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Important Questions for Class 12 History Chapter 12 Colonial Cities: Urbanisation, Planning and Architecture

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Important Questions for Class 12 History Chapter 12 Colonial Cities: Urbanisation, Planning and Architecture

Important Questions for Class 12 History Chapter 12 – 2 Marks Questions

Question 1.
Name the fortification of East India Company in Madras. Mention any one feature of it. (All India 2016)
or
Mention two features of fort St. George of white town where most of the Europeans lived. (All India 2010)
Answer:
Madras was an important part. The English East India Company built its factories and fortified the city for protection. They built Fort St George in Madras. Two features of fort St. George are as follows:

  • Fort St. George was nucleus of the white town.
  • Walls and bastions made this a unique enclave. Colour and religion determined who was allowed to live within the fort i.e., only British could live. Indian merchants, artisans and other workers who had economic dealings with European merchants lived outside these forts in settlements of their own.

Question 2.
Point out one supportive and one conservative view on the opportunities provided to the Indian women in colonial cities. (Delhi 2015)
Answer:
The two views on the opportunities provided to the Indian women in colonial cities are as follows:

  1. Cities provided new opportunities for women. Middle class women express themselves through medium of journals autobiographies and books. They started working as domestic and factory worker teachers, and theatre and film accesses.
  2. Conservative feared that education and working of women would alter the basis of the entire social order.

Question 3.
Analyse how did the introduction of the railways by the British prove advantageous for Indian in the late 19th century. (HOTS; Delhi 2015)
Answer:
The introduction of railways in 1853 meant a change in the fortunes of Indian cities. The advantages due to this change were:

  • Economic activity gradually shifted away from traditional towns to new cities. Every railway station became a collect depot for raw materials and a distribution point for imported goods. Thus, the activity of the old city decreased and new cities got importance.
  • With the expansion of the railway network, railway workshops and railway colonies were established. New railway towns like Jamalpur, Waltair and Bareilly developed.

Question 4.
Explain how the conversion of census data into convenient statistical data by the Britishers in India riddled with ambiguities in late 19th century. (Delhi 2015)
Answer:
The conversion of census data into convenient statistical data by the Britishers in India riddled with ambiguities in late 19th Century. The reasons behind this were:

  • Often Indians refused to cooperate or gave evasive answers to the census officials. People were suspicious of census operation and believed that it would impose new taxes.
  • Upper caste people were unwilling to give any information regarding women folk in their family as women were not subjected to public enquiry.
  • People often claimed that they were associated with higher status, so their real identities were not revealed.
  • The figures of mortality and diseases were difficult to collect as these were not registered.

Question 5.
Name the region where the Lottery Committee initiated town planning during 18th century. Mention any one feature of it. (All India 2015)
Answer:
Lottery Committee initiated town planning in the Calcutta. Main function of the Lottery Committee in the context of colonial Calcutta was that the Lottery Committee built roads and cleared the river bank of encroachments. The Committee removed many huts and displace the labouring poor of the city.

Question 6.
Why was colonial government keen on mapping from the early years? Mention two reasons. Delhi 2014
Answer:
From the early years, the colonial government was keen on mapping due to the following reasons:

  • The colonial government felt that good maps were necessary to understand the landscape and know the topography.
  • With the useful information through maps, the government can easily control the region.

Question 7.
How did changes occur in building pattern of colonial cities after the revolt of 1857? Cite two examples.
(HOTS; All India 2014)
Answer:
Nature and design of colonial city changed after the revolt of 1857. Because after the revolt of 1857, British attitude in India were shaped by constant fear of rebellion. The two examples of this change were as follows:

  1. The British felt the need of security for white elites. So, they developed Civil lines, which were secure and segregated enclaves, away from the threat of the natives.
  2. Another example is cantonments places where Indian troops under European command were stationed were also developed as safe enclaves.

Question 8.
How were the town often defined in opposition to rural areas during pre-colonial times? Give any two point of difference. (HOTS; Delhi 2012)
Answer:
Generally, there was a wide difference between towns and rural areas:

  • The towns developed as the representatives of specific forms of economic activities and cultures. But cultivating land, foraging in the forest or rearing animals were the main sources of income for the rural people.
  • Another difference between the rural areas and the towns was that the towns and the cities were generally fortified, whereas the village did not have any kind of fortification.

Question 9.
How were the hill stations a distinctive feature of colonial urban development? Give two reasons. (All India 2012)
or
How did Indian hill stations beacamc racial enclave for the European in the 19th century? Explain two reasons. (Delhi 2016)
Answer:
British Government initialed the development of hill stations because of the following reasons:

  • Hill stations developed as a place of strategic importance for guarding frontiers, billeting of troops and launching operations against enemy.
  • Hill stations have cool climate. These areas are free from the disease of tropical regions, so these areas w’ere developed as sanitarium and place for recreation for the European elites.
  • As climate of hill stations were similar to climate of Europe, they became an attractive destination for rulers.

Question 10.
Why were the hill stations important for the colonial economy? Give any two reasons. (Delhi 2010)
or
How were hill stations important for the colonial economy? Give one example. (All India 2012)
Answer:
Hill stations were important for the colonial economy because:

  • With the setting up of tea and coffee plantations in the adjoining areas, influx of immigrant labour from the plains began which proved beneficial for the colonial economy. For e.g. Tea plantations of Darjeeling. With this cheap labour, British Company made huge profits.
  • The introduction of railways made hill stations more accesssible to Indians, upper and middle class Indian started to go there and it helped to improve the overall economic development of the hill stations.

Question 11.
Give any two features of colonial cities after 1850 in India. (Delhi 2008)
Answer:
Two features of colonial cities after 1850 in India were as follows:

  1. Institutions were set up to regulate economic activity and demonstrate authority of a new ruler.
  2. The colonial cities grew as new commercial and administrative centres.

Important Questions for Class 12 History Chapter 12 – 4 Marks Questions

Question 12.
“A careful study of census reveals some fascinating trends of urbanisation in 19th century”. Support the statement with facts. (HOTS; All India 2013)
or
The pace of urbanisation and city development was slow after 1800 CE. Justify this statement with suitable examples. (HOTS)
Answer:
The trends of urbanisation in 19th century can be discussed through the following facts:

Stagnant Growth of Urban Population:
The growth of urban population in proportion to the total population of India was almost stagnant. From 1900 to 1940, the urban population increased from 10 to 13 percent.

Significant Variations in the Patterns of Urban Development: In this period the smaller town had little opportunity to grow economically. But Culcutta, Bombay, Madras grew rapidly and became large cities. Gradually, these three cites became the new commercial and administrative centres.

Clear Difference between Colonical Cities and Traditional Towns: The large cities functioned as collection depots for the export of Indian manufactures like cotton textiles. After the Industrial Revolution in England, these cities became the entry point for British manufactured goods and for the export of Indian raw materials. This nature of economic activities sharply differentiated these colonial cities from traditional towns.

Changes due to the Introduction of Railways: The introduction of railways further deregulated the pace and uniformity of urban development. Economic activities were accelerated in Calcutta, Bombay and Madras after the introduction of Railways. The traditional towns were lagged behind in growth. For e.g. Mirzapur was a famous collection centre of cotton and cotton goods from the Deccan. When a rail link was made to Bombay, this town witnessed a decline. On the other hand, with the expansion of the railway network, railway workshops and railway colonies were established, for e.g. Jamalpur, Waltair and Bareilly.

Important Questions for Class 12 History Chapter 12 – 8 Marks Questions

Question 13.
“The architecture in colonial Bombay represent ideas of imperial power, nationalism and religious glory.”
Support statement with example. (All India 2015)
or
Taking the examples of Bombay (Mumbai), explain how the imperial vision of the British was realised through town planning. (Delhi 2008)
or
Describe the characteristics of the public buildings built in the new classical style with special reference to ‘Town Hall’ of Bombay.
Answer:
With the expansion of economy the . architecture or town planning of Bombay changed a lot. These changes are as follows:
1. In the mid-nineteenth century; many new building were constructed in Bombay and the architectural style was European. The British wanted to create a familiar landscape in an alien country by adopting this architectural style. They also thought that European styles symbolised their superiority, authority and power. European style of buildings marked the difference between the buildings of colonial masters and their Indian subject.

2. Gradually, Indians got used to European architecture and accepted these as their own. British in turn adapted some Indian styles e.g. bungalow which was derived from . Bengali hut. Bungalow was used by
government officers in Bombay and all over India. The surrourding veranda kept the bungalow cool in summer. These ensured privacy without daily social contact with Indians.

3. For public bindings, three broad architectural styles were adapted in India. The Town Hall in Bombay was built in new classical style. Elphinstone Circle or Horniman Circle was inspired from models in Italy. Neo-Gothic structure was also adapted for making building like the secretariat, University of Bombay and Eight Court, etc. Victoria Terminus is the most spectacular example of the Neo-gothic style. The Gateway of India, Taj Mahal hotel built by Jamsetji Tata were followed the Indo-Saracenic style.
Taj Mahal hotel became a challenge to the racially exclusive club and hotel maintained by the British.

4. The increasing population due to migration and lack of space in Bombay led to a type of building unique to it. This was known as chawl, the multi-storeyed single-room apartment with long open corridor built around a courtyard. Such buildings housed a large number of families sharing common spaces and this type of building helped in the growth of neighbourhood identity and solidarity.

Question 14.
Explain the changes reflected in the history of urban centres in India during the 18th century with special reference to network of trade. (Delhi 2012)
or
Describe the changes which occurred in the cities during the 18th century.
Answer:
The towns underwent many significant changes during the 18th century which were as follows:

Decline of Old Towns and Emergence of New Towns:

The political and commercial reorganisation, the old towns headed towards decline and new towns were developed in the 18th century. With the decline of the Mughal power, the towns associated with their adminisration also started declining. Delhi and Agra, the Mughal capital cities started losing their political importance.

Increasing Importance of Regional Capitals:

The regional capitals such as Lucknow, Hyderabad, Seringapatam, Pune, Baroda and Tanjore started gaining importance. The traders, administrators, artisans and other people in search of work and patronage began to reside in new capitals leaving the old Mughal centres. Some local notables and officials used new urban settlements like qasbah and ganj. The effects of political decentralisation were uneven. Some areas became strong for economic activity and in some areas political uncertainty led to economic decline.

Changes in Trading Networks:

The changes taking place in the networks of the trade also influenced the history of the urban centres. The European commercial companies had established their settlement at various places of the country during the Mughal period. These were Panaji by the Portuguese, Masulipatnam by the Dutch, Madras by the British and Puducherry by the French. By the end of the eighteenth century, the land based empires in Asia were replaced by the powerful sea-based European empires. The nature of society was regulated by forces of international trade, mercantilism and capitalism.

Decline of Renowned Trading Centres:

A new phase of change came into existence by the middle of the 18 th century .The renowned trading centres like Surat, Masulipatnam and Dhaka which had developed during 17th century, started heading towards decline and trading activities started concentrating at other places.

Impact of Battle of Plassey:

The Battle of Plassey in 1757, established British dominance in Bengal. As a result, there was a significant increase in the territories of the Company. Now many colonial port cities like Madras, Calcutta and Bombay started developing rapidly into new economic capitals.
Soon these cities grew into important centres of colonial power and administration. The people were so much attracted to these cities that these became the biggest cities in India in terms of population of about 1800.

Question 15.
Explain any three broad architectural styles used by the British for the public buildings in the colonial states with examples. (All India 2012)
Answer:
The British used three broad architectural styles for the public buildings in the colonial state. They have been described as under:

Neo-Classical /New-Classical:
It comprised construction of geometrical structures fronted with lofty pillars. It was derived from the Roman style and got popular during the European Renaissance. It was considered appropriate during the British times, since the British imagined that a style that embodied the glory of imperial Rome could now’ be made to express the glory of imperial India. The Mediterranean origins of this architecture was also thought to be suitable for tropical weather. For e.g., the Town Hall of Bombay (1883), Elphinstone Circle, etc.

Neo-Gothic:

The style was characterised by high pitched roofs, pointed arches and detailed decoration. The Gothic originated from Churches built in Northen Europe during the medieval period.

The neo-gothic style revived in the mid-nineteenth century in England. This • was the time when the government in Bombay was building its infrastructure and this style was adopted for Bombay. The most spectacular example of the neo-gothic style is the Victoria Terminus, the station and headquarters of the great Indian Peninsular Railway Company. The University Hall, the University Library, Clock Tower are other examples of this style of building.

Indo-Saracenic:

It was a fusion of Indian and the European style which evolved towards the beginning of the 20th century.
The inspiration for this style was medievel buildings in India with their domes, chhatris, jalis and arches. The Indo-Saracenic style became a medium of expression by the British that they were the legitimate rulers of India. For e.g. the Gateway of India, the Taj Mahal Hotel built by industrialist Jamsetji Tata.

Question 16.
Explain why hill stations were a distinctive feature of colonial urban development. (Delhi 2008)
or
Describe the aims of establishing hill stations and cantonments in India by the British.
Answer:
Hill stations were a distinctive feature of the colonial urban development because of the following reasons:

Connection with the Needs of the British Army:

The founding and setting of hill stations was initially connected with the needs of the British army. For instance,
Simla, Mount Abu and Darjeeling were established for special purposes. Hill stations were made as cantonments for the army.

Strategic Places for Harbouring Troops:

Hill stations became the strategic places Un-accommodating troops, guarding frontiers and launching campaigns against enemy rulers.

Pleasant Climate:

The temperate and cool . climate of the Indian hills was seen as an advantage and these were free from epidemics like cholera and malaria.

Served as Sanitariums:

The hill stations where developed as sanitariums as the places where soldiers could be sent for rest and recovery from illness.

Attractive Destinations for the British:

Since the hill stations experienced the same kind of climate as of Europe, they became an attractive destination for the new rulers. It became practice of the Viceroys to move to hill stations during summer months.

Reflected Settlements Native to the British:

The hill stations w’ere characterised by settlements that were reminiscent of homes of the British and other Europeans.The buildings were deliberately built in the European style.

Linked to Diverse Groups:

The introduction of railways made the hill stations more accessible to a wide range of people including upper and middle class Indians like maharajas, lawyers and merchants.

Economic Significance:

The hill stations were important for the colonial economy in the sense that they were the important centres for tea and coffee plantations.

Important Questions for Class 12 History Chapter 12  Source Based Question

Question 17.
A rural city:
Read this excerpt on Madras from the Imperial Gazetteer, 1908:
…the better European residences are built in the midst of compounds which almost attain the dignity of parks, and rice-fields frequently wind in and out between these in almost rural fashion. Even in the most thickly peopled native quarters such as Black Town and Triplicane, there is little of the crowding found in many other towns…

  1. Where and why were better European residences built?
  2. Explain the condition of black towns.
  3. State the meaning of gradual urbanisation of Madras.

Answer:
1. The better European residences were built in the midst of compounds which almost like attain the dignity of parks. These were rice field in almost rural fashion.

2. Indian merchants, artisans and other workers who had economic dealing with Europeans merchants lived outside the St. George fort in settlements known as “Black Town”. These towns laid out in straight lines. Later a new Black Town developed and resembled like a traditional Indian town with living quarters built around its own temple and bazaars.

3. The gradual urbanisation of Madras meant that the areas between the villages were brought within the city. As a result, Madras had a semi-rural air about it.

Important Questions for Class 12 History Chapter 12 Value Based Questions

Question 18.

  1. “The colonical cities provided new opportunities for women during 19th century.” Give two examples.
  2. Explain any two values which encouraged women for their empowerment. (Delhi 2014)

Answer:
1.  The colonial cities provided opportunities for women and these were evident by following examples:

  • Women from middle class sought to express themselves through medium of journals, autobiographies and books.
  • Over the period of time women became more visible in public as they entered new profession in the city as domestic and factory worker, teacher and theatre actress.

2. women gained confidence and inspiration to do something for her own self. They developed social awareness. These confidence and awareness of women encouraged them for empowerment.

Question 20.

  1. Explain the importance of census data for the holistic development of a country.
  2. When was the first All India Census attempted? Mention two of its early aims. (Delhi 2009)

Answer:
1. The first attempt of All India Census was made in 1872. Thereafter, decennial census became a regular feature. Undoubtedly census data are of immense significance for the holistic development of a country. The census data are essential for the all round development and plannings.

2. The first attempt of All India Census was made in 1872. Its two early aims were:

  • To study urbanisation in India.
  • To know about the total number of white and non-white people living in various towns.

Important Questions for Class 12 History

The post Important Questions for Class 12 History Chapter 12 Colonial Cities: Urbanisation, Planning and Architecture appeared first on Learn CBSE.


NIOS D.El.Ed Study Material 2019 | Download Official Subject Wise Study Material PDF’s Here

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NIOS D.El.Ed Study Material: National Institute for open schooling also known as NIOS helps students the facility of securing the diploma in education and teaching for the in-service teachers. The main purpose of the Institute is to provide the training to the teachers that are untrained for primary as well as upper classes. The diploma has an exam to crack and a course duration. Teachers that are able to crack the exams successfully become liable for being called as trained professionals.

NIOS is the conducting body that is assigned by the human resource department ministry. The NIOS D.El.Ed syllabus includes the study for the understanding of students at various levels. Also, it includes the history of the Indian education sector, the difficulty of the subjects based on their level, trends in the education sector, etc. In this article, we are going to provide you with the official PDFs that are released by the NIOS official website which nios.ac.in. Also, you can refer these PDFs to understand the concepts that you are going to apply and study them during the diploma sessions.

The human resource department recently released a report that states that for every in-service, it is compulsory to have a D.El.Ed certificate in 2019. It is also compulsory to have at least 50% in the 12th class for teachers belonging to the general category. Teachers from other categories like ST/SC/OBC can have a relaxation of up to 5%. Also, D.El.Ed certificate will be issued to the candidates that secure at least 50% marks.

NIOS D.El.Ed Subject Material – Subject Wise PDF’s

NIOS D.El.Ed syllabus is a combination of various cultural and social subjects. These subjects are also designed with the main focus on education and the academic scenario of India. Because of the training, an in-service teacher can have access to the knowledge of the history of the education system, drawbacks, and benefits, etc. The main idea behind designing the program is to make teachers understand that there are many challenges that they can ace successfully.

NIOS D.El.Ed Study Material for Code 501

Elementary education in India: A Socio-Cultural Perspective

  1. Elementary Education in India – A Retrospective
  2. Elementary Education in India terms of Contemporary Context- I
  3. Elementary Education in India terms of Contemporary Context- II
  4. Assignment 501

NIOS D.El.Ed Study Material for Code 502

Pedagogical processes in Elementary Schools

  1. Teaching and Learning Process
  2. Management of Learning- Teaching Processes
  3. Emerging Issues in the Classroom learning
  4. Learning Assessment
  5. Assignment 502

NIOS D.El.Ed Study Material for Code 503

Learning languages at the elementary level

  1. Understanding Language
  2. Skills that are associated with language learning
  3. Language learning in a class
  4. Assignment 503

NIOS D.El.Ed Study Material for Code 504

Learning Mathematics at Elementary Level

  1. Importance of Learning Mathematics at the Elementary Stage of Schooling
  2. Enriching Candidates and Methodology
  3. Learner Assessment in Mathematics
  4. Assignment 504

NIOS D.El.Ed Study Material for Code 505

Learning Environmental Studies at Primary Level

  1. Importance of Teaching-Learning of EVS at primary level
  2. Curriculum and Pedagogy of EVS
  3. Assessment of learning in EVS
  4. Assignment 505

NIOS D.El.Ed Study Material for Code 506

Understanding Children in Inclusive Context

  1. Development Basics and Child Growth
  2. Personality development of Children
  3. Inclusive Education
  4. Girl Child and Child Right

NIOS D.El.Ed Study Material for Code 507

Community and Elementary Education

  1. Community, Society, and School
  2. School System
  3. Managing School Community Interface

NIOS D.El.Ed Study Material for Code 508

Learning in Art, Health and Physical and Work Education at Elementary Level

  1. Art Education
  2. Health and Physical Education
  3. Work Education

NIOS D.El.Ed Study Material for Code 509

Learning Social Science at Upper Primary Level

  1. Understanding Social Science in terms of a Discipline
  2. Concepts and Social Subjects
  3. Issues in Pedagogy of Social Sciences

NIOS D.El.Ed Study Material for Code 510

Learning Science at Upper Primary Level

  1. Understanding Science
  2. Measuring and Managing Science Learning

As you can notice that NIOS D.El.Ed study material js a combination of various academic and society structure-based concepts. The main idea behind launching this course is to provide enough training to the teachers so that the quality of education is improved. Also, it is expected that the training will be helpful for teachers in understanding the challenges in a professional way.

The post NIOS D.El.Ed Study Material 2019 | Download Official Subject Wise Study Material PDF’s Here appeared first on Learn CBSE.

Important Questions for Class 12 History Chapter 14 Understanding Partition (Politics, Memories, Experiences)

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Important Questions for Class 12 History Chapter 14 Understanding Partition (Politics, Memories, Experiences)

Important Questions for Class 12 History Chapter 14 – 2 Marks Questions

Question 1.
“The relationship between India and Pakistan has been profoundly shaped by the legacy of partition.” Explain any two consequences of it. (All India 2017)
Answer:
It is true that the relationship between India and Pakistan has been profoundly shaped by the legacy of partition.
The two consequences of this are:

  • Indian haters in Pakistan and Pakistan haters in India are both products of partition. Some people spread false ideas in India that Muslims are cruel, bigoted, unclean and descendants of invaders, while Hindus are kind, liberal and pure. They even wrongly believed that Indian Muslims are supporters of Pakistan.
  • Similarly in Pakistan, feeling was spread that Muslims are fair, brave, monotheists and meat-eaters, while Hindus are dark, cowardly, polytheist and vegetarians.

Question 2.
Why did the Congress not accept the proposal to form a joint government with the Muslim League in the United Provinces? Give any two reasons. (HOTS; Delhi 2012)
Answer:
The two reasons were as follows:

  • The Congress had won an absolute majority in the province.
  • The Muslim League supported landlordism whereas the Congress wanted to abolish it.

Question 3.
Was the demand of the league reasonble? Comment. (Delhi 2011)
Answer:
No, the demand of league was not reasonable.
It clearly demanded the partition of India on the basis of religion.

Important Questions for Class 12 History Chapter 12 – 4 Marks Questions

Question 4.
Examine the events that took place during 1920s and 1930s which consolidated the communal identities in the country.
(All India 2017)
Answer:
Between 1920 and 1930, many incidents took place which created tensions. Frequent riots took place. Many Hindu organisations were formed. They carried out purification movement and started playing communal cards. Hindu identity was defined against Muslim identity by the Hindu Mahasabha. Music playing before the mosque became frequent.

Hindi became the language of the Hindus and Urdu became the language of Muslims. There were increased communal feelings within Hindus and the Muslims. Hindus were angered by the rapid spread of ‘tabligh’ (propaganda) and ‘tanzim’ (organisation) after 1923. Gaps between Hindus and Muslims widened due to these deliberate actions.

In the 1937 elections, the communal parties, the Hindu Mahasabha and the Muslim League fared poorly. Apprehensive of their survival, both the parties began to make use of religion to secure the support of the masses. The British encouraged the Muslim League and when the Congress resigned in 1939, they were invited to form the government in the provinces.

Question 5.
Analyse the provisions of Cabinet Mission of 1946. (HOTS; Delhi 2015)
or
Examine the recommendations of the Cabinet Mission and explain the reasons for rejecting the plan suggested by Cabinet Mission in 1946 by both Congress and the Muslim League.
Answer:
Recommendations of the Cabinet Mission were as follows:

  • A loose three-tier confederation of United India.
  • A weak Central Government controlling foreign affairs, defence and communications.
  • Provincial assemblies were grouped into three sections, i.e. section A Hindu majority provinces, section B and section C Muslim-Majority provinces of the North-West and the North-East India.
  • The provinces would have their own executives and legislatures.
    Reasons of rejecting the plans were as follows:
    • The league wanted the right to secede from the Union to sections B and C.
    • The Congress wanted that the provinces should have the right to join any group.
      Therefore, neither the league nor the Congress agreed to the proposal.

Question 6.
Analyse the impact of partition of India on Punjab and Bengal. (HOTS; All India 2015)
Am. Impact of partition was horrendous in the Punjab.

  • There was carnage on both sides across the border. Near total displacement of Hindus and Sikhs took place from West Punjab to Eastward into India and similarly Punjabi Muslims were displaced from Indian Punjab to Westward into Pakistan.
  • There was massacre in Amritsar, thousands of people were killed in Punjab, hundreds of women were raped and abducted. Properties were looted and immovable property were captured.
  • Many women killed themselves in order to protect themselves from being dishonoured by violent mob. This total displacement of Hindu, Sikh and Punjabi-speaking Muslims happened in a relatively short period between 1946 and 1948.

In Bengal situation was more or less similar but process of migration was more protracted.

  • There was no total displacement of population in Bengal as many Hindus remained in East Bengal and many Muslims in West Bengal. But in Bengal people were also murdered in large numbers, women were raped and abducted.
  • Finally Bengali Muslims of East Pakistan broke away from Pakistan and created independent country Bangladesh in 1971-72.

Question 7.
Why did the Congress reject the offer of the Muslim League to form a joint government? Explain. (All India 2009)
Answer:
There were various reasons due to which the Congress rejected the offer of the Muslim League to form a joint government which were:

Absolute Majority of Congress:
In the United Provinces, the Muslim League ‘ wanted to form a joint government with the Congress. The Congress had won an absolute majority in the province so it rejected the offer.

Ill effects of the Rejection:
It is argued that this rejection convinced the league that if India remained united, then Muslim would find it difficult to gain political power because they would remain a minority.

Muslim League to be the Sole Spokesman of Muslims:
The league assumed that only a Muslim party could represent Muslim interests and the Congress was essentially a Hindu party. Jinnah’s insistence that the League be recongised as the sole spokesman of Muslims could convince few at the time.

Muslim League’s Support to Landlordism:
Congress Party rejected the Muslim League proposal for a coalition government because the League tended to support landlordism, which the Congress wished to abolish.

Question 8.
Examine the views of Gandhiji against the partition of India. (Delhi 2008)
Answer:
Gandhiji W’as against the partition of India. He worked for Hindu-Muslim unity. He rejected religion as the basis of forming a “ nation. He belived in Hindu-Muslim unity- based on sharing of a common culture and a single nation. Gandhiji believed that the Pakistan demand had put forward by the Muslim League was un-Islamic.

He called it as sinful. According to him Islam stood for the unity and brotherhood of mankind, not for disrupting the oneness of the human family.

Gandhiji was saddened by the partition and the migrations taking place. He said, “What could be more shameful for us”, during a speech, “than the fact that not a single Muslim could be found in Chandni Chowk?” Gandhiji continued to be in Delhi, fighting the mentality of those who wished to drive out • every Muslim from the city, seeing them as Pakistani.

Important Questions for Class 12 History Chapter 12 – 8 Marks Questions

Question 9.
Examine various events that led to the partition of British India. (Delhi 2016)
Answer:
There are a number of events which led to the partition of British India whether directly or indirectly:

Politicisation of religion:
Politicisation of religion started with separate electorate in 1909, and was further strengthened by the colonial government of India in 1919. Thus, communal riots were taking place and deepened the difference between communities. Yet it would be incorrect to see partition as outcome of communal tensions.

The Provincial elections of 1937 and the Congress ministries:
In 1937, for the first time provincial elections were held. In this Congress won with majority. In United Provinces, Muslim League wanted to form government with Congress but Congress rejected the idea as it had absolute majority. League believed that as they are minority they would not get political power.
The league also believed that only Muslim party can represent Muslim and Congress is Hindu Party. Growth of RSS and Hindu Mahasabha also played an important role in widening the difference between Hindus and Muslims.

The ‘Pakistan’ Resolution:
On 23rd March, 1940, League passed a resolution demanding a measure of autonomy for Muslim majority areas of the sub-continent. This resolution never mentioned partition or a separate state. First demand of Pakistan was made by the urdu poet Mohammad Iqbal.

Post-War Development:
In 1946, again provincial elections were held. In this election Congress swept general constituencies and league succeed in gaining large majority of Muslim vote. Therefore, in 1946 League established itself as a dominant party among Muslims.

A possible alternative to Partition:
In March 1946, Cabinet Mission came in India to make a suitable political framework for India. But both the parties i.e. the Muslim League and the Congress did not agree to the Cabinet Mission’s proposal and talks failed. Congress sensed after this failure that partition become inevitable and took it as tragic but unavoidable.

Towards Partition:
After withdrawal from Cabinet Mission, Muslim League decided on direct action for winning its Pakistan demand. It announced 16th August, 1946 as ‘direct action day’. Initially riots broke out in Calcutta and gradually spread to other parts of Northern India. In March 1947, Congress accepted the partition.
Thus, these events led to the partition of British India.

Question 10.
Analyse the distinctive aspects of the oral testimonies to understand the history of the partition of British India. (Delhi 2016,15)
or
Describe the strengths and weaknesses of oral history. Mention any four sources from which the history of partition has been constructed. (All India 2014,09)
Answer:
The strengths of oral history are:

  • It helps us to grasp experiences and memories in detail.
  • It also enables historians to write richly textured, vivid accounts of what happened to people during partition.
  • Government sources cannot provide this kind of information.
  • Oral history also facilitates historians to broaden the boundaries of their discipline by making them aware about the lived experiences of the poor and the powerless.

The weaknesses of oral history are:

  • Oral data lack concreteness and the chronology they yield may not be very precise.
  • Historians argue that the uniqueness of personal experience makes generalisation difficult. A larger picture cannot be built from such micro evidence and one witness is not sufficient for the whole analysis.
  • Oral accounts are concerned with tangential issues in the sense that they provide an indirect evidence of the event.
  • The small individual experiences which remain in memory are irrelevant to the unfolding of larger processes of history.

Following are the four sources from which the history of partition has been constructed:
Diaries:
It helps to understand experiences and memories in detail and enables historians to write richly textured, vivid accounts of what happened. It is impossible to extract this kind of information from government documents. The later deal with policy and party matters and various state sponsored schemes in case of partition. Government reports and files and personal writings of high level functionaries throw sufficient light on negotiations between the British and the major political parties about the future of India or rehabilitation of refugees.

Memoirs and Experiences:
Partition has been generally viewed in terms of suffering and challenges of the times. It was not merely a constitutional division or just the party politics of the Muslim League,
Congress and others.
For the common people, it meant unexpected changes in life as it unfolded between 1950 and beyond requiring psychological, emotional and social adjustments. Memoirs and experiences shape the reality of an event.

Oral Narration:
Oral history allows historians to broaden the boundaries of their discipline by making them aware of the live experiences of the poor and the powerless. The oral history of partition has succeeded in exploring the experiences of those men and women whose experience has hitherto been ignored, taken for granted or mentioned only in passing in mainstream history.

Family Histories:
The accounts of family histories tell us about the pains and trauma faced by members, of their families during partition.

Question 11.
Explain how Indian partition was a culmination of communal politics that started developing in the opening decades of the 20th century. (Delhi 2015)
or
Explain how communal identities were consolidated by host of developments other than political in the early 20th century. (Delhi 2013)
Answer:
The rise and role of various communal parties was a significant factor for the partition of India. The Muslim League was founded in 1906 to work for the interest and favour of the Muslims. The demand for more and more political rights by the Muslims awakened the Hindus and as a result of it Hindu Mahasabha was founded in 1915. They demanded more political rights and representation in different government organisations for the Hindus.

Later on, the Sikh League was founded and the Akah Dal also raised voice for the Sikhs. Undoubtedly, these parties increased communal feelings and brought a feeling of esparatism between different communities especially the Hindus and the Muslims.

Between 1920 and 1930, many incidents took place which created tensions. Frequent riots took place. Many Hindu organisations were formed. They carried out purification movement (Shuddhi Karan) and started playing communal cards. Hindu identity was defined against Muslim identity by the Hindu Mahasabha, Music playing before the mosque became frequent.

Hindi became the language of the Hindus and Urdu became the language of Muslims. There were increased communal feelings within Hindus and the Muslims. Hindus were angered by the rapid spread of ‘tabligh’ (propaganda) and ‘tanzim’ (organisation) after 1923. Gaps between Hindus and Muslims widened due to these deliberate actions by these religious communities.

In the 1937 elections, the communal parties, the Hindu Mahasabha and the Muslim League fared poorly.
Apprehensive of their survival, both the parties began to make use of religion to secure the support of the masses. The British encouraged the Muslim League and when the Congress resigned in 1939, they were invited to form the government in the provinces.

The British seeking to weaken the National Movement now recognised the Muslim League as the sole spokes person of the Muslim and they were given the power to veto any political settlement. The Muslim League realised that all communal demands conceded by the British would hardly give the Muslim League anything when the country became free. The Muslim League demanded a separate state since, they feared domination by the Hindu majority when India would become free.

Question 12.
Describe the harrowing experiences of ordinary people during the period of partition of India. (Delhi 2014)
Answer:
The harrowing experiences were as follows:
Massive Killings and Bloodshed:
Ordinary, people underwent traumatic experiences during the period of partition. There were massive killings and bloodshed in the North-Western part of India due to collapse of law and order. Communal riots were frequent. Due to complete breakdown of authority, people were left helpless in that situation.

Rape, Abduction and Selling of Women:
Women were raped, abducted and sold frequently and forced to live with strangers in unknown circumstances. Some began to develop new7 family bonds in the changed circumstances and faced insensitivity of the Indian and Pakistani government. There were also instances of honour killings and women were forced to preserve community honour by getting killed by their family members if they felt that they have been violated by the enemy. The incidence of Thoa Khalsa, Rawalpindi district where ninety Sikh women were said to have ‘voluntarily’ jumped into a well rather than falling into ‘enemy’ hands indicates this fact.

Large-Scale Migration of People:
Apart from this, there were large-scale displacement of Hindus and Sikhs from West Punjab to Eastern part of the country. In Bengal too, there were large-scale migration, though it was protracted with people moving across a porous border. In both Punjab and Bengal, women became prime targets of persecution. Thus to sum up, ordinary people especially women experienced painful turmoils during the course of partition.

Question 13.
Explain the reason why the plan,
suggested by the Cabinet Mission was finally not accepted by the Congress and the Muslim League. (All India 2010)
Answer:
The reasons for which the plan suggested by the Cabinet Mission not accepted by the Congress and Muslim League were as follows:
United India as per Cabinet Mission Plan:
The Cabinet Mission toured the country for three months and recommended a loose three-tier confederation, India was to remain united.

According to the Cabinet Mission Plan:

  • India was to have a weak Central Government controlling only foreign affairs, defence and communications with the existing provincial assemblies.
  • Three sections while electing the Constituent Assembly. Section A was for the Hindu-majority provinces and Sections B and C for the Muslim-majority provinces of the North-West and the North-East (including Assam) respectively
  • The sections or groups of provinces would comprise various regional units. These would have the power to set up intermediate-level executives and legislatures of their own.

Opposition by the Congress and League:
The Congress wanted that provinces should be given the right to join a group. It was not satisfied with the cabinet mission’s clarification that grouping would be compulsory at first, but provinces would have the right to opt out after the constitution had been finalised and new elections were held in accordance with it. League wanted that these grouping should be compulsory and provinces should be given the right to secede from union.

Rejected by both Congress and the League:
Finally, neither the League nor the Congress agreed to the Cabinet Mission proposal. This was the most crucial juncture because after this partition became more or less inevitable.

Important Questions for Class 12 History Chapter 12  Sourse Based Question

Question 14.
A small basket of grapes
This is what Khushdeva Singh writes about his experience during one of his visits to Karachi in 1949.
My friends took me to a room at the airport where we all sat down and talked….(and) had lunch together. I had to travel from Karachi to London…at 2.30 am… At 5.00 pm…I told my friends that they had given me so generously of their time, I thought it would be too much for them to wait the whole night and suggested they must spare themselves the trouble. But nobody left until it was dinner time… Then they said they were leaving and that I must have a little rest before emplaning …I got up at about 1.45 am and when I opened the door,
I saw that all of them were still there…

They all accompanied me to the plane, and before parting, presented me gratitude for the overwhelming affection with which I was treated and the happiness this stopover had given me.

  1. Give a brief introduction of Khushdeva Singh.
  2. How did his friends show their affection for him?
  3. Explain how Khushdeva Singh was seen as a symbol of humanity and harmony.
  4. How does oral history help historians in reconstructing events of the past? (All India 2012)

Answer:
1. Khushdeva Singh was a Sikh doctor. He had specialisation in the treatment of tuberculosis. He was posted at Dharampur.

2. Friends of Khushdeva Singh showed their affection by having lunch together. Dr Khushdeva Singh was highly impressed by their affection.

3. Dr Khushdeva Singh was seen as a symbol of humanity and mutual harmony. He was a workaholic. He provided immense help to several migrants.

4. Oral history persists historians in reconstructing events of the past. It helps historians to write clear description of what happened to people during the past events. Oral history enables historians to broaden the limits of their discipline.

Question 15.
“No no! You can never be ours”
This is the third story the researcher related.
I still vividly remember a man I met in Lahore in 1992. He mistook me to be a Pakistani studying abroad. For some reason he liked me. He urged me to return home after completing my studies to serve the qaum (nation). I told him I shall do so but, at some stage in the conversation, I added that my citizenship happens to be Indian.

All of a sudden his tone changed as much as he was restraining himself., he blurred out.

“Oh Indian! I had thought you were Pakistani”. I tried my best to impress upon him that I always see myself as South Asian. “No, no! You can never be ours. Your people wiped out my entire village in 1947, we are sworn enemies and shall always remain so”.

  1. What did the person advice the researcher who met him in Lahore in 1992? Why did he say like this? Explain.
  2. How did the person react on knowing that the researcher was an Indian?
  3. What did the Indian try to explain?
  4. Who was right and why? Explain. (Delhi 2011)

Answer:
1. The person urged the researcher to return home after completing his studies to serve the nation.
2. As soon as the person knew that researcher was an Indian, his voice and tone changed. He restrained himself as much as he could but he blurred out.
3. The Indian tried to convince him that he always looked himself as South Asian.
4. The researcher was right because he did not show any particular association with any particular region and associated himself to a more broader context which unified both India and Pakistan.

Question 16.
“A Voice in the Wilderness”
Mahatma Gandhi knew that he was “a voice in the wildeness” but he nevertheless continued to oppose the idea of Partition.
But what a tragic change we see today. I wish the day may come again when Hindus and Muslims will do nothing without mutual consultation. I am day and night tormented by the question what I can do to hasten the coming of that day.
I appeal to the League not to regard any India as its enemy. Hindus and Muslims are born of the same soil. They have the same blood, eat the same food, drink the same water and speak the same language.

Speech at Prayer meeting, 7th September, 1946, CWMG, VOL. 92. P. 139

“But I am firmly convinced that the Pakistan demand as put forward by the Muslim League is un-Islamic and I have not hesitated to call it sinful. Islam stands for the unity and brotherhood of mankind, nor for disrupting the oneness of the human family.

Therefore, those who want to divide India into possible warring groups are enemies alike of Islam and India. They may cut me into pieces but they cannot make me subscribe to something which I consider to be wrong.”

  1. Explain what did Gandhiji wish to see again.
  2. Explain how the demand for Pakistan was un-Islamic.
  3. Why did Mahatma Gandhi say that his voice was in the wilderness? Explain. (All India 2017)

Answer:
1. Mahatma Gandhi wished to see again a day when the Hindus and Muslims would unite and do nothing without mutual consulation.

2. The demand for Pakistan was un-Islamic since Islam stands for unity and brotherhood of mankind, not for disrupting the unity of the human family.

3. Mahatma Gandhi said that his voice was in wilderness because he knew that he was in his fight against the partition of India when both dominant parties Congress and league are determined to partition of the country after failure of the Cabinet Mission.

Question 17.
“Without a shot being fired”
For over twenty-four hours riotous mobs were allowed to rage through this great commercial city unchallenged and unchecked. The finest bazaars were burnt to the ground without a shot being fired to disperse the incendiaries (i.e., those who
stirred up conflict). The District
Magistrate marched his (large police) force into the city and marched it out again without making any effective use of it at all.

  •  To which event does this source refer to? Describe what the mobs were doing.
  • Why did Amritsar become the scene of bloodshed later in 1947?
  • What was the attitude of the soldiers and policemen towards the mob?
  • Give one example to show how Gandhiji tried to bring about communal harmony. (All India 2008)

Answer:
1. This event refers to riot in Amritsar in March 1947. Mobs were doing riot and they were razing the city, burning bazaars, killings were there and Amritsar become the scene of bloodshed.

2. In Amritsar there was complete breakdown of authority. British officials did not know how to handle the situation. Indian officials were fearing for their own life and property. Senior leaders were busy in negotiating the Independence, so they are also not able to present there to handle the situation.

3. District magistrate was reluctant to control mob, he even feared for his life.
By riots, soldier and policemen were also affected. Few of them even forgot their professional commitment and sided with their co-religionist.

4. Gandhiji moved from village to village in Noakhali district, then Bihar, Calcutta to Delhi. He tried to console the people and told the people- to assure the safety of minorities.

Important Questions for Class 12 History Chapter 12  Map Based Question

Question 18.
On the outline map of India three centres related to the Indian National movement have been marked as A, B, C. Identify them and write their correct names on the lines drawn near them. (Delhi 2016)
Important Questions for Class 12 History Chapter 14 Understanding Partition (Politics, Memories, Experiences) Q18
Answer:
Important Questions for Class 12 History Chapter 14 Understanding Partition (Politics, Memories, Experiences) Q18(i)

Question 19.
On the given political outline map of India, locate and label the following with appropriate symbols. (All India 2016)

  1. The place where Gandhiji called off Non-cooperation movement.
  2. The imperial capital of Mughal.

Important Questions for Class 12 History Chapter 14 Understanding Partition (Politics, Memories, Experiences) Q19
Answer:
Important Questions for Class 12 History Chapter 14 Understanding Partition (Politics, Memories, Experiences) Q19(i)
Question 20.
On the outline map of India three places related to the Indian National Movement have been marked as A, B and C. Identify them and write their correct names on thelines drawn near them. (Delhi 2015)
Important Questions for Class 12 History Chapter 14 Understanding Partition (Politics, Memories, Experiences) Q20
Answer:
Important Questions for Class 12 History Chapter 14 Understanding Partition (Politics, Memories, Experiences) Q20(i)

Question 21.
On the map three centres of National Movement have been marked as 1, 2 and 3. Identify them and write their names 6n the line drawn near them. (All India 2010)
Important Questions for Class 12 History Chapter 14 Understanding Partition (Politics, Memories, Experiences) Q21
Answer:
Important Questions for Class 12 History Chapter 14 Understanding Partition (Politics, Memories, Experiences) Q21(i)

Question 22.
On the given political outline map of India locate and label the following. (All India 2009)
1. Dandi
2. Bardoli
Answer:
Important Questions for Class 12 History Chapter 14 Understanding Partition (Politics, Memories, Experiences) Q22

Question 23.
On the given political outline map of India, locate and label the following. (All India 2008)
1. Delhi
2. Goa
Answer:
Important Questions for Class 12 History Chapter 14 Understanding Partition (Politics, Memories, Experiences) Q23

Important Questions for Class 12 History Chapter 12 Value Based Questions

Question 24.
Read the following passage and answer the question that follows.
Arya Samaj, a North Indian Hindu reform organisation of the late 19th and early 20th centuries particularly active in Punjab (tried to bring back Hindus who had converted to some other religion) Which sought to revive vedic learning and combine it with modern education in the sciences.
1. Illustate how the values integrated with the rich Indian literature paved way for the scientific development of modern India. (All India 2016)
Answer:
1. The two values were as follows:

  • Rational and analytical values which are present in the ancient literature helps in developing scientific temperament in Indian people.
  • The scientific and analytical values can revive vedic learning and combine it with modern education in the science.

Question 25.
Read the following passage and answer the question that follows.
Dr Khushdeva Singh describes his work as “humble efforts I made to discharge my duty as a human being to fellow human beings”.
1. “Love is stronger than hate”. How true is this value which was proved at the time of the partition of India? What are the values one needs to instill and nurture to avoid hatred? Explain.
(All India 2015)
Answer:
1. “Love is stronger than hate”. This value is proved by the act of Dr Khushdeva Singh and an act of lady in Jammu which saved the life of man. To avoid hatred and build feeling of love, there should be feeling of brotherhood, tolerance and empathy towards each other. There is a need to shed discriminations on the ground of trivial issues, and on the personal faith, belief, etc.

Question 26.
Read the following ‘value-based’ passage and answer the questions that follows.
The 77-year-old Gandhiji decided to stake his all in a bid to vindicate his lifelong principle of non-violence, and his conviction that people’s hearts could be changed. He moved from the villages of Noakhali in East Bengal (present-day Bangladesh) to the villages of Bihar and then to the riot-torn slums of Calcutta and Delhi, in a heroic effort to stop Hindus and Muslims kill each other, careful everywhere to reassure the minority community. In October 1946, Muslims in East Bengal targeted Hindus. Gandhiji visited the area, toured the villages on foot, and persuaded the local Muslims to guarantee the safety of Hindus. Similarly, in other places such as Delhi he tried to build a spirit of mutual trust and confidence between the two communities.

  1. Explain hoW the spirit of mutual trust and confidence building was initiated by Gandhiji.
  2. Explain the value mentioned in this passage. (All India 2013)

Answer:
1. Gandhiji toured these riot and violence . affected areas. He told crowd to assure the
safety of their fellow minority community. Trust and confidence between communities had completely broken due to riots and they started to see each other as enemy but when one community would take responsibility to protect and assure safety of other community then, again feeling of trust and confidence will start to develop among the communities.
2. In this passage value of non-violence, peaceful co-existence have been emphasised. Passage suggests that Gandhiji believed that people’s heart can be changed.
To foster the feeling of unity, to develop a mutual trust and confidence between the community, Gandhiji told communities to protect and assure the safety of minority community in that area. Gandhiji tried to impress upon the value of tolerance, respect for each other, and feeling of brotherhood between different religious communities.

Question 27.

  1. What is communalism? Explain.
  2. How can we weed out the virus of communalism from our society. (All India 2012)

Answer:
1. Communalism is a particular kind of politics based on the basis of religious identity. It is an ideology that seeks to promote conflict between religious communities. Communalism nurture a politics of hatred for people of other religion and believe that their interest are not common and sometimes even antogonist. During the partition period extreme level of communalism was there when people believed that peace-coexistence between Hindus and Muslims is not possible.

2. Communalism or communal politics can be countered by secularism or secular politics. People should adopt policy of tolerance and respect for other religions. Principle of peaceful co-existence should be followed by every community and sense of brotherhood should be spread.

Question 28.

  1. “Some scholars see partition as a culmination of communal politics.” Examine the statement.
  2. “Secularism is the best solution to the problems of communalism”. Comment. (Delhi 2011)

Answer:
1. The policy of direct action started by Muslim League led to loss of life and property. The fanatics of the Muslim League received further encouragement from the leaders of the Muslim League. A retaliation started in some of the Hindu provinces. Partition was considered better than the murder of the innocent people. Therefore, it may be called as the culmination of the communal politics. Scholars suggest that separate electorates for Muslims by government crucially shaped the nature of communal politics. Further, religious identities acquired a functional use within modern political system. Sadly communal identities came to mean active opposition and hostilities between communities.

2. Secularism is a principle that involves two basis propositions:

  • The first is the strict separation of the state from religious institutions,
  • The second is that people of different religious and beliefs are equal before the law. Secularism has a number of advantages.

These are:

  • Separation of religion from state which is the foundation of secularism.
  • It protects both believers and non-believers.
  • It gives religious freedom to every single person.
  • It is about democracy and fairness, where all citizens are equal before law.
  • It provides equal access to public services.
  • It provides free speech and expression.

Thus, we can say that secularism is the best solution to the problems of communalism.

Question 29.

  1. Why are the historians sceptical of oral history? Explain briefly.
  2. To what extent oral narratives, memoirs, diaries, family histories, first hand accounts help us to understand the trails and tribulations of ordinary people during partition of country? Explain with examples. (Delhi 2010)

Answer:
1. The historians are sceptical of oral history because:

  • Oral data are not concrete and the chronology they yield may be imprecise.
  • Historians argue that the uniqueness of personal experience makes generalisations difficult. A large picture cannot be built from such micro-evidence and one witness is not sufficient enough to paint the larger picture.
  • Oral accounts are not directly related to the concerned issues.
  • The small individual experiences, which remain in memory are irrelevant to the unfolding of larger processes of history.

2. Oral history presents the story of poor and powerless, it helps in exploring the experiences of men and women who were earlier ignored or taken for granted. Oral history enabled historians to write about pain and anguish of people during the time of partition.

Question 30.
“Amidst all turmoil following March 1947, Gandhiji valiant efforts bore fruit to bring harmony among the people”. Justify the statement. (All India 2011)
Answer:
The following efforts of Gandhiji bore fruit to bring harmony among the people at the time of partition:

  • Gandhiji decided to stake everything to support his life long principle of non-violence and his conviction that people’s heart could be changed.
  • He moved from the village of Naokhali in East Bengal to the villages of Bihar and then to the riot affected slums of Calcutta and Delhi in a heroic effort to stop communal fights between Hindus and Muslims.
  • In October 1946, when Muslims in East Bengal targeted Hindus, Gandhiji visited the area, toured the villages on foot and convinced the Muslims to guarantee the safety of Hindus.
  • In places like Delhi, he tried to build a spirit of mutual trust and confidence between Hindus and Muslims.
  • On 28th November, 1947 when Gandhiji noticed that there were no Muslims on the Chandni Chowk road, he was very disappointed. He condenmed the mentality of those who wished to drive out every Muslim from the city, seeing them as Pakistani.

Question 31.
The India haters in Pakistan and Pakistan haters in India are the products of partition. Explain. (All India 2009)
Answer:
The India haters in Pakistan and Pakistan haters in India are the products of partition because of the following reasons:

  • It has been mistakenly believed that the loyalties of Indian Muslim lie with Pakistan.
  • Apart from this belief of pan-Islamic loyalties, other highly objectionable ideas come such as Muslims are cruel, bigoted, unclean, descendants of invaders. On the other hand, Hindus are believed to be kind, liberal, pure, children of the invaded.
  • According to RM Murphy, similar stereotypes has proliferated in Pakistan. For instance, some Pakistanis believe that Muslims are fair, brave, monotheists and non-vegetarians while Hindus are dark, cowardly, polythesists and vegetarian.
  • Some of these stereotypes predate partition but there is no doubt that they were immensely strengthened because of 1947.
  • Every myth of these constructions has been systematically criticised by historians. However in both the countries voices of hatred are still persisting.

Important Questions for Class 12 History

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CBSE Class 12 Sanskrit सङ्केताधारितम् अनुच्छेदलेखनम्

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CBSE Class 12 Sanskrit सङ्केताधारितम् अनुच्छेदलेखनम्

(क) चित्रमधिकृत्य
प्रियच्छात्रा:! भवद्भिः पूर्वम् दशमकक्षापर्यन्तं चित्रम् अधिकृत्य निर्दिष्टशब्दसूचीसाहाय्येन च चित्रवर्णनस्य अभ्यास: कृतः। तत्र केवलं पञ्चवाक्यानि एवं संस्कृतेन लिखित्वा कार्य पूर्णम् अभवत् किन्तु अधुना अनुबन्ध-लेखनम् अपेक्षितम्। अनुबन्ध: निबन्धस्य एव लघुरूप: भवति। अयम् अनुबन्ध: कथात्मकोऽपि भवितुम् शक्यते। अत्र अस्मिन् पुस्तके उदाहरणरूपेण दश अनुबन्धाः प्रस्तूयन्ते। तत्र चित्राणि अपि सन्ति, निर्दिष्टा शब्दसूची अपि वर्तते। अनुबन्धलेखने कल्पनायाः आश्रयः अनिवार्यः अस्ति। केवलं चित्रं दृष्ट्वा अथवा शब्दसूचीसहायतया एव अपि यदि सङ्केतः वर्तते तदापि भवन्तः कल्पनाया: प्रयोगेण अनुबन्धलेखने सिद्धा: भवन्तु इति अपेक्ष्यते।

प्रश्न:
अधोलिखितं चित्रम् दृष्ट्वा शब्द-सूचीसहायता अनुच्छेद लिखत
CBSE Class 12 Sanskrit सङ्केताधारितम् अनुच्छेदलेखनम् 1
शब्दसूची – समुद्रतटम्। कन्याकुमारी। स्वामिविवेकानन्दस्मारकम्। दर्शनीयम्। मण्डपम् नौकया प्राप्यम्। अतीव रम्यम् भव्यम्। ध्यानमण्डपम्। शिलाभि: निर्मितम्। शिलायाम् भगवद्ध्वज: विराजते। समुद्रे शिला: शोभन्ते। जनाः समुद्रे स्नानं कुर्वन्ति।
उत्तरम्:
अनुच्छेदः- भारते बहूनि दर्शनीयानि तीर्थस्थानानि सन्ति। दक्षिणदिशायाम् समुद्रमध्ये शिलाभि: निर्मितम् श्रीविवेकानन्द स्मारकम् तेषु अन्यतमम् अस्ति। पुरा स्वामी विवेकानन्द: समुद्र तीत्व अस्यामेव शिलायां ध्यानावस्थितः आसीत्। तदनु स: विदेशं गत्वा हिन्दूधर्मस्य संस्कृतेश्च प्रचार कर्तुम् निश्चयम् अकरोत्। स्वामी विवेकानन्द: लोक-कल्याणाय प्रवृत्त: आसीत्। अत: समुद्रस्य नातिदूरे शिलातले तस्य स्मारकम् विनिर्मितम्। एतत् भण्डपं नौकया प्राप्यं भवति। सम्पूर्ण भवनम् अतीव रम्यं त्र्यं च अस्ति। तत्र गत्वा चतुर्दिक्षु समुद्रतरङ्गा: कल्लोलं कुर्वन्ति। जनाः स्मारकम् अवतीर्य स्नानं कुर्वन्ति, पुष्पाणि गृहीत्वा पूजाकायं कुर्वन्ति। तत्र मध्ये ध्यानमण्डपम् अस्ति। एकस्मिन् भवने भगवद्ध्वजः विराजते। अस्माभिः एतत्स्मारकं प्राप्य स्वामिविवेकानन्द महाभागानां सन्देशा: ग्रहीतव्याः।

प्रश्न:
अधोलिखितं चित्रम् दृष्ट्वा शब्द-सूचीसहायता अनुच्छेद लिखते
CBSE Class 12 Sanskrit सङ्केताधारितम् अनुच्छेदलेखनम् 2
शब्दसूची – वसन्तस्य दृश्यम्। पुष्पाणाम् विकासः। सौरभाणां प्रसारः। वृक्षा: लता: पादपाः पुष्पिता:। कुसुमाकरः। विविधवर्णनं। पुष्पाणि। पुष्पेषु भ्रमराः। मधूनि पिबन्ति। पिक: कूजति। बालकाः। नदीतटे गायन्ति। अनुकुर्वन्ति। पल्लवाः। आमेषु मजर्यः। खगाः।
उत्तरम्:
अनुच्छेद : चित्रे वयं वसन्तस्य रम्यं मनोरम दृश्यम् अवलोकयाम:। वसन्त ऋतुराजः अस्ति। अयं कुसुमाकरः इति कथ्यते। सर्वत्र पुष्पाणां विकास:, सौरभाणां प्रसार:, भ्रमराणां गुञ्जनम् च अतीव आनन्दं यच्छन्ति। वने उपवने नानावर्णानि पुष्पाणि शोभन्ते। पुष्पाणाम् उपरि भ्रमरा: गुञ्जन्ति। ते च सानन्दं मधूनि पिबन्ति। मधुना मत्तः पिकः अपि मधुरं कूजति, बालकाः तस्य ध्वनिम् अनुकुर्वन्ति। जनाः उद्याने नदीतटेषु च प्रकृतिशोभा पश्यन्तः गायन्ति। इतस्तत: च भ्रमणं कुर्वन्ति। आम्रमञ्जरीषु मधुनः लोलुपा: चञ्चरीका: तदुपरि उड्डीयन्ते। सर्वतः पीत वसनं दधानः सर्षपः मनोरम दृश्यते। जनाः अपि पीतानि वस्त्राणि धारयन्ति। प्रकृति: नर्तकीव नृत्यन्ती प्रतीयते।

CBSE Class 12 Sanskrit सङ्केताधारितम् अनुच्छेदलेखनम् 3
शब्दसूची – महाभारत – प्रसङ्गः। यक्ष-युधिष्ठिर-वार्तालापः। तृषार्ताः पाण्डवाः। सरोवर। गतवन्तः। बकरूपधारी यक्षः। प्रश्नान् करोति। उत्तरं न दत्त्वा जलपानं चेष्टन्ते। चत्वार: चेतनारहिताः। अन्ते युधिष्ठिर: उत्तरं दत्वा यक्ष प्रसीदति।
उत्तरम्:
अनुच्छेद : प्रस्तुतं चित्रम् महाभारतस्य वनपर्वण: प्रसङ्गं स्मारयति। एकदा वने तृषार्ताः पाण्डवाः युधिष्ठिरस्य आज्ञया एकैकं कृत्वा जलस्य अन्वेषणे एकस्य सरोवरस्य तटम् अगच्छन्। तत्र स्थितस्य वृक्षस्य पृष्ठत: तै: बकरूपधारिण: यक्षस्य ध्वनिः श्रूयते-“तिष्ठ! मम प्रश्नानाम् उत्तरं दत्त्वैव जलं पिब। अन्यथा गतप्राणः भविष्यसि।” सहदेवः, नकुलः, भीमः अर्जुनः च यक्षस्य आज्ञां शृण्वन्ति किन्तु ते यक्षस्य प्रश्नानाम् उत्तरदानेन विना एवं जलं पातुं चेष्टन्ते। परिणामत: ते चत्वारः नि:संज्ञाः जाताः। यदा चतुषु भातृषु कोऽपि न अपरावर्तत, तदा युधिष्ठिरः स्वयं तत्र गतवान्। सः यक्षस्य प्रश्नानाम् सन्तोष-प्रदम् उत्तरं दत्तवान्। अनेन निजज्ञानबलात् बुद्धिकौशलात् च यक्ष प्रसन्नं कृत्वा युधिष्ठिरः स्वानुजान् अपि पुनर्जीवितान् अकरोत्। अयं यक्षयुधिष्ठिरसंवाद: अतीव ज्ञानप्रदः अस्ति।

CBSE Class 12 Sanskrit सङ्केताधारितम् अनुच्छेदलेखनम् 4
शब्दसूची – तपोवनम्। वृक्षस्य पृष्ठतः। राजा। चिन्तनमुद्रा। बालकः। सिंहस्य मुखम् उद्घाटयति। दन्तान् गणयति। द्वे महिले। एका वर्णयति। अन्या निवारयति सर्वदमन भरतः। दुष्यन्तस्य पुत्रः।
उत्तरम्:
अनुच्छेद : चित्रे वयम् तपोवनस्य दृश्यम् अवलोकयामः एकतः पुष्पिता: पादपाः तपोवनस्य रम्यतां वर्धयन्ति। वृक्षस्य पृष्ठतः आश्चर्यान्वितः राजा तिष्ठति। तस्य हस्ते खड्गः अस्ति। तत्र एकः अश्वोऽपि वर्तते। मध्ये एकः बालकः सिंहस्य मुखम् उद्घाटयति। द्वे महिले तं पश्यतः। बालकः सिंहशावकस्य दन्तान् गणयितुं चेष्टते। एका महिला तं निवारयति। सा कथयति-अलं क्रीडया, अन्यथा केसरिणी त्वां लङद्धयिष्यति। हठी वीरबालकः साहसं न परित्यजति। बालकं दृष्ट्वा राज्ञः हृदयं बालकं प्रति वात्सल्यभावेन पूर्णं भवति। वस्तुत: अयं बालकः तस्य एव पुत्रः अस्ति। बालकस्य नाम सर्वदमनः अस्ति। राज्ञः दुष्यन्तस्य अयं पुत्रः कालान्तरे ‘ भरतः’ इति नाम्ना चक्रवर्ती सम्राट् अभवत्।

CBSE Class 12 Sanskrit सङ्केताधारितम् अनुच्छेदलेखनम् 5
शब्दसूची – आदिकवि: वाल्मीकिः। तमसा नदी। व्याधेन। विद्धम्। क्रौञ्चम्। अपश्यत्। खगः अध: पतितः। क्रौञ्च्याः रोदनेन। द्रवीभूतम्। ऋषे: शोकः। श्लोकत्वम् आगतः। मा निषाद प्रतिष्ठां त्वमगम: शाश्वती समाः।
उत्तरम्:
अनुच्छेद : चित्रे वाल्मीके: रामायणस्य रचनायाः मूले वर्तमान: प्रसङ्गः चित्रितः। एकदा शिष्यैः सह स्नानार्थम् ऋषिः वाल्मीकिः तमसा नदीम् अगच्छत्। मार्गे सः व्याधेन विद्धम् एकं क्रौञ्चपक्षिण्म् अपश्यत्। व्याधः बाणेन काममोहितं कौञ्चयुग्मं प्राहरत्। तेन नरक्रौञ्चः भूमौ पतितः। सहचरस्य वियोगेन क्रौञ्ची व्याकुला अभवत्। सा च उच्चैः अरुदत्। तस्याः तीव्र रोदनं श्रुत्वा दु:खं च विलोक्य ऋषिः द्रवीभूतः। क्रौञ्च्याः क्रन्दनात् कवेः करुणाईहृदये जातः शोकः निम्नश्लोकत्वम् आगत :

मा निषाद प्रतिष्ठां त्वमगम: शाश्वतीः समाः।
यत्क्रौञ्चमिथुनादेकमवधी: काममोहितम्।।

अयं श्लोक: लोकसाहित्यस्य प्रथमा काव्यरचना। तदाधृत्य एव वाल्मीकिः आदिकविः तस्य च रचना आदिकाव्यं रामायणं रूपे प्रसिद्धे जाते।

CBSE Class 12 Sanskrit सङ्केताधारितम् अनुच्छेदलेखनम् 6
शब्दसूची – विपणेः दृश्यम्। फलविक्रेता। भूमौ। तस्य अग्रत: । विविधेषु पात्रेषु विविधानि फलानि। क्वचित्। कदलीफलानि, आम्रफलानि, हृढबीजफलानि फलविक्रेतुः हस्तयो:। फलानि। द्वौ बालकौ मूल्यं साधयत:। जना: रूप्यकाणि दत्त्वा फलानि क्रीणन्ति। कश्चित् नारङ्गफलानि इच्छति।
उत्तरम्:
अनुच्छेद : इदम् चित्रं फलविपणे: दृश्यम् उपस्थापयति। फलविक्रेता भूमौ तिष्ठति। तस्य अग्रे विविधपात्रेषु विविधानि फलानि सन्ति यथा-कदलीफलानि, नारङ्गफलानि, आम्रफलानि, दृढबीजफलानि इत्यादीनि। क्रेता प्रथम फलानि सम्यक्तया द्रष्टुम् इच्छति। विक्रेता हस्तयो: अभीष्टानि, फलानि गृहीत्वा दर्शयति। तदनन्तरं मूल्यं कथ्यते। विक्रेता कदाचित् अधिकं मूल्यम् वदति। क्रेता न्यूनमूल्येन क्रेतुम् इच्छति। अन्तत: तयोः मध्ये मूल्य स्थिरं भवति। क्रेता तदनु फलानां सङ्ख्यां वदति। सः स्वरुच्यनुसारं सुन्दराणि पक्वानि मधुराणि च फलानि अवचिनोति। मात्रानुसारं अवचितानां फलानां मूल्यं दत्त्वा क्रेता गच्छति, अन्ये क्रेतारः अपि तथैव कुर्वन्ति। तत्र क्वचित् बालिकाः, क्वचित् काऽपि महिला तिष्ठति। इत्थं फलानां क्रयणं विक्रयणं च नित्यं प्रचलतः।

CBSE Class 12 Sanskrit सङ्केताधारितम् अनुच्छेदलेखनम् 7
शब्दसूची – नगरे। नृपः। पशवः। वानरः। प्रियः। सुप्तः। अवीजयत्। मक्षिका। वारं वारं न्यवारयत्। पुन:। तत्रैव। अतिष्ठित्। खड्गप्रहार उड्डयते। नृपनासिका छिन्ना। मूर्खण मित्रता। नोचिता।
उत्तरम्:
अनुच्छेद : एकस्मिन् नगरे एकः नृपः आसीत्। तस्य भवने बहवः पशवः आसन्। ते तस्य सेवाम् अकुर्वन्। तेषु पशुषु एक: वानरः तस्य प्रियमित्रम् अभवत्। एकदा वानरः व्यजनेन नृपम् अवीजयत्। एका मक्षिका नृपस्य नासिकायाम् उपाविशत्। वानरः वारं वारं ताम् व्यजनेन निवारयति स्म। तथापि सा मक्षिका पुनः पुनः आगत्य तत्रैव अतिष्ठत्। तेन सः वानरः अतीव क्रुद्धः अभवत्। क्रोधेन विवेकः नश्यति। तां मक्षिकां हन्तुं वानरः तस्यामुपरि खड्ग प्रहारं कृतवान्। मक्षिका तु उड्डीय दूरम् अगच्छत् किन्तु नृपस्य नासिका छिन्ना अभवत्। सत्यमुक्तम्- मूर्खमित्रैः सह मित्रता नोचिता।

CBSE Class 12 Sanskrit सङ्केताधारितम् अनुच्छेदलेखनम् 8
शब्दसूची – चित्रे एका चतुर्दशवर्षीय बालकः। परितः केचन जनाः। तेषु भटै: बद्धः ग्रामाधिकारी। सः अधिकारी विधवां कारणं विना अपीडयत्। विधवा पृष्ठतः तिष्ठति। बद्धः पुरुषः बालकेन पृष्टम् उत्तरं न ददाति। बाल: अकथयत् ये जनान् पीडयन्ति ते मम शत्रवः। भो भटाः, इमं दण्डयत।
उत्तरम्:
अनुच्छेद :
चित्रे एक: उष्णीषवेष्टित: एकया अङ्गुल्या निर्दिशन् कोऽपि चतुर्दशवर्षीय बालकः शिलायाः उपरि उपविष्टः वर्तते। तस्य पृष्ठतः काचित् स्त्री असहाया इव करबद्धा तिष्ठति। बालकस्य समक्षे त्रिभिः भटै: बद्धः एकः ग्रामाधिकृत: दृश्यते। बद्धः पुरुषः एव तां विधवा स्त्री कारणं विना भृशम् अपीडयत्। सा स्त्री न्यायं वाञ्छति। बालकः भटानां स्वामी अस्ति। ग्रामाधिकृत: कुमारात् न बिभेति। कुमार: अपृच्छत्- भो:, किम् एतत् सत्यम् यत् त्वया असहाया स्त्री पीडिता। बद्धः पुरुषः तूष्णीं तिष्ठति। अन्ततोगत्वा कुमार: आदिशति- भो भटाः, इमं पुरुषं दण्डयत। ये जनान् पीडयन्ति ते मम शत्रवः। भटाः बालकस्य नीतिं प्रशंसन्ति, बद्धं पुरुषं दण्डयन्ति। धन्योऽयं नीतिमान् बालः!

