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Aunt Jennifer’s Tigers Important Questions Class 12 English

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Aunt Jennifer’s Tigers Important Questions CBSE Class 12 English

Aunt Jennifer’s Tigers Important Questions Short Answer Type Questions (3-4 Marks)

Question 1.
Describe the tigers created by Aunt Jennifer. (Delhi 2009)
Answer:
The poet describes Aunt Jennifer’s tigers as ‘bright topaz denizens’ of the forest. They are fearless and ferocious in sharp contrast to their creator, Aunt Jennifer’s nervousness and timidity. Gallant and confident, they are sure of their purpose and move ahead undeterred by any kind of hindrance or obstruction.

Question 2.
Why did Aunt Jennifer choose to embroider tigers on the panel? (All India 2009)
Answer:
Aunt Jennifer chose to embroider tigers on the panel because of the nature of the tigers. They symbolise strength and splendour which was in sharp contrast to her own meek nature. The massive weight of the wedding band that sits heavily on her finger symbolises the ordeals and hardships of her married life so she creates tigers as they are a striking contrast to the frail, meek old lady who created them.

Question 3.
Read the extract given below and answer the questions that follow: (Delhi 2010)
Aunt Jennifer’s tigers prance across a screen,
Bright topaz denizens of a world of green.
They do not fear the men beneath the tree;
They pace in sleek chivalric certainty.

  1. How are aunt Jennifer’s tigers described?
  2. Why are they described as denizens of a world of green?
  3. Why are they not afraid of the men?

Answer:
1. Aunt Jennifer’s tigers are described as powerful, strong and fearless.
2. The tigers are inhabitants of the dense green forests so they are described as dwellers of a world of green.
3. Their courage and fearlessness gives them a confidence due to which they are not afraid of men.

Question 4.
What will happen to Aunt Jennifer’s tigers when she is dead? (All India 2010)
Answer:
Aunt Jennifer’s tigers will survive even after she is dead. She has created the tigers in a panel out of wool. These objects of art are immortal. They will continue prancing, proudly and fearlessly. To express her desire for freedom she had created the chivalrous tigers who will survive long after her death but her own longing for freedom will remain unfulfilled.

Question 5.
How has Aunt Jennifer created her tigers? What traits of tigers do they reveal? (All India 2010)
Answer:
Aunt Jennifer has created shining topaz yellow- coloured tigers who are denizens of a dense, green forest. They are fierce, unafraid and fearless and pace in ‘sleek’ and ‘chivalric’ certainty.

Question 6.
Why are Aunt Jennifer’s hands fluttering through her wool? (Comptt. Delhi 2010)
Answer:
Aunt Jennifer is a victim of gender oppression at the hands of her husband. She lives a life of total domination and constant fear. So she feels nervous and terrified that the hands shake and flutter through her wool as she sits down to knit.

Question 7.
Describe the contrast between Aunt Jennifer ‘ and her creation, the tigers. (Comptt. All India 2010)
Answer:
Aunt Jennifer is totally victimised and suffers from oppression by her male counterpart. So she creates an alternate world of freedom in her art. The tigers she creates go on prancing menacingly, exhibiting their pride and fearlessness of any social group or gender conflicts.

Question 8.
How do ‘denizens’ and ‘chivalric’ add to our understanding of the tigers’ attitude? (Delhi 2011)
Answer:
‘Denizens’ means that the tigers inhabit a green world. They live in the forests where they are free from constraints. ‘Chivalric’ means they are brave and fearless creatures. This helps us to understand that bravery and fearlessness are the basic nature of the tigers.

Question 9.
Why do you think Aunt Jennifer created animals that are so different from her own character? (All India 2011)
Answer:
Aunt Jennifer’s tigers possessed all the qualities that Aunt Jennifer did not have. The tigers were free, fearless, confident and proud whereas Aunt Jennifer was meek, submissive and without any identity. She was a rather indecisive woman unlike the confident tigers she had created.

Question 10.
What do the symbols, ‘tigers’, ‘fingers’ and ‘ring’ stand for in the poem, ‘Aunt Jennifer’s Tigers’? (Comptt. Delhi 2011)
Answer:
The ‘tigers’ are symbols of bravery and courage and also of Aunt Jennifer’s desire for freedom. The ‘fingers’ are symbolic of the nervousness and fear experienced by Aunt Jennifer and the ‘ring’ symbolises a binding marriage that is full of oppression and curtails one’s freedom.

Question 11.
Describe the tigers created by Aunt Jennifer. (Comptt. All India 2011)
Answer:
The poet describes Aunt Jennifer’s tigers as ‘bright topaz denizens’ of the forest. They are fearless and ferocious in sharp contrast to their creator, Aunt Jennifer’s nervousness and timidity. Gallant and confident, they are sure of their purpose and move ahead undeterred by any kind of hindrance or obstruction.

Question 12.
Read the extract given below and answer the questions that follow: (Comptt. All India 2012)
Bright topaz denizens of a world of green.
They do not fear the men beneath the tree;
They pace in sleek chivalric certainty.

  1. Who are ‘They’? Where are ‘They’?
  2. Why are They’ not afraid of men?

Answer:
1. ‘They’ refers to the tigers that Aunt Jennifer has knitted on the panel. They are prancing jerkily in the forest (across a screen).
2. The tigers are not afraid of men because they are gallant and fearless creatures who are undeterred by any obstacles or hindrances and thus are not afraid of the men.

Question 13.
Why did Aunt Jennifer choose to embroider tigers on the panel? (Delhi 2012)
Answer:
Aunt Jennifer chose to embroider tigers on the panel because of the nature of the tigers. They symbolise strength and splendour which was in sharp contrast to her own meek nature. The massive weight of the wedding band that sits heavily on her finger symbolises the ordeals and hardships of her married life so she creates tigers as they are a striking contrast to the frail, meek old lady who created them.

Question 14.
How do the words, “denizens’ and ‘chivalric’ add to our understanding of Aunt Jennifer’s tigers? (All India 2012)
Answer:
Aunt Jennifer chose to embroider tigers on the panel because of the nature of the tigers. They symbolise strength and splendour which was sharp contrast to her own meek nature. The massive weight of the wedding band that sits heavily on her finger symbolises the ordeals and hardships of her married life so she creates tigers as they are a striking contrast to the frail, meek old lady who created them.

Question 15.
What kind of married life did Aunt Jennifer lead? (Comptt. Delhi 2012)
Answer:
Aunt Jennifer’s wedding band lies heavily on her hand. It reminds her of her unhappy married life. It is symbolic of male authority and power of her husband who had suppressed her and made her a nervous wreck. He had dominated over her for so long that she had lost her identity.

Question 16.
What will happen to Aunt Jennifer’s tigers when she is dead? (Delhi 2013)
Answer:
Aunt Jennifer’s tigers will survive even after she is dead. She has created the tigers in a panel out of wool. These objects of art are immortal. They will continue prancing, proudly and fearlessly. To express her desire for freedom she had created the chivalrous tigers who will survive long after her death but her own longing for freedom will remain unfulfilled.

Question 17.
What lies heavily on Aunt Jennifer’s hand? How is it associated with her husband? (All India 2013)
Answer:
Aunt Jennifer’s wedding band lies heavily on her hand. It reminds her of her unhappy married life. It is symbolic of male authority and power of her husband who had suppressed her and made her a nervous wreck. He had dominated over her for so long that she had lost her identity.

Question 18.
Why has Aunt Jennifer made ‘prancing, proud and unafraid’ tigers? (Comptt. Delhi 2013)
Answer:
Aunt Jennifer chose to embroider tigers on the panel because of the nature of the tigers. They symbolise strength and splendour which was sharp contrast to her own meek nature. The massive weight of the wedding band that sits heavily on her finger symbolises the ordeals and hardships of her married life so she creates tigers as they are a striking contrast to the frail, meek old lady who created them.

Question 19.
What is the meaning of the phrase, ‘massive weight of uncle’s wedding band’? (Comptt. All India 2013)
Answer:
Aunt Jennifer’s wedding band lies heavily on her fingers as she has been a victim of gender oppression at the hands of her husband. She has been so physically and mentally trapped for so many years that she lives in a perpetual state of mental fear which she has never been able to overcome.

Question 20.
What are the difficulties that aunt Jennifer faced in her life? (Delhi 2014)
Answer:
Aunt Jennifer faced great hardships in her married life. She led a terrifying and oppressed life wherein she had never been free but a helpless victim of male chauvinism. Dominated and terrorised by her husband, Aunt Jennifer struggled for an existence within the deep conflicts of slavery.

Question 21.
How are Aunt Jennifer’s tigers different from her? (All India 2014)
Answer:
Aunt Jennifer’s tigers possessed all the qualities that Aunt Jennifer did not have. The tigers were free, fearless, confident and proud whereas Aunt Jennifer was meek, submissive and without any identity. She was a rather indecisive woman unlike the confident tigers she had created.

Question 22.
How does Aunt Jennifer express her bitter-ness and anger against male dominance? (Comptt. Delhi 2014)
Answer:
To express her bitterness and anger against male dominance, Aunt Jennifer chooses to embroider tigers on the panel. The nature of tigers symbolizes strength, fearlessness and splendour which is in sharp contrast to her own meek nature because of which she has suffered endlessly.

Question 23.
Read the extract given below and answer the questions that follow: (Comptt. Delhi 2015)
Aunt Jennifer’s tigers prance across a screen,
Bright topaz denizens of a world of green.
They do not fear the men beneath the tree;
They pace in sleek chivalric certainty.

  1. Why are the tigers called ‘Aunt Jennifer’s tigers’?
  2. What does the phrase,’ a world of green’ mean?
  3. How are the tigers different from their creator?
  4. Why are the tigers not afraid of the men beneath the trees?

Answer:
1. The tigers are called ‘Aunt Jennifer’s tigers’ because they are her creation. She has knitted (embroidered) the tigers on a screen.
2. The phrase ‘a world of green’ means ‘the green forest to which the tigers belong.’
3. The tigers are brave, chivalric, confident and strong unlike their creator who is weak, timid, frightened and meek.
4. The tigers are brave and fearless by nature. They are ferocious wild beasts so they are not afraid of the men beneath the trees.

Question 24.
What is suggested by the phrase, ‘massive weight of Uncle’s wedding band’? (Delhi 2015)
Answer:
Aunt Jennifer’s wedding band lies heavily on her fingers as she has been a victim of gender oppression at the hands of her husband. She has been so physically and mentally trapped for so many years that she lives in a perpetual state of mental fear which she has never been able to overcome.

Question 25.
Why does Aunt Jennifer create animals that are so different from her own character? (Delhi 2015)
Answer:
Aunt Jennifer’s tigers possessed all the qualities that Aunt Jennifer did not have. The tigers were free, fearless, confident and proud whereas Aunt Jennifer was meek, submissive and without any identity. She was a rather indecisive woman unlike the confident tigers she had created.

Question 26.
Aunt Jennifer’s efforts to get rid of her fear proved to be futile. Comment. (Delhi 2016)
Answer:
Aunt Jennifer has been a victim of oppression by the overbearing dominance of her husband. Completely terrorised by her husband she struggled for an existence and was so victimised that even after her death she will not be able to liberate her mind and spirit from the fear of male-dominance.

Question 27.
What picture of male chauvinism (tyranny) do we find in the poem, ‘Aunt Jennifer’s Tigers’? (All India 2016)
Answer:
Aunt Jennifer faced great hardships in her married life. She led a terrifying and oppressed life wherein she had never been free but a helpless victim of male chauvinism. Dominated and terrorised by her husband, Aunt Jennifer struggled for an existence within the deep conflicts of slavery.

Question 28.
Read the extract given below and answer the questions that follow: (Comptt. Delhi 2016)
When Aunt is dead, her terrified hands will lie
Still ringed with ordeals she was mastered by.
The tigers in the panel that she made
Will go on prancing, proud and unafraid.

  1. Name the poem and the poet.
  2. What was the aunt’s ordeal?
  3. Why did she ‘make’ tigers?
  4. How were the tigers different from her?

Answer:
1. The poem is ‘Aunt Jennifer’s Tigers’ by Adrienne Rich.
2. The aunt’s ordeal was that she was dominated by her husband and was denied freedom.
3. Aunt Jennifer made tigers to give expression to her desire for freedom.
4. Aunt Jennifer was meek and submissive whereas the tigers she embroidered were strong and courageous.

Question 29.
Read the extract given below and answer the questions that follow :(Comptt. All India 2016)
Aunt Jennifer’s fingers fluttering through her wool
Find even the ivory needle hard to pull.
The massive weight of Uncle’s wedding band
Sits heavily upon Aunt Jennifer’s hand.

  1. Name the poem and the poet.
  2. What is Aunt Jennifer doing with the wool?
  3. Why are her fingers fluttering?
  4. What does ‘wedding band’ mean?

Answer:
1. The poem is ‘Aunt Jennifer’s Tigers’ by Adrienne Rich.
2. Aunt Jennifer is embroidering tigers on a canvas with wool.
3. She has been tormented and dominated by her husband all her life, so her fingers are fluttering due to nervousness.
4. ‘Wedding band’ means a wedding ring.

Question 30.
Read the extract given below and answer the questions that follow: (Delhi 2017)
Aunt Jennifer’s tigers prance across a screen,
Bright topaz denizens of a world of green.
They do not fear the men beneath the tree;
They pace in sleek chivalric certainty.

  1. Why are the tigers called Aunt Jennifer’s tigers?
  2. How are they described here?
  3. How are they different from Aunt Jennifer?
  4. What does the word ‘chivalric’ mean?

Answer:
1. The tigers are called Aunt Jennifer’s tigers because they have been created by her, she has embroidered a panel of prancing tigers.
2. They are described here as yellowish brown (topaz) coloured inhabitants of the jungle. They are fearless, ferocious and brave creatures.
3. Aunt Jennifer is a timid and terrified old woman whose nature is in stark contrast to the fearless and chivalrous tigers she has created.
4. The word ‘chivalric’ means ‘brave’/ respectful towards women.

Question 31.
Read the extract given below and answer the questions that follow: (All India 2017 )
Aunt Jennifer’s fingers fluttering through her wool
Find even the ivory needle hard to pull.
The massive weight of Uncle’s wedding band
Sits heavily upon Aunt Jennifer’s hand.

  1. What is Aunt Jennifer doing with her wool?
  2. Why does she find it difficult to pull her ivory needle?
  3. What does ‘wedding band’ stand for ?
  4. Describe the irony in the third line.

Answer:
1. Aunt Jennifer is embroidering the tigers on a panel with her wool.
2. Suppressed under male domination, Aunt Jennifer has become a nervous wreck. As a result, her fingers flutter and she finds it difficult to pull her ivory needle.
3. ‘Wedding band’ is a symbol of male authority and power. The band symbolizes her unhappy marriage, her husband and the patriarchal society that limits the freedom of women.
4. Even though Aunt Jennifer wears the wedding band, it is ironical that the poet describes it as belonging to uncle.