CBSE Class 12 Sanskrit सङ्केताधारितम् अनुच्छेदलेखनम् 9
शब्दसूची – नद्याः दृश्यम्। मेघवर्षणम्। जलधारा: नदीं प्रति आगच्छन्ति। उपरि पर्वतीयभाग: वर्तते। तत्रापि जलबिन्दवः पतन्ति। नद्यः पवर्तेभ्यः एव प्रभवन्ति। ग्रीष्मे पर्वतोपरि स्थितं हिमम् प्रतीभूय जलरूपेण प्रवहति। नदीषु जनाः तरन्ति। तत्र मत्स्याः , मकराश्च अपि विद्यन्ते। नदीनां बहवः लाभाः।
उत्तरम्:
अनुच्छेद : इदं पर्वतीय-नद्याः दृश्यम् अस्ति। भारते बहवः गिरयः सन्ति। ते गिरयः मेधानां कलापं धृत्वा वृष्टिं सम्पादयन्ति। अनेकाः नद्यः प्रवहन्ति। शीतर्ती पर्वतोपरि पतितं हिमम् अपि ग्रीष्मकाले द्रवीभूय जलरूपेण प्रवहति। अनेकाः उपनद्यः महानदीं प्रविशन्ति। गङ्गा, यमुना, सरस्वती प्रभृतयः काश्चन अतीव प्रसिद्धा: नद्यः सन्ति। वर्षाकाले वृष्ट्या नद्या: जलेन कृषिक्षेत्राणि प्लावितानि सन्ति। कृषकाः कृषिकार्यं कृत्वा अन्नोत्पादनं कुर्वन्ति। नद्याः जलं प्रतिरुध्य सेतुं बद्धवा महाजलाशयान् निर्माय च जलसञ्चयः क्रियते। तेभ्य: जलाशयेभ्यः क्षेत्रेषु जलं प्रेष्यते। नद्यां बन्धं निर्माय विद्युत् उत्पाद्यते। एवं नदीनां बहवः लाभाः सन्ति।

CBSE Class 12 Sanskrit सङ्केताधारितम् अनुच्छेदलेखनम् 10
शब्दसूची – समवायो हि दुर्जयः। नीतिकथा। वृक्षे। चटका। प्रमत्त: गजः। चटकाया नीड भग्नम्। काष्ठकूट: खग: चटका दु:खं शृणोति। तां मणिकासमीपं नयति। तया: मण्डूक: मित्रम्। गजकर्णे मक्षिका शब्दं करोति। काष्ठकूट:
गजनयने स्फोटयति। अन्धः तृषार्त: गजः जलाशयं गत्वा गर्ते पतति।
उत्तरम्:
अनुच्छेद : अत्र चित्रे वृक्षे चटका दृश्यते। उन्मत्तेन गजेन तस्याः नीड त्रोटयित्वा तस्याः विशीर्णानि अण्डानि अधः पातितानि। चटका दु:खिता वर्तते। तस्य एव वृक्षस्य कोटरे काष्ठकूटः नाम खगः आसीत्। सः अपि चटकाया: दु:खेन दु:खितो भवति। काष्ठकूट: चटकां मक्षिकायाः समीपं नयति। ते सर्वे मिलित्वा मण्डूकस्य समीपं गच्छन्ति। तत्र ते गजस्य वधस्य उपायं कुर्वन्ति। प्रथमं मक्षिका गजस्य कर्णे शब्दं करोति। तेन गजस्य नयने निमीलिते भवतः। काष्ठकूट: चञ्च्वा गजनयने स्फोटयति। सम्प्रति अन्धः गजः जलाशयं गच्छति। मार्गे महान् गर्तः अस्ति। तदा मण्डूकः शब्दं करोति। तेन गजः गर्त जलाशयं मत्वा तस्मिन्नेव गर्ते पतति म्रियते च एवं सर्वे मिलित्वा कार्यं साधयन्ति। सत्युमुक्तम्- बहूनामप्यसाराणां समवायो हि दुर्जयः।

CBSE Class 12 Sanskrit सङ्केताधारितम् अनुच्छेदलेखनम् 11
शब्दसूची – नृपः रघुः। सैनिकाः। सुवर्णपात्राणि। वटुः। कौत्सः। निजपात्रे चतुर्दश-कोटी: स्वर्णमुद्रा: स्वीकरोति। रघुः सर्वं स्वर्णं दातुम् इच्छति किन्तु कौत्स: अधिकं न नयति।
उत्तरम्:
अनुच्छेदः- चित्रे वयं नृपं रघु पश्यामः। सः उभौ करौ प्रसार्य वटुकं कौत्सं कथयति यत् सुवर्णमुद्राभिः पूर्णानि अवशिष्टानि पात्राणि अपि स्वीकरोतु। किन्तु कौत्सः प्राक् चतुर्दशकोटी: स्वर्णमुद्राः स्वकरोति स्म। ततः अधिकाम् एकामपि सुवर्णमुद्रां नेतुं कौत्सः तत्परः न भवति। नृपः रघु: वारं वारं निवेदयति यत् एतत् सर्वं धनं तस्मै एव तेन कुबेरस्य भाण्डागारात् प्राप्तम्। वस्तुत: कौत्सस्य गुरुभक्त्या प्रभावितः रघुः कौत्सस्य कृते चतुर्दशकोटी: मुद्रा: प्राप्तुम् कुबेरम् आक्रान्तुं सङ्कल्पं कृतवान्। परन्तु महापराक्रमिण: महादानिनश्च रघो: आक्रमणात् पूर्वम् एव कुबेर: रात्रौ एव रघो: भाण्डागारे प्रभूतां स्वर्णवृष्टिम् अकरोत्। अन्ततोगत्वा रघुः चतुर्दशकोटी: स्वर्णमुद्राः एव प्रदाय वटुकौत्सं सत्कृतवान्। धन्या कौत्सस्य गुरुभक्तिः धन्या च रघोः उदारता।

(ख) निर्दिष्टशब्दसूची-साहाय्येन

प्रश्न:
अधोलिखितायाः शब्दसूचेः सहायतया निर्दिष्टविषये अनुबन्धम् लिखत।

1. मम प्रियः कविः ( भर्तृहरिः)
शब्दसूची – भर्तृहरिः। नीतिशतकम्। श्रृंगारशतकम्। वैराग्यशतकम्। विक्रमादित्यस्य अग्रजः। इत्सिंगस्य मतम्। गृहस्थ-वानप्रस्थयो: वारंवारं भ्रमणम्। सरसपदावली। नीति: वैराग्यम्। शृङ्गारादिविषयाः।।
उत्तरम्
मम प्रियः कविः (महाकविः भर्तृहरिः)

साहित्यसंगीतकलाविहीनः साक्षात्पशुपुच्छविषाणहीनः।
तृणं न खादन्नपि जीवमानस्तद्भागधेयं परमं पशूनाम्।।

महाकविः भर्तृहरिः संस्कृत-गीतिकाव्यस्य अद्वितीयः अलंकारः अस्ति। सः नीतिशतकम् श्रृंगारशतकम्, वैराग्यशतकम् च रचितवान्। सः महाराजविक्रमादित्यस्य अग्रजः आसीत् इति जनश्रुतिः अस्ति। तस्य स्थितिकालः षष्ठीशताब्द्याः उत्तरार्धः अस्ति । इत्सिग-मतानुसारं भर्तृहरिः बारम्बारं गृहस्थाश्रमस्य वानप्रस्थाश्रमस्य च मध्ये इतस्तत: पर्यभ्रमत्। शतकेषु वर्णिता: प्रसङ्गाः इत्सिंगमतस्य पुष्टिं कुर्वन्ति। एकैके शतके शतसंख्यकाः पद्याः सन्ति। नीतिशतके विद्या-वीरता-साहस-सत्संगति-मैत्री-परोपकार-धर्म-राजधर्म-इत्यादि वृत्तीनां सरसपदावल्यां वर्णनम् अस्ति। शृङ्गारशतके मधुरशैल्यां शृङ्गारस्य प्रमुख: विषयः वर्तते परं शनैः शनैः जीवनस्य अस्थिरतां प्रदर्य शान्तरसस्य तुलनायां तस्य तुच्छता प्रकटीक्रियते। वैराग्यशतके संसारस्य निस्सारताम्, मोक्षसोपानरूपस्य वैराग्यस्य च महत्त्वं प्रदर्यते।

2. सदाचारः
शब्दसूची – यद्यदाचरति श्रेष्ठः। संताम् आचरणम्। परमोधर्मः। शीलम्। शिष्टः। सभ्य: विनीत: नम्रः। मातृवत् परदारेषु। परद्रव्येषु लोष्ठवत्। श्रीराम: वाक्संयमी। मधुरभाषी।
उत्तरम्:
सदाचारः। यद्यदाचरति श्रेष्ठ स्वत्तदेवेतरो जनः।
स यत्प्रमाणं कुरुते लोकस्तदनुवर्तते।।

श्रेष्ठजनानाम् आचरणम् एव सदाचार: शीलं वा कथ्यते सताम् आचारः सदाचारः। सत्पुरुषाः शास्त्रविहितं प्रशस्तं कार्य कुर्वते, इतरे जनाः तदेव अनुसरन्ति। ‘आचार: परमो धर्म:’ इति शास्त्रविहितं मतम्। शीलमेव जीवस्य परमं भूषणम् अस्ति। सदाचारी जन: शिष्टः, सभ्यः, विनीतः, विनम्रः च भवति। सः परदारेषु मातृवत् परधनेषु लोष्ठवत्, सर्वभूतेषु च आत्मवत् पश्यति। वाक्यसंयतो भूत्वा स कामक्रोधादीन् विकारान् निगृह्य कर्मसु प्रवर्तते। सदाचारस्य प्रतिपालनेन एव भगवान् श्रीरामः मर्यादापुरुषोत्तमः इति विश्रुतः भवति। वयं तादृशं सुशीलं कामयामहे येन जगत् विनम्र भवेत्।

3. श्रीमद्भगवद्गीता
शब्दसूची – महाभारते दीपस्तम्भः। श्रीकृष्णार्जुनसंवादः। नष्टो मोहः कर्मोपदेशः। निष्काम-कर्म। कर्मण्येवाधिकारस्ते। असक्तः कुरु कर्माणि सङ्ग त्यक्त्वा धनञ्जय। कृष्णस्य विराट् रूपं गीतायाम् उपलभ्यते।
उत्तरम्:
श्रीमद्भगवद्गीता
संस्कृतभाषायां महर्षिवेदव्यासेन विरचितस्य महाभारते श्रीमद्भगवद्गीता दीपस्तम्भ: इव स्थिता प्रतीयते। धर्मक्षेत्रे कुरुक्षेत्रे समवेतानां युयुत्सुकां पाण्डवानां कौरवाणां च मध्ये युद्धविरतम् अर्जुनम् अभिलक्ष्य भगवान् श्रीकृष्णस्य उपदेशेन तस्य मोहम् अनश्यत् तत्सर्वं श्रीमद्भगवद्गीतायां निबद्धम् अस्ति। श्रीमद्भगवद्गीतायां कर्मयोगस्य उपदेशः वर्तते। अर्जुनस्य विषाद: जीवन-संग्रामात् निवर्तकस्य जनसामान्यस्य विषाद: अस्ति। कृष्णस्य उपदेशैः, भगवत्कृपया, विराट्रूपदर्शनेन च सः विषाद नष्टः। सङ्ग्रहितो भूत्वा, कर्मफलेषु आसक्ति त्यक्त्वा कर्माणि कुर्वन् जनः मोक्षं प्राप्नोति, यथा गीतायाम् उपदिष्टम् अस्ति– असक्तः कुरु कर्माणि, सङ्गं त्यक्त्वा धनञ्जय।

4. कालिदासः
शब्दसूची – महाकविः। संस्कृतसाहित्यस्य सर्वश्रेष्ठः कविः। नाटककारः। महाकाव्यद्वयम्-रघुवंशम्, कुमारसम्भवम्। खण्डकाव्यद्वयम्-मेघदूतम्, ऋतुसंहारः। नाटकत्रयम्-अभिज्ञानशाकुन्तलम्, विक्रमोर्वशीयम्, मालविकाग्निमित्रम्। शृङ्गाररसस्य विशेष: परिपाकः। तपोवनवर्णनम्। भारतीयसंस्कृतेः तत्त्वानां चित्रणम्।।
उत्तरम्:
कालिदासः
महाकविः कालिदासः संस्कृतवाङ्मयस्य सर्वश्रेष्ठ: महाकविः अस्ति। असौ नाटककारः, महाकाव्यनिर्माता, गीतिकाव्यविर चयिता च आसीत्। कवेः नवनवोन्मेषशालिनी प्रतिभा सर्वतोमुखी आसीत्।। महाकविः कालिदासः विक्रमादित्यस्य नवरत्नेषु एकं रत्नम् आसीत् इति प्रसिद्धिः अस्ति। विक्रमस्य राजधानी उज्जयिनी आसीत्। महाकालस्य अस्याः नगर्याः भव्यं वर्णनं महाकविना स्वगीतिकाव्ये मेघदूते कृतं वर्तते। विक्रमादित्यस्य सभायां कालिदासस्य नाटकानां मञ्चनं सञ्जातम्। विक्रमोर्वशीयनाटके कालिदासः परोक्षतया विक्रमादित्यस्य कीर्तिम् चिरस्थायिनी चकार। विक्रमादित्यः खीष्टाब्दात् 57 वर्ष-पूर्वं उज्जयिन्यां नवसंवत्सरं प्रारब्धवान् इति प्रसिद्धिः वर्तते।।

5. दीपावली
शब्दसूची – कार्तिकमासस्य अमावस्यायाम्। सुधालिप्तानि भवनानि। भगवतः रामस्य गृहागमने दीपमालिकायाः परम्परा। शुद्धवस्त्रधाारणम्। मिष्टान्नसेवनम्। स्फोटकानाम् आस्फोटनम्। दीपानां, मोमवर्तिकानां प्रयोगः। होमादिकम्। लक्ष्मीपूजनम्। नव-बही-खाता-आरम्भः। सर्वातिशय: उत्साहः।
उत्तरम्
दीपावली भारतीयजनानां दीपावलि महोत्सवः तैः सर्वातिशयेन उत्साहन मान्यते। अयम् उत्सवः कार्तिकमासस्य अमावस्यायाः दिने भवति। श्रुयते यत् अस्मिन् दिने भगवान् श्रीरामचन्द्रः रावणं हत्वा अयोध्या निवृत्तः आसीत्। तदा च तस्य स्वागतं लोकैः रात्रौ दीपमालां कृत्वा कृतम् आसीत्। ततः प्रभृति जनाः तस्मिन् एव दिने प्रतिवर्ष स्वगृहेषु दीपमालां कुर्वन्ति। अस्मिन् अवसरे रात्रौ लक्ष्मीपूजनम् क्रियते। केचित् गृहेषु होमादि अपि सम्पाद्यते। वणिजः प्रायेण अस्मात् दिनात् एव ‘बही-खाता’ इत्यस्य आरम्भं कुर्वन्ति, अत: ते ‘बही-खाता’–पूजनं कुर्वन्ति। उद्योगिनः यन्त्रपूजनं कुर्वन्ति।

6. जननी जन्मभूमिश्च स्वर्गादपि गरीयसी
शब्दसूची – जननी यस्याः उदरात् (कुक्षेः) जन्म गृह्यते। सा पूज्यतमा। माता सन्ततेः आजीवनम् सुख-समृद्धेः चिन्तनम् करोति। स्वर्गसुखम् मातुः सुखात् अल्पतरम् इत्थम् स: देशः, तत् राष्ट्रम्, सः समाजः अपि मातृवत् भवति। जन्मभूमिः मातृभूमिः कथ्यते। रामेण उक्तम्-अपि सुवर्णमयी लङका न में लक्ष्मण रोचते। जननी जन्मभूमिश्च स्वर्गादपि गरीयसी।
उत्तरम्
जननी जन्मभूमिश्च स्वर्गादपि गरीयसी अस्याः सूक्ते: अयम् अर्थः यत् जननी अर्थात् माता, जन्मभूमिश्च अर्थात् स: देश: यत्र जनस्य जन्म भवति, एते उभे स्वर्गात् स्वर्गस्थानात् अपि गरीयस्यौ अर्थात् गुरुतरे स्तः। लोके इदं विश्रुतं भवति यत् स्वर्गलोके दु:खानां सर्वविधक्लेशानां च निराकरणं भवति, तत्र सर्वसुखानाम् उद्भवो भवति। दु:खस्थानं नरकं विहाय सर्वे जनाः सुखस्य परमोच्चं स्थानं स्वर्ग कामयन्ते। परम् अत्र अन्यत् एवं प्रतिपादितम्। स्वर्गतः अपि द्वे महत्तरे स्तः, एका जन्मदात्री जननी, अपरा च जन्मभूमिः एताभ्यां जनः स्वर्गादपि अधिकं सुखं प्राप्नोति इति अभिप्राय:।।

7. मम प्रियं पुस्तकम् ( उत्तररामचरितम्)
शब्दसूची – उत्तररामचरितम्। भवभूतेः नाटकम्। करुणरसप्राधान्यम्। पत्नी परित्यागस्य दोषः निराकृतः। छायासीता-अङ्कः । तमसा-मुरला इत्यादि नदीपात्राणि। वनदेवता। एकः रस: करुण: एव। भवभूति: तादृशः कविः यं ब्राह्मणं वाग् वश्येव अनुवर्तते। सप्त अङ्का । चित्रदर्शनम्। गर्भाङकनाटकम्।
उत्तरम्:
मम प्रियं पुस्तकम्
मम प्रियं पुस्तकम् उत्तररामचरितम् अस्ति। इदं नाटकं महाकवि-भवभूते: रचना अस्ति। अत्र करुणरसस्य प्राधान्यं वर्तते। ‘एको रसः करुण एव’ इति अत्र प्रतिपादितम् अन्यरसानां समावेशः करुणे एव भवति। भवभूतिः तादृशः कविः अस्ति यं ब्रह्माणमियं देवी वाग्वश्येवानुवर्तते। उत्तररामचरिते भाषा प्रौढता, औदार्यम् अर्थगौरवम्-सर्वे गुणाः समाविष्टाः सन्ति। अस्मिन् नाटके सप्त अङ्का सन्ति। तत्र चित्रदर्शनं, छायासीता, गर्भाकनाटकम् च इति दृश्यानि कवेः स्वकल्पनाप्रसूतानि सन्ति ।

8. महाकविः वाल्मीकिः
शब्दसूची – आदिकविः। रामायणस्य प्रणेताः आश्रमे सीतायाः, लव-कुशयोश्च परिपालकः। शस्त्रास्त्र-शिक्षणं, रामायण कथाज्ञानम् च लवकुशकृते अकारयत्। महर्षिः। भगवान्। रामायणम् उपजीव्यकाव्यम्। नीतीनां, मर्यादा पूर्णव्यवहारस्य आदर्शः।
उत्तरम्:
महाकविः वाल्मीकिः आदिकवि: वाल्मीकिः एवं मम प्रियः कविः अस्ति। तस्य जन्म नाम रत्नाकरः आसीत्। तपश्चर्यां कुर्वतः तस्य उपरि पुत्तिकाभिः मृद्-आवरणम् (वल्मीकम्) अकरोत्। ततः अयं लोके वाल्मीकिः इति विश्रुतः अभवत्। एकदा तमसातीरे परिभ्रमता अनेन व्याधेन विध्यमानम् क्रौञ्चमेकं दृष्टम्। तत् करुणं दृश्यं दृष्ट्वा सः निम्नपद्यमाध्यमेन व्याधं शापग्रस्तं कृतवान्

मा निषाद प्रतिष्ठां त्वमगम: शाश्वतीः समाः।
यत्क्रौञ्चमिथुनादेकमवधी: काममोहितम्।।

9. कोऽपि महापुरुषः ( श्री कृष्णः)
शब्दसूची – वसुदेवसुतः देवकीनन्दनः वासुदेवः श्रीकृष्णः मम आदर्शः। कारावासे जन्म। नन्दगृहे रक्षणम्। द्वैमातुरः। यशोधानन्दनः। असुरविनाशं। कंस-जरासन्ध-शिशुपालादीनां वधं। महाभारते। सारथिकार्यग्रहणं। न्यायपक्षस्य विजयः। मोहग्रस्ताय अर्जुनाय निष्कामकर्मोपदेशः। श्रीमद्भगवद्गीतायाः गायकः।
उत्तरम्:
कोऽपि महापुरुषः ( श्री कृष्णः) भगवान् श्रीकृष्णः षोडश-कला-सम्पूर्णः आसीत्। भीष्मपितामहः अपि तस्य महिमानम् अजानात्। युधिष्ठिरस्य राजसूययज्ञे श्रीकृष्णस्य एवं प्रथमपूजा सर्वैः स्वीकृता। शिशुपालः तस्य विराधम् अकरोत्। यदा तस्य विरोधस्य पराकाष्ठा जाता तदा श्रीकृष्णः तस्य शिरच्छेदनं कृतवान्। महाभारते पाण्डवानां विजयस्य एकमात्रं कारणम् श्रीकृष्णः एव आसीत्। सः कर्मयोगी आसीत्। सः जरासन्धस्य वधं कृत्वा तत्र बन्दीगृहे निर्दोषं स्थापितान् संख्यातीत् जनान्, तस्य अन्तः पुरे च स्थिताः महिलाश्च मुक्ताः कारयति स्म।

10. गणतन्त्र दिवसः
शब्दसूची – जनवरीमासस्य षड्विंशतितमे दिवसे। संविधानस्य आरम्भ दिवसः। 1950 तमे वर्षे। नवदिल्लयाम्। महान् |
संरम्भ: विविधाः दृश्यावल्याः। सैनिकै: राष्ट्रपति सम्माने। प्रणामाः। स्वतन्त्रतायाः आन्दोलनकाले लवपुरे 1924 तमे वर्षे रावीतटे नेहरूमहाभागैः ‘पूर्णस्वतन्त्रताप्राप्तिः’ इति स्वतन्त्रता-आन्दोलनस्य लक्ष्य निर्धारितम्। अस्माभिः संविधानस्य परिपालनार्थं सङ्कल्पः करणीयः।
उत्तरम्:
गणतन्त्रदिवस समारोहः
जनवरीमासस्य षड्-विंशति-तमे दिवसे भारते गणतन्त्रदिवसः प्रतिवर्ष सम्पद्यते। अस्य दिवसस्य इव अस्याः तिथेः अपि महत्त्वं वर्तते। स्वतन्त्रतायाः आन्दोलनकाले 1929 मते ईसवीये वर्षे लवपुरे जवाहरलालनेहरूमहोदयस्य सभापतित्वे कांग्रेसदलस्य अधिवेशने जनवरीमासस्य 26 तमे दिवसे ‘पूर्णस्वतन्त्रताप्राप्तिः एव अस्माकं लक्ष्यम् अस्ति, इति घोषणा सञ्जाता। अतः सः एव दिवस: संविधानस्य अपि आरम्भदिवसः स्वीकृतः नेतृभिः।

11. होलिका-महोत्सवः
शब्दसूची – फाल्गुनमासस्य पूर्णिमायाम्। ‘होली’ इति नाम्ना प्रसिद्धम्। द्विदिनात्मकः महोत्सवः प्रथमदिने होलिका-पूजन | दहनं च। द्वितीयदिने परस्परं गुलाल, रागम् आदाय अन्योन्यमुखानि कपोलानि लिम्पन्ति। पिचकारी-माध्यमेन रागोऽपि क्षिप्यते। जनाः विविधाः विनोदपूर्णाः संगीतमयाः च चेष्टाः कुर्वन्ति।
उत्तरम्:
होलिकोत्सवः
होलिका महोत्सवः फाल्गुनमासस्य पूर्णिमायां समायोज्यते। इदं पर्व ‘होली’ इत्यभिधानेन प्रसिद्धमस्ति। परम्परया अयमुत्सवः वसन्तोत्सवः अपि कथ्यते यतः इदं पर्व अपि वसन्तर्ती समायोज्यते। होलिकादहनं कृत्वा जनाः अग्रिमे दिने परस्परं ‘गुलाल’ इति रागम् आदाय अन्योन्यमुखानि लिम्पन्ति, पिचकारी माध्यमेन द्रवं रागमपि क्षिपन्ति, विविधाश्च विनोदपूर्णाः चेष्टाः कुर्वन्ति, इत्यनेन रङ्गक्रीडां विधाय सायंकाले मिलित्वा मिष्टान्नसेवनमापि कुर्वन्ति।

NCERT Solutions for Class 12 Sanskrit

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Important Questions for Class 12 History Chapter 15 Framing the Constitution (The Beginning of a New Era)

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Important Questions for Class 12 History Chapter 15 Framing the Constitution (The Beginning of a New Era)

Important Questions for Class 12 History Chapter 15 – 2 Marks Questions

Question 1.
Why is ‘Objective Resolution’ of Nehru considered as momentous resolution? Give two reasons? (HOTS; Delhi 2013)
Answer:
Objective resolution was considered as momentous resolution because:

  • It outlined the defining ideals of Constitution of Independent India and provided framework within which constitution making was to be proceeded.
  • It proclaimed India to be an “Independent Sovereign Republic”.

Question 2.
Mention any two arguments given by Balakrishna Sharma for greater power to the centre. (All India 2013)
Answer:
Balakrishna Sharma said following things in favour for greater power to the centre:

  • He said strong centre could plan for well-being of the country and it can mobilise the available economic resources of the country.
  • Strong centre can establish proper administration and defend the country against foreign invasion.

Important Questions for Class 12 History Chapter 15 – 4 Marks Questions

Question 3.
Describe the different arguments made in favour of protection on of depressed class in the Constituent Assembly. (All India 2017)
Answer:
The following arguments were made in favour of protection of depressed classes in the Constituent Assembly:
1. It was realised that the depressed classes especially tribals and untouchables needed special attention and safeguards to raise their status in society and provide them equality. But some members of the depressed class emphasised that the problem of the “Untouchables” could not be resolved through protection and safeguards alone.

2. These members believed that the disabilities of the depressed class were caused by the social norms and the moral values of caste divided society. The depressed class had been left in isolation with this belief that they are not born to be fit in the civil society.

Their suffering was due to their systematic marginalisation. They had no aceess to education and also had no share in the administration.
Thus, in the Constituent Assembly many recognised that social discrimination could not solve only through constitutional legislation, there had to be a change in the attitudes within society.

Question 4.
“The discussions within the Constituent Assembly were also influenced by the opinions expressed by the public”.
Examine the statement. (HOTS; Delhi 2012)
Answer:
The public opinion had a considerable effect on the discussions of the Constituent Assembly that were:

  • There was public debate on all the resolutions.
  • The newspapers reported the arguments presented by different members on any issue.
  • Criticisms and counter criticism in the press shaped the nature of the consensus that was ultimately reached on specific issues.
  • Suggestions from the public was also welcomed which created a sense of collective participation.
  • Many linguistic minorities demanded protection of their mother tongue. Religious minorities asked for special safeguards.
  • The groups low caste or dalits demanded an end to ill-treatments by upper caste people and reservation of separate seats on the basis of their population in legislatures.
  • Important issues of cultural rights and social justice raised in the public discussions were debated in the Assembly.
  • In the same way, groups of religious minorities came forward and asked for special safeguards.

Question 5.
“A communist member Somnath Lahiri saw the dark hand of British . imperialism hanging over the deliberations of the Constituent Assembly”. Examine the statement and give your own views in support of your answer. (All India 2012)
Answer:
The statement implies that Somnath Lahiri saw the influence of the British imperialism over the deliberations of the Constituent Assembly.
As a result, he urged the members to completely free themselves from the influences of imperial rule. During the winter of 1946-47, when the assembly was constituted, the British were still in India.

An interim administration headed by • Jawaharlal Nehru was in place, but it could only operate under the directions of the viceroy and the British Government in London. Lahiri exhorted his colleagues to realise that the Constituent Assembly was British made and was working on the British plans as the British should like it to be worked out.

Important Questions for Class 12 History Chapter 15 – 8 Marks Questions

Question 6.
“Within the Constituent Assembly of India the language issue was intensely debated”. Examine the views put forward by members of the assembly on the issue. (All India 2016)
Answer:
The language issue was intensely debated in the Constituent Assembly. R.V. Dhulekar, Shrimati G. Durgabai, Shri Shankarrao Deo and T.A. Ramalingam Chettiar were prominent members of the Constituent Assembly who gave their remarkable views on language.

R.V. Dhulekar, a Congressman from the United Provinces, made a strong plea that Hindi must be used as the language of constitution making. He stated! “People who are present in this house to fashion a constitution for India and do not know Hindustani are not worthy to be member of this Assembly. They better leave”. Many members of the Assembly became agitated and the controversy regarding language continued over the next three years.

After three years, the Language Committee of the Constituent Assembly had produced its report. The committee tried to give a compromise formula to resolve the dead lock between those who advocated Hindi as the national language and those who opposed it. The committee suggested Hindi in the Devanagari script would be the official language along with English. But this solution could not satisfy members like Dhulekar who wanted to see Hindi as the national language of India.

Shrimati G. Durgabai from Madras expressed her worry that this controversy made the non-Hindi speaking people to think that other powerful languages of India would be neglected and it was an obstacle for the composite culture of our nation. She informed the House that the opposition in the South against Hindi was very strong. She said “The opponents feel perhaps justly that this propaganda for Hindi cuts at the very root of the provincial languages”.

She along with many others had obeyed the call of Mahatma Gandhi and carried on Hindi propaganda in the South. She accepted Hindustani as the language of the people. But its character was changed as it took many Urdu words and regional vocabulary. Durgabai believed this composite character of Hindustani was bound to create anxieties and fears among different language groups.

Shri Shankarrao Deo, a member from Bombay, a Congressman and a follower of Mahatma Gandhi accepted Hindustani as a language of the nation. But he warned “If you want my whole-hearted support (for Hindi) you must not do now any thing which may arise my suspicions and which will strengthen my fears”.

T. A. Ramalingam Chettiar from Madras suggested that whatever was done had to be done with caution. Because the cause of Hindi would not be helped if it was pushed too aggressively. There would be fear and bitter feelings among people if Hindi was applied forcefully, although the people might be unjustified. So he believed that to form a united nation “there should be mutual adjustment and no question of forcing things on people.”In this way different members of the Constituent Assembly expressed their views regarding the controversy.

Question 7.
How did Constituent Assembly of India protected the powers of the Central government? Explain. (All India 2016)
Answer:
India achieved its independence on 15 th August, 1947 and was also divided into two parts i.e. India and Pakistan. Before the partition, the Constituent Assembly did not communicated itself in commendation of a strong Central Government, but after the declaration of partition on 3rd June, 1947, Constituent Assembly considered itself free from all restrictions inflicted by Cabinet Mission and political pressures. Constituent Assembly decided to opt for a federation alongwith strong centre. There were arguments in favour of strong provinces which evoked powerful reactions from the leaders who preferred strong centre.

Dr BR Ambedkar and Jawaharalal Nehru propounded a strong Central Government for India. They mentioned to the riots’and violences that were fearing the nation apart and stated that only a strong centre can stop the communal disharmony. Balakrishna Sharma focussed on length of the nation and stated that only a centre, which was powerful could plan for the well-being of the country. Strong centre would help in mobilising available economic resources and proper administration was possible only through strong centre only.

In spite of arguments of the centre has likely to break or inefficiency of the centre, the rights of the states were most impressively defended by K Santhanam from Madras. Also the decision of the Constituent Assembly to have a strong centre was occasioned by the situations in which it was taken. Most of the members felt that strong centre was the need of the hour. It was necessary to ensure peace, prosperity and political stability, and hence, Gopalaswami Ayyangar requested to make centre as strong as possible.

Important Questions for Class 12 History Chapter 15  Source Based Question

Question 8.
“There cannot be any divided loyalty”
Govind Ballabh Pant argued that in order to become loyal citizens people had to stop focusing only on the community and the self.
For the success of democracy one must train himself in the art of self discipline.

In Democracies one should care less for himself and more for others. There cannot be any divided loyalty. All loyalties must exclusively be centred round the state. If in a democracy, you create rival loyalties, or you create a system in which any individual or group, instead of suppressing his extravagance, cares nought for larger or other interests, then democracy is doomed.

  1. Why did Govind Ballabh Pant lay more stress on the art of self-discipline?
  2. What was considered important for the success of democracy?
  3. ‘In Democracies one should care less for himself and more for other.’ Give your views on this philosopy. (Delhi 2015)

Answer:
1. Govind Ballabh Pant suggested that to make democracy successful, one should be self disciplined. Individual should care less for personal gain and focus more on collective benefit or for others gain in democracy. So a trait of sacrifice should be present in every citizen and this character of sacrifice can be learned through discipline.

2. For success of democracy, there should not be divided loyality and it must be centred round the state and citizens
should care less for themselves and more for fellow citizens.

3. This philosophy of democracy suggests that one should be considerate towards other, nothing should be done for personal gain which can harm the interest of other person or large section of people. This philosophy promotes the feeling of people centric benefits instead of individual centric.

Question 9.
‘We are not just going to copy’
We say that it is our firm and solemn resolve to have an independent sovereign republic. India is bound to be sovereign, it is bound to be independent and it is bound to be a republic….Now, some friends have raised the question “Why have you not put in the word ‘democratic’ here.?” Well, I told them that it is conceivable of course, that a republic may not be democratic but the whole of our past is witness to this fact that we stand for democratic institutions.

Obviously, we are aiming at democracy and nothing less than a democracy. What form of democracy, what shape it might take is another matter. The democracies of the present day, many of them in Europe and elsewhere, have played a great part in the world’s progress. Yet it may be doubtful if those democracies may not have to change their shape somewhat before long if they have to remain completely democratic. We are not going just to copy, I hope, a certain democratic procedure or an institution of a so-called democratic country.

We may improve upon it. In any event whatever system of government we may establish here must fit in with the temper of our people and be acceptable to them. We stand for democracy. It will be for this House to determine what shape to be given to that democracy, the fullest democracy, I hope the House will notice that in this resolution, although we have not used the word “democratic” because we thought it is obvious that the word “republic” contains that word and we have done something much more than using the word.

We have given the content of democracy in this resolution and not only the content of democracy but the context, also, if I may say so of economic democracy in this resolution. Others might take objection to this Resolution on the grounds that we have not said that it should be a Socialist State.

Well, I stand for Socialism and, I hope, India will stand for Socialism and that India will go towards the Constitution of a Socialist State and I do believe that the whole world will have to go that way.

  1. Explain why Nehru did not mention the word democratic in the resolution.
  2. Mention the three basic features of the constitution given in the above passage.
  3. On what kind of socialism did Nehru give stress to? (Delhi 2014)

Answer”
1. The explanation given by Jawaharlal Nehru for not using the term ‘Democratic’ in the objective resolution is as follows:
(a) It was thought by the makers of the constitution that the word ‘republic’ contains that word.
(b) They did not want to use unnecessary and redundant words.
(c) They had given the content of democracy in the resolution especially democracy.

2. Three basic features of the constitution given in above passage are independent, sovereign, republic.

3. Nehru was supporter of Socialism and he said that India would stand for socialism, where every citizen would be provided equal opportunities for growth and development. There would be economic democracy and economic justice.

Question 10.
“British element is gone but they have left the mischief behind”
Sardar Vallabh Bhai Patel said
It is no use saying that we ask for separate electorates, because it is good for us. We have heard it long enough. We have heard it for years, and as a result of this agitation we are now a separate nation… Can you show me one free country where there are separate electorates? If so, I shall be prepared to accept it. But in this unfortunate country if this separate electorate is going to be persisted in, even after the division of the country, woe betide the country; it is not worth living in. Therefore, I say, it is not for my good alone, it is for your own good that I say it, forget the past.

One day, we may be united… The British element is gone, but they have left the mischief behind. We do not want to perpetuate that mischief. (Hear, hear).

When the British introduced this element they had not expected that they have to go so soon. They wanted it for their easy administration. That is all right. But they have left the legacy behind. Are we to get out of it or not?

  1. Why are separate electorates considered as a mischief?
  2. State the arguments given by Sardar Vallabh Bhai Patel for building political unity and forging a nation.
  3. How did the philosophy of separate electorates result in a separate nation? (All India 2015)

or

  1. Explain Sardar Vallabh Bhai Patel’s views on the issue of separate electorate system.
  2. In what ways did Sardar Patel explain that “The British element is gone, but they have left the mischief behind”?
  3. Mention the reasons behind Sardar Patel urging the assembly members to get rid of separate electorate.
    (All India 2014)

Answer:
1. Separate electorate was considered as a mischief because in the name of giving representation to minorities and making the administration easy, Britishers divided two major communities of India politically. Later, this issue of separate electorate played an important role in partition of the country.

2. Patel said in an assembly that there was no provision of separate electorate in any free country. He further said that separate electorate could not deliver any good, so it was better to forget it. For political unity he said, this electorate had to go. British introduced the policy of divide and rule. After the British we should reject.it for the sake of the unity of our nation.
3. Philosophy of separate electorate saw Hindus and Muslims as separate political identity. It believed that interest of Hindus and Muslims were not common, so to represent Muslims there should be a Muslim only, similarly for Hindu only Hindu should represent. This policy separated the people on the basis of religion and started to keep one community isolated from another politically. It was there to divide Indians on the basis of religion.
or
Answer:
1. According to Sardar Vallabh Bhai Patel, separate electorates would be suicidal to the minorities and would do tremendous harm to them and the whole society. It was a demand that had turned one community against another, divided the nation, caused , bloodshed and led to the tragic
partition of the country. He argued that it would permanently isolate the minorities, make them vulnerable and deprive them of any effective say within the government.

2. Sardar Patel said that British policy of separate electorate created a division in the people of India and divided them on the basis of religion. This division culminated with partition of the country Britishers have left the country but negative consequence of that policy still haunted Indians.

3. Sardar Patel was urging for no separate electorates because it may harm the unity of the country as such no country is in the world having separate electorates.

Question 11.
“That is Very Good, Sir-Bold Words, Noble Words”
Somnath Lahiri said: Well, Sir, I must congratulate Pandit Nehru for the fine expression he gave to the spirit of the Indian people when he said that no imposition from the British will be accepted by the Indian people.
Imposition would be resented and objected to he said and he added that if need be we will walk to the valley of struggle. That is very good, Sir-bold words, noble words.
But the point is to see when and how are you going to apply that challenge. Well Sir the point is that the imposition is here right now.

Not only has the British plan made any future Constitution dependent on a treaty satisfactory to the Britisher but it suggests that for every little difference you will have to run to the Federal Court or dance attendance there in England or to call on the British Prime Minister Clement Attlee or someone else.

Not only is it a fact that this Constituent Assembly, whatever plans we may be hatching, we are under the shadow of British guns. British Army, their economic and financial stranglehold-which means that the final power is still in the British hands and the question of power has not yet been finally decided which means the future is not yet completely in our hands. Not only that, but the statements made by Attlee and others recently have made it clear that if need be, they will even threaten you with division entirely.

This means Sir there is no freedom in this country. As Sardar Vallabh Bhai Patel put in some days ago, we have freedom only to fight among ourselves. That is the only freedom we have got…therefore, our humble suggestion is that it is not a question of getting something by working out this plan but to declare independence here and now and call upon the Interim government and call upon the people of India to stop fratricidal warfare and look out against its enemy, which still had the whip in hand of the British imperialism and go together to fight it and then resolve our claim afterward when we will be free.

  1. Why did Somnath Lahiri congratulate Pandit Nehru?
  2. Explain why Somnath feels that the absence of constitution will mean dependence on the British.
  3. How did he feel that final power was still in hands of the British?
  4. Explain the views of Sardar Vallabh Bhai Patel.
  5. Explain the intentions of the British in not framing the Constitution beforehand. What did they want?
    (Delhi 2012, 2010)

Answer:
1. Somnath Lahiri congratulated Pandit Nehru for his fine expression that gave to the Indian people, when he said no imposition from the British would be accepted by India which was his true spirit for India and its free people.

2. Somnath felt that in the absence of Constitution for every basic law and rule, there would be need to refer to British government. British would obviously want control over the governance. So he felt that Indians should draft their own constitution according to their will and will of people, so it could be truly, independent and free.

3. Somnath Lahiri feels that although we have made our constitution but still we are not free. We are under British Army, British economic and financial stranglehold and this means that final power is still in the hands of British.

4. According to Sardar Vallabh Bhai Patel, we had freedom to fight only among ourselves, there was as such no freedom in our country.

5. The intentions of the British in not framing the constitution beforehand can be explained in the context of their convenience. It suggested that for every little difference, one would have to run to the federal court or act on the rule of the Government of England or to call on the British Prime Minister Clement Attlee.

Question 12.
We are not Just Going to Copy’
This is what Jawaharlal Nehru said in his famous speech of 13th December, 1946.
My mind goes back to the various Constituent Assemblies that have gone before and of what took place at the making of the great American nation when the father of that nation met and fashioned out a Constitution which has stood the test of so many years, more than a century and a half and of the great nation which has resulted, which has been built up on the basis of that construction.

My mind goes back to that mighty revolution, which took place also over 150 years ago and to that Constituent Assembly that met in that gracious and lovely city or Paris which has fought so many battles for freedom, to the difficulties than that Constituent Assembly had and to now the king and other authorities came in its way and still it continued.

The house will remember that when these difficulties came and even the room for a meeting was denied to the Constituent Assembly, they took themselves to an open tennis court and met there and took the oath, which is called the Oath of the Tennis Court that they continued meeting inspite of kings, inspite of the others and did not disperse till they had finished the task they had undertaken.

Well I trust that it is in that solemn spirit that we too are meeting here and that we too whether we meet in this chamber or other chambers or in the fields or in the market place will go on meeting and continue our work till we have finished it.

  1. How was the American Constitution finalised and explain its results?
  2. What does Nehru’s determination to pass the Constitution show? Explain any two such difficulties that were faced by the Constituent Assembly. (Delhi 2010)

Answer:
1. American Constitution was finalised when fathers of that Constitution met and fashioned out a Constitution. As a result, this Constitution stood the test of so many years.

2. Nehru’s determination to pass ‘The Constitution’ shows that our leaders were determined to establish parliamentary democracy in India and they were not ready to copy from other Constitution. But dedication and spirit were things that they wanted to copy. Two such difficulties faced by Constituent Assembly were:

  • There was no hall for meeting.
  • Kings and other authorities were not too willing and thus, posed hurdles in the making of Constitution.

Question 13.
“The Real Minorities are the Masses Of this country”
Welcoming the Objectives Resolution introduced by Jawaharlal Nehru, NG Ranga said. Sir, there is a lot of talk about minorities. Who are the real minorities? Not the Hindus in the so-called Pakistan provinces, not the Sikhs, not even the Muslims. No, the real minorities are the masses of this country.

These people are so depressed and oppressed and suppressed till now that they are not able to take advantage of the ordinary civil rights. What is the position? You go to the tribal areas. According to law, their own traditional . law, their tribal law, their lands, cannot be alienated.

Yet our merchants go there and in the so-called free market they are able to snatch their lands. Thus, even though the law goes against this snatching away of their lands, still the merchants are able to turn the tribal people into veritable slaves by various kinds of bonds and make them hereditary bond-slaves.

Let us go to the ordinary villagers. There goes the money lender with his money and he is able to get the villagers in his pocket. There is the landlord himself, the zamindar and the malguzar and there are the various other people who are able to exploit these poor villagers. There is no elementary education even among these people. These are the real minorities that need protection and assurances of protection. In order to give them the necessary protection, we will need much more than this Resolution…

  1. How is the notion of minority defined by NG Ranga?
    or
    Who are the real minorities according to Shri NG Ranga and why?
  2. Do you agree with Ranga? If not, mention who are real minorities according to you and why?
  3. Explain the conditions of ordinary villagers.
  4. Describe the living condition of the tribals. (All India 2010)

Answer:
1. According to Shri NG Ranga, the real minorities were the poor and downtrodden, especially the tribals because these people are so depressed, oppressed and suppressed till now that they are not able to take advantage of the ordinary civil rights.

2. I agree with Ranga that masses of this country are real minorities because these people have been depressed, suppressed and oppressed from the very long time.

3. The conditions of ordinary villagers

  • According to NG Ranga the life of ordinary villagers is miserable since they remain subjugated and exploited in the hands of the landlords, zamindars and the malguzars.
  • Secondly, there is no elementary education even among these people.
  • According to NG Ranga, these are the real minorities that need protection and assurances.

4. According to Professor NG Ranga, the living conditions of the tribals can be described as follows:

  • According to tribal law, the tribals cannot be alienated from their lands in their own areas but they were being alienated.
  • When the merchants go to the lands of tribals, they snatch their lands. The merchants were able to turn the tribal people into veritable slaves by various kinds of bonds and make them hereditary bond slaves.

Question 14.
“I Believe Separate Electorates will be Suicidal to the Minorities”
During the debate on 27th August, 1947, Govind Ballabh Pant said. I believe separate electorates will be suicidal to the minorities and will do them tremendous harm. If they are isolated forever, they can never convert themselves into a majority and the feeling of frustrations will cripple them even from the very beginning. What is that you desire and what is our ultimate objective? Do the minorities always want to remain as minorities or do they ever expect to form an integral part of a great nation and as such to guide and control its destinies?

If they do can they ever achieve that aspiration and that ideal if they are isolated from the rest of the community? I think it would be extremely dangerous for them if they were segregated from the rest of the community and kept aloof in an air-tight compartment where they would have to rely on others even for the air they breathe. The minorities if they are returned by separate electorates can never have any effective voice.

  1. Why were some Muslims, like Begum Aizaz Rasul against it?
  2. What are separate electorates?
  3. Why did GB Pant feel separate electorate would be suicidal for the minorities? (Delhi 2008)
  4. Do you think that seats should be reserved for Muslims and other minorities in educational institutions? Give one argument for or against it.
    (Delhi 2008)

Answer:
1. Not all Muslims supported the demand for separate electorates. Begum Aizaz Rasul felt that separate electorates were self-destructive, since they isolated the minorities from the majority.

2. It was political arrangements where the seats were reserved for minority candidates. Means Muslim candidates in the election would be voted by only Muslim members. This was done to give representation to Muslims in the governance of the country.

3. Views put forward by GB Pant against the system of separate electorates were:

  • It would be suicidal to minorities and would tremendously harm them.
  • It would be difficult for them to be an integral part of a nation and as such guide and control their destinies.
  • They would have to always be dependent upon others.
  • In this way, if they were returned by separate electorates, they could never have an effective voice.

4. In my opinion, there should not be any reservation for Muslims and other minorities because it will make the minorities being isolated which will cripple them from the beginning.

Important Questions for Class 12 History Chapter 15 Value Based Questions

Question 15.
Read the following passage and answer the question that follows. (Delhi 2016)
Every citizen in a free state should be treated in a manner that satisfied not only his material wants but also his spiritual sense of the self respect and the majority community has an obligation to try and understand the problems of the minorities and empathise with their aspirations.

  1. How could a citizen of a free nation express his imbibed value of equality and social justice while dealing with the members of the minority community? Explain.

Answer:
1. Majority community should understand the problem of their brothers from minority community and . empathise with their aspirations.

Constitution has granted its every citizen right to equality. Majority community should protect the rights of minorities and prevent them from discrimination. Majority community should keep in mind that culture of the majoritarian community should not dominate over the minorities and minorities culture and customs should be protected.

Question 16.

  1. Why did Mahatma Gandhi think Hindustani should be the national language?
  2. How did the Constituent Assembly seek to resolve the language controversy? (All India 2010)

Answer:
1. Mahatma Gandhi thought that Hindustani should be the national language because of following reasons:
Easily Understood:
The common people could easily understand it.

Blend of Diverse Cultures:
Hindustani was a blend of Hindi and Urdu and was a composite language enriched by interaction of diverse cultures.

Ideal Language of Communication between Diverse Cultures:
It would help to unify Hindus and Muslims and the people from North and South.

2. The language committee of the Constituent Assembly had submitted its report and thought of a composite formula to resolve the deadlock between those who advocated Hindi as the national language and those who opposed it.

Question 17.

  1. How was the concept of separate electorates erratic? Explain by giving examples from the Constituent Assembly.
  2. “I believed separate electorates will be suicidal to the minorities” How far these view of GB Pant are relevant in contemporary India? (Delhi 2008)

Answer:
1. Intense debate took place in the Constituent Assembly on the issue of separate electorate. Some were in favour of it and many nationalist leaders saw this system as a tool to divide people on the basis of religion and they believed that this idea finally culminated in partition of the country.

Sarder Patel strongly declared that separate electorate was a poison that has entered the body of politics of our contry and instigate one community against other caused bloodshed, riots and partition. So, for peace we need to remove separate electorate.

2. According to GB Pant, separate electorates would be suicidal to the minorities. By doing this, there would be tremendous harm to the minority sections, they would remain isolated forever. The feelings of isolation and frustration would cripple them from the very beginning.
Isolation would be extremely dangerous for them if they were segregated from the rest of the community. Through separate electorates, they could never have any effective voice.

Important Questions for Class 12 History

The post Important Questions for Class 12 History Chapter 15 Framing the Constitution (The Beginning of a New Era) appeared first on Learn CBSE.

The Last Lesson Important Questions Class 12 English

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The Last Lesson Important Questions CBSE Class 12 English

The Last Lesson Important Questions Short Answer Type Questions (3-4 Marks)

Question 1.
What was the mood in the classroom when M. Hamel gave his last French lesson? (Delhi 2009)
Answer:
When M.Hamel was giving his last French ; lesson, the mood in the classroom was solemn and sombre. When he announced that this was their last French lesson everyone present in the classroom suddenly developed patriotic feelings for their native language and genuinely regretted ignoring their mother tongue.

Question 2.
What had the narrator counted on to enter the school, unnoticed? (Delhi 2010)
Answer:
The narrator (Franz) had counted on the hustle and bustle that was usual when the school began in order to enter there unnoticed. He had thought he could depend on the commotion to get to his desk without anybody noticing that he was late for the class.

Question 3.
“This is your last French lesson.” How did Franz react to this declaration of M.Hamel? (Delhi 2010)
Answer:
When M.Hamel declared that it was their last French lesson a grim realisation dawned on Franz that he had so much more yet to learn. He felt sorry for whiling away his time and skipping his lessons. Now he did not want to part with his books, which he had earlier consi¬dered a nuisance. He was also deeply pained at the thought that M.Hamel was leaving.

Question 4.
“What a thunder clap these words were to me!” Which were the words that shocked and surprised little Franz? (Delhi 2010)
Answer:
When M.Hamel mounted on the chair and announced that he was there to teach his last French lesson that day, Franz was shocked and surprised. He felt very guilty for deliberately ignoring to learn his native language and he suddenly developed a strange fascination for his language and his school.

Question 5.
What changes came over little Franz after he heard M.Hamel’s announcement? (All India 2010)
Answer:
When M.Hamel announced that this was to be their last French class a grim realisation dawned on Franz that he could hardly write his language and now he was being deprived of the opportunity to learn it. He felt very guilty of neglecting his classes and escaping school. The thought of losing his teacher, M. Hamel, also pained him.

Question 6.
What was tempting Franz to keep away from school That morning’? (Comptt. Delhi 2010)
Answer:
Franz was very late for school ‘that morning’. M. Hamel was to question the students on participles and Franz knew nothing and feared a scolding. The bright weather, woods, fields and chirping of birds tempted him to spend the day outdoors.

Question 7.
What was unusual about M. Hamel’s dress and behaviour on the day of his last French lesson? (Comptt. Delhi 2010)
Answer:
On the day of his last French lesson M. Hamel was unexpectantly kind. He had put on his beautiful green coat, frilled shirt and an embroidered black silk cap that he normally wore only on inspection and prize days. In his last lesson he explains everything with unusual patience as if he wants to teach them everything he has ever learned in their last French class.

Question 8.
Why had the bulletin-board become a centre of attention during the last two years? (Comptt. All India 2010)
Answer:
The bulletin-board had become a centre of attention because for the last two years all the bad news had come from there. It was the source of all bad news-the lost battles, the draft and the orders of the commanding officers.

Question 9.
What was Franz expected to be prepared with for the school that day? (Delhi 2011)
Answer:
Franz had started very late for school that morning. He feared facing his class-teacher, M. Hamel. Because he was expected to be prepared with his grammar lesson on participles and he did not know even the first word about them. So he feared a scolding.

Question 10.
What had been put up on the bulletin board? (Delhi 2011)
Answer:
A notice had been put up on the bulletin board informing the people about the order from Berlin which declared that only German was to be taught in the schools of Alsace and Lorraine. They had also been informed that the new German teacher was coming on the following day.

Question 11.
Who did M.Hamel blame for the neglect of learning on the part of boys like Franz? (Delhi 2011)
Answer:
Mr. Hamel blamed various people for different reasons for neglect of learning on the part of boys like Franz. First he blamed the parents for not being anxious to have their children learn and rather engaging them into household work. He blamed the boys themselves for preferring to work on mills and farms. He also blamed himself for sending boys on errands for his convenience.

Question 12.
Franz thinks, “Will they make them sing in German, even the pigeons?” What does this tell us about the attitude of the Frenchmen?
(All India 2011)
Answer:
During his last French class Franz hears the
pigeons cooing and wonders if the Prussians would make them sing in German too. His thought reveals the Frenchmen’s fear of complete domination by the Germans whereby they would try to enslave even their minds.

Question 13.
“What a thunderclap these words were to me!”—Which were the words that shocked and surprised Franz? (All India 2011)
Answer:
When M.Hamel mounted on the chair and announced that he was there to teach his last French lesson that day, Franz was shocked and surprised. He felt very guilty for deliberately ignoring to learn his native language and he suddenly developed a strange fascination for his language and his school.

Question 14.
“What announcement did M. Hamel make? What was the impact of this on Franz? (All India 2011)
Answer:
When M.Hamel declared that it was their last French lesson a grim realisation dawned on Franz that he had so much more yet to learn. He felt sorry for whiling away his time and skipping his lessons. Now he did not want to part with his books, which he had earlier consi¬dered a nuisance. He was also deeply pained at the thought that M.Hamel was leaving.

Question 15.
What did M. Hamel tell the people in the class about French language? What did he ask them to do and why? (Comptt. Delhi 2011)
Answer:
M. Hamel told the people in his class that French was the most beautiful, the clearest and the most logical language in the world. He advised them to guard it among themselves j and never to forget it because it would prove to [ be the key to their prison when they were I enslaved.

Question 16.
What was the mood in the classroom when M.Hamel gave his last French lesson? (Comptt. All India 2011)
Answer:
When M.Hamel was giving his last French; lesson, the mood in the classroom was solemn and sombre. When he announced that this was their last French lesson everyone present in the classroom suddenly developed patriotic feelings for their native language and genuinely regretted ignoring their mother tongue.

Question 17.
What changes did the order from Berlin cause in the school? (Delhi 2012)
Answer:
The shocking order from Berlin that only German would be taught in the schools of Alsace and Lorraine saddened and hurt the pride of the French people. So as a gesture to express their solidarity for their language the village elders also attended school for the last French lesson. The usual hustle and bustle was missing from school and it was unusually quiet. M. Hamel was also dressed in his formal suit which he wore only on special occasions. Question 56. How different did M.Hamel look that particu¬lar morning in ‘The Last Lesson’?
(Comptt. All India) Answer: In honour of his last French lesson in the school, M. Hamel had put on his fine Sunday clothes. He had on his green coat, his frilled shirt and the little black, all embroidered silk cap that he never wore except on inspection and prize days.

Question 18.
How did M.Hamel say farewell to his students and the people of the town? (All India 2012)
Answer:
To bid farewell to his students and the people of the town M.Hamel wore his formal suit on the day he gave his last French lesson. He assumed the role of a mentor and advised the villagers to safeguard their language, which was the most beautiful and logical language in the world. Being rimmed up with emotions, he wrote “Vive La France” which mean Tong live France’ in order to inculcate a feeling of patriotism within everyone.

Question 19.
What were Franz’s regrets after M. Hamel’s announcement of his last lesson? (Comptt. Delhi 2012)
Answer:
When M. Hamel announced that it was their last French lesson, Franz realised he had so much more to learn. He felt sorry for whiling away his time and skipping his lessons. He was also deeply pained at the thought of M. Hamel leaving the school.

Question 20.
What change was there in Franz’ attitude towards M. Hamel? (Comptt. All India 2012)
Answer:
Franz’s attitude towards M. Hamel underwent a complete change on the day of the last French lesson. He developed a strange fascination for his school and the French language. When M. Hamel read out a grammar lesson to the class, Franz was amazed at how well he understood it. All M. Hamel said seemed so easy to Franz and he genuinely regretted ignoring M. Hamel’s classes and lessons.

Question 21.
Why did M. Hamel not scold Franz when the latter fumbled with the rule for the participle? (Comptt. All India 2012)
Answer:
Since it was M. Hamel’s last class in Alsace, he did not scold Franz when the latter fumbled with the rule for the participle. He told Franz that he would not scold him as he knew that Franz was already feeling terrible enough and reproaching himself a great deal.

Question 22.
How did Franz react to the declaration that it was their last French lesson? (Delhi 2013)
Answer:
When M.Hamel declared that it was their last French lesson a grim realisation dawned on Franz that he had so much more yet to learn. He felt sorry for whiling away his time and skipping his lessons. Now he did not want to part with his books, which he had earlier considered a nuisance. He was also deeply pained at the though that M.Hamel was leaving.

Question 23.
“What a thunderclap these words were to me !” What were the words that shocked and surprised the narrator? (Delhi, Comptt. AT 2013)
Answer:
When M.Hamel mounted on the chair and announced that he was there to teach his last French lesson that day, Franz was shocked and surprised. He felt very guilty for deliberately ignoring to learn his native language and he suddenly developed a strange fascination for his language and his school.

Question 24.
What did M. Hamel tell them about the French language? What did he ask them to do and why? (Delhi 2013)
Answer:
M. Hamel told the people in his class that French was the most beautiful, the clearest and the most logical language in the world. He advised them to guard it among themselves j and never to forget it because it would prove to [ be the key to their prison when they were I enslaved.

Question 25.
What was the order from Berlin and what changes did it cause in the school? (All India 2013)
Answer:
The shocking order from Berlin that only German would be taught in the schools of Alsace and Lorraine saddened and hurt the pride of the French people. So as a gesture to express their solidarity for their language the village elders also attended school for the last French lesson. The usual hustle and bustle was missing from school and it was unusually quiet. M. Hamel was also dressed in his formal suit which he wore only on special occasions. Question 56. How different did M.Hamel look that particu¬lar morning in ‘The Last Lesson’?
(Comptt. All India) Answer: In honour of his last French lesson in the school, M. Hamel had put on his fine Sunday clothes. He had on his green coat, his frilled shirt and the little black, all embroidered silk cap that he never wore except on inspection and prize days.

Question 26.
Why did Franz think of running away from the school that morning? (Comptt. Delhi 2013)
Answer:
Franz was tempted to run away and spend the day outside instead of attending school as it was a warm and bright day and he wanted to watch the Prussian soldiers’ drill. Moreover he had not done his lesson on participles from which his teacher, M. Hamel was going to question and feared a scolding from his teacher.

Question 27.
What did Franz see when he passed the town hall? (Comptt. Delhi 2013)
Answer:
When Franz passed the town hall he saw there was a crowd in front of the bulletin board. For the last two years all the bad news of lost battles, the draft etc. had come from there and Franz thought to himself what the matter was then and what bad news awaited them.

Question 28.
When Franz reached his school, what unusual situation did he observe? (Comptt. Delhi 2013)
Answer:
The usual hustle and bustle of the morning hours could not be seen on the day of the last lesson. It was all very still and quiet as it used to be on a Sunday morning. The din of opening and closing of desks and the rapping of the teacher’s ruler on the table could not be heard. The students had already taken their places and even the village elders had gathered there.

Question 29.
After sitting down at his desk what unusual things did Franz observe about M. Hamel?
(Comptt. All India)
Answer:
Franz realised that M. Hamel was not his usual cranky self and said nothing to him even though he was late for the class. Moreover their teacher was wearing his beautiful green coat, his frilled shirt and the little black silk cap, all embroidered, that he wore only on special days in school.

Question 30.
“But the thing that surprised me most was to see ” What surprised Franz most in the class? (Comptt. All India 2013)
Answer:
The thing that surprised Franz the most when he entered the classroom was to see the village people sitting as quietly as the students on the back benches of the class that were always empty. This included old Hauser, the former mayor, the former postmaster and several others.

Question 31.
What tempted Franz to stay away from school? (Delhi 2014)
Answer:
Little Franz was tempted to run away and spend the day outside instead of attending school as it was a warm and bright day with the birds chirping merrily. Moreover, he was tempted to watch the Prussian soldiers’ drill. All these things appeared rather tempting considering he did not know his participles and feared the French teacher M. Hamel’s scolding on that day.

Question 32.
Why were the elders of the village sitting in the classroom? (All India 2014)
Answer:
The elders of the village were sitting in the classroom as a mark of respect which they wanted to show for their language and their country. They felt sorry for not having learnt their language and wanted to thank M. Hamel for his forty years of faithful service.

Question 33.
What words did M. Hamel write on the black¬board before dismissing the last class? What did they mean? (Comptt. Delhi 2014)
Answer:
Before dismissing the last French class, M. Hamel turned to the blackboard, took a piece of chalk and wrote as large as he could -“Vive la France!” These words meant “Long Live France”. He then made a gesture with his hand that the class was dismissed.

Question 34.
Who were sitting on the back benches during M. Hamel’s last lesson? Why? (Comptt. Delhi 2014)
Answer:
The village elders were sitting on the back benches during M. Hamel’s last lesson. They included old Hauser, the former mayor, the former postmaster and several others. They were there to show their respect for their language and their country.

Question 35.
Why is the order from Berlin called a thunderclap by Franz? (Comptt. Delhi 2014)
Answer:
The order from Berlin is called a thunderclap by Franz, because it was a complete shock for him. He had never thought that he would be deprived of the right of learning his native language. He realised he had so much more yet to learn and was going to lose the opportunity.

Question 36.
Why does M. Hamel reproach himself for his students’ unsatisfactory progress in studies? (Comptt. All India 2014)
Answer:
M. Hamel reproaches himself for his students’ unsatisfactory progress in studies. He had often sent them to water his flowers instead of learning their lessons. And when he wanted to go fishing, he gave them a holiday.

Question 37.
What made M. Hamel cry towards the end of his last lesson? (Comptt. All India 2014)
Answer:
Towards the end of his last French lesson, as the church clock struck twelve and the trumpets of the Prussians sounded under the window, M. Hamel was completely overwhelmed with emotions. Feeling choked, he was unable to talk further and dismissed the class by making a gesture with his hand.

Question 38.
What was unusual about M. Hamel’s dress on his last day in the school? (Comptt. All India 2014)
Answer:
In honour of his last French lesson in the school, M. Hamel had put on his fine Sunday clothes. He had on his green coat, his frilled shirt and the little black, all embroidered silk cap that he never wore except on inspection and prize days.

Question 39.
Who occupied the back benches in the classroom on the day of the last lesson? Why? (Delhi 2015)
Answer:
The village people occupied the backbenches in the classroom on the day of the last lesson. They included the old hauser, the former mayor, the former postmaster and several others. They had come there to thank their master for his valuable service and to show their respect for their language and country which was no longer theirs.

Question 40.
Why did M. Hamel write ‘Vive La France’ on the blackboard? (Delhi 2015 )
Answer:
Before dismissing the class, M. Hamel writes ‘Vive La France’ (Long Live France) on the blackboard. He was extremely patriotic and he does this to make the Alsacians conscious that they had to safeguard their language and identity among themselves as this was the key to their prison.

Question 41.
“We’ve all a great deal to reproach ourselves with”, said M. Hamel. Comment. (Delhi 2015)
Answer:
M. Hamel blamed everyone for putting off learning till the next time and thus losing out on the opportunity to learn their native language, French. The parents preferred to put their children to work on a farm or at the mills so that they could earn more money. M. Hamel too sent them to water his flowers and gave them a holiday when he wanted to go for fishing.

Question 42.
How did the order from Berlin change the situation in the school? (All India 2015)
Answer:
The shocking order from Berlin after the defeat of France shocked the French people. It declared that all the schools in the two French districts of Alsace and Lorraine would now teach only German. This not only saddened but also hurt the pride of the people of France. On the day of the French teacher M.Hamel’s last lesson, even the village elders came to show their respect to him for serving the community faithfully for forty years. The usual hustle and bustle too was missing from school and Franz found it unusually quiet as it used to be on a Sunday morning. M.Hamel was dressed in his formal suit. Even though Franz reached late M.Hamel did not scold him. He then went on to praising the French language at length by referring to it as the most beautiful language. He called upon the French people to safeguard their language among themselves and not to forget it ever.

Question 43.
Whom did M.Hamel blame for Franz’s inabil¬ity to answer his questions? (Comptt. Delhi 2015)
Answer:
Hamel blamed the parents who sent the children to work instead of sending them to school. He blamed himself for making them water the plants and giving them a holiday when he wanted to go fishing. He also blamed the students themselves for delaying the learning of their lessons thinking that there was plenty of time.

Question 44.
How did M.Hamel display his love for the French language? (Comptt. Delhi 2015)
Answer:
M. Hamel had great love for the French language which he had been teaching for forty years. For him it was the most beautiful, most logical and clearest language. He wanted the French people to preserve it among themselves and this would enable them to be free from foreign rule.

Question 45.
What was the bulletin board news that caused a change in the school? (Comptt. All India 2015)
Answer:
The bulletin board news that caused a change in the school was that, that day was going to be the last French lesson for the schools of Alsace and Lorraine. The order had come from Berlin to teach only German in the schools and their new German teachers would come the next day.

Question 46.
Who were sitting on the backbenches on the day of the last lesson? Why? (Comptt. AI 2016)
Answer:
The village people occupied the backbenches in the classroom on the day of the last lesson. They included the old hauser, the former mayor, the former postmaster and several others. They had come there to thank their master for his valuable service and to show their respect for their language and country which was no longer theirs.

Question 47.
Why were people standing in front of the bulletin board in ‘The Last Lesson’? Why did Franz not join the crowd? (Comptt. All India 2016)
Answer:
For the past two years the bulletin board had been the bearer of all bad news—the lost battles, drafts, orders from the commanding officer. That day an order had come from Berlin that from the next day a new teacher would teach German instead of French in the school. Franz could not join the crowd as he was already late for his class and could not afford further delay.

Question 48.
Why did Franz not want to go to school that day? (Delhi 2016)
Answer:
Franz was tempted to run away and spend the day outside instead of attending school as it was a warm and bright day and he wanted to watch the Prussian soldiers’ drill. Moreover he had not done his lesson on participles from which his teacher, M. Hamel was going to question and feared a scolding from his teacher.

Question 49.
Why was Franz not scolded for reaching the school late that day? (Delhi 2016)
Answer:
It was to be their last French lesson that day. The order from Berlin allowed only German language to be taught in the schools of Alsace and Lorraine. Hence their teacher, M. Hamel, was in a sad and sombre mood. So he did not scold Franz for reaching school late that day.

Question 50.
What did Franz wonder about when he entered the class that day? (Delhi 2016)
Answer:
Franz wondered about a number of things when he entered the class that day. He wondered as to why he was not given a scolding by his teacher, M. Hamel for being late, why his teacher was wearing his formal clothes which he wore only on inspections and prize days and why the village elders were sitting on the backbenches in the classroom.

Question 51.
Why were some elderly persons occupying the backbenches that day? (All India 2016)
Answer:
The elders of the village were sitting in the classroom as a mark of respect which they wanted to show for their language and their country. They felt sorry for not having learnt their language and wanted to thank M. Hamel for his forty years of faithful service.

Question 52.
“What a thunderclap these words were to me!” (Franz). What were those words and what was their effect on Franz? (All India 2016)
Answer:
When M.Hamel mounted on the chair and announced that he was there to teach his last French lesson that day, Franz was shocked and surprised. He felt very guilty for deliberately ignoring to learn his native language and he suddenly developed a strange fascination for his language and his school.

Question 53.
How were the parents and M. Hamel respon¬sible for the children’s neglect of the French language? (All India 2016)
Answer:
M. Hamel blamed everyone for putting off learning till the next time and thus losing out on the opportunity to learn their native language, French. The parents preferred to put their children to work on a farm or at the mills so that they could earn more money. M. Hamel too sent them to water his flowers and gave them a holiday when he wanted to go for fishing.

Question 54.
How is the title ‘The Last Lesson’ appropriate? (Comptt. Delhi 2016)
Answer:
The title ‘The Last Lesson’ is appropriate because it was M. Hamel’s last lesson and also the last French lesson in the school. During the last lesson, the teacher, as well as all those who were attending it, were feeling nostalgic and repentant.

Question 55.
What changes did the order from Berlin cause on the day of the last lesson? (Comptt. Delhi 2016)
Answer:
The shocking order from Berlin that only German would be taught in the schools of Alsace and Lorraine saddened and hurt the pride of the French people. So as a gesture to express their solidarity for their language the village elders also attended school for the last French lesson. The usual hustle and bustle was missing from school and it was unusually quiet. M. Hamel was also dressed in his formal suit which he wore only on special occasions.

Question 56.
How different did M.Hamel look that particu¬lar morning in ‘The Last Lesson’? (Comptt. All India)
Answer:
In honour of his last French lesson in the school, M. Hamel had put on his fine Sunday clothes. He had on his green coat, his frilled shirt and the little black, all embroidered silk cap that he never wore except on inspection and prize days.

Question 57.
What was M.Hamel’s regret on the day of the last lesson? (Comptt. All India 2016)
Answer:
M. Hamel’s regret on the last day of the French lesson was that they all had put off learning their language till the next day and now most of them who claimed to be Frenchmen could neither speak nor write their language. For this they all, including M. Hamel himself, had a great deal to reproach themselves with.

Question 58.
How different was the scene in the classroom on the day of the last lesson? (Comptt. All India 2016)
Answer:
The usual hustle and bustle of the morning hours could not be seen on the day of the last lesson. It was all very still and quiet as it used to be on a Sunday morning. The din of opening and closing of desks and the rapping of the teacher’s ruler on the table could not be heard. The students had already taken their places and even the village elders had gathered there.

The Last Lesson Important Questions Long Answer Type Questions (5-6 marks)

Question 59.
What did the French teacher tell his students in his last French lesson? What impact did it have on them? Why? (All India 2009)
Answer:
M. Hamel told his students that a new order from Berlin has declared that all schools of Alsace and Lorraine would teach only German so this was going to be their last French lesson. This new order aroused patriotic feelings in him and he, in turn, wanted to arouse similar patriotism in his students and the village elders. He made them conscious of the glory and value of the French language and told them to safeguard it among themselves and keep it alive at all costs as it was the key to their unity and liberation. Everyone listened to him sadly but with rapt attention and respect. Even little Franz listened to his teacher’s words with a new-found interest. He felt sorry that he had neglected learning French. Everyone from the village assembled in the class to thank Mr. Hamel for his forty years of faithful service to the community.

Question 60.
How different from usual was the atmosphere at school on the day of the last lesson? (All India 2015)
Answer:
Acquiring power over the Alsacians made the Prussians so dominating that they even imposed their language on them. This way the Prussians intended to dominate the hearts and minds of the Alsacians and wanted them to even think in their language and thereby lose their complete identity. An order had been received from Berlin that only German would be taught in schools of Alsace and Lorraine. So there was something unusual about the school on the last day of the French lesson. The usual hustle and bustle was missing. Everything was ‘strange and solemn’ as on a Sunday morning. The village elders were seated on the back desks.

M.Hamel, who had been teaching French at the school for the last forty years, was wearing his formal suit in honour of the last French lesson. While delivering the last lesson, he called upon his students and the village elders to guard the French language among themselves and never forget it, declaring French to be the most beautiful language in the world. Franz developed a sudden fascination for school and the French language and a sudden respect for M.Hamel. He wanted his teacher to stay and felt very guilty for having neglected his French lessons as now he was being deprived of the opportunity of learning his language.

Question 61.
Everybody during the last lesson is filled with regret. Comment. (All India 2015)
Answer:
The one common feeling that fills each and every person who is present in the last French lesson is an acute sense of regret. M. Hamel reproaches himself for putting off his students’ learning till the next day and sending them to water his flowers instead of learning their lessons. He also gave his students a holiday when he wanted to go fishing. Franz felt sorry for not learning his lessons and escaping school. He wished he had attended his classes more often and even the thought of losing his teacher saddened him. The village elders occupied the back benches of the class to atone for their guilt and express their regret for not having attended school regularly. They were now showing their respect for the country that was theirs no more.

Question 62.
Our language is part of our culture and we are proud of it. Describe how regretful M.Hamel and the village elders are for having neglected their native language, French. (Delhi 2016)
Answer:
The feeling of regretfulness for having neglected their native language, French comes quite late to M. Hamel and the village elders. They realise rather late that their language is part of their culture and they should be proud of it. It is only after they have been deprived of learning their language that they understand its value. The imposition of German language made them suddenly realise the authority of their captors and they felt a loss of freedom. So on the day of the last French lesson the village elders are seated on the back desks and M. Hamel, who had been teaching French at the school for the last forty years, was wearing his formal suit as a mark of respect for the last French lesson. M. Hamel expressed how they all had a great deal to reproach themselves for as most of the people of Alsace could neither speak nor write French. Parents preferred to put their children to work on farms or mills.

M. Hamel regretfully said how he himself sometimes sent his students to water his flowers instead of learning their language lessons. While delivering the last lesson M. Hamel called upon his students and the village elders to guard the French language among themselves, declaring it to be the most beautiful language tn the world. Each one of them felt guilty for having ignored the French lessons.

Question 63.
Our native language is part of our culture and we are proud of it. How does the presence of village elders in the classroom and M.Hamel’s last lesson show their love for French? (All India 2016)
Answer:
Acquiring power over the French made the Prussians so dominating that they decided to impose even their language on them. So on the day of the last French lesson the village elders were seated on the back seats of the classroom. They felt sad and sorry for not having gone to school more often. Their presence was also their way of thanking their master, M.Hamel for his forty years of faithful service. M.Hamel himself had put on his fine Sunday clothes in honour of the last French lesson. Assuming the role of mentor he advised the class to safeguard and preserve their language among themselves and never forget it. He called French the most beautiful and logical language in the world. By the end of the class M.Hamel was so choked with emotion and could not speak so he took a piece of chalk and wrote on the blackboard “Vive la France!” (Long Live France) and then dismissed the class.

Question 64.
Describe the atmosphere in the class on the day of the last lesson. (Comptt. Delhi 2016)
Answer:
On the day of the last French lesson the atmosphere in the class was strange and solemn. The school was as quiet as if it was a Sunday morning. The teacher (M.FIamel) moving in the class with his ruler under his arm was wearing his special dress for the last class. The elders of the village were sitting on the backbenches of the classroom. All the students were studying with complete attention and the teacher was teaching with full dedication. All this was due to an order from Berlin that from the next day German would be taught instead of French by a new teacher. While delivering the last lesson, the teacher called upon his students and the village elders to guard the French language among themselves and never forget it, declaring French to be the most beautiful language in the world.

Important Questions for Class 12 English

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Lost Spring Important Questions Class 12 English

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Lost Spring Important Questions CBSE Class 12 English

Lost Spring Important Questions Short Answer Type Questions (3-4 marks)

Question 1.
What does the writer mean when she says, ‘Saheb is no longer his own master’? (Delhi 2000)
Answer:
Since Saheb now works in a tea-stall, he is now bound to his master and feels burdened. The steel canister he carries is very heavy as compared to his light plastic bag. The bag was his own and the canister belongs to his master whose orders he now has to follow. So he is no longer his own master.

Question 2.
Is it possible for Mukesh to realize his dream? Justify your answer? (All India 2000)
Answer:
Mukesh’s determination is going to prove instrumental in helping him to realize his dream. His dream can become a reality only if he is able to find a garage where he can be taken in as an apprentice and then he will have to learn how to drive a car. He will then be able to graduate himself to be a good mechanic.

Question 3.
Do you think Saheb was happy to work at the tea stall? Answer giving reasons. (All India 2000)
Answer:
Since Saheb now works in a tea-stall, he is now bound to his master and feels burdened. The steel canister he carries is very heavy as compared to his light plastic bag. The bag was his own and the canister belongs to his master whose orders he now has to follow. So he is no longer his own master.

Question 4.
What does the title, ‘Lost Spring’ convey? (All India 2000)
Answer:
The title ‘Lost Spring’ conveys how millions of children in India lose out on living the ‘spring’ of their lives, that is their childhood. The best phase of life is lost in the hardships involved to earn their livelihood. Poverty forces these young children to work in the most inhuman conditions as a result of which they miss out on the fun of childhood which hampers their growth.

Question 5.
Why does the author say that the bangle makers are caught in a vicious web?(All India 2010)
Answer:
The author says that the bangle makers are caught in a vicious web which starts from poverty, to indifferences, then to greed and finally to injustice. Mind-numbing toil kills their hopes and dreams. They cannot organise themselves into cooperatives and have fallen into a vicious circle of ‘sahukars’, middlemen and the police so they get condemned to poverty and perpetual exploitation.

Question 6.
What does the writer mean when she says, ‘Saheb is no longer his own master’? (Comptt. All India 2011)
Answer:
Since Saheb now works in a tea-stall, he is now bound to his master and feels burdened. The steel canister he carries is very heavy as compared to his light plastic bag. The bag was his own and the canister belongs to his master whose orders he now has to follow. So he is no longer his own master.

Question 7.
Who is Mukesh? What is his dream? (Delhi 2012)
Answer:
Mukesh is a child labourer in a glass factory in Firozabad. Belonging to a family of bangle makers, he shows no fascination towards bangle-making and insists on being his own master. He dreams of becoming a motor mechanic. He desires to go to a garage and get the required training for this job.

Question 8.
Is Saheb happy working at the tea stall? Why/ Why not? (Delhi 2012)
Answer:
No, Saheb is not happy working at the tea stall. Even though, he now gets a fixed income of ?800 alongwith all his meals, he has lost his freedom and his carefree days. He is no longer his own master and is bound and burdened by the steel canister he now has to carry.

Question 9.
Why could the bangle-makers not organise themselves into a co-operative? (All India 2012)
Answer:
The bangle-makers are caught in a vicious web which starts from poverty to indifferences then to greed and finally to injustice. Mind-numbing toil kills their hopes and dreams.
The bangle makers of Ferozabad were not able to organise themselves into a cooperative because they had got trapped in a vicious circle j of the sahukars, the middlemen, the policemen, j the bureaucrats and the politicians. Together they had imposed a baggage on these people 1 which they could not put down.

Question 10.
Mention any two problems faced by the bangle sellers.(Comptt. All India 2012)
Answer:
The bangle makers had to work in the glass furnaces with high temperatures, in dingy cells without air and light. They are exposed to various health hazards like losing their eyesight as they work in abysmal conditions in dark and dingy cells. They were also caught in a web of poverty, burdened by the stigma of caste in which they were born and also caught in a vicious circle of sahukars, middlemen and policeman.

Question 11.
Garbage has two different meanings—one for the children and another for the adults. Comment. (Comptt. All India 2012)
Answer:
For the children garbage has a different meaning from what it means for the adults. For the children it is wrapped in wonder, their eyes light-up when they find a rupee or a ten-rupee note in it. They search the garbage excitedly with the hope of finding something more. But for the elders it is a means of survival.

Question 12.
Why didn’t the bangle makers of Ferozabad organise themselves into a cooperative? (Comptt. All India 2012)
Answer:
The bangle-makers are caught in a vicious web which starts from poverty to indifferences then to greed and finally to injustice. Mind-numbing toil kills their hopes and dreams.
The bangle makers of Ferozabad were not able to organise themselves into a cooperative because they had got trapped in a vicious circle j of the sahukars, the middlemen, the policemen, j the bureaucrats and the politicians. Together they had imposed a baggage on these people 1 which they could not put down.

Question 13.
How is Mukesh’s attitude towards life different from that of his family? (Comptt. Delhi 2013)
Answer:
Unlike his family Mukesh insists on being his own master. He dreams to be a motor mechanic which in itself is a daring thought because he wants to break away from the family’s work of making bangles wherein his forefathers have spent generations working around furnaces.

Question 14.
Why can’t the bangle makers of Ferozabad organize themselves into a cooperative? (Comptt. Delhi 2013)
Answer:
The bangle-makers are caught in a vicious web which starts from poverty to indifferences then to greed and finally to injustice. Mind-numbing toil kills their hopes and dreams.
The bangle makers of Ferozabad were not able to organise themselves into a cooperative because they had got trapped in a vicious circle j of the sahukars, the middlemen, the policemen, j the bureaucrats and the politicians. Together they had imposed a baggage on these people 1 which they could not put down.

Question 15.
Why is Saheb unhappy working at the tea i stall? (Comptt. Delhi 2013)
Answer:
Saheb was unhappy while working at the tea- stall because he was no longer the master of his own life. He lost his freedom and carefree look. He had to live and work under the instructions of the owner of the tea-stall. He was not at liberty to go out and spend time with his friends.

Question 16.
Survival in Seemapuri means rag-picking. , Comment. (Comptt. All India 2013)
Answer:
Survival in Seemapuri means rag-picking. Over the years it has acquired the proportions of a fine art. For the slum dwellers of Seemapuri, rag-picking is their daily bread, it gives them the roof over their heads and is the very means for their survival.

Question 17.
It is ‘a tradition to stay barefoot ‘ What is the attitude of the rag-pickers of Seemapuri towards wearing shoes? (Comptt. All India 2013)
Answer:
The rag-pickers of Seemapuri have different attitudes towards wearing shoes. One boy does not feel like wearing shoes. Another boy who has never owned a pair of shoes all his life wants them. But the author feels it its not lack of money but a tradition to stay barefoot for these poor people.

Question 18.
A young man in Ferozabad is burdened under the baggage of two worlds. What are they? (Comptt. All India)
Answer:
The two worlds that burden a young man in Ferozabad include one of the family, caught in the web of poverty, burdened by the stigma of ” caste in which they are born; the other a vicious circle of the sahukars, the middlemen, the policemen, the keepers of law, the bureaucrats and the politicians.

Question 19.
How is Mukesh different from the other bangle makers of Firozabad? (Delhi 2014)
Answer:
Mukesh was different from other bangle makers because he wanted to be his own master. He had a dream of becoming a motor mechanic whereas other bangle makers did not even dare to dream but had accepted their fate.

Question 20.
What job did Saheb take up? Was he happy? (Delhi 2014)
Answer:
Saheb took up a job in a tea stall. Though he gets 800 rupees and all his meals, he is not happy and his face has lost the carefree look. He is bound and burdened as he now has to follow the orders of his master and is no longer his own master.