Question 32.
Read the extract given below and answer the questions that follow: (Comptt. Delhi 2017)
Aunt Jennifer’s fingers fluttering through her
wool.
Find even the ivory needle hard to pull.
The massive weight of Uncle’s wedding band
Sits heavily upon Aunt Jennifer’s hand.

  1. What is Aunt Jennifer’s mood?
  2. Why are her fingers fluttering?
  3. What is Uncle’s wedding band?
  4. Why is it heavy?

Answer:
1. Aunt Jennifer is in a state of anxiety and nervousness.
2. Aunt Jennifer’s fingers are fluttering as she is nervous because of her husband.
3. Uncle’s wedding band is the wedding ring that Aunt Jennifer wears in one of her fingers.
4. It is metaphorically heavy because it is a symbol of her subjugation and oppression by her husband.

Question 33.
Read the extract given below and answer the questions that follow: (Comptt. All India 2017)
‘Bright topaz denizens of a world of green.
They do not fear the men beneath the tree;
They pace in sleek chivalric certainty.’

  1. Who are ‘bright topaz denizens’?
  2. Where do you find them?
  3. Why are ‘they’ not afraid of the men?
  4. What does the word ‘sleek’ mean?

Answer:
1. The tigers embroidered on the panel by Aunt Jennifer are referred to as ‘bright topaz denizens’.
2. They are found in the jungle, the world of green on a screen.
3. They are fearless and bold creatures so they are not afraid of the men.
4. The word ‘sleek’ means ‘elegant’ or ‘glossy’.

Important Questions for Class 12 English

The post Aunt Jennifer’s Tigers Important Questions Class 12 English appeared first on Learn CBSE.


The Third Level Important Questions Class 12 English

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The Third Level Important Questions CBSE Class 12 English

1. Answer each of the following questions in about 30-40 words:

Question 1.
What does the third level refer to? What is the significance of the third level? (2001; 2004, Delhi)
Answer:
The third level is a medium of escape through which man yearns to be away from life’s harsh realities. Modem life is devoid of peace and tranquility so man in his quest to seek solace escapes to a place where his aim is to seek the realization of his dreams and unfulfilled wishes of his subconscious mind.

Question 2.
What convinced Charley that he had reached the third level at Grand Central Station and not the second level? (2010 Delhi)
Answer:
The general layout of the third level was different from that of the second level. It had comparatively smaller rooms, fewer ticket windows and lesser train gates. The infor¬mation booth in the centre was made of wood and looked old. The place with its brass spittoons did not look very bright. So Charley was convinced it was not the second level.

Question 3.
How does Charley, the narrator describe the third level at Grand Central Station? (2013 Delhi)
Answer:
Charley says that the rooms on the third level were smaller than that of the second level. There were fewer ticket windows and train gates and the information booth in the centre was wood and old looking. There were open- flame gaslights and brass spittoons on the floor. Everyone at the station was dressed in nineteenth century dresses.

Question 4.
How did Charley make sure that he was not in the present time? (2002 Delhi)
Answer:
To make sure that he was not in the present time, Charley did a reality check. He looked at the newspapers which were on sale at a kiosk and found a copy of the newspaper ‘The World’, which carried the main story on President Cleveland. Then he confirmed from the Public Library files that the newspaper he had seen was dated 11th June, 1894.

Question 5.
How did Charley often get lost on the Grand Central Station? (2010 Delhi)
Answer:
The Grand Central Station was growing like a tree pushing out endless corridors, doorways and stairs like roots. It had intricate and tangled pathways. The network of passages was so complicated that instead of reaching his destination, one did tend to move up and down to look for entries and exits. So, Charley often got lost on this station.

Question 6.
Why did Charley suspect that Sam had gone to Galesburg? (2011 Outside Delhi)
Answer:
When Sam disappeared all of a sudden and no one knew about his whereabouts, Charley suspected he had gone to Galesburg as Sam was a city boy and liked Galesburg very much. Then Charley found an envelope mailed to Sam by his grandfather from his home in Galesburg and so it confirmed that Sam was indeed in Galesburg.

Question 7.
How does Charley describe Galesburg as it used to be in 1894? (2013 Comptt. Outside Delhi)
Answer:
Charley describes Galesburg as a quiet, simple and peaceful place with big old frame houses, huge lawns and tremendous trees. The summer evenings were rather long and people sat out on their lawns in a peaceful world, men smoking cigars and women waving palm-leaf fans.

Question 8.
What did Charley learn about Sam from the stamp and coin store? (2012 Outside Delhi)
Answer:
From the stamp and coin store Charley gets to know that Sam had bought old style currency worth eight hundred dollars. This money was sufficient to set him up in a little hay, feed and grain business in Galesburg.

Question 9.
How did Sam reach Galesburg? What did he advise Charley to do? (2012 Outside Delhi)
Answer:
Sam was fascinated by Charley’s description of Galesburg. He was so burdened by the tensions and stress of modem life that he thought of escaping to the peaceful world of Galesburg. His advice to Charley is that, he (Charley) and his wife, Louisa should come over to Galesburg through the medium of the ‘third level’.

Question 10.
Why did the booking clerk refuse to accept the money? (2010 Delhi)
Answer:
The booking clerk refuses to accept the money because the notes Charley had given him were of old style. He did not pay in the currency notes that were in circulation in 1894. So the clerk stared at him and told him, “That ain’t money, Mister”. He thought Charley was trying to cheat him and even threatened to get him arrested.

Question 11.
Why did Charley rush back from the third level? (2012 Outside Delhi)
Answer:
When Charley took out the modem currency to pay for the two tickets to Galesburg, the ticket clerk accused him of trying to cheat him. He threatened to hand Charley over to the police. Charley was frightened and he decided to rush back from the third level, lest he was arrested and put into prison.

2. Answer each of the following questions in about 125-150 words.

Question 12.
How did Charley reach the third level of Grand Central? How was it different from the other levels? (2009 Delhi; 2012 Comptt. Delhi)
Answer:
One night Charley worked till late at the office. Then he was in a hurry to get back to his apartment. So he decided to take the subway from Grand Central. He went down the steps and came to the first level. Then he walked down to the second level from where the suburban trains left. He ducked into an arched doorway that headed to the subway. Then he got lost. Knowing that he was going wrong he continued to walk downward. The tunnel turned a sharp left and then taking a short flight of stairs he came out on the third level at the Grand Central Station. Here he saw many unusual things. There were very few ticket windows and train gates that were old-looking and made of wood. Dim gaslights flickered and men wore derby hats and four-button suits. It was a rather strange world of sideburns, beards and fancy moustaches.

Question 13.
Do you think that the third level was a medium of escape for Charley? Why? (2005; 2008 Delhi)
Answer:
The fears, anxieties and insecurities of the modem world are taking a toll on man’s mind. He feels helpless and frustrated and seeks temporary respite from life’s harsh realities. Charley too was unable to cope up with his fastpaced and stressful life so his flight to the third level was undoubtedly a medium of escape for him. It is nothing but a creation of Charley’s own mind. He wants to escape from the modern world’s insecurity, fear, worries and stress and so seeks an exit, a medium to get away into the world of dreams and fancies.

Question 14.
What made Charley believe that the was actually standing at the third level? (2010 Comptt. Delhi)
Answer:
One night Charley worked late at the office. He was in a hurry to get to his apartment. So he decided to take the subway from Grand Central. He ducked into an arched doorway and then he got lost. He walked down the steps to the second level, turned left and kept on walking. He came out on the third level at the Grand Central Station. This was a different, old and romantic world. So he was convinced that he was actually standing at the third level. There were fewer ticket windows there which were made of wood and were old-looking. There were open flame gaslights. He saw people with beards, sideburns and fancy moustaches. Then he caught a glimpse of an old locomotive and also saw an 1894 issue of ‘The World’ newspaper. Perhaps Charley is under pressure to escape from the harsh world of realities. He would like to escape to the peaceful world of 1894.

Question 15.
What kind of people did Charley ‘See’ at the third level? (2011 Outside Delhi, 2010 Comptt. Outside Delhi)
Answer:
Having worked late at the office Charley decided to take a train back home. So he came to Grand Central Station and from the second level he got lost while ducking into an arched doorway and found himself inside a tunnel. This tunnel took him to another light of stairs and he found himself on the third level of the station. As compared to the second level, the third level had smaller rooms, fewer ticket windows and train gates. Everyone there was dressed in ‘eighteen-ninety-something’. Charley came across men and women wearing 19th century dresses. Men sported fancy moustaches, beards and sideburns. Tiny lapels, four-button suits, derby hats and pocket gold watches seemed to be in vogue. Women were wearing fancy cut sleeves, long skirts and high-buttoned shoes. Charley was confused to see people sporting old-fashioned clothes and hair styles at the third level.

Question 16.
How does Charley make his description of the third level very realistic? (2013 Comptt. Delhi)
Answer:
To make his description of the third level very realistic, Charley describes its minute details, vividly comparing it to the second level of the Grand Central station. He says the rooms here were smaller. There were fewer ticket windows and train gates, and the information booth was wooden and old-looking. He also gives a detailed description about the people he saw at the third level and their dresses. He says the people wore nineteenth century dresses; many men had beards, sideburns and fancy moustaches. He also buys tickets to Galesburg, Illinois thus making the reader believe that he was actually at the third level.

Question 17.
What is being inferred from Sam’s letter to Charley? (2003 Delhi)
Answer:
Sam’s letter to Charley is dated 18th July, 1894. It is written from Galesburg, Illinois. In response to Charley’s claim of having visited the third level, Sam who is equally insecure wishes the entire episode is true, as he too believes in the existence of the third level. There are some inferences made by the letter. The introductory part of the letter confirms Charley’s belief in the existence of the third level. It also suggests that those who find the third level can travel across to Galesburg and enjoy the festivities, songs, music and peaceful world of the 1890s. So the author uses Sam’s letter as a unique combination of the real and fantasy world.

Important Questions for Class 12 English

The post The Third Level Important Questions Class 12 English appeared first on Learn CBSE.

AP ePASS 2019-20 | Online Scholarship Platform for all Students in AP

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AP ePASS: The scholarships in Andhra Pradesh state are intended for learners who are a member of the education system. The main aims of Epass scholarship in Andhra Pradesh are, to help the worthy students who cannot afford learning and the government will pay the expenditure fee designated by the colleges in the form of scholarships.

The Electronic Payment and Application System of Scholarships (Epass) is essentially an online scholarship payment system which was conceived by Andhra Pradesh government to assure the faster disbursal of scholarships to the economically underprivileged students. It is a very good action taken by the administration of Andhra Pradesh to award the scholarship to the students whose families are not able of giving their children for study for obtaining higher knowledge or the candidates those belong to SC/ ST / BC class.

It attracts the scholarships schemes for students who belong to SC, ST, BC, EBC, Minority and Disabled categories. Some of the famous scholarships schemes are listed on the Andhra Pradesh ePASS website which includes post-matric scholarships, pre-matric scholarships, overseas scholarships, skill up-gradation services, corporate admissions, etc. Read the article moreover to get complete information about the scholarships, their eligibility, awards, application procedure and more.

AP ePASS List of Scholarship

Let us discuss here, the list of AP Epasss, who provide these scholarships, kind of financial assistance is given to it when to apply for it, etc. in this section. Check the table below for further details.

  • Pre matric Scholarships For SC/ST/BC and Disabled Welfare
  • Corporate Admission Scholarship
  • GRE, TOEFL, IELTS Coaching Registration
  • Overseas Scholarship for Kapu Students
  • Overseas Scholarship for BC Students
  • Overseas Scholarship for EBC Students

Which Students are Eligible for AP ePASS?

  • Students who belong to the categories of SC, ST whose annual parental income is Rs. 200000 or below.
  • Students who belong to BC, EBC, Disabled Welfare Students whose parental income is RS. 100000 or below.
  • Students whose attendance is 75% at the end of each quarter.

Note: The following category students are not eligible for AP Epass.

  • Students belonging to the categories other than SC, ST, BC, EBC, and DW(Disabled).
  • SC, ST Students whose annual parental income is more than Rs. Two Lakhs and BC, EBC, Disabled Students whose parental income is more than Rupees One lakh.
  • All Students pursuing part-time courses, online courses.
  • Students admitted under Sponsored seats, Management Quota seats.
  • Students drawing the stipend more than the scholarship amount in aggregate per annum.
  • Students of BC, EBC and DW students studying the Courses offered by open universities, distant mode, category B seats in MBBS, BDS.
  • EBC students studying Intermediate or equivalent courses

AP ePASS Eligible Courses

Post Matric Courses certified by the concerned University/Board having a term of 1 year and above:

Group -IProfessional Courses (Degree and PG Courses) in:

Medicine
Engineering Technology
Management
Agriculture
Veterinary and Allied Sciences
Business Finance
Business Administration
Computer Applications/ Science
Commercial pilot License course

Group -IIOther professional and technical graduate and PG including (M.Phil, Ph.D. and post-doctoral research) level courses not covered in Group-I.
C.A./I.C.W.A./C.S./ etc., courses, all postgraduate, graduate-level diploma courses, all certificate level courses.
Group -IIIDegree courses(not covered in group I & II)
NIFT (National Institute of Fashion Technology) courses
Group -IVIntermediate
ITI/ITCs
Vocational courses (Intermediate level)

AP ePASS Eligible Colleges

  • All Post Matric Colleges in Andhra Pradesh approved by Government of Andhra Pradesh/Competent Authority.
  • The list of colleges is communicated by Administrative Departments (Departments of Higher Education, Technical Education, School Education, Health Medical and Family Welfare, Employment and Training) to the Commissioner of Social Welfare.

AP ePASS Timeline

It is necessary to submit the application form online within one month from the date of admission. The principal of the Colleges or schools should issue the bonafide certificate on the same day of submission of the application form (hard copy) in the college.

Field Officer/ASWO: Verification would be done physically twice in a year.

  • Within one month of re-opening of college.
  • Within one month from the last date of closing of Admissions.

An officer from any of the Government Department verifies the Certificates and confirms their candidature in the Starting day of AP Epass.

If any candidate missed the process will have to face many problems. Hence, the Government has introduced the Biometric Process of Verification. The school or college can conduct this at any time. The overall process should be completed within one month of submission of application.

AP ePASS Status

Candidates, who have successfully applied for the AP Epass and filled the application form along with the needed documents can check the status of the application for the scholarship provided by Andhra PradeshBoard, time to time, on the university website.

With this, they will be able to trace the progress of their application and can get an approach if they will get the scholarship or not. All they have to do is, use their registration ID and password and submit the details to check the application status. Follow the below steps to check the status of Andhra PradeshState Scholarships for different schemes.