Question 21.
Why did Saheb’s parents leave Dhaka and migrate to India? (Comptt. Delhi 2014)
Answer:
Saheb’s home was set amidst the green fields of Dhaka. His mother told him that many storms had swept away their fields and homes. For this reason his parents were forced to leave Dhaka and migrate to India, looking for gold in the big city where they now live.

Question 22.
What is Mukesh’s dream? Do you think he will be able to fulfil his dream? Why? Why not? (Comptt. Delhi 2014)
Answer:
Mukesh’s dream is to learn to drive a car and become a motor mechanic. His dream is likely to be fulfilled because one can sense a kind of determination in him to ensure the fulfillment of his dream. Though the garage is a long way from his home he is willing to walk to learn despite the odds against him.

Question 23.
In what sense is garbage gold to the ragpickers? (Comptt. All India 2014)
Answer:
The rag-pickers of Seemapuri consider garbage as nothing less than gold. For the elders it is their only means of survival as it provides them with their daily bread. For the innocent chil¬dren it is wrapped in wonder as they marvel at the chance of finding a coin, a currency note or a curio that really thrills them and gives them a hope of finding more elusive notes.

Question 24.
Whom does Anees Jung blame for the sorry plight of the bangle makers’?(Comptt. All India 2014)
Answer:
Anees Jung blames the family of the bangle makers, who are caught in a web of poverty, burdened by the stigma of the caste in which they are born and the vicious circle of the sahukars, the middlemen, the policemen, the keepers of law, the bureaucrats and the politicians for the sorry plight of the bangle makers.

Question 25.
To which country did Saheb’s parents originally belong? Why did they come to India? (Comptt. All India 2014)
Answer:
Saheb’s parents originally belonged to Dhaka in Bangladesh. His home, which was set amidst the green fields of Dhaka, had been swept away due to the storms and that was when his parents had left their native place and come to the big city in search of livelihood.

Question 26.
Most of us do not raise our voice against injustice in our society and tend to remain
mute spectators. Anees Jung in her article, l “Lost Childhood” vividly highlights the I miserable life of street children and bangle makers of Firozabad. She wants us to act. Which qualities does she want the children to develop? (Comptt. Delhi 2014)
Answer:
Anees Jung feels that there is dire need to provide these poverty-stricken children a life of dignity and respect. This can mainly be done j through the medium of education, which will further provide them with opportunities wherein they will be able to pursue their dreams. There is utter lack of compassion and concern for unfortunate children like Saheb and Mukesh. They are caught in a vicious circle of poverty and exploitation. The author wants all i children to become aware of their basic rights which will empower them and enable them to j organise themselves into cooperatives whereby j they will not be ruthlessly exploited.

Question 27.
What does Saheb look for in the garbage 1 dumps? (All India 2015)
Answer:
According to the author Saheb scrounges for ‘gold’ in the garbage dumps. ‘Gold’ here infers . to items that are valuable to him like used clothes, shoes, plastic scrap, stray coins or 1 currency notes. For children like Saheb, i garbage is ‘wrapped in wonder’.

Question 28.
What did garbage mean to the children of Seemapuri and to their parents? (All India 2015)
Answer:
The rag-pickers of Seemapuri consider garbage as nothing less than gold. For the elders it is their only means of survival as it provides them with their daily bread. For the innocent chil¬dren it is wrapped in wonder as they marvel at the chance of finding a coin, a currency note or a curio that really thrills them and gives them a hope of finding more elusive notes.

Question 29.
“It is his karam, his destiny.” What is Mukesh’s family’s attitude towards their situation? (All India 2015)
Answer:
Mukesh’s family have accepted their misery and impoverished condition as factors that have been ordained by destiny. Years of depravation and suffering has made them accept their condition passively in the name of fate or destiny. They feel that a God-given lineage can never be broken and have accepted bangle making as his destiny.

Question 30.
Describe the irony in Saheb’s name. (Delhi 2016)
Answer:
Saheb’s full name is Saheb-e-Alam which means ‘Lord of the Universe’. But ironically Saheb is a poverty-stricken ragpicker who scrounges the garbage dumps to earn his livelihood. His name is in complete contrast to his miserable existence.

Question 31.
What does the reference to chappals in ‘Lost Spring7 tell us about the economic condition of the rag pickers? (All India 2016)
Answer:
The reference to chappals in ‘Lost Spring’ tells us that the ragpickers were poverty-stricken. The fact that they are not able to buy chappals reflects their extreme state of poverty because of which they are unable to buy basic things.

Question 32.
How was Mukesh different from other bangle makers? (Comptt. Delhi 2016)
Answer:
Mukesh was different from other bangle makers because he wanted to be his own master. He had a dream of becoming a motor mechanic whereas other bangle makers did not even dare to dream but had accepted their fate.

Question 33.
Why was Saheb unhappy while working at the tea-stall? (Comptt. Delhi 2016)
Answer:
Saheb was unhappy while working at the tea- stall because he was no longer the master of his own life. He lost his freedom and carefree look. He had to live and work under the instructions of the owner of the tea-stall. He was not at liberty to go out and spend time with his friends.

Question 34.
Which industry was a boon and also bane for the people of Firozabad? How? (Comptt. Delhi 2017)
Answer:
The glass-bangles making industry was a boon and also bane for the people of Firozabad. The industry has given them a means of livelihood but the hazardous working conditions in the hot furnaces take a toll on their physical health.

Question 35.
How are Saheb and Mukesh different from each other? (Comptt. All India 2017)
Answer:
Mukesh’s attitude towards life was different from that of Saheb. Unlike Saheb he was optimistic about his future and so he dared to dream. He wanted to become a motor mechanic and also wanted to learn to drive a car. Saheb lacked determination so he harboured no dreams or ambitions about his future.

Lost Spring Important Questions Long Answer Type Questions (5-6 marks)

Question 36.
The bangle-makers of Ferozabad make bea-utiful bangles and make everyone happy but they live and die in squalor. Elaborate. (Delhi 2010)
Answer:
Through the story of the bangle-makers of Ferozabad, the author expresses her concern over their exploitation in the hazardous job of bangle-making. Extreme poverty, hard work and dismal working conditions result in the loss of the childhood of children who are in this profession. The working conditions of all bangle-makers are pathetic and miserable. They work in high temperature, badly lit and poorly ventilated glass furnaces due to which child workers especially are at risk of losing their eyesight at an early age and get prone to other health hazards. The stinking lanes of Ferozabad are choked with garbage and humans and animals live together in these hovels. There is no development or progress in their lives with the passage of time. They have no choice but to work in these inhuman conditions. Mind-numbing toil kills their dreams and hopes. They are condemned to live and die in squalor, subjected to a life of poverty and perpetual exploitation.

Question 37.
Why did Saheb become a ragpicker? What did j he look for in the garbage dumps? (Comptt. Delhi 2010)
Answer:
Saheb belongs to a Bangladeshi refugee family that migrated to Delhi from Dhaka in the wake of the 1971 Indo-Pak war. Here he finds himself in a vicious circle of social stigma, poverty and exploitation. He represents a growing number of refugee migrant poor population who are forced to lead a life of penury. This migratory population settle on the periphery of big cities and start doing menial jobs to earn a living. Due to extreme poverty Saheb’s parents are unable to provide for him and so he ends up picking rags for his own survival as well as to support his parents. The author says that Saheb scrounges for ‘gold’ in the garbage. Anything valuable like used clothes, shoes, bits of metal, plastic scrap, stray coins and currency notes can be termed as gold for them.

Question 38.
What change did Anees Jung see in Saheb when she saw him standing by the gate of the neighbourhood club? (Comptt. Delhi 2010)
Answer:
Saheb was a poor ragpicker who later takes up a job at a tea-stall in an attempt to be a master of his own destiny. But unfortunately this move further enslaves him. His new job replaces his light polythene bag with a heavy steel canister.
It even deprives him of roaming around with his friends without a care in the world. Earlier, though he did not have the security of a regular income, he had his freedom, and later he did have an assured income at the end of the month but he had lost his freedom. He was no more a free bird and his own master. He appeared burdened and forlorn. He was now a bonded labour who had surrendered his freedom. From being a spirited free bird who was not answerable to anyone he has become bound. He lives in a society where there is utter lack of compassion and commitment for the upliftment of these unfortunate children.

Question 39.
Describe the life of squatters at Seemapuri. (Comptt. All India 2010)
Answer:
Most of the squatters at Seemapuri were refugees from Bangladesh.
Also:
Most of the people like Saheb-e-Alam settled in Seemapuri were refugees from Bangladesh
who had fled their country and migrated to Delhi from Dhaka in the wake of the 1971 Indo- Pak war. Their dwellings were structures of mud, tin and tarpaulin with no sewage, drainage or running water. Picking garbage and rags helped them to earn their daily bread, gave them a roof over their heads and was their only means of livelihood and survival. Though these squatters of Seemapuri have no identity but they do have valid ration cards that enable them to buy grain. Living in Seemapuri, which is on the periphery of Delhi, is like living in hell. Children here grow up to become partners in survival to their parents. An army of barefoot children appears every morning, carrying their plastic bags on their shoulders and disappear by noon. They are forced to live a life of abject poverty that results in the loss of childhood innocence.
Saheb, a ragpicker, roamed in the streets, scrounging for garbage, barefoot and deprived of education. Later he starts working in a tea stall but he loses his freedom and carefree life as he is no longer his own master.

Question 40.
What does Anees Jung tell us about life at Mukesh’s home in Ferozabad? (Comptt. AT 2010)
Answer:
Mukesh’s father represents the underpaid, over exploited bangle makers of Ferozabad who is a victim of his own caste and is caught in a vicious circle of Sahnkars, middlemen, politi¬cians and policemen. He leads a hand-to-mouth existence in a shack with his family which includes two elderly parents, two sons and a daughter-in-law. Lack of education and awareness, the stigma of caste and a vicious nexus of people who exploit them have killed all initiative and drive in the young and the old. Time seems to stand still in Mukesh’s home in Ferozabad. There is no progress and no development despite years of mind- numbing toil. All the labourers of Ferozabad are victims of middlemen and touts. Their desire to dream and dare is snubbed in their childhood. They have no choice but to accept their subservience silently as their spirit is broken and their initiative dormant.

Question 41.
Give a brief account of the life and activities of the people like Saheb-e-Alam settled in Seemapuri. (Delhi 2011 )
Answer: Refer to Question 49, Page 166

Question 42.
‘Lost Spring’ explains the grinding poverty and traditions that condemn thousands of people to a life of abject poverty. Do you agree? Why/Why not? (All India 2011)
Answer:
‘Lost Spring’ does indeed highlight the miserable plight of thousands of poor people whose life is completely marred by abject poverty and thoughtless traditions. They work extremely hard in the most pathetic conditions and accept poverty and exploitation as their destiny. Through the lives of Saheb-e-Alam, a ragpicker, and Mukesh, a bangle maker, the author highlights the vicious circle of social stigma and poverty which these people are subjected to. Saheb and Mukesh also represent a growing number of refugee migrants and people who are forced to live a life of penury. No one shows any kind of compassion or sensitivity to their pathetic plight and there is also no initiative or commitment for the upliftment of these downtrodden people. Acute poverty, no education and no infrastructural development has drained their energy and willpower and they have no choice but to accept their destiny of inevitable poverty.

Question 43.
What circumstances forced Mukesh not to pursue his family business of bangle making? Instead, what did he decide to do?
(Comptt. Delhi 2012)
Answer:
Mukesh dares to dream of a different life and decides not to pursue his family business of bangle-making. He does not want to accept his life of misery in the name of destiny. Though he is born in a poverty-ridden family in the caste of bangle makers he dreams of a better future. He wants to break free from the vicious circle of sahukars and middlemen and carve a new beginning for himself by becoming a motor- mechanic. He knows what it is like to work in glass furnaces that are neither well-lit nor well- ventilated. They are dingy hovels with high temperatures. He has seen that the youngsters are weighed down by the baggage of generations of subservience and have forgotten to dream of an alternative world. So Mukesh’s dream of going to a garage and learning to be a motor-mechanic is an attempt to break free off the mind-numbing toil.

Question 44.
In 1971 Bangladeshi migrants came to Delhi ‘looking for gold in the big city’. What kind of life are they living in Seemapuri now?
(Comptt. Delhi 2012)
Answer:
Most of the people like Saheb-e-Alam settled in Seemapuri were refugees from Bangladesh
who had fled their country and migrated to Delhi from Dhaka in the wake of the 1971 Indo- Pak war. Their dwellings were structures of mud, tin and tarpaulin with no sewage, drainage or running water. Picking garbage and rags helped them to earn their daily bread, gave them a roof over their heads and was their only means of livelihood and survival. Though these squatters of Seemapuri have no identity but they do have valid ration cards that enable them to buy grain. Living in Seemapuri, which is on the periphery of Delhi, is like living in hell. Children here grow up to become partners in survival to their parents. An army of barefoot children appears every morning, carrying their plastic bags on their shoulders and disappear by noon. They are forced to live a life of abject poverty that results in the loss of childhood innocence.
Saheb, a ragpicker, roamed in the streets, scrounging for garbage, barefoot and deprived of education. Later he starts working in a tea stall but he loses his freedom and carefree life as he is no longer his own master.

Question 45.
Describe the difficulties the bangle makers of Firozabad have to face in their lives. (Delhi 2015 2012)
Answer:
Through the story of the bangle-makers of Ferozabad, the author expresses her concern over their exploitation in the hazardous job of bangle-making. Extreme poverty, hard work and dismal working conditions result in the loss of the childhood of children who are in this profession. The working conditions of all bangle-makers are pathetic and miserable. They work in high temperature, badly lit and poorly ventilated glass furnaces due to which child workers especially are at risk of losing their eyesight at an early age and get prone to other health hazards. The stinking lanes of Ferozabad are choked with garbage and humans and animals live together in these hovels. There is no development or progress in their lives with the passage of time. They have no choice but to work in these inhuman conditions. Mind-numbing toil kills their dreams and hopes. They are condemned to live and die in squalor, subjected to a life of poverty and perpetual exploitation.

Question 46.
Describe the circumstances which keep the workers in the bangle industry in poverty. (Delhi 2015)
Answer:
Through the story of the bangle makers of Firozabad, the author expresses concern over their exploitation in the hazardous job of bangle making and addresses the circumstances which keep the workers in poverty. They live in stinking lanes, choked with garbage in homes. Their houses are hovels with families of humans and animals coexisting in a primeval state. They cannot organise themselves into cooperatives. Their families are caught in a web of poverty and in a vicious circle of the sahukars, the middlemen, the keepers of law, the policemen, the bureaucrats and the politicians who impose on them a baggage which they cannot put down. They move in a spiral from poverty, to apathy, to greed and to injustice.

Question 47.
How is Mukesh’s attitude towards his situation different from that of Saheb? Why? (Delhi 2015)
Answer:
Mukesh’s attitude towards his situation is different from that of Saheb. Mukesh is more of a rebel who dares to be different and wishes to become a motor mechanic. Though, Mukesh too, like his community, is working in back breaking, mind-numbing glass industry but unlike his peers, the spark in him has not extinguished. He wants to break free from the vicious circle which his community has been caught in. Saheb, on the other hand, has enslaved himself. By taking up work in the tea stall he is no longer his own master. The difference in their attitude towards their situation can be attributed to the fact that Saheb is a rootless migrant from Bangladesh and Mukesh is a citizen of India. Moreover, Mukesh dares to dream. The author too senses a flash of daring in Mukesh and this is what makes his attitude a little more aggressive than Saheb.

Question 48.
“It is his karam, his destiny” that made Mukesh’s grandfather go blind. How did Mukesh disprove this belief by choosing a new vocation and making his own destiny?
(Comptt. Delhi 2015)
Answer:
Mukesh disproved the belief of “his karam, his destiny” by choosing a new vocation in an attempt to make his own destiny. He decided to become a motor mechanic. He mustered the courage to break free from the family lineage of bangle making. He had seen his parents and others suffering because of the vicious circle of poverty and exploitation that they were caught in. Mukesh was ready to walk a long distance to reach the garage to learn the vocation of car mechanic. His determination is going to prove instrumental in helping him to realize his dream. Mukesh dares to dream of a different life and decides not to pursue his family business of bangle-making. He does not want to accept his life of misery in the name of destiny. Though he is born in a poverty-ridden family in the caste of bangle makers he dreams of a better future. He wants to break free from the vicious circle of sahukars and middlemen and carve a new beginning for himself by becoming a motor-mechanic. He knows what it is like to work in glass furnaces that are neither well-lit nor well-ventilated. They are dingy hovels with high temperatures. He has seen that the youngsters are weighed down by the baggage of generations of subservience and have forgotten to dream of an alternative world. So Mukesh’s dream of going to a garage and learning to be a motor-mechanic is an attempt to break-free off the mind-numbing toil.

Question 49.
What kind of life did Saheb lead at Seemapuri? (Comptt. All India 2016 )
Answer:
Most of the people like Saheb-e-Alam settled in Seemapuri were refugees from Bangladesh
who had fled their country and migrated to Delhi from Dhaka in the wake of the 1971 Indo- Pak war. Their dwellings were structures of mud, tin and tarpaulin with no sewage, drainage or running water. Picking garbage and rags helped them to earn their daily bread, gave them a roof over their heads and was their only means of livelihood and survival. Though these squatters of Seemapuri have no identity but they do have valid ration cards that enable them to buy grain. Living in Seemapuri, which is on the periphery of Delhi, is like living in hell. Children here grow up to become partners in survival to their parents. An army of barefoot children appears every morning, carrying their plastic bags on their shoulders and disappear by noon. They are forced to live a life of abject poverty that results in the loss of childhood innocence.
Saheb, a ragpicker, roamed in the streets, scrounging for garbage, barefoot and deprived of education. Later he starts working in a tea stall but he loses his freedom and carefree life as he is no longer his own master.

Question 50.
Describe the living conditions prevailing in Firozabad. Mention why the bangle making business does not give bangle makers a comfortable life. (Comptt. All India 2016)
Answer:
Through the story of the bangle-makers of Ferozabad, the author expresses her concern over their exploitation in the hazardous job of bangle-making. Extreme poverty, hard work and dismal working conditions result in the loss of the childhood of children who are in this profession. The working conditions of all bangle-makers are pathetic and miserable. They work in high temperature, badly lit and poorly ventilated glass furnaces due to which child workers especially are at risk of losing their eyesight at an early age and get prone to other health hazards. The stinking lanes of Ferozabad are choked with garbage and humans and animals live together in these hovels. There is no development or progress in their lives with the passage of time. They have no choice but to work in these inhuman conditions. Mind-numbing toil kills their dreams and hopes. They are condemned to live and die in squalor, subjected to a life of poverty and perpetual exploitation.

Question 51.
‘Garbage to them is gold’. How do ragpickers of Seemapuri survive? (Delhi 2017)
Answer:
Garbage, for the ragpickers of Seemapuri, is considered to be nothing less than gold. Picking garbage helps them to earn their daily bread, gives them a roof over their head and is their only means of livelihood and survival. For the innocent children garbage is wrapped in wonder as they marvel at the prospect of finding a coin, a currency note or a curio that sustains their hope. These children grow up to become partners in survival with their parents. Seemapuri houses around 10,000 ragpickers mostly Bangladeshi refugees, who have lived there for more than thirty years without an identity and without permits. These people live in mud structures with roofs made of tin and tarpaulin. The ration cards that enable them to get their names on voters’ lists and enable them to buy grain and garbage are their means of survival. Food is more important to them for survival than an identity.

Question 52.
“For the children it is wrapped in wonder, for the elders it is a means of survival.” What kind of life do the rag-pickers of Seemapuri lead? (Comptt. All India 2017)
Answer:
Garbage, for the ragpickers of Seemapuri, is considered to be nothing less than gold. Picking garbage helps them to earn their daily bread, gives them a roof over their head and is their only means of livelihood and survival. For the innocent children garbage is wrapped in wonder as they marvel at the prospect of finding a coin, a currency note or a curio that sustains their hope. These children grow up to become partners in survival with their parents. Seemapuri houses around 10,000 ragpickers mostly Bangladeshi refugees, who have lived there for more than thirty years without an identity and without permits. These people live in mud structures with roofs made of tin and tarpaulin. The ration cards that enable them to get their names on voters’ lists and enable them to buy grain and garbage are their means of survival. Food is more important to them for survival than an identity.

Important Questions for Class 12 English

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Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions

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Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions

The Solid State Class 12 Important Questions Very Short Answer Type

Question 1.
Which point defect in crystals does not alter the density of the relevant solid? (Delhi) 2009
Answer:
Frenkel defect.

Question 2.
Which point defect in its crystal units alters the density of a solid? (Delhi) 2009
Answer:
Schottky defect.

Question 3.
Which point defect in its crystal units increases the density of a solid? (Delhi) 2009
Answer:
Metal excess defect increases the density of a solid. It is due to presence of extra cations in the interstitial sites.

Question 4.
How do metallic and ionic substances differ in conducting electricity? (All India) 2009
Answer:
The electrical conductivity in metallic substances is due to free electrons while in ionic substances the conductivity is due to presence of ions.

Question 5.
Which point defect of its crystals decreases the density of a solid? (Delhi & All India) 2009
Answer:
Schottky defect.

Question 6.
What is the number of atoms in a unit cell of a face-centred cubic crystal? (All India) 2009
Answer:
The number of atoms in a unit cell of fcc-crysta! is 4 atoms.

Question 7.
Write a feature which will distinguish a metallic solid from an ionic solid. (Delhi) 2010
Answer:
The electrical conductivity in metallic solid is due to free electrons while in ionic solid the conductivity is due to presence of ions.

Question 8.
Which point defect in crystals of a solid does not change the density of the solid? (Delhi) 2010
Answer:
Frenkel defect.

Question 9.
Which point defect in crystals of a solid decreases the density of the solid? (Delhi) 2010
Answer:
Schottky defect.

Question 10.
What type of interactions hold the molecules together in a polar molecular solid? (All India) 2010
Answer:
Dipole-dipole forces of attractions hold the molecules together in a polar molecular solid.

Question 11.
What type of semiconductor is obtained when silicon is doped with arsenic? (All India) 2010
Answer:
n-type semiconductor.

Question 12.
Write a distinguishing feature of metallic solids. (All India) 2010
Answer:
Metallic solids possess high electrical and thermal conductivity due to presence of free electrons.

Question 13.
‘Crystalline solids are anisotropic in nature.’ What does this statement mean? (Delhi) 2011
Answer:
It means that crystalline solids show different values of their some properties like electrical conductivity, refractive index, thermal expansion etc. in different directions.

Question 14.
Which stoichiometric defect in crystals increases the density of a solid? (Delhi) 2011
Answer:
Interstitial defect in crystals increases the density of a solid.

Question 15.
What is meant by ‘doping’ in a semiconductor? (Delhi) 2012
Answer:
Addition of a suitable impurity to a semiconductor to increase its conductivity is called doping.

Question 16.
Write a point of distinction between a metallic solid and an ionic solid other than metallic lustre. (Delhi) 2012
Answer:
Metallic solid conducts electricity in solid state but ionic solids do so only in molten state or in solution or metals conduct electricity through electrons and ionic substances through ions. Metallic solids are malleable and ductile while ionic solids are hard and brittle.

Question 17.
How may the conductivity of an intrinsic semiconductor be increased? (All India) 2012
Answer:
The conductivity is increased by adding an appropriate amount of suitable impurity. This process is called as intrinsic doping.

Question 18.
Which stoichiometric defect increases the density of a solid? (All India) 2012
Answer:
Interstitial defect increases the density of a solid.

Question 19.
What are n-type semiconductors? (All India) 2012
Answer:
n-type semiconductor : They are obtained by doping silicon with an element of group15, like P, As etc.

Question 20.
What type of stoichmetric defect is shown by AgBr and Agl ? (Comptt. All India) 2012
Answer:
AgBr shows both Frenkel defect and Schottky defect whereas Agl shows Frenkel defect.

Question 21.
What type of defect can arise when a solid is heated ? (Comptt. All India) 2012
Answer:
Vacancy defects can arise when a solid is heated.

Question 22.
Why does LiCl acquire pink colour when heated in Li vapours? (Comptt. All India) 2012
Answer:
This is due to metal excess defect due to anionic vacancies in which the anionic sites are occupied by unpaired electrons (F-centres).

Question 23.
How many atoms constitute one unit cell of a face-centered cubic crystal? (Delhi) 2013
Answer:
4 atoms constitute one unit cell of a fee crystal.

Question 24.
What type of stoichiometric defect is shown by AgCl? (Delhi) 2013
Answer:
Frenkel defect is shown by AgCl.

Question 25.
What type of substances would make better Permanent Magnets: Ferromagnetic or Ferrimagnetic? (Delhi) 2013
Answer:
Ferromagnetic substances would make better I permanent magnets
Example : Fe, Co, Ni etc.

Question 26.
Calculate the number of atoms in a face centred cubic unit cell. (Comptt. Delhi) 2013
Answer:
In face centered cubic arrangement, number of lattice points are : 8 + 6.
∴ Lattice points per unit cell = 8 \times \frac{1}{8}+6 \times \frac{1}{2} = 4

Question 27.
On heating a crystal of KC1 in potassium vapour, the crystal starts exhibiting a violet colour. What is this due to? (Comptt. Delhi) 2013
Answer:
The Cl ions diffuse to the surface and combine j with atoms which get ionized by losing electrons. ! These electrons are trapped in anions vacancies j and act as F-centre which imparts violet colour to the crystal.

Question 28.
Which type of ionic substances show Schottky defect in solids? (Comptt. Delhi) 2013
Answer:
Highly ionic compounds with high coordination rjuniber and small difference in size of cations and anions show schottky defect.

Question 29.
How many atoms per unit cell (z) are present in bcc unit cell? (Comptt. Delhi) 2014
Answer:
Number of atoms in a unit cell of a body centred cubic structure :
Contribution by 8 atoms on the corners
= \frac{1}{8} × 8 = 1
Contribution by the atom presents within the body = 1
∴ Total number of atoms present in the unit cell = 1 + 1 = 2 atoms

Question 30.
What type of stoichiometric defect is shown by NaCl? (Comptt. Delhi) 2014
Answer:
Schottky defect is shown by NaCl.

Question 31.
Write a distinguishing feature between a metallic solid and an ionic solid. (Comptt. Delhi) 2014
Answer:
The electrical conductivity in metallic substances is due to free electrons while in ionic substances the conductivity is due to presence of ions.

Question 32.
Why are crystalline solids anisotropic? (Comptt. All India) 2014
Answer:
Crystalline solids show different values of their some properties like electrical conductivity, refractive index, thermal expansion etc. in different directions.

Question 33.
What is meant by ‘antiferromagnetism’? (Comptt. All India) 2014
Answer:
Antiferromagnetism : These substances possess zero net magnetic moment because of presence of equal number of electrons with opposite spins.

Question 34.
Write a distinguishing feature of a metallic solid compared to an ionic solid. (Comptt. All India) 2014
Answer:
Metallic solid conducts electricity in solid state but ionic solids do so only in molten state or in solution or metals conduct electricity through electrons and ionic substances through ions. Metallic solids are malleable and ductile while ionic solids are hard and brittle.

Question 35.
What is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy 1/3rd of tetrahedral voids? (Delhi) 2015
Answer:
Formula is X2Y3.

Question 36.
What is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy 2/3rd of tetrahedral voids? (All India) 2015
Answer:
Y atoms are N (No. of tetrahedral voids are 2N), No. of tetrahedral voids occupied by X are
\frac{2}{3} × 2N = \frac{4 \mathrm{N}}{3}
X : Y = 4N : 3N
Formula : X4Y3

Question 37.
What is the no. of atoms per unit cell (z) in a body-centred cubic structure? (Comptt. Delhi) 2015
Answer:
Contribution by the atoms present at eight comers = 8 × \frac{1}{8} = 1
Contribution by the atoms present at centre = 1
Total number of atoms present in unit cell = 1 + 1 = 2

Question 38.
What type of stoichiometric defect is shown by AgCl? (Comptt. All India) 2015
Answer:
AgCl shows Frenkel defect.

Question 39.
What type of magnetism is shown by a substance if magnetic moments of domains are arranged in same direction? (Delhi) 2016
Answer:
Ferromagnetism is shown by a substance if magnetic moments of domains are arranged in same direction.
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 1

Question 40.
Give an example each of a molecular solid and an ionic solid. (All India) 2016
Answer:
Molecular solid → Iodine (I2)
Ionic solid → Sodium chloride (NaCl)

Question 41.
A metallic element crystallises into a lattice having a pattern of AB AB … and packing of spheres leaves out voids in the lattice. What type of structure is formed by this arrangement? (Comptt Delhi) 2017
Answer:
Tetrahedral void is formed in AB AB … pattern. The hexagonal close packing (hep) is formed in this arrangement.

Question 42.
A metallic element crystallises into a lattice having a ABC ABC … pattern and packing of spheres leaves out voids in the lattice. What type of structure is formed by this arrangement? (Comptt. Delhi) 2017
Answer:
Octahedral voids are formed in ABC ABC … pattern. The cubic close packing (ccp) is formed in this arrangement.

Question 43.
What type of Stoichiometric defect is shown by AgCl? (Comptt. Delhi) 2017
Answer:
Frenkel defect.

Question 44.
What type of stoichiometric defect is shown by NaCl? (Comptt. All India) 2017
Answer:
Schottky defect is shown by NaCl.

Question 45.
Which ionic compound shows both Frenkel and Schottky defects? (Comptt. All India) 2017
Answer:
Silver bromide (AgBr) shows both Schottky and Frenkel defect.

The Solid State Class 12 Important Questions Short Answer Type – I (SA – 1)

Question 46.
Explain how you can determine the atomic mass of an unknown metal if you know its mass density and the dimensions of unit cell of its crystal. (All India) 2011
Answer:
Suppose edge of the unit cell = a pm
Number of atoms present per unit cell = Z
∴ Volume of unit cell = (a pm)3
= (a × 10-10cm)3 = a3 × 10-30 cm3
Density of unit cell = \frac{\text { Mass of unit cell }}{\text { Volume of unit cell }} ……………… (i)
Mass of unit cell = Number of atoms in the unit cell × mass of each atom
= Z × m
Mass of each atom = \frac{\text { Atomic mass }}{\text { Avogadro's no. }}=\frac{M}{N_{0}}
Substituting these values in equation (i), we get
Density of unit cell = \frac{Z \times M}{a^{3} \times 10^{-30} \times N_{0}}
If a is in cm, d = \frac{\mathrm{Z} \times \mathrm{M}}{a^{3} \times \mathrm{N}_{0}} g/cm3
∴ Molar mass can be calculated as
M = \frac{d \times a^{3} \times N_{0}}{Z}

Question 47.
Calculate the packing efficiency of a metal crystal for a simple cubic lattice. (All India) 2011
Answer:
Percentage efficiency of packing of simple cubic lattice = 52.4%.

Question 48.
Define the following terms in relation to crystalline solids :
(i) Unit cell (ii) Coordination number Give one example in each case. (All India) 2011
Answer:
(i) Unit cell : The smallest three dimensional portion of a complete space lattice which when repeated over and again in different directions produces the complete space lattice is called the unit cell.
Example: Cubic unit cell, Hexagonal unit cell etc.
(ii) Coordination number : The number of nearest spheres with which a particular sphere is in contact is called co-ordination number.
Example : Co-ordination number of hexagonal (hep) structures is 12.

Question 49.
The unit cell of an element of atomic mass 108 u and density 10.5 g cm-3 is a cube with edge length, 409 pm. Find the type of unit cell of the crystal. [Given : Avogadro’s constant = 6.023 × 1023 mol-1] (Comptt. Delhi) 2012
Answer:
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 2
So it forms cubic- closed packed (ccp) lattice or fee structure.

Question 50.
Explain the following terms with suitable examples : Ferromagnetism and Ferrimagnetism (Comptt. Delhi) 2012
Answer:
Ferromagnetic solids : The solids which are strongly attracted by external magnetic field and do not lose their magnetism when the external field is removed are called ferromagnetic solids. The property, thus exhibited, is termed as ferromagnetism.
Example: Fe, Co and Ni show ferromagnetism at room temperature.

Ferrimagnetic solids : The solids which are expected to show large magnetism due to the presence of unpaired electrons but in fact have small net magnetic moment, are called ferrimagnetic solids.
Example : Fe3O4 and ferrites.

Question 51.
An element X crystallizes in f.c.c structure. 208 g of it has 4.2832 × 1024 atoms. Calculate the edge of the unit cell, if density of X is 7.2 g cm-3. (Comptt. Delhi) 2012
Answer:
Z = 4(fcc) A = 7,2 g/cm3 a = ?
4.2832 × 1024 atoms have mass = 208 g
6.022 × 1023 atoms have mass
= \frac{208}{4.2832 \times 10^{24}} × 6.022 × 1023 = 29.24 (at. mass)
a3 = \frac{\mathrm{Z} \times \mathrm{M}}{d \times \mathrm{N}_{\mathrm{A}}}=\frac{4 \times 29.24}{7.2 \times 6.022 \times 10^{23}}
= 269.6 x 10-24 cm3
∴ a = 6.46 x 10-8 cm = 6.46 Å

Question 52.
What is a semiconductor? Describe the two main types of semiconductors. (Comptt. Delhi) 2012
Answer:
Semiconductor : The solid materials whose electrical conductivity lies between those of the typical metallic conductors and insulators are termed as semiconductors. The semiconductors possess conductivity in the range of 102 to 10-9 ohm-1 cm-1.
These are of two types :
(a) n-type semiconductors : Doping of higher group element impurity forms n-type semiconductors, e.g. when ‘As’ is doped in ‘Ge’.
(b) p-type semicondctors : Impurity of lower groups forms electron deficient bond in the structure. Electron deficiency develops to p-hole.

Question 53.
Account for the following:
(i) Schottky defects lower the density of related solids.
(ii) Conductivity of silicon increases on doping it with phosphorus. (All India) 2013
Answer:
(i) Schottky defect produced due to missing of equal number of cation and anion from lattice as a result of which the density of the lattice solid decreases.
(ii) The conductivity of silicon increases due to negatively charged extra electron of doped pentavalent phosphorus.

Question 54.
Aluminium crystallizes in an fee structure. Atomic radius of the metal is 125 pm. What is the length of the side of the unit cell of the metal? (All India) 2013
Answer:
For fee, Formula : r = \frac{a}{2 \sqrt{2}}
Given: r = 125 pm
∴ a = 2 \sqrt{2} r ÷ a = 2 \sqrt{2} × 125
⇒ a = 2 × 1.414 × 125 = 353.5 pm

Question 55.
(a) Why does presence of excess of lithium makes LiCl crystals pink?
(b) A solid with cubic crystal is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? (All India) 2013
Answer:
(a) This is due to metal excess defect due to anionic vacancies in which the anionic sites are occupied by unpaired electrons (F-centres).
(b) As atoms of Q are present at the 8 centres of the cube, therefore, number of atoms of Q in the unit cell = \frac{1}{8} × 8 = 1
The atom P is at the body centre .-. Number of atoms = 1
Ratio of atoms P : Q = 1 : 1
Hence, the formula of the compound is PQ.

Question 56.
(a) What change occurs when AgCl is doped with CdCl2?
(b) What type of semiconductor is produced when silicon is doped with boron? (All India) 2013
Answer:
(a) Impurity defect of ionic solids is produced when AgCl is doped with CdCl2. Due to this defect vacancies are created that result in higher electrical conductivity of the solid.
(b) p-type semi-conductor is obtained when silicon is doped with boron.

Question 57.
If NaCl is doped with 10-3 mole percent SrCl2, what will be the concentration of cation vacancies? (NA = 6.02 × 1023 mol-1) (Comptt. All India) 2013
Answer:
10-3 mol percent means 100 moles of NaCl are doped with 10-3 moles of SrCl2
∴ 1 mole of NaCl is doped with SrCl2
= \frac{10^{-3}}{100} = 10-5 mole
Since each Sr2+ ion introduces one cation vacancy
∴ Concentration of cation vacancies
= 10-5 mol/mol of NaCl
= 10-5 × 6.02 × 1023 mol-1
= 6.02 × 1018 mol-1

Question 58.
What is a semiconductor? Describe the tivo main types of semiconductors and contrast their conduction mechanism. (Comptt. All India) 2013
Answer:
Semiconductor : The solids which have intermediate conductivities between metals and non-metals i,e. between 10-6 to 104 π-1 m-31 are called semiconductors.
Example : Germanium and Silicon.
Main types of semiconductors are of two types :
(i) Intrinsic semiconductor : These are insulators at room temperature and become semiconductors when temperature is raised
(ii) Extrinsic semiconductor :
p-type semiconductor
n-type semiconductor
These are formed by dopping impurity of lower or higher group.
These are subdivided into two types :
• p-type semiconductor : When a silicon crystal is doped with atoms of group-13 elements like B, Al, Ga etc., the atom forms only 3 covalent bonds with the Si atom and 4th missing electron creates a hole which conducts electricity.
• n-type semiconductor : When a silicon crystal is doped with atoms of group-15 elements like P, As etc., then only four of the five valence electrons of each impurity atom, participate in 4 covalent bond formation and 5th e conducts electricity.

Question 59.
A compound forms hep structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids? (Comptt. All India) 2013
Answer:
No. of atoms in the hep = 0.5 × 6.022 × 1023
= 3.011 × 1023
No. of octahedral voids
= No. of atoms in packing = 3.011 × 1023
No. of tetrahedral voids
= 2 × No. of atoms in packing
= 2 × 3.011 × 1023 = 6.022 × 1023
∴ Total no. of voids
= 3.011 × 1023 + 6.022 × 1023 = 9.033 × 1023

Question 60.
An element crystallizes in a structure having fee unit cell of an edge 200 pm. Calculate the density if 200 g of this element contains 24 × 1023 atoms. (Comptt. All India) 2013
Answer:
24 × 1023 atoms of an element have mass = 200 g
∴ 6.022 × 1023 atoms of an element have mass
= \frac{200}{24 \times 10^{23}} × 6.022 × 1023 = 50.18 g
Given : a = 200 pm = 200 × 10-12 cm,
Z = 4 (For fee), M = 50.18 g
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 3

Question 61.
An element with density 11.2 g cm-3 forms a f.c.c. lattice with edge length of 4 × 10-8 cm. Calculate the atomic mass of the element. (Given : NA = 6.022 × 1023 mol-1) (Delhi) 2014
Answer:
Given : p = 11.2 g cm-3, a = 4 × 10-8 cm
For fee lattice, Z = 4
Using formula,
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 4

Question 62.
Examine the given defective crystal 2014
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 5
Answer the following questions :
(i) What type of stoichiometric defect is shown by the crystal?
(ii) How is the density of the crystal affected by this defect?
(tii) What type of ionic substances show such defect? (Delhi)
Answer:
(i) Schottky defect
(ii) Density of the crystal decreases
(iii) NaCl (Ionic solids having approximate equal size of cations and anions)

Question 63.
An element with density 2.8 g cm-3 forms a f.c.c. unit cell with edge length 4 × 10-8 cm. Calculate the molar mass of the element.
(Given : NA = 6.022 × 1023mol-1) (All India) 2014
Answer:
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 6

Question 64.
(i) What type of non-stoichiometric point
defect is responsible for the pink colour of LiCl?
(ii) What type of stoichiometric defect is shown by NaCl? (All India) 2014
Answer:
(i) This is due to metal excess defect due to anionic vacancies in which the anionic sites are occupied by unpaired electrons (F-centres).
(ii) Schottky defect is shown by NaCl.

Question 65.
How will you distinguish between the following pairs of terms :
(i) Tetrahedral and octahedral voids
(ii) Crystal lattice and unit cell (All India) 2014
Answer:

Tetrahedral voidsOctahedral voids
1.It is much smaller than the size of spheres in  the packing.Size is much larger than tetrahedral voids.
2.Each tetrahedral void is surrounded by 4 spheres. Hence, co­ordination no. is 4.Each octahedral void is surrounded by 6 spheres. Hence, its co­ordination no. is 6.

(ii) A regular arrangement of the constituent particles of a crystal in a three dimensional space is called crystal lattice.
The smallest three dimensional portion of a complete crystal lattice, which when repeated over and again in different directions produces the complete crystal lattice is called the unit cell.

Question 66.
(i) Write the type of magnetism observed when the magnetic moments are oppositively aligned and cancel out each other.
(ii) Which stoichiometric defect does not change » the density of the crystal? (All India) 2014
Answer:

  1. Diamagnetism is observed when the magnetic moments are oppositively aligned and cancel out each other.
  2. Frenkel defect does not change the density of the crystal.

Question 67.
(i) Write the type of magnetism observed when the magnetic moments are aligned in parallel and anti-parallel directions in unequal numbers.
(ii) Which stoichiometric defect decreases the density of the crystal? (All India) 2014
Answer:

  1. Ferrimagnetism is observed.
  2. Schottky defect decreases the density of the crystal.

68. Define the following terms: (Comptt. Delhi) (2016)
(i) n-type semiconductor
(ii) Ferrimagnetism
Answer:
(i) n-type semiconductor : When Si/Ge is doped with group 15 element.
(ii) Ferrimagnetism : When magnetic domains are aligned in parallel and anti-parallel directions in unequal numbers.

Question 69.
Explain the following terms with suitable examples : (Comptt. All India) (2016)
(i) Frenkel defect (ii) F-centres
Answer:
(i) Frenkel defect : The defect in which the smaller ion/cation is dislocated to a nearby interstitial site.
Example : Silver halides, ZnS.
(ii) F-centres : The anion vacancy occupied by an electron is called F-centre in Alkali metal halides.
Example : NaCl, KC1, Li Cl.

Question 70.
Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in a face-centered cubic (f.c.c.) structure. (Atomic mass of Al = 27 g mol-1) (Comptt. All India) 2017
Answer:
1 mole of Aluminium = 27 g = 6.022 × 1023
Hence, No. of atoms present in 27 g of Al
= \frac{6.022 \times 10^{23}}{27}
As f.c.c. unit cell contains 4 atoms
∴ No. of f.c.c. unit cells present
= \frac{6.022 \times 10^{23} \times 8.1}{27 \times 4}
= 0.45165 × 1023 = 4.5165 × 1022

The Solid State Class 12 Important Questions Short Answer Type – II [SA II]

Question 71.
Iron has a body centred cubic unit cell with a cell edge of 286.65 pm. The density of iron is 7.87 g cm-3. Use this information to calculate Avogadro’s number (At. mass of Fe = 56 g mol-1). (Delhi & All India) 2009
Answer:
Given :
a = 286.65 pm = 286.65 × 10-10,
d = 7.87 g cm-3, M = 56 g mol-1
Z = 2 NA = ?
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 7
∴ Avogadro’s number NA = 6.022 × 1023

Question 72.
Silver crystallises with face-centred cubic unit cells. Each side of the unit cell has a length of 409 pm. What is the radius of an atom of silver? (Assume that each face atom is touching the four comer atoms.) (All India) 2009
Answer:
Given : a = 409 pm r = ?
For fee unit cell, the formula is a
r = \frac{a}{2 \sqrt{2}}
or r = \frac{409}{2 \sqrt{2}}=\frac{409}{2 \times 1.414}=\frac{409}{2.828}
∴ r = 144.62
∴ Radius of an atom of silver = 144.62 pm

Question 73.
The well known mineral fluorite is chemically calcium fluoride. It is known that in one unit cell of this mineral there are 4 Ca2+ ions and 8 F ions and that Ca2+ ions are arranged in a fee lattice. The F ions fill all the tetrahedral holes in the face centred cubic lattice of Ca2+ ions. The edge of the unit cell is 5.46 × 10-8 cm in length. The density of the solid is 3.18 g cm-3. Use this information to calculate Avogadro’s number (Molar mass of CaF2 = 78.08 g mol-1). (Delhi) 2010
Answer:
Given
Edge of the unit cell (a) = 5.46 × 10-8 cm
Density (P) = 3.18 g cm-3
According to the formula ;
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 8

Question 74.
The density of copper metal is 8.95 g cm-3. If the radius of copper atom is 127.8 pm, is the copper unit cell a simple cubic, a body-centred cubic or a face centred cubic structure?
(Given : At. mass of Cu = 63.54 g mol-1 and NA = 6.02 × 1023 mol-1) (Delhi & All India) 2010
Answer:
If copper atom were simple cubic :
a = 2 × r = 2 × 127.8 pm = 255.6 pm
= 255.6 pm = 255.6 × 10-10 cm
Z = 1
p = \frac{\mathrm{Z} \times \mathrm{M}}{a^{3} \times \mathrm{N}_{\mathrm{A}}}=\frac{1 \times 63.54}{\left(255.6 \times 10^{-10}\right)^{3} \times\left(6.02 \times 10^{23}\right)}
∴ P = 6.34 g cm-3
Actual density = 8.95 g cm-3
Hence copper atom is not simple cubic.
If copper atom were body-centred :
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 9
∴ P = 8.21 g cm-3
Hence, copper atom is not body centered
If copper atom were face-centered
a = 2 \sqrt{2}
or a = 2 × 1.414 × 127.8 pm
= 361.4 pm = 361.4 × 10-10 cm
P = \frac{\mathrm{Z} \times \mathrm{M}}{a^{3} \times \mathrm{N}_{\mathrm{A}}}=\frac{4 \times 63.54}{\left(361.4 \times 10^{-10}\right)^{3} \times 6.02 \times 10^{23}}
∴ P = 8.94 cm-3
Hence, copper is face-centred cubic.

Question 75.
Silver crystallises in face-centred cubic unit cells. Each side of the unit cell has a length of 409 pm. What is the radius of silver atom? (All India) 2010
Answer:
Refer to Q. 72, Page 9
Given : a = 409 pm r = ?
For fee unit cell, the formula is a
r = \frac{a}{2 \sqrt{2}}
or r = \frac{409}{2 \sqrt{2}}=\frac{409}{2 \times 1.414}=\frac{409}{2.828}
∴ r = 144.62
∴ Radius of an atom of silver = 144.62 pm

Question 76.
Silver crystallizes in face-centered cubic unit cell. Each side of this unit cell has a length of 400 pm. Calculate the radius of the silver atom. (Assume the atoms just touch each other on the diagonal across the face of the unit cell. That is each face atom is touching the four comer atoms.) (Delhi) 2011
Answer:
Given : a = 400 pm, r = ?
For fee unit cell : r = \frac{a}{2 \sqrt{2}}
or r = \frac{400}{2 \times 1.4142}=\frac{400}{2.828} ∴ r = 141.4
∴ Radius of the silver, r = 141.4 pm

Question 77.
The density of lead is 11.35 g cm-3 and the metal crystallizes with fee unit cell. Estimate the radius of lead atom.
(At. Mass of lead = 207 g mol-1 and NA = 6.02 × 1023 mol-1) (Delhi) 2011
Answer:
Given : d = 11.35 g cm-3
According to the formula
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 10

Question 78.
Tungsten crystallizes in body centred cubic unit cell. If the edge of the unit cell is 316.5 pm, what is the radius of tungsten atom? (Delhi) 2012
Answer:
For bee, unit cell : radius, r = \frac{\sqrt{3} a}{4} (∵ a = 316.5 pm)
= \frac{\sqrt{3} \times 316.5 \mathrm{pm}}{4}=\frac{1.732 \times 316.5 \mathrm{pm}}{4} = 137 pm

Question 79.
Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm-3. Use this information to calculate Avogadro’s number. (At mass of Fe = 55.845 u) (Delhi) 2012
Answer:
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 11
∴ Avogadro’s number, NA = 6.02 × 1023 mol-1

Question 80.
Copper crystallises with face centred cubic unit cell. If the radius of copper atom is 127.8 pm, calculate the density of copper metal.
(Atomic mass of Cu = 63.55 u and Avogadro’s number NA = 6.02 × 10223 mol-1) (All India) 2012
Answer:
In fee, lattice, Z = 4 atoms [fee = face centred cubic]
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 12

Question 81.
Iron has a body centred cubic unit cell with the cell dimension of 286.65 pm. Density of iron is 7.87 g cm-3. Use this information to calculate Avogadro’s number. (Atomic mass of Fe = 56.0 u) (All India) 2012
Answer:
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 13

Question 82.
(a) Some of the glass objects recovered from ancient monuments look milky instead of being transparent. Why?
(b) Iron (II) oxide has a cubic structure and each side of the unit cell is 5Å. If density of the oxide is 4 g cm-3, calculate the number of Fe2+ and O2- ions present in each unit cell. [Atomic mass : Fe = 56 u, O = 16 u; Avagadro’s number = 6.023 × 1023 mol-1]
(Comptt. All India) 2012
Answer:
(a) Some of the glass objects found from ancient monuments look to be milky in appearance because of crystallisation of glass.
(b) Volume of unit cell = a3 = (5 A) = (5 × 10-8)3
= 1.25 × 10-22 cm
Density of FeO = 4g cm-3
Mass of unit cell = Volume × Density
= 1.25 × 10-22 × 4 g
= 5 × 10-22 g
Mass of FeO molecule per unit cell
= \frac{5 \times 10^{-22} g}{1.195 \times 10^{-22} g} = 4.19 ≈ 4
Thus 4Fe+2, 4O-2 will be present in each unit cell.

Question 83.
(a) What are intrinsic semi-conductors? Give an example.
(b) What is the distance between Na+ and Cl ions in NaCl crystal if its density is 2.165 g cm-3? [Atomic Mass of Na = 23u, Cl = 35.5u; Avogadro’s number = 6.023 × 1023] (Comptt. All India) 2012
Answer:
(a) Intrinsic semi-conductors : These are insulators at room temperature and become semi-conductors when temperature is raised, Example : silicon and germanium.
(b) Applying the formula
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 14

Question 84.
(a) What type of semiconductor is obtained when silicon is doped with boron?
(b) What type of magnetism is shown in the following alignment of magnetic moments?
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 15
(c) What type of point defect is produced when AgCl is doped with CdCl2? (Delhi) 2013
Answer:
(a) p-type semi-conductor is obtained when silicon is doped with boron.
(b) Ferromagnetism is shown when the alignment of magnetic movements will be
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 16
(c) Impurity defect of inonic solids is produced when AgCl is doped with CdCl2. Due to this defect vacancies are created that result in higher electrical conductivity of the solid.

Question 85.
An element occurs in bcc structure. It has a cell edge length of 250 pm. Calculate the molar mass if its density is 8.0 g cm-3. Also calculate the radius of an atom of this element. (Comptt. Delhi) 2013
Answer:
Given : For bcc structure, Z = 2
Edge of the unit cell, a = 250 pm
Density of the element, p = 8.0 g/ cm3
M = ? r = ?
Using the formula
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 17

Question 86.
Iron (II) oxide has a cubic structure and each unit cell has a size of 5 Å. If density of this oxide is 4 g cm-3, calculate the number of Fe2+ and O2- ions present in each unit cell.
(Atomic mass of Fe = 56, O = 16, NA = 6.023 × 1023 and 1 Å = 10-8 cm) (Comptt. Delhi) 2013
Answer:
Given : p = 4g cm-3
a = 5Å = 5 × 10-8 cm M = 72 g/’mol, Z = ?
Using the formula for cubic crystals
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 18
There are four formula units of FeO present per unit cell. Hence it has face-centred cubic lattice where each Fe2+ and O2- are four in number.

Question 87.
Niobium crystallizes in body-centred cubic structure. If its density is 8.55 g cm-3, calculate atomic radius of niobium, given its atomic mass 93u. (Comptt. Delhi) 2013
Answer:
Given : p = 8.55 g cm3,
M = 93 g/mol-1
Z = 2 (For bcc) a = ?
Using formula
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 19
a = (36.1)1/3 × 102 ρm
∴ a = 3.304 × 102 ρm = 330.4 ρm
For bcc r1 = \frac{\sqrt{3}}{4} a = \frac{\sqrt{3}}{4} × 330.4 = 143.1 ρm

Question 88.
The density of copper is 8.95 g cm-3. It has a face centred cubic structure. What is the radius of copper atom?
(Atomic mass Cu = 63.5 g mol-1, NA = 6.02 × 1023 mol-1) (Comptt. Delhi) 2014
Answer:
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 20

Question 89.
Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm-3. Use this information to calculate Avogadro’s number.
(Atomic mass of Fe = 55.84 g mol-1) (Comptt. Delhi) 2014
Answer:
Given :
a = 286.65 pm = 286.65 × 10-10,
d = 7.87 g cm-3, M = 56 g mol-1
Z = 2 NA =?
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 21
∴ Avogadro’s number NA = 6.022 × 1023

Question 90.
Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm-3. Use this information to calculate Avogadro’s number.
(Gram atomic mass of Fe = 55.84 g mol-1). (Comptt. All India) 2014
Answer:
Given :
a = 286.65 pm = 286.65 × 10-10,
d = 7.87 g cm-3, M = 56 g mol-1
Z = 2 NA = ?
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 7
∴ Avogadro’s number NA = 6.022 × 1023

Question 91.
An element with molar mass 27 g mol-1 forms a cubic unit cell with edge length 4.05 × 10-8 cm. If its density is 2.7 g cm-3, what is the nature of the cubic unit cell? (Delhi) 2015
Answer:
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 22
= 6.022 × 10-2 × 66.43 = 4.0004 = 4
It has Face centred cubic cell/fee.

Question 92.
Examine the given defective crystal :
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 23
Answer the following questions :
(i) Is the above defect stoichiometric or non- stoichiometric?
(ii) Write the term used for this type of defect. Give an example of the compound which shows this type of defect.
(iii) How does this defect affect the density of the crystal? (All India) 2015
Answer:
(i) It is stoichiometric defect.
(ii) Schottky defect, e.g. NaCl.
(iii) Density of crystal decreases.

Question 93.
Define the following :
(i) Schottky defect
(ii) Frenkel defect
(iii) F-centre (Comptt. Delhi) 2015
Answer:
(i) Schottky defect: If in an ionic crystal of type A B , equal number of cations and anions are missing from their lattice sites so that the electrical neutrality is maintained, it is called Schottky defect.
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 24
(ii) Frenkel defect : If an ion leaves its site from its lattice site and occupies the interstitial site and maintains electrical neutrality, then it is called Frenkel defect.
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 25
(iii) F-centre : The centres which are created by trapping of electrons in anionic vacancies
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 26
and which are responsible for imparting colour to the crystals are called F-centres. (F = Fabre)

Question 94.
Silver crystallises in fee lattice. If edge length of the unit cell is 4.077 × 10-8 cm, then calculate the radius of silver atom. (Comptt. All India) 2015
Answer:
Given : a = 4.077 × 10-8 cm r = ? for fee lattice
Using formula,
Radius (r) = \frac{a}{2 \sqrt{2}}
r = \frac{4.077 \times 10^{-8}}{2 \times 1.414} \mathrm{cm}=\frac{4.077 \times 10^{-8}}{2.828}
∴ r = 1.441 × 10-8 cm

Question 95.
An element crystallizes in a f.c.c. lattice with cell edge of 250 pm. Calculate the density if 300 g of this element contains 2 × 1024 atoms. (Delhi) 2015
Answer:
Given: a = 250 pm = 250 × 10-10 cm
z = 4 (for fee)
M = ? d = ?
Using formula : d = \frac{z \times M}{a^{3} N_{A}}
∵ 2 × 1024 atoms of an element have mass = 300g
∴ 6.022 × 1023 atoms of an element have mass
= \frac{300 \times 6.022 \times 10^{23}}{2 \times 10^{24}} = 90.33 g
Now M = 90.33 g

Question 96.
An element crystallizes in a b.c.c. lattice with cell edge of 500pm. The density of the element is 7.5g cm-3. How many atoms are present in 300 g of the element? (All India) 2015
Answer:
Given: For b.c.c. structure, z = 2
Edge of the unit cell, a = 500 pm = 500 × 1010 cm
Density d = 7.5 g cm-3
Using the formula,
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 27
∵ 282.28 g of the element contains = 6.022 × 1023 atoms
∴ 300 g of the element contains
= \frac{6.022 \times 10^{23}}{282.28} × 300
= \frac{1806.6}{282.28} \times 10^{23} × 1023 = 6.40 × 1023 atoms

Question 97.
If NaCl is doped with 10-3 mol % of SrCl2, what is the concentration of cation vacancies? (Comptt. Delhi) 2015
Answer:
Concentration of SrCl2 = 10-3 mol% = 10-3/100 mol = 10-5 mol
1 mol of NaCl on doping procuces = 6.022 × 1023 cation vacancies
Therefore, 10-5 mol of NaCl on doping produces = 6.022 × 1023 × 10-5 = 6.022 × 1018 cation vacancies

Question 98.
Silver crystallises in f.c.c. lattice. If edge length of the cell is 4.07 × 10-8 cm and density is 10.5 g cm-3, calculate the atomic mass of silver.  (Comptt. All India) 2015
Answer:
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 28

Question 99.
(a) Based on the nature of intermolecular forces, classify the following solids: Silicon carbide, Argon
(b) ZnO turns yellow on heating. Why?
(c) What is meant by groups 12-16 compounds? Give an example. (All India) 2017
Answer:
(a) Silicon carbide is a covalent or network solid while Argon is a non-polar molecular solid.
(b) ZnO shows metal excess defect due to presence of extra cations, i.e., Zn2+ ions in interstitial sites which on heating changes into yellow due to loss of oxygen.
\mathrm{ZnO} \stackrel{\Delta}{\longrightarrow} \mathrm{Zn}^{2+}+\frac{1}{2} \mathrm{O}_{2}+2 \mathrm{e}^{-}
(c) Group 12-16 compounds are imperfect covalent compounds in which the ionic character depends on the electronegativities of the two elements, e.g., ZnS, CdS, etc.

Question 100.
(a) Based on the nature of intermolecular forces, classify the following solids: Benzene, Silver
(b) AgCl shows Frenkel defect while NaCl does not. Give reason.
(c) What type of semiconductor is formed when Ge is doped with Al? (All India) 2017
Answer:
(a) Benzene — Molecular solid (non-polar)
Silver — Metallic solid
(b) Due to intermediate radius of AgCl, the size of Ag+ is smaller than larger Na+ ion of NaCl so it can easily occupy interstitial spaces and shows Frenkel defect.
(c) p-type semiconductor is formed when Ge is doped with Al.

Question 101.
(a) Based on the nature of intermolecular forces, classify the following solids: Sodium sulphate, Hydrogen
(b) What happens when CdCl2 is doped with AgCl?
(c) Why do ferrimagnetic substances show better magnetism than antiferromagnetic substances? (All India) 2017
Answer:
(a) Sodium sulphate — Ionic solid
Hydrogen — Molecular solid (non-polar)
(ib) Cd2+ ion is dipositive and therefore addition of one Cd2+ ion results in the loss of two Ag+ ions from the lattice. But out of 2 holes obtained, one is occupied by Cd2+ ion and one left empty. Hence, addition of CdCl2 results in an impurity defect with cation vacancy.
(c) In ferrimagnetism, domains /magnetic moments are aligned in opposite direction in unequal numbers while in antiferromagnetic substances, the domains align in opposite direction in equal numbers so they cancel magnetic moments completely, i.e., net magnetism is zero.

Question 102.
An element crystallises in b.c.c. lattice with cell edge of 400 pm. Calculate its density if 500 g of this element contains 2.5 × 1024 atoms. (Comptt. Delhi) 2017
Answer:
Given : a = 400 pm = 400 × 10-10 cm
Z = 2 (for bcc) M = ? d = ?
Using formula, d = \frac{\mathrm{Z} \times \mathrm{M}}{a^{3} \times \mathrm{N}_{\mathrm{A}}}
∵ 2.5 × 1024 atoms of an element have mass = 500 g
∴ 6.022 × 2023atoms of an element have mass
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 29
= \frac{240.88}{6.4 \times 6.022}=\frac{240.88}{38.5408}
∴ d = 6.25 g cm-3

Question 103.
An element crystallises in fee lattice with cell edge of 400 pm. Calculate its density if 250 g of this element contain 2.5 × 1024 atoms. (Comptt. Delhi) 2017
Answer:
Given : a = 400 pm = 400 × 10-10 cm
Z = 4 (for fee), M = ?, d = ?
Using formula, d = \frac{\mathrm{Z} \times \mathrm{M}}{a^{3} \times \mathrm{N}_{\mathrm{A}}}
∵ 2.5 × 1024 atoms of an element have mass = 250 g
∴ 6.022 × 1023 atoms of an element have mass
= \frac{250 \times 6.022 \times 10^{23}}{2.5 \times 10^{24}}
∴ M = 60.22 g
Substituting all values in formula :
d = \frac{4 \times 60.22}{\left(400 \times 10^{-10}\right)^{3} \times 6.022 \times 10^{23}}=\frac{240.88}{38.5408}
∴ d = 6.25 g cm-3

Question 104.
An element crystallises in bcc lattice with cell edge of 400 pm. Calculate its density if 250 g of this element contains 2.5 × 1024 atoms.
(Comptt. Delhi) 2017
Answer:
Given : a = 400 pm = 400 × 10-10 cm
Z = 2 (for bcc), M = ?, d = ?
Using formula, d = \frac{\mathrm{Z} \times \mathrm{M}}{a^{3} \times \mathrm{N}_{\mathrm{A}}}
∵ 2.5 × 1024 atoms of an element have mass = 250g
∴ 6.022 × 1023 atoms of an element have mass
= \frac{250 \times 6.022 \times 10^{23}}{2.5 \times 10^{24}}
∴ M = 60.22 g
Substituting all values in formula : .
d = \frac{2 \times 60.22}{\left(400 \times 10^{-10}\right)^{3} \times 6.022 \times 10^{23}}=\frac{120.44}{38.5408}
∴ d = 3.125 g cm-3

Question 105.
An element exists in bcc lattice with a cell edge of 288 pm. Calculate its molar mass if its density is 7.2 g/cm3. (Comptt. All India) 2017
Answer:
Given : Cell edge, a = 288 pm = 288 × 10-10 cm
Density, d = 7.2 g/cm3
For bcc formula, units per cell Z = 2, M = ?
Using formula and substituting values,
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 30

The Solid State Class 12 Important Questions Long Answer Type (LA)

Question 106.
(a) An element has an atomic mass 93 g mol-1 and density 11.5 g cm-3. If the edge length of its unit cell is 300 pm, identify the type of unit cell.
(b) Write any two differences between amorphous solids and crystalline solids. (Delhi) (2017)
Answer:
(a) Given:
M = 93 g mol-1; ρ = 11.5 g cm-3;
a = 300 pm = 300 × 10-10 cm = 3 × 10-8 cm
Using formula,
Z = \frac{\rho \times a^{3} \times N_{A}}{M}
= \frac{11.5 \times\left(3 \times 10^{-8}\right)^{3} \times 6.022 \times 10^{23}}{93}
= 2.01 (approx.)
As the number of atoms present in given unit cells are coming nearly equal to 2, hence the given units cell is body centered cubic unit cell (BCC).

Amorphous solidsCrystalline solids
(i) They are  isotropic, i.e., they will show same value of physical perties in directions.They are aniso­tropic, i.e.,  tropic, i.e., value   of physical properties will be different when measured along different directions.
(ii) They have short range order.(ii) They have long range order.

Question 107.
(a) Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in a f.c.c. structure. (Atomic mass of Al = 27 g mol-1)
(b) Give reasons:
(i) In stoichiometric defects, NaCl exhibits Schottky defect and not Frenkel defect.
(ii) Silicon on doping with Phosphorus form n-type semiconductor.
(iii) Ferrimagnetic substances show better magnetism than antiferromagnetic substances. (Delhi) 2017
(a) Given:
Mass of Al = 8.1,
Atomic mass of Al = 27 g mol-1
No. of atoms = η × 6.022 × 1023
= \frac{8.1}{27} × 6.022 × 1023
= 0.3 × 6.022 × 1023
= 1.8066 × 1023
Since one f.c.c. unit cell has 4 atoms
∴ No. of unit cells = \frac{1.8066 \times 10^{23}}{4}
= 4.5 × 1022 unit cells
(b) (i) Schottky defect is shown by the ionic solids having very small difference in their cationic and anionic radius whereas Frenkel defect is shown by ionic solids having large difference in their cationic and anionic radius. NaCl exhibits Schottky defect because radius of both Na+ and Cl have very small difference.
(ii) Phosphorus is pentavalent that is it has 5 valence electrons, an extra electron results in the formation of n-type semi conductors on doping with Silicon. The conductivity is due to presence of extra electrons.
(iii) In antiferromagnetic substances the magnetic moments of domains are half aligned in one direction and remaining half in opposite direction in the presence of magnetic field so magnetic moment will be zero while in ferrimagnetic substances the magnetic moments of domains are aligned in parallel and anti-parallel directions in unequal numbers, hence shows some value of magnetic moment.
Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions 31

Important Questions for Class 12 Chemistry

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Deep Water Important Questions Class 12 English

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Deep Water Important Questions CBSE Class 12 English

Deep Water Important Questions Short Answer Type Questions (3-4 Marks)

Question 1.
How did Douglas finally get rid of the fear he had of water? (Delhi 2009)
Answer:
The terror that seized Douglas because of his misadventure with water twice was so intense that he sought professional help to overcome this fear. He engaged a swimming instructor who gave him intensive training for six months to ensure he overcame his fear of water which he eventually did.

Question 2.
How did the incident at the YMCA pool affect Douglas? (All India 2009)
Answer:
Douglas, a ten year old boy, was standing alone at the YMCA pool when a big bully of a boy picked him up and tossed him into the deep end and at once he was at the bottom of the pool. Though he did manage to come up with extreme difficulty, he could never again go back to the pool. He started fearing and avoiding water. Whenever he went near water a haunting terror would seize him.

Question 3.
Why was Douglas keen to overcome his fear of water? (All India 2009)
Answer:
Douglas was determined to overcome his fear of water because this phobia had ruined his fishing trips. Moreover, he had also never been able to enjoy water sports like canoeing, boating, and swimming. He was determined to get an instructor and learn swimming to get over his fear of water.

Question 4.
When Douglas realised that he was sinking, how did he plan to save himself? (Delhi 2010)
Answer:
When Douglas realised he was sinking he was frightened out of his wits and it was then that he decided to make a big jump and come up to the surface. He thought of lying flat on water for some time and then to paddle to the edge of the pool.

Question 5.
What did Douglas experience as he went down to the bottom of the pool for the first time? (Delhi 2010)
Answer:
When Douglas is pushed into the pool, he at once goes to the bottom of the pool. The nine feet deep pool appears like ninety feet to him. He feels a sense of acute uneasiness and as if his lungs are ready to burst. Despite feeling absolutely suffocated he makes desperate efforts for survival.

Question 6.
What sort of terror seized Douglas as he went down the water with a yellow glow? How could he feel that he was still alive? (Delhi 2010)
Answer:
When Douglas went down the water with a yellow glow it was a nightmarish experience for him. His legs were almost paralysed, his lungs were aching and his head was throbbing. He felt the thumping of his heart and the pounding in his head and these made him realize that he was still alive.

Question 7.
Why did William Douglas develop aversion to the water when he was three or four years old? (Comptt. Delhi 2010)
Answer:
William Douglas developed an aversion to water when he was three or four years old. He stood at a beach with his father when a wave Swept over him and knocked him down. He was buried under water and became breathless. He was petrified and developed a permanent fear for water.

Question 8.
What misadventure did William Douglas experience at the YMCA pool? (Comptt. All India 2010)
Answer:
A misadventure at the YMCA pool wherein Douglas was thrown into the deeper end of the pool by a big boy made Douglas afraid of water. He went down into the water three times but failed to come up. Though he was ultimately saved, a terror of water developed in him as his lungs filled with water. His head throbbed and his legs felt paralysed thus making him fear water permanently.

Question 9.
Why was Douglas determined to get over his fear of water? (Delhi 2011)
Answer:
Douglas was determined to overcome his fear of water because this phobia had ruined his fishing trips. Moreover, he had also never been able to enjoy water sports like canoeing, boating, and swimming. He was determined to get an instructor and learn swimming to get over his fear of water.

Question 10.
Why did Douglas go to Lake Wentworth in New Hampshire? (Delhi 2011)
Answer:
Douglas went to Lake Wentworth in New Hampshire to test whether he still had any fear of water. After his vigorous swimming practice he was still not very sure if his terror for water had left him. So he wanted to try out his swimming skills at Lake Wentworth. He dived into the lake and only after swimming across the shore and back did he feel certain his terror of water had fled.

Question 11.
Which factors led Douglas to decide in favour of YMCA pool? (All India 2011)
Answer:
When Douglas decided to overcome his childhood fear of water he preferred to go to YMCA swimming pool to learn swimming because it was safe. It was only two or three feet deep at the shallow end; and although it was nine feet at the other end, the drop was quite gradual.

Question 12.
What did Douglas learn from his experience at the YMCA pool? (Comptt. Delhi 2011)
Answer:
His experience of drowning at the YMCA pool had a very deep impact on Douglas. He became extremely terrorised and fearful of death. He had experienced both the sensation of dying and the terror that fear of it can produce. So his will to live grew in intensity. He learnt slowly to become a swimmer brushing aside his fear gradually.

Question 13.
Why was Douglas determined to get over his fear of water? (Delhi 2012)
Answer:
Douglas was determined to overcome his fear of water because this phobia had ruined his fishing trips. Moreover, he had also never been able to enjoy water sports like canoeing, boating, and swimming. He was determined to get an instructor and learn swimming to get over his fear of water.

Question 14.
Why did Douglas go to Lake Wentworth in New Hampshire? How did he make his terror flee? (All India 2012)
Answer:
Douglas went to Lake Wentworth in New Hampshire to test whether he still had any fear of water. After his vigorous swimming practice he was still not very sure if his terror for water had left him. So he wanted to try out his swimming skills at Lake Wentworth. He dived into the lake and only after swimming across the shore and back did he feel certain his terror of water had fled.

Question 15.
What efforts did Douglas make to get over his fear of water? (All India 2012)
Answer:
Douglas was determined to get over his fear of water. He engaged a professional instructor who understood the intensity of his fear and decided to not just teach him how to swim but ‘build’ a swimmer out of him slowly and steadily.

Question 16.
Which two frightening experiences did Douglas have in water in his childhood? (Comptt. Delhi 2012)
Answer:
Douglas’ first frightening experience in water was when he was three or four years old. He was knocked down by waves while surfing at California beach. He had been terror-struck at that time. Years later, his experience at the YMCA pool revived unpleasant memories when an eighteen year old boy had hurled Douglas into the deep end of the pool. Both these experiences led Douglas to develop a fear of water.

Question 17.
How did the instructor turn Douglas into a swimmer? (Delhi 2013)
Answer:
To ‘build a swimmer’ out of Douglas, the instructor began his training with extreme caution. He attached a rope to a belt and put it around Douglas. The rope went through a pulley that ran on an overhead cable. Supported by the cable, they went to and fro in the pool and practiced for weeks together. The instructor taught Douglas to put his face underwater and exhale and to raise his nose and inhale. He then taught him to kick in water for many weeks. Finally after seven months, the instructor told him to swim the length of the pool and Douglas’ persistent fear started fading.

Question 18.
How did Douglas make sure that he conquered the old terror? (All India 2013)
Answer:
Douglas went to Lake Wentworth in New Hampshire to test whether he still had any fear of water. After his vigorous swimming practice he was still not very sure if his terror for water had left him. So he wanted to try out his swimming skills at Lake Wentworth. He dived into the lake and only after swimming across the shore and back did he feel certain his terror of water had fled.

Question 19.
How did William Douglas’s aversion to water begin? (Comptt. Delhi 2013)
Answer:
William Douglas developed an aversion to water when he was three or four years old. He stood at a beach with his father when a wave Swept over him and knocked him down. He was buried under water and became breathless. He was petrified and developed a permanent fear for water.

Question 20.
Why did William Douglas use the YMCA pool and not Yakima river to learn swimming? (Comptt. Delhi 2013)
Answer:
Douglas used the YMCA pool and not the Yakima river to learn swimming because the river was treacherous His mother continually warned him against it and kept the details of each drowning in the river fresh in his mind. On the other hand the YMCA pool was safe, only two or three feet deep at the shallow end.

Question 21.
What happened when ‘a big bruiser of a boy’ tossed Douglas into the YMCA pool? How did Douglas plan to come out? (Comptt. Delhi 2013)
Answer:
Douglas landed inside the pool in a sitting position, swallowed water and went at once to the bottom. He was frightened but on the way down he planned how he would come out. When his feet would hit the bottom, he would make a big jump, come to the surface, lie flat on it and paddle to the edge of the pool.

Question 22.
Which two incidents in Douglas’ early life made him scared of water? (Comptt. Delhi 2014)
Answer:
Douglas’ first frightening experience in water was when he was three or four years old. He was knocked down by waves while surfing at California beach. He had been terror-struck at that time. Years later, his experience at the YMCA pool revived unpleasant memories when an eighteen year old boy had hurled Douglas into the deep end of the pool. Both these experiences led Douglas to develop a fear of water.

Question 23.
Why did Douglas prefer to go to YMCA swimming pool to learn swimming? (Comptt. Delhi 2014)
Answer:
When Douglas decided to overcome his childhood fear of water he preferred to go to YMCA swimming pool to learn swimming because it was safe. It was only two or three feet deep at the shallow end; and although it was nine feet at the other end, the drop was quite gradual.

Question 24.
What did Douglas feel and do when he was pushed into the swimming pool? (Comptt. All India 2014)
Answer:
Douglas landed in the pool in a sitting position, swallowed water and went at once to the bottom. Though he was frightened, on the way down he planned that when his feet would hit the bottom he would make a big jump and come to the surface.

Question 25.
How did his swimming instructor ‘build a swimmer’ out of Douglas? (Comptt. All India 2014)
Answer:
To ‘build a swimmer’ out of Douglas, the instructor began his training with extreme caution. He attached a rope to a belt and put it around Douglas. The rope went through a pulley that ran on an overhead cable. Supported by the cable, they went to and fro in the pool and practiced for weeks together. The instructor taught Douglas to put his face underwater and exhale and to raise his nose and inhale. He then taught him to kick in water for many weeks. Finally after seven months, the instructor told him to swim the length of the pool and Douglas’ persistent fear started fading.

Question 26.
Why did Douglas’ mother recommend that he should learn swimming at the YMCA swimming pool? (Delhi 2015)
Answer:
Douglas’ mother recommended that he should
learn swimming at the YMCA pool because it was safe. It was only two to three feet deep at the shallow end; and while it was nine feet deep at the other end, the drop was gradual.

Question 27.
How did Douglas remove his residual doubts about his fear of water? (Delhi 2015)
Answer:
To remove his residual doubts about his fear of water, Douglas went up the Tieton to Conrad Meadows, up the Conrad Creek Trail to Meade Glacier. He camped in the high meadow by the side of the warm lake. There he dove into the warm lake, swam across to the other shore and back just as Doug Corpron used to do.

Question 28.
How did Douglas’s experience at the YMCA pool affect him? (Delhi 2015)
Answer:
The drowning experience left Douglas weak and trembling. He was unable to eat that night and was haunted by extreme fear for days. The slightest exertion would upset him, making his knees wobble and his stomach sick. He developed a complete aversion to the swimming pool and this aversion stayed with him for years.