  • Visit the official website epass.apcfss.in
  • Click on the option “Scholarship status tab”
  • Next, you will be redirected to the new page.
  • Now, fill the required details in this page such as SSC hall ticket number, exam category, the application number, academic year, date of birth.
  • Click on the “Get Status” button to know the current status of your application for the scholarship of AP ePASS. Student can also take the print of the application/scholarship status.

AP ePASS Application Procedure

As you have already given the information on the list of AP Epass scholarship and their eligibility criteria, now it’s time to assemble the information about its application process. Let us now examine how to apply for AP Epas by following these steps given below.

  • Visit the official website of the AP Epass, https://epass.apcfss.in/.
  • Select the Course for which you want to apply.
  • Now, go to the option of pre-matric or post matric as per your demand.
  • The students who belong to the scheduled tribe, scheduled caste, BC and Disabled can apply for the pre-matric scholarship program.
  • Click on the “Registration button”,  to get admission or registration form.
  • Provide your Aadhaar and ration card details and all the necessary details.
  • Also give details of your parents/guardian, as per directed.
  • Provide details of your school or college as well as the course.
  • Mention the State bank account number clearly with IFSC code.
  • Mention your caste and income certificate details.
  • Upload the required documents, enter the captcha and click on Submit button.
  • Your application is submitted now.
  • You can take the print out of the application form for future reference.

Documents Required for AP ePASS

  • Domicile Certificate
  • Aadhar Card Number
  • Latest Passport Size Photograph
  • Qualifying Exam Mark Sheet
  • Category Certificate
  • Copy Family Income Certificate
  • Copy Bank Passbook
  • Fee Receipt Number
  • Annual Non Refundable Amount
  • Enrollment Number
  • Student ID Proof

Reason for Rejection of AP ePASS Application

Students application form for scholarships can be rejected for the following condition;

  • If the student is not bonafide.
  • Incorrect caste or annual Income details.
  • Non-submission of caste, Income certificates.
  • Non-receiving of renewal proposal.
  • Claiming scholarship for same level courses.
  • Not recommended by Field Officer.
  • Student admitted under management quota .
  • Incorrect course & year of study.
  • Non-submission of enclosures.
  • Incorrect course & year of study.
  • Non-submission of hard copy of the application.
  • If the student is not physically present.
  • Discontinued/Detained students in case of renewals.
  • Claiming scholarship for same level courses.
  • Previous sanction verification for renewal.
  • Non-receiving of hard copy for Fresh.

Important Instructions for AP ePASS

  • AP Epass is fairly open for all the candidates belonging to any of the categories such as General, OBC, SC, ST.
  • It is important for students to attach a hardcopy of the photocopy of the SBI bank passbook with the application form.
  • Candidates must provide their own effective email ID and mobile phone number.
  • All the details and documents produced by the applicant must be genuine and authentic. If the given information is obtained wrong, then the application and data collected will be rejected.
  • Suspect applications will not be sent for scholarship and will be discarded at the same time.
  • Failed students are not eligible and should not apply for the scholarship.
  • Candidates who have applied for the scholarship or willing to apply are suggested to not share their individual details, bank account number, IFSC code, class 10th/12th roll number, password, and other delicate information to anyone.
  • Candidates can review their application status from time to time on the official website.
  • Candidates are required their ID and password produced at the time of application to see the scholarship status.
  • Candidates, those who are already enrolled with portal have to renew the account by adding the updates. No new registration is required.
  • All updates regarding the scholarship are updated on the official portal in a regular interval of time. Candidates are requested to stay in contact for regular updates.
  • Candidates are suggested to check the scholarship updates on the portal on a regular basis. They must obey all the instructions timely and carefully so that they cannot miss the chance to get the scholarship.

FAQ’s

Question 1.
How much reimbursement, a student can get?

Answer:
The reimbursement off tuition will be dependent on the course you pursue.  While most of the courses are eligible for 100% of the tuition fee as fixed by the government. Self-financed courses are eligible only for a maximum of Rs. 20,000 or the actual fee charged by the college, whichever is less.

Question 2.
Can I obtain a scholarship for the previous year?

Answer:
No, the scholarship can be claimed only for the current year. Scholarships for the previous year can not be claimed under any circumstance.

Question 3.
How do I know my verification officer?

Answer:
The District Collector appoints verification officers for one or more colleges depending on the number of students in the college. The verification process is a two-step verification process:

  • Verification by the College Principle: All applications to be verified individually by the college and signed by the principal of the college.
  • Verification by the verification officer: The verification officer will have to verify all the student in the college an appointed date and time.

The details of the verification officer can be viewed on this website by clicking the verification officer details given on the right side of the web page.

Question 4.
What is the meaning of verification in the scholarship process?

Answer:
The process of verification is essentially meant to check whether the particulars given in the scholarship form are correct as per the documents enclosed.  The verification is done in two steps namely college verification and Independent verification.

  • College Verification:  In this verification, the college has been mandated to verify the documents furnished by each student with the entry made in the application form.  Once the verification is completed and all entries are found correct, he would finally sign the same and send it to the department for verification by the verification officer appointed by the District Collector
  • Verification by the Verification Officer:  the verification officer appointed by the District Collector will conduct physical and documentary verification and submit his report either accepting or rejecting the scholarship application.
  • Scrutiny by the Welfare Officer: before each and every application is taken up for sanction it is the responsibility of the welfare officer to satisfy himself of the verification by the college principal and the verification officer and finally sanction the scholarship.

Contact Details

ePASS
Project Monitoring Unit
The Director of Social Welfare
Service Road Opposite to Manipal Hospital,
TG Plaza, Tadepalli,
Vijayawada – 520001.

The post AP ePASS 2019-20 | Online Scholarship Platform for all Students in AP appeared first on Learn CBSE.

Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions

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Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions

Electrochemistry Class 12 Important Questions Very Short Answer Type

Question 1.
What is meant by ‘limiting molar conductivity’? (All India 2010)
Answer:
The molar conductivity of a solution at infinite dilution is called limiting molar conductivity and is represented by the symbol Λm.

Question 2.
Express the relation between conductivity and molar conductivity of a solution held in a cell. (Delhi 2011)
Answer:
Λm = \frac{\mathrm{K}}{\mathrm{C}}=\frac{\text { Conductivity }}{\text { Concentration }}

Question 3.
What is the effect of catalyst on:
(i) Gibbs energy (ΔG) and
(ii) activation energy of a reaction? (Delhi 2017)
Answer:
(i) There will be no effect of catalyst on Gibbs .energy.
(ii) The catalyst provides an alternative pathway by decreasing the activation energy of a reaction.

Question 4.
What is the effect of adding a catalyst on
(a) Activation energy (Ea), and
(b) Gibbs energy (AG) of a reaction? (All India 2017)
Answer:
(a) On adding catalyst in a reaction, the activation energy reduces and rate of reaction is fastened.
(b) A catalyst does not alter Gibbs energy (AG) of a reaction.

Electrochemistry Class 12 Important Questions Short Answer Type -I [SA – I]

Question 5.
Two half cell reactions of an electrochemical cell are given below :
MnO4(aq) + 8H+ (aq) + 5e → Mn2+ (aq) + 4H2O (I), E° = + 1.51 V
Sn2+ (aq) → 4 Sn4+ (aq) + 2e, E° = + 0.15 V
Construct the redox equation from the two half cell reactions and predict if this reaction favours formation of reactants or product shown in the equation. (All India 2009)
Answer:
The reactions can be represented at anode and at cathode in the following ways :
At anode (oxidation) :
Sn2+ → = Sn4+ (aq) + 2e ] × 5 E° = + 0.15 V
At cathode (reduction) :
MnO4(aq) + 8H+ (aq) + 5e → Mn2+ (aq) + 4H2O (I)] × 2 E° = + 1.51 V
The Net R × M = 2MnO4(aq) + 16H+ + 5Sn2+ → 2Mn2+ + 5Sn4+ + 8H2O
Now E°cell = E°cathode – E°anode
= 1.51 – 0.15 = + 1.36 V
∴ Positive value of E°cell favours formation of product.

Question 6.
Express the relation among the cell constant, the resistance of the solution in the cell and the conductivity of the solution. How is the conductivity of a solution related to its molar conductivity? (All India 2010)
Answer:
\frac{1}{\mathrm{R}} \times \frac{1}{a} = Conductance (C) × Cell constant
Molar conductance : (Λm) = \frac{\mathrm{K} \times 1000}{\mathrm{c}}.

Question 7.
Given that the standard electrode potentials (E°) of metals are :
K+/K = -2.93 V, Ag+/Ag = 0.80 V, Cu2+/Cu = 0.34 V,
Mg2+/Mg = -2.37 V, Cr3+/Cr = -0.74 V, Fe2+/Fe = -0.44 V.
Arrange these metals in increasing order of their reducing power. (All India 2010)
Answer:
Ag+/Ag < Cu2+/Cu < Fe2+/Fe < Cr3+/Cr < Mg2+/ Mg < K+/K
More negative the value of standard electrode potentials of metals is, more will be the reducing power.

Question 8.
Two half-reactions of an electrochemical cell are given below :
MnO4 (aq) + 8H+ (aq) + 5e → Mn2+ (aq) + 4H2O (I), E° = 1.51 V
Sn2+ (aq) → Sn4+ (aq) + 2e, E° = + 0.15 V.
Construct the redox reaction equation from the two half-reactions and calculate the cell potential from the standard potentials and predict if the reaction is reactant or product favoured. (All India 2010)
Answer:
The reactions can be represented at anode and at cathode in the following ways :
At anode (oxidation) :
Sn2+ → Sn4+ (aq) + 2e ] × 5 E° = + 0.15 V
Af cathode (reduction) :
MnO4(aq) + 8H+ (aq) + 5e → Mn2+ (aq) + 4H2O (I)] × 2 E° = + 1.51 V
The Net R × M = 2MnO4(aq) + 16H+ + 5Sn2+ → 2Mn2+ + 5Sn4+ + 8H2O
Now E°cell = E°cathode – E°anode
= 1.51 – 0.15 = + 1.36 V
∴ Positive value of E°cell favours formation of product.

Question 9.
The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere. (Delhi 2011)
Answer:
The mechanism of corrosion is explained on the basis of electrochemical theory. By taking example of rusting of iron, we Refer tothe formation of small electrochemical cells on the surface of iron.
The redox reaction involves
At anode : Fe(S) → Fe2+ (aq) + 2e
At cathode : H2O + CO2 ⇌ H2CO3 (Carbonic acid)
H2CO3 ⇌2H+ + CO22-
H2O ⇌ H+ + OH
H+ + e → H
4H + O2 → 2H2O
Then net resultant Redox reaction is
2Fe(s) + O2 (g) + 4H+ → 2Fe2+ + 2H2O

Question 10.
Determine the values of equilibrium constant (Kc) and ΔG° for the following reaction :
Ni(s) + 2Ag+ (aq) → Ni2+ (aq) + 2Ag(s),
E° = 1.05 V
(1F = 96500 C mol-1) (Delhi 2011)
Answer:
According to the formula
ΔG° = -nFE° = – 2 × 96500 ×1.05
or ΔG° = -202650 J mol-1 = -202.65 KJ mol-1
Now ΔG° ⇒ -202650 J Mol-1
R = 8.314 J/Mol/K, T = 298 K
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 49

Question 11.
Two half-reactions of an electrochemical cell are given below :
MnO4 (aq) + 8H+ (aq) + 5e → Mn2+ (aq) + 4H2O (I), E° = 1.51 V
Sn2+ (aq) → Sn4+ (aq) + 2e, E° = + 0.15 V.
Construct the redox equation from the standard potential of the cell and predict if the reaction is reactant favoured or product favoured. (Delhi 2011)
Answer:
The reactions can be represented at anode and at cathode in the following ways :
At anode (oxidation) :
Sn2+ → = Sn4+ (aq) + 2e ] × 5 E° = + 0.15 V
At cathode (reduction) :
MnO4(aq) + 8H+ (aq) + 5e → Mn2+ (aq) + 4H2O (I)] × 2 E° = + 1.51 V
The Net R × M = 2MnO4(aq) + 16H+ + 5Sn2+ → 2Mn2+ + 5Sn4+ + 8H2O
Now E°cell = E°cathode – E°anode
= 1.51 – 0.15 = + 1.36 V
∴ Positive value of E°cell favours formation of product.

Question 12.
Express the relation among cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to its conductivity? (All India 2012
Answer:
GG* = K
where Q is conductance;
G * is cell constant;
K is conductivity
G* × \frac{1}{\mathrm{R}} = K ⇒ G* = RK
∴ Λm = \frac{\mathbf{K} \times 1000}{\mathbf{C}} S cm2 mol-1

Question 13.
The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2 mol-1. Calculate the conductivity of this solution. (All India 2012)
Answer:
C = 1.5 M, Λm = 138.9 S cm2 mol-1
Λm = \frac{\mathrm{K} \times 1000}{\mathrm{c}}
∴K = \frac{\Lambda_{m} \times C}{1000}=\frac{138.9 \times 1.5}{1000} = 0.20835 S cm-1

Question 14.
A zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential.
[ E°Zn2+ /Zn = – 0.76 V] (Comptt. Delhi 2012)
Answer:
The electode reaction is given as
Zn+2 + 2e → Zn
Using Nernest Equation
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 1

Question 15.
Write the reactions taking place at cathode and anode in lead storage battery when the battery is in use. What happens on charging the battery ? (Comptt. All India 2012)
Answer:
At Anode: Pb + SO4-2 → PbSO4 + 2e
at Cathode : PbO2 + SO4-2 + 4H+ + 2e → PbSO4 + 2H2O
On charging the battery, the reaction is reversed and PbSO4 on anode and cathode is converted into Pb and PbO2 respectively.