Question 29.
What lesson did Douglas learn when he got rid of his fear of water? (Comptt. Delhi 2015)
Answer:
After getting rid of his fear of water Douglas realized that ‘what one has to fear is fear itself’ and if he is able to overcome that fear he can achieve anything he wants.

Question 30.
How did the instructor turn Douglas into a swimmer?
Answer:
To ‘build a swimmer’ out of Douglas, the instructor began his training with extreme caution. He attached a rope to a belt and put it around Douglas. The rope went through a pulley that ran on an overhead cable. Supported by the cable, they went to and fro in the pool and practiced for weeks together. The instructor taught Douglas to put his face underwater and exhale and to raise his nose and inhale. He then taught him to kick in water for many weeks. Finally after seven months, the instructor told him to swim the length of the pool and Douglas’ persistent fear started fading.

Question 31.
How did his experience at the YMCA swimming pool affect Douglas? (Delhi 2016)
Answer:
The drowning experience left Douglas weak and trembling. He was unable to eat that night and was haunted by extreme fear for days. The slightest exertion would upset him, making his knees wobble and his stomach sick. He developed a complete aversion to the swimming pool and this aversion stayed with him for years.

Question 32.
What deep meaning did his experience at the YMCA swimming pool have for Douglas? (All India 2016)
Answer:
After his experience at the YMCA there was a haunting fear in Douglas’ heart. The experience of fear and death and its conquest made him live intensely. Conquering fear made him realise the true value of life and helped him enjoy every moment.

Question 33.
‘All we have to fear is fear itself.’ When did Douglas learn this lesson? (All India 2016)
Answer:
Douglas learnt this lesson after he had conquered his fear of water completely. He went to Lake Wentworth, dived into the warm lake, and swam across to the other shore and back. He shouted with joy at finally having con¬quered his fear of water and realized the meaning of Roosevelt’s words.

Question 34.
When did Douglas first become afraid of water? (Comptt. Delhi 2016)
Answer:
Douglas first became afraid of water when he was three years old and had gone to the California beach with his father. He went under a wave for a few seconds and though he was not in any kind of danger, yet he developed a fear of water.

Question 35.
When did Douglas’ dislike of water first begin? (Comptt. All India 2016)
Answer:
Douglas’ dislike of water first began when he was three years old and had gone to the Cali-fornia beach with his father. He was knocked down by the waves which swept over him and he was buried under the water. His breath was gone and he developed a fear of water.

Question 36.
How did his experience at the YMCA pool make Douglas feel scared of water? (Comptt. All India 2016)
Answer:
A misadventure at the YMCA pool wherein Douglas was thrown into the deeper end of the pool by a big boy made Douglas afraid of water. He went down into the water three times but failed to come up. Though he was ultimately saved, a terror of water developed in him as his lungs filled with water. His head throbbed and his legs felt paralysed thus making him fear water permanently.

Question 37.
What shocking experience did Douglas have at YMCA pool? (Delhi 2017)
Answer:
A misadventure at the YMCA pool wherein Douglas was thrown into the deeper end of the pool by a big boy made Douglas afraid of water. He went down into the water three times but failed to come up. Though he was ultimately saved, a terror of water developed in him as his lungs filled with water. His head throbbed and his legs felt paralysed thus making him fear water permanently.

Question 38.
Why did Douglas fail to come to the surface of the pool as he hoped to? (All India 2017)
Answer:
Douglas had hoped that when his feet hit the bottom of the pool, he would make a big jump and come to the surface but before he touched the bottom his lungs were ready to burst. Then when his feet hit the bottom and he summoned up all his strength to spring upwards, he came up slowly and saw nothing but water.

Question 39.
How did Douglas’ introduction to YMCA pool revive his childhood fear of water? (All India 2017)
Answer:
Douglas’ introduction to the YMCA swimming pool revived unpleasant memories and stirred his childish fears when he was knocked down and swept over by the waves in a beach in Cali¬fornia and had been buried in water. Thereafter he had an aversion to water when he was near it.

Question 40.
When did Douglas first start fearing water? (Comptt. All India 2017)
Answer:
Douglas first became afraid of water when he was three years old and had gone to the California beach with his father. He went under a wave for a few seconds and though he was not in any kind of danger, yet he developed a fear of water.

Question 41.
What was the deep fear in William Douglas’ mind? How did he get over it? (Comptt. AI 2017)
Answer:
William Douglas had a deep fear of water. Douglas had to resort to professional assistance to overcome his fear of water. He employed an instructor to teach him how to swim. He practiced five days a week, an hour each day with the instructor and piece by piece the instructor built a swimmer out of Douglas.

Question 42.
What happened to Douglas at the YMCA pool?
Answer: As Douglas was sitting on the side of the YMCA pool waiting for others, a boy who was around eighteen years old came there and yelled, ‘Hi, Skinny! How’d you like to be ducked?” saying this, he picked up Douglas and tossed him into the deep end of the pool.

Deep Water Important Questions Long Answer Type Questions (5-6 MARKS)

Question 43.
” There was terror in my heart at the overpowering force of the waves.” When did Douglas start fearing water? Which experience had further strengthened its hold on his mind and personality? (All India 2010)
Answer:
As a child of three or four years, Douglas had been knocked down by the waves at California beach. The waves swept over him and he was buried under them. The overpowering force of water terrorized him and he developed an aversion for water. This aversion to water resurfaced once again when Douglas was eleven years old and further strengthened its hold on his mind and personality. An eighteen years old boy tossed little Douglas into the deep end of the pool. This downward journey into water for the second time was a nightmarish experience. His legs were almost paralysed, his lungs ached, his head throbbed and he felt suffocated. Keeping his wits intact, he pushed himself up to the surface of the swimming pool but all his efforts proved futile. Finally when he came to his senses he was lying on his stomach beside the pool. Now the haunting fear of water gripped his heart and he continued to be mortally scared of water for a very long time.

Question 44.
How did Douglas develop an aversion to water? (Delhi 2011)
Answer:
When Douglas was three or four years old, his father had taken him to the beach in California. As he and his father had stood together in the surf, the waves had knocked him down and swept over him. He was buried in water. His breath was gone and he was frightened. Then, when he was about ten or eleven years old and had decided to learn to swim, he had gone to the YMCA pool. There an eighteen year old boy picked him up and tossed him into the deep end of the pool. After this incident he never went back to the pool. He developed a fear of water and avoided it whenever he could. Even when he went wading or boating in water the terror that had seized him during these experi¬ences would come back and take possession of him completely. His legs would become paralysed and icy horror would grab his heart.

Question 45.
How did Douglas try to save himself from drowning in the YMCA pool? (Delhi 2011)
Answer:
When his feet hit the bottom of the pool Douglas summoned up all his strength and made a great spring upwards thinking he would bob to the surface like a cork. But when nothing like that happened Douglas tried to yell but no sound came out. Now a great force was pulling him under. He was paralysed under water stiff and rigid with fear. Then in the midst of the terror came a touch of reason that he should remember to jump when he hit the bottom. As soon as he felt the tiles under him he reached out his toes towards them and jumped again with all his strength. Yet again the jump did not make any difference. The water was still around him. Stark terror took an even deeper hold on him and he shook and trembled with fright. He could not move his arms and legs. He tried to call for help but nothing happened. Finally he ceased all his efforts and decided to relax as blackness swept over his brain.

Question 46.
How did the instructor make Douglas a good swimmer? (All India 2011)
Answer:
The instructor put in serious efforts to ‘build a swimmer’ out of Douglas. He understood Douglas’ mortal water-phobia and practiced five days a week, an hour each day, with him. He devised a unique way to teach him how to swim. He attached a rope to Douglas’ belt that went through a pulley which ran over an overhead cable. Holding the end of the rope in his hand, he made Douglas move back and forth in the pool without causing him much fear. Douglas was taught how to exhale under water and raise his nose to inhale.

This exercise was repeated numerous times and they went to and fro across the pool week after week. The instructor then taught Douglas to kick with his legs. At first his legs would not work but finally he was able to control and command them. Finally he was transformed into quite a perfect swimmer by his instructor.

Question 47.
What horrific experiences did Douglas have in his childhood? What impact did they have on him? (Comptt. All India 2011)
Answer:
When Douglas was three or four years old, his father had taken him to the beach in California. As he and his father had stood together in the surf, the waves had knocked him down and swept over him. He was buried in water. His breath was gone and he was frightened. Then, when he was about ten or eleven years old and had decided to learn to swim, he had gone to the YMCA pool. There an eighteen year old boy picked him up and tossed him into the deep end of the pool. After this incident he never went back to the pool. He developed a fear of water and avoided it whenever he could. Even when he went wading or boating in water the terror that had seized him during these experi¬ences would come back and take possession of him completely. His legs would become paralysed and icy horror would grab his heart.

Question 48.
What misadventure did Douglas experience at the YMCA swimming pool? (Comptt. All India 2013)
Answer:
As the timid Douglas sat alone at the side of the YMCA swimming pool waiting for other people to come, a big bruiser of a boy, probably eighteen years old came there. He asked Douglas whether he would like to be ducked. Saying this he picked up Douglas and tossed him into the deep end of the pool. Douglas landed inside the pool in a sitting position, swallowed water and went at once to the bottom of the pool. Though Douglas was extremely frightened he had his wits intact so on his way down he started to plan. He decided that when his feet would hit the bottom he would make a big jump, come to the surface, lie flat on it and then paddle to the edge of the pool. At that moment the nine feet deep pool seemed like ninety feet to him and before he touched the bottom he felt his lungs were ready to burst.

Question 49.
Describe the efforts made by Douglas to save himself from drowning in the YMCA swimming pool. (Comptt. All India 2013)
Answer:
Douglas was picked up and tossed into the deep end of the YMCA swimming pool. At that time those nine feet seemed a long way down. As his feet hit the bottom he summoned all his strength and made an upward spring. He came up slowly, opened his eyes and saw only water. He reached up as if to grab a rope and his hands clutched only at water. He flailed at the surface of the water, swallowed and choked. He tried to bring up his legs but they hung as if paralysed. He again started on a journey back to the bottom of the pool.

Then he remembered the strategy —he would spring from the bottom of the pool and come like a cork to the surface. He would lie flat on the water, strike out with his arms and thrash with his legs. Then he would get to the edge of the pool and be safe. Yet again the jump made no difference and finally Douglas ceased all efforts and relaxed as blockness swept over his brain.

Question 50.
How did Douglas’s experience at the YMCA pool affect him and how did he overcome his fear of water? (Comptt. All India 2013)
Answer:
The big bully of a boy found Douglas alone at the pool. He picked him up and tossed him into the deep end of the pool. Douglas was thrown at the bottom of the pool and feared that he would be drowned. This ‘misadventure’ caused Douglas a lot of trouble and agony. He developed an aversion to water as he experienced a series of fears and emotions. Icy horror grabbed his heart and made him panicky. This experience had a lasting effect on Douglas. It deprived him of the joys of boating and swimming. This fear of water ruined his fishing trips. He never went back to the pool. This fear of water stayed with him as the years rolled by.

Whenever he tried to enter water, he was seized by fear. Wherever he went his joys of fishing, boating and swimming were ruined. This fear of deep waters stayed with him for years and firmly held him in its grip. Douglas had to resort to professional assistance to overcome his fear of water. He employed an instructor to teach him how to swim. He practiced five days a week, an hour each day with the instructor and piece by piece the instructor built a swimmer out of Douglas.

Question 51.
The story “Deep Water” has made you realize that with determination and perseverance one can accomplish the impossible. Write a paragraph in about 100 words on how a positive attitude and courage will aid you to achieve success in life. (Delhi 2014)
Answer:
Douglas was afraid of water from a very young age. His misadventure at the YMCA pool further increased his water phobia. He was unable to enjoy fishing and boating trips. Finally he decided he had to overcome his fear. He could not do it without seeking professional help, so he appointed an instructor who gradually made him an excellent swimmer. Still Douglas was not satisfied. He made use of every opportunity to swim and dive in water thus challenging his fear. He was able to overcome his fear completely and this led him to make the statement that what one is afraid of is fear itself and if we are able to overcome that fear then we can achieve anything in life.

Question 52.
Desire, determination and diligence lead to success. Explain the value of these qualities in the light of Douglas’ experience in “Deep Water”. (Comptt. All India 2014)
Answer:
Determination and perseverance is a combination of attributes and abilities that drive people to set goals for themselves and then to take the initiative to achieve these goals. Douglas was able to overcome his fear of water by the values of positive attitude and courage. Initially he was afraid of water but his grit and determination made him get an instructor and overcome his fear. Determination today leads to our success tomorrow. It is that innate quality in our soul, which comes to surface when an incident irks it. It is a reflection of our values taught to us by society and
circumstances and enables us to overcome all obstacles.

There is always admiration for heroes like William Douglas who face challenges courageously and finally emerge successful. For years the fear of water haunted him. It deprived him of the joy of canoeing, swimming, fishing and boating. However, it was deliberate, planned and continuous efforts that enabled Douglas to get over his fear. He was absolutely determined to get rid of his fear and it was due to his perseverance and resoluteness that he emerged victorious.

Question 53.
“All we have to fear is fear itself.” Describe Douglas’s experiences which led to the making of this statement. (Comptt. Delhi 2015)
Answer:
Douglas was afraid of water from a very young age. His misadventure at the YMCA pool further increased his water phobia. He was unable to enjoy fishing and boating trips. Finally he decided he had to overcome his fear. He could not do it without seeking professional help, so he appointed an instructor who gradually made him an excellent swimmer.

Still Douglas was not satisfied. He made use of every opportunity to swim and dive in water thus challenging his fear. He was able to overcome his fear completely and this led him to make the statement that what one is afraid of is fear itself and if we are able to overcome that fear then we can achieve anything in life.

Question 54.
With the help of courage one can achieve a lot. How did Douglas overcome his fear of water? (Comptt. Delhi 2016)
Answer:
Douglas was afraid of water from a very young age. His misadventure at the YMCA pool further increased his water phobia. He was unable to enjoy fishing and boating trips. Finally he decided he had to overcome his fear. He could not do it without seeking professional help, so he appointed an instructor who gradually made him an excellent swimmer. Still, Douglas was not satisfied. He made use of every opportunity to swim and dive in water thus challenging his fear. He was able to overcome his fear completely and this led him to make the statement that what one is afraid of is fear itself and if we are able to overcome that fear then we can achieve anything in life.

Question 55.
We always admire those as heroes who face challenges bravely in different phases of life and emerge successfully. Elaborate on this statement with reference to William Douglas. (Comptt. Delhi 2017)
Answer:
Determination and perseverance is a combination of attributes and abilities that drive people to set goals for themselves and then to take the initiative to achieve these goals. Douglas was able to overcome his fear of water by the values of positive attitude and courage. Initially he was afraid of water but his grit and determination made him get an instructor and overcome his fear. Determination today leads to our success tomorrow. It is that innate quality in our soul, which comes to surface when an incident irks it. It is a reflection of our values taught to us by society and
circumstances and enables us to overcome all obstacles.

There is always admiration for heroes like William Douglas who face challenges courageously and finally emerge successful. For years the fear of water haunted him. It deprived him of the joy of canoeing, swimming, fishing and boating. However, it was deliberate, planned and continuous efforts that enabled Douglas to get over his fear. He was absolutely determined to get rid of his fear and it was due to his perseverance and resoluteness that he emerged victorious.

Important Questions for Class 12 English

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The Rattrap Important Questions Class 12 English

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The Rattrap Important Questions CBSE Class 12 English

The Rattrap Important Questions Short Answer Type Questions (3-4 Marks)

Question 1.
Why did the peddler sign himself as Captain von Stahle? (All India 2009)
Answer:
Edla Willmansson had been rather nice to the peddler and had treated him with the honour that was due to a Captain. The peddler, through this mistaken identity, got an opportunity to raise himself and get above the petty temptations of the world. So he signed himself as Captain von Stahle.

Question 2.
Why did the peddler think that the world was a rattrap? (All India 2009)
Answer:
The peddler considered the whole world as a big rattrap, its sole purpose being to set baits for people. The joys and riches of this world are nothing but tempting baits and anyone who is tempted by them was captured by the rattrap which completely closed in on him.

Question 3.
Why was Edla happy to see the gift left by the peddler? (All India 2010)
Answer:
Edla was happy to see the gift left by the peddler as he had respected her faith in him. Edla had retained him in her house even after knowing his real identity and he, in turn, had shown her that the guest she had honoured was as honourable as the Captain. The latent goodness of his heart had been awakened and he had been able to overcome the bait of the rattrap.

Question 4.
Which act of the crofter surprised the peddler? Why? (Comptt. Delhi 2010)
Answer:
The peddler was always shunned away wherever he went. No one treated him with kindness and so he had lost all hope of being shown any kind of warmth. But when he approached the crofter’s roadside cottage he was surprised by the latter’s warm welcome and generous hospitality.

Question 5.
How was the peddler treated at the crofter’s cottage? (Comptt. All India 2010)
Answer:
At the crofter’s cottage the peddler was welcomed warmly and received generous hospitality. The crofter was an old and lonely man and the prospect of getting the peddler’s company overjoyed him. So he poured all his warmth and friendly courtesy on the peddler.

Question 6.
Why was the crofter so talkative and friendly with the peddler? (Delhi 2011)
Answer:
The crofter was lonely. He lived alone in his cottage without a wife, a child or any companion. Since he suffered from acute loneliness he was extremely happy when he got the peddler’s company. That is why he was so talkative and friendly with the peddler.

Question 7.
Why was the peddler surprised when he knocked on the door of the cottage? (All India 2011)
Answer:
At the crofter’s cottage the peddler was welcomed warmly and received generous hospitality. The crofter was an old and lonely man and the prospect of getting the peddler’s company overjoyed him. So he poured all his warmth and friendly courtesy on the peddler.

Question 8.
What made the peddler finally change his ways? (All India 2011)
Answer:
The peddler was touched by Edla’s kind treatment. She treated him like a Captain in spite of knowing his real identity. This awakened the latent goodness of his heart because he wanted to show Edla he was worthy of the honour she had given him. So he finally changed his ways.

Question 9.
Why did the peddler keep to the woods after leaving the crofter’s cottage? How did he feel? (All India 2011)
Answer:
The peddler discards the public highway and keeps to the woods after leaving the crofter’s cottage because he wants to avoid being caught with the thirty kronors that he had stolen from the crofter’s house. He walks through mazes of forest paths but lands nowhere. When he realizes he has been trapped he feels extremely tired and sinks to the ground in despair.

Question 10.
Did the stranger agree to go to the ironmaster’s house? Why or why not? (Comptt. Delhi 2011)
Answer:
Initially the stranger declined the ironmaster’s invitation. He had the stolen thirty kronors on him and thought it was like going into the lion’s den. But then he accepted the ironmaster’s invitation because Edla’s sympathy and compassion allayed his fears and her friendly manner made him to have confidence in her.

Question ll.
Why did the peddler decline the invitation of the ironmaster? (Delhi 2012)
Answer:
The peddler declined the ironmaster’s invitation because he had the stolen thirty kronors on him. He feared that he would be caught there for stealing the crofter’s money. For the peddler going to the manor house of the ironmaster would be like throwing himself voluntarily into the lion’s den.

Question 12.
Why was the crofter so friendly and talkative with the peddler? (Delhi 2012)
Answer:
The crofter was lonely. He lived alone in his cottage without a wife, a child or any companion. Since he suffered from acute loneliness he was extremely happy when he got the peddler’s company. That is why he was so talkative and friendly with the peddler:

Question 13.
Who was the owner of Ramsjo iron mills?
Why did he visit the mills at night? (All India 2012)
Answer:
The ironmaster was the owner of the Ramsjo iron mills. He was very particular about the quality of his products. So he made nightly rounds of inspection to his mill to inspect the quality control.

Question 14.
How did the ironmaster react on seeing the stranger lying close to the furnace? (All India 2012)
Answer:
When the iron master saw the stranger lying close to the furnace he walked up to him and looked him over carefully. Mistaking him for an old acquaintance, a comrade from his regiment, he got very excited and invited him to come to his cottage.

Question 15.
Why didn’t the stranger tell the ironmaster that he was not Nils Olof? (All India)
Answer:
When the ironmaster mistakes the stranger for Captain Nils Olof, an old regimental comrade, the stranger decides not to correct him as hopes to get a couple of kronors from the ironmaster. So he does not want to undeceive him all at once.

Question 16.
Why was the crofter happy when the peddler knocked on his door? (Comptt. All India 2013)
Answer:
The crofter was very lonely. He lived alone in his cottage without a wife, a child or any companion. Since he was suffering from acute loneliness he felt very happy when the peddler knocked on his door. He treated the peddler in the most kind and hospitable manner.

Question 17.
How was the crofter ‘generous with his confidences’ when he spoke to the peddler? (Comptt. All India 2013)
Answer:
The crofter was generous not just with his porridge and tobacco but also with his confidences with the peddler. He informed the peddler that he had been a crofter but now his
cow supported him. She would give milk for the creamery every day, and last month he had even received thirty kronors as payment.

Question 18.
What did the peddler do to keep his body and soul together? (Comptt. All India 2013)
Answer:
The peddler made and sold rattraps but his business was not really profitable so he had to resort to both begging and petty thievery to keep his body and soul together.

Question 19.
Why did the peddler derive pleasure from his idea of the world as a rattrap? (Delhi 2014)
Answer:
The world had never been kind to the peddler so it gave him unwanted joy to think ill of it by considering it as a rattrap. It became his cherished pastime to think of people he knew who had let themselves be caught in the dangerous snare and of others who were still circling around the bait in the rattrap.

Question 20.
Why did Edla plead with her father not to send the vagabond away? (All India 2014)
Answer:
Edla pleaded with her father not to send the vagabond away as it was Christmas eve and she wanted to keep the spirit of Christmas alive. She wanted to provide the vagabond with a day of comfort and solace. She justified that they had invited him against his will and since he was lonely she wanted to do something special for him on Christmas eve.

Question 21.
In what sense was the world a big rattrap ac¬cording to the peddler? (Comptt. Delhi 2014)
Answer:
According to the peddler the whole world with its lands, seas, cities and villages was nothing but a big rattrap. It only existed to set baits for people. If offered riches and joys, shelter, food and clothing as the rattrap offered cheese and pork and as soon as anyone let himself be tempted to touch the bait, it closed on him and then everything came to an end.

Question 22.
Why did the peddler knock on the cottage by the roadside? How was he treated by the owner of the cottage? (Comptt. Delhi 2014)
Answer:
The peddler knocked on the cottage by the roadside to ask shelter for the night. The owner, who was an old man, without a wife or child, greeted him warmly, served him supper and played cards with him as he was happy to get someone to talk to in his loneliness.

Question 23.
What conclusion did the ironmaster reach when he heard that the crofter had been robbed by the peddler? (Comptt. All India 2014)
Answer:
When the ironmaster heard that the crofter had been robbed by a man who went around selling rat-traps, he sarcastically remarked to his
daughter that she had let a fine man into the house and was wondering as to how many silver spoons were left in their cupboard by that time.

Question 24.
What were the contents of the package left by the peddler as a Christmas gift for Edla Willmansson? (Cornptt. All India 2014)
Answer:
The package that the peddler left as a Christmas gift for Edla Willmansson contained a small rattrap which had a letter he had written in large, jagged characters and in it also lay three wrinkled ten kronor notes.

Question 25.
How was the peddler received by the crofter? (Comptt. Delhi 2015)
Answer:
The crofter was a lonely person who received the peddler warmly, gave him supper and tobacco to fill his pipe and played a game of cards with him. He was very friendly with the peddler. He shared the details of his life with him and showed him the thirty kronor notes he had received as payment.

Question 26.
What brought about a change in the life of the peddler? (Comptt. All India 2015)
Answer:
Edla’s warmth, sympathy and compassion brought about a change in the life of the peddler. He is touched by the kind treatment Edla gives him despite knowing his real identity. The latent goodness of his heart is awakened and he actually behaves like a true Captain.

Question 27.
At the crofter’s home, why did the peddler feel very happy? (Comptt. All India 2015)
Answer:
The peddler was received very warmly and received generous hospitality at the crofter’s home. The crofter, an old and lonely man, served him porridge, treated him kindly and the two smoked and played cards. This made the peddler feel very happy.

Question 28.
Why was the peddler amused at the idea of the world being a rattrap? (Delhi 2016)
Answer:
The world had never been very kind to the peddler. So it gave him unwanted joy to think of the world with its lands and seas, cities and villages as nothing but a big rattrap that sets baits for people in the form of riches, joys and other necessities, and as soon as one got tempted, it closed in on him.

Question 29.
What hospitality did the peddler receive from the crofter? (Delhi 2016)
Answer:
Instead of the sour faces which normally met the peddler, the crofter who was an old and lonely man received the peddler most warmly and offered him generous hospitality. He gave him porridge for supper and the two smoked tobacco and played cards. The crofter also shared his confidences with the peddler.

Question 30.
What do we learn about the crofter’s nature from the story, ‘The Rattrap’? (All India 2016)
Answer:
The crofter was an old man who was very lonely as he had no family. He was very happy when the peddler knocked on his door as he got someone to talk to in his loneliness. He treated the peddler most courteously and offered him food and tobacco. The crofter was as generous with his confidences as he was with his hospitality.

Question 31.
Why did the crofter show the thirty kroner to the peddler? (All India 2016)
Answer:
The crofter was too happy to get someone to talk to in his loneliness so he was generous with his confidences with the peddler. The stranger must have seemed doubtful, for the crofter took down a leather pouch hanging on a nail near the window and showed the thirty kroner notes to the peddler.

Question 32.
Why did the ironmaster speak kindly to the peddler and invite him home? (All India 2016)
Answer:
The ironmaster mistook the peddler to be an old acquaintance Nils Olof, his old regimental comrade, so he spoke kindly to the peddler and invited him to his house. He and his daughter Edla did not have any company for Christmas so he wanted the peddler to join them for Christmas dinner.

Question 33.
Why did Edla invite the peddler? (Compff. Delhi 2016)
Answer:
Edla invited the peddler because she wanted to celebrate Christmas in the true spirit of the festival by having a guest over for Christmas supper. Moreover, she wanted to give the peddler a day of peace.

Question 34.
Describe the crofter’s meeting with the rattrap peddler. (Comptt. All India 2016)
Answer:
The crofter gave the rattrap peddler shelter for a night. He was very hospitable with the peddler. He gave the peddler supper, tobacco to smoke, played cards with him and even shared his confidences with him.

The Rattrap Important Questions Long Answer Type Questions (5-6 MARKS)

Question 35.
Describe how the story, ‘The Rattrap’ shows that basic human goodness can be brought out by understanding and love. (Delhi 2006)
Answer:
The theme of the story ‘The Rattrap’ is that most human beings are prone to fall into the trap of material gains. However, love and understanding can transform a person and bring out his essential human goodness. The peddler had been treated very cruelly by the world. So even though the old crofter was kind and hospitable to him, he betrayed his trust and stole thirty kronors from him. He was not impressed by the iron-master’s invitation also. But Edla Willmansson’s compassion and understanding brought about a transformation in his nature. Her human qualities helped in raising him to be a gentleman. He was easily able to overcome petty temptations. The peddler who always considered the whole world to be a rattrap finally felt released from this rattrap due to the sympathetic, kind, loving and generous treatment of Edla Willmansson that was able to bring out his basic human goodness.

Question 36.
Give examples from the story, “The Rattrap” to show how the iron master is different from his daughter. (Delhi 2006)
Answer:
The character of Edla Willmansson and that of the iron master are in stark contrast to each other. Despite being young, the daughter displays a better sense of maturity than her father who acts impulsively and behaves . casually. He jumps to conclusions without thinking. First he mistakes the peddler to be an old regimental comrade and without confirming his identity he instantly invites him to the manor house and again on realising his mistake he refers the matter to the sheriff thoughtlessly. Edla, on the other hand, displays a keen sense of observation. She rightly judges that her guest is a tramp and has a sympathetic attitude towards him. She persuades her father to allow the guest to stay, leads him courteously to the dining table and makes him eat despite her father’s protest. It is because of her compassion and generosity that the peddler undergoes a change of heart and redeems himself from dishonesty. He leaves behind thirty kronors to be given back to the old crofter and a Christmas present for Edla.

Question 37.
The story ‘The Rattrap’ focuses on human loneliness and the need to bond with others. Explain. (Delhi 2010)
Answer:
The main focus of the story ‘The Rattrap’ is on human loneliness. All the characters, whether it is the peddler, the crofter, the ironmaster or his daughter, suffer from loneliness. The peddler is a lonely man who has always been shunned by society’s cold and unkind words. When he knocks at the door of the old crofter’s cottage he does not expect hospitality but the crofter welcomes him as he is too happy to get someone to talk to after being alone for so long. By serving the peddler the crofter is in fact serving himself. He serves the peddler with supper, gives him tobacco and plays ‘mjolis’ with him. He is a very good host. The iron¬master and his daughter too miss company and this makes them all the more lonely on the occasion of Christmas. So the ironmaster, who mistakes the peddler for his old regimental comrade, invites him to his manor house for Christmas. The ironmaster’s daughter, Edla, extends this invitation again and tells the peddler he can leave any time after Christmas. Thus the need to bond is the main focus of the story ‘The Rattrap’.

Question 38.
Describe how the story, ‘The Rattrap’ shows that basic human goodness can be brought out by understanding and love. (Comptt. All India 2011)
Answer:
The theme of the story ‘The Rattrap’ is that most human beings are prone to fall into the trap of material gains. However, love and understanding can transform a person and bring out his essential human goodness. The peddler had been treated very cruelly by the world. So even though the old crofter was kind and hospitable to him, he betrayed his trust and stole thirty kronors from him. He was not impressed by the iron-master’s invitation also. But Edla Willmansson’s compassion and understanding brought about a transformation in his nature. Her human qualities helped in raising him to be a gentleman. He was easily able to overcome petty temptations. The peddler who always considered the whole world to be a rattrap finally felt released from this rattrap due to the sympathetic, kind, loving and generous treatment of Edla Willmansson that was able to bring out his basic human goodness.

Question 39.
How are the attitudes of the ironmaster and his daughter different? Support your answer from the text. (Delhi 2011)
Answer:
The character of Edla Willmansson and that of the iron master are in stark contrast to each other. Despite being young, the daughter displays a better sense of maturity than her father who acts impulsively and behaves . casually. He jumps to conclusions without thinking. First he mistakes the peddler to be an old regimental comrade and without confirming his identity he instantly invites him to the manor house and again on realising his mistake he refers the matter to the sheriff thoughtlessly. Edla, on the other hand, displays a keen sense of observation. She rightly judges that her guest is a tramp and has a sympathetic attitude towards him. She persuades her father to allow the guest to stay, leads him courteously to the dining table and makes him eat despite her father’s protest. It is because of her compassion and generosity that the peddler undergoes a change of heart and redeems himself from dishonesty. He leaves behind thirty kronors to be given back to the old crofter and a Christmas present for Edla.

Question 40.
Describe the peddler’s interaction with the ironmaster’s daughter. To what extent was he influenced by her? (Comptt. Delhi 2011)
Answer:
The peddler first meets Edla Wilmansson, the ironmaster’s daughter when, on her fathers insistence, she comes to invite him to their manor house for Christmas. He was so touched by the sincerity in her voice that he could not refuse her invitation. Later, she comes to know of the peddler’s real identity but it does not change her warmth, friendliness and hospitality towards him. She continues to treat him like a Captain and the peddler quite spontaneously, starts behaving like a real Captain. He leaves a rattrap as a Christmas gift for Edla and encloses a letter of thanks and a note of confession in it. He leaves behind the stolen money to be restored to its rightful owner, the crofter, thus redeeming himself from his dishonest ways. Edla Wilmansson’s sympathy, compassion and understanding give the peddler an opportunity to redeem and reform himself.

Question 41.
Describe the crofter’s interaction with the peddler. How did the latter get tempted? (Comptt. Delhi 2011)
Answer:
When the peddler knocked on the door of the crofter’s cottage he was greeted by the lonely old man who was just too happy to get someone to talk to. He served the peddler with extreme hospitality and even played cards with him. He also shared his confidences with the peddler telling him that he had been a crofter at Ramsjo Ironworks during his days of prosperity and now his cow supported him. Last month he had received thirty kronors in payment for the cow milk he had sold. He even showed the peddler the leather pouch on the window where he had kept the thirty kronors thus tempting the peddler. The next day the peddler after leaving the crofter’s cottage came back there again, smashed the window pane, stuck in his hand and got hold of the pouch that contained the thirty kronors. Then hanging the leather pouch back very carefully, he went away.

Question 42.
‘The Rattrap’ highlights the impact of compassion and understanding on the hidden goodness in human beings. Substantiate with evidence from the story. (Comptt. All India 2011)
Answer:
The theme of the story ‘The Rattrap’ is that most human beings are prone to fall into the trap of material gains. However, love and understanding can transform a person and bring out his essential human goodness. The peddler had been treated very cruelly by the world. So even though the old crofter was kind and hospitable to him, he betrayed his trust and stole thirty kronors from him. He was not impressed by the iron-master’s invitation also. But Edla Willmansson’s compassion and understanding brought about a transformation in his nature. Her human qualities helped in raising him to be a gentleman. He was easily able to overcome petty temptations. The peddler who always considered the whole world to be a rattrap finally felt released from this rattrap due to the sympathetic, kind, loving and generous treatment of Edla Willmansson that was able to bring out his basic human goodness.

Question 43.
Given his temperament, Edla’s father would have failed in reforming the peddler. How did Edla succeed? (Comptt. All India 2011)
Answer:
Unlike her father Edla is a compassionate, sympathetic and understanding girl and because of these virtues she succeeded in reforming the peddler. The ironmaster, on the other hand, is impulsive and whimsical. He invites the stranger to his house without confirming the latters identity and as soon as the peddler’s true identity is revealed he decides to refer the whole matter to the sheriff. But Edla continues to be nice and hospitable to him and does not turn him out of their house on Christmas eve. She treats the peddler like a real captain and he too behaves like one. She thus awakened the potential goodness of his heart and before leaving the manor house he leaves behind a rattrap as a Christmas gift for Edla and the money he had stolen from the crofter’s cottage alongwith a letter in which he confesses his crime and requests Edla to return the thirty kronors to the old crofter.

Question 44.
How did the peddler feel after robbing the crofter? What course did he adopt and how did he react to the new situation? What does his reaction reveal? (Delhi 2013)
Answer:
After robbing the crofter the peddler felt quite pleased with his smartness. He immediately realised that he could not dare to continue with his journey on the public highway so he turned off the road, into the woods. During the first few hours his decision caused him no difficulty but later it became worse for he had gotten into a big and confusing forest. He continued to walk and when he came to the end of the forest he realized that he had been walking around in the same part of the forest. Then he recalled his thoughts about the world and the rattrap and knew his own turn had come. He had let himself be fooled by a bait of thirty kronors and had been caught. His reaction reveals his gloom and despair as he realized the forest had closed in open upon him like an impenetrable prison from which he thought he could never escape. It also reveals that he was basically a good person at heart and was repentant of his folly.

Question 45.
There is a saying, ‘Kindness pays, rudeness never’. In the story, ‘The Rattrap’ Edla’s attitude towards men and matters is different from her fathers attitude. How are the values of concern and compassion brought out in the story, ‘The Rattrap’? (All India 2013)
Answer:
Edla Willmansson displays qualities of compassion and understanding that transform the peddler and brings out his essential human goodness. Her human qualities help in raising him to the level of a gentleman and he is able to overcome petty temptations. The peddler, who always considered the whole world to be a rattrap, was finally able to release himself from this rattrap due to the sympathetic, kind and generous treatment of Edla and thus redeems himself from his dishonest ways. Despite knowing his real identity Edla continues to treat him like a Captain and the peddler quite spontaneously starts behaving like a real captain. Edla’s compassion and the peddler’s reformation arouses our optimism and belief in the essential goodness of man and other human values.

Question 46.
How does the peddler respond to the hospitality shown to him by the crofter? (Comptt. Delhi 2013)
Answer:
The peddler responds to the crofter’s hospitality by betraying him. One dark evening
when the peddler was walking along the road he knocked on the door of a cottage to seek shelter for the night. To his surprise, he was welcomed by an old man, the crofter, who lived alone in the cottage. The lonely crofter was happy to find a man whom he could talk to. He served the peddler supper, gave him tobacco and played a game of cards with him. The old crofter then went to the window and took down a leather pouch. He counted three ten kronor notes and put them into the pouch. This provided a big bait for the peddler who was tempted to steal the notes. The peddler unwillingly allowed himself to be tempted to touch the bait and was thus trapped in. He stole the money and thus committed a breach of trust. So he betrayed the confidence reposed in him by his host.

Question 47.
How did the seller of rattraps realize that he himself was caught up in a rattrap after he left the crofter’s cottage? (Comptt. Delhi 2013)
Answer:
After robbing the crofter the peddler felt quite pleased with his smartness. He immediately realised that he could not dare to continue with his journey on the public highway so he turned off the road, into the woods. During the first few hours his decision caused him no difficulty but later it became worse for he had gotten into a big and confusing forest. He continued to walk and when he came to the end of the forest he realized that he had been walking around in the same part of the forest. Then he recalled his thoughts about the world and the rattrap and knew his own turn had come. He had let himself be fooled by a bait of thirty kronors and had been caught. His reaction reveals his gloom and despair as he realized the forest had closed in open upon him like an impenetrable prison from which he thought he could never escape. It also reveals that he was basically a good person at heart and was repentant of his folly.

Question 48.
Why did the peddler accept Edla’s invitation when he had already declined the iron¬master’s to go home with him? (Comptt, Delhi 2013)
Answer:
The peddler first meets Edla Willmannson, the ironmaster’s daughter, when on her father’s insistence she comes to invite the peddler to their manor house for Christmas. The peddler who had earlier refused to accept the ironmaster’s invitation could not decline Edla’s invitation. She looked at him with compassion because she immediately noticed that he was afraid. So she spoke to him in such a friendly and warm manner that the peddler felt a kind of confidence in her. Infact the peddler even tells her that it would never have occurred to him that she would bother herself over him and he would come with her immediately. So he accepted the fur coat which Edla’s valet handed him and followed the young lady out to the carriage.

Question 49.
The peddler declined the invitation of the ironmaster but accepted the one from Edla. Why? (Delhi 2015)
Answer:
The peddler first meets Edla Willmannson, the ironmaster’s daughter, when on her father’s insistence she comes to invite the peddler to their manor house for Christmas. The peddler who had earlier refused to accept the ironmaster’s invitation could not decline Edla’s invitation. She looked at him with compassion because she immediately noticed that he was afraid. So she spoke to him in such a friendly and warm manner that the peddler felt a kind of confidence in her. Infact the peddler even tells her that it would never have occurred to him that she would bother herself over him and he would come with her immediately. So he accepted the fur coat which Edla’s valet handed him and followed the young lady out to the carriage.

Question 50.
Edla proved to be much more persuasive than her father while dealing with the peddler. Comment. (Delhi 2015)
Answer:
Edla does prove to be much more persuasive than her father while dealing with the peddler. When the ironmaster invites the peddler to his manor house the latter vehemently refuses to go. Despite trying to allay the peddlers’s fears, the ironmaster is unable to convince him to give them company for Christmas. He finally gives up after the peddler’s repeated refusals and goes away. Later Edla arrives, approaches the peddler and extends the invitation once again. She assures him he can leave after Christmas anytime. The peddler is touched by Edla’s warmth, friendly and sympathetic nature. So he agrees to accompany Edla.

Question 51.
Why did the crofter repose confidence in the peddler? How did the peddler betray that and with what consequences? (Delhi 2015)
Answer:
One dark evening when the peddler was walking along the road he knocked on the door of a cottage to seek shelter for the night. To his surprise, he was welcomed by an old man, the crofter, who lived alone in the cottage. The lonely crofter was happy to find a man whom he could talk to. He served the peddler supper, gave him tobacco and played a game of cards with him. The old crofter then went to the window and took down a leather pouch. He counted three ten kronor notes and put them into the pouch. This provided a big bait for the peddler who was tempted to steal the notes. The peddler unwillingly allowed himself to be tempted to touch the bait and was thus trapped in. He stole the money and thus committed a breach of trust. So he betrayed the confidence reposed in him by his host. After stealing the crofter’s money he turned off the road, and ventured into the woods. There he got completely lost in the big and confusing forest. Later the whole forest seemed to close in upon him like an impenetrable prison from which he could just not escape.

Question 52.
To be grateful is a great virtue of a gentleman. How did the peddler show his gratitude to Edla? (Comptt. Delhi 2015)
Answer:
Edla knew that her father was mistaken when he invited the peddler home thinking he was his long lost friend. Later it was revealed that he was a complete stranger and not a straight forward man. Despite this Edla begged her father not to send him away on Christmas eve. She invited him home and gave him food, shelter and clothes. Her kindness, compassion and sympathy brings out the goodness in the tramp. He leaves a packet for her as a Christmas gift which contains a rattrap and three ten kronor notes stolen from the crofter. It also contains a letter in which he signs himself as captain. Edla’s care and concern changes the peddler into a dignified gentleman.
Shiv Das Chapterwise Board Solutions (English (core) XII)

Question 53.
How did Edla treat the peddler? (Compff. Delhi 2016)
Answer:
Edla was the daughter of the ironmaster. She was a warm, sensitive and compassionate person. She understood in the first meeting that the peddler was scared of someone and he had not had a single day of peace. So she decided to let him spend Christmas with them in their house. She assured him that he could leave anytime he wanted after celebrating Christmas with them. Edla woke him up to eat food only. Otherwise she let him sleep for two days. She showed genuine goodness which brought about a positive transformation in the peddler. She continues to treat him like a captain and the peddler, quite spontaneously, starts behaving like a real captain.

Question 54.
The peddler thinks that the whole world is a rattrap. This view of life is true only of himself and of no one else in the story. Comment. (Delhi 2017)
Answer:
The peddler had naturally been thinking of his rattraps when he was struck by the idea that the whole world was nothing but a big rattrap. It only existed to set baits for people. When someone let himself be tempted to touch the bait, it closed on him, and then everything came to an end. But this view of life is true only of the peddler himself and of no one else in the story. The three ten kronor notes of the old crofter provide a bait for the peddler and after he steals the money, he gets lost in the big and confusing forest. He then recalls his thoughts about the world and the rattrap and knew his . turn had come. He had let himself be tempted by a bait and had been caught in the rattrap. He realized the forest had closed upon him like an impenetrable prison from which he thought
he could never escape. The theme of the story of ‘The Rattrap’ is that most human beings are prone to fall into the trap of material benefit. However, love and understanding can transform a person and bring out his essential human goodness. So even though the old crofter was kind and hospitable to him, he betrayed his trust and stole thirty kronors from him. He was not impressed by the iron¬master’s invitation also. But Edla Willmansson’s compassion brought out a transformation in his nature. Human qualities helped in raising him to be a gentleman.

Question 55.
The peddler believed that the whole world is a rattrap. How did he himself get caught in the same? (All India 2017)
Answer:
The peddler had naturally been thinking of his rattraps when he was struck by the idea that the whole world was nothing but a big rattrap. It only existed to set baits for people. When someone let himself be tempted to touch the bait, it closed on him, and then everything came to an end. But this view of life is true only of the peddler himself and of no one else in the story. The three ten kronor notes of the old crofter provide a bait for the peddler and after he steals the money, he gets lost in the big and confusing forest. He then recalls his thoughts about the world and the rattrap and knew his . turn had come. He had let himself be tempted by a bait and had been caught in the rattrap. He realized the forest had closed upon him like an impenetrable prison from which he thought
he could never escape. The theme of the story of ‘The Rattrap’ is that most human beings are prone to fall into the trap of material benefit. However, love and understanding can transform a person and bring out his essential human goodness. So even though the old crofter was kind and hospitable to him, he betrayed his trust and stole thirty kronors from him. He was not impressed by the iron¬master’s invitation also. But Edla Willmansson’s compassion brought out a transformation in his nature. Human qualities helped in raising him to be a gentleman.

Question 56.
The people we meet in life leave an impression on us. How is the rattrap peddler affected by meeting the crofter and Edla? (Comptt. All India 2017)
Answer:
The good people we meet sometimes leave an impression on us. Compassion and understanding can transform a person and bring out his essential human goodness. As is the case with the rattrap peddler whose meeting with the crofter and Edla bring about a positive transformation in his nature. The lonely old crofter was extremely kind to the peddler. Despite his hospitality the peddler stole his money and committed a breach of trust. Edla too treated him nicely, even after she came to know his true identity. The peddler who had always considered the world to be a rattrap that enclosed upon people finally felt released from this rattrap due to the kind, generous and sympathetic treatment of the crofter and Edla. He leaves behind a letter of thanks for Edla with a Christmas gift and the money he had stolen from the crofter, to be restored to its rightful owner.

Important Questions for Class 12 English

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CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण

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CBSE Class 12 Sanskrit व्याकरणम् सन्धि – प्रकरण

पाठ्यक्रम में पाँच अंश हैं – (i) सन्धि, (ii) समास, (ii) प्रत्यय, (iv) अन्विति (v) उपपदविभक्ति।
समास के अंतर्गत अव्ययीभाव, तत्पुरुष, द्विगु, कर्मधारय, द्वन्द्व तथा बहुव्रीहि समास हैं। प्रत्यय के अंतर्गत कृदन्त (क्त, क्तवतु, क्त्वा, ल्यप्, तुमुन्, तव्यत्, अनीयर्, यत्, क्तिन्,) तथा तद्धित (मतुप्, इन्, ठक्, ठञ्, त्व तथा तल्) आते हैं। सन्धि, समास तथा उपपद विभक्ति के संपूर्ण उदाहरण पाठ्यपुस्तक पर आधारित होंगे। इन सब बातों को ध्यान में रखते हुए आगे आने वाले पृष्ठों में व्याकरण के नियम बताकर वे सभी उदाहरण दिए गए हैं जो पाठ्यपुस्तक में आए हैं।

  1. सन्धि-प्रकरण
  2. समास-प्रकरण
  3. प्रकृति-प्रत्यय-विभाग
  4. अन्विति-प्रकरण
  5. उपपदविभक्ति प्रयोग

1. सन्धि-प्रकरण
अत्यंत समीपवर्ती दो वर्गों के मेल को सन्धि कहते हैं; जैसे- ‘यमुनाभम्’ में ‘यमुना’ व ‘आभम्’ पदों में दो समीपवर्ती आ, आ वर्गों का मेल होकर एक ‘आ’ वर्ण हो गया है। सन्धि को संहिता भी कहते हैं। सन्धि के नियमअत्यन्त निकट होने के कारण दो वर्षों में कभी सुविधा से, तो कभी शीघ्रता के परिणामस्वरूप परिवर्तन होता है। यह परिवर्तन निम्न प्रकार से होता है; जैसे

  1. कभी दोनों वर्गों के स्थान पर एक नया वर्ण बन जाता है; जैसे- गङ्गा + इव = गङ्गेव । यहाँ ‘आ’ तथा ‘इ’ के मेल से नया वर्ण ‘ए’ बन गया है।
  2. कभी पूर्व वर्ण में परिवर्तन हो जाता है; जैसे- इति + एषः = इत्येषः। यहाँ इति के अंतिम वर्ण ‘इ’ को ‘ए’ परे होने पर ‘य’ हो गया है।
  3. कभी उत्तरवर्ती (परवर्ती) वर्ण का लोप हो जाता है; जैसे- बाले + अस्मिन् = बालेऽस्मिन्। यहाँ अस्मिन् के ‘अ’ का लोप दिखाने के लिए अवग्रह 2019-08-05 17_17_52-Extracted pages from Ch 1.pdf - Foxit PhantomPDFका चिह्न अंकित किया गया है।
  4. कभी दो वर्षों के बीच में एक नया वर्ण आ जाता है; जैसे- तरु + छाया = तरुच्छाया। यहाँ ‘उ’ और ‘छ’ के बीच ‘च्’ का आगमन हुआ है।
  5. कभी वर्ण द्वित्व हो जाता है; जैसे- पिबन् + इव = पिबन्निव।

सन्धि के भेद –
(क) स्वर सन्धिः
(ख) व्यञ्जन सन्धिः
(ग) विसर्ग सन्धिः

(क) स्वर सन्धि : दो समीपस्थ स्वरों में परिवर्तन हो तो उसे स्वर सन्धि कहते हैं; जैसे- केन + अपि = केनापि।
स्वर संन्धि के प्रकार – दीर्घ, गुण, वृद्धि, यण, अयादि, पूर्वरूप।

(i) दीर्घ सन्धि- समान वर्ण परे होने पर अक् (अ, इ, उ) को दीर्घ हो जाता है; जैसे
CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण 1

उदाहरण :
CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण 2

(i) गुण सन्धि – ‘अ’ या ‘आ’ के अनन्तर ह्रस्व या दीर्घ इ, उ, ऋ, लू हों तो वे क्रमशः ए, ओ, अर्, अल् हो जाते । है; जैसे –
CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण 3
CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण 4

(iii) वृद्धि सन्धि – ‘अ’ या ‘आ’ के बाद ए, ऐ तथा ओ, औ होने पर दोनों के मेल से ऐ, औ हो जाते हैं –
अ, आ + ए, ऐ = ऐ                  अ, आ + ओ, औ = औ ।
उदाहरण :
CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण 5

(iv) यण् सन्धि – इक् अर्थात् ह्रस्व या दीर्घ इ, ई, उ, ऊ, ऋ, ऋ, लू के अनन्तर कोई असवर्ण (असमान) स्वर आए तो इ, उ, ऋ, लू के स्थान पर क्रमश: य्, व, र, ल्, (यण्) हो जाते हैं तथा परवर्ती स्वर इन वर्गों के साथ मिल जाते हैं; जैसे –
CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण 6

(v) अयादि सन्धि – ए, ओ, ऐ, औ (एच्) के अनन्तर कोई भी स्वर हो तो ‘ए’ को अय्, ‘ओ’ को ‘अ’, ‘ऐ’ को ‘आय्’ तथा ‘औ’ को ‘आव्’ हो जाते हैं।
उदाहरण:
रात्रौ + अपि = रात्रावपि

(vi) पूर्वरूप सन्धि- यदि पदान्त (पद के अंत) में ‘ए’ या ‘ओ’ हो तथा बाद में ह्रस्व ‘अ’ हो तो अयादि संधि का । नियम लागू नहीं होता अपितु परवर्ती स्वर ‘अ’ पूर्ववर्ती स्वर में बिना परिवर्तन के मिल जाता है तथा उसके स्थान पर () पूर्वरूप यह चिह्न हो जाती है; जैसे –
CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण 7

(ख) व्यञ्जन सन्धि – किसी व्यंजन का किसी व्यंजन या स्वर के साथ मेल होने पर व्यंजन में जो परिवर्तन होता है। उसे व्यंजन सन्धि कहते हैं; जैसे –

(i) न् को द्वित्व – ‘न’ से पहले ह्रस्व अ, इ, उ हो तथा ‘न्’ के बाद कोई स्वर हो तो ‘न्’ को द्वित्व हो जाता है।
पिबन्     + इव    = पिबन्निव
पीडयन्, + अङ्गः = पीडयन्नङ्गः
ध्यक्षन्    + इव    = ध्यक्षन्निव
तस्मिन्   + एव    = तस्मिन्नेव
खादन्    + अपि  = खादन्नपि
गच्छन्   + एव     = गच्छन्नेव। इत्यादि
कुर्वन्    + अस्ति = कुर्वन्नस्ति

(ii) त्- च् (च परे होने पर)
अचिरात् + च = अचिराच्च यत् + च = यच्च
तत् + श्रुत्वा = तच्छुत्वा (श् परे होने पर त् को च् तथा श् को छू हो जाता है)

(iii) त् – ज् (ज परे होने पर)।
वशात् + जनः = वशाज्जनः

(iv) हल् सन्धिः (जश्त्व सन्धि) – पूर्वपद के अन्त में क्, च्, ट्, त्, ए होते हैं और बाद में कोई वर्ण होता है। पूर्वपद के स्थान पर क्रमशः वर्ग का तीसरा वर्ण होता है; जैसे – क् → ग्। च् → ज् ट्। → ड्। त् → दू। तथा प् → ब्।
उदाहरण – त् → द्
CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण 9

(v) अनुस्वार सन्धि : ‘म्’ के स्थान पर अनुस्वार हो जाता है।
CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण 10

(ग) विसर्ग सन्धि-विसर्ग से परे स्वर या व्यंजन होने पर विसर्ग में होनेवाले परिवर्तन को विसर्ग संधि कहते हैं।

(i) सत्व > विसर्ग को स् – (‘त’ परे होने पर)
CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण 11

(ii) शत्व > विसर्ग को श्- (‘च’ परे होने पर)
CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण 12

(iii) उत्व > विसर्ग को उ- (पूर्ववर्ती ‘अ’ के साथ मिलकर ‘ओ’)
CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण 13
CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण 14

(vi) रत्व > विसर्ग को र्- (परवर्ती स्वर, वर्ग का 3, 4, 5, य्, र, ल, व्, हु, होने पर)
उदाहरण :
CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण 15

(v) विसर्ग लोपः (एषः एवं सः का संयोग यदि किसी भी स्वर व व्यंजन के साथ हो तो विसर्ग का लोप ही जाता है)
CBSE Class 12 Sanskrit व्याकरणम् सन्धि-प्रकरण 16

अभ्यासार्थ

निम्न वाक्येषु स्थूलशब्देषु सन्धिच्छेदं कुरुत –

1. तृणानि भूमिरुदकम्।
उत्तर:
भूमिः + उदकम्

2. एतान्यपि सतां गेहे।
उत्तर:
एतानि + अपि

3. नोच्छिद्यन्ते कदाचन।
उत्तर:
न + उच्छिद्यन्ते

4. अहिंसया च भूतात्मा।
उत्तर:
भूत + आत्मा

5. सत्यमेव जयति नानृतम्।
उत्तर:
न + अनृतम्

6. सत्येन पन्था विततोदेवयानः।
उत्तर:
विततः + देवयानः

7. येनाक्रमन्त्य॒षयो ह्याप्तकामाः।
उत्तर:
येन + आक्रमन्ति + ऋषयः, हि + आप्तकामाः

8. यस्तु सर्वाणि भूतान्यात्मन्येवानुपश्यति।
उत्तर:
यः + तुः, भूतानि + आत्मनि + एव + अनुपश्यति

9. सर्वभूतेषु चात्मानम् ततो न विजुगुप्सते।
उत्तर:
च + आत्मानम्, ततः + न

10. दुर्गं पथस्तत् कवयोवदन्ति।
उत्तर:
पथः + तत्, कवयः + वदन्ति

11. प्रारम्भे अपि सूर्योदयस्य रम्यम् वर्णनम् उपलभ्यते।
उत्तर:
सूर्य + उदयस्य

12. भगवतोमरीचिमालिनः।
उत्तर:
भगवतः + मरीचिमालिन:

13. एष भगवान् मणिराकाश-मण्डलस्य
उत्तर:
मणिः + आकाश

14. इनश्च दिनस्य।
उत्तर:
इन: + च।

15. अयम् एव कारणं षण्णाम् ऋतूणाम्।।
उत्तर:
षट् + नाम्

16. एष एव अङ्गीकरोति उत्तरं दक्षिणं चायनम्।
उत्तर:
एषः + एव, अम् + गीकरोति, च + अयनम्।

17. भो: भो: प्रासादाधिकृताः पुरुषा:।
उत्तर:
प्रसाद + अधिकृताः

18. सुगाङ्गप्रासादस्य उपरि स्थिताः प्रदेशाः संस्क्रियन्ताम्।
उत्तर:
सुगाम् + ग, सम् + क्रियन्ताम्

19. कौमुदी-महोत्सवः प्रतिषिद्धः?
उत्तर:
महा + उत्सवः

20. आर्य! अथ अस्मद्वचनात् कौमुदीमहोत्सवः आघोषित:?
उत्तर:
अस्मत् + वचनात्

21. अपहृतः पेक्षकाणाम् चक्षुषोविषयः?
उत्तर:
चक्षुषः + विषय:

22. अहो राजाधिराजमन्त्रिणो विभूतिः।
उत्तर:
राजा + अधिराजमन्त्रिणः + विभूति:

23. तदुपविशतु आर्यः।।
उत्तर:
तत् + उपविशतु

24. न प्रयोजनम् अन्तरा चाणक्यः स्वप्नेऽपिचेष्टते।
उत्तर:
स्वप्ने + अपि

25. वृषल! किम् अस्थाने महान् प्रजा-धनापव्यय:?
उत्तर:
धन + अपव्ययः

26. पितृवधात् क्रुद्धः राक्षसोपदेशप्रवणः।
उत्तर:
राक्षस + उपदेशप्रवणः

27. अतः इदानीं दुर्गसंस्कारः प्रारब्धव्यः।
उत्तर:
प्र + आरब्धव्यः

28. प्रत्युत्पन्नमतिः शीघ्रमेव निर्णयं कृत्वा आत्मरक्षां करोति।
उत्तर:
प्रति + उत्पन्नमतिः

29. कस्मिंश्चित् जलाशये त्रयोमत्स्याः प्रतिवसन्ति स्म।
उत्तर:
कस्मिन् + चित्, त्रयः + मत्स्याः

30. गच्छभिः मतस्यजीविभिः उक्तम्।
उत्तर:
गच्छद् + भिः

31. अहो, बहुमत्स्योऽयं सरः।
उत्तर:
बहुमत्स्यः + अयम्

32. अस्माभिः कदापि न अन्वेषितः।
उत्तर:
अनु + एषितः

33. अशक्तैर्बलिनः शत्रो: प्रपलायनं कर्तव्यम्।
उत्तर:
अशक्तैः + बलिनेः

34. आश्रितव्योऽथवा दुर्गः।
उत्तर:
आश्रितव्यः + अथवा

35. नान्या तेषां गतिर्भवेत्।
उत्तर:
न + अन्या, गति: + भवेत्

36. तन्न साम्प्रतं क्षणमप्यत्र अवस्थातुं युक्तम्।
उत्तर:
तत् + न, क्षणम् + अपि + अत्र

37. ते विद्वांसः देहभङ्ग कुलक्षयम् न पश्यन्ति।
उत्तर:
देहभम् + गम्

38. तदाकर्ण्य प्रत्युत्पन्नमतिः प्राह।
उत्तर:
तत् + आकर्त्य, प्रति + उत्पन्नमतिः

39. ममापि अभीष्टमेतत्।
उत्तर:
मम + अपि

40. तदन्यत्र गम्यताम् इति।
उत्तर:
तत् + अन्यत्र

41. किं वाङ्मात्रेणापि एतत् सर: त्यक्तुं युज्यते?
उत्तर:
वाक् + मात्रेण + अपि

42. जीवत्यनाथोऽपि वने विसर्जितः।
उत्तर:
जीवति + अनाथः + अपि

43. अनागतविधता प्रत्युत्पन्नमतिश्च निष्क्रान्तौ।
उत्तर:
प्रत्युत्पन्नमतिः + च

44. तैः मत्स्यजीविभि: जालैस्तं जलाशयम् निर्मत्स्यतां नीतम्।
उत्तर:
जालैः + तं ।

45. अहो! कीदृशीयं हिमानी राजते।
उत्तर:
कीदृशी + इयं

46. किम् एतानि पर्वतारोहाणस्य चित्राणि सन्ति?
उत्तर:
पर्वत + आरोहणस्य

47. इदम् अभियानम् अतीव रोचकम् साहसिकं चासीत्।
उत्तर:
च + आसीत्।

48. नास्ति सन्देहः।।
उत्तर:
न + अस्ति

49. किं लद्दाख-शब्दस्य कश्चिद् विशिष्टोऽर्थः?
उत्तर:
कः + चित्, विशिष्टः + अर्थ:

50. लद्दाखमार्गेणैव तिब्बत क्षेत्रे बौद्धधर्मस्य प्रवेशोऽभवत्।
उत्तर:
लद्दाखमार्गेण + एव, प्रवेशः + अभवत्

51. उपत्यकाया चित्रेऽस्मिन् या रेखा प्रतिभाति।
उत्तर:
चित्रे + अस्मिन्

52. यत: बालुका उड्डीय सर्वांभूमिम् आवृणोति।
उत्तर:
उत् + डीय, सर्वाम् + भूमिम्

53. इदं ‘लेह’ इत्यभिधानेन प्रसिद्ध पर्यटन स्थलम्।
उत्तर:
इति + अभिधानेन

54. एषः बौद्धधर्मस्य प्रसिद्धः प्राचीनश्च श्वेतस्तूपः।
उत्तर:
प्राचीनः + च

55. “स्टाक पैलेस” इत्याख्यः प्रासादोऽपि अत्रैव वर्तते।
उत्तर:
इति + आख्यः, प्रासादः + अपि, अत्र + एव

56. स्टाक पैलेस संग्रहालयोवर्तते।
उत्तर:
संग्रह + आलयः + वर्तते

57. किं तेऽपि उत्सवप्रिया:?
उत्तर:
ते + अपि

58. मानवः स्वभावादेव उत्सवप्रियाः।
उत्तर:
स्वभावात् + एव ।

59. बौद्धानां ‘गम्पा’ नाम वार्षिकोत्सवः शीते आयाति।
उत्तर:
वार्षिक + उत्सवः

60. सः प्रदेश: अत्रैव अस्ति।
उत्तर:
अत्र + एव

61. ग्रीष्मे ऋतौ पर्वतारोहिणोऽत्र प्रायः दृश्यन्ते।
उत्तर:
पर्वत + आरोहिणः + अत्र

62. एको हि दोषो गुणसन्निपाते।
उत्तर:
एकः + हि, दोषः + गुणसन्निपाते

63. निमज्जतीन्दोः किरणेष्विवाङ्कः।
उत्तर:
निमज्जति + इन्दोः, किरणेषु + इव + अङ्कः

64. जीवनस्य प्रत्येक क्षेत्रे सुभाषितानि साहित्ये सुलभानि सन्ति।
उत्तर:
प्रति + एकं

65. सन्मार्ग च प्रदर्शयन्ति।
उत्तर:
सत् + मार्ग

66. अस्मिन् पाठे केषाञ्चित् मधुरवचनानां सङ्कलनं प्रस्तूयते।
उत्तर:
केषाम् + चित्

67. सदयं हृदयं सुधामुचोवाचः।
उत्तर:
सुधामुचः + वाचः

68. येषां करणं परोपकरणं ते केषां न वन्द्याः ।
उत्तर:
पर + उपकरणं

69. हुतं च दत्तं च सदैव तिष्ठति।
उत्तर:
सदा + एव

70. यथा चतुर्भिः कनकम् परीक्ष्यते।
उत्तर:
परि + ईक्ष्यते

71. निर्घषणच्छेदन तापताडनैः।
उत्तर:
निर्घषण + छेदन

72. प्रविश्य हि छनयन्ति शठस्तथाविधाः।
उत्तर:
शठः + तथाविधा:

73. नसंवृत्ताङ्गान्निशिता इवेषवः।
उत्तर:
न + सम् + वृत्त + अङ्गात् + निशिता, इव + इषवः

74. स किंसखा।
उत्तर:
किम् + सखा

75. साधु न शास्ति योऽधिपम्।
उत्तर:
यः + अधिपम्

76. हितान्न यः संशृणुते स: किंप्रभुः।
उत्तर:
हितात् + न

77. सदानुकूलेषु हि कुर्वते रतिम्।
उत्तर:
सदा + अनुकूलेषु

78. नृपेष्वमात्येषु च सर्वसम्पदः।
उत्तर:
नृपेषु + अमात्येषु

79. लोभश्चेदगुणेन किम्?
उत्तर:
लोभः + चेत् + अगुणेन

80. पिशुनता यद्यस्ति किं पातकै:?
उत्तर:
यदि + अस्ति

81. सत्यंचेत्तपसा च किम्?
उत्तर:
सत्यम् + चेत् + तपसा

82. शुचिमनोयद्यस्ति तीर्थेन किम्?
उत्तर:
शुचिमनः + यदि + अस्ति

83. सदविद्या यदि किम्?
उत्तर:
सत् + विद्या

84. धनैरपयशो यद्यस्ति किं मृत्युना?
उत्तर:
धनैः + अपयशः

85. ‘चारुदत्तं’ नाटकस्य प्रथमाङ्कात् सङ्कलितः।
उत्तर:
प्रथम + अम् + कात्, सम् + कलितः

86. सः उदारतावशदानकारणात् च शीघ्रं दरिद्रो जातः।
उत्तर:
दरिद्रः + जातः

87. किन्तु दैन्येऽपि तस्य मनः भ्रष्टं न भवति।
उत्तर:
दैन्ये + अपि

88. दरिद्रावस्थायां मित्राणाम् उपेक्षया कटुः अनुभवः भवति।
उत्तर:
दरिद्र + अवस्थायां

89. नान्द्यन्ते ततः प्रविशति सूत्रधारः।
उत्तर:
नान्दी + अन्ते

90. किन्नु खलु अद्य प्रत्यूष एव गेहान्निष्क्रान्तस्य।
उत्तर:
किम् + नु, गेहात् + निष्क्रान्तस्य

91. चञ्चलायेते इव मेऽक्षिणी।
उत्तर:
मे + अक्षिणी

92. किन्नु खलु संविधा विहिता न वेति।
उत्तर:
वा + इति

93. आर्ये! इतः+तावत्।
उत्तर:
इतस्तावत्

94. आर्य! दिष्ट्या खलु आगतोऽसि।
उत्तर:
आगतः + असि

95. किम् अस्त्यस्माकं गेहे कोऽपि प्रातराश:?
उत्तर:
अस्ति + अस्माकं

96. घृतं गुडो दधि तण्डुलाश्च सर्वमस्ति।
उत्तर:
तण्डुलाः + च

97. नहि नहि, अन्तरापणे।
उत्तर:
अन्तर + आपणे

98. आर्य! अद्य ममोपवासः अस्ति।
उत्तर:
मम + उपवासः

99. यदि आर्यस्यानुग्रहः स्यात्।
उत्तर:
आर्यस्य + अनुग्रह

100. भणामि कार्यान्तरे व्यस्तः।
उत्तर:
कार्य + अन्तरे

101. अहम् अन्यत्र भुक्त्वा तस्यावासमेव गच्छामि।
उत्तर:
तस्य + आवासम् + एव

102. मयापि मैत्रेयेण परस्य आमन्त्रणकानि अभिलषणीयानि।
उत्तर:
मया + अपि

103. पुनरपि सन्तुष्टोऽहम्।
उत्तर:
पुनः + अपि

104. तदैव तत्रभवत: आर्यचारूदत्तस्य देवकार्यकारणात् गृहीतानि।
उत्तर:
तदा + एव

105. एष चारुदत्त: गृहदैवतानि अर्चयन् इतएवागच्छति।
उत्तर:
इतः + एव + आगच्छति

106. यावएनमुपसर्पामि।
उत्तर:
यावत् + एनम्।

107. भो: दारिद्र्यं खलु पुरुषस्य सोच्छ्वासं मरणम्।
उत्तर:
स + उत् + श्वास

108. न खल्वहं नष्टां श्रियम् अनुशोचामि।
उत्तर:
खलु + अहम्

109. सुखं हि दुःखान्यनुभूय शोभते।
उत्तर:
दुःखानि + अनुभूय

110. यथान्धकारादिव दीपदर्शनम्।
उत्तर:
यथा + अन्धकारात् + इव

111. इति प्रत्ययादेव ममार्थाः क्षीणाः जाताः।।
उत्तर:
प्रत्ययात् + एव, मम + अर्थाः

112. भाग्यक्रमेण हि धनानि पुनर्भवन्ति।
उत्तर:
पुन: + भवन्ति

113. एतत्तु मां दहति नष्टधनश्रियो मे।
उत्तर:
माम् + दहति, नष्टधनश्रियः +मे

114. सुहृदः स्फीता: भवन्त्यापदः।
उत्तर:
भवन्ति + आपद:

115. पापं कर्म च यत् परैरपि कृतं तत्तस्य संभाव्यते।।
उत्तर:
परैः + अपि

116. अङ्कुराद् अङ्कुराः निस्सरन्ति।
उत्तर:
अम् + कुरात्, अम् + कुराः

117. तथैव धनविनाशदु:खस्य।
उत्तर:
तथा + एव

118. चिन्त्यमानस्य नानाविधा: चिन्ताकुराः प्रादुर्भवन्ति।
उत्तर:
चिन्ता + अङ्कुराः, प्रादुः + भवन्ति

119. तदलं भवतः सन्तापेन।।
उत्तर:
तत् + अलं

120. विभवानुवशी भार्या दरिद्रेषु दुर्लभा।
उत्तर:
विभव + अनुवशा

121. सभागारस्य दृश्यम् अत्र वर्तते।
उत्तर:
सभा + आगारस्य

122. नमः सभाभ्य: सभापतिभ्यश्च।
उत्तर:
सभापतिभ्यः + च

123. अद्य अस्माकं मध्ये ग्रीष्मावकाश परियोजनाकायें।
उत्तर:
अस्माकम् + मध्ये, ग्रीष्म + अवकाश

124. सर्वोत्तमान् अङ्कान् लब्धवन्तः छात्राः समुपस्थिताः।
उत्तर:
सर्व + उत्तमान्, अम् + कान्

125. एते स्वाध्यायस्य विशिष्टांशान् अत्र प्रस्तोष्यन्ति।
उत्तर:
स्व + अध्यायस्य, विशिष्ट + अंशान्।

126. भागत्रयं भवेदस्य त्रिपुरस्य यथाक्रमम्।।
उत्तर:
भवेत् + अस्य

127. द्वितीयभागस्सञ्चारो जलस्यान्तर्बहिः क्रमात्।
उत्तर:
द्वितीयभागः + सञ्चारं:, जलस्य + अन्तर्बहिः

128. तृतीयभागस्सञ्चारस्त्वन्तरिक्षं भवेत् स्वतः।।
उत्तर:
तृतीयभागः + सञ्चारः + तु + अन्तरिक्ष

129. यदि जम्बूवृक्षस्य प्राग्वल्मीको समीपस्थ: भवेत्।
उत्तर:
प्राक् + वल्मीकः

130. तस्माद् दक्षिणपावें स्वादु सलिलं पुरुषद्वये भवष्यिति।
उत्तर:
तस्मात् + दक्षिणपाश्र्वे

131. तत्र तु एकं शून्यञ्च द्वे एव संख्ये महत्त्वपूर्ण स्तः।
उत्तर:
शून्यम् + च

132. परं समयाभावात् तस्याः सर्वस्याः प्रस्तुतिः अत्र न भविष्यति।
उत्तर:
समय + अभावात्

133. आधुनिकै: वैज्ञानिकैरपि तथैव मन्यते।।
उत्तर:
वैज्ञानिकैः + अपि, तथा + एव

134. सूर्य प्रति पूर्वाभिमुखा पृथिवी 365.25 वारं प्रतिवर्ष भ्रमति।
उत्तर:
पूर्व + अभिमुखा

135. तथैव नक्षत्रादयः पश्चिमं प्रति धावन्तः प्रतीयन्ते।।
उत्तर:
नक्षत्र + आद्यः

136. नातप्तं लोहं लोहेन सन्धत्ते।
उत्तर:
न + अतप्तं

137. सम्यग् वर्णितम् त्वया।
उत्तर:
सम्यक् + वर्णितम्

138. ओइम् द्यौः शान्तिरन्तरिक्षइम् शान्तिः।
उत्तर:
शान्तिः + अन्तरिक्षम्

139. सर्वे मिलित्वा उच्चरन्ति।।
उत्तर:
उत् + चरन्ति

140. तत् रात्रावपि किञ्चित् निकटं सरः गम्यताम्।
उत्तर:
रात्रौ + अपि, किम् + चित्

141. सः चन्द्रगुप्तस्य आदेशस्य उल्लङ्घनम् कर्तुम् उत्सहते स्म।।
उत्तर:
उत् + लङ्घनम्

142. तत् कुलिशपातोपमं वचः श्रुत्वा अनागत विधाता अवदत्।।
उत्तर:
कुलिशपात + उपमं

143. भगवतोबुद्धस्य सप्तदशशताब्द्याः मूर्तिः आकर्षणकेन्द्रम् अस्ति।
उत्तर:
भगवतः + बुद्धस्य

144. मन्ये उत्कीर्णा लेखा भित्तिलेखाश्च तिब्बतशैल्याः परिचायकाः।
उत्तर:
उत्कीर्णाः + लेखाः, भित्तिलेखाः + च

145. पर्वतारोहणाय ‘लिकिर’ ‘स्टाक’ नाम्नी स्थले उपयुक्ते स्तः।।
उत्तर:
पर्वत + आरोहणाय

146. संस्कृतवाङ्मयं सहस्रशः सुमधुरवचनैः सम्यग् अलङ्कृतं वर्तते।
उत्तर:
संस्कृतवाक् + मयं, सम्यक् + अलङ्कृतं

147. गुरुवासरे अर्धावकाशानन्तरं सभागारे एका संङ्गोष्ठी भविष्यति।।
उत्तर:
अर्ध + अवकाश + अनन्तरं

148. छात्रैः कृतस्य विशिष्टाध्ययनस्य परिचयः अस्मभ्यं भविष्यति।
उत्तर:
विशिष्ट + अध्ययनस्य

149. वराहमिहिरेण स्वग्रन्थे वृक्षायुर्वेदः, वास्तुविज्ञानं, ज्योतिषं इत्यादयः विषयाः वर्णिताः।
उत्तर:
वृक्ष + आयुवेदः, इति + आदयः

NCERT Solutions for Class 12 Sanskrit

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Important Questions for Class 10 Maths Chapter 6 Triangles

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Important Questions for Class 10 Maths Chapter 6 Triangles

Triangles Class 10 Important Questions Very Short Answer (1 Mark)

Question 1.
If ∆ABC ~ ∆PQR, perimeter of ∆ABC = 32 cm, perimeter of ∆PQR = 48 cm and PR = 6 cm, then find the length of AC. (2012)
Solution:
∆ABC ~ ∆PQR …[Given
Important Questions for Class 10 Maths Chapter 6 Triangles 1

Question 2.
∆ABC ~ ∆DEF. If AB = 4 cm, BC = 3.5 cm, CA = 2.5 cm and DF = 7.5 cm, find the perimeter of ∆DEF. (2012, 2017D)
Solution:
∆ABC – ∆DEF …[Given
Important Questions for Class 10 Maths Chapter 6 Triangles 2
Important Questions for Class 10 Maths Chapter 6 Triangles 3

Question 3.
If ∆ABC ~ ∆RPQ, AB = 3 cm, BC = 5 cm, AC = 6 cm, RP = 6 cm and PQ = 10, then find QR. (2014)
Solution:
∆ABC ~ ∆RPQ …[Given
Important Questions for Class 10 Maths Chapter 6 Triangles 4
∴ QR = 12 cm

Question 4.
In ∆DEW, AB || EW. If AD = 4 cm, DE = 12 cm and DW = 24 cm, then find the value of DB. (2015)
Solution:
Let BD = x cm
then BW = (24 – x) cm, AE = 12 – 4 = 8 cm
In ∆DEW, AB || EW
Important Questions for Class 10 Maths Chapter 6 Triangles 5