Question 16.
The conductivity of 0.20 M solution of KCl at 298 K is 0.025 S cm-1. Calculate its molar conductivity. (Delhi 2013)
Answer:
Molar conductivity Λm = \frac{1000 \times \kappa}{M}
Given : K = 0.025 S cm-1, M = 0.20 M
Hence, Λm = \frac{0.025 \times 1000}{0.20} ∴ Λm = 125 S cm2 mol-1

Question 17.
The standard electrode potential (E°) for Daniel cell is +1.1 V. Calculate the ΔG° for the reaction
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
(1 F = 96500 C mol-1). (All India 2013)
Answer:
We know, ΔG° = -nF E°cell
Given : E°cell = 1.1 volt
∴ ΔG° = -2 × 96500 C mol-1 × 1.1 volt
= -212300 CV mol-1
= -212300 J mol-1 = -212.3 KJ mol-1

Question 18.
The conductivity of 0.001 M acetic acid is 4 × 10-5 S/cm. Calculate the dissociation constant of acetic acid, if molar conductivity at infinite dilution for acetic acid is 390 S cm2/mol. (Comptt. Delhi 2013)
Answer:
Given : K = 4 × 10-5 S/cm, M = 0.001 M
Λ°m = 390 S cm2/mol, k = ?
Using the formula
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 2

Question 19.
The standard electrode potential for Daniell cell is 1.1 V. Calculate the standard Gibbs energy for the cell reaction. (F = 96,500 C mol-1) (Comptt. Delhi 2013)
Answer:
Given : E° = 1.1V, F = 96,500 C mol-5, n = 2
Zn + Cu2 ⇌ Cu + Zn2+
Using ΔG° = -nFE° = -2 × 96500 × 1.1
= 212,300 CV mol-1

Question 20.
State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution? (All India 2014)
Answer:
Kohlrausch law of independent migration of ions: The limiting molar conductivity of an electrolyte (i.e. molar conductivity at infinite dilution) is the sum of the limiting ionic conductivities of the cation and the anion each multiplied with the number of ions present in one formula unit of the electrolyte
Λ°m for AxBy = xλ°+ + yλ°
For acetic acid Λ° (CH3COOH) = λ°CH3COO + λ°H+
Λ°(CH3COOH) = Λ° (CH3COOK) + Λ° (HCl) – Λ° (KCl)

Question 21.
Define the following terms :
(i) Fuel cell
(ii) Limiting molar conductivity (Λ°m) (All India 2014)
Answer:
(i) Fuel cells : These cells are the devices which convert the energy produced during combustion of fuels like H2, CH4, etc. directly into electrical energy.
(ii) The molar conductivity of a solution at infinite dilution is called limiting molar conductivity and is represented by the symbol Λ°m.

Question 22.
Define the following terms :
(i) Molar conductivity (Λm)
(ii) Secondary batteries (All India 2014)
Answer:
Molar conductivity: Molar conductivity of a solution at a given concentration is the conductance of the volume ‘V’ of a solution containing one mole of electrolyte kept between two electrodes with area of cross section ‘A’ and distance of unit length. It is represented by Λm (lamda).
Λm = \frac{\mathrm{KA}}{l} I = 1 and A = V
∴ Λm = KV Unit = S cm2 mol-1
Secondary batteries : Those batteries which can be recharged by passing an electric current through them and can be used again and again are called secondary batteries.

Question 23.
Define the following terms :
(i) Rate constant (k)
(ii) Activation energy (Ea) (Comptt. Delhi 2014)
Answer:
(i) Rate constant (k): It is a proportionality constant and is equal to the rate of reaction when the molar concentration of each of the reactants is unity.
(ii) Activation energy (Ea): The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex is called activation energy.

Question 24.
Set up Nemst equation for the standard dry cell. Using this equation show that the voltage of a dry cell has to decrease with use. (Comptt. All India 2014)
Answer:
Cell reaction of a dry cell can be represented as
Zn + Hg2+ → Zn2+ + Hg (n = 2)
Nemst equation
Ecell = E°cell\frac{0.0591}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\mathrm{Hg}^{2+}}
The voltage of dry cell has to decrease because the concentration of electrolyte decreases in the reactions.

Question 25.
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their Variation with change in temperature. (Comptt. All India 2014)
Answer:
Conductivity: Conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section. It is represented by K.
Its unit is S cm-1
Molar conductivity : Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm3 of the solution when the electrodes are 1 cm apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is represented by Λm.
Its unit is S cm2 mol-1
Conductivity and molar conductivity of electrolytes increase with increasing temperature.

Question 26.
(a) Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution :
Ag+(aq) + e → Ag(s) E° = +0.80 V
H+(aq) + e\frac{1}{2}H2(g) E° = 0.00 V
On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?
(b) Define limiting molar conductivity. Why conductivity of an electrolyte solution decreases with the decrease in concentration? (Delhi 2015)
Answer:
(a) At the cathode Ag+ (aq) + e → Ag(s)
reaction is feasible, because Ag+ ion has higher reduction potential i.e. higher E° value.
(b) Limiting molar conductivity or the molar conductivity of solution at infinite dilution is the sum of molar conductivity cations and anions.
Conductivity of an electrolyte solution decreases on dilution because number of ions per unit volume decreases.

Question 27.
Calculate the time to deposit 1.27 g of copper at cathode when a current of 2A was passed through the solution of CuSO4.
(Molar mass of Cu = 63.5 g mol-1,1 F = 96500 C mol-1) (All India 2015)
Answer:
CuSO4 → Cu+ + SO42-
Cu2+ + 2e → Cu
63.5 gram of copper is deposited = 2 × 96500 C
1.27 gram of Cu is deposited = \frac{2 \times 96500}{63.5} × 1.27
= I × t (Q = I × t)
t = \frac{2 \times 96500 \times 1.27}{63.5 \times 2} = 1930 seconds

Question 28.
From the given cells: Lead storage cell, Mercury cell, Fuel cell and Dry cell
Answer the following:
(i) Which cell is used in hearing aids?
(ii) Which cell was used in Apollo Space Programme?
(iii) Which cell is used in automobiles and inverters?
(iv) Which cell does not have long life?(Delhi 2016)
Answer:
(i) Mercury cell is used in hearing aids.
(ii) Fuel cell was used in the Apollo Space Programme.
(iii) Lead storage cell is used in automobiles and inverters.
(iv) Dry cell does not have a long life.

Question 29.
Calculate the degree of dissociation (a) of acetic acid if its molar conductivity (Λm) is 39.05 S cm2 mol-1.
Given: λ°(H+) = 349.6 S cm2 mol-1 and
λ°(CH3COO) = 40.9 S cm2 mol-1 (Delhi 2017)
Answer:
Λ°m(HAC) = λ°H+ + λ°AC
= λ°CH3COOH = λ°H+ + λ°CH3COO
= 349.6 S cm2 mol-1 + 40.9 S cm2 mol-1
= 390.5 S cm2 mol-1
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 50

Question 30.
Write the name of the cell which is generally used in hearing aids. Write the reactions taking place at the anode and the cathode of this cell. (All India 2017)
Answer:
Mercury cells are used in hearing aids.
Reaction at anode:
Zn (Hg) + 2OH → ZnO (s) + H2O + 2e
Reaction at cathode:
HgO + H2O + 2e → Hg (l) + 2OH

Question 31.
Write the name of the cell which is generally used in transistors. Write the reactions taking place at the anode and the cathode of this cell. (All India 2017)
Answer:
Leclanche cells (Dry cell) is used in transistors.
Reaction at Anode:
Zn(s) → Zn2+ + 2e
At Cathode:
MnO2 + \mathrm{NH}_{4}^{+} + e → MnO(OH) + NH3

Question 32.
Write the name of the cell which is generally used in inverters. Write the reactions taking place at the anode and the cathode of this cell. (Delhi 2017)
Answer:
Lead storage battery is used in inverters.
At Anode:
Pb(s) + \mathrm{SO}_{4}^{2-}(\mathrm{aq}) → PbSO4 (s) + 2e
At Cathode:
PbO2(s) + \mathrm{SO}_{4}^{2-}(\mathrm{aq}) + 4H+ (aq) + 2e
PbSO4 (s) + 2H2O

Question 33.
Following reactions can occur at cathode during the electrolysis of aqueous silver nitrate solution using Pt electrodes :
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 3
On the basis of their standard electrode potential values, which reaction is feasible at cathode and why? (Comptt. All India 2017)
Answer:
As the standard electrode potential of silver is greater than that of the other hydrogen electrode, so reduction of silver takes place and reaction (i) will be feasible
i.e., \mathrm{Ag}_{(\mathrm{aq})}^{+} + e → Ag(s)

Question 34.
Following reactions may occur at cathode during the electrolysis of aqueous silver nitrate solution using silver electrodes :
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 4
On the basis of their standard electrode potential values, which reaction is feasible at cathode and why? (Comptt. All India 2017)
Answer:
The value of \mathrm{E}_{\mathrm{cell}}^{0} is positive due to higher standard electrode potential (E° = +0.80 V) of Ag+ than that of H+ so reaction (i) will be feasible at cathode
i.e. \mathrm{Ag}_{(\mathrm{aq})}^{+} + e → Ag(s). Silver has higher reduction potential.

Question 35.
Following reactions may occur at cathode during the electrolysis of aqueous CuCl2 solution using Pt electrodes:
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 5
On the basis of their standard electrode potential values, which reaction is feasible at cathode and why? (Comptt. All India 2017)
Answer:
Since the standard electrode potential of Cu2+ is greater than that of H+, so reaction (i) will be feasible at cathod
i.e. \mathrm{Cu}_{(\mathrm{aq})}^{2+} + 2e → Cu
Cu2+ has higher reduction potential.

Electrochemistry Class 12 Important Questions Short Answer Type -II [SA – II]

Question 36.
A copper-silver cell is set up. The copper ion concentration in it is 0.10 M. The concentration of silver ion is not known. The cell potential is measured 0,422 V. Determine the concentration of silver ion in the cell.
Given : E°Ag+/Ag = + 0.80 V, E° Cu2+/Cu = + 0.34 V. (All India 2009)
Answer:
The reaction takes place at anode and cathode in the following ways :
At anode (oxidation) :
Cu(s) → Cu2+(aq) + 2e
At cathode (reduction) :
Cu(s) + 2Ag2+(aq) → Cu2+(aq) + 2Ag(s)
The complete cell reaction is
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 6

Question 37.
The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 103 ohm. Calculate its resistivity, conductivity and molar conductivity.(All India 2012)
Answer:
A = πr2 = 3.14 × (0.5)2 = 0.785 cm2, l = 50 cm
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 7

Question 38.
A voltaic cell is set up at 25°C with the following half cells :
Al/Al3+ (0.001 M) and Ni/Ni2+ (0.50 M)
Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{0}=-0.25 \mathrm{V} \text { and } E_{\mathrm{Al}^{3+} / \mathrm{Al}}^{0}=-1.66 \mathrm{V}
(Log 8 × 10-6 = -0.54) (All India 2012)
Answer:
Half cell reactions and overall cell reaction are
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 8
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 9

Question 39.
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
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 10
∴ Molecular mass, MB = 34 g/mol.

Question 40.
What is corrosion? Explain the electrochemical theory of rusting of iron and write the reactions involved in the rusting of iron. (Comptt. Delhi 2012)
Answer:
Corrosion: Corrosion is defined as the deterioration of a substance because of its reaction with its environment. Corrosion is an electrochemical phenomenon. At a particular spot of an object made of iron, oxidation takes place and that spot behaves as anode and the reaction is
At Anode : 2Fe → 2Fe+2 + 4e
Electrons released at anodic spot move through the metal and go to another spot on the metal and reduce oxygen in presence of H+. This spot behaves as cathode
At Cathode : O2 + 4H+ + 4e-
Overall reaction : 2Fe + O2 + 4H+ → 2Fe+2 + 2H2O

Question 41.
When a certain conductance cell was filled with 0.1 M KCl, it has a resistance of 85 ohms at 25°C. When the same cell was filled with an aqueous solution of 0.052 M unknown electrolyte, the resistance was 96 ohms. Calculate the molar conductance of the electrolyte at this concentration.
[Specific conductance of 0.1 M KCl = 1.29 × 10-2 ohm-1 cm-1] (Comptt. All India 2012)
Answer:
Cell contant = Conductivity × Resistance
G* = K × R
= 1.29 × 10-2-1 × 85
= 109.65 × 10-2-1
= 1.0965 cm-1
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 11

Question 42.
(a) How many coulombs are required to reduce 1 mole Cr2O72- to Cr3+?
(b) The conductivity of 0.001 M acetic acid is 4 × 10-5 S/m. Calculate the dissociation constant of acetic acid if \Lambda_{\mathrm{m}}^{0},for acetic acid is 390 S cm2 mol-1. (Comptt. All India 2012)
Answer:
(a) Cr2O7-2 + 14H+ + 6e → 2Cr-3 + 7H2O
∴ 6 Faraday of charge is required
(b) Conductivity (K) = 4 × 10-5 S cm-1
Concentration (C) = 0.001M
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 12

Question 43.
The cell in which the following reaction occurs :
2Fe3+ (aq) + 2I (aq) → 2Fe2+ (aq) + I2 (s) has E0cell = 0.236V at 298K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
(Antilog of 6.5 = 3.162 × 106; of 8.0 = 10 × 108; of 8.5 = 3.162 × 108) (Comptt. All India 2012)
Answer:
log KC = \frac{n E_{\text { cell }}^{0}}{0.0591}=\frac{2 \times 0.236}{0.0591} = 8
KC = antilog 8 = 1 × 108
ΔG° = -nFE0cell = -2 × 96500 × 0.236
= -45548 J/mol-1
= -45.548 kJ/mol-1

Question 44.
Calculate the emf of the following cell at 298 K: Fe(s) | Fe2+ (0.001 M) || H+ (1M) | H2(g) (1 bar), Pt(s) (Given E°cell = +0.44V) (Delhi 2013)
Answer:
As Fe + 2H+ → Fe2+ + H2 (n = 2)
According to Nernst equation
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 13

Question 45.
Calculate the emf of the following cell at 25°C : Ag(s) | Ag+ (10-3 M) || Cu2+ (10-1 M) | Cu(s) Given E0cell = +0.46 V and log 10n = n. (All India 2013)
Answer:
Given: Cell notation is incorrect. Correct cell formula is
Cu2+ (10-1 m) | Cu(5) || Ag+ (10-3 M) | Ag(s)
According to Nernst equation
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 14
∴Ecell = 0.46 V – 0.14775 = 031 V

Question 46.
(a) What are fuel cells? Explain the electrode reactions involved in the working of H2 – O2 fuel cell.
(b) Represent the galvanic cell in which the reaction
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s) takes place. (Comptt. Delhi, Comptt. All India 2013)
Answer:
(a) Fuel cells : These cells are the devices which convert the energy produced during combustion of fuels like H2, CH4, etc. directly into electrical energy.
The electrode reaction for H2 – O2 fuel cell :
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 15
(b) The galvanic cell may be represented as Zn (s) | Zn2+ (1M) || Cu2+ (1M) | Cu (s)

Question 47.
(a) State and explain Kohlrausch law.
(b) How much electricity in terms of Faradays is required to produce 20 g of calcium from molten CaCl2? (Comptt. Delhi 2013)
Answer:
(a) Corrosion is an electrochemical phenomenon. At a particular spot of iron j oxidation it takes place and the spot behaves j as anode and the reaction will be: i
At anode :
2Fe(s) → 2Fe2+ + 4e
Electrons released above move through the metal to another spot and reduce oxygen in presence of H+ and the spot behaves as cathode with the reaction.
At cathode :
O2(g) + 4H+ (aq) + 4e → 2H2O(I)
Overall reaction :
2Fe(s) + O2(g) + H+(aq) → 2Fe2+(aq) + 2H2O (l)
The ferrous ions are further oxidised by atmospheric oxygen to form hydrated ferric oxide (Fe2O3 xH2O) as rust.

(b) Ca22+ + 2e → Ca
Thus, 1 mole of Ca i.e. 40 g of Ca requires electricity = 2F
∴ 20 g of Ca will require electricity = 1F

Question 48.
Silver is uniformly electro-deposited on a metallic vessel of surface area of 900 cm2 by passing a current of 0.5 ampere for 2 hours. Calculate the thickness of silver deposited.
[Given: the density of silver is 10.5 g cm-3 and atomic mass of Ag = 108 amu.] (Comptt. All India 2013)
Answer:Quantity of electricity passed
= 0.5 × 2 × 60 × 60 = 3600 c
Ag+ + e → Ag
96500 c deposits Ag = 107.92 g
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 16

Question 49.
A current was passed for 5 hours through two electrolytic cells connected in series. The first cell contains AuCl3 and second cell CuSO4 solution. If 9.85 g of gold was deposited in the first cell, what amount of copper gets deposited in the second cell? Also calculate magnitude of current in ampere.
Given: Atomic mass of Au = 197 amu and Cu = 63.5 amu. (Comptt. All India 2013)
Answer:
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 17

Question 50.
(a) Calculate ΔrG0 for the reaction
Mg (s) + Cu2+ (aq) → Mg2+ (aq) + Cu (s)
Given : E0cell = + 2.71 V, 1 F = 96500 C mol-1
(b) Name the type of cell which was used in Apollo space programme for providing electrical power. (All India 2014)
Answer:
(a) ΔrG0 = – nFE0
= -2 × 96500 × 2.71 (∵ n = 2)
= -523,030 J mol-1 = -523.03 KJ mol-1
(b) Fuel cell was used in Apollo space programme for providing electrical power.

Question 51.
The resistance of 0.01 M NaCl solution at 25° C is 200 Ω. The cell constant of the conductivity cell used is unity. Calculate the molar conductivity of the solution.(Comptt. All India 2014)
Answer:
For 0.01 M NaCl solution,
R = 200 Ω, cell constant is unity.
∴ Conductivity (K) = \frac{\text { Cell constant }}{\text { Resistance }}
= \frac{1}{200} = 0.005 Sm-1
Concentration of solution = 0.01 M = 0.01 mol L-1
= 0.01 × 103 mol m-3 = 10 mol m-3
Molar conductivity = \frac{K}{C_{m}}=\frac{0.005}{10}
5 × 10-4 Sm2 mol-1

Question 52.
Calculate emf of the following cell at 25°C :
Fe | Fe2+ (0.001 M) || H+ (0.01 M) | H2(g) (1 bar) | Pt(s)
E0(Fe2+ | Fe) = -0.44 V E0(H+ | H2) = 0.00V (Delhi 2015)
Answer:
Fe | Fe2+ (0.001 M) || H+ (0.01 M) | H2(g) (1 bar) | Pt(s)
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 18
= 0.4105 Volts

Question 53.
Conductivity of 2.5 × 10-4 M methanoic acid is 5.25 × 10-5 S cm-1. Calculate its molar conductivity and degree of dissociation.
Given : λ0(H+) = 349.5 Scm2 mol-1
and λ0(HCOO) = 50.5 Scm2 mol-1. (All India 2015)
Answer:
Concentration is 2.5 x 10-4 M
K = 5.25 × 10-5 Scm-1.
\Lambda_{m}^{c}=\frac{\mathrm{K} \times 1000}{\text { Concentration }}
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 19

Question 54.
Calculate e.m.f. of the following cell at 298 K: 2Cr(s) + 3Fe2+ (0.1 M) → 2Cr3+ (0.01 M) + 3 Fe(s)
Given: E0(Cr3+| Cr) = -0.74 V E0(Fe2+ | Fe)
= -0.44 V (Delhi 2016)
Answer:
Cell reaction: 2Cr(s) + 3Fe2+ (0.1 M) → 2Cr3+ (0.01 M) + 3Fe(s)
Given: E0(Cr3+/Cr) = -0.74
E0(Fe2+/Fe) = -0.44 V
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 20

Question 55.
(i) Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO3 for 15 minutes.
[Given: Molar mass of Ag = 108 g mol-1 1F = 96,500 C mol-1)
(ii) Define fuel cell. (Delhi 2017)
Answer:
(i) Q = I × t …(Charge = Current ∝ Time)
. = 2 × 15 × 60 = 1800 C
∵ 96500 C deposit Ag = 108 g
∴ 1800 C deposit Ag = \frac{108}{96500} × 1800
= 2.0145 g
(ii) Cells that convert the energy of combustion of fuels like hydrogen, methanol, methane, etc directly into electrical energy are called fuel cells.

Question 56.
(a) The cell in which the following reaction occurs:
2Fe3+ (aq) + 2I(aq) → 2Fe2+ (aq) + I2(s)
has \mathbf{E}_{\text { Cell }}^{\mathrm{o}} = 0.236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction. (Given: 1F = 96,500 C mol-1)
(b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given: 1F = 96,500 C mol-1) (All India 2017)
Answer:
(a) 2Fe3+ (aq) + 2I (aq) → 2Fe2+ (aq) + I2 (s)
For the given reaction, n = 2, E° = 0.236 V
Using formula
ΔG° = -nF E°cell
= -2 × 96,500 C mol-1 × 0.236 V
∴ ΔG° =-45.55 kj mol-1

(b) Given:
I = 0.5 A
t = 2 hrs. = 2 × 60 ×60 s = 7,200 s
Q = I × t = 0.5 × 7200 = 3,600 C
96,500 C electricity flows to produce
= 6.022 × 1023 electrons
∴ 1 C electricity flows to produce
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 51

Question 57.
Why an electrochemical cell stop working after some time? The reduction potential of an electrode depends upon the concentration of solution with which it is in contact. (All India 2017)
Answer:
As the cell works, the concentration of reactants decrease. Then according to Le chatelier’s pri¬nciple it will shift the equilibrium in backward direction. On the other hand if the concentration is more on the reactant side then it will shift the equilibrium in forward direction. When cell works concentration in anodic compartment in cathodic compartment decreases and hence E° cathode will decrease. Now EMF of cell is
E0cell = E0cathode – E0anode
A decrease in E°cathode and a corresponding increase in E°anode will mean that EMF of the cell will decrease and will ultimately become zero i.e., cell slops working after some time.

Question 58.
Calculate ΔrG° and log Kc for the following reaction at 298 K.
2 \mathrm{Cr}_{(\mathrm{s})}+3 \mathrm{Cd}_{(\mathrm{aq})}^{2+} \longrightarrow 2 \mathrm{Cr}_{\mathrm{raq}}^{3+}+3 \mathrm{Cd}_{(\mathrm{s})}
[Given : \mathbf{E}_{\mathrm{Cell}}^{\mathbf{0}} = +0.34 V, IF = 96500 C mol-1] (Comptt. All India 2017)
Answer:
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 21

Question 59.
Calculate ΔrG° and log K. for the following reaction at 298 K. (Comptt. All India 2017)
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 22
Answer:
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 23

Electrochemistry Class 12 Important Questions long Answer Type (LA)

Question 60.
(a) Define molar conductivity of a substance and describe how for weak and strong electrolytes, molar conductivity changes with concentration of solute. How is such change explained?
(b) A voltaic cell is set up at 25°C with the following half cells :
Ag+ (0.001 M) | Ag and Cu2+ (0.10 M) | Cu What would be the voltage of this cell? (E0cell = 0.46 V) (Delhi 2009)
Answer:
(a) Molar conductivity: Molar conductivity of a solution at a given concentration is tire conductance of the volume ‘V’ of a solution containing one mole of electrolyte kept between two electrodes with area of cross section ‘A’ and distance of unit length. It is represented by Λm (lamda).
Λm = \frac{K A}{l} ∴ l = 1 and A = V
∴ Λm= KV Unit = S cm2 mol-1
Effect of change of concentrations on molar conductivity. In case of strong electrolytes there is a small increase in conductance with dilution because a strong electrolyte is completely dissociated in solution and the number of ions remains constant. Moreover there will be greater inter-ionic attractions at higher concentrations which retards the motion of ions and conductance decreases. In case of weak electrolytes there is increase in conductance with decrease in concentration due to the increase in the number of ions in the solution.
The graph between Λm and concentration also rectifies the above statement.
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 24
(b) The reaction takes place in cell as
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 25
= 0.46 – \frac{0.059}{2} × 5
= 0.46 – 0.1475 = 0.3125

Question 61.
(a) State the relationship amongst cell constant of a cell, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solute related to conductivity of its solution?
(b) A voltaic cell is set up at 25°C with the following half-cells :
Al | Al3+ (0.001 M) and Ni | Ni2+ (0.50 M)
Calculate the cell voltage
E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{0}=-0.25 \mathrm{V}, E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{0}=-1.66 \mathrm{V}
(Delhi 2009)
Answer:
(a) The relationship between the cell constant of a cell (G*), resistance of the solution in the cell (R) and conductivity (K) is given by
K = \frac{\text { Cell constant }}{\mathrm{R}}=\frac{\mathrm{G}^{\star}}{\mathrm{R}}
The relationship between molar conductivity (Λm) and conductivity of the solution (K) is given by
Λm = \frac{\mathrm{K}}{\mathrm{C}}

(b) The cell may be represented as
Al | Al3+ || Ni2+| Ni
E0cell = E0R – E0L
E0cell = (-0-25) – (-1.66)
∴ E0 = -0.25 + 1.66 = 1.41 V

Question 62.
(a) What type of a cell is the lead storage battery? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery while operating.
(Delhi, All India 2009)
(b) A voltaic cell is set up at 25 °C with the half-cells, Al | Al3+ (0.001 M) and Ni | Ni2+ (0.50 M). Write the equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
(Given : E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{\circ}=-0.25 \mathrm{V}, E_{\mathrm{Al}^{3+} / \mathrm{Al}}^{\circ}=-1.66 \mathrm{V}) (Delhi 2009)
Answer:
(a) The lead storage battery is a secondary cell (rechargeable). During discharging the electrode reaction occurs as follows :
At anode :
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 26

(b) The cell may be represented as
Al | Al3+ || Ni2+| Ni
E0cell = E0R – E0L
E0cell = (-0-25) – (-1.66)
∴ E0 = -0.25 + 1.66 = 1.41 V

Question 63.
(a) Express the relationship amongst cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solute related to conductivity of its solution?
(b) Calculate the equilibrium constant for the reaction
Fe(s) + Cd2+(aq) ⇌ Fe2+(aq) + Cd(s)
(Given : E_{\mathrm{Cd}^{2+} / \mathrm{Cd}}^{0} = -0.40 V,
E_{\mathrm{Cd}^{2+} / \mathrm{Cd}}^{0} = -0.44 V). (Delhi 2009)
Answer:
(a) The relationship between the cell constant of a cell (G*), resistance of the solution in the cell (R) and conductivity (K) is given by
K = \frac{\text { Cell constant }}{\mathrm{R}}=\frac{\mathrm{G}^{\star}}{\mathrm{R}}
The relationship between molar conductivity (Λm) and conductivity of the solution (K) is given by
Λm = \frac{\mathrm{K}}{\mathrm{C}}

Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 27

Question 64.
(a) Define the term molar conductivity. How is it related to conductivity of the related solution?
(b) One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration. Its other half-cell consists of a zinc electrode dipping in 1.0 M solution of Zn(NO3)2. A voltage of 1.48 V is measured for this cell. Use this information to calculate the concentration of silver nitrate solution used.
(E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0}=-0.76 \mathrm{V}, E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0} = + 0.80 V) (Delhi 2009)
Answer:
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 23

(b) Half cell reactions :
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 28
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 29

Question 65.
(a) Corrosion is essentially an electrochemical phenomenon. Explain the reactions ; occurring during corrosion of iron kept in j an open atmosphere.
(b) Calculate the equilibrium constant for the equilibrium reaction
Fe(s) + Cd2+(aq) ⇌ Fe2+(aq) + Cd(s)
(Given : E_{\mathrm{Cd}^{2+} / \mathrm{Cd}}^{0} = – 0.40 V,
E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{0} = -0-44V) (Delhi 2009)
Answer:
(a) Corrosion is an electrochemical i phenomenon. At a particular spot of iron j oxidation it takes place and the spot behaves j as anode and the reaction will be : i
At anode :
2Fe(s) → 2Fe2+ + 4e
Electrons released above move through the metal to another spot and reduce oxygen in presence of H+ and the spot behaves as cathode with the reaction.
At cathode :
O2(g) + 4H+ (aq) + 4e → 2H2O(I)
Overall reaction :
2Fe(s) + O2(g) + H+(aq) → 2Fe2+(aq) + 2H2O (l)
The ferrous ions are further oxidized by atmospheric oxygen to form hydrated ferric oxide (Fe2O3 xH2O) as rust

Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 27

Question 66.
(a) State Kohlrausch law of independent migration of ions. Write an expression for the molar conductivity of acetic acid at infinite dilution according to Kohlrausch law.
(b) Calculate Λ°m for acetic acid.
Given that Λ°m (HCl) = 426 S cm2 mol-1
Λ°m (NaCl) = 126 S cm2 mol-1
Λ°m (CH3COONa) = 91 S cm2 mol-1 (Delhi 2010)
Answer:
(a) Kohlrausch law of independent migration of ions : The limiting molar conductivity of an electrolyte (i.e. molar conductivity at infinite dilution) is the sum of the limiting ionic conductivities of the cation and the anion each multiplied with the number of ions present in one formula unit of the electrolyte
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 30

Question 67.
(a) Write the anode and cathode reactions and the overall reaction occuring in a lead storage battery.
(b) A copper-silver cell is set up. The copper ion concentration is 0.10 M. The concen¬tration of silver ion is not known. The cell potential when measured was 0.422 V. Determine the concentration of silver ions in the cell. (Delhi 2010)
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 31
Answer:
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 32

(b) The reaction takes place at anode and cathode in the following ways :
At anode (oxidation) :
Cu(s) → Cu2+(aq) + 2e
At cathode (reduction) :
Cu(s) + 2Ag2+(aq) → Cu2+(aq) + 2Ag(s)
The complete cell reaction is
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 6

Question 68.
(a) What type of a battery is lead storage battery? Write the anode and cathode reactions and the overall cell reaction occuring in the operation of a lead storage battery.
(b) Calculate the potential for half-cell containing
0.10 M K2Cr2O7 (aq), 0.20 M Cr3+ (aq) and 1.0 × 10-4 M H+ (aq).
The half-cell reaction is
Cr2O72-(aq) + 14 H+ (aq) + 6e → 2 Cr3+ (aq) + 7 H2O (l)
and the standard electrode potential is given as E0 = 1.33 V. (All India 2011)
Answer:
(a) The lead storage battery is a secondary cell (rechargeable). During discharging the electrode reaction occurs as follows :
At anode :
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 26

(b) E = ? E0 = 1.33 V
E0cell = 1.33 – \frac{0.0591}{6} \log \frac{\left[\mathrm{Cr}^{+3}\right]^{2}}{\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]\left[\mathrm{H}^{+}\right]^{14}}
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 33

Question 69.
(a) How many moles of mercury will be produced by electrolysing 1.0 M Hg(NO3)2 solution with a current of 2.00 A for 3 hours?
[Hg(NO3)2 = 200.6 g mol-1]
(b) A voltaic cell is set up at 25 °C with the following half-cells Al2+ (0.001 M) and Ni2+ (0.50 M). Write an equation for the reaction that occurs when the cell generates an electric current and determine the cell potential.
(Given : E_{\mathrm{Ni}^{2+} / \mathrm{Ni}}^{0} = – 0.25 V, E_{\mathrm{Al}^{3+} / \mathrm{Al}}^{0} = – 1.66 V) (All India 2011)
Answer:
(a) Quantity of electricity (Q) = I × t
= 2 × 3 × 60 × 60
= 21600 C
Hg2+ + 2e → Hg
Thus 2F i.e. 2 × 96500 C deposists Hg = 1 mole
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 34

Question 70.
(a) What type of a battery is the lead storage battery? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery when current is drawn from it
(b) In the button cell, widely used in watches, the following reaction takes place
Zn(s) + Ag2O(s) → Zn2+(aq) + 2Ag(s) + 2OHO(aq)
Determine E0 and ΔG0 for the reaction.
(Given E_{\mathrm{A} \mathrm{g}^{+} / \mathrm{Ag}}^{0} = +0.80V, E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0} = -0.76 V) (Delhi 2011)
Answer:
(a) It is a secondary cell.
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 35

Question 71.
(a) Define molar conductivity of a solution and explain how molar conductivity changes with change in concentration of solution for a weak and a strong electrolyte.
(b) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10-3 S cm-1 ? (Delhi 2011)
Answer:
(a) Molar conductivity: Conductivity of 1 M electrolytic solution placed between two electrodes 1 cm apart and have enough area of cross-section to hold the entire volume is known as molar conductivity or conductivity observed for one molar solution of electrolyte. Molar condctivity increases with decrease in concentration of solute for both weak and strong electrolytes.