Question 5.
In ∆ABC, DE || BC, find the value of x. (2015)
Important Questions for Class 10 Maths Chapter 6 Triangles 6
Solution:
In ∆ABC, DE || BC …[Given
Important Questions for Class 10 Maths Chapter 6 Triangles 7
x(x + 5) = (x + 3)(x + 1)
x2 + 5x = x2 + 3x + x + 3
x2 + 5x – x2 – 3x – x = 3
∴ x = 3 cm

Question 6.
In the given figure, if DE || BC, AE = 8 cm, EC = 2 cm and BC = 6 cm, then find DE. (2014)
Important Questions for Class 10 Maths Chapter 6 Triangles 8
Solution:
In ∆ADE and ∆ABC,
∠DAE = ∠BAC …Common
∠ADE – ∠ABC … [Corresponding angles
∆ADE – ∆ΑΒC …[AA corollary
Important Questions for Class 10 Maths Chapter 6 Triangles 9

Question 7.
In the given figure, XY || QR, \frac{P Q}{X Q}=\frac{7}{3} and PR = 6.3 cm, find YR. (2017OD)
Important Questions for Class 10 Maths Chapter 6 Triangles 10
Solution:
Let YR = x
\frac{\mathrm{PQ}}{\mathrm{XQ}}=\frac{\mathrm{PR}}{\mathrm{YR}} … [Thales’ theorem
Important Questions for Class 10 Maths Chapter 6 Triangles 11

Question 8.
The lengths of the diagonals of a rhombus are 24 cm and 32 cm. Calculate the length of the altitude of the rhombus. (2013)
Solution:
Diagonals of a rhombus are ⊥ bisectors of each other.
∴ AC ⊥ BD,
OA = OC = \frac{A C}{2} \Rightarrow \frac{24}{2} = 12 cm
OB = OD = \frac{B D}{2} \Rightarrow \frac{32}{2} = 16 cm
In rt. ∆BOC,
Important Questions for Class 10 Maths Chapter 6 Triangles 12

Question 9.
If PQR is an equilateral triangle and PX ⊥ QR, find the value of PX2. (2013)
Solution:
Altitude of an equilateral ∆,
Important Questions for Class 10 Maths Chapter 6 Triangles 13

Triangles Class 10 Important Questions Short Answer-I (2 Marks)

Question 10.
The sides AB and AC and the perimeter P, of ∆ABC are respectively three times the corresponding sides DE and DF and the perimeter P, of ∆DEF. Are the two triangles similar? If yes, find \frac { ar\left( \triangle ABC \right) }{ ar\left( \triangle DEF \right) } (2012)
Solution:
Given: AB = 3DE and AC = 3DF
Important Questions for Class 10 Maths Chapter 6 Triangles 14
…[∵ The ratio of the areas of two similar ∆s is equal to the ratio of the squares of their corresponding sides

Question 11.
In the figure, EF || AC, BC = 10 cm, AB = 13 cm and EC = 2 cm, find AF. (2014)
Important Questions for Class 10 Maths Chapter 6 Triangles 15
Solution:
BE = BC – EC = 10 – 2 = 8 cm
Let AF = x cm, then BF = (13 – x) cm
In ∆ABC, EF || AC … [Given
Important Questions for Class 10 Maths Chapter 6 Triangles 16

Question 12.
X and Y are points on the sides AB and AC respectively of a triangle ABC such that \frac{\mathbf{A X}}{\mathbf{A B}}=\frac{1}{4}, AY = 2 cm and YC = 6 cm. Find whether XY || BC or not. (2015)
Solution:
Given: \frac{A X}{A B}=\frac{1}{4}
Important Questions for Class 10 Maths Chapter 6 Triangles 17
AX = 1K, AB = 4K
∴ BX = AB – AX
= 4K – 1K = 3K
Important Questions for Class 10 Maths Chapter 6 Triangles 18
∴ XY || BC … [By converse of Thales’ theorem

Question 13.
In the given figure, ∠A = 90°, AD ⊥ BC. If BD = 2 cm and CD = 8 cm, find AD. (2012; 2017D)
Important Questions for Class 10 Maths Chapter 6 Triangles 19
Solution:
∆ADB ~ ∆CDA …[If a perpendicular is drawn from the vertex of the right angle of a rt. ∆ to the hypotenuse then As on both sides of the ⊥ are similar to the whole D and to each other
\frac{B D}{A D}=\frac{A D}{C D} …[∵ Sides are proportional
AD2 = BD , DC
AD2 = (2) (8) = 16 ⇒ AD = 4 cm

Question 14.
In ∆ABC, ∠BAC = 90° and AD ⊥ BC. Prove that AD\frac{B D}{A D}=\frac{A D}{C D} = BD × DC. (2013)
Solution:
In 1t. ∆BDA, ∠1 + ∠5 = 90°
In rt. ∆BAC, ∠1 + ∠4 = 90° …(ii)
∠1 + ∠5 = ∠1 + ∠4 …[From (i) & (ii)
.. ∠5 = ∠4 …(iii)
In ∆BDA and ∆ADC,
Important Questions for Class 10 Maths Chapter 6 Triangles 20
∠5 = 24 … [From (iii)
∠2 = ∠3 …[Each 90°
∴ ∆BDA ~ ∆ADC…[AA similarity
\frac{B D}{A D}=\frac{A D}{C D}
… [In ~ As corresponding BA sides are proportional
∴ AD2 = BD × DC

Question 15.
A 6.5 m long ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall. Find the height of the wall where the top of the ladder touches it. (2015)
Solution:
Let AC be the ladder and AB be the wall.
Important Questions for Class 10 Maths Chapter 6 Triangles 21
∴Required height, AB = 6 m

Question 16.
In the figure ABC and DBC are two right triangles. Prove that AP × PC = BP × PD. (2013)
Important Questions for Class 10 Maths Chapter 6 Triangles 22
Solution:
Important Questions for Class 10 Maths Chapter 6 Triangles 23
In ∆APB and ∆DPC,
∠1 = ∠4 … [Each = 90°
∠2 = ∠3 …[Vertically opp. ∠s
∴ ∆APB ~ ∆DPC …[AA corollary
\frac{\mathrm{BP}}{\mathrm{PC}}=\frac{\mathrm{AP}}{\mathrm{PD}} … [Sides are proportional
∴ AP × PC = BP × PD

Question 17.
In the given figure, QA ⊥ AB and PB ⊥ AB. If AO = 20 cm, BO = 12 cm, PB = 18 cm, find AQuestion (2017OD)
Important Questions for Class 10 Maths Chapter 6 Triangles 24
Solution:
Important Questions for Class 10 Maths Chapter 6 Triangles 25
In ∆OAQ and ∆OBP,
∠OAQ = ∠OBP … [Each 90°
∠AOQ = ∠BOP … [vertically opposite angles
Important Questions for Class 10 Maths Chapter 6 Triangles 26

Triangles Class 10 Important Questions Short Answer-II (3 Marks)

Question 18.
In the given figure, CD || LA and DE || AC. Find the length of CL if BE = 4 cm and EC = 2 cm. (2012)
Important Questions for Class 10 Maths Chapter 6 Triangles 27
Solution:
In ∆ABL, CD || LA
Important Questions for Class 10 Maths Chapter 6 Triangles 28

Question 19.
If a line segment intersects sides AB and AC of a ∆ABC at D and E respectively and is parallel to BC, prove that \frac{A D}{A B}=\frac{A E}{A C}. (2013)
Solution:
Given. In ∆ABC, DE || BC
Important Questions for Class 10 Maths Chapter 6 Triangles 29
To prove. \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}
Proof.
In ∆ADE and ∆ABC
∠1 = ∠1 … Common
∠2 = ∠3 … [Corresponding angles
∆ADE ~ ∆ABC …[AA similarity
\frac{\mathbf{A D}}{\mathbf{A B}}=\frac{\mathbf{A} \mathbf{E}}{\mathbf{A C}}
…[In ~∆s corresponding sides are proportional

Question 20.
In a ∆ABC, DE || BC with D on AB and E on AC. If \frac{A D}{D B}=\frac{3}{4} , find \frac{\mathbf{B} C}{\mathbf{D} \mathbf{E}}. (2013)
Solution:
Given: In a ∆ABC, DE || BC with D on AB and E
on AC and \frac{A D}{D B}=\frac{3}{4}
To find: \frac{\mathrm{BC}}{\mathrm{DE}}
Proof. Let AD = 3k,
Important Questions for Class 10 Maths Chapter 6 Triangles 30
DB = 4k
∴ AB = 3k + 4k = 7k
In ∆ADE and ∆ABC,
∠1 = ∠1 …[Common
∠2 = ∠3 … [Corresponding angles
∴ ∆ADE ~ ∆ABC …[AA similarity
Important Questions for Class 10 Maths Chapter 6 Triangles 31

Question 21.
In the figure, if DE || OB and EF || BC, then prove that DF || OC. (2014)
Important Questions for Class 10 Maths Chapter 6 Triangles 32
Solution:
Given. In ∆ABC, DE || OB and EF || BC
To prove. DF || OC
Proof. In ∆AOB, DE || OB … [Given
Important Questions for Class 10 Maths Chapter 6 Triangles 33

Question 22.
If the perimeters of two similar triangles ABC and DEF are 50 cm and 70 cm respectively and one side of ∆ABC = 20 cm, then find the corresponding side of ∆DEF. (2014)
Solution:
Important Questions for Class 10 Maths Chapter 6 Triangles 34
Given. ∆ABC ~ ∆DEF,
Perimeter(∆ABC) = 50 cm
Perimeter(∆DEF) = 70 cm
One side of ∆ABC = 20 cm
To Find. Corresponding side of ∆DEF (i.e.,) DE. ∆ABC ~ ∆DEF …[Given
Important Questions for Class 10 Maths Chapter 6 Triangles 35
∴ The corresponding side of ADEF = 28 cm

Question 23.
A vertical pole of length 8 m casts a shadow 6 cm long on the ground and at the same time a tower casts a shadow 30 m long. Find the height of tower. (2014)
Solution:
Important Questions for Class 10 Maths Chapter 6 Triangles 36
Let BC be the pole and EF be the tower Shadow AB = 6 m and DE = 30 m.
In ∆ABC and ∆DEF,
∠2 = ∠4 … [Each 90°
∠1 = ∠3 … [Sun’s angle of elevation at the same time
∆ABC ~ ∆DEF …[AA similarity
\frac{A B}{D E}=\frac{B C}{E F} … [In -As corresponding sides are proportional
\frac{6}{30}=\frac{8}{\mathrm{EF}} ∴ EF = 40 m

Question 24.
In given figure, EB ⊥ AC, BG ⊥ AE and CF ⊥ AE (2015)
Prove that:
(a) ∆ABG ~ ∆DCB
(b) \frac{\mathbf{B C}}{\mathbf{B D}}=\frac{\mathbf{B E}}{\mathbf{B A}}
Important Questions for Class 10 Maths Chapter 6 Triangles 37
Solution:
Important Questions for Class 10 Maths Chapter 6 Triangles 38
Given: EB ⊥ AC, BG ⊥ AE and CF ⊥ AE.
To prove: (a) ∆ABG – ∆DCB,
(b) \frac{B C}{B D}=\frac{B E}{B A}
Proof: (a) In ∆ABG and ∆DCB,
∠2 = ∠5 … [each 90°
∠6 = ∠4 … [corresponding angles
∴ ∆ABG ~ ∆DCB … [By AA similarity
(Hence Proved)
∴ ∠1 = ∠3 …(CPCT … [In ~∆s, corresponding angles are equal

(b) In ∆ABE and ∆DBC,
∠1 = ∠3 …(proved above
∠ABE = ∠5 … [each is 90°, EB ⊥ AC (Given)
∆ABE ~ ∆DBC … [By AA similarity
\frac{B C}{B D}=\frac{B E}{B A}
… [In ~∆s, corresponding sides are proportional
\frac{B C}{B D}=\frac{B E}{B A} (Hence Proved)

Question 25.
∆ABC ~ ∆PQR. AD is the median to BC and PM is the median to QR. Prove that \frac{\mathbf{A B}}{\mathbf{P Q}}=\frac{\mathbf{A D}}{\mathbf{P M}}. (2017D)
Solution:
Important Questions for Class 10 Maths Chapter 6 Triangles 39
∆ABC ~ ∆PQR … [Given
∠1 = ∠2 … [In ~∆s corresponding angles are equal
Important Questions for Class 10 Maths Chapter 6 Triangles 40

Question 26.
State whether the given pairs of triangles are similar or not. In case of similarity mention the criterion. (2015)
Important Questions for Class 10 Maths Chapter 6 Triangles 41
Solution:
Important Questions for Class 10 Maths Chapter 6 Triangles 42
(b) In ∆PQR, ∠P + ∠Q + ∠ZR = 180° …[Angle-Sum Property of a ∆
45° + 78° + ∠R = 180°
∠R = 180° – 45° – 78° = 57°
In ∆LMN, ∠L + ∠M + ∠N = 180° …[Angle-Sum Property of a ∆
57° + 45° + ∠N = 180°
∠N = 180° – 57 – 45° = 78°
∠P = ∠M … (each = 45°
∠Q = ∠N … (each = 78°
∠R = ∠L …(each = 57°
∴ ∆PQR – ∆MNL …[By AAA similarity theorem

Question 27.
In the figure of ∆ABC, D divides CA in the ratio 4 : 3. If DE || BC, then find ar (BCDE) : ar (∆ABC). (2015)
Important Questions for Class 10 Maths Chapter 6 Triangles 43
Solution:
Given:
D divides CA in 4 : 3
CD = 4K
DA = 3K
DE || BC …[Given
Important Questions for Class 10 Maths Chapter 6 Triangles 44
In ∆AED and ∆ABC,
∠1 = ∠1 …[common
∠2 = ∠3 … corresponding angles
∴ ∆AED – ∆ABC …(AA similarity
\frac { ar\left( \triangle AED \right) }{ ar\left( \triangle ABC \right) } =\left( \frac { AD }{ AC } \right) ^{ 2 }
… [The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides
\\frac { \left( 3K \right) ^{ 2 } }{ \left( 7K \right) ^{ 2 } } =\frac { { 9K }^{ 2 } }{ { 49K }^{ 2 } } =\frac { ar\left( \triangle AED \right) }{ ar\left( \triangle ABC \right) } =\frac { 9 }{ 49 }
Let ar(∆AED) = 9p
and ar(∆ABC) = 49p
ar(BCDE) = ar (∆ABC) – ar (∆ADE)
= 49p – 9p = 40p
\frac { ar\left( BCDE \right) }{ ar\left( \triangle ABC \right) } =\frac { 40p }{ 49p }
∴ ar (BCDE) : ar(AABC) = 40 : 49

Question 28.
In the given figure, DE || BC and AD : DB = 7 : 5, find \frac { ar\left( \triangle DEF \right) }{ ar\left( \triangle CFB \right) } [/latex] (2017OD)
Important Questions for Class 10 Maths Chapter 6 Triangles 45
Solution:
Given: In ∆ABC, DE || BC and AD : DB = 7 : 5.
To find: \frac { ar\left( \triangle DEF \right) }{ ar\left( \triangle CFB \right) } = ?
Important Questions for Class 10 Maths Chapter 6 Triangles 46
Proof: Let AD = 7k
and BD = 5k then
AB = 7k + 5k = 12k
In ∆ADE and ∆ABC,
∠1 = ∠1 …(Common
∠2 = ∠ABC … [Corresponding angles
Important Questions for Class 10 Maths Chapter 6 Triangles 47

Question 29.
In the given figure, the line segment XY is parallel to the side AC of ∆ABC and it divides the triangle into two parts of equal areas. Find the ratio \frac{\mathbf{A} \mathbf{X}}{\mathbf{A B}}. (2017OD)
Important Questions for Class 10 Maths Chapter 6 Triangles 48
Solution:
We have XY || AC … [Given
So, ∠BXY = ∠A and ∠BYX = ∠C …[Corresponding angles
∴ ∆ABC ~ ∆XBY …[AA similarity criterion
Important Questions for Class 10 Maths Chapter 6 Triangles 49

Question 30.
In the given figure, AD ⊥ BC and BD = \frac{1}{3}CD. Prove that 2AC2 = 2AB2 + BC2. (2012)
Important Questions for Class 10 Maths Chapter 6 Triangles 50
Solution:
BC = BD + DC = BD + 3BD = 4BD
\frac{\mathrm{BC}}{4} = BD
In rt. ∆ADB, AD2 = AB2 – BD2 ….(ii)
In rt. ∆ADC, AD2 = AC2 – CD2 …(iii)
From (ii) and (iii), we get
AC2 – CD2 = AB2 – BD2
AC2 = AB2 – BD2 + CD2
Important Questions for Class 10 Maths Chapter 6 Triangles 51
∴ 2AC2 = 2AB2 + BC2 (Hence proved)

Question 31.
In the given figure, ∆ABC is right-angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE. (2012, 2017D)
Important Questions for Class 10 Maths Chapter 6 Triangles 52
Solution:
Given: ∆ABC is rt. ∠ed at C and DE ⊥ AB.
AD = 3 cm, DC = 2 cm, BC = 12 cm
To prove:
(i) ∆ABC ~ ∆ADE; (ii) AE = ? and DE = ?
Proof. (i) In ∆ABC and ∆ADE,
∠ACB = ∠AED … [Each 90°
∠BAC = ∠DAE …(Common .
∴ ∆ABC ~ ∆ADE …[AA Similarity Criterion

(ii) ∴ \frac{A B}{A D}=\frac{B C}{D E}=\frac{A C}{A E} … [side are proportional
\frac{A B}{3}=\frac{12}{D E}=\frac{3+2}{A E}
…..[In rt. ∆ACB, … AB2 = AC2 + BC2 (By Pythagoras’ theorem)
= (5)2 + (12)2 = 169
∴ AB = 13 cm
Important Questions for Class 10 Maths Chapter 6 Triangles 53

Question 32.
In ∆ABC, if AP ⊥ BC and AC2 = BC2 – AB2, then prove that PA2 = PB × CP. (2015)
Solution:
Important Questions for Class 10 Maths Chapter 6 Triangles 54
AC2 = BC2 – AB2 …Given
AC2 + AB2 = BC2
∴ ∠BAC = 90° … [By converse of Pythagoras’ theorem
∆APB ~ ∆CPA
[If a perpendicular is drawn from the vertex of the right angle of a triangle to the hypotenuse then As on both sides of the perpendicular are similar to the whole triangle and to each other.
\frac{\mathrm{AP}}{\mathrm{CP}}=\frac{\mathrm{PB}}{\mathrm{PA}} … [In ~∆s, corresponding sides are proportional
∴ PA2 = PB. CP (Hence Proved)

Question 33.
ABCD is a rhombus. Prove that AB2 + BC2 + CD2 + DA2 = AC2 + BD2. (2013)
Solution:
Given. In rhombus ABCD, diagonals AC and BD intersect at O.
Important Questions for Class 10 Maths Chapter 6 Triangles 55
To prove: AB2 + BC2 + CD2 + DA2 = AC2 + BD2
Proof: AC ⊥ BD [∵ Diagonals of a rhombus bisect each other at right angles
∴ OA = OC and
OB = OD
In rt. ∆AOB,
AB2 = OA2 + OB2 … [Pythagoras’ theorem
AB2 = \left(\frac{A C}{2}\right)^{2}+\left(\frac{B D}{2}\right)^{2}
AB2 = \left(\frac{A C}{2}\right)^{2}+\left(\frac{B D}{2}\right)^{2}
4AB2 = AC2 + BD2
AB2 + AB2 + AB2 + AB2 = AC2 + BD2
∴ AB2 + BC2 + CD2 + DA2 = AC2 + BD2
…[∵ In a rhombus, all sides are equal

Question 34.
The diagonals of trapezium ABCD intersect each other at point o. If AB = 2CD, find the ratio of area of the ∆AOB to area of ∆COD. (2013)
Solution:
In ∆AOB and ∆COD, … [Alternate int. ∠s
∠1 = ∆3
∠2 = ∠4
Important Questions for Class 10 Maths Chapter 6 Triangles 56

Question 35.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that \frac{A O}{B O}=\frac{C O}{D O}. Show that ABCD is a trapezium. (2014)
Solution:
1st method.
Given: Quadrilateral ABCD in which
AC and BD intersect each other at 0.
Such that \frac{A O}{B O}=\frac{C O}{D O}
To prove: ABCD is a trapezium
Const.: From O, draw OE || CD.
Important Questions for Class 10 Maths Chapter 6 Triangles 57
Important Questions for Class 10 Maths Chapter 6 Triangles 58
But these are alternate interior angles
∴ AB || DC Quad. ABCD is a trapezium.

Triangles Class 10 Important Questions Long Answer (4 Marks).

Question 36.
In a rectangle ABCD, E is middle point of AD. If AD = 40 m and AB = 48 m, then find EB. (2014D)
Solution:
Important Questions for Class 10 Maths Chapter 6 Triangles 59
E is the mid-point of AD …[Given
AE = \frac{40}{2} = 20 m
∠A = 90° …[Angle of a rectangle
In rt. ∆BAE,
EB2 = AB2 + AE2 …[Pythagoras’ theorem
= (48)2 + (20)2
= 2304 + 400 = 2704
∴ EB = \sqrt{2704} = 52 m

Question 37.
Let ABC be a triangle and D and E be two points on side AB such that AD = BE. If DP || BC and EQ || AC, then prove that PQ || AB. (2013)
Solution:
Important Questions for Class 10 Maths Chapter 6 Triangles 60
In ∆ABC,
DP || BC
and EQ || AC … [Given
Important Questions for Class 10 Maths Chapter 6 Triangles 61
Now, in ∆ABC, P and Q divide sides CA and CB respectively in the same ratio.
∴ PQ || AB

Question 38.
In the figure, ∠BED = ∠BDE & E divides BC in the ratio 2 : 1.
Prove that AF × BE = 2 AD × CF. (2015)
Important Questions for Class 10 Maths Chapter 6 Triangles 62
Solution:
Construction:
Draw CG || DF
Proof: E divides
BC in 2 : 1.
\frac{B E}{E C}=\frac{2}{1} …(i)
Important Questions for Class 10 Maths Chapter 6 Triangles 63
Important Questions for Class 10 Maths Chapter 6 Triangles 64

Question 39.
In the given figure, AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm, then find the pair of parallel lines and hence their lengths. (2015)
Important Questions for Class 10 Maths Chapter 6 Triangles 65
Solution:
Important Questions for Class 10 Maths Chapter 6 Triangles 66
Important Questions for Class 10 Maths Chapter 6 Triangles 67

Question 40.
If sides AB, BC and median AD of AABC are proportional to the corresponding sides PQ, QR and median PM of PQR, show that ∆ABC ~ ∆PQR. (2017OD)
Solution:
Important Questions for Class 10 Maths Chapter 6 Triangles 68

Question 41.
Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. (2012)
Solution:
Given: ∆ABC ~ ∆DEF
Important Questions for Class 10 Maths Chapter 6 Triangles 69
Important Questions for Class 10 Maths Chapter 6 Triangles 70
Important Questions for Class 10 Maths Chapter 6 Triangles 71

Question 42.
State and prove converse of Pythagoras theorem. Using the above theorem, solve the following: In ∆ABC, AB = 6\sqrt{3} cm, BC = 6 cm and AC = 12 cm, find ∠B. (2015)
Solution:
Part I:
Statement: Prove that, in a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
Important Questions for Class 10 Maths Chapter 6 Triangles 72
To prove: ∠ABC = 90°
Const.: Draw a right angle ∆DEF in which DE = BC and EF = AB.
Proof: In rt. ∆ABC,
AB2 + BC2 = AC2 …(i) Given
In rt. ∆DEF
DE2 + EF2 = DF2 … [By Pythagoras’ theorem
BC2 + AB2 = DF2…(ii)…[∵ DE = BC; EF = AB
From (i) and (ii), we get
AC2 = DF2 = AC = DF
Now, DE = BC …[By construction
EF = AB …[By construction
DF = AC … [Proved above :
∴ ∆DEF = ∆ABC … (SSS congruence :
∴ ∠DEF = ∠ABC …[c.p.c.t.
∵ ∠DEF = 90° ∴ ∠ABC = 90°
Given: In rt. ∆ABC,
AB2 + BC2 = AC2
AB2 + BC2 = (6\sqrt{3})2 + (6)2
= 108 + 36 = 144 = (12)2
AB2 + BC2 = AC2 ∴ ∠B = 90° … [Above theorem

Question 43.
In the given figure, BL and CM are medians of a triangle ABC, right angled at A. Prove that: 4(BL2 + CM2) = 5BC2 (2012)
Important Questions for Class 10 Maths Chapter 6 Triangles 73
Solution:
Given: BL and CM are medians of ∆ABC, right angled at A.
To prove: 4(BL2 + CM2) = 5 BC2
Proof: In ∆ABC, BC2 = BA2 + CA2 …(i)
In ∆BAL,
BL2 = BA2 + AL2 …[Pythagoras’ theorem
BL2 = BA2 + \left(\frac{\mathrm{CA}}{2}\right)^{2}
BL2 = BA2+ \frac{\mathrm{CA}^{2}}{4}
⇒ 4BL2 = 4BA2 + CA2 …(ii)
Now, In ∆MCA,
MC2 = CA2 + MA2 …[Pythagoras’ theorem
MC2 = CA22 + \left(\frac{\mathrm{BA}}{2}\right)^{2}
MC2 = CA2 + \frac{\mathrm{BA}^{2}}{4}
4MC2 = 4CA2 + BA2
Adding (ii) and (iii), we get
4BL2 + 4MC2 = 4BA2 + CA2 + 4CA2+ BA2 …[From (ii) & (iii)
4(BL2 + MC2) = 5BA2 + 5CA2
4(BL2 + MC2) = 5(BA2 + CA2)
∴ 4(BL2 + MC2) = 5BC2 … [Using (1)
Hence proved.

Question 44.
In the given figure, AD is median of ∆ABC and AE ⊥ BC. (2013)
Prove that b2 + c2 = 2p2 + \frac{1}{2} a2.
Important Questions for Class 10 Maths Chapter 6 Triangles 74
Solution:
Important Questions for Class 10 Maths Chapter 6 Triangles 75
Proof. Let ED = x
BD = DC = \frac{B C}{2}=\frac{a}{2} = …[∵ AD is the median
In rt. ∆AEC, AC2 = AE2 + EC2 …..[By Pythagoras’ theorem
b2 = h2 + (ED + DC)2
b2 = (p2 – x2) + (x = \frac{a}{2})2
…[∵ In rt. ∆AED, x2 + h2 = p2 ⇒ h2 = p2 – x2 …(i)
b2 = p2 – x2 + x2 + \left(\frac{a}{2}\right)^{2}2+ 2(x)\left(\frac{a}{2}\right)
b2 = p2 + ax + \frac{a^{2}}{4} …(ii)
In rt. ∆AEB, AB2 = AE2 + BE2 … [By Pythagoras’ theorem
Important Questions for Class 10 Maths Chapter 6 Triangles 76

Question 45.
In a ∆ABC, the perpendicular from A on the side BC of a AABC intersects BC at D such that DB = 3 CD. Prove that 2 AB2 = 2 AC2 + BC2. (2013; 2017OD)
Solution:
In rt. ∆ADB,
AD2 = AB2 – BD2 …(i) [Pythagoras’ theorem
In rt. ∆ADC,
AD2 = AC2 – DC2 …(ii) [Pythagoras’ theorem
Important Questions for Class 10 Maths Chapter 6 Triangles 77
From (i) and (ii), we get
AB2 – BD2 = AC2 – DC2
AB2 = AC2 + BD2 – DC2
Now, BC = BD + DC
= 3CD + CD = 4 CD …[∵ BD = 3CD (Given)
⇒ BC2 = 16 CD2 …(iv) [Squaring
Now, AB2 = AC2 + BD2 – DC2 …[From (iii)
= AC2 + 9 DC2 – DC2 ….[∵ BD = 3 CD ⇒ BD2 = 9 CD2
= AC2 + 8 DC2
= AC2 + \frac{16 \mathrm{DC}^{2}}{2}
= AC2 + \frac{B C^{2}}{2} … [From (iv)
∴ 2AB2 = 2AC2 + BC2 … [Proved

Question 46.
In ∆ABC, altitudes AD and CE intersect each other at the point P. Prove that: (2014)
(i) ∆APE ~ ∆CPD
(ii) AP × PD = CP × PE
(iii) ∆ADB ~ ∆CEB
(iv) AB × CE = BC × AD
Solution:
Important Questions for Class 10 Maths Chapter 6 Triangles 78
Given. In ∆ABC, AD ⊥ BC & CE ⊥ AB.
To prove. (i) ∆APE ~ ∆CPD
(ii) AP × PD = CP × PE
(iii) ∆ADB ~ ∆CEB
(iv) AB × CE = BC × AD
Proof: (i) In ∆APE and ∆CPD,
∠1 = ∠4 …[Each 90°
∠2 = ∠3 …[Vertically opposite angles
∴ ∆APE ~ ∆CPD …[AA similarity
(ii) \frac{\mathrm{AP}}{\mathrm{CP}}=\frac{\mathrm{PE}}{\mathrm{PD}} … [In ~ ∆s corresponding sides are proportional
∴ AP × PD = CP × PE
(iii) In ∆ADB and ∆CEB,
∠5 = ∠7 …[Each 90°
∠6 = ∠6 …(Common
∴ ∆ADB ~ ∆CEB …[AA similarity
(iv) ∴ \frac{A B}{C B}=\frac{A D}{C E} … [In ~ ∆s corresponding sides are proportional
∴ AB × CE = BC × AD

Question 47.
In the figure, PQR and QST are two right triangles, right angled at R and T resepctively. Prove that QR × QS = QP × QT. (2014)
Important Questions for Class 10 Maths Chapter 6 Triangles 79
Solution:
Given: Two rt. ∆’s PQR and QST.
Important Questions for Class 10 Maths Chapter 6 Triangles 80
To prove: QR × QS = QP × QT
Proof: In ∆PRQ and ∆STQ,
∠1 = ∠1 … [Common
∠2 = ∠3 … [Each 90°
∆PRQ ~ ∆STO …(AA similarity
\frac{Q R}{Q T}=\frac{Q P}{Q S} ..[In -∆s corresponding sides are proportional
∴ QR × QS = QP × QT (Hence proved)

Question 48.
In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \frac { ar\left( ABC \right) }{ ar\left( DBC \right) } =\frac { AO }{ DO } . (2012)
Important Questions for Class 10 Maths Chapter 6 Triangles 81
Solution:
Given: ABC and DBC are two As on the same base BC. AD intersects BC at O.
To prove:
Important Questions for Class 10 Maths Chapter 6 Triangles 82
Important Questions for Class 10 Maths Chapter 6 Triangles 83
Important Questions for Class 10 Maths Chapter 6 Triangles 84

Question 49.
Hypotenuse of a right triangle is 25 cm and out of the remaining two sides, one is longer than the other by 5 cm. Find the lengths of the other
two sides. (2013)
Solution:
Let Base, AB = x cm
Then altitude, BC = (x + 5) cm
In rt. ∆,
By Pythagoras’ theorem
Important Questions for Class 10 Maths Chapter 6 Triangles 85
AB2 + BC2 = AC2
⇒ (x)2 + (x + 5)2 = 252
⇒ x22 + x2 + 10x + 25 – 625 = 0
⇒ 2x2 + 10x – 600 = 0
⇒ x2 + 5x – 300 = 0 … [Dividing both sides by 2
⇒ x2 + 20x – 15x – 300 = 0
⇒ x(x + 20) – 15(x + 20) = 0
(x – 15)(x + 20) = 0
x – 15 = 0 or x + 20 = 0
x = 15 or x = -20
Base cannot be -ve
∴ x = 15 cm
∴ Length of the other side = 15 + 5 = 20 cm
Two sides are = 15 cm and 20 cm

Question 50.
In Figure, AB ⊥ BC, FG ⊥ BC and DE ⊥ AC. Prove that ∆ADE ~ ∆GCF. (2016 OD)
Important Questions for Class 10 Maths Chapter 6 Triangles 86
Solution:
In rt. ∆ABC,
∠A + ∠C = 90° …(i)
In rt. ∆AED,
∠A + ∠2 = 90°
From (i) and (ii), ∠C = ∠2
Similarly, ∠A = ∠1
Now in ∆ADE & ∆GCF
∠A = 1 … [Proved
∠C = 2 … [Proved
∠AED = ∠GFC … [rt. ∠s
∴ ∆ADE – ∆GCF …(Hence Proved)

Important Questions for Class 10 Maths

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Going Places Important Questions Class 12 English

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Going Places Important Questions CBSE Class 12 English

Going Places Important Questions Short Answer Type Questions (3-4 MARKS)

Question 1.
“Damn that Geoff, this was a Geoff thing not a Jansie thing.” Why did Sophie say so? (Delhi 2010)
Answer:
Sophie had cooked up the story of her meeting with Danny Casey to impress Geoff who was a great fan of the football prodigy. She knew this news would be of great interest to her brother. She had also thought her brother would keep her secret. She did not want Jansie, who was ‘nosey’, to know about this. So when Geoff told Jansie, Sophie cursed him because she knew Jansie would spread her story in the entire neighbourhood.

Question 2.
Why did Sophie like her brother Geoff more than any other person? (All India 2010)
Answer:
For Sophie, her brother Geoff was the only person who listened to her fantasies and long cherished dreams. She only trusted Geoff to keep all her secrets to himself.

Question 3.
Why didn’t Sophie want Jansie to know about her story with Danny? (All India 2010)
Answer:
Jansie told Sophie that Geoff had given her the news of her (Sophie’s) meeting with Danny Casey. Sophie cursed Geoff for telling Jansie this because it was meant to be something special between Geoff and her. ‘It wasn’t the jansie kind of thing at all’. Jansie was ‘nosey’ and now the whole neighbourhood would come to know her story.

Question 4.
What did Sophie tell Geoff about her meeting with Danny Casey? (All India 2010)
Answer:
Sophie told Geoff that she had met Danny Casey, the wonder footballer. As she was looking at the clothes in a shop’s window Casey came and stood beside her. Geoff did not believe her. Sophie then went on to tell him that Casey had green and gentle eyes and he was not very tall.

Question 5.
Why didn’t Sophie want Jansie to know about her story with Danny? (Delhi 2011)
Answer:
Sophie did not want Jansie to know anything about her meeting with Danny Casey because she knew Jansie was good at spreading rumours. Telling anything to Jansie was like telling the whole town. Sophie had made up this story only for her family and she did not want this story to spread in the town.

Question 6.
Did Geoff keep his promise to Sophie? How do you know? (Delhi 2011)
Answer:
No, Geoff did not keep his promise to Sophie. He told Sophie’s secret to Jansie’s brother Frank who in turn told Jansie about it. We know this because Jansie questions Sophie about her meeting with footballer, Danny Casey and Sophie is startled that Jansie knows about this meeting.

Question 7.
How did Sophie’s father react when Geoff told him about her meeting with Danny Casey? (Delhi 2011)
Answer:
When Geoff told their father about Sophie’s meeting with Danny Casey he did not believe it and suspected it was one of Sophie’s wild stories. He was filled with disdain and did not want to encourage her in any way. So he warned her to keep herself out of any kind of trouble.

Question 8.
What thoughts came to Sophie’s mind as she sat by the canal? (All India 2011)
Answer:
Sophie considers the spot along the canal the best place for a date, as it is away from the rush of the city. As she sits by the canal she is lost in the world of her dreams. She imagines Casey coming along the river and her own excitement thereafter.

Question 9.
Which was the only occasion when Sophie got to see Danny Casey in person? (All India 2011)
Answer:
The only occasion when Sophie got to see Danny Casey in person was when she had gone to see the football match with her family. Sitting among the spectators, she saw Danny Casey from a distance.

Question 10.
Why did Jansie discourage Sophie from enter¬taining thoughts about the sports-star, Danny Casey? (All India 2011)
Answer:
Jansie is a practical girl who lives in the real world. She knows that Danny Casey, a sports- star is far beyond Sophie’s dreams and imagination. So she discourages Sophie from entertaining any kind of wild thoughts about him.

Question 11.
Did Geoff keep his promise? How do you know? (Comptt. Delhi 2011)
Answer:
No, Geoff did not keep his promise. He told Frank about Sophie’s meeting with Danny Casey though he had promised her he would not disclose it. Even Jansie came to know about it. Sophie was happy to know that Geoff had at least not revealed the date of her meeting with Casey.

Question 12.
Why did Sophie not want Jansie to know any¬thing about her meeting with Danny Casey?(All India 2012)
Answer:
Sophie did not want Jansie to know anything about her meeting with Danny Casey because she knew Jansie was good at spreading rumours. Telling anything to Jansie was like telling the whole town. Sophie had made up this story only for her family and she did not want this story to spread in the town.

Question 13.
What do you think of Sophie’s dream about her career? (Comptt. Delhi 2012)
Answer:
Sophie’s dreams of her future life are far removed from her reality. She dreams of owning a boutique, of becoming an actress but the fact was that she was earmarked for the biscuit factory. Sophie drifts into a world of fantasy and wants to live her dreams by stepping out of her middle-class status.

Question 14.
Though Sophie and Jansie were good friends, how were they basically different from each other? (Comptt. All India 2012)
Answer:
Although Sophie and Jansie were good friends they were completely different from each other.
Sophie lived in a world of dreams and fantasy, a world which was far removed from reality. Jansie, on the other hand, was a sensible and realistic girl who was grounded to her reality.

Question 15.
Why did Sophie want to keep her story with Danny a secret from Jansie? (Comptt. All India 2012)
Answer:
Sophie did not want Jansie to know anything about her meeting with Danny Casey because she knew Jansie was good at spreading rumours. Telling anything to Jansie was like telling the whole town. Sophie had made up this story only for her family and she did not want this story to spread in the town.

Question 16.
Why did Sophie like her brother, Geoff more than any other person? (Delhi 2013)
Answer:
For Sophie, her brother Geoff was the only person who listened to her fantasies and long cherished dreams. She only trusted Geoff to keep all her secrets to himself.

Question 17.
Why was Sophie jealous of Geoff’s silence? (All India 2013)
Answer:
Sophie’s brother’s hallmark is his silence. Sophie feels “words have to be prized out of him like stone out of the ground.” She is jealous of his silence and is very keen to enter the recesses of his mind. There is so much about her brother that Sophie only suspects because she has never really known anything about him.

Question 18.
Did Geoff keep up his promise? How do you know? (All India 2013)
Answer:
No, Geoff did not keep his promise. He told Frank about Sophie’s meeting with Danny Casey though he had promised her he would not disclose it. Even Jansie came to know about it. Sophie was happy to know that Geoff had at least not revealed the date of her meeting with Casey.

Question 19.
What is unrealistic about Sophie’s dreams of her future life? (Comptt. Delhi 2013)
Answer:
Sophie’s dreams of her future life are far removed from her reality. She dreams of owning a boutique, of becoming an actress but the fact was that she was earmarked for the biscuit factory. Sophie drifts into a world of fantasy and wants to live her dreams by stepping out of her middle-class status.

Question 20.
Why does Jansie discourage Sophie from living in a world of fantasy? (Comptt. Delhi 2013)
Answer:
Jansie was a realistic and practical girl who knew both she and Sophie were earmarked for the biscuit factory. So she discouraged Sophie from living in a world of fantasy. Whenever Sophie expressed her wishes aloud Jansie became melancholic and told Sophie to behave sensibly.

Question 21.
Why did Sophie not want Jansie to know about her story with Danny? (Comptt. Delhi 2013)
Answer:
Sophie did not want Jansie to know anything about her meeting with Danny Casey because she knew Jansie was good at spreading rumours. Telling anything to Jansie was like telling the whole town. Sophie had made up this story only for her family and she did not want this story to spread in the town.

Question 22.
How do we know that Sophie’s family lived in poor circumstances? (Comptt. All India 2013)
Answer:
We know that Sophie’s family lived in poor circumstances as her father worked hard as a manual labourer. Her mother’s back had become crooked due to the burden of household work. Their house was rather small, easily suffocated with steam from the stove.

Question 23.
Why did Sophie want to be admitted into Geoff’s affections? (Comptt. All India 2013)
Answer:
Geoff was an apprentice mechanic who traveled to the far side of the city every day. Sophie wished she could be admitted into Geoff’s affections so that someday he might take her with him to meet those exotic and interesting people whom he never spoke about.

Question 24.
Whom did Sophie like more than anyone else in the story? Why? (Comptt. All India 2013)
Answer:
Sophie liked her brother Geoff more than anyone else. She was most fascinated by his evasiveness and wanted to travel with him to his enigmatic world. Her brother’s lost and inquisitive eyes captivated her and his silence intrigued her.

Question 25.
Why did Sophie long for her brother’s affection? (All India 2014)
Answer:
Sophie longed for her brother’s affection because she trusted him more than any other member of her family. He symbolised freedom and she wanted to be a part of his exotic and adventurous world. He was the only one to whom she could look for approval as her father was critical and disapproving and she hero- worshipped him.

Question 26.
How are Jansie and Sophie different from each other? (Comptt. Delhi 2014)
Answer:
Although Jansie and Sophie are classmates and friends, their approach to life is completely different from each other. While Jansie is practical and grounded to reality, Sophie lives in a make-believe world of her own, which she ‘invents’ in her imagination.

Question 27.
Why is Sophie attracted to Danny Casey? (Comptt. Delhi 2014)
Answer:
Danny Casey is an Irish football prodigy and Sophie is attracted to him because like most youngsters she idolizes and hero-worships him. So she envisions her meetings with the football star and also gives vivid details of their meetings.

Question 28.
What was Sophie’s ambition in life? How did she hope to achieve that? (Delhi 2017)
Answer:
Sophie’s ambition in life was either to have her own boutique and be a fashion designer or to be an actress. She wanted to do something a bit sophisticated. She hoped to achieve her ambition by being a manager at a boutique to start with and then save enough money to have her own boutique.

Question 29.
Why did Jansie discourage Sophie from hav-ing dreams? (All India 2017)
Answer:
Jansie was a realistic and practical girl who knew both she and Sophie were earmarked for the biscuit factory. So she discouraged Sophie from living in a world of fantasy. Whenever Sophie expressed her wishes aloud Jansie became melancholic and told Sophie to behave sensibly.

Going Places Important Questions Long Answer Type Questions (5-6 MARKS)

Question 30.
Contrast Sophie’s real world with her fantasies. (Delhi 2009)
Answer:
Sophie belongs to a lower middle class family and lives a humble life with her parents and elder brother. But her dreams far supercede the reality in which she is living. Her dreams are far beyond her reach. So she wishes to open a boutique, entertains the idea of being an actress and also aspires to be a fashion designer. Her dreamy disposition and romantic allusions lead her to hero-worship the wonder-footballer, Danny Casey towards whom she develops a romantic fascination. Though she sees him only once in person she sits for hours imagining Danny Casey coming to her. The incurable dreamer in her remains an escapist who wants to remain away from her real world.

Question 31.
Describe the character of Sophie’s father and the role played by him. (Delhi 2009)
Answer:
Sophie’s father is a happy-go-lucky and carefree man. He does not appear to be either soft or even sophisticated. He is a heavy breathing man. He usually sits in his vest at the table. Sophie, it appears, fears him. He does not believe in his daughter’s ‘wild stories’ and so he ignores her completely and prefers to go and watch television than listen to her. Even when his son Geoff tells him that Sophie had met the Irish prodigy Danny Casey, he completely ignores this news. He is extremely interested in football and, like all his children, he also adores Danny Casey. He is a middle-class man who goes to the pub on his bicycle to celebrate his team’s victory and the fact that Casey had scored a second goal. He is a rather dominating person and a typical representation of the lower middle-class family of that time.

Question 32.
Jansie is just as old as Sophie but she is very different from her. Bring out the contrast bet¬ween the two friends citing relevant instances from the story, “Going Places”. (All India 2009)
Answer:
Sophie and Jansie were classmates as well as friends. They both belonged to lower middle- class families. But that is where their similarity ends. There is a striking contrast between their characters. Sophie is a day-dreamer and Jansie is practical. Sophie lives in a world of dreams and does not want to come out of this fairyland. She is an incurable escapist and dreams of having a boutique, becoming an actress or a fashion designer. Jansie, on the other hand, is very grounded. Jansie has her feet firmly planted on the ground and knows they are both ‘earmarked for the biscuit factory’. She knows big things require big money and experience which they lack desperately. She advises Sophie to be sensible and not entertain wild dreams. Sophie and Jansie’s temperaments differ greatly. While Sophie shares her dreams only with one person, her brother Geoff, Jansie on the other hand is nosey. She takes an interest in learning new things about others and can spread stories in the whole neighbourhood.

Question 33.
Contrast Sophie’s real world with her fantasies. (Comptt. All India 2011)
Answer:
Sophie belongs to a lower middle class family and lives a humble life with her parents and elder brother. But her dreams far supercede the reality in which she is living. Her dreams are far beyond her reach. So she wishes to open a boutique, entertains the idea of being an actress and also aspires to be a fashion designer. Her dreamy disposition and romantic allusions lead her to hero-worship the wonder-footballer, Danny Casey towards whom she develops a romantic fascination. Though she sees him only once in person she sits for hours imagining Danny Casey coming to her. The incurable dreamer in her remains an escapist who wants to remain away from her real world.

Question 34.
Compare and contrast Sophie and Jansie high¬lighting their temperament and aspirations. (Delhi 2012)
Answer:
Sophie and Jansie were classmates as well as friends. They both belonged to lower middle- class families. But that is where their similarity ends. There is a striking contrast between their characters. Sophie is a day-dreamer and Jansie is practical. Sophie lives in a world of dreams and does not want to come out of this fairyland. She is an incurable escapist and dreams of having a boutique, becoming an actress or a fashion designer. Jansie, on the other hand, is very grounded. Jansie has her feet firmly planted on the ground and knows they are both ‘earmarked for the biscuit factory’. She knows big things require big money and experience which they lack desperately. She advises Sophie to be sensible and not entertain wild dreams. Sophie and Jansie’s temperaments differ greatly. While Sophie shares her dreams only with one person, her brother Geoff, Jansie on the other hand is nosey. She takes an interest in learning new things about others and can spread stories in the whole neighbourhood.

Question 35.
Attempt a character sketch of Sophie as a woman who lives in her dreams. (Delhi 2012)
Answer:
Sophie belongs to a lower middle class family and lives a humble life with her parents and elder brother. But her dreams far supercede the reality in which she is living. Her dreams are far beyond her reach. So she wishes to open a boutique, entertains the idea of being an actress and also aspires to be a fashion designer. Her dreamy disposition and romantic allusions lead her to hero-worship the wonder-footballer, Danny Casey towards whom she develops a romantic fascination. Though she sees him only once in person she sits for hours imagining Danny coming to her. The incurable dreamer in her remains an escapist who wants to remain away from her real world.

Question 36.
Why did Sophie enjoy living in a world of dreams? Describe some of her dreams. (All India 2012)
Answer:
Sophie, like most girls of her age, enjoyed living in her world of dreams which provided to her the refuge she needed from her middle – class reality. Through her dream world she visualised for herself a life that she would like to lead. Her imagination drew into her life all the things she desired and people she idolised but could never make a part of her real life. Sophie dreams of having her own boutique after she leaves school, a boutique which will be the most amazing in the city. Becoming an actress or a fashion designer is also her dream. She was conscious of a vast world which waited for her and she was very sure that she would adjust easily in that world. In her dream world, Sophie dreams of meeting the sensational footballer Danny Casey. She even fixes a date with him and actually travels to that place and waits for him to come. In her world of fantasy, Sophie moves rapidly from one dream to another through the leaps of her mind.

Question 37.
Has Sophie met Danny Casey? What details of her meeting with Danny Casey did she narrate to her brother? (Delhi 2014)
Answer:
No, Sophie had never met Danny Casey. Her story is merely a part of her fantasy. Her fertile mind made up the story of her encounter with him. She told her brother that she had met Danny Casey in the arcade. When she was looking at the clothes in Royce’s window someone came and stood beside her and she looked around and saw Danny. She also told Geoff that Danny Casey has gentle, green eyes and he is not as tall as one would think him to be. She also asked Danny for his autograph but could not get it as none of them had a paper or pen with them. Sophie’s dream world makes her travel into those aspects of life which she may never be able to achieve in reality.

Question 38.
What were Sophie’s plans for her future? Why would you call her dreams unrealistic? (Delhi 2014)
Answer:
Sophie had rather unrealistic plans for her future. She wanted to own a boutique and have the most amazing shop in the city. She also dreamed of being an actress or a fashion designer. Considering her meagre resources and family’s position, her dreams are quite unrealistic, in fact they are in sharp contrast to her reality. Her father works hard for a living and so she is never likely to be a part of the sophisticated world. Her mother bears the back-breaking burden of household chores. I Ier brother works as an apprentice mechanic. Considering her circumstances she would actually have to work in the biscuit factory. So her dreams would just remain dreams and never become a reality.

Question 39.
Sophie lives in a world full of dreams which she does not know she cannot realise. Comment. (All India 2015)
Answer:
Sophie, like most girls of her age, loves to fantasize and live in a world full of dreams which is far away from reality. She envisions a life for herself which she would like to lead not knowing that she might never be able to realise her dreams. She dreams of owning a boutique which again she might never be able to fulfil. The best part of her fantasy is that she is unaware of the fact that her dreams may never materialise. Though her friend Jansie tries her best to keep her grounded, Sophie is undeterred. Sophie’s dream world makes her traverse into those spheres of life which she may never be able to achieve in reality and this does not cause her any anxiety. She blissfully ignores the fact that she comes from a lower middle class family and continues to dream on.

Question 40.
How different is Jansie from Sophie?(All India 2015)
Answer:
Sophie and Jansie were classmates as well as friends. They both belonged to lower middle- class families. But that is where their similarity ends. There is a striking contrast between their characters. Sophie is a day-dreamer and Jansie is practical. Sophie lives in a world of dreams and does not want to come out of this fairyland. She is an incurable escapist and dreams of having a boutique, becoming an actress or a fashion designer. Jansie, on the other hand, is very grounded. Jansie has her feet firmly planted on the ground and knows they are both ‘earmarked for the biscuit factory’. She knows big things require big money and experience which they lack desperately. She advises Sophie to be sensible and not entertain wild dreams. Sophie and Jansie’s temperaments differ greatly. While Sophie shares her dreams only with one person, her brother Geoff, Jansie on the other hand is nosey. She takes an interest in learning new things about others and can spread stories in the whole neighbourhood.

Question 41.
It is not unusual for a lower middle class girl to dream big. How unrealistic were Sophie’s dreams? (All India 2015)
Answer:
Sophie belongs to a lower middle class family and lives a humble life with her parents and elder brother. But her dreams far supercede the reality in which she is living. Her dreams are far beyond her reach. So she wishes to open a boutique, entertains the idea of being an actress and also aspires to be a fashion designer. Her dreamy disposition and romantic allusions lead – her to hero-worship the wonder-footballer, Danny Casey towards whom she develops a romantic fascination. Though she sees him only once in person she sits for hours imagining Danny Casey coming to her. The incurable dreamer in her remains an escapist who wants to remain away from her real world.
Her friend Jansie continued to ground her to reality by stating that they were earmarked for the biscuit factory. Her father also stated that if she ever had enough money she would first buy them a decent house to live in. Sophie’s little brother Derek too feels that his sister thinks money grows on trees.

Question 42.
Every teenager must dream big. Yet the dream should also be rooted to the ground. Write a character sketch of Sophie in the light of this remark. (Comptt. All India 2015)
Answer:
Every teenager must dream big. Yet the dream should also be rooted to the ground. This statement is so true for Sophie whose dreams are larger than life and far supercede her real life. Sophie had rather unrealistic plans for her future. She wanted to own a boutique and have the most amazing shop in the city. She also dreamed of being an actress or a fashion designer. Considering her meagre resources and family’s position, her dreams are quite unrealistic, in fact they are in sharp contrast to her reality. Her father works hard for a living and is never likely to be a part of the sophisticated world. Her mother bears the back-breaking burden of household chores. Her brother works as an apprentice mechanic. Considering her circumstances she would actually have to work in the biscuit factory. So her dreams would just remain dreams and never become a reality.

Question 43.
Teachers always advise their students to dream big. Yet, the same teachers in your classrooms find fault with Sophie when she dreams. What is wrong with Sophie’s dreams? (Delhi 2016)
Answer:
There is nothing wrong with having big dreams. Infact it is an advice given to us by our teachers. But then our dreams must not be in j complete contrast to our reality. This is the fault with Sophie’s dreams. Her dreams are not ! only over-ambitious but also most impractical. She is an incurable escapist and lives in the world of her dreams. She drifts into the world of her fantasy and wants to make her dreams her reality. Dreaming the unachievable can have a negative impact on one’s personality. It can lead to depression. Sophie blissfully ignores the fact that she comes from a lower middle class family and is ear-marked for the biscuit factory. She dreams of owning a boutique, becoming a fashion designer or an actress, dreams which she may never be able to fulfill. Sophie not only dreams big but concocts stories about her imaginary meeting with the Irish prodigy, Danny Casey. She envisions her meeting with the football star and also gives vivid details of the meeting. Such unrealistic dreams need to be discouraged.

Question 44.
Every teenager has a hero/heroine to admire. So many times they become role models for them. What is wrong if Sophie fantasizes about Danny Casey and is ambitious in life? (All India 2016)
Answer:
Dreams and fantasies are an integral part of every teenager’s life. Their dreams motivate them to achieve their goals. Every teenager has an idol to admire and these idols become role models for them. Same is the case with Sophie who idolizes Danny Casey. The only thing wrong with Sophie’s fantasy is that she nurtures unrealistic dreams which are in sharp contrast to her reality. Her ambitions too are way beyond the confines of her lower middle class status. Sophie has never met Danny Casey but she envisions her meeting with the football star and even invents vivid details of this meeting. Her over-imaginative mind concocts stories and she starts living in her make-believe world. Through her dreams Sophie escapes from the harsh realities of life and such people find it difficult to cope with reality.

Question 45.
Describe the fantasies Sophie had about Danny Casey. (Comptt. All India 2016)
Answer:
Sophie had never met Danny Casey. Her story is merely a part of her fantasy. Her fertile mind made up the story of her encounter with him. She told her brother that she had met Danny Casey in the arcade. When she was looking at the clothes in Royce’s window someone came and stood beside her and she looked around and saw Danny. She also told Geoff that Danny Casey has gentle, green eyes and he is not as tall as one would think him to be. She also asked Danny for his autograph but could not get it as none of them had a paper or pen with them. Sophie’s dream world makes her travel into those aspects of life which she may never be able to achieve in reality. Sophie imagines Danny Casey coming to her, her own excitement and subsequent disappointment.

Question 46.
Geoff and Sophie are different from each other, though they belong to the same family. Com¬ment with examples from the text. (Comptt. All India 2016)
Answer:
Though Geoff and Sophie belong to the same family they are very different from each other. Geoff, Sophie’s elder brother is an apprentice mechanic who travelled to work to the far side of the city everyday. He lived in reality and spoke little. In fact Sophie felt that words had to be prized out of him like stone out of the ground. He had a life of his own and though he listened to Sophie’s fantasies, he did not believe her. Sophie, on the other hand, lived in a dream world. She dreams of having a boutique, becoming an actress or a fashion designer. She is a day-dreamer and an incurable escapist. She imagines meeting the football prodigy Danny Casey. Geoff is the most important member of her family for Sophie who wishes to be admitted into her brother’s affections so that some day he might take her with him to meet the exotic and interesting people whom he never spoke about.

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Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions

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Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions

Moving Charges and Magnetism Class 12 Important Questions Very Short Answer Type

Question 1.
What is the direction of the force acting on a charged particle q, moving with a velocity \overrightarrow{\mathbf{v}} in a uniform magnetic field B? (Delhi)
Answer:
The direction of the force acting on a charged particle q, moving with a velocity \overrightarrow{\mathbf{v}} in a uniform
magnetic field \overrightarrow{\mathbf{B}} is perpendicular to the plane of vectors \overrightarrow{\mathbf{v}} and \overrightarrow{\mathbf{B}}
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 1
So, force is perpendicular to both \overrightarrow{\mathbf{v}} and \overrightarrow{\mathbf{B}}. From equation (i), we can also say that the force \overrightarrow{\mathbf{F}} acts in the direction of the vectors \overrightarrow{\mathbf{v}} and \overrightarrow{\mathbf{B}}

Question 2.
Why should the spring/suspension wire in a moving coil galvanometer have low torsional constant? (All India 2008)
Answer:
Low torsional constant is basically required to increase the current/charge sensitivity in a moving coil ballistic galvanometer.

Question 3.
Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why? (Delhi 2008)
Answer:
At the edges of the solenoid, the field lines get diverged due to other fields and/or non-availability of dipole loops, while in toroids the dipoles (in loops) orient continuously.

Question 4.
An electron does not suffer any deflection while passing through a region of uniform magnetic field. What is the direction of the magnetic field? (All India 2009)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 2
∴Magnetic field will be in the line of the velocity of electron.

Question 5.
A beam of a particles projected along +x-axis, experiences a force due to a magnetic field along the +y-axis. What is the direction of the magnetic field? (All India 2009)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 3
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 4
Direction of the magnetic field is towards negative direction of z-axis.

Question 6.
A beam of electrons projected along +x-axis, experiences a force due to a magnetic field along the +y/-axis. What is the direction of the magnetic field? (All India 2010)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 1
Answer:
Direction of the magnetic field is F = q (v × B) towards positive direction of z-axis.

Question 7.
A beam of protons, projected along + x-axis, experiences a force due to a magnetic field along the – y-axis. What is the direction of the magnetic field? (All India 2010)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 220
Answer:
The direction of the magnetic field is towards positive direction of z-axis.

Question 8.
Depict the trajectory of a charged particle moving with velocity v as it enters a uniform magnetic field perpendicular to the direction of its motion. (Comptt. All India 2012)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 5
The force acting on the charge particle will be perpendicular to both v and S and therefore will describe a circular path.

Question 9.
Write the expression in vector form, for the magnetic force \overrightarrow{\mathrm{F}} acting on a charged particle moving with velocity \overrightarrow{\mathrm{V}} in the presence of a magnetic field B. (Comptt. All India 2012)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 6

Question 10.
An ammeter of resistance 0.6 Ω can measure current upto 1.0 A. Calculate
(i) The shunt resistance required to enable the ammeter to measure current upto 5.0 A
(ii) The combined resistance of the ammeter and the shunt. (Delhi 2013)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 7
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 8

Question 11.
Write the expression, in a vector form, for the Lorentz magnetic force \overrightarrow{\mathrm{F}} due to a charge moving with velocity \overrightarrow{\mathrm{V}} in a magnetic field \overrightarrow{\mathrm{B}}. What is the direction of the magnetic force? (Delhi 2013)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 9
… [q is the magnitude of the moving charges)
This force is normal to both the directions of velocity \overrightarrow{\mathrm{V}} and magnetic field \overrightarrow{\mathrm{B}} .

Question 12.
Using the concept of force between two infinitely long parallel current carrying conductors, define one ampere of current. (All India 2013)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 10
“One ampere of current is the value of steady current, which when maintained in each of the two very long, straight, parallel conductors of negligible cross-section; and placed one metre apart in vacuum, would produce on each of these conductors a force of equal to 2 × 10-7 newtons per metre (Nm-1) of length. ”

Question 13.
Write the condition under which an electron will move undeflected in the presence of crossed electric and magnetic fields.(Comptt. All India 2013)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 11
and electric and magnetic fields are mutually perpendicular.

Question 14.
Why do the electrostatic field lines not form closed loops? (All India 2015)
Answer:
Electric field lines do not form closed loops because the direction of an electric field is from positive to negative charge. So one can regard a line of force starting from a positive charge and ending on a negative charge. This indicates that electric field lines do not form closed loops.

Question 15.
A particle of mass ‘m’ and charge ‘q’ moving with velocity V enters the region of uniform magnetic field at right angle to the direction of its motion. How does its kinetic energy get affected? (Comptt. Delhi 2015)
Answer:
Kinetic energy will NOT be affected.
*(When \vec{v} is perpendicular to \vec{B}, then magnetic field provides necessary centripetal force)

Question 16.
Write the underlying principle of a moving coil galvanometer. (Delhi 2015)
Answer:
Principle of a galvanometer : “A current carrying coil, in the presence of magnetic field, experiences a torque which produces proportionate deflection”.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 12

Question 17.
A coil, of area A, carrying a steady current I, has a magnetic moment, \vec{m}, associated with it. Write the relation between \vec{m}, I and A in vector form. (Comptt Delhi 2015)
Answer:
Relation for magnetic moment = \vec{m} = I\vec{A}

Moving Charges and Magnetism Class 12 Important Questions Short Answer Type SA II

Question 18.
Using Ampere’s circuital law, obtain an expression for the magnetic field along the axis of a current carrying solenoid of length l and having N number of turns. (All India 2008)
Answer:
Magnetic field due to Solenoid Let length of solenoid = L
Total number of turns in solenoid = N
No. of turns per unit length = \frac{N}{L} = n
ABCD is an Ampere’s loop
AB, DC are very large
BC is in a region of \overrightarrow{\mathrm{B}} = 0
AD is a long axis
Length of AD = x
Current in one turn = I0
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 13
Applying Ampere’s circuital loop — | B .dl = go I ’
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 14
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 15
No. of turns in x length = nx,
Current in turns nx, I = nx I0
According to Ampere’s circuital law
Bx = µ0 I => Bx = µ0 nx I0
∴ B = µ0nI0

Question 19.
A charge ‘q’ moving B along the X-axis with a velocity v is subjected to a uniform magnetic field B acting along the Z-axis as it crosses the origin O. (Delhi 2009)
(i) Trace its trajectory.
(ii) Does the charge gain kinetic energy as it enters the magnetic field? Justify your answer.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 16
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 17
(ii) K.E does not change irrespective of the direction of the charge as
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 18

Question 20.
State Biot-Savart law.
A current I flows in a conductor placed perpendicular to the plane of the paper. Indicate the direction of the magnetic field due to a small element \overrightarrow{d \vec{l}} at point P situated at a distance \vec{r} from the element as shown in the figure.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 19
Answer:
Biot-Savart law and its applications :
Biot-Savart law states that “the magnitude of the magnetic field dB at any point due to a small current element dl is given by
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 20

Question 21.
(a) In what respect is a toroid different from a solenoid? Draw and compare the pattern of the magnetic field lines in the two cases.
(b) How is the magnetic field inside a given’ solenoid made strong? (All India 2011)
Answer:
(a) Solenoid consists of a long wire wound in the form of a helix where the neighbouring turns are closely spaced, whereas, the toroid is a hollow circular ring on which a large number of turns of a wire is closely wound.
(b) Magnetic field inside a given solenoid is made strong by putting a soft iron core inside it. It is strengthened by increasing the amount of current through it.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 21

Question 22.
Write the expression for Lorentz magnetic force on a particle of charge ‘q’ moving with velocity \vec{v} in a magnetic field \vec{B}. Show that no work is done by this force on the charged particle. (All India 2011)
Answer:
Expression for Lorentz magnetic force on a particle of charge ‘q’ moving with velocity \vec{v} in a magnetic field \vec{B} is \vec{F} = q(\vec{E} + \vec{v} × \vec{B})
Work done by a magnetic force on a charged particle :
The magnetic force \overrightarrow{\mathrm{F}}=q(\overrightarrow{\mathrm{E}}+\vec{v} \times \overrightarrow{\mathrm{B}}) always acts perpendicular to the velocity \vec{v} on the direction of motion of charge q.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 22

Question 23.
A steady current (I1) flows through a long straight wire. Another wire carrying steady current (I2) in the same direction is kept close and parallel to the first wire. Show with the help of a diagram how the magnetic field due to the current I1 exerts a magnetic force on the second wire. Write the expression for this force. (All India 2011)
Answer:
Consider two infinitely long parallel conductors carrying current I1 and I2 in the same direction.
Let d be the distance of separation between these two conductors.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 51
Hence, force is attractive in nature.
Ampere : Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10-7 N on each metre of the other wire.
Then current flowing is 1A
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 52
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 53

Question 24.
Using Ampere’s circuital law, obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis. (All India 2011)
Answer:
Magnetic field due to Solenoid Let length of solenoid = L
Total number of turns in solenoid = N
No. of turns per unit length = \frac{N}{L} = n
ABCD is an Ampere’s loop
AB, DC are very large
BC is in a region of \overrightarrow{\mathrm{B}} = 0
AD is a long axis
Length of AD = x
Current in one turn = I0
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 13
Applying Ampere’s circuital loop — | B .dl = go I ’
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 14
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 15
No. of turns in x length = nx,
Current in turns nx, I = nx I0
According to Ampere’s circuital law
Bx = µ0 I => Bx = µ0 nx I0
∴ B = µ0nI0

Question 25.
Two identical circular wires P and Q each of radius R and carrying current ‘I’ are kept in perpendicular planes such that they have a common centre as shown in the figure. Find the magnitude and direction of the net magnetic field at the common centre of the two coils. (Delhi 2011)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 221
Answer:
Magnetic field produced by the two coils at their common centre are:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 23
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 222
The net magnetic field is directed at an angle of 45° with either of the fields.

Question 26.
Two identical circular loops, P and Q, each of radius r and carrying current I and 21 respectively are lying in parallel planes such that they have a common axis. The direction of current in both the loops is clockwise as seen from O which is equidistant from both the loops. Find the magnitude of the net magnetic field at point O. (Delhi 2011)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 24
Answer:
When the currents are in the same direction, the resultant field at point O is,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 25

Question 27.
Two identical circular loops, P and Q, each of radius r and carrying equal currents are kept in the parallel planes having a common axis passing through O. The direction of current in P is clockwise and in Q is anti-clockwise as seen from O which is equidistant from the loops P and Q. Find the magnitude of the net magnetic field at O. (Delhi 2011)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 223
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 26

Question 28.
A circular coil of closely wound N turns and radius r carries a current I. Write the expressions for the following :
(i) the magnetic field at its centre
(ii) the magnetic moment of this coil (All India 2011)
Answer:
(i) The magnetic field at the centre of a circular coil of N turns and radius r carrying a current, I is
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 27
(iii) Magnetic moment, M = NIA = NIπr2

Question 29.
A proton and a deuteron, each moving with velocity \vec{v} enter simultaneously in the region of magnetic field \vec{B} acting normal to the direction of velocity. Trace their trajectories establishing the relationship between the two. (Comptt. Delhi 2011)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 28

Question 30.
A particle of mass 10-3 kg and charge 5 pC enters into a uniform electric field of 2 × 105 NC-1, moving with a velocity of 20 ms-1 in a direction opposite to that of the field. Calculate the distance it would travel before coming to rest. (Comptt. Delhi 2011)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 29

Question 31.
A particle of mass 2 x 10-3 kg and charge 2 µC enters into a uniform electric field of 5 × 105 NC-1, moving with a velocity of 10 ms-1 in a direction opposite to that of the field. Calculate the distance it would travel before coming to rest. (Comptt. Delhi 2011)
Answer:
Force applied on the charged particle, f = qE
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 30

Question 32.
A particle of mass 5 × 10-3 kg and charge 4 µC enters into a uniform electric field of 2 × 105 NC-1, moving with a velocity of 30 ms-1 in a direction opposite to that of the field. Calculate the distance it would travel before coming to rest. (Comptt. Delhi 2011)
Answer:
Force applied on the charged particle,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 31
Acceleration exerted on the charged particle when it enters in electric field.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 2
Distance travelled by charged particle before coming to rest will be
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 33

Question 33.
An ammeter of resistance 0.80 Ω can measure current upto 1.0 A.
(i) What must be the value of shunt resistance to enable the ammeter to measure current upto 5.0A?
(ii) What is the combined resistance of the ammeter and the shunt? (Delhi 2013)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 34

Question 34.
(a) How is a toroid different from a solenoid?
(b) Use Ampere’s circuital law to obtain the magnetic field inside a toroid.
(c) Show that in an ideal toroid, the magnetic field
(i) inside the toroid and
(ii) outside the toroid at any point in the open space is zero. (Comptt. All India 2014)
Answer:
(a) A toroid is essentially a solenoid which has been bent into a circular shape to close on itself.
(b)

(b) A toroid is a solenoid bent to form a ring shape.
Let N number of turns per unit length of toroid and I be current flowing in it.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 183
Consider a loop (region II) of radius r passes through the centre of the toroid.
Let (region II) \overrightarrow{\mathrm{B}} be magnetic field along the loop is
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 184
Let (region I) B1 be magnetic field outside toroid in open space. Draw an amperian loop L2 of radius r2 through point Q.
Now applying ampere’s law :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 185
As I = 0, because the circular turn current coming out of plane of paper is cancelled exactly by current going into it, so net I = 0, equation (i) becomes
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 186

(c) For the loop 1, Ampere’s circuital law gives,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 35
Thus the magnetic field, in the open space inside the toroid is zero.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 36
Also at point Q, we have
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 37
But from the sectional cut, we refer to that the current coming out of the plane of the paper is cancelled exactly by the current going into it
Hence Ienclosed = 0
∴ B3 = 0

Question 35.
Derive an expression for the magnetic moment (\vec{\mu}) of an electron revolving around the nucleus in terms of its angular momentum ( \vec{\l} ). What is the direction of* the magnetic moment of the electron with respect to its angular momentum? (Comptt. All India 2014)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 38
[ ∴ electron has a negative charge
The direction of (\vec{\mu}) is opposite to that of (\vec{\l}), because of the negative charge of the electron.

Question 36.
State the underlying principle of a cyclotron. Write briefly how this machine is used to accelerate charged particles to high energies.
(Delhi 2014)
Answer:

Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 151
Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D1 to be negative. It gets accelerated towards dee D1. As it enters the dee D1, it does not experience any electric field due to shielding
effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D1. It finds dee D1 positive and dee D2 negative. It now gets accelerated towards dee D2. It moves faster through dee D2 describing a larger semicircle than
before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 152

Question 37.
Draw the magnetic field lines due to a current passing through a long solenoid. Use Ampere’s circuital law, to obtain the expression for the magnetic field due to the current I in a long solenoid having n number of turns per unit length. (Comptt. Delhi 2014)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 39
(ii) Expression for magnetic field :
Magnetic field due to Solenoid Let length of solenoid = L
Total number of turns in solenoid = N
No. of turns per unit length = \frac{N}{L} = n
ABCD is an Ampere’s loop
AB, DC are very large
BC is in a region of \overrightarrow{\mathrm{B}} = 0
AD is a long axis
Length of AD = x
Current in one turn = I0
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 13
Applying Ampere’s circuital loop — | B .dl = go I ’
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 14
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 15
No. of turns in x length = nx,
Current in turns nx, I = nx I0
According to Ampere’s circuital law
Bx = µ0 I => Bx = µ0 nx I0
∴ B = µ0nI0

Question 38.
A rectangular coil of sides ‘V and ‘b’ carrying a current I is subjected to a uniform magnetic field \overrightarrow{\mathbf{B}} acting perpendicular to its plane. Obtain the expression for the torque acting on it.
(Comptt. Delhi 2014)
Answer:
(a) Torque on a rectangular current loop in a uniform magnetic field:
Let I = current through the coil
a, b – sides of the rectangular loop
A = ab = area of the loop
n = Number of turns in the loop
B = Magnetic field
θ = angle between magnetic field
\overrightarrow{\mathrm{B}} and area vector \overrightarrow{\mathrm{A}}
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 176
Force exerted on the arm DA inward
F1 = I b B …[∵ F = ILB]
Force exerted on the arm BC outward
F2 = I b B ∴ F2 = F1
Thus net force on the loop is zero
∴ Two equal and opposite forces form a couple which exerts a torque
∴ Magnitude of the torque on the loop is,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 177
If loop has n turns then M = n I A
∴ τ = nIAB sin θ
When θ = 90° then \tau_{\max }=n I A B
When θ = 0° then τ = 0
(b) Since the momentum and the charge on both the proton and deutron are the same, the particle will follow a circular path with radius 1:1.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 178

Question 39.
(i) State Biot – Savart law in vector form expressing the magnetic field due to an \overrightarrow{\mathbf{B}} element \overrightarrow{\mathbf{dl}} carrying current I at a distance \overrightarrow{\mathbf{r}} from the element.
(ii) Write the expression for the magnitude of the magnetic field at the centre of a circular loop of radius r carrying a steady current I. Draw the field lines due to the current loop. (Comptt. All India 2014)
Answer:
(i) According to Biot-Savart’s law, “magnetic field acting at a particular point due to current carrying element is proportional to the division of cross product of current element and position vector of point where the field is to be calculated from the current element to the cube of the distance between current element and the point where the field is to be calculated”.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 157
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 158
Magnetic field on the axis of circular current loop :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 159
As in a special case we may obtain the field at the centre of the loop. Here x = 0, and we obtain
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 160
In a current loop, both the opposite faces behave as opposite poles, making it a magnetic dipole. One side of the current carrying coil behaves like the N-pole and the other side as the S-pole of a magnet.

Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 40

Question 40.
A square loop of side 20 cm carrying current of 1A is kept near an infinite long straight wire carrying a current of 2A in the same plane as shown in the figure.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 41
Calculate the magnitude and direction of the net force exerted on the loop due to the current carrying conductor. (Comptt. All India)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 42
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 43
The direction of force is towards the infinitely long straight wire.

Question 41.
A square shaped plane coil of area 100 cm2 of 200 turns carries a steady current of 5A. It is placed in a uniform magnetic field of 0.2 T acting perpendicular to the plane of the coil. Calculate the torque on the coil when its plane makes an angle of 60° with the direction of the field. In which orientation will the coil be in stable equilibrium? (Comptt. All India 2014)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 44
The coil will be in stable equilibrium when it is parallel to the magnetic field.