(b) R = ρ (l/a)
Cell constant \frac{l}{a}=\frac{R}{\rho} = RK
= (1500 Ω) × 0.146 × 10-3 S cm-1 = 0.219 cm-1

Question 72.
(a) Define the following terms :
(i) Limiting molar conductivity (ii) Fuel cell
(b) Resistance of a conductivity cell filled with 0.1 mol L-1 KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L-1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L-1 KCl solution. The conductivity of 0.1 mol L-1 KCl solution is 1.29 × 10-2-1 cm-1. (Delhi 2014)
Answer:
(a) (i) Fuel cells : These cells are the devices which convert the energy produced during combustion of fuels like H2, CH4, etc. directly into electrical energy.
(ii) The molar conductivity of a solution at infinite dilution is called limiting molar conductivity and is represented by the symbol Λ°m.

(b) For 0.1 M KCl solution
R = 100 Q, K = 1.29 × 10-2-1 cm-1
Formula : Cell constant
= Conductivity × Resistance
= 1.29 × 10-2 × 100
= 1.29 cm-1 or 129 m-1
For 0.2 M KCl solution, conductivity
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 36

Question 73.
(a) State Faraday’s first law of electrolysis. How much charge in terms of Faraday is required for the reduction of 1 mol of Cu2+ to Cu.
(b) Calculate emf of the following cell at 298 K : Mg(s) | Mg2+ (0.1 M) || Cu2+ (0.01) | Cu (s)
[Given E0cell = +2.71 V, 1 F = 96500 C mol-1] (Delhi 2014)
Answer:
According to first law of Faraday’s “the amount of chemical reaction and hence the mass of any substance deposited/liberated at any electrode is directly proportional to the quantity of electricity passed through the electrolyte.”
The quantity of charge required for reduction of 1 mol of Cu2+
= 2 faradays (∵ Cu2+ + 2e → Cu)
= 2 × 96500 C = 193000 C
Cell reaction : Mg + Cu2+ Mg2+ + Cu(n = 2)
Using Nernst equation,
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 37

Question 74.
(a) Define the terms conductivity and molar conductivity for the solution of an electrolyte. Comment on their variation with temperature.
The measured resistance of a conductance cell was 100 ohms. Calculate (i) the specific conductance and (ii) the molar conductance of the solution.
(KC1 = 74.5 g mol-1 and cell constant = 1.25 cm-1 ) (Comptt. Delhi 2014)
Answer:
(a) Conductivity : Reciprocal of resistivity is called conductivity
k = \frac{1}{R} \times \frac{l}{A}
Molar conductivity : It is defined as the conductivity of solution containing 1 mole solute dissolved per litre when placed between two electrodes of unit area separated by 1 cm.
Molar conductivity and conductivity of solution increase on increasing the temperature.

(b) Given : R = 100 Ω Cell constant = 1.25 cm-1
Molarity = 74.5 g mol-1
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 38

Question 75.
(a) Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with platinum electrodes.
(ii) An aqueous solution of H2SO4 with platinum electrodes.
(b) Estimate the minimum potential difference needed to reduce Al2O3 at 500°C. The Gibbs energy change for the decomposition reaction \frac{2}{3} Al2O3\frac{4}{3} Al + O2 is 960 kJ
(F = 96500 C mol-1) (Comptt. Delhi 2014)
Answer:
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 39

Question 76.
(a) Define the following terms :
(i) Molar conductivity (Λm)
(ii) Secondary batteries
(iii) Fuel cell
(b) State the following laws :
(i) Faraday first law of electrolysis
(ii) Kohlrausch’s law of independent migration of ions (Comptt. Delhi 2014)
Answer:
(a) (i) Molar conductivity Λm): Molar conductivity can be defined as the conductance of the volume V of electrolytic solution kept between two electrodes of a conducting cell at distance of unit length but having area of cross section large enough to accomodate sufficient volume of solution that contains one mole of the electrolyte.
Λm = KV
(ii) Secondary batteries: Those cells which can be recharged on passing electric current through them in opposite direction and can be used again are called secondary batteries, e.g. Lead-acid storage cell.
(iii) Fuel cell : Galvanic cells that are designed to convert the chemical energy of combustion of fuels like hydrogen, methane etc. into electrical energy are called fuel cells, e.g. H2 – O2 fuel cell

(b) (i) Faraday first law of electrolysis : According to this law the mass of the substance deposited or liberated at any electrode during electrolysis is directly proportional to the quantity of charge passed through the electrolyte.
ω ∝ Q (∵ Q = I × t)
ω = = ZIt
(ii) Kohlrausch’s law of independent migration of ions : According to this law limiting molar conductivity of an electrolyte can be represented as the sum of the limiting ionic conductivities of the cation and the anion each multiplied with the number of ions present in one formula unit of electrolyte.
\Lambda_{m}^{\mathrm{o}} \text { for } \mathrm{A} x \mathrm{B} y=x \lambda_{+}^{\mathrm{o}}+y \lambda_{-}^{\mathrm{o}}

Question 77.
(a) Define the term degree of dissociation. Write an expression that relates the molar conductivity of a weak electrolyte to its degree of dissociation.
(b) For the cell reaction
Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)
Calculate the equilibrium constant at 25 °C. How much maximum work would be obtained by operation of this cell?
\mathrm{E}_{\mathrm{Ni}^{2} / \mathrm{Ni}}^{\mathrm{o}} = 0.25 V and \mathbf{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\mathrm{o}} = 0.80 V (Comptt. Delhi 2014)
Answer:
(a) Degree of dissociation: It is the measure of the extent to which an electrolyte gets dissociated into its constitutent ions.
Thus higher the degree of dissociation, higher will be its molar conductance. Mathematically it can be expressed as :
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 40
Maximum work = 106.150 KJ mol-1

Question 78.
Calculate ΔrG° and e.m.f. (E) that can be obtained from the following cell under the standard conditions at 25°C :
Zn (s) | Zn2+ (aq) || Sn2+ (aq) | Sn (s)
Answer:
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 41

Question 79.
(a) Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
(b) Calculate the standard cell potential of the galvanic cell in which the following reaction takes place :
Fe2+ (aq) + Ag+ (aq) → Fe3+ (aq) + Ag (s) Calculate the ΔrG° and equilibrium constant of the reaction also.
(\mathbf{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0} = 0.80 V; \mathbf{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{0} = 0.77 V)
(Comptt. All India 2014)
Answer:
(a) Conductivity : The conductance of the solution of an electrolyte enclosed in a cell between two electrodes of unit area of cross section separated by 1 cm. It is represented as K with unit ohm-1 cm-1
Molar conductivity : It is the conductance of the volume V of solution containing one mole of electrolyte kept between two electrodes with area of cross section A and distance of unit length.
Molar conductivity increases with decrease in concentration of solute for both weak and strong electrolytes.
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 42

Question 80.
(a) Calculate E0cell for the following reaction at 298 K:
2Al(s) + 3Cu2+ (0.01M) → 2Al2+ (0.01M) + 3Cu(s)
Given: Ecell = 1.98 V
(b) Using the E0 values of A and B, predict which is better for coating the surface of iron [E0(Fe2+/Fe) = -0.44 V] to prevent corrosion and why?
Given: E0(A2+/A) = -2.37 V; E0(B2+/B) = -0.14 V (All India 2016)
Answer:
(a) For the reaction
2Al(s) + 3Cu2+ (0.01M) → 2Al3+ (0.01M) + 3Cu(s)
Given: Ecell = 1.98 V E0cell = ?
Using Nernst equation
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 43
(b) Element A will be better for coating the surface of iron than element B because its E° value is more negative.

Question 81.
(a) The conductivity of 0.001 mol L-1 solution of CH3COOH is 3.905 × 10-5 S cm-1. Calculate its molar conductivity and degree of dissociation (α).
Given: λ0(H+) = 349.6 S cm2 mol-1 and λ0 (CH3COO) = 40.9 S cm2 mol-1
(b) Define electrochemical cell. What happens if external potential applied becomes greater than E0cell of electrochemical cell? (All India 2016)
Answer:
(a) Concentration = 0.001 mol L-1
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 44
(b) Electrochemical cell: It is a device which converts chemical energy into electrical energy i.e., produced as a result of redox reaction taking place in the electrolyte.
The reaction gets reversed and it becomes non-spontaneous. It starts acting as an electrolytic cell.

Question 82.
(a) Define the following :
(i) Molar conductivity
(ii) Fuel cell
(b) The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2 mol-1 Calculate the conductivity of the solution. (Comptt. Delhi 2016)
answer:
(a) (i) Molar conductivity Λm): Molar conductivity can be defined as the conductance of the volume V of electrolytic solution kept between two electrodes of a conducting cell at distance of unit length but having area of cross section large enough to accomodate sufficient volume of solution that contains one mole of the electrolyte.
Λm = KV
(ii) Secondary batteries: Those cells which can be recharged on passing electric current through them in opposite direction and can be used again are called secondary batteries, e.g. Lead-acid storage cell.
(iii) Fuel cell : Galvanic cells that are designed to convert the chemical energy of combustion of fuels like hydrogen, methane etc. into electrical energy are called fuel cells, e.g. H2 – O2 fuel cell
(b) Λm = \frac{k \times 1000}{c} S cm2 mol-1
138.9 = \frac{k \times 1000}{1.5}
k = 0.208 S cm-1

Question 83.
(a) Define the following terms :
(i) Primary batteries
(ii) Corrosion
(b) The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10-3 S cm-1? (Comptt. Delhi 2016)
Answer:
(a) (i) Primary batteries : Batteries which can’t be rechared/reused.
(ii) Corrosion : Loss of useful metals due to oxidation on exposure to atmospheric gases and moisture.
(b) k = \frac{1}{R}\left(\frac{l}{A}\right)
\frac{l}{\mathrm{A}} = k × R = 0.146 × 10-3 S cm-1 × 1500 Ω
\frac{l}{\mathrm{A}} = 0.219 cm-1

Question 84.
(a) What are the two classifications of batteries? What is the difference between them?
(b) The resistance of 0.01 M NaCl solution at 25°C is 200 Ω. The cell constant of the conductivity cell is unity. Calculate the molar conductivity of the solution. (Comptt. All India 2016)
Answer:
(a) Two types of batteries are Primary batteries and Secondary batteries. Primary batteries are non-chargeable whereas secondary batteries are rechargeable.
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 45

Question 85.
(a) What are fuel cells? Give an example of a fuel cell.
(b) Calculate the equilibrium constant (log Kc) and ΔrG° for the following reaction at 298 K.
Cu (s) + 2Ag+ (aq) ⇌ Cu2+ (aq) + 2Ag (s)
Given E0cell = 0.46 V and IF = 96500 C mol-1 (Comptt. All India 2016)
Answer:
(a) Cell which converts energy of combustion of fuel directly into electricity.
Example : H2 – O2 fuel cell.
Or
Those cells which convert fuel energy directly into electrical energy.
Example : H2 – O2 fuel cell
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 46

Question 86.
(a) When a bright silver object is placed in the solution of gold chloride, it acquires a golden tinge but nothing happens when it is placed in a solution of copper chloride. Explain this behaviour of silver.
[Given : \mathbf{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0} =+0.34V,\mathbf{E}_{\mathbf{A} \mathbf{g}^{+} / \mathbf{A g}}^{\mathbf{0}} =+0.80V, \mathrm{E}_{\mathrm{Au}^{3+} / \mathrm{Au}}^{0} = +1.40V]
(b) Consider the figure given and answer the following questions :
(i) What is the direction of flow of electrons?
(ii) Which is anode and which is cathode?
(iii) What will happen if the salt bridge is removed?
Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 47
(iv) How will concentration of Zn2+ and Ag+ ions be affected when the cell functions?
(v) How will concentration of these ions be affected when the cell becomes dead? (Comptt. Delhi 2017)
Answer:
(a) The standard electrode potential, E° for silver is 0.80 V and that of gold is 1.5 V, hence silver can displace gold from its solution. The replaced gold is deposited on silver object due to which golden tinge is obtained. On the other hand E° for Cu is 0.34 V which is lower than that of silver, thus silver cannot replace copper from its solution.

(b)

(i) Electrons flow from anode (Zinc plate) to cathode (Silver plate).
(ii) Zinc plate where oxidation occurs acts as anode and silver plate where reduction occurs acts as cathode.
(iii) If the salt bridge is removed then electrons from zinc electrode will flow to the silver electrode where they will neutralize some of Ag+ ions and the \mathrm{SO}_{4}^{2-} ions will be left and the solution will acquire a negative charge. Secondly the Zn2+ ions from zinc plate will enter into ZnSO4 solution producing positive charge. Thus due to accumulation of charges in two solutions, further flow of electrons will stop and hence the current stops flowing and the cell will stop functioning.

(iv) As silver from silver sulphate solution is deposited on the silver electrode and sulphate ions migrate to the other side, the concentration of AgSO4 solution decreases and of ZnSO4 solution increases as the cell operates.
(v) When the cell becomes dead, the concentration of these ions become equal due to attainment of equilibrium and zero EMF.