Question 42.
Find the condition under which the charged particles moving with different speeds in the presence of electric and magnetic field vectors can be used to select charged particles of a particular speed. (All India 2015)
Answer:
Condition: The velocity \vec{v} of the charged particles, and the \overrightarrow{\mathrm{E}} and \overrightarrow{\mathrm{B}} vectors, should be mutually perpendicular

It means that the forces on q, due to \overrightarrow{\mathrm{E}} and \overrightarrow{\mathrm{B}} must be oppositely directed.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 45
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 46

Question 43.
A charge q of mass m is moving with a velocity of v, at right angles to a uniform magnetic field B. Deduce the expression for the radius of the circular path it describes. (Comptt. Delhi 2015)
Answer:
Force experienced by charged particle in magnetic field.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 47
As v and B are perpendicular, F = qvB
This force is perpendicular to the direction of velocity and hence acts as
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 224

Question 44.
A proton and an alpha particle having the same kinetic energy are, in turn, passed through a region of uniform magnetic field, acting normal to the plane of the paper and travel in circular paths. Deduce the ratio of the radii of the circular paths described by them. (Comptt. Delhi 2015)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 48

Moving Charges and Magnetism Class 12 Important Questions Short Answer Type SA III

Question 45.
A circular coil of 200 turns and radius 10 cm is placed in a uniform magnetic field of 0.5 T, normal to the plane of the coil. If the current in the coil is 3.0 A, calculate the
(a) total torque on the coil.
(b) total force on the coil.
(c) average force on each electron in the coil, due to the magnetic field.
Assume the area of cross-section of the wire to be 10-5 m2 and the free electron density is 1029/m3. (All India 2015)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 49
(b) As the forces on different parts of the coil appears in pairs, equal in magnitude and opposite in direction, net force on the coil is zero.i.e., F = 0
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 3

Question 46.
An electron moves around the nucleus in a hydrogen atom of radius 0.51 A, with a velocity of 2 × 105 m/s. Calculate the following :
(i) the equivalent current due to orbital motion of electron
(ii) the magnetic field produced at the centre of the nucleus
(iii) the magnetic moment associated with the electron. (All India 2015)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 50

Question 47.
Derive the expression for force per unit length between two long straight parallel current carrying conductors. Hence define one ampere. (Delhi 2015)
Answer:
Consider two infinitely long parallel conductors carrying current I1 and I2 in the same direction.
Let d be the distance of separation between these two conductors.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 51
Hence, force is attractive in nature.
Ampere : Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10-7 N on each metre of the other wire.
Then current flowing is 1A
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 52
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 53

Question 48.
Explain the principle and working of a cyclotron with the help of a schematic diagram. Write the expression for cyclotron frequency. (Delhi 2015)
Answer:
Cyclotron :
Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 151
Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D1 to be negative. It gets accelerated towards dee D1. As it enters the dee D1, it does not experience any electric field due to shielding
effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D1. It finds dee D1 positive and dee D2 negative. It now gets accelerated towards dee D2. It moves faster through dee D2 describing a larger semicircle than
before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 152

Question 49.
Find the magnetic field at a point on the axis of a circular coil carrying current and hence find the magnetic field at the centre of the circular coil carrying current.
Answer:
Magnetic field at a point on the axis of a circular coil carrying current
Consider a circular coil of radius ‘a’ with centre ‘O’, carrying current I. Its plane is perpendicular to the plane of the loop. Suppose P is any point on the axis of the circular coil at a distance x from the centre, such that
OP = x
Consider two small elements of length dl at C and D at diametrically opposite current elements of the coil
PC = PD = r = \sqrt{a^{2}+x^{2}}
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 4
According to Biot Savart’s law, the magnitude of magnetic field at P due to current element dl at C is
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 55
The direction of \overrightarrow{d \overrightarrow{\mathrm{B}}} is perpendicular to \vec{r} in the plane paper i.e., along PQ.
Similarly, the magnitude of magnetic field at P due to current element dl at D is
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 56
Its direction is along PQ’
From (i) an (ii), we get, dB = dB’ = \frac{\mu_{0}}{4 \pi} \frac{\mathrm{I} d l}{\left(a^{2}+x^{2}\right)}
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 57
components :
(i) dB cos ϕ along PY and dB’ sin ϕ along PX
(ii) dB cos ϕ along PY and dB’ sin ϕ along PX
Since the components of the magnetic field along Y-axis are equal and opposite and cancel each other, the components along X-axis are in the same direction and are added up.
Hence the total magnetic field at point P is,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 58
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 59

Question 50.
Deduce the expression for the magnetic dipole moment of an electron orbiting around the central nucleus. (All India 2010)
Answer:
Consider an electron revolving around the nucleus of an atom. Electron is in uniform circular motion around the nucleus of charge + Ze. This constitutes a current.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 60
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 61
If ‘r’ is orbital radius of the electron and ‘V’ is orbital speed, then the time period is
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 62
Now putting the value of T in (i), we get
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 63
…where [Z is angular momentum of the electron.
According to Bohr hypothesis angular momentum we can have discrete values only.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 64

Question 51.
A long straight wire of a circular cross-section of radius ‘a’ carries a steady current ‘I’. The current is uniformly distributed across the cross-section. Apply Ampere’s circuital law to calculate the magnetic field at a point V in the region for
(i) r < a and (ii) r > a.
Answer:
Consider an infinite long thick wire of radius V with axis XY. Let I be the current flowing through the wire.
When the point P lies outside the wire :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 65
Let r be the perpendicular distance of point P from the axis of the cylinder, where r > a.
Here \overrightarrow{\mathrm{B}} and d \overrightarrow{\mathrm{l}} are acting in the same direction.
Applying Ampere’s circuital law, we have
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 66
When the point P lies inside the wire :
Here r ≤ a. We have two possibilities:
According to Ampere circuital law,
(i) “Whenever the current floras only through the surface of the wire, B = 0 as current in the closed path will be zero.”
(ii) “Wherever in the case when current is uniformly distributed through the cross-section of conductor, current through the closed path will be :
I’ = Current per unit area of the wire × area of the circle of radius r
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 67

Question 52.
State the underlying principle of working of a moving coil galvanometer. Write two reasons why a galvanometer can not be used as such to measure current in a given circuit. Name any two factors on which the current sensitivity of a galvanometer depends. (Delhi 2010)
Answer:
(i) Moving coil galvanometer works on the principle of a torque experienced by a current carrying coil placed in a magnetic field, whose magnitude is a function of current passing through the coil.
(ii) The galvanometer cannot be used to measure the value of the current in a given circuit due to the following two reasons:
(a) Galvanometer is a very sensitive device. It gives a full scale deflection for a small value of current.
(b) The galvanometer has to be connected in series for measuring currents and as it has a large resistance, this will change the value of the current in the circuit.
(iii)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 68
It depends on the number of turns N of the coil, torsion constant and the area A of the coil.

Question 53.
Write the expression for the magnetic moment (\overrightarrow{\mathbf{M}}) due to a planar square loop of side ‘l’ carrying a steady current I in a vector form. In the given figure this loop is placed in a horizontal plane near a long straight conductor carrying a steady current I, at a distance l as shown. Give reasons to explain that the loop will experience a net force but no torque. Write the expression for this force acting on the loop. (Delhi 2010)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 69
Answer:
(i) The magnetic moment (\overrightarrow{\mathrm{M}}) due to a planar square loop of side ‘l’ carrying a steady current I in a vector form is
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 70
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 5
(ii) CE will be attracted towards AB with a force F1 given by
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 71

Question 54.
Write the expression for the magnetic moment (\overrightarrow{\mathbf{M}}) due to a planar square loop of side ‘l’ carrying a steady current I in a vector form. In the given figure this loop is placed in a horizontal plane near a long straight conductor carrying a steady current I, at a distance l as shown. Give reasons to explain that the loop will experience a net force but no torque. Write the expression for this force acting on the loop. (Delhi 2010)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 69
Answer:
(i) The magnetic moment (\overrightarrow{\mathrm{M}}) due to a planar square loop of side ‘l’ carrying a steady current I in a vector form is
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 70
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 5
(ii) CE will be attracted towards AB with a force F1 given by
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 71

Question 55.
A rectangular loop of wire of size 4 cm × 10 cm carries a steady current of 2 A. A straight long wire carrying 5 A current is kept near the loop as shown. If the loop and the wire are coplanar, find
(i) the torque acting on the loop and
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 228
(ii) the magnitude and direction of the force on the loop due to the current carrying wire. (Delhi 2012)
Answer:
τ = IAB sin θ => τ = IAB sin θ (as θ = 0)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 85
(ii) Force acting on the loop on |F|
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 86
Direction : Towards conductor/Attractive Net force on the loop will act towards the long conductor (attractive) if the current in its closer side is in the same direction as the current in the long conductor, otherwise it will be repulsive.

Question 56.
A rectangular loop of wire of size 2 cm × 5 cm carries a steady current of 1 A. A straight long wire carrying 4 A current is kept near the loop as shown in the figure.
If the loop and the wire are coplanar, find
(i) the torque acting on the loop and
(ii) the magnitude and direction of the force on the loop due to the current carrying wire. (Delhi 2010)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 229
Answer:
(i) τ (Torque on the loop) = MB sin θ
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 87

Direction : Towards the conductor/Attractive
Net force on the loop will act towards the long conductor (attractive) if the current in its closer side is in the same direction as the
current in the long conductor, otherwise it will be repulsive.

Question 57.
Two identical coils, each of radius ‘R’ and number of turns ‘N’ are lying in perpendicular planes such that their centres coincide. Find the magnitude and direction of the resultant magnetic field at the centre of the coils, if they are carrying currents ‘I’ and √3I respectively.
(Comptt. Delhi 2010)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 88

Question 58.
Two identical coils, each of radius ‘R’ and number of turns ‘N’ are lying in perpendicular planes such that their centres coincide. Find the magnitude and direction of the resultant magnetic field at the centre of the coils, if they are carrying currents ‘I’ and √2I respectively.
(Comptt. Delhi 2010)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 89

Question 59.
Figure shows a rectangular loop conducting PQRS in which the arm PQ is free to move. A uniform magnetic field acts in the direction perpendicular to the plane of the loop. Arm PQ is moved with a velocity v towards the arm RS. Assuming that the arms QR, RS and SP have negligible resistances and the moving arm PQ has the resistance r, obtain the expression for
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 90
(i) the current in the loop
(ii) the force and
(iii) the power required to move the arm PQ. (Delhi 2010)
Answer:
Let the magnetic field acting on the loop be B and length of the rod PQ be l
The induced e.m.f. ε = Blv
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 91

Question 60.
A rectangular conductor LMNO is placed in a uniform magnetic field of 0.5 T. The field is directed perpendicular to the plane of the conductor. When the arm MN of length of 20 cm is moved towards left with a velocity of 10 ms-1, calculate the emf induced in the arm. Given the resistance of the arm to be 5Ω (assuming that other arms are of negligible resistance) find the value of the current in the arm. (All India 2010)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 92
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 93

Question 61.
A wire AB is carrying a steady current of 12A and is lying on the table. Another wire CD carrying 5A is held directly above AB at a height of 1 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB. [Take the value of g = 10 ms-2] (All India 2010)
Answer:
Given :
Current in the wire AB (I1) = 12 A,
Current in wire CD (I2) = 5 A
Separation between two wires (d) = 1 mm
= 10-3 m
Let ‘m’ be the mass of wire CD of length (L),
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 94
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 95
Direction of current in CD should be opposite to that of AB.

Question 62.
A wire AB is carrying a steady current of 10 A and is lying on the table. Another wire CD carrying 6 A is held directly above AB at a height of 2 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB. [Take the value of g = 10 ms-2] (All India 2010)
Answer:
Let AB has current in +ve x-direction
(I1) Current in wire AB = 10 A,
(I2) Current in wire CD = 6 A
Separation between the two wires = 2 mm
= 2 × 10-3m
To keep the wire CD suspended in its vertical position when left free. For this magnetic force on CD due to AB should balance mg due to its own weight.
Let m be the mass of the wire CD and L be its length
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 96
Direction of current in CD should be opposite to that AB.

Question 63.
(a) Define the current sensitivity of a galvanometer.
(b) The coil area of a galvanometer is 16 × 10-4 m2. It consists of 200 turns of a wire and is in a magnetic field of 0.2 T. The restoring torque constant of the suspension fibre is 10-6 Nm per degree. Assuming the magnetic field to be radial, calculate the maximum current that can be measured by the galva-nometer if the scale can accommodate 30° deflection. (Comptt. All India 2010)
Answer:
Sensitivity of a galvanometer: A galvanometer is said to be sensitive, if it gives a large deflection, even when a small current passes through it.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 97

Question 64.
(a) State Ampere’s circuital law, expressing it in the integral form.
(b) Two long coaxial insulated solenoids, S1 and S2 of equal lengths are wound one over the other as shown in the figure. A steady current “I” flows through the inner solenoid S1 to the other end B, which is connected to the outer solenoid S2 through which the same current “l” flows in the opposite direction so as to come out at end A. If n1 and n2 are the number of turns per unit length, find the magnitude and direction of the net magnetic field at a point
(i) inside on the axis and
(ii) outside the combined system. (Delhi 2014)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 98
Answer:
(a) According to Ampere’s Circuital law, the magnetic field B is related to steady current
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 99
(i) Inside the combined system : Magnetic field at a point on the axis.
Using Ampere’s Circuital law, the magnetic field due to inner solenoid S1 is given by
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 100
Similarly due to outer solenoid S2,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 230
Since these two magnetic fields are opposite in direction,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 101
in the upward direction.
(ii) Outside the combined system : At such a point, magnetic field is zero, because corresponding turns of the two halves of the solenoid produce equal and opposite magnetic fields.

Question 65.
Consider the motion of a charged particle of mass ‘m’ and charge ‘q’ moving with velocity \vec{v} in a magnetic field \vec{B}.
(a) If \vec{v} is perpendicular to \vec{B}, show that it describes a circular path having angular frequency ω = qB/m.
(b) If the velocity \vec{v} has a component parallel to the magnetic field \vec{B}, trace the path described by the particle. Justify your answer. (Comptt. Delhi 2014)
Answer:
(a) When a charged particle with charge q moves inside a magnetic field \overrightarrow{\mathrm{B}} with velocity v, it experiences a force, which is given by
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 196
Here, \overrightarrow{\mathrm{v}} is perpendicular to \overrightarrow{\mathrm{B}}, \overrightarrow{\mathrm{F}} is the force on the charged particle which acts as the centripetal force and makes it move along a circular path.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 197
Let m be the mass of the charged particle and r be the radius of the circular path.
Time period of circular motion of the charged particle can be calculated as shown below:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 198
Therefore, the frequency of the revolution of the charged particle is independent of the velocity or the energy of tire particle.
(b)

Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 151
Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D1 to be negative. It gets accelerated towards dee D1. As it enters the dee D1, it does not experience any electric field due to shielding
effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D1. It finds dee D1 positive and dee D2 negative. It now gets accelerated towards dee D2. It moves faster through dee D2 describing a larger semicircle than
before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 152

Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 102

Justification : Component of velocity \vec{v}, parallel to magnetic field, will make the particle move along the field.

Perpendicular component of velocity \vec{v} will cause the particle to move along a circular path in the plane perpendicular to the magnetic field Hence, the particle will follow a helical path, as shown above.

Question 66.
(a) Draw a schematic sketch of a moving coil galvanometer and describe briefly its working.
(b) “Increasing the current sensitivity of a galvanometer does not necessarily increase the voltage sensitivity.” Justify this statement. (Comptt. Delhi 2014)
Answer:
(a)
Principle : “If a current carrying coil is freely suspended/pivoted in a uniform magnetic field, it experiences a deflecting torque.”
Working: As the pivoted coil is placed in a radial magnetic field, hence on passing current I through it, a deflecting torque acts on the coil which is given by, τ = NAIB
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 204
The spring Sp attached to the coil provides the counter torque and in equilibrium state balances the deflecting torque. If φ is steady angular deflection then counter torque is kφ.
…where [k = torsional constant of the spring
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 205
In equilibrium state,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 206
Thus, deflection is directly proportional to the current flowing in the coil.
(a) (i) Uniform radial magnetic field. It keeps the magnetic field line normal to the area vector of the coil.
(ii) Soft iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.
(b) (i) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 207
(ii) Voltage sensitivity is defined as the deflection produced in the galvanometer when unit voltage is applied across the coil of the galvanometer.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 208
…where [R = Resistance of the coil
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 209
does not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

(b) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 103
may not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

Question 67.
A uniform magnetic field \overrightarrow{\mathrm{B}} is set up along the positive x-axis. A particle of charge ‘q’ and mass ‘m’ moving with a velocity v enters the field at the origin in X-Y plane such that it has velocity components both along and perpendicular to the magnetic field \overrightarrow{\mathrm{B}}. Trace, giving reason, the trajectory followed by the particle. Find out the expression for the distance moved by the particle along the magnetic field in one rotation. (All India 2014)
Answer:
Since the velocity of the particle is inclined to x-axis, thererfore, the velocity has a component along B, this component remains unchanged as the motion along the magnetic field will not be affected by the magnetic field. The motion in a plane perpendicular to B is as before a circular one, thereby producing a helical motion, which is its trajectory.

If r is the radius of the circular path of a particle, then a force of mv2/r, acts perpendicular to the path towards the centre of the circle and is called the centripetal force. If the velocity v is perpendicular to the magnetic field B, the magnetic force is perpendicular to both v and B and acts like a centripetal force. It has a magnitude qvB. Equating the two expressions (for centripetal force)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 104
mv2/r = qvB, which gives r = mv/qB …(i) for the radius of the circle described by the charged particle.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 106
There is a component of the velocity parallel to the magnetic field (denoted by v11), it will make the particle move along the field and the path of the particle would be a helical one.

The distance moved along the magnetic field in one rotation is called pitch p.
Using equation (ii), we have
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 107

Question 68.
Write the expression for the generalized form of Ampere’s circuital law. Discuss its significance and describe briefly how the concept of displacement current is explained through charging/discharging of a capacitor in an electric circuit. (All India 2014)
Answer:
Maxwell’s displacement current : According to Ampere’s circuital law, the magnetic field \overrightarrow{\mathrm{B}} is related to steady current I as,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 109
Maxwell showed that this relation is logically inconsistent. He accounted this inconsistency as follows :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 110
Ampere’s circuital laic for loop C, gives
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 111
Loop C2 lies in the region between the plates
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 112
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 113
which is logically inconsistent. So, Maxwell gave idea of displacement current.

Thus displacement current is that current which comes into play in the region in which the electric
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 114
where [ID is displacement current and
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 115
It is now called Ampere-Maxwell law. This is the generalization of Ampere’s Circuital law.

Question 69.
(a) Why is the magnetic field radial in a moving coil galvanometer? Explain how it is achieved.
(b) A galvanometer of resistance ‘G’ can be converted into a voltmeter of range (0 – V) volts by connecting a resistance ‘R’ in series with it. How much resistance will be required to change its range from 0 to V/2? (Comptt. All India 2014)
Answer:
(a) The magnetic field in a moving coil galvanometer is made ‘radial’ to keep the magnetic field ‘normal’ to the area vector of the coil. It is done by taking the cylindrical soft iron core. The torque acting on the coil is maximum (sin θ = 1, when, θ = 90°)
(b) Given : resistance of galvanomter = G Ω
Range of voltmeter (RL) = (0 – V) volts
Resistance to be connected in parallel = R
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 116
[ig is the maximum current which can flow through galvanometer]
From equation (i) and (ii), on solving we get
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 117

Question 70.
A closely wound solenoid of 2000 turns and cross sectional area 1.6 × 10-4 m2 carrying a current of 4.0 A is suspended through its centre allowing it to turn in a horizontal plane. Find
(i) the magnetic moment associated with the solenoid,
(ii) magnitude and direction of the torque on the solenoid if a horizontal magnetic field of 7.5 × 10-2 T is set up at an angle of 30° with the axis of the solenoid. (Comptt. All India 2014)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 118
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 119
(iii) Direction of torque is perpendicular to both the planes of the solenoid and the magnetic field.

Question 71.
(a) Write the expression for the magnetic force acting on a charged particle moving with velocity v in the presence of magnetic field B.
(b) A neutron, an electron and an alpha particle moving with equal velocities, enter a uniform magnetic field going into the plane of the paper as shown. Trace their paths in the field and justify your answer. (Delhi 2016)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 120
Answer:
(a) Expression for magnetic force :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 231
(b) Justification : Direction of force experienced by 01 the particle will be n according to Fleming’s Left hand rule.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 121

Question 72.
Two long straight parallel conductors carry steady current I1 and I2 separated by a distance d. If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere. (Delhi 2014)
Answer:

Consider two infinitely long parallel conductors carrying current I1 and I2 in the same direction.
Let d be the distance of separation between these two conductors.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 51
Hence, force is attractive in nature.
Ampere : Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10-7 N on each metre of the other wire.
Then current flowing is 1A
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 52
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 53

Question 73.
Use Biot-Savart law to derive the expression for the magnetic field on the axis of a current carrying circular loop of radius R.
Draw the magnetic field lines due to a circular . wire carrying current I. (All India 2014)
Answer:
(i)
According to Biot-Savart’s law, “magnetic field acting at a particular point due to current carrying element is proportional to the division of cross product of current element and position vector of point where the field is to be calculated from the current element to the cube of the distance between current element and the point where the field is to be calculated”.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 157
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 158
Magnetic field on the axis of circular current loop :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 159
As in a special case we may obtain the field at the centre of the loop. Here x = 0, and we obtain
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 160
In a current loop, both the opposite faces behave as opposite poles, making it a magnetic dipole. One side of the current carrying coil behaves like the N-pole and the other side as the S-pole of a magnet.
(ii)

(a) Magnetic field lines :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 195
(b) Moving coil galvanometer. It is a device used for the detection and measurement of small electric current.
Principle. The working is based on the fact that a current carrying coil suspended in a magnetic field experiences a torque.
Construction. It consists of a coil having a large number of turns of insulated copper wire wound on a metallic frame. The coil is suspended by means of a phosphor-bronze strip and is surrounded by a horse-shoe magnet NS. A hair spring is attached to lower end of the coil. The other end of the spring is attached to the scale through a pointer.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 7

Working. When current is passed, say along ABCD, the couple acts on it. Since the plane remains always parallel to the magnetic field in all positions of the coil (radial field), the force on the vertical arms always remains perpendicular to the place of the coil.

Let, I be the current flowing through coil,
B be magnetic field supposed to be uniform and always parallel to the coil, A be area of the coil
Deflecting torque acting on the coil is,
τ = nI BA sin 900

Due to deflecting torque, the coil rotates and suspension wire gets twisted. A restoring torque is set up in the suspension fibre. If <|) is angle through which the coil rotates and k is the restoring torque per unit angular twist, then restoring torque, τ = kϕ
In equilibrium,
Deflecting torque = Restoring torque
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 8
This provide a linear scale for the galvanometer.
Function of a radial magnetic field : Radial magnetic field being normal in all directions is formed to get maximum torque.
Function of Soft iron core, which not only makes the field radial but also increases the strength of the magnetic field.
(c) One uses a shunt resistance in parallel with the galvanometer, so that most of the current passes through the shunt. In the case of a voltmeter, a resistance of large value is used in series because it must draw a very small current, otherwise the voltage measurement will disturb the original set up by an amount which is very large.

Question 74.
Three long straight parallel wires are kept as shown in the figure. The wire (3) carries a current I
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 122
(i) The direction of flow of current I in wire (3), is such that the net force, on wire (1), due to the other two wires, is zero.
(ii) By reversing the direction of I, the net force, on wire (2), due to the other two wires, becomes zero. What will be the directions of current I, in the two cases? Also obtain the relation between the magnitudes of currents I1 I2 and I.
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 123
(i) Net force experienced by wire (1) can be zero only, when the current in wire (3) flows along – \hat{\mathrm{j}} i.e. downwards, it means that the forces acting on wire (1) due to wire (3) and wire (2) are equal and opposite.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 124
(ii) When direction of current in wire (3) is reversed then current should be along + \hat{\mathrm{j}} i.e. upwards.
For this case net force on wire (2) becomes zero, which means that the forces due to wire (1) and wire (3) are equal and opposite.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 125

Question 75.
A circular coil, having 100 turns of wire, of radius (nearly) 20 cm each, lies in the XY plane with its centre at the origin of co-ordinates. Find the magnetic field, at the point (0, 0, 20√3 cm), when this coil carries a current of (Comptt. Delhi 2016)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 126

Question 76.
Write the expression for the magnetic force \overrightarrow{\mathbf{F}} acting on a charged particle q moving with velocity \overrightarrow{\mathbf{F}} in the presence of the magnetic field \overrightarrow{\mathbf{B}} in a vector form. Show that no work is done and no change in the magnitude of the velocity of the particle is produced by this force. Hence define the unit of magnetic field. (Comptt. All India 2016)
Answer:
(i) The required expression is [latex]\overrightarrow{\mathbf{F}}=q(\vec{v} \times \overrightarrow{\mathbf{B}})
(ii) The magnetic force, at all instants, is, therefore, perpendicular to the instantaneous direction of \vec{v}, which is also the instantaneous direction of displacement (\overrightarrow{d s}).
Since, \overrightarrow{\mathrm{F}} is perpendicular to (\overrightarrow{d s}), at all instants, work done (=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{d s}) is zero
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 127
Hence, the magnetic field \overrightarrow{\mathrm{B}}, at a point equals one tesla if a charge of one coulomb, moving with a velocity 1/sec, along a direction perpendicular to the direction of \overrightarrow{\mathrm{B}}, experience a force of one newton.

Question 77.
A long straight wire, of circular cross section (radius = a) carries a current I which is uniformly distributed across the cross section of the wire.
Use Ampere's circuital law to calculate the magnetic field B(r), due to this wire, at a point distance r < a and r > a from its axis. Draw a graph showing the dependence of B(r) on r. (Comptt. All India 2016)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 128

Question 78.
Derive the expression for the torque τ acting on a rectangular current loop of area A placed in a uniform magnetic field B. Show that \vec{\tau}=\vec{m} \times \overrightarrow{\mathbf{B}} where \vec{m} is the magnetic moment of the current loop given by \vec{m}=\overrightarrow{\mathbf{I}} \overrightarrow{\mathbf{A}}. (Comptt. All India)
Answer:
(a) Torque on a rectangular current loop in a uniform magnetic field:
Let I = current through the coil
a, b - sides of the rectangular loop
A = ab = area of the loop
n = Number of turns in the loop
B = Magnetic field
θ = angle between magnetic field
\overrightarrow{\mathrm{B}} and area vector \overrightarrow{\mathrm{A}}
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 176
Force exerted on the arm DA inward
F1 = I b B ...[∵ F = ILB]
Force exerted on the arm BC outward
F2 = I b B ∴ F2 = F1
Thus net force on the loop is zero
∴ Two equal and opposite forces form a couple which exerts a torque
∴ Magnitude of the torque on the loop is,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 177
If loop has n turns then M = n I A
∴ τ = nIAB sin θ
When θ = 90° then \tau_{\max }=n I A B
When θ = 0° then τ = 0
(b) Since the momentum and the charge on both the proton and deutron are the same, the particle will follow a circular path with radius 1:1.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 178

Question 79.
(i) Obtain the expression for the cyclotron frequency.
(ii) A deuteron and a proton are accelerated by the cyclotron. Can both be accelerated with the same oscillator frequency? Give reason to justify your answer. (Delhi 2017)
Answer:
(i) Expression for cyclotron frequency : The magnetic field provides necessary centripetal force needed by the charged particle to move in a circular path.
m = mass of the charged particle,
v = velocity,
r = radius of the circular path
q = charge,
B = Magnetic field
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 129
Therefore, the frequency of revolution is independent of energy of the particle.
(ii) The mass of the two particles, i.e. deuteron and proton, is different. Since cyclotron frequency depends inversely on the mass, they cannot be accelerated by the same oscillator frequency.

Question 80.
Describe the working principle of a moving coil galvanometer. Why is it necessary to use
(i) a radial magnetic field and
(ii) a cylindrical soft iron core in a galvanometer? Write the expression for current sensitivity of the galvanometer.
Can a galvanometer as such be used for measuring the current? Explain. (Delhi 2017)
Answer:
Principle : “If a current carrying coil is freely suspended/pivoted in a uniform magnetic field, it experiences a deflecting torque."
Working: As the pivoted coil is placed in a radial magnetic field, hence on passing current I through it, a deflecting torque acts on the coil which is given by, τ = NAIB
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 204
The spring Sp attached to the coil provides the counter torque and in equilibrium state balances the deflecting torque. If φ is steady angular deflection then counter torque is kφ.
...where [k = torsional constant of the spring
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 205
In equilibrium state,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 206
Thus, deflection is directly proportional to the current flowing in the coil.
(a) (i) Uniform radial magnetic field. It keeps the magnetic field line normal to the area vector of the coil.
(ii) Soft iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.
(b) (i) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 207
(ii) Voltage sensitivity is defined as the deflection produced in the galvanometer when unit voltage is applied across the coil of the galvanometer.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 208
...where [R = Resistance of the coil
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 209
does not necessarily increase the voltage sensitivity. It may be affected by the resistance used.
No, the galvanometer cannot be used to measure current. It can only detect current but cannot measure as it is not calibrated. The galvanometer coil is likely to be damaged by currents in the (mA/A) range.

Question 81.
An electron of mass me revolves around a nucleus of charge +Ze. Show that it behaves like a tiny magnetic dipole. Hence prove that the magnetic moment associated with it is expressed as \vec{\mu}=-\frac{e}{2 m_{e}} \overrightarrow{\mathbf{L}} where L is the orbital angular momentum of the electron. Give the significance of negative sign. (Delhi 2017)
Answer:
(i) Electron, in circular motion around the nucleus, constitutes a current loop which behaves like a tiny magnetic dipole.
Current associated with the revolving electron :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 130
Magnetic moment of the loop, µ = IA
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 131
Negative sign signifies that the angular momentum of the revolving electron is opposite in direction to the magnetic moment associated with it.

Question 82.
(a) Write the expression for the force \overrightarrow{\mathbf{F}} acting on a particle of mass m and charge q moving with velocity \overrightarrow{\mathbf{v}} in a magnetic field \overrightarrow{\mathbf{B}}. Under what conditions will it move in
(i) a circular path and
(ii) a helical path?
(b) Show that the kinetic energy of the particle moving in magnetic field remains constant. (Delhi 2017)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 132
(i) When velocity of charged particle and magnetic field are perpendicular to each other, it will move in a circular path.
(ii) When velocity is neither parallel nor perpendicular to the magnetic field, it will move in helical path.
(b) The force experienced by the charged particle, is perpendicular to the instantaneous velocity \vec{v}, at all instants.
Hence the magnetic force cannot bring any change in the speed of the charged particle. Since speed remains constant, the kinetic energy also stays constant.

Question 83.
(a) State Biot-Savart law and express this law in vector form.
(b) Two identical circular coils, P and Q each of radius R, carrying currents 1 A and √3 A respectively, are placed concentrically and perpendicular to each other lying in the XY and YZ planes. Find the magnitude and direction of the net magnetic field at the centre of the coils. (All India 2017)
Answer:
(a) Biot-Savart law : It states that “the magnetic field dB due to a current element d l at any point P is:
(i) directly proportional to current dB ∝ I.
(ii) directly proportional to the length dl of the element d \vec{B} ∝ dl.
(iii) directly proportional to sin θ, where θ is the angle between d\overrightarrow{l} and \overrightarrow{r},
Therefore d \vec{B} ∝ sin θ
(iv) inversely proportional to the square of the distance r from the current element
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 190
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 6
Combining (i), (ii), (iii) and (iv), we get
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 191
The direction of \overrightarrow{d \mathrm{B}} is perpendicular to the plane of the vector d \vec{l} and \vec{r} given by Right Handed Screw Rule.
(b) Consider a circular coil of radis r, carrying current I. It consists of a large number of small current elements of length dl. According to Biot-Savart law, magnetic field at the centre O due to current element dl is,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 192
Magnetic field due to all such current elements will point into the plane of paper.
Hence total field at O is,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 193
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 194
(b) Given : RP = RQ = R, Ip = 1 A, IQ = √3 A
B = ? (Magnitude and direction)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 133
This net magnetic field B, is inclined to the field BP, at an angle θ, where
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 134

Question 84.
Two identical loops P and Q each of radius 5 cm are lying in perpendicular planes such that they have a common centre as shown in the figure. Find the magnitude and direction of the net magnetic field at the common centre of the two coils, if they carry currents equal to 3 A and 4 A respectively. (All India 2017)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 135
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 136

Question 85.
State the Lorentz's force and express it in vector form. Which pair of vectors are always perpendicular to each other? Derive the expression for the force acting on a current carrying conductor of length L in a uniform magnetic field 'B'. (Comptt. Delhi 2017)
Answer:
Lorentz's magnetic force is force experienced by a charged particle of charge 'q' moving in
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 137
perpendicular to each other Let us consider a conductor of uniform cross-sectional area A and length 'U having number density of electrons as V Total force on charge carriers in the conductor,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 138

Question 86.
Define the term magnetic moment of a current loop. Derive an expression for the magnetic field at any point along the axis of a solenoid of length 21, and radius «, and number of terms per unit length n. (Comptt. Delhi 2017)
Answer:
(i) Definition of magnetic moment: Magnetic moment of a current loop is equal to the product of current flowing in the loop and its area; and its direction is along area vector as per the right handed screw rule.
(ii) Magnetic field for solenoid :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 139
Using Ampere's circuital law
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 140

Question 87.
(a) Draw the pattern of magnetic field lines for a circular coil carrying current.
(b) Two identical circular loops X and Y of radius R and carrying the same current are kept in perpendicular planes such that they have a common centre at P as shown in the figure. Find the magnitude and direction of the net magnetic field at the point P due to the loops. (Comptt. All India 2017)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 232
Answer:
(a) Pattern of magnetic field lines for a circular coil carrying current :
(a) Magnetic field lines :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 195
(b) The magnetic field due to a circular coil at a point carrying current is given by
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 141
Since these two circular coils are identical and carrying the same current,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 142
Resultant magnetic field (BR)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 143

Question 88.
Define the term current sensitivity of a galvanometer. Write its SI unit. (Comptt. All India 2017)
Answer:
• Current sensitivity of a galvanometer is "deflection per unit current". It is defined as the ratio of deflection produced in the galvanometer to the current flowing through it.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 144
SI unit is radian per ampere.

Question 89.
A toroidal solenoid of mean radius 20 cm has 4000 turns of wire wound on a ferromagnetic core of relative permeability 800. Calculate the magnetic field in the core for a current of 3A , passing through the coil. How does the field change, when this core is replaced by a core of Bismuth? (Comptt. All India 2017)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 145
Since bismuth is diamagnetic, its μr < 1, therefore the magnetic field in the core will be very much reduced.

Moving Charges and Magnetism Class 12 Important Questions Long Answer Type

Question 90.
(a) Using Biot-Savart's law, derive an expression for the magnetic field at the centre of a circular coil of radius R, number of turns N, carrying current i.
(b) Two small identical circular coils marked 1, 2 carry equal currents and are placed with their geometric axes perpendicular to each other as shown in the figure. Derive an expression for the resultant magnetic field at O. (Delhi 2017)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 146
Answer:
(a) Consider a circular loop of wire of radius R carrying current I. The entire loop can be divided into a large number of small current elements.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 147
According to Biot-Savart's law, magnetic field due to current element 'Idl' at the centre O of a coil is
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 148
The direction of d \vec{l} is along the tangent
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 149
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 150

Question 91.
Draw a schematic diagram of a cyclotron. Explain its underlying principle and working, stating clearly the function of the electric and magnetic field applied on a charged particle. Deduce an expression for the period of revolution and show that it does not depend on the speed of the charged particle. (Delhi 2017)
Answer:
Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 151
Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D1 to be negative. It gets accelerated towards dee D1. As it enters the dee D1, it does not experience any electric field due to shielding
effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D1. It finds dee D1 positive and dee D2 negative. It now gets accelerated towards dee D2. It moves faster through dee D2 describing a larger semicircle than
before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 152

Question 92.
Draw a schematic sketch of a cyclotron. Explain briefly how it works and how it is used to accelerate the charged particles.
(i) Show that time period of ions in a cyclotron is independent of both the speed and radius of circular path.
(ii) What is resonance condition? How is it used to accelerate the charged particles? (All India 2017)
Answer:
(i) Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 151
Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D1 to be negative. It gets accelerated towards dee D1. As it enters the dee D1, it does not experience any electric field due to shielding
effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D1. It finds dee D1 positive and dee D2 negative. It now gets accelerated towards dee D2. It moves faster through dee D2 describing a larger semicircle than
before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 152
(ii) The frequency va of the applied voltage is adjusted so that the polarity of the dees is reversed in the same time that it takes the ions to complete one-half of the revolution. The requirement va = vc is called the resonance condition.
The phase of the supply is adjusted so that when the positive ions arrive at the edge of D1, D2 is at a lower potential and the ions are accelerated across the gap.

Question 93.
(a) Two straight long parallel conductors carry currents I1 and I2 in the same direction. Deduce the expression for the force per unit length between them.
Depict the pattern of magnetic field lines around them.
(b) A rectangular current carrying loop EFGH is kept in a uniform magnetic field as shown in the figure.
(i) What is the direction of the magnetic moment of the current loop?
(ii) When is the torque acting on the loop
(A) maximum,
(B) zero? (All India 2017)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 153
Answer:
Consider two infinitely long parallel conductors carrying current I1 and I2 in the same direction.
Let d be the distance of separation between these two conductors.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 51
Hence, force is attractive in nature.
Ampere : Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10-7 N on each metre of the other wire.
Then current flowing is 1A
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 52
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 53
(i) Magnetic moment will be out of the plane from the surface HEFG.
(ii) Torque
(A) Torque is maximum when MII B i.e., when it gets rotated by 90°.
(B) Torque is minimum when M and B are at 270° to each other.

Question 94.
(a) With the help of a diagram, explain the principle and working of a moving coil galvanometer.
(b) What is the importance of a radial magnetic field and how is it produced?
(c) Why is it that while using a moving coil galvanometer as a voltmeter a high resistance in series is required whereas in an ammeter a shunt is used? (All India)
Answer:
(a) Principle : “If a current carrying coil is freely suspended/pivoted in a uniform magnetic field, it experiences a deflecting torque."
Working: As the pivoted coil is placed in a radial magnetic field, hence on passing current I through it, a deflecting torque acts on the coil which is given by, τ = NAIB
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 204
The spring Sp attached to the coil provides the counter torque and in equilibrium state balances the deflecting torque. If φ is steady angular deflection then counter torque is kφ.
...where [k = torsional constant of the spring
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 205
In equilibrium state,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 206
Thus, deflection is directly proportional to the current flowing in the coil.
(a) (i) Uniform radial magnetic field. It keeps the magnetic field line normal to the area vector of the coil.
(ii) Soft iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.

(b) (i) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 207
(ii) Voltage sensitivity is defined as the deflection produced in the galvanometer when unit voltage is applied across the coil of the galvanometer.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 208
...where [R = Resistance of the coil
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 209
does not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

(b) For radial magnetic field, sin θ = 1,
so torque τ = NIAB.
Thus when radial magnetic field is used, the deflection of the coil is proportional to the current flowing through it. Hence a linear scale can be used to determine the deflection of the coil.

(c) A high resistance is joined in series with a galvanometer so that when the arrangement (voltmeter) is used in parallel with the selected section of the circuit, it should draw least amount of current. In case voltmeter draws appreciable amount of current, it will disturb the original value of potential difference by a good amount.

To convert a galvanometer into ammeter, a shunt is used in parallel with it so that when the arrangement is joined in series, the maximum current flows through the shunt, and thus the galvanometer is saved from its damage, when the current is passed through ammeter.

Question 95.
(a) Derive an expression for the force between two long parallel current carrying conductors.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 154
(b) Use this expression to define S.I. unit of current.
(c) A long straight wire AB carries a current I. A proton P travels with a speed v, parallel to the wire, at a distance d from it in a direction opposite to the current as shown in the figure. What is the force experienced by the proton and what is its direction? (All India)
Answer:
(a) For (a) and (b) :
Consider two infinitely long parallel conductors carrying current I1 and I2 in the same direction.
Let d be the distance of separation between these two conductors.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 51
Hence, force is attractive in nature.
Ampere : Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10-7 N on each metre of the other wire.
Then current flowing is 1A
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 52
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 53
(c) Force experienced by the proton,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 155
As magnetic field due to the current carrying wire is directed into the plane of the paper (θ = 90°)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 156
Force is directed away from the current carrying wire or in the right direction of observer.

Question 96.
State Biot-Savart law, giving the mathematical expression for it.
Use this law to derive the expression for the magnetic field due to a circular coil carrying current at a point along its axis.
How does a circular loop carrying current behave as a magnet? (Delhi 2011)
Answer:
According to Biot-Savart's law, "magnetic field acting at a particular point due to current carrying element is proportional to the division of cross product of current element and position vector of point where the field is to be calculated from the current element to the cube of the distance between current element and the point where the field is to be calculated".
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 157
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 158
Magnetic field on the axis of circular current loop :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 159
As in a special case we may obtain the field at the centre of the loop. Here x = 0, and we obtain
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 160
In a current loop, both the opposite faces behave as opposite poles, making it a magnetic dipole. One side of the current carrying coil behaves like the N-pole and the other side as the S-pole of a magnet.

Question 97.
With the help of a labelled diagram, state the underlying principle of a cyclotron. Explain clearly how it works to accelerate the charged particles. Show that cyclotron frequency is independent of energy of the particle. Is there an upper limit on the energy acquired by the particle? Give reason. (Delhi 2011)
Answer:

Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 151
Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D1 to be negative. It gets accelerated towards dee D1. As it enters the dee D1, it does not experience any electric field due to shielding
effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D1. It finds dee D1 positive and dee D2 negative. It now gets accelerated towards dee D2. It moves faster through dee D2 describing a larger semicircle than
before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 152

Yes, there is an upper limit. The increase in the kinetic energy of particles is qv. Therefore, the radius of their path goes on increasing each time, their kinetic energy increases. The lines are repeatedly accelerated across the dees, untill they have the required energy to have a radius approximately that of the dees. Hence, this is the upper limit on the energy required by the particles due to definite size of dees.

Question 98.
(a) State the principle of the working of a moving coil galvanometer, giving its labelled diagram.
(b) "Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity." Justify this statement
(c) Outline the necessary steps to convert a galvanometer of resistance RG into an ammeter of a given range. (All India 2011)
Answer:
(a)
Principle : “If a current carrying coil is freely suspended/pivoted in a uniform magnetic field, it experiences a deflecting torque."
Working: As the pivoted coil is placed in a radial magnetic field, hence on passing current I through it, a deflecting torque acts on the coil which is given by, τ = NAIB
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 204
The spring Sp attached to the coil provides the counter torque and in equilibrium state balances the deflecting torque. If φ is steady angular deflection then counter torque is kφ.
...where [k = torsional constant of the spring
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 205
In equilibrium state,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 206
Thus, deflection is directly proportional to the current flowing in the coil.
(a) (i) Uniform radial magnetic field. It keeps the magnetic field line normal to the area vector of the coil.
(ii) Soft iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.
(b) (i) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 207
(ii) Voltage sensitivity is defined as the deflection produced in the galvanometer when unit voltage is applied across the coil of the galvanometer.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 208
...where [R = Resistance of the coil
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 209
does not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

(b) Since \mathrm{v}_{\mathrm{s}}=\frac{\mathrm{I}_{\mathrm{s}}}{\mathrm{R}} increase in current sensitivity may not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

(c) Conversion of galvanometer into ammeter: By just connecting a low resistance known as shunt in parallel to the galvanometer, it can be converted into an ammeter.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 161
Let G = resistance of the galvanometer.
Ig = the current with which galvanometer gives full scale deflection.
S = shunt resistance
I - Ig = current through the shunt.
As the galvanometer and shunt are connected in parallel,
Potential difference across the galvanometer = Potential difference across the shunt
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 162

Question 99.
(a) Write the expression for the force, \overrightarrow{\mathbf{F}}, acting on a charged particle of charge ‘q’, moving with a velocity latex]\overrightarrow{\mathbf{v}}[/latex] in the presence of both electric field \overrightarrow{\mathrm{E}} and magnetic field \overrightarrow{\mathrm{B}}. Obtain the condition under which the particle moves undeflected through the fields.
(b) A rectangular loop of size l × b carrying a steady current I is placed in a uniform magnetic field \overrightarrow{\mathrm{B}}. Prove that the torque \vec{\tau} acting on the loop is given by \vec{\tau}=\vec{m} \times \overrightarrow{\mathrm{B}}, where \overrightarrow{\mathrm{m}} is the magnetic moment of the loop. (All India 2011)
Answer:
(a) A charge q in an electric field \overrightarrow{\mathrm{E}} experiences the electric force, \overrightarrow{\mathrm{F}}_{e}=q \overrightarrow{\mathrm{E}}

This force acts in the direction of field \overrightarrow{\mathrm{E}} and is independent of the velocity of the charge.

The magnetic force experienced by the charge q moving with velocity \overrightarrow{\mathrm{v}} in the magnetic field B is given by
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 163
This force acts perpendicular to the plane of \overrightarrow{\mathrm{V}} and \overrightarrow{\mathrm{B}} and depends on the velocity \overrightarrow{\mathrm{v}} of the charge.

The total force, or the Lorentz force, experienced by the charge q due to both electric and magnetic field is given by
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 164
Hence, A stationary charged particle does not experience any force in a magnetic field. (b) Torque on a current loop in a uniform magnetic field.
Let I = Current flowing through the coil PQRS
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 165
Its magnitude is, F3 = IaB sin(90° + 0)
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 166
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 167

Question 100.
(a) Explain, giving reasons, the basic difference in converting a galvanometer into
(i) a voltmeter and
(ii) an ammeter.
(b) Two long straight parallel conductors carrying steady currents I1 and I2 are separated by a distance’d’ Explain briefly, with the help of a suitable diagram, how the magnetic field due to one conductor acts on the other. Hence deduce the expression for the force acting between the two conductors. Mention the nature of this force.
Answer:
(a) (i) Voltmeter is connected in parallel with the circuit element across which the potential difference is intended to be measured.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 168
A galvanometer can be converted into a voltmeter by connecting a higher resistance in series with it. The value of this resistance is so adjusted that only current I which produces full scale deflection in the galvanometer, passes through the galvanometer.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 169
(ii) A galvanometer can be converted into an ammeter by connecting a low value
resistance in parallel with it.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 170
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 171
(b)
Consider two infinitely long parallel conductors carrying current I1 and I2 in the same direction.
Let d be the distance of separation between these two conductors.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 51
Hence, force is attractive in nature.
Ampere : Ampere is that current which is if maintained in two infinitely long parallel conductors of negligible cross-sectional area separated by 1 metre in vacuum causes a force of 2 × 10-7 N on each metre of the other wire.
Then current flowing is 1A
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 52
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 53

Question 101.
(a) Explain briefly with the help of a labelled diagram, the principle and working of a moving coil galvanometer.
(b) Define the term ‘current sensitivity’ of a galvanometer. How is it that increasing current sensitivity may not necessarily increase its voltage sensitivity? Explain. (Comptt. All India 2011)
Answer:
(a) Principle and working of a moving coil galvanometer:
Principle : “If a current carrying coil is freely suspended/pivoted in a uniform magnetic field, it experiences a deflecting torque.”
Working: As the pivoted coil is placed in a radial magnetic field, hence on passing current I through it, a deflecting torque acts on the coil which is given by, τ = NAIB
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 204
The spring Sp attached to the coil provides the counter torque and in equilibrium state balances the deflecting torque. If φ is steady angular deflection then counter torque is kφ.
…where [k = torsional constant of the spring
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 205
In equilibrium state,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 206
Thus, deflection is directly proportional to the current flowing in the coil.
(a) (i) Uniform radial magnetic field. It keeps the magnetic field line normal to the area vector of the coil.
(ii) Soft iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.
(b) (i) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 207
(ii) Voltage sensitivity is defined as the deflection produced in the galvanometer when unit voltage is applied across the coil of the galvanometer.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 208
…where [R = Resistance of the coil
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 209
does not necessarily increase the voltage sensitivity. It may be affected by the resistance used.
(b) Current sensitivity:

(a)
Principle : “If a current carrying coil is freely suspended/pivoted in a uniform magnetic field, it experiences a deflecting torque.”
Working: As the pivoted coil is placed in a radial magnetic field, hence on passing current I through it, a deflecting torque acts on the coil which is given by, τ = NAIB
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 204
The spring Sp attached to the coil provides the counter torque and in equilibrium state balances the deflecting torque. If φ is steady angular deflection then counter torque is kφ.
…where [k = torsional constant of the spring
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 205
In equilibrium state,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 206
Thus, deflection is directly proportional to the current flowing in the coil.
(a) (i) Uniform radial magnetic field. It keeps the magnetic field line normal to the area vector of the coil.
(ii) Soft iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.
(b) (i) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 207
(ii) Voltage sensitivity is defined as the deflection produced in the galvanometer when unit voltage is applied across the coil of the galvanometer.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 208
…where [R = Resistance of the coil
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 209
does not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

(b) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 103
may not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

Question 102.
(a) State Biot-Savart law. Deduce the expression for the magnetic field due to a circular current carrying loop at a point lying on its axis.
(b) Two long parallel wires carry currents I1 and I2 flowing in the same direction. When a third current carrying wire is placed parallel and coplanar in between the two, find the condition when the third wire would experience no force due to these two wires. (Comptt. All India 2011)
Answer:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 172
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 173
The components perpendicular to the axis of the loop will be equal and opposite to component along the axis of the loop and will cancel out. Their axial components will be in the same direction, i.e., along CP and get added up.
∴ Total magnetic field at point P in the direction CP is
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 174
The direction of this field is along the axis, in the sense given by the right hand (thumb) rule.
(b) The force \overrightarrow{f_{1}} on the third wire due to wire 1 is directed opposite to the force \overrightarrow{f_{2}} on the third wire due to wire 2. Hence, the net force on the third wire would become 0.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 175

Question 103.
(a) Derive the expression for the torque on a rectangular current carrying loop suspended in a uniform magnetic field.
(b) A proton and a deutron having equal momenta enter in a region of uniform magnetic field at right angle to the direction of the field. Depict their trajectories in the field. (Delhi 2013)
Answer:
(a) Torque on a rectangular current loop in a uniform magnetic field:
Let I = current through the coil
a, b – sides of the rectangular loop
A = ab = area of the loop
n = Number of turns in the loop
B = Magnetic field
θ = angle between magnetic field
\overrightarrow{\mathrm{B}} and area vector \overrightarrow{\mathrm{A}}
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 176
Force exerted on the arm DA inward
F1 = I b B …[∵ F = ILB]
Force exerted on the arm BC outward
F2 = I b B ∴ F2 = F1
Thus net force on the loop is zero
∴ Two equal and opposite forces form a couple which exerts a torque
∴ Magnitude of the torque on the loop is,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 177
If loop has n turns then M = n I A
∴ τ = nIAB sin θ
When θ = 90° then \tau_{\max }=n I A B
When θ = 0° then τ = 0
(b) Since the momentum and the charge on both the proton and deutron are the same, the particle will follow a circular path with radius 1:1.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 178

Question 104.
(a) Using Biot-Savart’s law, derive the expression for the magnetic field in the vector form at a point on the axis of a circular current loop.
(b) What does a toroid consist of? Find out the expression for the magnetic field inside a toroid for N turns of the coil having the average radius r and carrying a current I. Show that the magnetic field in the open space inside and exterior to the toroid is zero. (All India 2013)
Answer:
(a) Let P be the point on the axis of a circular loop or coil of radius a carrying current I. The distance of P from the centre of loop is x.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 179
According to Biot-Savart’s Law, magnetic field due to a small element XY (dl) at point P is
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 180
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 181
The perpendicular components of the magnetic field due to these elements being equal and opposite cancel each other. Hence the total contribution of perpendicular components, (i.e. dB cos θ) to the net magnetic field is zero.

On the other hand, dB sin θ component of magnetic field due to each element of the coil or loop is directed in the same direction.

Therefore, magnetic field at point P due to the whole coil or loop is equal to the sum of dB sin θ components of magnetic field due to each element
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 182
The right hand thumb rule can be used to find the direction of the field.
(b) A toroid is a solenoid bent to form a ring shape.
Let N number of turns per unit length of toroid and I be current flowing in it.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 183
Consider a loop (region II) of radius r passes through the centre of the toroid.
Let (region II) \overrightarrow{\mathrm{B}} be magnetic field along the loop is
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 184
Let (region I) B1 be magnetic field outside toroid in open space. Draw an amperian loop L2 of radius r2 through point Q.
Now applying ampere’s law :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 185
As I = 0, because the circular turn current coming out of plane of paper is cancelled exactly by current going into it, so net I = 0, equation (i) becomes
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 186

Question 105.
(a) Draw a schematic sketch of a cyclotron. Explain clearly the role of crossed electric and magnetic field in accelerating the charge. Hence derive the expression for the kinetic energy acquired by the particles.
(b) An α-particle and a proton are realeased from the centre of the cyclotron and made to accelerate.
(i) Can both be accelerated at the same cyclotron frequency? Give reason to justify your answer.
(ii) When they are accelerated in turn, which of the two will have higher velocity at the exit slit of the dees? (All India 2013)
Answer:
(a) Cyclotron :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 187
Role of crossed electric and magnetic field in cyclotron : The magnetic field makes the charged particle to cross the gap between the dees again and again by making it move along a circular path, while the oscillating electric field, applied across the dees, accelerates the charged particle again and again and hence increases its K.E.
Expression :
If the velocity v is perpendicular to the magnetic field B, the magnetic force is perpendicular to both v and B and acts like a centripetal force. It has a magnitude qvB. Equating the two expressions for centripetal force,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 188
No, both cannot be accelerated to same frequency because frequency depends upon mass and charge.
(ii) Velocity is given by the formula v = \frac{B q r}{m}
Velocity is also inversely proportional to mass and directly proportional to charge
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 189
Velocity of proton is higher than that α-particle.

Question 106.
State Biot-Savart law, expressing it in the vector form. Use it to obtain the expression for the magnetic field at an axial point, distance ‘d’ from the centre of a circular coil of radius V carrying current T. Also find the ratio of the magnitudes of the magnetic field of this coil at the centre and at an axial point for which
d = a√3. (Comptt. Delhi 2013)
Answer:
(a) Biot-Savart law : It states that “the magnetic field dB due to a current element d l at any point P is:
(i) directly proportional to current dB ∝ I.
(ii) directly proportional to the length dl of the element d \vec{B} ∝ dl.
(iii) directly proportional to sin θ, where θ is the angle between d\overrightarrow{l} and \overrightarrow{r},
Therefore d \vec{B} ∝ sin θ
(iv) inversely proportional to the square of the distance r from the current element
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 190
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 6
Combining (i), (ii), (iii) and (iv), we get
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 191
The direction of \overrightarrow{d \mathrm{B}} is perpendicular to the plane of the vector d \vec{l} and \vec{r} given by Right Handed Screw Rule.
(b) Consider a circular coil of radis r, carrying current I. It consists of a large number of small current elements of length dl. According to Biot-Savart law, magnetic field at the centre O due to current element dl is,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 192
Magnetic field due to all such current elements will point into the plane of paper.
Hence total field at O is,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 193
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 194
Question 107.
(a) Draw the magnetic field lines due to a current carrying loop.
(b) State using a suitable diagram, the working principle of a moving coil galvanometer. What is the function of a radial magnetic field and the soft iron core used in it?
(c) For converting a galvanometer into an ammeter, a shunt resistance of small value is used in parallel, whereas in the case of a voltmeter a resistance of large value is used in series. Explain why. (Comptt. Delhi 2011)
Answer:
(a) Magnetic field lines :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 195
(b) Moving coil galvanometer. It is a device used for the detection and measurement of small electric current.
Principle. The working is based on the fact that a current carrying coil suspended in a magnetic field experiences a torque.
Construction. It consists of a coil having a large number of turns of insulated copper wire wound on a metallic frame. The coil is suspended by means of a phosphor-bronze strip and is surrounded by a horse-shoe magnet NS. A hair spring is attached to lower end of the coil. The other end of the spring is attached to the scale through a pointer.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 7

Working. When current is passed, say along ABCD, the couple acts on it. Since the plane remains always parallel to the magnetic field in all positions of the coil (radial field), the force on the vertical arms always remains perpendicular to the place of the coil.

Let, I be the current flowing through coil,
B be magnetic field supposed to be uniform and always parallel to the coil, A be area of the coil
Deflecting torque acting on the coil is,
τ = nI BA sin 900

Due to deflecting torque, the coil rotates and suspension wire gets twisted. A restoring torque is set up in the suspension fibre. If <|) is angle through which the coil rotates and k is the restoring torque per unit angular twist, then restoring torque, τ = kϕ
In equilibrium,
Deflecting torque = Restoring torque
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 8
This provide a linear scale for the galvanometer.
Function of a radial magnetic field : Radial magnetic field being normal in all directions is formed to get maximum torque.
Function of Soft iron core, which not only makes the field radial but also increases the strength of the magnetic field.
(c) One uses a shunt resistance in parallel with the galvanometer, so that most of the current passes through the shunt. In the case of a voltmeter, a resistance of large value is used in series because it must draw a very small current, otherwise the voltage measurement will disturb the original set up by an amount which is very large.

Question 108.
(a) Deduce an expression for the frequency of revolution of a charged particle in a magnetic field and show that it is independent of velocity or energy of the particle.
(b) Draw a schematic sketch of a cyclotron. Explain, giving the essential details of its construction, how it is used to accelerate the charged particles. (All India 2011)
Answer:
(a) When a charged particle with charge q moves inside a magnetic field \overrightarrow{\mathrm{B}} with velocity v, it experiences a force, which is given by
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 196
Here, \overrightarrow{\mathrm{v}} is perpendicular to \overrightarrow{\mathrm{B}}, \overrightarrow{\mathrm{F}} is the force on the charged particle which acts as the centripetal force and makes it move along a circular path.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 197
Let m be the mass of the charged particle and r be the radius of the circular path.
Time period of circular motion of the charged particle can be calculated as shown below:
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 198
Therefore, the frequency of the revolution of the charged particle is independent of the velocity or the energy of tire particle.
(b) Cyclotron.

Principle : When a positively charged particle is made to move again and again in a high frequency electric field, it gets accelerated and acquires sufficiently large amount of energy.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 151
Working : Suppose a positive ion, say a proton, enters the gap between the two dees and finds dee D1 to be negative. It gets accelerated towards dee D1. As it enters the dee D1, it does not experience any electric field due to shielding
effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.

At the instant the proton comes out of dee D1. It finds dee D1 positive and dee D2 negative. It now gets accelerated towards dee D2. It moves faster through dee D2 describing a larger semicircle than
before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of the revolution of the proton, then everytime the proton reaches the gap between the two dees, the electric field is reversed and proton receives a push and finally it acquires very high energy. This proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.
Centripetal force is provided by magnetic field to charged particle to move in a circular back.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 152

Question 109.
(a) Draw a labelled diagram of a moving coil galvanometer. Describe briefly its principle and working.
(b) Answer the following :
(i) Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer?
(ii) Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. Explain, giving reason. (All India 2011)
Answer:
(a) Moving coil galvanometer.
Principle : “If a current carrying coil is freely suspended/pivoted in a uniform magnetic field, it experiences a deflecting torque.”
Working: As the pivoted coil is placed in a radial magnetic field, hence on passing current I through it, a deflecting torque acts on the coil which is given by, τ = NAIB
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 204
The spring Sp attached to the coil provides the counter torque and in equilibrium state balances the deflecting torque. If φ is steady angular deflection then counter torque is kφ.
…where [k = torsional constant of the spring
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 205
In equilibrium state,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 206
Thus, deflection is directly proportional to the current flowing in the coil.
(a) (i) Uniform radial magnetic field. It keeps the magnetic field line normal to the area vector of the coil.
(ii) Soft iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.
(b) (i) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 207
(ii) Voltage sensitivity is defined as the deflection produced in the galvanometer when unit voltage is applied across the coil of the galvanometer.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 208
…where [R = Resistance of the coil
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 209
does not necessarily increase the voltage sensitivity. It may be affected by the resistance used.
(b) (i) Iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.
(ii) Current sensitivity and voltage sensitivity.

(a)
Principle : “If a current carrying coil is freely suspended/pivoted in a uniform magnetic field, it experiences a deflecting torque.”
Working: As the pivoted coil is placed in a radial magnetic field, hence on passing current I through it, a deflecting torque acts on the coil which is given by, τ = NAIB
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 204
The spring Sp attached to the coil provides the counter torque and in equilibrium state balances the deflecting torque. If φ is steady angular deflection then counter torque is kφ.
…where [k = torsional constant of the spring
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 205
In equilibrium state,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 206
Thus, deflection is directly proportional to the current flowing in the coil.
(a) (i) Uniform radial magnetic field. It keeps the magnetic field line normal to the area vector of the coil.
(ii) Soft iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.
(b) (i) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 207
(ii) Voltage sensitivity is defined as the deflection produced in the galvanometer when unit voltage is applied across the coil of the galvanometer.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 208
…where [R = Resistance of the coil
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 209
does not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

(b) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 103
may not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

Question 110.
(a) State Ampere’s circuital law. Use this law to obtain the expression for the magnetic field inside an air cored toroid of average radius r, having ‘n’ turns per unit length and carrying a steady current I.
(b) Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 199
An observer to the left of a solenoid of N turns each of cross section area ‘A’ observes that a steady current I in it flows in the clockwise direction. Depict the magnetic field lines due to the solenoid specifying its polarity and show that it acts as a bar magnet of magnetic moment m = NIA. (Delhi 2011)
Answer:
(a) (i) Ampere’s Circuital Law. Line integral of magnetic field over a closed loop is equal to |i0 times the total current passing through the surface enclosed by the loop. Alternatively
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 200
(ii) Expression for magnetic field inside toroid
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 201
Let the current flowing through each turn of the toroid be I. The total number of turns equals n(2nr) where n is the number of turns per unit length. Applying Ampere’s circuital law, for the Amperian loop, for interior points.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 202
This is the expression for magnetic field inside air-cored toroid.
(b)
(i)Depiction of magnetic field for a solenoid.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 203
(ii) The solenoid contains N loops, each carrying a current I. Therefore, each loop acts as a magnetic dipole. The magnetic moment for a current I, flowing in loop of area (vector) A is given by, m = IA. The magnetic moments of all loops are aligned along the same direction.
Hence, net magnetic moment equals NIA.

Question 111.
Explain, using a labelled diagram, the principle and working of a moving coil galvanometer.
(a) What is the function of
(i) uniform radial magnetic field,
(ii) soft iron core?
(b)Define the terms
(i) current sensitivity and
(ii) voltage sensitivity of a galvanometer. Why does increasing the current sensitivity not necessarily increase voltage sensitivity? (All India 2011)
Answer:
Principle : “If a current carrying coil is freely suspended/pivoted in a uniform magnetic field, it experiences a deflecting torque.”
Working: As the pivoted coil is placed in a radial magnetic field, hence on passing current I through it, a deflecting torque acts on the coil which is given by, τ = NAIB
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 204
The spring Sp attached to the coil provides the counter torque and in equilibrium state balances the deflecting torque. If φ is steady angular deflection then counter torque is kφ.
…where [k = torsional constant of the spring
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 205
In equilibrium state,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 206
Thus, deflection is directly proportional to the current flowing in the coil.
(a) (i) Uniform radial magnetic field. It keeps the magnetic field line normal to the area vector of the coil.
(ii) Soft iron core in galvanometer. The cylindrical soft iron core, when placed inside the coil of a galvanometer, makes the magnetic field stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil may be, the magnetic field is always parallel to its plane.
(b) (i) Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 207
(ii) Voltage sensitivity is defined as the deflection produced in the galvanometer when unit voltage is applied across the coil of the galvanometer.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 208
…where [R = Resistance of the coil
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 209
does not necessarily increase the voltage sensitivity. It may be affected by the resistance used.

Question 112.
(a) Write, using Biot-Savart law, the expression for the magnetic field \overrightarrow{\mathrm{B}} due to an element d\overrightarrow{\mathrm{l}} carrying current I at a distance r from it in a vector form.
Hence derive the expression for the magnetic field due to a current carrying loop of radius R at a point P distant x from its centre along the axis of the loop.
(b) Explain how Biot-Savart law enables one to express the Ampere’s circuital law in the integral form, viz.,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 210 (All India 2011)
Answer:
According to Biot-Savart’s law, “magnetic field acting at a particular point due to current carrying element is proportional to the division of cross product of current element and position vector of point where the field is to be calculated from the current element to the cube of the distance between current element and the point where the field is to be calculated”.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 157
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 158
Magnetic field on the axis of circular current loop :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 159
As in a special case we may obtain the field at the centre of the loop. Here x = 0, and we obtain
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 160
In a current loop, both the opposite faces behave as opposite poles, making it a magnetic dipole. One side of the current carrying coil behaves like the N-pole and the other side as the S-pole of a magnet.

(b) Biot-Savart law can be expressed as Ampere’s Circuital law by considering the surface to be made up of a large number of loops. The sum of the tangential components of the magnetic field multiplied by the length of all such elements leads to integral. Ampere’s circuital law states that this’ integral is equal to p0 times the total current passing through that surface, i.e.,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 211

Question 113.
(a) Use Biot-Savart law to derive the expression for the magnetic field due to a circular coil of radius R having N turns at a point on the axis at a distance V from its centre. Draw the magnetic field lines due to this coil.
(b) A current ‘I’ enters a uniform circular loop of radius ‘R’ at point M and flows out at N as shown in the figure.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 212
Obtain the net magnetic field at the centre of the loop. (Comptt. Delhi 2011)
Answer:
(a) Biot-Savart Law to derive expression of magnetic field.
According to Biot-Savart’s law, “magnetic field acting at a particular point due to current carrying element is proportional to the division of cross product of current element and position vector of point where the field is to be calculated from the current element to the cube of the distance between current element and the point where the field is to be calculated”.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 157
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 158
Magnetic field on the axis of circular current loop :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 159
As in a special case we may obtain the field at the centre of the loop. Here x = 0, and we obtain
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 160
In a current loop, both the opposite faces behave as opposite poles, making it a magnetic dipole. One side of the current carrying coil behaves like the N-pole and the other side as the S-pole of a magnet.
Magnetic field lines due to circular coil.

(a) Magnetic field lines :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 195
(b) Moving coil galvanometer. It is a device used for the detection and measurement of small electric current.
Principle. The working is based on the fact that a current carrying coil suspended in a magnetic field experiences a torque.
Construction. It consists of a coil having a large number of turns of insulated copper wire wound on a metallic frame. The coil is suspended by means of a phosphor-bronze strip and is surrounded by a horse-shoe magnet NS. A hair spring is attached to lower end of the coil. The other end of the spring is attached to the scale through a pointer.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 7

Working. When current is passed, say along ABCD, the couple acts on it. Since the plane remains always parallel to the magnetic field in all positions of the coil (radial field), the force on the vertical arms always remains perpendicular to the place of the coil.

Let, I be the current flowing through coil,
B be magnetic field supposed to be uniform and always parallel to the coil, A be area of the coil
Deflecting torque acting on the coil is,
τ = nI BA sin 900

Due to deflecting torque, the coil rotates and suspension wire gets twisted. A restoring torque is set up in the suspension fibre. If <|) is angle through which the coil rotates and k is the restoring torque per unit angular twist, then restoring torque, τ = kϕ
In equilibrium,
Deflecting torque = Restoring torque
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 8
This provide a linear scale for the galvanometer.
Function of a radial magnetic field : Radial magnetic field being normal in all directions is formed to get maximum torque.
Function of Soft iron core, which not only makes the field radial but also increases the strength of the magnetic field.
(c) One uses a shunt resistance in parallel with the galvanometer, so that most of the current passes through the shunt. In the case of a voltmeter, a resistance of large value is used in series because it must draw a very small current, otherwise the voltage measurement will disturb the original set up by an amount which is very large.

(b) Let current I be divided at
point M into two parts I1 and I2; in bigger and smaller parts of the loop respectively. Magnetic field of current (clockwise) at point O :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 213
Magnetic field of current I2 (anticlockwise) at point O :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 214
Net magnetic field, \overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{B}_{1}}+\overrightarrow{\mathrm{B}_{2}}
But I1 = 3I2 (AS resistance of bigger part is three times that of the smaller part of the loop)
Substituting I1 = 3I2 in equation (i), we get
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 215
∴ Magnetic field at the centre of loop is zero.

Question 114.
(a) Show how Biot-Savart law can be alternatively expressed in the form of Ampere’s circuital law. Use this law to obtain the expression for the magnetic field inside a solenoid of length ‘l’, cross-sectional area ‘A’ having ‘N’ closely wound turns and carrying a steady current ‘I’.
Draw the magnetic field lines of a finite solenoid carrying current I.
(b) A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
Find the magnitude and direction of the magnetic field which should be set up in (Comptt. Delhi 2011)
Answer:
(a) (i) Bio-Savart law and Ampere’s Circuital law.

According to Biot-Savart’s law, “magnetic field acting at a particular point due to current carrying element is proportional to the division of cross product of current element and position vector of point where the field is to be calculated from the current element to the cube of the distance between current element and the point where the field is to be calculated”.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 157
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 158
Magnetic field on the axis of circular current loop :
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 159
As in a special case we may obtain the field at the centre of the loop. Here x = 0, and we obtain
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 160
In a current loop, both the opposite faces behave as opposite poles, making it a magnetic dipole. One side of the current carrying coil behaves like the N-pole and the other side as the S-pole of a magnet.

(b) Biot-Savart law can be expressed as Ampere’s Circuital law by considering the surface to be made up of a large number of loops. The sum of the tangential components of the magnetic field multiplied by the length of all such elements leads to integral. Ampere’s circuital law states that this’ integral is equal to p0 times the total current passing through that surface, i.e.,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 211

(ii) Expression for magnetic field inside a solenoid. Let ‘n’ be the number of turns per unit length. Then total number of turns in the length ‘h’ is nh.
Hence, total enclosed current = nhl
Using Ampere’s circuital law,
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 216
(iii) Magnetic field lines of a finite solenoid
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 217
(b) As per the figure given, magnetic field must be vertically inwards, to make tension zero.
Therefore, force on current carrying conductor and the weight of conductor are equal and opposite; and balance each other.
Important Questions for Class 12 Physics Chapter 4 Moving Charges and Magnetism Class 12 Important Questions 218
Direction : Perpendicular to the direction of

Important Questions for Class 12 Physics

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Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry

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Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry

Coordinate Geometry Class 10 Important Questions Very Short Answer (1 Mark)

Question 1.
Find the distance of the point (-3, 4) from the x-axis. (2012OD)
Solution:
B(-3, 0), A (-3, 4)
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 1

Question 2.
If the points A(x, 2), B(-3, 4) and C(7, -5) are collinear, then find the value of x. (2014D)
Solution:
When the points are collinear,
x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0
x(-4 – (-5)) + (-3)(-5 – 2) + 7(2 – (-4)) = 0
x(1) + 21 + 42 = 0
x + 63 = 0 ∴ x = -63

Question 3.
For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P.? (2016OD)
Solution:
As we know, a2 – a1 = a3 – a2
2k – 1 – (k + 9) = 2k + 7 – (2k – 1)
2k – 1 – k – 9 = 2k + 7 – 2k + 1
k – 10 = 8 ∴ k = 8 + 10 = 18

Question 4.
In which quadrant the point P that divides the line segment joining the points A(2, -5) and B(5,2) in the ratio 2 : 3 lies? (2011D)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 2

Question 5.
ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). Calculate the length of one of its diagonals. (2014OD)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 3
AB = 4 units
BC = 3 units
AC2 = AB2 + BC2 …[Pythagoras’ theorem
= (4)2 + (3)2
= 16 + 9 = 25
∴ AC = 5 cm

Question 6.
In the figure, calculate the area of triagle ABC (in sq. units). (2013D)
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 4
Solution:
Area of ∆ABC = \frac{1}{2} × base × corr, altitude
= \frac{1}{2} × 5 × 3 = 7.5 sq.units

Coordinate Geometry Class 10 Important Questions Short Answer-I (2 Marks)

Question 7.
Find a relation between x and y such that the point P(x, y) is equidistant from the points A (2, 5) and B (-3, 7). (2011D)
Solution:
Let P (x, y) be equidistant from the points A (2, 5) and B (-3, 7).
∴ AP = BP …[Given
AP2 = BP2 …[Squaring both sides
(x – 2)2 + (y – 5)2 = (x + 3)2 + (y – 7)2
⇒ x2 – 4x + 4 + y2 – 10y + 25
⇒ x2 + 6x + 9 + y2 – 14y + 49
⇒ -4x – 10y – 6x + 14y = 9 +49 – 4 – 25
⇒ -10x + 4y = 29
∴ 10x + 29 = 4y is the required relation.

Question 8.
Find the relation between x and y if the points A(x, y), B(-5, 7) and C(-4, 5) are collinear. (2015OD)
Solution: When points are collinear,
∴ Area of ∆ABC = 0
= (x1 (y2 – y3) + x2(y3 – y1) + x3(y1 – y2)) = 0
= x (7 – 5) – 5 (5 – y) -4 (y – 7) = 0
= 2x – 25 + 5y – 4y + 28 = 0
∴ 2x + y + 3 = 0 is the required relation.

Question 9.
Find the ratio in which the point P\left(\frac{3}{4}, \frac{5}{12}\right) divides the line segment joining the points A\left(\frac{1}{2}, \frac{3}{2}\right) and B(2, -5). (2015D)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 5
Let P divide AB in the ratio of K : 1.
Applying section formula,
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 6
∴ Required ratio = 1 : 5

Question 10.
Find the ratio in which y-axis divides the line segment joining the points A(5, -6), and B(-1, -4). Also find the coordinates of the point of division. (2016D)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 7
Let AC: CB = m : n = k : 1.
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 8

Question 11.
Let P and Q be the points of trisection of the line segment joining the points A (2, -2) and B (-7, 4) such that P is nearer to A. Find the coordinates of P and Question (2016D)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 9
AP : PB = 1 : 2
Coordinates of P are:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 10

Question 12.
Three vertices of a parallelogram taken in order are (-1, 0), (3, 1) and (2, 2) respectively. Find the coordinates of fourth vertex. (2011D)
Solution:
Let A(-1, 0), B(3, 1), C(2, 2) and D(x, y) be the vertices of a parallelogram ABCD taken in order. Since, the diagonals of a parallelogram bisect each other.
∴ Coordinates of the mid-point of AC = Coordinates of the mid-point of BD
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 11
Hence, coordinates of the fourth vertex, D(-2, 1).

Question 13.
A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, -5) is the mid point of PQ, then find the coordinates of P and Question (2017OD)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 12
We know that at y-axis coordinates of points are (0, y) and at x-axis (x, 0).
Let P(0, b) and Q(a,0)
Mid point of PQ = (2, -5) …[Given
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 13

Question 14.
Determine the ratio in which the line 3x + y – 9 = 0 divides the segment joining the points (1, 3) and (2, 7). (2011D)
Solution:
Suppose the line 3x + y – 9 = 0 divides the line segment joining A(1, 3) and B(2, 7) in the ratio k : 1 at point C.
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 14
⇒ 6k + 3 + 7k + 3 – 9k – 9 = 0
⇒ 4k – 3 = 0
⇒ 4k = 3 ∴ k = 8
So, the required ratio is 3 : 4 internally.