Question 87.
(a) What is limiting molar conductivity? Why there is steep rise in the molar conductivity of weak electrolyte on dilution?
(b) Calculate the emf of the following cell at 298 K :
Mg (s) | Mg2+ (0.1 M) || Cu2+ (1.0 × 10-3M) | Cu (s)
[Given = \mathbf{E}_{\text { Cell }}^{0} = 2.71 V]. (Comptt. Delhi 2017)
Answer:
(a) The molar conductivity of a solution at infinite dilution is called limiting molar conductivity and is represented by the
symbol \Lambda_{m}^{\circ}
There is steep rise in the molar conductivity of weak electrolyte on dilution because as the concentration of the weak electrolyte is reduced, more of it ionizes and thus increase in the number of ions in the solution.

Important Questions for Class 12 Chemistry Chapter 3 Electrochemistry Class 12 Important Questions 48

Important Questions for Class 12 Chemistry

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Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

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Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

Some Applications of Trigonometry Class 10 Important Questions Very Short Answer (1 Mark)

Question 1.
A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then calculate the height of the wall. (2013OD)
Solution:
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 1
∠BAC = 180° – 90° – 60o = 30°
sin 30° = \frac{B C}{A C}
\frac{1}{2}=\frac{B C}{15}
2BC = 15
BC = \frac{15}{2}m

Question 2.
In the given figure, a tower AB is 20 m high and BC, its shadow on the ground, is 20 \sqrt{3} m long. Find the Sun’s altitude. (2015OD)
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 2
Solution:
AB = 20 m, BC = 20 \sqrt{3} m,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 3
θ = ?
In ∆ABC,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 4

Question 3.
In the figure, AB is a 6 m high pole and CD is a ladder inclined at an angle of 60° to the horizontal and reaches up to a point D of pole. If AD = 2.54 m. Find the length of the ladder. (Use \sqrt{3} = 1.73) (2016D)
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 5
Solution:
BD = AB – AD = 6 – 2.54 = 3.46 m
In rt., ∆DBC, sin 60° = \frac{B D}{D C}
\frac{\sqrt{3}}{2}=\frac{3.46}{\mathrm{DC}}
\sqrt{3}DC = 3.46 x 2
∴ Length of the ladder, DC = \frac{6.92}{\sqrt{3}}=\frac{6.92}{1.73}
DC = 4 m

Question 4.
A ladder, leaning against a wall, makes anangle of 60° with the horizontal. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder. (2016OD)
Solution:
Let AC be the ladder
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 6
∴ Length of ladder, AC = 5 m 2.5 m

Question 5.
If a tower 30 m high, casts a shadow 10\sqrt{3} m long on the ground, then what is the angle of elevation of the sun? (2017OD)
Solution:
Let required angle be θ.
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 7
tan θ = \frac{30}{10 \sqrt{3}}
tan θ = \sqrt{3}
⇒ tan θ = tan 60° ∴ θ = 60°

Question 6.
The tops of two towers of height x and y, standing on level ground, subtend angles of 30° and 60° respectively at the centre of the line joining their feet, then find x : y. (2015D)
Solution:
When base is same for both towers and their heights are given, i.e., x and y respectively
Let the base of towers be k.
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 8

Some Applications of Trigonometry Class 10 Important Questions Short Answer-I (2 Marks)

Question 7.
The angles of depression of two ships from the top of a light house and on the same side of it are found to be 45° and 30°. If the ships are 200 m apart, find the height of the light house. (2012D)
Solution:
Let AB be the height of the light house,
D and C are two ships and DC = 200 m
Let BC = x m, AB = h m
In rt. ∆ABC,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 9
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 10
∴ Height of the light house = 273 m

Question 8.
The horizontal distance between two poles is 15 m. The angle of depression of the top of first pole as seen from the top of second pole is 30°. If the height of the second pole is 24 m, find the height of the first pole. [Use \sqrt{3} = 1.732] (2013D)
Solution:
Let AB be the 1st pole
and CE = 24 m be the 2nd pole
In ∆ADE,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 11
∴ Height of the first pole,
AB = CD = CE – DE
= 24 – 8.66 = 15.34 m

Question 9.
As observed from the top of a 60 m high light house from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the light-house, find the distance between the two ships. (Use \sqrt{3} = 1.732] (2013OD)
Solution:
Let AB = 60 m be the height of Light-house and C and D be the two ships.
In right ∆ABC,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 12
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 13
∴ Distance between the two ships,
CD = BD – BC = 103.92 – 60 = 43.92 m

Question 10.
Two ships are there in the sea on either side of a light house in such a way that the ships and the light house are in the same straight line. The angles of depression of two ships as observed from the top of the light house are 60° and 45°. If the height of the light house is 200 m, find the distance between the two ships. [Use \sqrt{3} = 1.73] (2014D)
Solution:
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 14
Let AB be the light house and C and D be the two ships,
In rt. ∆ABC, tan 45° = \frac{\mathrm{AB}}{\mathrm{BC}}
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 15

Question 11.
The angle of elevation of an aeroplane from a point on the ground is 60°. After a flight of 30 seconds the angle of elevation becomes 30°. If the aeroplane is flying at a constant height of 3000\sqrt{3} m, find the speed of the aeroplane. (2014OD)
Solution:
Let A be the point on the ground and C be the aeroplane.
In rt. ∆ABC,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 16
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 17

Question 12.
From the top of a 60 m high building, the angles of depression of the top and the bottom of a tower are 45° and 60° respectively. Find the height of the tower. (Take \sqrt{3} = 1.73] (2014OD)
Solution:
Let AC be the building & DE be the tower.
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 18
∴ Height of the tower, DE = BC
DE = AC – AB
DE = 60 – 20\sqrt{3} = 20(3 – \sqrt{3})
DE = 20(3 – 1.73) = 20(1.27)
DE = 25.4 m

Question 13.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 45°. If the tower is 30 m high, find the height of the building. (2015D)
Solution:
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 19
Let height of building be x and the distance between tower and building be y.
In ∆ABC,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 20
∴Height of the building is 10\sqrt{3} m.

Question 14.
The angle of elevation of an aeroplane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the aeroplane is flying at a constant height of 1500\sqrt{3} m, find the speed of the plane in km/hr. (2015OD)
Solution:
Let AL = x, BL = CM = 1500\sqrt{3}
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 21

Question 15.
The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building. [Use \sqrt{3} = 1.73]. (2016D)
Solution:
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 22
Let AE be the building and CD be the tower. Let height of the tower = h m
and, the horizontal distance between tower and building = x m …[Given
BD = AE = 50 m
∴ BC = CD – BD = (h – 50) m
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 23
From (i), x = h – 50
= 118.25 – 50 = 68.25 m
Height of the tower, h = 118.25 m
∴ Horizontal distance between tower and Building, x = 68.25 m

Question 16.
A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of hill as 30°. Find the distance of the hill from the ship and the height of the hill. (2016D)
Solution:
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 24
Let the man standing on the deck of a ship be at point A and let CE be the hill.
Here BC is the distance of hill from ship and CE be the height of hill.
In rt. ∠ABC, tan 30° = \frac{\mathrm{AB}}{\mathrm{BC}}
BC = 10\sqrt{3} m .(i)
BC = 10(1.73) = 17.3 m …[:: \sqrt{3} = 1.73
AD = BC = 10 \sqrt{3} m …(ii) [From (i)
In rt. ∆ADE, tan 60° = \frac{\mathrm{DE}}{\mathrm{AD}}
\sqrt{3}=\frac{\mathrm{DE}}{10 \sqrt{3}} … [From (ii)
⇒ DE = 10\sqrt{3} × \sqrt{3} = 30 m
∴ CE = CD + DE = 10 + 30 = 40 m
Hence, the distance of the hill from the ship is 17.3 m and the height of the hill is 40 m.

Some Applications of Trigonometry Class 10 Important Questions Long Answer (4 Marks)

Question 17.
The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 30°. Find the height of the tower. (2011OD)
Solution:
Let AC be the tower
Let AC = y m
Let DC = EB = x m
In rt. ∆ABE,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 25
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 26
\sqrt{3}.\sqrt{3} (y – 10) = y … [From (i)
3(y – 10) = y ⇒ 3y – 30 = y
3y – y = 30 ⇒ 2y = 30
∴ Height of the tower, y = 15 m

Question 18.
From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the cars. (Use \sqrt{3} = 1.732] (2011D)
Solution:
Let AB be the tower.
In rt. ∆ABC, tan 45° = \frac{\mathrm{AB}}{\mathrm{BC}}
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 27
∴ Distance between the cars, CD
= BD + BC
= 173 + 100
= 273 m

Question 19.
Two poles of equal heights are standing opposite to each other on either side of the road, which is 100 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles. (2011D)
Solution:
Let AB and DE be the two equal poles and C be the point on BD (road).
Let BC = x m
Then CD = (100 – x) m
Let AB = DE = y m
In rt. ∆ABC,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 28
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 29
⇒ 3y = 100\sqrt{3} – y
⇒ 4y = 100\sqrt{3}
∴ Height of the poles,
y = \frac{100 \sqrt{3}}{4} = 25\sqrt{3} m = 25(1.73) = 43.25 m

Question 20.
From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 10 m high building are 30° and 60° respectively. Find the height of the tower. (2011D)
Solution:
Let BC be the building and CD be the transmission tower. A be the point on the ground.
Let CD = y m
In rt. ∆ABC,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 30
∴ Height of transmission tower = 20 m

Question 21.
From the top of a vertical tower, the angles of depression of two cars, in the same straight line with the base of the tower, at an instant are found to be 45° and 60°. If the cars are 100 m apart and are on the same side of the tower, find the height of the tower. [Use \sqrt{3} = 1.732] (2011OD)
Solution:
Let AB be the tower
Let AB = h m,
and BC = x m
In rt. ∆ABC,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 31
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 32
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 33
∴ Height of the tower, h = 236.5 m

Question 22.
The angles of depression of the top and bottom of a 12 m tall building, from the top of a multi-storeyed building are 30° and 60° respectively. Find the height of the multistoreyed building. (2011OD)
Solution:
Let AC be the multi-storeyed building and DE be the other building.
Let AC = y m,
DC = EB = x m
In rt. ∆ABE,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 34
tan 30° = \frac{\mathrm{AB}}{\mathrm{BE}}
\frac{1}{\sqrt{3}}=\frac{y-12}{x}
x = \sqrt{3}(y – 12)
In rt. ∆ACD, tan
60° = \frac{\mathrm{AC}}{\mathrm{DC}}
\frac{\sqrt{3}}{1}=\frac{y}{x}
\sqrt{3}x = y
\sqrt{3}.\sqrt{3} (y – 12) = y … [From (i)
⇒ 3 (y – 12) = y
⇒ 3y – 36 = y
⇒ 3y – y = 36
⇒ 2y = 36
⇒ y = 18
∴ Height of the multi-storeyed building, y = 18 m

Question 23.
The angle of elevation of the top of a building from the foot of a tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. (2011OD)
Solution:
Let AB = 50 m be the tower
and CD be the building.
In rt. ∆ABC,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 35
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 36

Question 24.
The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of depression from the top of the tower of the foot of the hill is 30°. If the tower is 50 m high, find the height of the hill. (2012D)
Solution:
Let CD = H m, be the height of the hill
AB = 50 be the tower
and Distance between them, BC = x m
In rt. ∆ABC,
tan 30° = \frac{\mathrm{AB}}{\mathrm{BC}}
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 37
\frac{1}{\sqrt{3}}=\frac{50}{x}
x = 50\sqrt{3}
In rt. ∆BCD,
tan 60° = \frac{\mathrm{CD}}{\mathrm{BC}}
\sqrt{3}  = \frac{\mathrm{H}}{50 \sqrt{3}} ⇒ H = 50\sqrt{3} × \sqrt{3} = 150
∴ Height of the hill = 150 m

Question 25.
From the top of a hill, the angles of depression of two consecutive kilometre stones due east are found to be 30° and 45°. Find the height of the hill. (2012D)
Solution:
Let AB be the hill of height h km.
Let C and D be two stones due east of the hill at a distance of 1 km from each other.
Let BC = x km
In rt. ∆ABC,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 38
tan 45° = \frac{A B}{B C}
= 1 = \frac{h}{x}
= h = x … (1)
In rt. ∆ABD, tan 30° = \frac{AB}{BD}
\frac{1}{\sqrt{3}}=\frac{h}{x+1}
\sqrt{3}h = x + 1
\sqrt{3}x = x + 1 …[From (i)
\sqrt{3}x – x = 1
(\sqrt{3} – 1)x = 1
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 39
Hence, the height of the hill is 1.365 km.

Question 26.
The shadow of a tower standing on a level ground is found to be 20 m longer when the Sun’s altitude is 45° than when it is 60°. Find the height of the tower. (2012D)
Solution:
Let AB be the tower = h m
BC = x m, DC = 20 m
In rt. ∆ABC, tan 60° = \frac{AB}{BC}
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 40
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 41
∴ Height of the tower = 47.3 m

Question 27.
A kite is flying at a height of 45 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string assuming that there is no slack in the string. (2012D )
Solution:
Let AB = 45 m be the height of the kite from the ground,
In rt. ∆ABC, sin 60° = \frac{AB}{AC}
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 42

Question 28.
The angles of depression of the top and bottom of a tower as seen from the top of a 60\sqrt{3} m high cliff are 45° and 60° respectively. Find the height of the tower. (2012D)
Solution:
Let AB = 60\sqrt{3} be the cliff and CD be the tower.
In rt. ∆ABC,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 43
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 44

Question 29.
The angles of elevation and depression of the top and bottom of a light-house from the top of a 60 m high building are 30° and 60° respectively. Find:
(i) the difference between the heights of the light-house and the building.
(ii) the distance between the light-house and the building. (2012OD)
Solution:
Let AB = 60 m be the building
and CE be the light-house.
In rt. ∆ABC,
tan 60° = \frac{AB}{BC}
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 45
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 46
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 47
∴ (i) Difference between the heights = 20 m
and (ii) Distance = 34.64 m

Question 30.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high, find the height of the building. (2013D)
Solution:
Let AB = 60 m be the tower
and let CD be the building.
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 48
∴Height of the building, CD = 20 m

Question 31.
From a point P on the ground, the angle of elevation of the top of a 10m tall building is 30°. A flagstaff is fixed at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (Take \sqrt{3} = 1.73) (2013D)
Solution:
Let QR = 10 m be the building
and RS be the flagstaff.
In right ∆PQR,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 49
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 50
Length of the flagstaff, RS = 17.3 – 10
= 7.3 m

Question 32.
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distances of the point from the poles. (2013D)
Solution:
Let AB = DE = y m be the two poles such that
BD = 80 m, CD = x m and BC = 80 – x m
In rt. ∆CDE,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 51
Height of pole = 34.6 m
Puting the value of y in (i),
x = \sqrt{3} (20\sqrt{3}) ∴ x = 60
∴ CD, x = 60 m
Distance of poin from the pole
and BC = 80 – x = 80 – 60 = 20 m

Question 33.
The angle of elevation of the top of a building from the foot of a tower is 30°. The angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high, find the height of the building. (2013OD)
Solution:
Refer to Question 30, Page 143.