Question 15.
If A(4, 3), B(-1, y) and C(3, 4) are the vertices of a right triangle ABC, right-angled at A, then find the value of y. (2015OD)
Solution:
We have A(4, 3), B(-1, y) and C(3, 4). In right angled triangle ABC,
(BC)2 = (AB)2 + (AC)…. [Pythagoras theorem
⇒ (-1 – 3)2 + (y – 4)2 = (4 + 1)2 + (3 – y)2 + (4 – 3)2 + (3 – 4)2 …(using distance formula
⇒ (-4)2 + (y2 – 8y + 16)
⇒ (5)2 + (9 – 6y + y2) + (1)2 + (-1)2
⇒ y2 – 8y + 32 = y2 – 6y + 36 = 0
⇒ -8y + 6y + 32 – 36
⇒ -2y – 4 = 0 ⇒ -2y = 4
∴ y = -2

Question 16.
If A(5, 2), B(2, -2) and C(-2, t) are the vertices of a right angled triangle with ∠B = 90°, then find the value of t. (2015D)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 15
ABC is a right angled triangle,
∴ AC2 = BC2 + AB2 …(i)… (Pythagoras theorem
Using distance formula,
AB2 = (5 – 2)2 + (2 + 2)
= 25
BC2 = (2 + 2)2 + (t + 2)2
= 16 + (t + 2)2
AC2 = (5 + 2)2 + (2 – t)2
= 49 + (2 – t)2
Putting values of AB2, AC2 and BC2 in equation (i), we get
49 + (2 – t)2 = 16 + (t + 2)2 + 25
∴ 49 + (2 – t)2 = 41 + (t + 2)2
⇒ (t + 2)2 – (2 – t)2 = 8
⇒ (t2 + 4 + 4t – 4 – t2 + 4t) = 8
8t = 8 ⇒ t = 1

Question 17.
Find the area of the triangle whose vertices are (1, 2), (3, 7) and (5, 3).(2011OD)
Solution:
Area of Triangle
= \frac{1}{2} [x1(y2 – y3) + x2(y3 – y1) + x3 (y1 – y2)]
= \frac{1}{2}(1(7 – 3) + 3(3 – 2) + 5(2 – 7)]
= \frac{1}{2}[4 + 3 + 5(-5)] = \frac{18}{2} = 9 sq. unit

Question 18.
The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B. Find the value of p. (2015OD)
Solution:
Similar to Question 16, Page 102.

Question 19.
Prove that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle. (2013D)
Solution:
Let A (7, 10), B(-2, 5), C(3, -4) be the vertices of a triangle.
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 16
… [By converse of Pythagoras theorem
∆ABC is an isosceles right angled triangle. …(ii) From (i) & (ii), Points A, B, C are the vertices of an isosceles right triangle.

Coordinate Geometry Class 10 Important Questions Short Answer-II (3 Marks)

Question 20.
Find that value(s) of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units. (2011D)
Solution:
PQ = 10 …Given
PQ2 = 102 = 100 … [Squaring both sides
(9 – x)2 + (10 – 4)2 = 100…(using distance formula
(9 – x)2 + 36 = 100
(9 – x)2 = 100 – 36 = 64
(9 – x) = ± 8 …[Taking square-root on both sides
9 – x = 8 or 9 – x = -8
9 – 8 = x or 9+ 8 = x
x = 1 or x = 17

Question 21.
Find the value of y for which the distance between the points A (3,-1) and B (11, y) is 10 units. (2011OD)
Solution:
AB = 10 units … [Given
AB2 = 102 = 100 … [Squaring both sides
(11 – 3)2 + (y + 1)2 = 100
82 + (y + 1)2 = 100
(y + 1)2 = 100 – 64 = 36
y + 1 = ±6 … [Taking square-root on both sides
y = -1 ± 6 ∴ y = -7 or 5

Question 22.
The point A(3, y) is equidistant from the points P(6, 5) and Q(0, -3). Find the value of y. (2011D)
Solution:
PA = QA …[Given
PA2 = QA2 … [Squaring both sides
(3 – 6)2 + (y – 5)2 = (3 – 0)2 + (y + 3)2
9 + (y – 5)2 = 9 + (y + 3)2
(y – 5)2 = (y + 3)2
y – 5 = ±(y + 3) … [Taking sq. root of both sides
y – 5 = y + 3 y – 5 = -y – 3
0 = 8 … which is not possible ∴ y = 1

Question 23.
If a point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), then find the value of p. (2012D)
Solution:
Similar to Question 24, Page 103.

Question 24.
Find the value of k, if the point P(2, 4) is equidistant from the points A(5, k) and B(k, 7). (2012OD)
Solution:
Let P(2, 4), A(5, k) and B(k, 7).
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 17
PA = PB …[Given
PA2 = PB2 … [Squaring both sides
(5 – 2)2 + (k – 4)2 = (k – 2)2 + (7 – 4)2
9 + (k – 4)2 – (k – 2)2 = 9
(k – 4 + k – 2) (k – 4 – k + 2) = 0
(2k – 6)(-2) = 0
2k – 6 = 0
2k = 6 ∴ k = \frac{6}{2} = 3

Question 25.
If the point P(k – 1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k. (2014OD)
Solution:
PA = PB …Given
PA2 = PB2 … [Squaring both sides
⇒ (k – 1 – 3)2 + (2 – k)2 = (k – 1 – k)2 + (2 – 5)2
⇒ (k – 4)2 + (2 – k)2 = (-1)2 + (-3)2
k2 – 8k + 16 + 4 + k2 – 4k = 1 + 9
2k2 – 12k + 20 – 10 = 0
2k2 – 12k + 10 = 0
⇒ k2 – 6k + 5 = 0 …[Dividing by
⇒ k2 – 5k – k + 5 = 0
⇒ k(k – 5) – 1(k – 5) = 0
⇒ (k – 5) (k – 1) = 0
⇒ k – 5 = 0 or k – 1 = 0
∴ k = 5 or k = 1

Question 26.
Find a point P on the y-axis which is equidistant from the points A(4, 8) and B(-6, 6). Also find the distance AP. (2014OD)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 18
Let P(0, y) be any point on y-axis.
PA = PB … [Given
PA2 = PB2 … [Squaring both sides
⇒ (4 – 0)2 + (8 – y)2 = (-6 – 0)2 + (6 – y)2
16 + 64 – 16y + y2 = 36 + 36 + y2 – 12y
80 – 72 = -12y + 16y
8 = 4y ⇒ y = 2
∴ Point P (0, 2).
Now, AP = \sqrt{(4-0)^{2}+(8-2)^{2}}
∴ Distance, AP = \sqrt{16+36}=\sqrt{52}

Question 27.
If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b), prove that bx = ay. (2016OD)
Solution:
PA = PB … [Given
PA2 = PB2 … [Squaring both sides
⇒ [(a + b) – x]2 + [(b a) – y)2 = [(a – b) – x]2 + [(a + b) – y]2
⇒ (a + b)2 + x2 – 2(a + b)x + (b – a)2 + y2 – 2(b – a)y = (a – b)2 + x2 – 2(a – b)x + (a + b)2 + y2 – 2(a + b)y …[∵ (a – b) 2 = (b – a)2
⇒ -2(a + b)x + 2(a – b)x = -2(a + b)y + 2(b – a)y
⇒ 2x(-a – b + a – b) = 2y(-a – b + b – a)
⇒ -2bx = – 2ay
⇒ bx = ay (Hence proved)

Question 28.
If the point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. Also find the length of AB. (2014D)
Solution:
AB = AC … [Given
∴ AB2 = AC2 …[Squaring both sides
⇒ (3 – 0)2 + (p – 2)2= (p – 0)2 + (5 – 2)2
⇒ 9+ (p – 2)2 = p2 + 9
⇒ p2 – 4p + 4 – p2 = 0
⇒ -4p + 4 = 0
⇒ -4p = -4 ⇒ p = 1
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 19

Question 29.
If the point P(2, 2) is equidistant from the points A(-2, k) and B(-2k, -3), find k. Also find the length of AP. (2014D)
Solution:
PA = PB … [Given
PA2 = PB2 … [Squaring both sides
⇒ (2 + 2)2 + (2 – k)2 = (2 + 2k)2 + (2 + 3)2
16 + 4 + k2 – 4k = 4 + 4k2 + 8k + 25
4k2 + 8k + 25 – k2 + 4k – 16 = 0
3k2 + 12k + 9 = 0
⇒ k2 + 4k + 3 = 0 …[Dividing by 3
k2 + 3k + k + 3 = 0
k(k + 3) + 10k + 3) = 0
(k + 1) (k + 3) = 0
k + 1 = 0 ork + 3 = 0
i k = -1 or k = -3
AP = \sqrt{(2+2)^{2}+(2-k)^{2}}
When k = -1
AP =\sqrt{16+(2+1)^{2}}=\sqrt{16+9} = 5 units
When k = -3
AP = \sqrt{16+(2+3)^{2}}=\sqrt{16+25} units

Question 30.
Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle. (2016OD)
Solution:
Similar to Question 19, Page 102.

Question 31.
Prove that the points A(0, -1), B(-2, 3), C(6, 7) and D(8, 3) are the vertices of a rectangle ABCD. (2013D)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 20
Opposite sides BC = AD = 4 \sqrt{5}
Diag. AC = BD = 10
∴ ABCD is a rectangle.
… [∵ Opp. sides are equal & diagonals are also equal

Question 32.
Show that the points (-2, 3), (8, 3) and (6, 7) are the vertices of a right triangle. (2013D)
Solution:
Let A (-2, 3), B(8,3), C(6, 7).
(AB)2 = (8 + 2)2 + (3 – 3)2 = 102 + 02 = 100
(BC)2 = (6 – 8)2 + (7 – 3)2 = (-2)2 + 42 = 20
(AC)2 = (6 + 2)2 + (7 – 3)2 = 82 + 42 = 80
Now, (BC)2 + (AC)2= 20 + 80 = 100 = (AB)2
…[By converse of Pythagoras’ theorem
Therefore, Points A, B, C are the vertices of a right triangle.

Question 33.
Prove that the points A(2, 3), B(-2, 2), C(-1, -2) and D(3, -1) are the vertices of a square ABCD. (2013OD)
Solution:
Given: A(2, 3), B(-2, 2), C(-1, -2), D(3, -1)
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 21
AB = BC = CD = DA …[All four sides are equal
AC = BD …[Also diagonals are equal
∴ ABCD is a Square.

Question 34.
Prove that the points A(2, -1), B(3, 4), C(-2, 3) and D(-3, -2) are the vertices of a rhombus ABCD. Is ABCD a square? (2013OD)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 22
Given: A(2, -1), B(3, 4), C(-2,3), D(-3, -2)
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 23

Question 35.
Prove that the diagonals of a rectangle ABCD, with vertices A(2, -1), B(5, -1), C(5, 6) & D(2,6), are equal and bisect each other. (2014OD)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 24
∴ Diagonals also bisect each other.

Question 36.
Find that value of k for which point (0, 2), is equidistant from two points (3, k) and (k, 5). (2013OD)
Solution:
Similar to Question 29, Page 104.

Question 37.
If the points A(-2, 1), B(a, b) and C(4, -1) are collinear and a – b = 1, find the values of a and (2014D)
Solution:
Since A(-2, 1), B(a, b) and C(4, -1) are collinear.
∴ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0
⇒ -2[b – (-1)] + a(-1 – 1) + 4(1 – b) = 0
⇒ -2b – 2 – 2a + 4 – 4b = 0
⇒ -2a – 6b = -2
⇒ a + 3b = 1 … [Dividing by (-2)
⇒ a = 1 – 3
We have, a – b = 1 …[Given
(1 – 3b) – b = 1 …[From (i)
1 – 3b – b = 1
-4b = 1 – 1 = 0 ∴ b = \frac{0}{-4} = 0
From (i), a = 1 – 3(0) = 1 – 0 = 1
∴ a = 1, b = 0

Question 38.
If the points A(-1, -4), B(b, c) and C(5, -1) are collinear and 2b + c = 4, find the values of b and c. (2014D)
Solution:
A(-1, -4), B(b, c), C(5, -1) are collinear.
∴ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0
⇒ -1(c + 1) + b(-1 + 4) + 5(-4 – c) = 0
⇒ -c – 1 – b + 4b – 20 – 5c = 0
⇒ 3b – 6c = 21
⇒ – b – 2c = 7 …[Dividing by 3
⇒ b = 7 + 2c
We have, 2b + c = 4 … [Given
2(7 + 2c) + c = 4 … [From (i)
⇒ 14 + 4c + c = 4
⇒ 5c = 4 – 14 = -10 ⇒ c = -2
⇒ b = 7 + 2(-2) = 3 … [From (i)
∴ b = 3, c = -2

Question 39.
Find the area of the triangle ABC with A(1, 4) and mid-points of sides through A being (2, -1) and (0, -1). (2015D)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 25

Question 40.
Find the area of the triangle PQR with Q(3, 2) and the mid-points of the sides through Q being (2, -1) and (1, 2). (2015D)
Solution:
Let P(x, y), R(z, t).
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 26

Question 41.
Find the ratio in which the y-axis divides the line segment joining the points (-4, -6) and (10, 12). Also find the coordinates of the point of division. (2013D)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 27

Question 42.
Find the ratio in which point P(-1, y) lying on the line segment joining points A(-3, 10) and B(6, -8) divides it. Also find the value of y. (20130D)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 28

Question 43.
In what ratio does the point (14.4) divide the line segment joining the points P(2, -2) and Q(3, 7)? Also find the value of y. (2017OD)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 29

Question 44.
Find the coordinates of a point P on the line segment joining A(1, 2) and B(6, 7) such that AP = AB. (2015OD)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 30

Question 45.
Points A(-1, y) and B(5, 7) lie on a circle with centre 0(2, -3y). Find the values of y. Hence find the radius of the circle. (2014D)
Solution:
Join OA and OB. …[radii of a circle
∴ OA = OB OA2 = OB2 …[Squaring both sides
⇒ (2 + 1)2 + (-3y – y)2 = (5 – 2)2 + (7 + 3y)2
⇒ 9+ (-4y)2 = 9 + (7 + 3y)2
⇒ 16y2 = 49 + 42y + 9y2
⇒ 16y2 – 9y2 – 42y – 49 = 0
⇒ 7y2 – 42y – 49 = 0
⇒ y2 – 6y – 7 = 0 …[Dividing both sides by 7
⇒ y2 – 7y + y – 7 = 0
⇒ y(y – 7) + 1(y – 7) = 0
⇒ (y – 7) (y + 1) = 0
y – 7 = 0 or y + 1 = 0
y = 7 or y = -1
(i) Taking y = 7
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 31

Question 46.
If the points P(-3, 9), Q(a, b) and R(4, -5) are collinear and a + b = 1, find the values of a and b. (2014D)
Solution:
Similar to Question 37, Page 105.

Question 47.
The area of a triangle is 5 sq. units. Two of its vertices are (2, 1) and (3,-2). If the third vertex is \left(\frac{7}{2}, y\right), find the value of y. (2017D)
Solution:
Given: Area of ∆ = 5 sq. units
Vertices: (2, 1), (3, -2) and \left(\frac{7}{2}, y\right)
Area of ∆
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 32

Question 48.
Find the ratio in which the line segment joining the points A(3, -3) and B(-2, 7) is divided by x-axis. Also find the coordinates of the point of division. (2014OD)
Solution:
Similar to Question 41, Page 106.

Question 49.
In the Figure, ABC is a triangle coordinates of whose vertex A are (0, -1). D and E respectively are the mid-points of the sides AB and AC and their coordinates are (1, 0) and (0, 1) respectively. If F is the midpoint of BC, find the areas of ∆ABC and ∆DEF. (2016D)
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 33
Solution:
Let B (p, q), C (r, s)
and F(x, y)
Mid-point of AB = Coordinates of D
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 34
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 35
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 36

Coordinate Geometry Class 10 Important Questions Long Answer (4 Marks)

Question 50.
If the points A(x, y), B(3, 6) and C(-3, 4) are collinear, show that x – 3y + 15 = 0. (2012OD)
Solution:
Pts. A(x, y), B(3, 6), C(-3, 4) are collinear.
∴ Area of ∆ = 0
As area of ∆
= \frac{1}{2}[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
∴Area of ∆ABC
= x(6 – 4) + 3(4 – y) + (-3) (y – 6) = 0
= 2x + 12 – 3y – 3y + 18 = 0
= 2x – 6y + 30 = 0
∴ x – 3y + 15 = 0 … [Dividing both sides by 2

Question 51.
Find the values of k for which the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear. (2015OD)
Solution:
A (k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k).
When points are collinear, area of ∆ is 0.
∴ Area of triangle = 0
⇒ [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0
⇒ [(k + 1) (2k + 3 – 5k) + 3k (5k – 2k) + (5k – 1) (2k – 2k – 3)] = 0
⇒ [(k + 1) (3 – 3k) + 3k(3k) + (5k – 1)(-3)] = 0
⇒ [3k – 3k2 + 3 – 3k + 9k2 – 15k + 3) = 0
⇒ 6k2 – 15k + 6 = 0
⇒ 2k2 – 5k + 2 = 0
⇒ 2k2 – 4k – 1k + 2 = 0
⇒ 2k(k – 2) – 1(k – 2) = 0
⇒ (k – 2)(2k – 1) = 0
⇒ k – 2 = 0 or 2k – 1 = 0
⇒ k = 2 or k = \frac{1}{2}
We get, k = 2, \frac{1}{2}

Question 52.
If two vertices of an equilateral triangle are (3, 0) and (6, 0), find the third vertex. (2011D)
Solution:
Let A (3,0), B (6, 0), C (x, y).
∴ ∆ABC is an equilateral
∴ AB = BC = AC
AB2 = BC2 = AC2 …[Squaring throughout
(6 – 3)2 + (0 – 0)2 = (x – 6)2 + (1 – 0)2 = (x – 3)2 + (y – 0)2
9 = x2 – 12x + 36 + y2 = x2 – 6x + 9 + y2
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 37

Question 53.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5). (2011D)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 38

Question 54.
If the vertices of a triangle are (1, -3), (4, p) and (-9, 7) and its area is 15 sq. units, find the value(s) of p. (2012D)
Solution:
Area of ∆ = 15 sq. units
\frac{1}{2} [1(p – 7) + 4(7 + 3)) + (-9)(-3 – p)] = ±15
p – 7 + 40 + 27 + 9p = ±30
10p + 60 = ± 30
10p = -60 ± 30
p = \frac{-60 \pm 30}{10}
∴ Taking +ve sign, p = \frac{-60+30}{10}=\frac{-30}{10} = -3
Taking -ve sign, p = \frac{-60-30}{10}=\frac{-90}{10} = -9

Question 55.
For the triangle ABC formed by the points A(4, -6), B(3,-2) and C(5, 2), verify that median divides the triangle into two triangles of equal area. (2013OD)
Solution:
Let A(4, -6), B(3, -2) and C(5, 2) be the vertices of ∆ABC.
Since AD is the median
∴ D is the mid-point of BC.
⇒ D\left(\frac{3+5}{2}, \frac{-2+2}{2}\right) ⇒ D(4,0)
Area of ∆ABD
= \frac{1}{2} [4(-2 – 0) + 3(0 + 6) + 4(-6 + 2)]
= \frac{1}{2} [-8 + 18 – 16) = \frac{1}{2} [-6] = -3
But area of A cannot be negative.
∴ ar(∆ABD) = 3 sq.units …(i)
Area of ∆ADC
= \frac{1}{2} [4(0 – 2) + 4(2 + 6) + 5(-6 – 0)]
= \frac{1}{2}(-8 + 32 – 30] = \frac{1}{2} [-6] = -3
But area of ∆ cannot be negative.
∴ ar(∆ADC) = 3 sq.units
From (i) and (ii),
∴ Median AD of AABC divides it into two ∆s of equal area.

Question 56.
If A(4, 2), B(7,6) and C(1, 4) are the vertices of a AABC and AD is its median, prove that the median AD divides AABC into two triangles of equal areas. (2014OD)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 39
Area of ∆ABD
= \frac{1}{2} (x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= \frac{1}{2}[4(6 – 5) + 7(5 – 2) + 4(2 – 6))
= \frac{1}{2}(4 + 21 – 16) = \frac{9}{2} sq.units …(i)
Area of ∆ADC
= \frac{1}{2} [4(5 – 4) + 4(4 – 2) + 162 – 5)]
= \frac{1}{2}(4 + 8 – 3) = \frac{9}{2}sq.units
From (i) and (ii),
Area of ∆ABD = Area of ∆ADC
∴ Median AD divides ∆ABC into two triangles of equal area.

Question 57.
Find the values of k so that the area of the triangle with vertices (k + 1, 1), (4, -3) and (7, -k) is 6 sq. units. (2015OD)
Solution:
Let A(k + 1, 1), B(4, -3) and C(7, -k).
We have, Area of ∆ABC = 6 … [Given
6 = \frac{1}{2} [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
6 = \frac{1}{2}[(k + 1)(-3 + k) + 4(-k – 1) + 7(1 + 3)]
12 = (-3k + k2 – 3 + k – 4k – 4 + 28]
12 = [k2 – 6k + 21]
⇒ k2 – 6k + 21 – 12 = 0
⇒ k2 – 6k + 9 = 0
⇒ k2 – 3k – 3k + 9 = 0
⇒ k(k – 3) – 3(k – 3) = 0 =
⇒ (k – 3) (k – 3) = 0
⇒ k – 3 = 0 or k – 3 = 0
⇒ k = 3 or k = 3
Solving to get k = 3.

Question 58.
Prove that the area of a triangle with vertices (t, t – 2), (t + 2, t + 2) and (t + 3, t) is independent of t. (2016D)
Solution:
Let A(t, t – 2), B(t + 2, + + 2), C(t + 3, t).
Area of ∆ABC
= \frac{1}{2}[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= \frac{1}{2} [t(t + 2 – t) + (t + 2)(t – (t – 2)) +(t + 3)((t – 2) – (t + 2))]
= \frac{1}{2} [t(2) + (t + 2)(2) + (t + 3)(-4)]
= \frac{1}{2} (2+ + 2+ + 4 – 46 – 12] = \frac{1}{2} [-8] = -4
Area of ∆ is always positive.
∴ Area of ∆ = 4 sq. units, which is independent of t.

Question 59.
If the centroid of ∆ABC, in which A(a, b), B(b, c), C(c, a) is at the origin, then calculate the value of (a3 + b3 + c). (2012OD)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 40
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 41
a + b + c = 0
If a + b + c = 0
then, as we know
a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ac) ∴ a3 + b3 + c3 – 3abc = 0 … [Since a + b + c = 0
∴ a3 + b3 + c3 = 3abc …(Hence proved)

Question 60.
If the area of triangle ABC formed by A(x, y), B(1, 2) and C(2, 1) is 6 square units, then prove that x + y = 15 or x + y = -9. (2013D)
Solution:
Let A(x, y), B(1, 2), C(2, 1).
Area of ∆ABC = 6 sq. units …[Given
As \frac{1}{2} [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 6
∴ x(2 – 1) + 1(1 – y) + 2(y – 2) = ±12
x + 1 – y + 2y – 4 = ±12
Taking +ve sign
x + y = 12 + 4 – 1
∴ x + y = 15

Taking -ve sign
x + y = -12 + 4 – 1
∴ x + y = -9

Question 61.
In the given figure, the vertices of ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line segment DE is drawn to intersect the sides AB and AC at D and E respectively such that \frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{3}. Calculate the area of ∆ADE and compare it with area of ∆ABC. (2016OD)
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 42
Solution:
AD : AB = 1 : 3; AE : AC = 1 : 3
∴AD : DB = 1 : 2; AE : EC = 1 : 2
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 43
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 44

Question 62.
Point P(x, 4) lies on the line segment joining the points A(-5, 8) and B(4, -10). Find the ratio in which point P divides the line segment AB. Also find the value of x. (2011OD)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 45

Question 63.
A point P divides the line segment joining the points A(3, -5) and B(-4, 8) such that \frac{\mathbf{A P}}{\mathbf{P B}}=\frac{\mathbf{K}}{\mathbf{1}}. If P lies on the line x + y = 0, then find the value of K. (2012D)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 46
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 47

Question 64.
Find the ratio in which the point P(x, 2) divides the line segment joining the points A(12, 5) and B(4, -3). Also, find the value of x. (2014D)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 48

Question 65.
If (3, 3), (6, y), (x, 7) and (5, 6) are the vertices of a parallelogram taken in order, find the values of x and y. (2011D)
Solution:
Let A (3, 3), B (6, y), C (x, 7) and D (5, 6).
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 49

Question 66.
Find the area of the quadrilateral ABCD whose vertices are A(-3, -1), B(-2, -4), C(4, -1) and D(3, 4). (2011D, 2012OD)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 50
Area of ∆ = \frac{1}{2}[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
∴ Area of ∆ABC
= \frac{1}{2}[(-3)(4 + 1) + (-2){-1 – (-1)} +4{-1 – (- 4}}]
= \frac{1}{2}[9+ 0 + 12] = 21 sq. units …(i)
∴ Area of ∆ACD
= \frac{1}{2}[-3(-1 – 4) + 4{(4 – (-1)} + 34{-1 – (-1)}]
= \frac{1}{2} [15 + 20 + 0]
= \frac{1}{2}[15 + 20] = \frac{35}{2} sq. units
∴ ar(quad. ABCD) = ar (∆ABC) + ar(∆ACD)
= \frac{21}{2}+\frac{35}{2} …[From (i) & (ii)
= \frac{56}{2} = 28 sq. units

Question 67.
The three vertices of a parallelogram ABCD are A(3, 4), B(-1, -3) and C(-6, 2). Find the coordinates of vertex D and find the area of ABCD. (2013D)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 51
Area of ∆ = \frac{1}{2}[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
∴ Area of ∆ABC
= \frac{1}{2} [3(-3 – 2) + (-1)(2 + 4) + (-6)(+4+3)]
= \frac{1}{2} [-15 – 6 +6] = \frac{-15}{2}
= \frac{15}{2} sq. units …[∵ ar of A cannot be -ve.
Since diagram of a parallelogram divides it into 2 equal areas.
∴ ar(||gm ABCD) = 2(area of ∆ABC) = 2\left(\frac{15}{2}\right)
= 15 sq. units

Question 68.
If the points A(1, -2), B(2, 3), C(-3, 2) and D(-4, -3) are the vertices of parallelogram ABCD, then taking AB as the base, find the height of this parallelogram. (2013OD)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 52
1st method. A(1, -2), B(2, 3), D(-4, -3)
.. Area of ∆ABD
= \frac{1}{2} [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= \frac{1}{2} [1(3 + 3) + 2(-3 + 2) + (-4)(-2 – 3)]
= \frac{1}{2} [6 – 2 + 20) = \frac{24}{2} sq.units
ar(∆ABD) = 12 sq.units
= \frac{1}{2} × AB × DM = 12 sq.units …[Area of ∆ = 1/2 × Base × Altitude
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 53
Area of ∆ABC
= \frac{1}{2} [x2(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= \frac{1}{2}[1(3 – 2) + 2(2 + 2) + (-3)(-2 – 3)]
= \frac{1}{2}[1 + 8 + 15] = 12 sq. unit
Area of ∆ABC = 12
\frac{1}{2} × AB × CP = 12 …[Area of ∆ = 1/2 × Base × Height
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 54
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 55

Question 69.
Find the value of k, if the points P(5, 4), Q(7, k) and R(9, – 2) are collinear. (2011D)
Solution:
Given points are P(5, 4), Q(7, k) and R(9, -2).
x1 (y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0 …[∵ Points are collinear
∴ 5 (k + 2) + 7 (- 2 – 4) + 9 (4 – k) = 0
5k + 10 – 14 – 28 + 36 – 9k = 0
4 = 4k ∴ k=1

Question 70.
If P(2, 4) is equidistant from Q(7, 0) and R(x, 9), find the values of x. Also find the distance PQuestion (2011D)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 56
PQ = PR …[Given
PQ2 = PR2 … [Squaring both sides
∴ (7 – 2)2 + (0 – 4)2 = (x – 2)2 + (9 – 4)2
⇒ 25 + 16 = (x – 2)2 + 25
⇒ 16 = (x – 2)2
⇒ ±4 = x – 2 …[Taking sq. root of both sides
⇒ 2 ± 4 = x
⇒ x = 2 + 4 = 6 or x = 2 – 4 = -2
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 57

Question 71.
Find the point of y-axis which is equidistant from the points (-5, -2) and (3, 2). (2011D)
Solution:
Similar to Question 26, Page 103.

Question 72.
Point M(11, y) lies on the line segment joining the points P(15, 5) and Q(9, 20). Find the ratio in which point M divides the line segment PQuestion Also find the value of y. (2011OD)
Solution:
Similar to Question 62, Page 110.

Question 73.
Find the coordinates of a point P, which lies on the line segment joining the points A(-2, -2) and B(2, -4) such that AP = \frac{3}{7}AB. (2012OD)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 58
AP = \frac{3}{7}AB …(Given)
\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{3}{7}
Let AP = 3K, AB = 7K
BP = AB – AP = 7K – 3K = 4K
\frac{A P}{A B}=\frac{3 K}{4 K}=\frac{3}{4}
Using Section formula, we have
P\left(\frac{6-8}{3+4}, \frac{-12-8}{3+4}\right) \Rightarrow\left(\frac{-2}{7}, \frac{-20}{7}\right)

Question 74.
Find the value of k, for which the points A(6,-1), B(k, -6) and C(0, -7) are collinear. (2012OD)
Solution:
Similar to Question 69, Page 112.

Question 75. Find the value of p, if the points A(1, 2), B(3, p) and C(5, -4) are collinear. (2012OD)
Solution:
Similar to Question 69, Page 112.

Question 76. Find the value of x for which the points (x, -1), (2, 1) and (4, 5) are collinear. (2013D)
Solution:
Similar to Question 69, Page 112.

Question 77.
Find the area of a parallelogram ABCD if three of its vertices are A(2, 4), B(2 + \sqrt{3}, 5) and C(2,6). (2013OD)
Solution:
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 59
Area of ∆ABD
= \frac{1}{2} (x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
= \frac{1}{2} [2(5 – 6) + (2 + \sqrt{3})(6 – 4) + 2(4 – 5))
= \frac{1}{2} (-2 + 4 + 2\sqrt{3} – 2)
= \frac{1}{2} (2\sqrt{3}) = \sqrt{3}
Since diagonal of a ||gm divides it into two equal areas.
Area of ABCD (||gm) = 2(Area of ∆ABC)
= 2\sqrt{3}3 sq. units

Question 78.
If A(-3,5), B(-2, -7), C(1, -8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area. (2014OD0
Solution:
Similar to Question 66, Page 111.

Question 79.
A(4, -6), B(3, -2) and C(5, 2) are the vertices of a ∆ABC and AD is its median. Prove that the median AD divides AABC into two triangles of equal areas. (2014OD)
Solution:
Similar to Question 56, Page 109.

Question 80.
If A(-, 8), B(-3,-4), C(0, -5) and D(5, 6) are the vertices of a quadrilateral ABCD, find its area. (2015D)
Solution:
Similar to Question 66, Page 111.

Question 81.
If P(-5, -3), Q(-4, -6), R(2, -3) and S(1, 2) are the vertices of a quadrilateral PQRS, find its area. (2015D)
Solution:
Similar to Question 66, Page 111.

Question 82.
Find the values of k so that the area of the triangle with vertices (1, -1), (-4, 2k) and (-k, -5) is 24 sq. units. (2015OD)
Solution:
Similar to Question 57, Page 109.

Question 83.
The x-coordinate of a point P is twice its y coordinate. If P is equidistant from Q (2, -5) and R(-3, 6), find the coordinates of P. (2016D)
Solution:
Let the point P be (2k, k), Q(2,-5), R(-3, 6)
Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry 60
PQ = PR …Given
PQ2 = PR2 …[Squaring both sides
(2k – 2)2 + (k + 5)2 = (2k + 3)2 + (k – 6)2 …Given
4k2 + 4 – 8k + k2 + 10k + 25 = 4k2 + 9 + 12k + k2 – 12k + 36
⇒ 2k + 29 = 45
⇒ 2k = 45 – 29
⇒ 2k = 16
⇒ k = 8
Hence coordinates of point P are (16, 8).

Important Questions for Class 10 Maths

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Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions

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Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions

Solutions Class 12 Important Questions Very Short Answer Type

Question 1.
Differentiate between molarity and molality of a solution. (All India 2010)
Answer:
The distinction between molarity and molality.
Molarity : It is the number of moles of solute dissolved in 1 litre of solution. It is temperature dependent.
M = \frac{\omega \times 1000}{\text { mol.mass } \times V}
Molality : It is the number of moles of solute dissolved in 1 kg of the solvent.
m = \frac{\omega \times 1000}{M_{2} \times W}
The relationship between molarity and molality is
m = \frac{\mathrm{M}}{d-\frac{\mathrm{MM}_{2}}{1000}}
When molality = molarity, we get,
1 = \frac{1}{d-\frac{\mathrm{MM}_{2}}{1000}} or d – \frac{\dot{\mathrm{MM}}_{2}}{1000} = 1
∴ d = 1+\frac{\mathrm{MM}_{2}}{1000}
Molarity is temperature dependent while molarity is not.
For very dilute solution, the factor MM2/1000 can be neglected in comparison to 1.
Hence molality will be same to molarity when density d = 1.
Molality is independent of temperature, whereas molarity is a function of temperature because volume depends on temperature and mass does not.

Question 2.
What type of semiconductor is obtained when silicon is doped with arsenic? (Delhi 2010)
Answer:
n-type semiconductor.

Question 3.
What is meant by ‘reverse osmosis’? (All India 2011)
Answer:
If a pressure higher than the osmotic pressure is applied on the solution, the solvent will flow from the solution into the pure solvent through semipermeable membrane. This process is called reverse osmosis (R.O.).

Question 4.
What are isotonic solutions? (Delhi 2014)
Answer:
An isotonic solution is a kind of solution with the same salt concentration as blood and cells. Those solutions which are exerting same osmotic pressure under similar conditions (For example 0.9% NaCl solution by mass volume is Isotonic with human blood).

Question 5.
Some liquids on mixing form ‘azeotropes’. What are ‘azeotropes’? (Delhi 2014)
Answer:
The liquid mixture having a definite composition and boiling like a pure liquid without change in composition is called as azeotrope.

Question 6.
What type of intermolecular attractive interaction exists in the pair of methanol and acetone? (Delhi 2014)
Answer:
Solute-solvent dipolar interactions exist in the pair of methanol and acetone.

Question 7.
Out of BaCl2 and KCl, which one is more effective in causing coagulation of a negatively charged colloidal Sol? Give reason. (Delhi 2015)
Answer:
BaCl2 is more effective in causing coagulation because it has double +ve charge than K+.

Solutions Class 12 Important Questions Short Answer Type – I [SA-I]

Question 8.
Differentiate between molality and molarity of a solution. What is the effect of change in temperature of a solution on its molality and molarity? (Delhi 2009)
Answer:
Distinction between molarity and molality.
Molarity : It is the number of moles of solute dissolved in 1 litre of solution. It is temperature dependent.
M = \frac{\omega \times 1000}{\text { mol.mass } \times V}
Molality : It is the number of moles of solute dissolved in 1 kg of the solvent.
m = \frac{\omega \times 1000}{M_{2} \times W}
The relationship between molarity and molality is
m = \frac{\mathrm{M}}{d-\frac{\mathrm{MM}_{2}}{1000}}
When molality = molarity, we get,
1 = \frac{1}{d-\frac{\mathrm{MM}_{2}}{1000}} or d – \frac{\dot{\mathrm{MM}}_{2}}{1000} = 1
∴ d = 1+\frac{\mathrm{MM}_{2}}{1000}
Molarity is temperature dependent while molarity is not.
For very dilute solution, the factor MM2/1000 can be neglected in comparison to 1.
Hence molality will be same to molarity when density d = 1.
Molality is independent of temperature, whereas molarity is a function of temperature because volume depends on temperature and mass does not.

Question 9.
Non-ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused? Explain with one example for each type. (Delhi 2010)
Answer:
Non-ideal solutions exhibit Negative deviation from Raoult’s law : For any composition of the non-ideal solution, the partial vapour pressure of each component and total vapour pressure of the solution is less than expected from Raoult’s law. Such solutions show negative deviation.
Example : Mixture of CHCl3 and acetone.
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 1
Non-ideal solutions show positive deviations from Raoult’s law on mixing of two volatile components of the solution.
Example : Mixture of acetone and benzene solutions show positive deviation,

Question 10.
Define the terms, ‘osmosis’ and ‘osmotic pressure’.
What is the advantage of using osmotic pressure as compared to other colligative properties for the determination of molar masses of solutes in solutions? (All India 2010)
Answer:
Osmosis : The net spontaneous flow of the solvent molecules from the solvent to the solution or from a less concentrated solution to a more concentrated solution through a semipermeable membrane is called osmosis.
Osmotic pressure : The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semipermeable membrane is called the osmotic pressure.
The osmotic pressure method has the advantage that it uses molarities instead of molalities and it can be measured at room temperature.

Question 11.
A 1.00 molal aqueous solution of trichloroacetic acid (CCl3COOH) is heated to its boiling point. The solution has the boiling point of 100.18°C. Determine the van’t Hoff factor for trichloroacetic acid. (Kb for water = 0.512 K kg mol-1) (Delhi 2012)
Answer:
As ΔTb= iKbm
(100.18 – 100) °C = i × 0.512 K kg mol-1 × 1 m
0.18 K = i × 0.512 K kg mol-1 × 1 m
∴ i = 0.3

Question 12.
Define the following terms :
(i) Mole fraction
(ii) Isotonic solutions
(iii) van’t Hoff factor
(iv) Ideal solution (Delhi 2012)
Answer:
(i) Mole fraction : Mole fraction is the ratio of number of moles of one component to the total number of moles in a mixture.
(ii) Isotonic solution : Two solutions having same osmotic pressure at a given temperature are called Isotonic solutions.
(iii) van’t Hoff factor : van’t Hoff factor is expressed as :
i = \frac{\text { normal molar mass }}{\text { abnormal molar mass }}
(iv) Ideal solution : The solution which obeys Raoult’s law under all conditions is known as an ideal solution.

Question 13.
Explain why aquatic species are more comfortable in cold water rather than in warm water. (Comptt. Delhi 2012)
Answer:
Aquatic species need dissolved oxygen for breathing. As solubility of gases decreases with increase of temperature, less oxygen is available in summer in the lake. Hence the aquatic species feel more comfortable in winter (low temperature) when the solubility of oxygen is higher.

Question 14.
State Raoult’s law. How is it formulated for solutions of non-volatile solutes? (Comptt. Delhi 2012)
Answer:
Raoult’s Law : Raoult’s Law states that “for a solution of volatile liquids, the partial vapour of each component in the solution is directly proportional to its mole fraction”.
Thus for component 1 : p1 = p10 X1
where [p10 is vapour pressure of pure component 1]
For component 2 : p2 = P20 X2
According to Dalton’s law of partial pressure
PTotal = P2 + P2 ⇒ PT = p1 0X1 + p2 0X2
⇒ PT = p1 0(1 – X2) + p2 0X2
⇒ PT = p1 0+ (p2 0 – p1 0)X2

Question 15.
State Henry’s law and mention two of its important applications. (Comptt. All India 2012)
Answer:
Henry’s law : Henry’s law states that “The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution “, Applications of Henry’s law :

  • To increase the solubility of CO2 in soft drinks and soda water, the bottle is sealed under high pressure.
  • To avoid a dangerous medical condition called bends, scuba divers use oxygen diluted with less soluble helium gas.

Question 16.
Why do gases nearly always tend to be less soluble in liquids as the temperature is raised? (Comptt. All India 2012)
Answer:
This is because the dissolution of gas in liquid is an exothermic process. The solubility should decrease with increase in temperature.

Question 17.
18 g of glucose, C6H12O6 (Molar mass – 180 g mol-1) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil? (Kb for water = 0.52 K kg mol-1, boiling point of pure water = 373.15 K) (Delhi 2013)
Answer:
We know that :
Elevation of boiling point ∆Tb
\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}} \times \frac{100 \times \mathrm{K}_{\mathrm{b}}}{\mathrm{wt} . \text { of solvent }}
Given: WB = 18 g
MB = Formula of glucose is C6H12O6
= 6 × 12 + 12 + 6 × 16 = 180
Wt. of solvent = 1 kg or 1000 g,
Kb = 0.52 K kg mol-1
Hence, ∆Tb = \frac{18 g}{180} \times \frac{1000 \times 0.52}{1000 g} = 0.52 K
∴B.P of the solution = 373.15 + 0.052
= 373.202 K

Question 18.
An aqueous solution of sodium chloride freezes below 273 K. Explain the lowering in freezing points of water with the help of a suitable diagram. (Comptt. Delhi 2013)
Answer:
An aqueous solution of sodium chloride freezes below 273 K because vapour pressure g of the solution is g less than that of the pure solvent.
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 2

Question 19.
Derive expression for Raoult’s law when the solute is non-volatile. (Comptt. Delhi 2013)
Answer:
Raoult’s law : Raoult’s law states that “for a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction”.
Thus for component 1 : p1 = p10X1
where [p10 is vapour pressure of pure component 1]
For component 2 : p2 = p20X2
ccording to Dalton’s law of partial pressure
PTotal = P1 + P2 ⇒ PT = p10X1 + p20X2
⇒ PT = p10( 1 – X2) + p20X2
∴ PT = p10 + (p20 – p10)X2

Question 20.
Calculate the mass of compound (molar mass = 256 g mol-1) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K (Kf = 5.12 K kg mol-1). (Delhi 2014)
Answer:
Given : ∆Tf = 0.48 K, W1 = 75g,
M2 = 256 g mol-1 W2 =?
Using formula, W2 = \frac{\mathrm{M}_{2} \times \mathrm{W}_{1} \times \Delta \mathrm{T}_{f}}{1000 \times \mathrm{K}_{f}}
= \frac{256 \times 75 \times 0.48}{1000 \times 5.12} = 1.8 g

Question 21.
Define an ideal solution and write one of its characteristics. (Delhi 2014)
Answer:
Those solutions which are obeying Raoult’s law are called ideal solutions. An ideal solution is a solution in which no volume change and no enthalpy change takes place on mixing the solute and the solvent in any proportion.
Characteristic of an ideal solution :
There will be no change in enthalpy ∆Hmix = 0,
∆Vmix = 0, ∆Pmix = 0

Question 22.
State Henry’s law. What is the effect of temperature on the solubility of a gas in a liquid? (Delhi 2014)
Answer:
Henry’s law : Henry’s law states that, “The solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas in equilibrium with the liquid at that temperature.” Solubility of gas decreases with increase of temperature at the same pressure.

Question 23.
State Raoult’s law for the solution containing volatile components. What is the similarity between Raoult’s law and Henry’s law? (Delhi 2014)
Answer:
Raoult’s law : “In a solution, the vapour pressure of a component at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapour pressure of that component in pure state.”
Similarity between Raoult’s law and Henry’s law is that the partial pressure or vapour pressure of the volatile component (gas) is directly proportional to the mole fraction of that component in the solution.

Question 24.
How is the vapour pressure of a solvent affected when a non-volatile solute is dissolved in it? (Comptt. Delhi 2014)
Answer:
The vapour pressure of a solvent decreases when a non-volatile solute is dissolved in it because some solvent molecules are replaced by the molecules of solute.

Question 25.
Differentiate between molarity and molality of a solution. How can we change molality value of a solution into molarity value? (Comptt. Delhi 2014)
Answer:
The distinction between molarity and molality.
Molarity : It is the number of moles of solute dissolved in 1 litre of solution. It is temperature dependent.
M = \frac{\omega \times 1000}{\text { mol.mass } \times V}
Molality : It is the number of moles of solute dissolved in 1 kg of the solvent.
m = \frac{\omega \times 1000}{M_{2} \times W}
The relationship between molarity and molality is
m = \frac{\mathrm{M}}{d-\frac{\mathrm{MM}_{2}}{1000}}
When molality = molarity, we get,
1 = \frac{1}{d-\frac{\mathrm{MM}_{2}}{1000}} or d – \frac{\dot{\mathrm{MM}}_{2}}{1000} = 1
∴ d = 1+\frac{\mathrm{MM}_{2}}{1000}
Molarity is temperature dependent while molarity is not.
For very dilute solution, the factor MM2/1000 can be neglected in comparison to 1.
Hence molality will be same to molarity when density d = 1.
Molality is independent of temperature, whereas molarity is a function of temperature because volume depends on temperature and mass does not.
Molarity (M) and Molality (m) relationship :
∵ Molarity is M moles of solute present in 1000 mL solution
If density of solution is d g mL-1, then
Mass of solution = 1000 d g
Mass of solute = MM2
(M2 is molar mass of solute)
∴ Mass of solvent = 1000 d – MM2 g
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 3

Question 26.
What is meant by positive deviations from Raoult’s law? Give an example. What is the sign of ∆mixH for positive deviation? (Delhi 2015)
Answer:
In positive deviations, the partial vapour pressure of each component A and B of a solution and the total pressure of the solution is higher than the vapour pressure calculated from Raoult’s law. For example, Water and Ethanol.
In case of positive deviations, ∆mixH > 0 (Positive)

Question 27.
Define azeotropes. What type of azeotrope is formed by positive deviation from Raoult’s law? Given an example. (Delhi) 2015
Answer:
Azeotropes : Liquid mixture which distills without change in compositions are called azeotropic mixtures or Azeotropes.
In positive deviations from Raoult’s law, minimum boiling point azeotropic mixture is formed. For example, 95% ethanol + 5% water.

Question 28.
(i) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult’s law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y?
(ii) What happens when we place the blood cell in water (hypotonic solution)? Give reason. (All India 2015)
Answer:
(i) Volume decreases by mixing X and Y. It shows negative deviations from Raoult’s law. There will be rise in temperature. (∆Hmix < 0)
(ii) Blood cell will swell due to osmosis as water enters the cell.

Question 29.
Define osmotic pressure of a solution. How is the osmotic pressure related to the concentration of a solute in a solution? (Comptt. Delhi 2015)
Answer:
Osmotic pressure : It is the external pressure which is applied on the side solution which is sufficient to prevent the entry of the solvent through semi-permeable membrane.
According to the Boyle-van’t Hoff Law, the osmotic pressure (π) of a dilute solution is directly proportional to its molar concentration provided temperature is constant.
π ∝ C (At constant temperature)
π ∝ CT (At constant concentration)
π = CRT (R = Solution constant)
or, π = \frac{n}{v} \mathrm{RT}

Question 30.
Define the following terms :
(i) Mole fraction (x)
(ii) Molality of a solution (m) (Comptt. All India 2015)
Answer:
(i) Mole fraction : Mole fraction of a constituent is the fraction obtained by dividing number of moles of that constituent by the total number of moles of all the constituents present in the solution. It is denoted by ‘x’.
Example : x1 = \frac{\text { No. of moles of } x_{1}}{\text { Total no. of moles }}=\frac{n x_{1}}{n x_{1}+n x_{2}}

(ii) Molality of a solution : Molality of a solution is defined as the number of moles of the solute dissolved in 1000 grams (1 kg) of the solvent. It is denoted by’m’.
m = \frac{w \times 1000}{\mathrm{M} \times \mathrm{W}}
Where w = Weight of solute in grams
M = Molecular mass of solute
W = Weight of solvent in grams

Question 31.
(i) Gas (A) is more soluble in water than Gas
(B) at the same temperature. Which one of the two gases will have the higher value of KH (Henry’s constant) and why?
(ii) In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes? (All India 2016)
Answer:
(i) Gas (B) will have higher value of KH (Henry’s constant) than Gas (A) at the same temperature because lesser the solubility of a gas in a given solvent, higher will be the value of KH for a gas.
KH = \frac{\text { Partial pressure of gas }}{\text { Mole fraction of gas in the solution }}
(ii) Negative deviations from Raoult’s law show the formation of maximum boiling azeotropes.

Question 32.
What is osmotic pressure? Why it is a colligative property? (Comptt Delhi 2016)
Answer:
The excess pressure applied on solution side to stop the process of osmosis. Because it depends upon the number of solute particles but not on their nature.

Question 33.
Define osmotic pressure. How is osmotic pressure related to the concentration of a solute in a solution? (Comptt. All India 2016)
Answer:
Osmotic pressure is the measure of excess pressure applied on solution side to stop the process of osmosis. Osmotic pressure is directly proportional to the conentration of solute in solution π ∝ c

Question 34.
Define the following terms: (Delhi 2017)
(i) Colligative properties
(ii) Molality (m)
Answer:
(i) Colligative properties. All those properties which depend on the number of solute particles irrespective of the nature of solute are called as colligative properties.
(ii) Molality (m). Number of moles of solute dissolved per kg of the solvent.

Question 35.
Define the following terms:
(i) Abnormal molar mass
(ii) van’t Hoff factor (i)               (Delhi 2017)
Answer:
(i) Abnormal molar mass. If the molar mass calculated by using any of colligative properties tends to be different than theoretically expected molar mass, it is called abnormal molar mass.
(ii) van’t Hoff factor (i). Extent of dissociation or association or ratio of the observed colligative property to calculated colligative property.
i = \frac{\text { Observed colligative property }}{\text { Theoretical colligative property }}

Question 36.
Define the following terms:
(i) Ideal solution
(ii) Molarity (M)      (Delhi 2017)
Answer:
(i) Ideal solution: The solution that obeys Raoults ’ Law over the entire range of concentration.
(ii) Molarity is the number of moles of solute dissolved per litre of solution or
M = \frac{w_{b} \times 1000}{\mathrm{M}_{b} \times \mathrm{Volume}(\mathrm{mL})}

Question 37.
Explain why on addition of 1 mol of glucose to 1 litre of water, the boiling point of water increases. (Comptt. Delhi 2017)
Answer:
(a) Glucose is a non-volatile solute, therefore, addition of glucose to water lowers the vapour pressure of water as a result of which boiling point of water increases.
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 35
i.e. nH2O = 27.78 mol
No. of moles of CO2
\frac{n \mathrm{CO}_{2}}{27.78} = 1.515 × 10-3
i.e. nCO2 = 42.08 × 10-3 moles = 0.042 mol

Solutions Class 12 Important Questions Short Answer Type – II [SA – II]

Question 38.
100 mg of a protein is dissolved in just enough water to make 10.0 mL of solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar mass of the protein?
(R = 0.0821 L atm mol-1 K-1 and 760 mm Hg = 1 atm.) (Delhi & All India 2009)
Answer:
Given :
w = 100 mg = 0.100 = 0.1 g,
V = 10.0 mL = 0.01 L
π = 13.3 mm Hg = \frac{13.3}{760} atm,
T = 25°C = 25 + 273 = 298 K
R = 0.0821 L atm mol-1 K-1, M = ?
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 4
∴ Molar mass, M = 13980.4 g mol-1

Question 39.
Calculate the freezing point depression expected for 0.0711 m aqueous solution of Na2S04. If this solution actually freezes at – 0.320°C, what would be the value of Van’t Hoff factor? (Ky for water is 1.86°C mol-1) (Delhi 2009)
Answer:
Given : Molality, m = 0.0711 m
ΔTf = – 0.320°C Kf = 1.86°C f = ?
Substituting these values in the formula, we get
ΔTf = i Kf m or i = \frac{\Delta \mathrm{T}_{f}}{\mathrm{K}_{f} \mathrm{m}}
or i = \frac{-0.320}{1.86 \times 0.0711}=\frac{-0.320}{0.132246} = -2.4

Question 40.
A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31°C. Determine the molar mass of this compound. (B.P. of pure benzene = 80.10°C and Kb for benzene = 2.53°C kg mol-1)    (Delhi 2010)
Answer:
Given : w2 = 1.25 g, w1 = 99 g
ΔTb = 80.31 – 80.10°C = 0.21°C
Kb = 2.53°C kg mol-1
According to the formula :
M2 = \frac{1000 \mathrm{K}_{b} w_{2}}{w_{1} \Delta \mathrm{T}_{b}}
Substituting these values in the formula, we get
M2 = \frac{1000 \times 2.53 \times 1.25}{99 \times 0.21}=\frac{3162.5}{20.79}
= 152 g mol-1

Question 41.
A solution of glycerol (C3H8O3; molar mass = 92 g mol-1) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42 °C. What mass of glycerol was dissolved to make this solution? Kb for water = 0.512 K kg mol-1. (Delhi 2010)
Answer:
Given : M2 = 92 g mol-1 w1 = 500g
ΔTb = 100.42°C – 100°C = 0.42°C
Kb = 0.512 K kg mol-1
Substituting above values in the formula
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 5

Question 42.
What mass of NaCl (molar mass = 58.5 g mol-1) must be dissolved in 65 g of water to lower the freezing point by 7.5°C? The freezing point depression constant, Kf, for water is 1.86 K kg mol-1. Assume van’t Hoff factor for NaCl is 1.87. (All India 2010)
Answer:
Given : M2 = 58.5 g mol-1 w1 = 65 g
ΔTf = 7.5 °C K, = 1.86 Kf kg mol-1 i = 1.87
Substituting these values in the formula
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 6
∴ Mass of NaCl to be dissolved, w2 = 8.199 g

Question 43.
What mass of ethylene glycol (molar mass = 62.0 g mol-1) must be added to 5.50 kg of water to lower the freezing point of water from 0° C to -10.0° C? (K, for water = 1.86 K kg mol-1)? (All India 2010)
Answer:
According to the formula :
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 7
∴ Mass of ethylene glycol, w2 = 1.833 kg

Question 44.
15 g of an unknown molecular substance was dissolved in 450 g of water. The resulting solution freezes at -0.34° C. What is the molar mass of the substance?
(Kf for water = 1.86 K kg mol-1). (All India 2010)
Answer:
Given : w2 = 15 g, w1 = 450 g
ΔTf = -0.34°C CKf = 1.86 K kg mol-1 M2 =?
Substituting these values in the formula
M2 = \frac{1000 \times \mathrm{K}_{f} w_{2}}{w_{1} \Delta \mathrm{T}_{f}}
∴ M2 = \frac{1000 \times 1.86 \times 15}{450 \times 0.34}
or M2 = \frac{27900}{153} = 182.53
2 153
∴ Molar mass of the substance, M2 = 182.53 g

Question 45.
What mass of NaCl must be dissolved in 65.0 g of water to lower the freezing point of water by 7.5°C? The freezing point depression constant (Kf) for water is 1.86°C/m. Assume van’t Hoff factor for NaCl is 1.87. (Molar mass of NaCl = 58.5 g) (All India 2011)
Answer:
Given : M2 = 58.5 g mol-1 w1 = 65 g
ΔTf = 7.5 °C K, = 1.86 Kf kg mol-1 i = 1.87
Substituting these values in the formula
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 6
∴ Mass of NaCl to be dissolved, w2 = 8.199 g

Question 46.
Calculate the amount of KCl which must beadded to 1 kg of water so that the freezing point is depressed by 2K. (Kf for water = 1.86 K kg mol-1) (Delhi 2012)
Answer:
Since one mol of KCl gives 2 mole particles,
the value of i = 2, ΔTf, = 2K, Kf = 1.86 kg mol-1
Applying equation, ΔTf = iKf
m = \frac{\Delta \mathrm{T}_{f}}{i \mathrm{K}_{f}}=\frac{2}{2 \times 1.86}=\frac{1}{1.86} = 0.54 mol kg-1
∴ 0.54 mole of KC1 should be added to 1 kg of water
Molar mass of KCl = 39 + 35.5 = 74.5 g
∴ Amount of KCl = 0.54 × 74.5 g = 40.05 g

Question 47.
A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42 °C while pure water boils at 100 °C. What mass of glycerol was dissolved to make the solution? (Delhi 2012)
Answer:
ΔTb = (100.42 – 100)°C = 0.42°C or 0.42 K
ΔTb = Kb m
0.42 = 0.512 × \frac{\mathrm{W}_{2}}{92} \times \frac{1000}{500}
W2 = \frac{0.42 \times 92 \times 500}{0.512 \times 1000}=\frac{4.83}{0.128} = 37.7 g
where W2 is the weight of the solute.

Question 48.
15.0 g of an unknown molecular material was dissolved in 450 g of water. The resulting solution was found to freeze at -0.34 °C. What is the molar mass of this material? (Kf for water = 1.86 K kg mol-1) (Delhi 2012)
Answer:
As ΔTf = Kf × m
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 8
∴ Molar mass of material, M = 182 g mol-1

Question 49.
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further 18 g of water is added to this solution. The new vapour pressure becomes 2.9 kPa at 298 K. Calculate
(i) the molecular mass of solute and
(ii) vapour pressure of water at 298 K. (Comptt. Delhi 2012)
Answer:
For a very dilute solution
Since \frac{\mathrm{P}^{\circ}-\mathrm{P}}{\mathrm{P}^{\circ}}=\frac{\mathrm{W}_{\mathrm{B}} \times \mathrm{M}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 9
∴ Molecular mass, MB = 34 g/mol

Question 50.
If N2 gas is bubbled through water at 293K, how many millimoles of N2 gas would dissolve in 1 litre of water? Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henry’s law constant for N2 at 293K is 76.48 k bar. (Comptt. All India 2012)
Answer:
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 10
∴ In millimoles 7.16 × 10-4 × 1000 = 0.716 m mol

Question 51.
The partial pressure of ethane over a saturated solution containing 6.56 × 10-2 g of ethane is 1 bar. If the solution contains 5.0 × 10-2 g of ethane, then what will be the partial pressure of the gas? (Comptt. All India 2012)
Answer:
Applying the Henry’s law, m = KH × p
In first case, 6.56 × 10-2 = KH × 1
KH = 6.56 × 10-2 g bar-1
Putting the value of KH in the second case, we get
5 × 10-2 g = 6.56 × 10-2 g bar-1 × p
∴ p = \frac{5 \times 10^{-2}}{6.56 \times 10^{-2} \mathrm{g} \mathrm{bar}} = 0.762 bar

Question 52.
Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10-2 g of K2SO4 in 2L of water at 25° C, assuming that it is completely dissociated.
(R = 0.0821 L atm K-1 mol-1, Molar mass of K2SO4 = 174 g mol-1). (Delhi 2013)
Answer:
We know, π = iCRT ⇒ π = \frac{i \mathrm{n} \mathrm{RT}}{\mathrm{V}}
⇒ π = i × \frac{w}{M} \times \frac{1}{V} R T
Given : w = 2.5 × 10-2 g = 0.025 g
V = 2L, T = 25°C = 298 K
M = K2SO4 = 2 × 39 + 32 + 4 × 16 = 174 g mol-1
K2SO4 = dissociates completely as K2SO4 → 2K+ + SO42-
∴ Ions produced = 3 i.e., i = 3
Hence, π = 3 × \frac{0.025 \mathrm{g}}{174 \mathrm{g} \mathrm{mol}^{-1}} \times \frac{1}{2 \mathrm{L}} × 0.0821 L atm K-1 mol-1 × 298 K
∴ π = 527 × 10-3 atm

Question 53.
The partial pressure of ethane over a saturated solution containing 6.56 × 10-2 g of ethane is 1 bar. If the solution were to contain 5.0 × 10-2 g of ethane, then what will be the partial pressure of the gas? (Comptt. Delhi 2013)
Answer:
Applying the Henry’s law, m = KH × p
In first case, 6.56 × 10-2 = KH × 1
KH = 6.56 × 10-2 g bar-1
Putting the value of KH in the second case, we get
5 × 10-2 g = 6.56 × 10-2 g bar-1 × p
∴ p = \frac{5 \times 10^{-2}}{6.56 \times 10^{-2} \mathrm{g} \mathrm{bar}} = 0.762 bar

Question 54.
Some ethylene glycol, HOCH2CH2OH, is added to your car’s cooling system along with 5 kg of water. If the freezing point of a water-glycol solution is -15.0°C, what is the boiling point of the solution?
(Kb = 0.52 K kg mol-1 and Kf = 1.86 K kg mol-1 for water) (Comptt. Delhi 2014)
Answer:
Given : ΔTf = 0 – (-15) = +15° C, w1 = 5 kg = 5000 g
Kf = 1.86 K kg mol-1 Kb = 0.52 K kg mol-1
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 11

Question 55.
3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the Van’t Hoff factor and predict the nature of solute (associated or dissociated).
(Given : Molar mass of benzoic acid = 122 g mol-1, Kf for benzene = 4.9 K kg mol-1) (Delhi 2015)
Answer:
ΔTf = iKf × m
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 12
As i < 1, therefore solute gets associated

Question 56.
A solution is prepared by dissolving 10 g of non-volatile solute in 200 g of water. It has a vapour pressure of 31.84 mm Hg at 308 K. Calculate the molar mass of the solute.
(Vapour pressure of pure water at 308 K = 32 mm Hg) (All India 2015)
Answer:
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 13

Question 57.
45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. Calculate
(i) the freezing point depression and
(ii) the freezing point of the solution
(Given : Kf of water = 1.86 K kg mol-1) (Comptt. Delhi 2015)
Answer:
(i) Given : w = 45 g, W = 600 g,
Kf = 1.86 K kg mol-1, Δ Tf = ?
Using the formula for freezing point depression,
ΔTf = \mathrm{K}_{f} \frac{w \times 1000}{m \times \mathrm{W}}
m of ethylene glycol (C2H6O2)
= 2 × 12 + 6 × 1 + 2 × 16
= 24 + 6 + 32 = 62 g mol-1
Substituting above values in formula,
ΔTf = \frac{1.86 \mathrm{K} \mathrm{kg} \mathrm{mol}^{-1} \times 45 \mathrm{g} \times 1000 \mathrm{g} \mathrm{kg}^{-1}}{62 \mathrm{g} \mathrm{mol}^{-1} \times 600 \mathrm{g}}
= \frac{837}{372}
∴ ΔTf = 2.25 K
(ii) ΔTf = Tf0 – Tf
Where, Tf0 = Freezing point of pure water
⇒ Tf = 273.15 – 2.25 K
∴ Tf = 270.9 K (Freezing point of the solution)

Question 58.
A 5 percent solution (by mass) of cane-sugar (M.W. 342) is isotonic with 0.877% solution of substance X. Find the molecular weight of X. (Comptt. All India 2015)
Answer:
Given : W (mass) of cane-sugar = 5% means 5 g
Molar mass of cane-sugar (M) = 342 g mol-1
Mass of isotonic substance X
= 0.877% means 0.877 g
Molar mass of X = ?
Using formula,
\frac{\mathrm{W}_{\text { canesugar }}}{\mathrm{M}_{\text { cane sugar }}}=\frac{\mathrm{W}_{\mathrm{X}}}{\mathrm{M}_{\mathrm{X}}} \quad \Rightarrow \frac{5 \mathrm{g}}{342 \mathrm{g} \mathrm{mol}^{-1}}=\frac{0.877 \mathrm{g}}{\mathrm{M}_{\mathrm{X}}}
or Mx = \frac{0.877 \mathrm{g} \times 342 \mathrm{g} \mathrm{mol}^{-1}}{5 \mathrm{g}} \Rightarrow \frac{299.934 \mathrm{g} \mathrm{mol}^{-1}}{5 \mathrm{g}}
∴ Mx = 59.9 ≈ 60 g mol-1

Question 59.
Calculate the boiling point of solution when 4 g of MgSO4 (M =120 g mol-1) was dissolved in 100 g of water, assuming MgS04 undergoes complete ionization.
(Kb for water = 0.52 K kg mol-1) (All India 2016)
Answer:
Since MgS04 is an ionic compound, so undergoes complete ionisation in the following way:
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 14
Boiling point of water = 373.15 K
Tb= Tb° + ΔTb = 373.15 K + 0.346 K
= 373.496 K

Question 60.
Calculate the mass of a non-volatile solute (molecular mass 40) which should be dissolved in 114 g octane to reduce the vapour pressure to 80%. (Comptt. Delhi 2016)
Answer:
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 15

Question 61.
An aqueous solution of 2 percent non-volatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molecular mass of the solute?
[Vapour pressure of water = 1.013 bar] (Comptt. All India 2016)
Answer:
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 16
MB = \frac{2 \times 18 \times 1.013}{0.009 \times 98} g mol-1
MB = 41.3 g mol-1

Question 62.
A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K.
Given: (Molar mass of sucrose = 342 g mol-1) (Molar mass of glucose = 180 g mol-1) (Delhi 2017)
Answer:
Molality (m) = \frac{w \times 100}{\mathrm{W} \times \mathrm{M}}
Given:
Molar mass of sucrose
= C12H22O11 = 12 × 12 + 22 + 11 × 16 = 342
10% solution (by mass) of sucrose in water means 10 g of sucrose is present in (100 – 10)
= 90 g of water
10% solution of sucrose means, w = 10 g
Mass of water, W = 90 g
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 17
∴ ΔTf for glucose = 12.33 × 0.6166 = 7.60 K (approx.)
∴ Freezing point of 10% glucose solution
= (273.15 – 7.60) K = 265.55 K

Question 63.
The vapour pressure of pure liquids A and B at 400 K are 450 and 700 mmHg respectively. Find out the composition of liquid mixture if total vapour pressure at this temperature is 600 mmHg. (Comptt. Delhi 2017)
Answer:
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 18
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 19

Solutions Class 12 Important Questions Long Answer Type (LA)

Question 64.
(a) Define the following terms :
(i) Mole fraction (ii) Van’t Hoff factor
(b) 100 mg of a protein is dissolved in enough water to make 10.0 mL of a solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25°C, what is the molar mass of protein?
(R = 0.0821 L atm mol-1 K-1 and 760 mm Hg = 1 atm) (All India 2009)
Answer:
(a) (i) Mole fraction : Mole fraction of a constituent is the fraction obtained by dividing number of moles of that constituent by the total number of moles of all the constituents present in the solution. It is denoted by ‘x’.
Example :
xi = \frac{\text { No. of moles of } x_{1}}{\text { Total no. of moles }}=\frac{n x_{1}}{n x_{1}+n x_{2}}
(ii) Van’t Hoff factor : It is the ratio of the observed colligative property to the theoretical value. It is denoted by ‘i’.
i = \frac{\text { Observed colligative property }}{\text { Theoretical colligative property }}

(b) Given :
w = 100 mg = 0.100 = 0.1 g,
V = 10.0 mL = 0.01 L
π = 13.3 mm Hg = \frac{13.3}{760} atm,
T = 25°C = 25 + 273 = 298 K
R = 0.0821 L atm mol-1 K-1, M = ?
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 4
∴ Molar mass, M = 13980.4 g mol-1

Question 65.
(a) What is meant by :
(t) Colligative properties (ii) Molality of a solution (b) What concentration of nitrogen should be present in a glass of water at room temperature? Assume a temperature of 25° C, a total pressure of 1 atmosphere and mole fraction of nitrogen in air of 0.78.
[KH for nitrogen = 8.42 × 10-7 M/mm Hg] (All India 2009)
Answer:
(a) (i) Colligative properties : Those properties of ideal solutions which depend only on the number of particles of the solute dissolved in a definite amount of the solvent and do not depend on the nature of solute are called colligative properties.
(ii) Molality of a solution : Molality of a solution is defined as the number of moles of the solute dissolved in 1000 grams (1 kg) of the solvent. It is denoted by’m’.
m = \frac{w \times 1000}{\mathrm{M} \times \mathrm{W}}
Where
‘w = Weight of solute in grams
M = Molecular mass of solute
W = Weight of solvent in grams

(b) Given :
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 21

Question 66.
(a) Differentiate between molarity and molality for a solution.
How does a change in temperature influence their values?
(b) Calculate the freezing point of an acqueous solution containing 10.50 g of MgBr2 in 200 g of water. (Molar mass of MgBr2 = 184 g) (Kf for water = 1.86 K kg mol-1) (Delhi 2011)
Answer:
(a) Distinction between molarity and molality.
Molarity : It is the number of moles of solute dissolved in 1 litre of solution. It is temperature dependent.
M = \frac{\omega \times 1000}{\text { mol.mass } \times V}
Molality : It is the number of moles of solute dissolved in 1 kg of the solvent.
m = \frac{\omega \times 1000}{M_{2} \times W}
The relationship between molarity and molality is
m = \frac{\mathrm{M}}{d-\frac{\mathrm{MM}_{2}}{1000}}
When molality = molarity, we get,
1 = \frac{1}{d-\frac{\mathrm{MM}_{2}}{1000}} or d – \frac{\dot{\mathrm{MM}}_{2}}{1000} = 1
∴ d = 1+\frac{\mathrm{MM}_{2}}{1000}
Molarity is temperature dependent while molarity is not.
For very dilute solution, the factor MM2/1000 can be neglected in comparison to 1.
Hence molality will be same to molarity when density d = 1.
Molality is independent of temperature, whereas molarity is a function of temperature because volume depends on temperature and mass does not.

(b) Since MgBr2 is an ionic compound, so undergoes complete dissociation
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 22

Question 67.
(a) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain.
(b) Calculate the boiling point of a solution prepared by adding 15.00 g of NaCl to 250.0 g of water.
(Kb for water = 0.512 K kg mol-1, Molar mass of NaCl = 58.44 g) (Delhi 2011)
Answer:
Osmosis : The net spontaneous flow of the solvent molecules from the solvent to the solution or from a less concentrated solution to a more concentrated solution through a semipermeable membrane is called osmosis.
Osmotic pressure : The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semipermeable membrane is called the osmotic pressure.
The osmotic pressure method has the advantage that it uses molarities instead of molalities and it can be measured at room temperature.

(b) Since NaCl is an ionic compound so undergoes complete dissociation.
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 23

Question 68.
(a) State the following :
(i) Henry’s law about partial pressure of a gas in a mixture.
(ii) Raoult’s law in its general form in reference to solutions.
(b) A solution prepared by dissolving 8.95 mg of a gene fragment in 35.0 mL of water has an osmotic pressure of 0.335 torr at 25°C. Assuming the gene fragment is a non-electrolyte, determine its molar mass. (All India 2011)
Answer:
(a) (i) Henry’s law : “The solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas in equilibrium with the liquid at that temperature.”
Applications of Henry’s law :
• In the production of carbonated beverages which are prepared under high pressure.
• Deep sea divers depend upon compressed air for their oxygen supply.

(ii) Raoult’s law : For a solution of volatile liquids the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
P = P°x
Non-ideal solution shows positive and negative deviations from Raoult’s law.
Positive deviation from Raoult’s law : The total vapour pressure for any solution is greater than the corresponding ideal solution of same composition. Such behaviour is called positive deviation.
Example : Mixtures of ethanol + cyclohexane
Mixture of acetone + carbon disulphide
Negative deviation from Raoult’s law: When the total vapour pressure will be less than corresponding vapour pressure, then it is termed as negative deviation.
Example : Chloroform + Benzene Chloroform + Diethylether

(b) Given : w2 = 8.95 mg = 8.95 × 10-3 g
V = 35 mLπ = 0.335 torr
T = 25°C = 298 K M2 = ?
Substituting all the values in the given formula
π = CRT
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 24
∴ Molar mass, M2 = 1.42 × 104 g mol-1

Question 69.
(a) Differentiate between molarity and molality in a solution. What is the effect of temperature change on molarity and molality in a solution?
(b) What would be the molar mass of a compound if 6.21 g of it dissolved in 24.0 g of chloroform form a solution that has a boiling point of 68.04°C. The boiling point of pure chloroform is 61.7°C and the boiling point elevation constant, Kb for chloroform is 3.63°C/m. (Delhi 2011)
Answer:
(a) Distinction between molarity and molality.
Molarity : It is the number of moles of solute dissolved in 1 litre of solution. It is temperature dependent.
M = \frac{\omega \times 1000}{\text { mol.mass } \times V}
Molality : It is the number of moles of solute dissolved in 1 kg of the solvent.
m = \frac{\omega \times 1000}{M_{2} \times W}
The relationship between molarity and molality is
m = \frac{\mathrm{M}}{d-\frac{\mathrm{MM}_{2}}{1000}}
When molality = molarity, we get,
1 = \frac{1}{d-\frac{\mathrm{MM}_{2}}{1000}} or d – \frac{\dot{\mathrm{MM}}_{2}}{1000} = 1
∴ d = 1+\frac{\mathrm{MM}_{2}}{1000}
Molarity is temperature dependent while molarity is not.
For very dilute solution, the factor MM2/1000 can be neglected in comparison to 1.
Hence molality will be same to molarity when density d = 1.
Molality is independent of temperature, whereas molarity is a function of temperature because volume depends on temperature and mass does not.

(b) Given : w2 = 6.21 g, w1 = 24 g
ΔTb = Tb – T°b = 68.04 – 61.7 °C
= 6.34 °C
Kb = 3.63 °C/m, M2 = ?
Substituting all these values in formula given below
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 25
∴ Molar mass of the compound, M = 148.14 g mol-1

Question 70.
(a) Define the following terms :
(i) Mole fraction (ii) Ideal solution
(b) 15.0 g of an unknown molecular material is dissolved in 450 g of water. The resulting solution freezes at – 0.34°C. What is the molar mass of the material?
(Kf for water = 1.86 K kg mol-1) (All India 2011)
Answer:
(a) (i) Mole fraction : Mole fraction is the ratio of number of moles of one component to the total number of moles in a mixture.
XA = \frac{n_{\mathrm{A}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}, XB = \frac{n_{\mathrm{B}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}
(ii) Ideal solution : The solution which obeys Raoult’s law under all conditions is known as an ideal solution.

(b) ΔTf = Kf m
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 26

Question 71.
(a) Explain the following :
(i) Henry’s law about dissolution of a gas in a liquid
(ii) Boiling point elevation constant for a solvent
(b) A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C. What mass of glycerol was dissolved to make this solution?
(Kb for water = 0.512 K kg mol-1) (All India 2011)
Answer:
(a) (i) Henry’s law : The law states “that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.”
(ii) Boiling point elevation constant for a solvent or molal elevation constant may be defined as the elevation in the boiling point when the molality of the solution is unity.

(b) ΔTb = (100.42 – 100)°C = 0.42°C or 0.42 K
ΔTb = Kb m
0.42 = 0.512 × \frac{\mathrm{W}_{2}}{92} \times \frac{1000}{500}
W2 = \frac{0.42 \times 92 \times 500}{0.512 \times 1000}=\frac{4.83}{0.128} = 37.7 g
where W2 is the weight of the solute.

Question 72.
(a) State Raoult’s law for a solution containing volatile components. How does Raoult’s law become a special case of Henry’s law?
(b) 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. Find the molar mass of the solute. (Kf for benzene = 5.12 K kg mol-1) (All India 2013)
Answer:
Raoult’s law: For a solution of volatile liquids the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
P = P°x
Non-ideal solution shows positive and negative deviations from Raoult’s law.
(i) Positive deviation from Raoult’s law : The total vapour pressure for any solution is greater than the corresponding ideal solution of same composition. Such behaviour is called positive deviation.
Example : Mixtures of ethanol + cyclohexane Mixture of acetone + carbon disulphide
(ii) Negative deviation from Raoult’s law : When the total vapour pressure will be less than corresponding vapour pressure, then it is termed as negative deviation.
Example : Chloroform + Benzene
Chloroform + Diethvlether
According to Raoult’s law PA = PA° × xA
According to Henry’s law PA = KH × xA
Thus both laws are identical and differ by their proportionality constants.

(b) We know that M2 = \frac{1000 \mathrm{K}_{f} w_{2}}{w_{1} \mathrm{DT}_{f}}
Given : w2 = 1.0 g, w1 = 50 g, ΔTf = 0.40
Kf = 5.12 K kg mol-1
∴ M2 = \frac{1000 g \mathrm{kg}^{-1} \times 5.12 \mathrm{kg} \mathrm{mol}^{-1} \times 1.0 \mathrm{g}}{50 \mathrm{g} \times 0.40 \mathrm{K}}
= 256 g mol-1

Question 73.
(a) Define the following terms :
(i) Ideal solution (ii) Azeotrope
(iii) Osmotic pressure
(b) A solution of glucose (C6H12O6) in water is labelled as 10% by weight. What would be the molality of the solution?
(Molar mass of glucose = 180 g mol-1) (All India 2013)
Answer:
(a) (i) Ideal solution : An ideal solution is that which obeys Raoult’s law and in which the intermolecular interactions between the different components are of same magnitude as that is found in pure components.
(ii) Azeotrope : It is a type of liquid mixture having a definite composition and boiling like a pure liquid, (distills without change in compositions)
(iii) Osmotic pressure : The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through semi- permeable membrane is called osmotic pressure.

(b) 10% of glucose means 10 g of solute in 100 g of solvent
∴ Mass of solute = 10 g
Mass of solvent = 100 – 10 = 90 g
= \frac{90}{1000} kg
Molar mass of glucose = 180 g mol-1
No. of moles of \frac{10}{100}=\frac{1}{18} mole
∴ Molarity = \frac{\text { No. of moles of solute }}{\text { mass of solvent in } \mathrm{kg}}
= \frac{1}{18} \times \frac{1000}{90}=\frac{100}{162}
= 0.67 mol kg-1 = 0.6 m

Question 74.
(a) The vapour pressures of benzene and toluene at 293 K are 75 mm Hg and 22 mm Hg respectively. 23.4 g of benzene and 64.4 g of toluene are mixed. If the two form an ideal solution, calculate the mole fraction of benzene in the vapour phase assuming that the vapour pressures are in equilibrium with the liquid mixture at this temperature. (b) What is meant by +ve and -ve deviations from Raoult’s law and how is the sign of AH solution related to +ve and -ve deviations from Raoult’s law?
(Comptt. All India 2013)
Answer:
(a) Given : Mass of benzene = 23.4 g
Molar mass of benzene = C6H6
= 12 × 6 + 6 = 78g mol-1
Mass of toluene = 64.4 g
Molar mass of toluene = C6H5CH3
= 12 × 7 + 8 = 92g mol-1
Moles of Benzene = \frac{23.4}{78} = 0.3 mole
Moles of toluene = \frac{64.4}{92} = 0.7 mole
Vapour pressure of benzene, PB = xB × P
= 0.3 × 75 = 22.5 mm
Vapour pressure of toluene, PT = xT × P
= 0.7 × 22 = 15.4 mm
Total vapour pressure = 22.5 + 15.4
= 37.9 mm
∴ Mole fraction of benzene = \frac{22.5}{37.9} = 0.59

(b) +ve and -ve deviations :
Raoult’s law: For a solution of volatile liquids the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
P = P°x
Non-ideal solution shows positive and negative deviations from Raoult’s law.
(i) Positive deviation from Raoult’s law : The total vapour pressure for any solution is greater than the corresponding ideal solution of same composition. Such behaviour is called positive deviation.
Example : Mixtures of ethanol + cyclohexane Mixture of acetone + carbon disulphide
(ii) Negative deviation from Raoult’s law : When the total vapour pressure will be less than corresponding vapour pressure, then it is termed as negative deviation.
Example : Chloroform + Benzene
Chloroform + Diethvlether
According to Raoult’s law PA = PA° × xA
According to Henry’s law PA = KH × xA
Thus both laws are identical and differ by their proportionality constants.
If it is higher, the solution exhibits positive deviation and if it is low, it exhibits negative deviation from Raoult’s law.
For positive deviation ΔmixH = +ve
For negative devation ΔmixH = -ve

Question 75.
(a) A 5% solution (by mass) of cane-sugar in water has freezing point of 271 K. Calculate the freezing point of 5% solution (by mass) of glucose in water if the freezing point of pure water is 273.15 K.
[Molecular masses : Glucose C6H12O6 : 180 amu; Cane-sugar C12H22C11 : 342 amu]
State Henry’s law and mention two of its important applications. (Comptt. All India 2013)
Answer:
Molality of sugar solution
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 27

(b) Henry’s law : Henry’s law states that, “The solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas in equilibrium with the liquid at that temperature.”
Solubility of gas decreases with increase of temperature at the same pressure.

Question 76.
(a) Define the following terms :
(i) Molarity
(ii) Molal elevation constant (Kb)
(b) A solution containing 15 g urea (molar mass = 60 g mol-1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol-1) in water. Calculate the mass of glucose present in one litre of its solution. (All India 2014)
Answer:
(a) (i) Molarity is the number of moles of solute dissolved in one litre of solution.
(ii) Molal elevation constant may be defined as the elevation in boiling point when the molality of solution is unity i.e. 1 mole of solute is dissolved in 1 kg of the solvent.

(b) For urea, concentration = \frac{15}{60} moles/lt.
For glucose, concentration = \frac{w}{180} moles/lt.
∵ Solutions are isotonic
\frac{w}{180}=\frac{15}{60} ∴ w = \frac{15 \times 180}{60} = 45 g

Question 77.
(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) A solution of glucose (molar mass = 180 g mol-1) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution?
(Density of solution = 1.2 g mL-1) (All India 2014)
Answer:
(a) Since acetone is nearly non-polar in nature and ethanol is polar in nature therefore, no interaction occurs between acetone and ethanol, the number of molecules increases, which shows positive deviation.

(b) 10% glucose means 10 g in 100 g solution or, 90 g of water = 0.090 kg of water
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 28

Question 78.
(a) What is van’t Hoff factor? What types of values can it have if in forming the solution, the solute molecules undergo
(i) Dissociation? (ii) Association?
(b) How many mL of a 0.1 M HCl solution are required to react completely with 1 g of a mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?
(Molar mass : Na2CO3 = 106 g, NaHCO3 = 84 g) (Comptt. All India 2014)
Answer:
(a) (i) Van’t Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property .
i = \frac{\text { Experimental value }}{\text { Calculated value }}
If there is dissociation of the solute in the solution, the Van’t Hoff factor T’ will be greater than one i.e. i > 1.
It means observed colligative property will be greater than calculated value.
(ii) Association : If there is association of solute in the solution, the Van’t Hoff factor ‘f’ will be less than one i.e. i < 1. Thus, observed colligative property will be less than the calculated value.

(b) Calculation of no. of moles of the components in the mixture
Suppose Na2CO3 in the mixture = x g
∴ NaHCO3 in the mixture = (1 – x) g
Molar mass of Na2CO3 = 106 g mol-1
Molar mass of NaHCO3 = 84 g mol-1
∴ Moles of Na2CO3 = \frac{x}{106}
and Moles of NaHCO3 = \frac{1-x}{84}
As the mixture contains equimolar amounts of both
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 29
No. of moles of HCl required
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
NaHCO3 + HCl → NaCl + H2O + CO2
1 mole of Na2CO3 requires HC1 = 2 moles
∴ 0.00526 mole of Na2CO3 requires HCl
= 0.00526 × 2 = 0.01052 mole
Similarly, 0.00526 mole of NaHCO3 requires
HCl = 0.00526
∴ Total HCl required = 0.01052 + 0.00526 = 0.01578 moles
Thus 0.1 mole of HCl is present in 1000 mL of HCl
∴ 0.01578 mole of HCl present in 1000 mL
= \frac{1000}{0.1} × 0.01578 = 157.8 mL

Question 79.
(a) Define
(i) Mole fraction (Hi) Raoult’s law
(b) Assuming complete dissociation, calculate the expected freezing point of a solution prepared by dissolving 6.00 g of Glauber’s salt, Na2SO4.10H2O in 0.100 kg of water. (Kf for water = 1.86 K kg mol-1, Atomic masses : Na = 23, S = 32, O = 16, H = 1) (Comptt. All India 2014)
Answer:
(a) (i) Mole fraction : Mole fraction is the ratio of number of moles of one component to the total number of moles in a mixture.
(ii) Molality of a solution : Molality of a solution is defined as the number of moles of the solute dissolved in 1000 grams (1 kg) of the solvent. It is denoted by’m’.
m = \frac{w \times 1000}{\mathrm{M} \times \mathrm{W}}
Where
‘w = Weight of solute in grams
M = Molecular mass of solute
W = Weight of solvent in grams

(ii) Raoult’s law : For a solution of volatile liquids the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution.
P = P°x
Non-ideal solution shows positive and negative deviations from Raoult’s law.
Positive deviation from Raoult’s law : The total vapour pressure for any solution is greater than the corresponding ideal solution of same composition. Such behaviour is called positive deviation.
Example : Mixtures of ethanol + cyclohexane
Mixture of acetone + carbon disulphide
Negative deviation from Raoult’s law: When the total vapour pressure will be less than corresponding vapour pressure, then it is termed as negative deviation.
Example : Chloroform + Benzene Chloroform + Diethylether

(b) Since Na2SO4.10H2O is an ionic compound, so undergoes complete dissociation.
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 30
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 31

Question 80.
Calculate the freezing point of solution when 1.9 g of MgCl2 (M = 95 g mol-1) was dissolved in 50 g of water, assuming MgCl2 undergoes complete ionization.
(Kf for water = 1.86 K kg mol-1)
(b) (i) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?
(ii) What happens when the external pressure applied becomes more than the osmotic pressure of solution? (Delhi 2016)
Answer:
(a) Since MgCl2 is an ionic compound, so it undergoes complete dissociation.
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 32

(b)
(i) 2M glucose will have a higher boiling point than 1M glucose because elevation in boiling point is a colligative property which depends upon the number of particles in the solution which is more in the case of 2M glucose solution.
(ii) When the external pressure applied becomes more than the osmotic pressure of the solution, then the solvent will flow from the solution into the pure solvent through the semi-permeable membrane. The process is called reverse osmosis (RO).

Question 81.
(a) When 2.56 g of sulphur was dissolved in 100 g of CS2, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (Sx).
(Kf for CS2 = 3.83 K kg mol-1, Atomic mass of Sulphur = 32 g mol-1)
(b) Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing
(i) 1.2% sodium chloride solution?
(ii) 0.4% sodium chloride solution?(Delhi 2016)
Answer:
(a) Given: wb = 2.56 g wa = 100 g = 0.1 kg
ΔTf = 0.383 K Kf= 3.83 K kg mol-1
Atomic mass of sulphur = 32 g mol-1
Mb = ?
Using formula,
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 33
∴ Mb = 256 g mol-1
Hence the no. of atoms present in one molecule of sulphur = \frac{256}{32} = 8
∴ the formula is S8.

(b) (i) If RBCs are placed in contact with 1.2% NaCl solution, then the osmotic pressure of 1.2% NaCl becomes higher than that of RBCs due to which water present inside the cells moves into the NaCl solution which results in shrinkage of RBCs.
(ii) Reverse process will take place if RBCs are kept in contact with 0.4% NaCl solution which has less osmotic pressure ’ due to which water moves into RBCs and they will swell.

Question 82.
(a) A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K.
Given: (Molar mass of sucrose = 342 g mol-1) (Molar mass of glucose = 180 g mol-1)
(b) Define the following terms:
(i) Molality (m)
(ii) Abnormal molar mass (All India 2017)
Answer:
(a) Molality (m) = \frac{w \times 100}{\mathrm{W} \times \mathrm{M}}
Given:
Molar mass of sucrose
= C12H22O11 = 12 × 12 + 22 + 11 × 16 = 342
10% solution (by mass) of sucrose in water means 10 g of sucrose is present in (100 – 10)
= 90 g of water
10% solution of sucrose means, w = 10 g
Mass of water, W = 90 g
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 17
∴ ΔTf for glucose = 12.33 × 0.6166 = 7.60 K (approx.)
∴ Freezing point of 10% glucose solution
= (273.15 – 7.60) K = 265.55 K

(b) (i) Molality (m): Number of moles of solute dissolved per kg of the solvent.
(ii) Abnormal molar mass: If the molar mass calculated by using any of the colligative properties comes to be different than theoretically expected molar mass.

Question 83.
(a) 30 g of urea (M = 60 g mol-1) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.
(b) Write two differences between ideal solutions and non-ideal solutions. (All India 2017)
Answer:
(a) Given:
Urea (W) = 30 g; H2O (W) = 846 g
Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 34

(b)

Ideal SolutionNon- Ideal solution
(i) They obey Raoult’s law over the enitre range of concentration.(i) They do no obey Raoult’s law over the entire range of concentration
(ii) Neither the heat is evolved or absorbed during dissolution(ii) Heat is evolved or absorbed during dissolution
(iii) Δmix H = 0
Δmix V = 0
(iii) Δmix H is not equal to 0.
ΔmixV is not equal to 0.

Question 84.
(a) Explain why on addition of 1 mol glucose to 1 litre water the boiling point of water increases.
(b) Henry’s law constant for CO2 in water is 1.67 × 108 Pa at 298 K. Calculate the number of moles of CO2 in 500 ml of soda water when packed under 2.53 × 105 Pa at the same temperature. (Comptt. All India 2017)
Answer:
(a) Glucose is a non-volatile solute, therefore, addition of glucose to water lowers the vapour pressure of water as a result of which boiling point of water increases.

Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 35
i.e. nH2O = 27.78 mol
No. of moles of CO2
\frac{n \mathrm{CO}_{2}}{27.78} = 1.515 × 10-3
i.e. nCO2 = 42.08 × 10-3 moles = 0.042 mol

Question 85.
(a) Define the following terms :
(i) Ideal solution (ii) Osmotic pressure
(b) Calculate the boiling point elevation for a solution prepared by adding 10 g CaCl2 to 200 g of water, assuming that CaCl2 is completely dissociated.
(Kb) for water = 0.512 K kg mol-1; Molar mass of CaCl2 = 111 g mol-1) (Comptt. All India 2017)
Answer:
(a) (i) Ideal solution : The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions.
(ii) The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through the semipermeable membrane is called the osmotic pressure.

Important Questions for Class 12 Chemistry Chapter 2 Solutions Class 12 Important Questions 36

Important Questions for Class 12 Chemistry

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Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions

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Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions

Magnetism and Matter Class 12 Important Questions Very Short Answer Type VSA

Questions 1.
The permeability of a magnetic material is 0.9983. Name the type of magnetic materials it represents. (Delhi 2011)
Answer:
It represents diamagnetic materials.

Question 2.
The susceptibility of a magnetic material is 1.9 × 10-5. Name the type of magnetic materials it represents. (Delhi 2011)
Answer:
It represents Paramagnetic substance.

Question 3.
The susceptibility of a magnetic material is – 4.2 × 10-6. Name the type of magnetic materials it represents. (Delhi 2011)
Answer:
It represents diamagnetic substances.

Question 4.
Where on the surface of Earth is the angle of dip 90°? (All India 2011)
Answer:
At the magnetic poles, the angle of dip is 90° on the surface of Earth.

Question 5.
Where on the surface of Earth is the angle of dip zero? (All India 2011)
Answer:
At the magnetic equator, the angle of dip is 0°.

Question 6.
Where on the surface of Earth is the vertical component of Earth’s magnetic field zero? (All India 2011)
Answer:
At the Magnetic equator the vertical component of Earth’s magnetic field is zero.

Question 7.
The horizontal component of the earth’s magnetic field at a place is B and angle of dip is 60°. What is the value of vertical component of earth’s magnetic field at equator? (Delhi 2011)
Answer:
BH = B cos δ
BV = BH tan δ = B tan 60° = B × √3 = √3B
∴ At equator, BV = 0 (zero).

Question 8.
Current flows through a circular loop. Depict the north and south pole of its equivalent magnetic dipole. (Comptt. Delhi 2012)
Answer:
Direction of the magnetic field lines is given by right hand thumb rule.
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 1

Question 9.
A straight wire extending from east to west falls with a speed v at right angles to the horizontal component of the Earth’s magnetic field. Which end of the wire would be at the higher electrical potential and why? (Comptt. Delhi 2012)
Answer:
West end of the wire must be at higher electric potential. According to Fleming’s Right Hand rule, “the direction of induced emf is from West to East”.

Question 10.
What are permanent magnets? Give one example. (Delhi 2013)
Answer:
Substances which at room temperature retain their ferromagnetic property for a long period of time are called permanent magnets.
Example: Steel, alinco

Question 11.
Which of the following substances are diamagnetic?
Bi, Al, Na, Cu, Ca and Ni (Delhi 2013)
Answer:
Bi and Cu

Question 12.
Which of the following substances are para-magnetic ?
Bi, Al, Cu, Ca, Pb, Ni (Delhi 2013)
Answer:
Al and Ca are para-magnetic.

Question 13.
Is the steady electric current the only source of magnetic field? Justify your answer. (Comptt. Delhi 2013)
Answer:
No. Steady current is not the only source of magnetic field. Magnets are also source of magnetic field. Unsteady current will also be source of varying magnetic field.

Question 14.
Where on the surface of Earth is the vertical com-ponent of Earth’s magnetic field zero? (Comptt. Delhi 2013)
Answer:
At the Equator the vertical component of the Earth’s magnetic field is zero.

Question 15.
Where on the surface of Earth is the horizontal component of Earth’s magnetic field zero? (Comptt. Delhi 2013)
Answer:
At poles of Earth the horizontal component of Earth’s magnetic field is zero.

Question 16.
Where on the surface of Earth is the Earth’s magnetic field perpendicular to the surface of the Earth? (Comptt. Delhi 2013)
Answer:
At poles of the Earth. The Earth’s magnetic field is perpendicular to the surface of the Earth.

Question 17.
The motion of copper plate is damped when it is allowed to oscillate between the two poles of a magnet. What is the cause of this damping? (All India 2013)
Answer:
The cause of this damping is eddy current.

Question 18.
Relative permeability of a material, µr = 0.5. Identify the nature of the magnetic material and write its relation to magnetic susceptibility. (Comptt. Delhi 2014)
Answer:

  1. Diamagnetic material
  2. µr = 1 + Xm

Question 19.
Relative permeability of a material µr = 400. Identify the nature of the magnetic material (Comptt. Delhi 2014)
Answer:
It is Ferromagnetic.

Question 20.
Relative permeability (µr) of a material has a value lying 1 < µr < 1 + ε (where ε is a small quantity). Identify the nature of the magnetic material. (Comptt. Delhi 2014)
Answer:
Substance : Paramagnetic

Question 21.
In what way is the behaviour of a diamagnetic material different from that of a paramagnetic, when kept in an external magnetic field? (All India 2016)
Answer:

  1. A diamagnetic specimen would move towards the weaker region of the field; while a paramagnetic specimen would move towards the stronger region.
  2. A diamagnetic specimen is repelled by a magnet while a paramagnetic specimen moves towards the magnet.
  3. The paramagnetic gets aligned along the field and the diamagnetic perpendicular to the field.

Question 22.
At a place, the horizontal component of earth’s magnetic field is B and angle of dip is 60°. What is the value of horizontal component of the earth’s magnetic field at the equator? (Delhi 2017)
Answer:
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 2

Magnetism and Matter Class 12 Important Questions Short Answer Type SA – I

Question 23.
Define magnetic susceptibility of a material. Name two elements, one having positive susceptibility and the other having negative susceptibility. What does negative susceptibility signify? (Delhi 2008)
Answer:
(i) Magnetic susceptibility \left(\chi_{m}\right) : It is the property of a material which determines how easily it can be magnetised when kept in a magnetising field.

Also, it is the ratio of intensity of magnetisation (I) produced in the material to the intensity of magnetising field (H)
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 3

(ii) Positive susceptibility : para-magnetic material
Example: Al, Ca.
Negative susceptibility : diamagnetic material
Example: Bi, Cu.

(iii) Negative susceptibility signifies that the material is diamagnetic in nature.

Question 24.
The figure shows the variation of intensity of magnetisation versus the applied magnetic field intensity, H, for two magnetic materials A and B :
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 4
(a) Identify the materials A and B.
(b) Why does the material B, has a larger susceptibility than A, for a given field at constant temperature? (All India 2008)
Answer:
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 5
Slope of the line gives magnetic susce¬ptibilities.
For magnetic material B, it is giving higher +ve value.
So material is ‘ferromagnetic’.
For magnetic material A, it is giving lesser +ve value than ‘B’.
So material is ‘paramagnetic’.

(b) Larger susceptibility is due to characteristic ‘domain structure’. More number of mag¬netic moments get aligned in the direction of magnetising field in comparision to that for paramagnetic materials for the same value of magnetising field.

Question 25.
(i) Write two characteristics of a material used for making permanent magnets.
(ii) Why is core of an electromagnet made of ferromagnetic materials? (Delhi)
Answer:
(i) Two characteristics of a material used for making permanent magnets are :
(a) High retentivity so that it produces a strong magnetic field.
(b) High coercivity so that its magnetisation is not destroyed by strong magnetic fields, temperature variations or minor mechanical damage.

(ii) The core of electromagnet is made of ferromagnetic materials because they have
high initial permeability so that magnetisation is large even for a small magnetising field and low resistivity to reduce losses due to eddy currents.

Question 26.
Draw magnetic field lines when a
(i) diamagnetic,
(ii) paramagnetic substance is placed in an external magnetic field.
Which magnetic property distinguishes this behaviour of the field lines due to the two substances?
Answer:
(i) When a diamagnetic material is placed in an external magnetic field.
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 6

(ii) When a paramagnetic material is placed in an external magnetic field.
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 7
Magnetic susceptibility distinguishes this behaviour of the field lines due to the two substances.

Question 27.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip down at 60° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.4 G. Determine the magnitude of the earth’s magnetic field at the place. (Delhi 2011)
Answer:
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 8

Question 28.
The susceptibility of a magnetic material is – 2.6 × 10-5. Identify the type of magnetic material and state its two properties. (Delhi 2011)
Answer:
Magnetic material is diamagnetic, because susceptibility of a magnetic material is in negative.
Properties are :

  1. In a non-uniform magnetic field, it tends to move slowly from stronger to weaker parts of the field.
  2. A freely suspended diamagnetic rod aligns itself perpendicular to the field.
  3. They expel magnetic field lines.
  4. Such substances are repelled by a magnet. [any two]

Question 29.
The susceptibility of a magnetic material is 2.6 × 10-5. Identify the type of magnetic material and state its two properties. (Delhi 2012)
Answer:
The material is paramagnetic.
Its two properties are :

  1. They are feebly attracted by magnets.
  2. In a non-uniform magnetic field, they tend to move slowly from weaker to stronger parts of the field.

Question 30.
The relative magnetic permeability of a magnetic material is 800. Identify the nature of magnetic material and state its two properties. (Delhi 2012)
Answer:
Substance is ferromagnetic.
Its properties are :

  1. They are strongly attracted by magnets.
  2. In a non-uniform magnetic field, they tend to move quickly from weaker to stronger parts of the field.

Question 31.
A circular coil of N turns and radius R carries a current I. It is unwound and rewound to make another coil of radius R/2, current I remaining the same. Calculate the ratio of the magnetic moments of the new coil and the original coil. (All India 2012)
Answer:
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 9

Question 32.
A circular coil of N turns and diameter ‘d’ carries a current ‘I’. It is unwound and rewound to make another coil of diameter ‘2d’, current T remaining the same. Calculate the ratio of the magnetic moments of the new coil and the original coil. (All India 2012)
Answer:
Magnetic moment of the coil is given by M = NIA
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 10
But as given, Ist coil is rewound to make new coil.
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 11

Question 33.
(a) How does a diamagnetic material behave when it is cooled to very low temperatures?
(b) Why does a paramagnetic sample display greater magnetisation when cooled? Explain. (Comptt. Delhi 2012)
Answer:
(a) When diamagnetic material is cooled to very low temperature then it exhibits both perfect conductivity and perfect diamagnetism.
(b) This is because at lower temperature, the tendency to disrupt the alignment of dipoles (due to magnetising field) decreases on account of reduced random thermal motion.

Question 34.
State two characteristic properties distinguishing the behaviour of paramagnetic and diamagnetic materials. (Comptt. All India 2012)
Answer:

ParamagneticDiamagnetic
(i)Susceptibility is small and positive, i.e., 0 < \chi_{m} < ε (where e is a small number) for paramagneticSusceptibility is small and negative, i.e., -1 ≤ \chi_{m}  for diamagnetic.
(ii)Paramagnetic materials are feebly attracted by magnetsDiamagnetic materials are feebly repelled by magnets

Question 35.
State two characteristic properties distinguish¬ing the behaviour of diamagnetic and ferromagnetic materials. (Comptt. All India 2013)
Answer:

Diamagnetic materialFerromagnetic material
(i) Relative magnetic permeability of dia-magnetic substances is always less than unity, i.e.,\mu_{r} < 1.

(ii) The susceptibility of diamagnetic substances has a small -ve value :

\mu_{r} < 1 ⇒ -1 ≤ \chi_{m} ≤ 0

Relative magnetic per­meability of ferro­magnetic materials is very large
(= 103 to 105).The susceptibility of ferromagnetic materi­als is very large.
\chi_{m} > 1

Question 36.
Write two characteristic properties each to select materials suitable for
(i) permanent magnets and
(ii) electromagnets. (Comptt. All India 2013)
Answer:
Properties of a material—
(a) For making a permanent magnet:

  1. High retentivity
  2. High coercivity
  3. High permeability

(b) For making an electromagnet:

  1. High permeability .
  2. Low retentivity
  3. Low coercivity

Question 37.
A coil of ‘N’ turns and radius ‘R’ carries a current ‘I’. It is unwound and rewound to make a square coil of side ‘a’ having same number of turns (N). Keeping the current ‘I’ same, find the ratio of the magnetic moments of the square coil and the circular coil. (Comptt. Delhi 2013)
Answer:
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 14

Question 38.
Depict the behaviour of magnetic field lines when
(i) a diamagnetic material and
(ii) a paramagnetic material is placed in an external magnetic field. Mention briefly the properties of these materials which explain this distinguishing behaviour. (Comptt. All India 2013)
Answer:
Diamagnetic materials. Diamagnetic materials are those which have tendency to move from stronger to the weaker part of the external magnetic field.
Examples. Bismuth, copper, lead and silicon.
Properties:
(i) When a rod of diamagnetic material is sus-pended inside a magnetic field, it slowly sets itself at right angles to the direction of field.
(ii) When a diamagnetic material is placed inside a magnetic field, the magnetic field lines become slightly less dense in the diamagnetic material.
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 15
(iii) For diamagnetic material :
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 16
Paramagnetic materials. Paramagnetic materials are those which get weakly magnetised when placed in an external magnetic field. They have tendency to move from a region of weak magnetic field to strong magnetic field.
Examples. Aluminium, sodium, calcium and oxygen.

Properties :
(i) When a rod of paramagnetic material is suspended inside a magnetic field, it slowly sets itself parallel to the direction of the magnetic field.

(ii) When a paramagnetic material is placed inside a magnetic field, the magnetic field lines become slightly more dense in the paramagnetic material.
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 17

(iii) The magnetic susceptability ‘\chi_{m}‘ of a paramagnetic material has a small positive value, ie. 0 < \chi_{m} < ε

Question 39.
Out of the two magnetic materials, ‘A’ has relative permeability slightly greater than unity while ‘B’ has less than unity. Identify the nature of the materials ‘A’ and ‘B’. Will their susceptibilities be positive or negative? (Delhi 2013)
Answer:
‘A’ is paramagnetic
‘B’ is diamagnetic
The susceptibility of material ‘A’ is positive while of ‘B’ is negative.

Question 40.
Show diagrammmatically the behaviour of magnetic field lines in the presence of
(i) paramagnetic and
(ii) diamagnetic substances.
How does one explain this distinguishing feature? (All India 2013)
Answer:
(i) Paramagnetic substance
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 18
(ii) Diamagnetic substance
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 19
The range of relative magnetic permeability (µr) of paramagnetic substance is µr > 1 . while for diamagnetic substance, it is µr < 1.

Question 41.
(a) How is an electromagnet different from a permanent magnet?
(b) Write tivo properties of a material which make it suitable for making electromagnets. (Comptt. All India 2013)
Answer:
(a) A permanent magnet is prepared from a ferromagnetic material, which retains magnetic properties for a long time at room temperature while

  • an electromagnet consists of a core made of a ferromagnetic material placed inside a solenoid. It behaves like a permanent magnet as long as a current flows through it.

(b) Properties of material :

  1. high permeability
  2. low retentivity
  3. low coercivity (any two)

Question 42.
Write two properties of a material suitable for making
(a) a permanent magnet, and
(b) an electromagnet. (All India 2017)
Answer:

Properties of a material—
(a) For making a permanent magnet:

  1. High retentivity
  2. High coercivity
  3. High permeability

(b) For making an electromagnet:

  1. High permeability .
  2. Low retentivity
  3. Low coercivity

Question 43.
Depict the behaviour of magnetic field lines near
(i) diamagnetic and
(ii) paramagnetic substances. Justify, giving reasons. (Comptt. Delhi 2017)
Answer:
Behaviour of magnetic lines of force near
(i) diamagnetic substances
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 20
(ii) paramagnetic substances
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 21
Justification : The field lines are repelled or expelled and the field inside the material is reduced near diamagnetic substances.

In the presence of magnetic field, the individual atomic dipoles can get aligned in the direction of the applied magnetic field. Therefore, field lines get concentrated inside the material and the field inside is enhanced near paramagnetic substances.

Magnetism and Matter Class 12 Important Questions Short Answer Type SA – II

Question 44.
Define the following using suitable diagrams :
(i) magnetic declination and
(ii) angle of dip. In what direction will a compass needle point when kept at the
(i) poles and
(ii) equator? (Comptt. Delhi 2015)
Answer:
Magnetic declination:
Angle between magnetic meridian and geographical meridian
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 22

Angle of dip : It is the angle which the magnetic needle makes with the horizontal in the magnetic meridian.
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 23

  1. Direction of compass needle is vertical to the earth’s surface at poles.
  2. Parallel to the earth’s surface at equator.

Magnetism and Matter Class 12 Important Questions Long Answer Type

Question 45.
(a) A small compass needle of magnetic moment ‘m’ is free to turn about an axis perpendicular to the direction of uniform magnetic field ‘B’. The moment of inertia of the needle about the axis is ‘I’. The needle is slightly disturbed from its stable position and then released. Prove that it executes simple harmonic motion. Hence deduce the expression for its time period.

(b) A compass needle, free to turn in a vertical plane orients itself with its axis vertical at a certain place on the earth. Find out the values of

  1. horizontal component of earth’s magnetic field and
  2. angle of dip at the place. (Delhi 2013)

Answer:
(a) This is done by placing a small compass needle of known magnetic moment m and moment of inertia I and allowing it to
Important Questions for Class 12 Physics Chapter 5 Magnetism and Matter Class 12 Important Questions 24
(b) Since, the compass needle is oriented vertically

  1. Horizontal component of earth’s magnetic field will be zero.
  2. The value of angle of dip at that place will be 90°.

Important Questions for Class 12 Physics

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Keeping Quiet Important Questions Class 12 English

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Keeping Quiet Important Questions CBSE Class 12 English

Keeping Quiet Important Questions Short Answer Type Questions (3-4 Marks)

Question 1.
Read the extract given below and answer the questions that follow: (All India 2009)
Now we will count to twelve
and we will all keep still.
For once on the face of the Earth
let’s not speak in any language,
let’s stop for one second,
and not move our arms so much.

  1. How long does the poet want to stay still?
  2. What does he hope to achieve by keeping quiet?
  3. What does the poet mean by ‘not move our arms so much’?

Answer:
1. The poet wants to count upto twelve and stay still during the time we are counting to twelve.
2. By keeping quiet the poet hopes that he will be able to live a fascinating moment and then realise the value of calm reflection and quiet introspection. The ! poet believes that by keeping quiet, we will be able to hear the voice of our conscience in these moments of silence.
3. Whenever man has used his arms he has caused undue harm to others. So the poet uses the phrase, ‘not to move our arms so much’ to emphasise the importance of physical inactivity to lessen destruction caused by man in the world.

Question 2.
What is the exotic moment the poet Pablo Neruda wishes for? (Delhi 2009)
Answer:
Pablo Neruda wishes for that exotic moment when mankind will be free from greed, cruelty and harmful actions. Unnecessary rush and noise have caused unpleasantness and troubles. The poet wishes the noise of engines and machines should cease and peace and tranquility should prevail.

Question 3.
Read the extract given below and answer the questions that follow: (Delhi 2010)
It would he an exotic moment
without rush, without engines,
we would all be together
in a sudden strangeness.

  1. What does ‘TP’ refer to?
  2. Who is the poet speaking to?
  3. What would be the moment like?

Answer:
1. TP refers to that moment of silence and quietness when peace and calmness will prevail.
2. The poet is addressing mankind who has no time for quiet introspection in its over-hectic, busy life.
3. The moment would be exotic and unusual, a new experience in which mankind will experience a sense of bonding and togetherness with each other.

Question 4.
According to the poet, what is it that human beings can learn from Nature? (All India 2010)
Answer:
Earth can teach a lesson to mankind on the preservation and resurrection of life and how new life is believed to rise from the ashes of the dead remains. The Earth never attains total inactivity. Nature remains at work under apparent stillness and keeps the Earth alive. Life never ceases on earth when everything appears to be dead only one thing remains alive and that is earth itself.

Question 5.
Why is Pablo Neruda against ‘total inactivity’? (Comptt. Delhi 2010)
Answer:
Pablo Neruda is against ‘total inactivity’ because that would amount to death and he has no association with death. Life is a continuous and on-going process and so is all about being on the move. Life should not come to a standstill under any circumstance.

Question 6.
Which sadness is Pablo Neruda worried about in his poem? (Comptt. All India 2010)
Answer:
The sadness that Pablo Neruda is worried about in his poem is the sadness of isolation that has made modern man its victim. This sadness has made man self-centered and uncaring to the needs of his fellowmen. He has sacrificed the needs of his emotional self in favour of materialism.

Question 7.
Do you think the poet, Pablo Neruda advocates total inactivity and death? Why/ Why not? (Delhi 2011)
Answer:
No, the poet Pablo Neruda does not advocate total inactivity and death. He wants to give mankind an opportunity of quiet introspection to know and realize how he has been harming himself and others. He makes it clear that stillness should not be confused with inactivity. Stillness means halting of harmful human activities. He also wants mankind to understand that life is about sprouting out of seemingly dead surroundings.

Question 8.
What is the sadness that the poet, Pablo Neruda refers to in the poem ‘Keeping Quiet’? (All India 2011)
Answer:
The sadness that Pablo Neruda is worried about in his poem is the sadness of isolation that has made modern man its victim. This sadness has made man self-centered and uncaring to the needs of his fellowmen. He has sacrificed the needs of his emotional self in favour of materialism.

Question 9.
Do you think the poet advocates total inactivity and death in the poem, ‘Keeping Quiet’? Give reasons. (Comptt. Delhi 2011)
Answer:
No, the poet especially says that keeping quiet should not be confused with total inactivity. He wants no association with death. He wants to keep quiet as this will prepare mankind for introspection and regeneration.

Question 10.
What is the exotic moment the poet Pablo Neruda wishes for? (Comptt. All India 2011)
Answer:
Pablo Neruda wishes for that exotic moment when mankind will be free from greed, cruelty and harmful actions. Unnecessary rush and noise have caused unpleasantness and troubles. The poet wishes the noise of engines and machines should cease and peace and tranquility should prevail.

Question 11.
Read the extract given below and answer the questions that follow: (Delhi 2012)
For once on the face of the Earth
let’s not speak in any language,
let’s stop for one second,
and not move our arms so much.

  1. Why does the poet want us to keep quiet?
  2. What does he want us to do for one second?
  3. What does he mean by ‘not move our arms’?

Answer:
1. The poet wants us to keep quiet because this moment of quietness will give us the time to meditate and introspect upon the kind of turmoil we have created on earth through our ill-actions. Too much activity and rush has only brought misfortunes to mankind, so it is better to be quiet and still.
2. The poet wants us to do nothing and remain inactive for one second.
3. The phrase ‘not move our arms’ implies that no physical activity should be carried out for one second. Man has used his arms only to kill and destroy others. Therefore, let him not move his arms to harm others.

Question 12.
How can suspension of activities help? (All India 2012)
Answer:
Suspension of activities will give us time to introspect and reflect on our follies. It will give us the much needed time to forget our differences, wither away our discriminations and enter into a world of peace.

Question 13.
How does stopping for a second help us, according to the poet, Pablo Neruda? (Comptt. Delhi 2012)
Answer:
According to Pablo Neruda, when we stop for a second we get an opportunity to introspect on our follies. During this moment we will forget our differences and experience a strength of togetherness which will give us a moment of bliss. By indulging in such moments of inacti-vity, we will realise the harm we are causing to the world with our words and actions.

Question 14.
What is the exotic moment referred to in the poem, ‘Keeping Quiet’? What makes it exotic? (Comptt. All India 2012)
Answer:
When people sit still without speaking any language in quiet introspection, that moment, according to the poet will be an exotic one because this moment will help us to experience a strength of togetherness and a strange relationship with which humanity will bind itself.

Question 15.
Read the extract given below and answer the questions that follow: (Delhi 2013)
Perhaps the Earth can teach us
as when everything seems dead
and later proves to be alive.
Now I’ll count upto twelve
and you keep quiet and I will go.

  1. What does the Earth teach us?
  2. What does the poet mean to achieve by counting upto twelve?
  3. What is the significance of ‘keeping quiet’?

Answer:
1. The Earth teaches us how to nurture life under apparent quietitude and how to work silently. It teaches us the importance of sometimes taking a break from all activities.
2. The poet counts up to twelve in an attempt to remain calm and still. He wants to make mankind realise the futility of their mindless words and actions and encourage people to introspect for their own good.
3. By keeping quiet and calm man will realise his follies and refrain from harmful destructive activities. It will also help in creating a feeling of natural understanding and introspection among human beings.

Question 16.
Read the extract given below and answer the questions that follow: (Comptt. All India 2013)
Perhaps the Earth can teach us
as when everything seems dead
and later proves to be alive.

  1. What can Earth teach us?
  2. How can we achieve the state of ‘seems dead’?
  3. Give one example of how the seeming to be dead things become alive.

Answer:
1. The Earth can teach us a lesson in pro¬tection and resurrection of life. It teaches us that life and living never cease to exist.
2. The state of ‘seems dead’ symbolizes the doomsday when all life of earth will come to an end.
3. New life emerges from the dead remains like a seed gives life to a new plant.

Question 17.
What are the different kinds of wars mentioned in the poem? What is Neruda’s attitude towards these wars? (All India 2013)
Answer:
The different wars Neruda mentions in the poem are green wars, wars with gas, wars with fire. Neruda feels that these chemical and nuclear wars and the war that man wages against nature will lead him towards his doom.

Question 18.
Which sadness is Pablo Neruda referring to? (Comptt. Delhi 2013)
Answer:
The sadness that the poet refers to in the poem ‘Keeping Quiet’ is the sadness of never understanding oneself and nature. Human beings in a frenzy of activities, have no time for introspection and thus threaten themselves with death or destruction. This darkens their ways with distress and wretchedness.

Question 19.
Read the extract given below and answer the questions that follow: (Comptt. All India 2014)
If we were not so single-minded
about keeping our lives moving,
and for once could do nothing,
perhaps a huge silence
might interrupt this sadness
of never understanding ourselves
and of threatening ourselves with death.

  1. Whom does ‘we’ refer to in the above lines?
  2. Why does the poet want us to ‘do nothing’ for once?
  3. What is the ‘sadness’ that the poet refers to in the poem?

Answer:
1. We’ in the above lines refers to mankind.
2. The poet wants us to ‘do nothing’ for once because during this time of inactivity man will realize the strength of humanity and become aware of universal brotherhood wherein he will be able to give mankind a healing touch. The poet wants that for once we should not single-mindedly focus on keeping our lives moving and do some introspection and spend our time in silence doing nothing. This way we can understand ourselves better and escape from the calls of death.
3. The ‘sadness’ that the poet refers to in the poem is the sadness of isolation that has made modern man its victim. This sadness is the selfishness of the modern man due to which he has put his own life and the life of his fellowmen in danger.

Question 20.
Which is the exotic moment that the poet refers to in ‘Keeping Quiet’? (Delhi 2014)
Answer:
Pablo Neruda wishes for that exotic moment when mankind will be free from greed, cruelty and harmful actions. Unnecessary rush and noise have caused unpleasantness and troubles. The poet wishes the noise of engines and machines should cease and peace and tranquility should prevail.

Question 21.
What is the sadness the poet refers to in the poem ‘Keeping Quiet’? (All India 2014)
Answer:
The sadness that the poet refers to in the poem ‘Keeping Quiet’ is the sadness of never understanding oneself and nature. Human beings in a frenzy of activities, have no time for introspection and thus threaten themselves with death or destruction. This darkens their ways with distress and wretchedness.

Question 22.
How, according to Neruda, can keeping quiet change our attitude to life? (Comptt. Delhi 2015)
Answer:
According to Neruda, when people on earth think of keeping quiet for sometime, that will be an exotic moment when they will be able to forget their differences and a feeling of brotherhood will prevail among them. Most of the evil thoughts disappear when man becomes silent for a while. By keeping quiet, man will realize his follies and refrain from harmful and destructive activities.

Question 23.
Which images in the poem, “Keeping Quiet” show that the poet condemns violence? (Comptt. All India 2015)
Answer:
The images in the poem, ‘Keeping Quiet’, which show that the poet condemns violence are, that he is totally against alienation among communities, races and violence. The poet is against chemical and nuclear wars which leave no survivors. He wants to bring all the destructive activities to a standstill.

Question 24.
Read the extract given below and answer the questions that follow: (Comptt. Delhi 2015)
‘ It would be an exotic moment
without rush, without engines,
we would all be together
in a sudden strangeness.’

  1. Which exotic moment is referred to in these lines?
  2. Why would that moment be strange?
  3. What does the poet advocate in the poem?
  4. What does the poet mean by the word, ‘engines’?

Answer:
1. The ‘exotic moment’ referred to in these lines is that moment when everyone keeps quiet and there is no movement.
2. That moment would be strange because there will be no rush or engines and it will bring the whole of humanity together for the first time.
3. The poet advocates the need to introspect and think before you act in the poem.
4. By the word ‘engines’ the poet means ‘automobiles’ or ‘machines.’

Question 25.
What will counting up to twelve and keeping still help us achieve? (Delhi 2015)
Answer:
Counting up to twelve and keeping still will help us to introspect upon the kind of turmoil we have created in this world with our mind-less actions. This will be an exotic moment during which we will experience a sense of bonding with one another. It will give us an opportunity to understand each other and save our¬selves from death.

Question 26.
Which symbol from nature does the poet invoke that there can be life under apparent stillness? (Delhi 2015)
Answer:
The poet uses the Earth as a symbol from nature to convey that there can be life under apparent stillness. Earth teaches us that despite a dreary stillness, nature continues to work and everything comes to life again.

Question 27.
‘Life is what it is all about;…’ How is keeping quiet related to life? (All India 2015)
Answer:
The poet advocates keeping quiet and inactivity but he does not want these to be confused with death. Life is an ongoing and continuous process, so it should not come to a standstill under any circumstances. People pursue their goals single-mindedly and their lives keep on moving.

Question 28.
Why does one feel ‘a sudden strangeness’ on counting to twelve and keeping quiet? (All India 2015)
Answer:
One feels ‘a sudden strangeness’ on counting to twelve and keeping quiet because this is an exotic moment which builds a sudden and strange feeling of universal brotherhood. This feeling of togetherness and brotherhood is rather new and inexplicable.

Question 29.
How will ‘keeping quiet’ protect our environ¬ment? (All India 2015)
Answer:
When we keep quiet and refrain from activity, we will not indulge in chemical and nuclear wars that cause total destruction by releasing toxic gases in the environment. These will de-stroy all greenery and life on earth making it a dead planet.

Question 30.
What does the poem, ‘Keeping Quiet’ teach us? (Comptt. All India 2015)
Answer:
The poem ‘Keeping Quiet’ emphasises the need for quiet introspection and creating a feeling of mutual understanding among human beings. It conveys to us the poet’s philosophy of an exotic moment of silence which will be an antidote to violence, hatred and war.

Question 31.
Read the extract given below and answer the questions that follow: (All India 2016)
Noiv we will count to twelve
and we will all keep still.
For once on the face of the Earth
let’s not speak in any language,
let’s stop for one second,
and not move our arms so much.

  1. What is the significance of the number ‘twelve’?
  2. Which two activities does the poet want us to stop?
  3. What does the poet mean by ‘let’s not speak in any language’?
  4. Describe the pun on the word, ‘arms’.

Answer:
1. The number ‘twelve’ signifies the twelve divisions in the clock (measure of time).
2. The poet wants the fishermen to stop gathering salt from the sea and he also wants the chemical and nuclear wars to stop.
3. The poet means we should be silent and introspect and enter into a world which is quiet and peaceful.
4. The word ‘arms’ means not only our own human arms but also refers to the arms and ammunition that are used to harm others.

Question 32.
What does the poet want us to do in the poem, ‘Keeping Quiet’? (Comptt. Delhi 2016)
Answer:
The poet wants us to keep quiet and still and not move our arms and legs too much in the poem, Keeping Quiet. This will give us the time to introspect over our actions and also help to create a feeling of mutual understanding among human beings.

Question 33.
Read the extract given below and answer the questions that follow: (Comptt. All India 2016)
‘Perhaps the Earth can teach us
as when everything seems dead
and later proves to be alive’.

  1. Name the poem and the poet.
  2. What does the earth teach us?
  3. When do things seem to be dead?
  4. Write the antonym of ‘teach’.

Answer:
1. The poem is ‘Keeping Quiet’ by Pablo Neruda.
2. The earth teaches us to continue doing constructive work quietly and the positive results will be evident at a later stage.
3. Things seem to be dead during the winter season.
4. The antonym for ‘teach’ is ‘learn’.

Question 34.
How can ‘mighty dead’ be things of beauty? (Delhi 2017)
Answer:
The ‘mighty dead’ can be things of beauty because of their glorious deeds and achievements. Their achievements make their lives extraordinary so they continue to be a source of inspiration for others.

Question 35.
How would keeping quiet affect life in and around the sea? (All India 2017)
Answer:
Fishermen who are responsible for endangering marine life would get a chance to realize that they are harming the whales who are on the verge of extinction. Men who gather salt from the sea would also get a chance to reflect on the pain they are causing to their hands.

Question 36.
How is total inactivity on the Earth in the winter months full of life? (All India 2017)
Answer:
In the winter months when there appears to be total inactivity on the Earth, life never ceases on it. The Earth teaches us that despite a dreary stillness, nature remains at work and keeps the Earth alive and new life rises from the ashes of the dead remains.

Question 37.
Why does Pablo Neruda want us to count till twelve and keep still? (Comptt. Delhi 2017)
Answer:
Counting up to twelve and keeping still will help us to introspect upon the kind of turmoil we have created in this world with our mind-less actions. This will be an exotic moment during which we will experience a sense of bonding with one another. It will give us an opportunity to understand each other and save ourselves from death.

Question 38.
What will keeping still help us achieve? (Comptt. Delhi 2017)
Answer:
Counting up to twelve and keeping still will help us to introspect upon the kind of turmoil we have created in this world with our mind-less actions. This will be an exotic moment during which we will experience a sense of bonding with one another. It will give us an opportunity to understand each other and save our¬selves from death.

Question 39.
Read the extract given below and answer the questions that follow: (Comptt. All India)
‘It would be an exotic moment
without rush, without engines,
we would all be together
in a sudden strangeness.’

  1. What would be an exotic moment?
  2. Why would it be exotic?
  3. What would be the result of all being together?
  4. Explain: ‘Sudden strangeness’.

Answer:
1. The exotic moment would be that time when silence, peace and calmness will prevail.
2. It would be exotic because we will not be in a rush so it will give us time for quiet introspection, to reflect upon the futility of our mindless actions.
3. The result of being together would result in a new experience in which man will get a chance to have a sense of bonding and togetherness with each other.
4. ‘Sudden strangeness’ is a strange and unusual feeling of universal brotherhood.

Important Questions for Class 12 English

The post Keeping Quiet Important Questions Class 12 English appeared first on Learn CBSE.

A Thing of Beauty Important Questions Class 12 English

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A Thing of Beauty Important Questions CBSE Class 12 English

A Thing of Beauty Important Questions Short Answer Type Questions (3-4 Marks)

Question 1.
Read the extract given below and answer the questions that follow: (Delhi 2000)
Therefore, on every morrow, are we wreathing
A flowery band to bind us to the earth,
Spite of despondence, of the inhuman dearth
Of noble natures, of the gloomy days,
Of all the unhealthy and o’er-darkened ways
Made for our searching:

  1. What are the flowery bands that bind us to the earth?
  2. What message do the above lines convey?

Answer:
1. The flowery band that binds us to earth is beauty in one shape or the other. It removes all sufferings and sorrow that covers our mind and spirit. There is disappointment and dejection all around but the presence of some objects of beauty removes this sadness from our hearts.
2. There are many things that bring us troubles and sufferings. The message conveyed in these lines is that the natural beauty of objects around us takes away the suffering from our sorrowful hearts.
Some beautiful shape or any object of beauty removes the pall of gloom from our mind and spirit.

Question 2.
According to Keats, what makes man love life in spite of all its problems and miseries? (All India 2000)
Answer:
In spite of all the problems and miseries that make man’s life gloomy and cause him suffering and pain, he does not cease to love life because a thing of beauty removes all the sadness that covers his spirit. The beautiful things of nature make life sweet and happy.

Question 3.
What image does Keats use to describe the beautiful bounty of the earth? (Delhi 2010)
Answer:
Keats uses the image of a perennial fountain that constantly pours forth bounties on the earth in the form of an immortal drink from the heavens into our hearts. The beauty of the sun, the moon, the trees, the daffodils and clear rivers are reflections of the beautiful bounties God has blessed us with.

Question 4.
What makes human beings love life inspite of troubles and sufferings? (All India 2010)
Answer:
It is the occasional phases of joy and happiness that make life beautiful and make human beings love life in spite of troubles and sufferings. Natural beauty in its various forms, like the clear rivers, the gurgling brooks and forest vegetation, motivates us to live life and moves away the pall from our dark spirits.

Question 5.
Mention any two ‘things of beauty’ that Keats has described in his poem. How do they make us joyful? (Comptt. Delhi 2010)
Answer:
According to Keats every small or big thing of nature is a thing of beauty and a source of pleasure. The sun, the moon, the trees and daffodil flowers are all things of beauty. So are the small streams with clear water, mass of ferns and the blooming musk roses.

Question 6.
Mention any two things which, according to Keats, give us pain and suffering. (Comptt. All India 2010)
Answer:
According to Keats man suffers from pain and suffering due to the inhuman dearth of noble natures on earth and due to the inhuman and hostile attitude that makes our days sad and darkens our ways with distress and wretchedness.

Question 7.
Read the extract given below and answer the questions that follow: (Comptt. Delhi 2011)
A thing of beauty is a joy forever
Its loveliness increases, it will never
Pass into nothingness; but will keep
A bower quiet for us, and a sleep
Full of sweet dreams, and health, and quiet
breathing

  1. List any tivo things of beauty mentioned above.
  2. What does the phrase, ‘pass into nothingness’ mean?
  3. What are the effects of beautiful things on man’s spirit?

Answer:
1. The things of beauty mentioned above include a quiet bower, peaceful sleep and quiet breathing.
2. The phrase ‘pass into nothingness’ means it will never cease to exist but will continue to have a lasting impression.
3. Beautiful things leave a lasting impression on man’s spirit. They give him great joy and happiness and also help him to gain inner peace and calm.

Question 8.
Read the extract given below and answer the questions that follow: (Comptt. All India 2011)
Therefore, on every morrow, are we wreathing
A flowery band to bind us to the earth,
Spite of despondence, of the inhuman dearth
Of noble natures, of the gloomy days,
Of all the unhealthy and o’er-darkened ways
Made for our searching:

  1. What are the flowery bands that bind us to the earth?
  2. What message do the above lines convey?

Answer:
1. The flowery band that binds us to earth is beauty in one shape or the other. It removes all sufferings and sorrow that covers our mind and spirit. There is disappointment and dejection all around but the presence of some objects of beauty removes this sadness from our hearts.
2. There are many things that bring us troubles and sufferings. The message conveyed in these lines is that the natural beauty of objects around us takes away the suffering from our sorrowful hearts.
Some beautiful shape or any object of beauty removes the pall of gloom from our mind and spirit.

Question 9.
Why and how is grandeur associated with the mighty dead? (Delhi 2011)
Answer:
Grandeur is associated with the mighty dead because of their grand deeds and achieve-ments. Their achievements make their lives extraordinary. Therefore, the grandeur of the mighty dead is a thing of beauty that is a source of inspiration for other.

Question 10.
What is the message of the poem, ‘A Thing of Beauty’? (All India 2011)
Answer:
John Keats was a worshipper of beauty and he saw it as an everlasting source of joy and happiness. Through his poem he conveys that a thing of beauty removes the pall of sadness and sorrow and gives us joy and pleasure. The beauty of a thing goes on increasing and never passes into nothingness.

Question 11.
How is a thing of beauty a joy forever? (Delhi 2012)
Answer:
A thing of beauty is a joy forever because it gives us eternal and everlasting happiness and leaves such an impact in our mind that we are able to relive the wonderful feeling we get from it each time we think about it. It never fades into nothingness, in fact its loveliness increases with each passing moment.

Question 12.
Why is ‘grandeur’ associated with the ‘mighty dead’? (All India 2012)
Answer:
Grandeur is associated with the mighty dead because of their grand deeds and achieve-ments. Their achievements make their lives extraordinary. Therefore, the grandeur of the mighty dead is a thing of beauty that is a source of inspiration for other.

Question 13.
What makes human beings love life in spite of troubles and sufferings? (All India 2012)
Answer:
It is the occasional phases of joy and happiness that make life beautiful and make human beings love life in spite of troubles and sufferings. Natural beauty in its various forms, like the clear rivers, the gurgling brooks and forest vegetation, motivates us to live life and moves away the pall from our dark spirits.

Question 14.
Describe any three things of beauty mentioned in the poem, ‘A Thing of Beauty’. (All India 2012)
Answer:
The poet sees beauty in various natural things. He sees simple and scenic beauty in the image of the sun, the moon, the trees, the sheep, the green pastures and the clear water of the small streams. All these things of beauty are a constant source of joy for us.

Question 15.
Why does a thing of beauty never pass into nothingness? (Comptt. Delhi 2012)
Answer:
The joy provided by a thing of beauty is ever-lasting. It leaves an indelible imprint on our mind. Its loveliness never fades away and so it does not pass into nothingness. It increases manifold each time we think about it and thus we are forever able to relive the joyful experience.

Question 16.
Mention any two things of beauty that Keats talks of in his poem and explain how they influence us. (Comptt. All India 2012)
Answer:
The two things of beauty that influence us are the lushgreen surroundings of meadows and pastures that provide life to all living beings and the simple lambs and sheep that Keats envisions as the embodiments of serene and divine beauty.

Question 17.
Read the extract given below and answer the questions that follow: (Delhi 2013)
A flowery band to bind us to the Earth,
Spite of despondence, of the inhuman dearth
Of noble natures, of the gloomy days,
Of all the unhealthy and o’er-darkened ways
Made for our searching:

  1. What are we doing everyday?
  2. Which evil things do we possess and suffer from?
  3. What are the circumstances that contribute towards making humans unhappy and disillusioned with life?

Answer:
1. Every day we are wreathing a flowery band which binds us to the earth and enables us to live life despite the dejection that surrounds us. We are looking for lovely things around us and establishing a close bond with the earth and nature.
2. We suffer from selfishness and self-centredness due to which there is dearth of noble souls on earth.
3. Man becomes unhappy and disillusioned because he suffers from pain and hopelessness at various junctures in life. Also the lack of nobility in human beings and gloomy days make him unhappy.

Question 18.
Read the extract given below and answer the questions that follow: (Comptt. All India 2013)
Therefore, on every morrow, are we wreathing
A flowery band to bind us to the earth,
Spite of despondence, of the inhuman dearth Of noble natures,

  1. What is ‘morrow’?
  2. Why do we need ‘a flowery band’?
  3. What is inhuman in life?

Answer:
1. ‘Morrow’ means the next day.
2. Everyday we are wreathing a flowery band because it binds us to earth and makes us live our life despite the dejection that surrounds us.
3. Man’s self-centred nature and his inability to rise above pettiness is inhuman in life.

Question 19.
What does Keats consider an endless fountain of immortal drink and why does he call its drink immortal? (All India 2013)
Answer:
Keats considers beauty, in all its forms, which is God’s greatest gift to man as an endless fountain of immortal drink. He calls it immortal because the beauty bestowed by God is everlasting and perennial and men can bask in its glory forever.

Question 20.
According to Keats, what spreads the pall of despondence over our dark spirits? How is it removed? (All India 2013)
Answer:
According to Keats, suffering and pain caused by man’s malice and his evil ways spreads the pall of despondence over our dark spirits. Man lacks noble qualities and his hostile and inhuman nature makes the world gloomy. This can be removed by some shape of beauty that is a source of constant joy.

Question 21.
How does a thing of beauty provide shelter and comfort? (All India 2013)
Answer:
A thing of beauty will always provide a pleasant shelter, a place under the shade of a tree, where we will always remain peaceful and comfortable. Whenever we are under nature’s shelter, we will have a quiet and peaceful sleep which will give us serenity and comfort.

Question 22.
Describe any two things mentioned by Keats in “A Thing of Beauty” which cause suffering and pain. (Comptt. Delhi 2013)
Answer:
According to Keats man suffers from pain and suffering due to the inhuman dearth of noble natures on earth and due to the inhuman and hostile attitude that makes our days sad and darkens our ways with distress and wretchedness.

Question 23.
Read the extract given below and answer the questions that follow: (Delhi 2014)
All lovely tales that we have heard or read;
An endless fountain of immortal drink.
Pouring unto us from the heaven’s brink.

  1. Name the poem and the poet.
  2. What is the thing of beauty mentioned in these lines?
  3. What image does the poet use in these lines?

Answer:
1. The poem is ‘A Thing of Beauty’ and the poet is John Keats.
2. The thing of beauty mentioned in these lines are the tales we have heard or read describing the glorified sacrifices of mighty warriors.
3. Using the divine image the poet tells us that beauty is God’s greatest gift to man. This beauty is eternal and everlasting in whose glory man can bask and it is an endless fountain from where he can drink the immortal elixir of life.

Question 24.
Read the extract given below and answer the questions that follow: (All India 2014)
Spite of despondence, of the inhuman dearth
Of noble natures, of the gloomy days,
Of all the unhealthy and o’er-darkened ways
Made for our searching: yes in spite of all,
Some shape of beauty moves away the pall
From our dark spirits.

  1. Name the poem and the poet.
  2. Why are we ‘despondent’?
  3. What removes ‘the pall from our dark spirits’?

Answer:
1. The poem is ‘A Thing of Beauty’ and the poet is John Keats.
2. We are ‘despondent’ due to lack of noble people, because of gloomy days and because of unhealthy and over darkened ways.
3. Any shape or manifestation of beauty removes ‘the pall from our dark spirits’.

Question 25.
Read the extract given below and answer the questions that follow: (Comptt. All India 2014)
Some shape of beauty moves away the pall
From our dark spirits.

  1. How does beauty help us when we are burdened with grief?
  2. Explain: “Some shape of beauty.”
  3. Identify the figure of speech used in the above lines.

Answer:
1. Beauty, in some shape or form, will help to drive away the sadness and despair from the dark enclosures of our spirit when we are burdened with grief.
2. ‘Some shape of beauty’ means some object of beauty bestowed on us by nature.
3. The figure of speech used in the above lines is alliteration.

Question 26.
Read the extract given below and answer the questions that follow: (Comptt. Delhi 2015)
And such too is the grandeur of the dooms
We have imagined for the mighty dead;
All lovely tales that we have heard or read;
An endless fountain of immortal drink,
Pouring unto us from the heaven’s brink.

  1. Name the poem.
  2. Who are the ‘mighty dead’ referred to here?
  3. What is the endless fountain of immortal drink?
  4. What does the word, ‘brink’ mean?

Answer:
1. The poem is ‘A Thing of Beauty’.
2. The ‘mighty dead’ referred to here are our ancestors, the great people of the world who are worthy of respect.
3. The endless fountain of immortal drink are all the things of beauty that sustain mankind.
4. The word ‘brink’ means ‘an edge at the top/ the point of onset’.

Question 27.
Read the extract given below and answer the questions that follow: (Comptt. All India 2015)
Spite of despondence, of the inhuman dearth
Of noble natures, of the gloomy days,
Of all the unhealthy and o’er-darkened ways
Made for our searching : yes, in spite of all,
Some shape of beauty moves away the pall
From our dark spirits.

  1. Name the poem.
  2. Give one cause of human suffering.
  3. What moves away the pall from our lives?
  4. What does the word, ‘gloomy’ mean?

Answer:
1. The poem is ‘A Thing of Beauty.’
2. Man lacks noble qualities and his hostile and inhuman nature is the cause of human suffering.
3. Nature’s beauty moves away the pall from our lives.
4. The word ‘gloomy’ means ‘dull or depressed.

Question 28.
‘ What does a thing of beauty do for us? (Delhi 2015)
Answer:
A thing of beauty gives us eternal and everlasting happiness and leaves an indelible impression on our mind that we are able to relive the wonderful feeling we get from it each time we think about it. It never fades into nothingness and its loveliness increases with each passing moment.

Question 29.
What makes human beings love life in spite of all the troubles they face? (Delhi 2015)
Answer:
It is the occasional phases of joy and happiness that make life beautiful and make human beings love life in spite of troubles and sufferings. Natural beauty in its various forms, like the clear rivers, the gurgling brooks and forest vegetation, motivates us to live life and moves away the pall from our dark spirits.

Question 30.
Mention any four things of beauty that add joy to our life. (All India 2015)
Answer:
The poet sees beauty in various things of nature that add joy to our life. These include the sun, the moon, the trees, the sheep, the green pastures and the clear water of the small streams. These remove the sadness from our dark spirits.

Question 31.
Mention any two things which cause pain and suffering. (All India 2015)
Answer:
Pain and suffering is caused by man’s malice and unhealthy and evil ways. Man lacks noble qualities and his hostile and inhumane nature makes the world gloomy and depressing and brings misery and suffering in his life, thus distressing him with sorrow and despair.

Question 32.
Which objects of nature does Keats mention as sources of joy in his poem, ‘A Thing of Beauty’? (All India 2015)
Answer:
The poet sees beauty in various natural things. He sees simple and scenic beauty in the image of the sun, the moon, the trees, the sheep, the green pastures and the clear water of the small streams. All these things of beauty are a constant source of joy for us.

Question 33.
Read the extract given below and answer the questions that follow: (Delhi 2016)
Its loveliness increases, it will never
Pass into nothingness; but will keep
A bcnver quiet for us, and a sleep
Full of sweet dreams, and health, and quiet
breathing.

  1. Whose loveliness will keep on increasing?
  2. Identify the phrase which says that ‘it’ is immortal.
  3. What is a ‘bower’?
  4. Why do we need sweet dreams, health and quiet breathing in our lives?

Answer:
1. The loveliness of a thing of beauty will keep increasing.
2. The phrase which implies its immortality is ‘it will never pass into nothingness’.
3. A ‘bower’ is a pleasant shady place under the trees.
4. We need sweet dreams, health and quiet breathing in our lives to refresh and relax our minds and drive away our anxieties and restlessness. They help us to bear the problems of life and remove, our sufferings.

Question 34.
How do beautiful things influence our lives? (Comptt. Delhi 2016)
Answer:
Beautiful things take away sadness and gloom from our spirits. They give us hope and joy. Whenever we think about a thing of beauty it brings a smile on our lips and is a source of joy forever.

Question 35.
Read the extract given below and answer the questions that follow :(Comptt. Delhi 2016)
…….and clear rills
That for themselves a cooling covert make
Gainst the hot season; the mid forest brake,
Rich with a sprinkling of musk-rose blooms;
…………………………

  1. Name the poem and the poet.
  2. How do the rills protect themselves from the heat?
  3. What makes the mid forest brake rich?
  4. What does the word, ‘sprinkling7 mean?

Answer:
1. The poem is ‘A Thing of Beauty’ by John Keats.
2. The rills protect themselves from the heat by making a cooling covert overhead.
3. The sprinkling of musk-rose blossoms makes the mid forest brake rich.
4. The word ‘sprinkling’ means spreading of fragrance.

Question 36.
In the hot season, how do man and beast get comfort? (Delhi 2017)
Answer:
Man and beast get comforted in the hot season by walking past the rivers and small streams which are a natural source of water and are considered as the vital potions of life. They relax under a bower and under the shade of trees.

Question 37.
How is the Earth a source of life when all seems dead on it? (All India 2017)
Answer:
The Earth, which is a source of life, can teach us how new life emerges from the dead remains of the ashes. Life is an eternal and continuous process and new life is believed to rise from the dead remains.

Question 38.
Read the extract given below and answer the questions that follow: (Comptt. Delhi 2017)
Answer:
‘A thing of beauty is a joy forever
Its loveliness increases, it will never
Pass into nothingness; but will keep
A bower quiet for us, and a sleep
Full of sweet dreams, and health and quiet breathing.

  1. How does a thing of beauty give us everlasting joy?
  2. What is the effect of increase in its loveliness?
  3. Which one example of the beauty of nature has the poet given here?
  4. What kind of joy do we get from a quiet bower?

Answer:
1.  A thing of beauty gives us eternal joy as its loveliness never dies out or fades away but keeps on increasing.
2. The effect of increase in its loveliness is that a thing of beauty will never cease to give us joy.
3. The example of beauty of nature that the poet describes is of a quiet bower, a shelter under the shade of trees which protects us from the hot sunrays.
4. The quiet bower gives us a place where we can sleep soundly and peacefully and enjoy sweet dreams.

Question 39.
Mention two things of beauty that Keats refers to in his poem. How do they influence us? (Comptt. All India 2017)
Answer:
The two things of beauty that influence us are the lushgreen surroundings of meadows and pastures that provide life to all living beings and the simple lambs and sheep that Keats envisions as the embodiments of serene and divine beauty.

Important Questions for Class 12 English

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A Roadside Stand Important Questions Class 12 English

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A Roadside Stand Important Questions CBSE Class 12 English

1. Answer each of the following questions in about 30-40 words:

Question 1.
Why do the people who run the roadside stand wait for the squeal of brakes so eagerly? (2003 Delhi)
Answer:
The “squealing of brakes” means that a car has stopped at their roadside stand. It raises their hopes that the city-folk have stopped there to buy something from their roadside stand and some city money will come into their hands.

Question 2.
Explain: “soothe them out of them wits” with reference to the poem The Roadside Stand’. (2005 Delhi)
Answer:
The powerful men approach the country folk with false promises of providing them with better living conditions and a better life. These innocent and simple rustics repose blind faith in their false claims and feel soothed and satisfied. They fail to see through their crookedness and selfishness.

Question 3.
Why does Robert Frost sympathise with the rural poor? (2009 Delhi; 2011 Comptt. Outside Delhi)
Answer:
Robert Frost feels an unbearable agony at the plight of the rural poor who are ignored and neglected by the rich politicians. The Government and the party in power are indifferent to their welfare. They fool them by making false promises and then fully exploit them to suit their own selfish interests.

Question 4.
What was the plea of the folk who had put up the roadside stand? (2008 Delhi; 2011 Delhi; 2013 Delhi)
Answer:
The folk who had put up the roadside stand pleaded to the city dwellers to stop and buy their wares so as to enable them to earn some extra money for a decent living. They wanted that the rich people who passed from there in their cars should stop there and buy some goods from them. The money that these folks would earn from the rich people would help them to lead a better life.

Question 5.
What is the ‘childish longing’ of the folk who had put up the roadside stand? Why is it ‘in vain’? (2011 Comptt. Delhi)
Answer:
The ‘childish longing’, the poet refers to, is the dreams and desires of the rural folk who have a child-like longing for a better life that they hope to live with the help from the city dwellers. Their longing is in vain because the city folk are not willing to help them and so their ‘childish longings’ are not likely to be fulfilled.

Question 6.
Why didn’t the ‘polished traffic’ stop at the roadside stand? (2012 Delhi)
Answer:
The ‘polished traffic’ conveniently overlook the roadside stand and do not stop there as their mind is focussed only on their destination. Moreover, they were critical of the poor decor of the stand, its artless interior and paint.

Question 7.
What news in the poem ‘A Roadside Stand’ is making its round in the village? (2013 Outside Delhi)
Answer:
The news making its round is about the resettlement of the poor, rural people who will be resettled in the villages, next to the theatre and the store. They would be close to the cities and will not have to worry about themselves any more.

Question 8.
Why do people at the roadside stand ask for city money? (2013 Comptt. Delhi)
Answer:
The rural people running the roadside stand are poor and deprived, unlike the people of the city. They thus ask for city money so that they too can lead a life of happiness and prosperity. This much-needed city money can give them the life that had been promised to them by the party in power.

Question 9.
What does Frost himself feel about the roadside stand? (2011 Comptt. Outside Delhi)
Answer:
The poet is distressed to see the interminable wait on the part of the shed owners for their prospective buyers. He is agonised at the ‘childish longing in vain’ of the people who have put up the roadside stand.

2. Read the extract and answer the questions that follow:

Question 10.
The little old house was out with a little new shed
In front at the edge of the road where the traffic sped,
A roadside stand that too pathetically pled,
It would not be fair to say for a dole of bread,
But for some of the money, the cash, whose flow
supports
The flower of cities from sinking and withering faint.

  1. Where was the new shed put up? What was its purpose?
  2. Why does the poet use the word ‘pathetic’?
  3. Explain: ‘too pathetically pled’
  4. Who are referred to as ‘the flower of cities’? (2009 Outside Delhi; 2010 Comptt. Delhi; 2012 Comptt. Delhi)

Answer:
1. A little house at one side of the road was extended and a shed was added to it to put up a road stand. It was set up to attract passersby to buy things from them so that they could earn some money.
2. By using the word ‘pathetic’ the poet emphasizes on the fact that the condition of the shed was most humble and that it presented a rather pitiable sight.
3. It was as if by putting up the shed the owner was desperately pleading to the rich city folks to stop by at his roadside stand and buy things from there so that they could earn some extra money.
4. ‘The flower of the cities’ here refers to the rich and wealthy city-dwellers who can afford the best things.

Question 11.
The polished traffic passed with a mind ahead,
Or if ever aside a moment, then out of sorts
At having the landscape marred with the artless paint
Of signs that with N turned wrong and S turned
wrong
Offered for sale wild berries in wooden quarts,

  1. What does the poet mean by ‘with a mind ahead?
  2. What are N and S signs?
  3. Why have these sings turned wrong? (2010 Comptt. Outside Delhi)

Answer:
1. The phrase ‘with a mind ahead’ suggests that the people who pass the roadside stand in their polished cars conveniently overlook the roadside stand as their mind is focussed only on their destination.
2. The N and S signs stand for the North and the South direction.
3. These signs have turned wrong because they have been painted in the wrong way and so these signboards are wrongly presented.

Question 12.
Or beauty rest in a beautiful mountain scene,
You have the money, but if you want to be mean,
Why keep your money (this crossly) and go along.
The hurt to the scenery wouldn’t be my complaint
So much as the trusting sorrow of what is unsaid

  1. What attraction does the place offer?
  2. What should one do if one wants to be mean?
  3. What does the poet not complain about?
  4. What do you think is the real worry of the poet? (2010 Outside Delhi)

Answer:
1. The place offers a scenic view of the beautiful mountains.
2. If one wants to be mean he can keep his money and move on ahead.
3. The poet does not complain about the landscape which has been spoilt because of the artless painting done on the building.
4. The poet’s real worry is the unexpressed sorrow of the people who have put up the roadside stand.

Question 13.
It is in the news that all these pitiful kin
Are to be bought out and mercifully gathered in
To live in villages, next to the theatre and the store,
Where they won’t have to think for themselves
anymore,
While greedy good-doers, beneficent beasts of prey,
(2000; 2007, Delhi)

  1. Name the poem and the poet.
  2. Explain why merciful have been called ‘greedy good-doers’ and ‘beneficent beasts of prey’?
  3. Why won’t these poor people have to think for themselves any more?

Answer:
1. The poem is ‘A Roadside Stand’ by Robert Frost.
2. The merciful are the crooked politicians, greedy people pretending to be good, who only pose as beneficiaries. These powerful men are actually beasts of prey in the guise of beneficiaries who ruthlessly exploit the common people.
3. These poor people are now in the hands of the so-called ‘merciful beneficiaries’, who will actually do them more harm than any good, so they will not have to think about themselves any more.

Question 14.
Sometimes 1 feel myself I can hardly bear
The thought of so much childish longing in vain,
The sadness that lurks near the open window there,
That waits all day in almost open prayer
For the squeal of brakes, the sound of a stopping car,
Of all the thousand selfish cars that pass.

  1. What cannot be borne by the poet and why?
  2. What is the ‘childish longing7?
  3. Why the longing has been termed as ‘vain’?
  4. Why do the people driving in the cars stop sometimes? (2004 Delhi; 2011 Outside Delhi)

Answer:
1. The poet cannot bear the thought of how these country folks are lured with false promises which are never going to be fulfilled because he feels genuinely sad about so much deprivation to these innocent people.
2. Like children, these country folk have many unfulfilled wishes and desires. So they keep their windows open expecting some prospective customers to turn up so that some good fortune can fall into their share.
3. The longing has been termed as ‘vain’ because it will never be fulfilled.
4. The people driving in the car stop sometimes either to just enquire about the way to their destination or to ask for a gallon of gas if they ran short of it.

Question 15.
Sometimes I feel myself I can hardly bear
The thought of so much childish longing in vain,
The sadness that lurks near the open window there,

  1. Why is the longing called childish?
  2. Where is the window?
  3. Why does sadness lurk there? (2012 Comptt. Outside Delhi)

Answer:
1. Like children, these rural folk nurture many unfulfilled dreams and desires which might never be satisfied. They crave in vain like children waiting for their wishes to be fulfilled.
2. The window is a part of their roadside stand where they wait expectantly.
3. Sadness lurks there because no car halts there to buy anything from their roadside stand and the rural folk are unable to earn some extra money.

Question 16.
The sadness that lurks near the open window there, That waits all day in almost open prayer For the squeal of brakes, the sound of a stopping car, Of all the thousand selfish cars that pass,
Just one to inquire a farmer’s prices are.

  1. Which open window is referred to? Why does sadness lurk there?
  2. What does the farmer pray for?
  3. Is the farmer’s prayer ever granted? How do you know? (2012 Outside Delhi)

Answer:
1. The open window is that of the roadside stand where they wait expectantly for a car to stop by. Sadness lurks there because no city dweller halts there and thus the hopes of the country folk are belied as no customer stops there.
2. The farmer prays that the city folks apply the brakes of the car and halt at their roadside stand to buy something from there.
3. The farmers’ prayers are not granted. The poet tells us that even if city folk do stop at the roadside stand it is only to enquire about the prices of the goods.

Important Questions for Class 12 English

The post A Roadside Stand Important Questions Class 12 English appeared first on Learn CBSE.

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