Question 34.
The angles of elevation and depression of the top and the bottom of a tower from the top of a building, 60 m high, are 30° and 60° respectively. Find the difference between the heights of the building and the tower and the distance between them. (2014D)
Solution:
Let AB be the building and CE be the tower.
In rt. ∆ABC, tan 60° = \frac{AB}{AC}
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 52
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 53
∴ Difference between the heights of the building and the tower, DE = 20 m
Distance between them,
BC= 20\sqrt{3}m = 20(1.73) = 34.6 m

Question 35.
A peacock is sitting on the top of a pillar, which is 9 m high. From a point 27 m away from the bottom of the pillar, a snake is coming to its hole at the base of the pillar. Seeing the snake the peacock pounces on it. If their speeds are equal, at what distance from the hole is the snake caught? (2014D)
Solution:
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 54
Lęt PH be the pillar. Let the distance from the hole to the place where snake is caught = x m
Let P be the top of the pillar and S be the point where the snake is
∴ SC = (27 – x)m
SC = PC = (27 – x)m …[∵ Their speeds are equal
In rt. ∆PHC
PH2 + CH2 = PC2 …[Pythagoras’ theorem
92 + x2 = (27 – x)2
81 + x2 = 729 – 54x + x2
54x = 729 – 81 = 648
x = \frac{648}{54} = 12 m
Hence, required distance, x =12 m

Question 36.
The angle of elevation of the top of a tower at a distance of 120 m from a point A on the ground is 45°. If the angle of elevation of the top of a flagstaff fixed at the top of the tower, at A is 60°, then find the height of the flagstaff. (Use \sqrt{3} = 1.732] (2014OD)
Solution:
Let BC be the tower
and CD be the flagstaff.
In rt. ∆ABC, tan 45° = \frac{BC}{AB}
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 55
1 = \frac{BC}{120}
BC = 120 m …(i)
In rt. ∆ABD,
tan 60° = \frac{BD}{AB}
\sqrt{3} = \frac{BD}{120}
BD = 120 \sqrt{3} …(ii)
Height of the flagstaff,
CD = BD – BC
= 120\sqrt{3} – 120
= 120(\sqrt{3} – 1)
= 120(1.73 – 1)
= 120(0.73) = 87.6 m

Question 37.
From a point P on the ground the angle of elevation of the top of a tower is 30° and that of the top of a flag staff fixed on the top of the tower, is 60°. If the length of the flag staff is 5m, find the height of the tower. (2015D)
Solution:
In ABCD,
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 56
(i) \frac{x}{y} = tan 30° = \frac{1}{\sqrt{3}}
y = \sqrt{3}x ..(i)
In ∆ABC,
(ii) \frac{x+5}{y} = tan 60°
tan 60° = \sqrt{3}
\frac{x+5}{\sqrt{3} x} = \sqrt{3} …[From (i)
⇒ 3x = x + 5 or x = 2.5
∴ Height of Tower = x = 2.5 m

Question 38.
At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30°. The angle of depression of the reflection of the cloud in the lake, at A is 60°. Find the distance of the cloud from A. (2015OD)
Solution:
In ∆ABC
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 57
Putting the value of h in equation (i),
∴ x = 20\sqrt{3} m
Using pythagoras’ theorem,
AC = \sqrt{(h)^{2}+(x)^{2}}
∴ AC = \sqrt{(20)^{2}+(20 \sqrt{3})^{2}} = 40 m
Distance of the cloud from A = 40 m.

Question 39.
The angle of elevation of a cloud from a point 60 m above the surface of the water of a lake is 30° and the angle of depression of its shadow in water of lake is 60°. Find the height of the cloud from the surface of water. (2017D)
Solution:
Let the height of the cloud from the surface of lake EC = H m = EF (image in water)
and A be the point, the distance between A and perpendicular of cloud AD = x m
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 58
⇒ 3H – 180 = H + 60
⇒ 3H – H = 60 = 180
⇒ 2H = 240
⇒ H = 120
∴ Height of the cloud = 120 m

Question 40.
A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such a way that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Find the speed of flying of the bird. (Take \sqrt{3} = 1.732) (2016D)
Solution:
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 59
Let BC be the tree
In rt. ∆ABC, tan 45° = \frac{BC}{AB}
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 60

Question 41.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are 60° and 30° respectively. Find the height of the tower. (2016D)
Solution:
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 61
Let the height of the tower, AB = h m
In ∆ABC, tan 60° = \frac{AB}{AC}
\sqrt{3} = \frac{h}{4} ⇒ h = 4\sqrt{3}
Hence, height of the tower = 473 m

Question 42.
The angle of elevation of the top of a vertical tower PQ from a point X on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of tower is 45°. Find the height of the tower PQ and the distance PX. (Use \sqrt{3} = 1.732) (2016 OD)
Solution:
Let QR = x m
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 62
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 63
Hence, height of the tower PQ is 96.54 m and the distance PX is 109.3 m.

Question 43.
As observed from the top of a light house, 100 m high above sea level, the angles of depression of a ship, sailing directly towards it, changes from 30° to 60°. Find the distance travelled by the ship during the period of observation. (Use \sqrt{3} = 1.732) (2016OD)
Solution:
Let AB = 100 m be the light-house and B, C be the two positions of the ship and CD be the distance travelled by ship.
In rt. ∆ABC, tan 60° = \frac{AB}{BC}
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 64
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 65
∴ Distance travelled by ship is 115.46 m

Question 44.
An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45o and 60° respectively. Find the width of the river. [Use \sqrt{3} = 1.732] (2017OD)
Solution:
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 66
Let CD be height of an aeroplane flying above the ground and AB be the two banks of the river.
In rt. ∆BCD, tan 60° = \frac{CD}{BC}
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 67
The width of the river, AB = AC – BC
= 300 – 173.2
= 126.8 m

Question 45.
From a point on the ground, the angle of elevation of the top of a tower is observed to be 60°. From a point 40 m vertically above the first 300 AC point of observation, the angle of elevation of the top of the tower is 30°. Find the height of the tower and its horizontal distance from the point of observation. (2016OD)
Solution:
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 68
Let BD be the tower, i.e., x m.
In rt. ∆ABD, tan 60° = \frac{BD}{AB}
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 69
Important Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry 70
From (i) x = \sqrt{3} (20\sqrt{3}) = 60 m
∴Height of the tower, r = 60 m
Required horizontal distance, y = 20\sqrt{3}m
y = 20(1.73) = 34.6 m …[∵ \sqrt{3} = 1.73

Important Questions for Class 10 Maths

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Bihar Scholarship 2019 | List, Eligibility, Application, Timeline

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Bihar Scholarship 2019: Bihar Scholarship gives a vast range of educational opportunities for the candidates who are residing in the Bihar state permanently. The Bihar Government and different private organizations collectively offer plenty of scholarships for worthy and impoverished scholars of the state. The key purpose of each scholarship is to assist and prompt deserving learners to pursue farther studies despite any financial limitation.

The article provides here all the highlights of Bihar scholarships presented by the government of Bihar and other firms. Get details of the list of scholarship, eligibility criteria, the detailed application process, application timeline, and award details.

List of Bihar Scholarship

Here you will get the details of a number of Bihar scholarships are available,  time duration to apply for the scholarship. The table given here contains the entire Bihar scholarship list, with their provider details and tentative application period. This will present you with an insight into scholarships which are prepared exclusively for the candidates of Bihar. Check for the scholarship that suits your requirements and fills the respective application, specified for them.

Name of the ScholarshipProvided ByApplication Duration
Post Matric Scholarship (PMS) For OBC/SC/ST Students, BiharBackward and Extremely Backward Class Welfare Department, Government of BiharJanuary – March
Pre-Matric Scholarship for OBC StudentsBackward and Extremely Backward Class Welfare Department, Government of BiharNot Available
Chief Minister Medhavi Yojana for EBC/BC Students, BiharBackward and Extremely Backward Class Welfare Department, Government of BiharNot Available
BTSE Bihar Talent Search ExaminationMployd Edutech Pvt. Ltd. in association with Lok Kalyan ParishadFebruary – August
Combined Counselling Board (CCB) Scholarship, BiharCombined Counselling BoardOctober – April

Eligibility for Bihar Scholarship

Now, as you are informed of all the list of Bihar Scholarship, let us know further the respective eligibility criteria. Every scholarship has its own eligibility criteria but there is one criterion, which each scholarship has stated, which is, candidates need to be a permanent resident of Bihar state and no candidate is eligible who reside outside the state.

Now let us check the academic qualifications and maximum family income criteria for all the category students required as per the eligibility defined by the conducting Bihar Board.

Post Matric Scholarship (PMS), Bihar for OBC/SC/ST Students

  • The students studying at post-matriculation level (class 11 to postgraduation level) can apply for this scholarship.
  • They must have passed the last qualifying examination.
  • The annual income of the family should be less than INR 2.50 Lakh (for SC/ST candidates) and INR 1 Lakh (for OBC candidates).
  • They must not be in receipt of any other scholarship.

Pre-Matric Scholarship, Bihar, for OBC Students

  • The students studying in class 1 to 10 at a government or government-recognized school can apply for this scholarship.

Chief Minister Medhavi Yojana, Bihar, for EBC/BC Students

  • The students who have passed their class 10 examination organized by Bihar School Examination Board with first-class (over 60% marks) can apply for this scholarship.
  • The annual income of the family for BC students should be less than INR 1.50 Lakh.

BTSE Bihar Talent Search Examination, Bihar (for all candidates)

  • This scholarship examination is open for students of class 7 to 10.

Combined Counselling Board (CCB) Scholarship, Bihar (for all candidates)

  • The students studying at college/university level (pursuing diploma level, degree level or postgraduate level courses) can apply for this scholarship.
  • They must have passed the last qualifying examination with at least 40% to 50% marks.

How To Apply for Bihar Scholarships?

The application process varies for different Bihar scholarships. We are providing the details here for individual scholarships and how to apply for them. Check the details in the below-given table and proceed with the application procedure as per your requirement.

Post Matric Scholarship (PMS) For OBC/SC/ST StudentsCandidates have to submit online application through the National Scholarship Portal (NSP).
Pre-Matric Scholarship for OBC StudentsCandidates have to submit an offline application through respective institutions/District Education Officer (DEO).
Chief Minister Medhavi Yojana for EBC/BC Students,Candidates have to submit an offline application through the Education Department of the state.
BTSE (Bihar Talent Search Examination)Candidates have to submit an online application of Bihar Talent Search Examination.
Combined Counselling Board (CCB) ScholarshipCandidates have to submit an online application of Combined Counselling Board.

Bihar Scholarship – Application Procedure of Post Matric Scholarship

Interested candidates may apply online through the official website of National Scholarship Portal from Feb-2019 to 05-March-2019. To apply for the scholarship follow the below-given steps:

  • Visit the National Scholarship Portal, https://scholarships.gov.in/
  • Click on New User/Register.
  • Read the guidelines properly.
  • Continue to register.
  • Once registered, take note of student application ID and continue to apply.
  • The applicants will receive an OTP on their registered mobile number. Confirm the OTP.
  • Change the password.
  • Log in with your registration number, date of birth and password.
  • You will be directed to a webpage comprising important instructions about form filling.
  • Read all instructions thoroughly and tick the box given at the end of the page and click “Proceed”.
  • The minute you click on the “Proceed” button, you will be directed to the user dashboard.
  • Click on “Fill in application form” part.
  • Fill in additional details in the scholarship application form.
  • Click on submit button.
  • Once you click on submit after finishing the application form, you need to upload your photograph and other supporting documents.
  • Before heading towards the final submission of the application form, the candidates are suggested to go through every detail filled carefully to avoid any sort of error afterward. Also, there is no requirement of making changes to the information filled by the candidate, once finally submit the application form.
  • After final submission of the application form, the candidates are asked to take the print out of the form and submit it along with other supporting documents to their respective educational institutions.

Bihar Scholarship Documents Required for Application

  • Last Qualifying Exam Mark Sheet
  • Category Certificate
  • Family Income Certificate
  • Bank Passbook
  • Fee Receipt Number
  • Annual Non Refundable Amount
  • Enrollment Number
  • Student ID Proof
  • Domicile Certificate
  • Aadhar Card Number
  • Latest Passport Size Photograph

Bihar Scholarship Reward Details

Get the details of the scholarship amount that each Bihar scholarship will offer and the number of students who will be benefitted through these scholarships. The prize money or reward amount of scholarship and the number of students receiving benefits from them differs from scholarship to scholarship. Moreover, the government scholarships transfer the amount directly into the bank account of the receivers through Direct Benefit Transfer (DBT).

Post Matric Scholarship (PMS) For OBC/SC/ST Students

  • There are 3,00,000 scholarships available in this category
  • Maintenance allowance of up to Rs.1200 per month (for hostellers) and INR 550 per month (for day scholars) for 10 months
  • An additional allowance of up to Rs.240 per month for students with disabilities
  • Reimbursement of compulsory non-refundable fees
  • Study tour charges of up to Rs.1600 per annum
  • Thesis typing and printing charges of up to Rs.1600 for research scholars

Pre-Matric Scholarship for OBC Students

  • There are 3,00,000 scholarships available in this category
  • For day-scholars of class 1 to 4 – Rs.50 per month
  • For day scholars of class 5 to 6 – Rs.100 per month
  • For day scholars of class 7 to 10 – Rs.150 per month
  • For hostellers of class 1 to 10 – Rs.250 per month

Chief Minister Medhavi Yojana for EBC/BC Students

  • There are 1,55,000 scholarships available in this category
  • One-time financial assistance of Rs.10,000 to each scholar

BTSE Bihar Talent Search Examination

  • Rs.20,000 and a laptop for students scoring from 91% to 100% in BTSE
  • Rs.15,000 and a notepad for students scoring from 81% to 90% in BTSE
  • Rs.10,000 and a tab for students scoring from 71% to 80% in BTSE
  • Rs.5,000 and a smartwatch for students scoring from 61% to 70% in BTSE
  • Gift hampers for students scoring from 51% to 60% in BTSE

Combined Counselling Board (CCB) Scholarship

  • Scholarship from Rs.1 Lakh to Rs.3 Lakh

Important Instructions for Bihar Scholars

  • Fields provided with a red asterisk (*) mark are mandatory fields.
  • Candidates should separately inform the mistakes detected by them to the Institute/District/Region/State. The software provides the facility at the level of the Institute & State to edit & correct limited information.
  • The Fields which can be edited are, Gender, Religion, Category, Profession, Annual Income, Aadhar Number, Disability, Day Scholar/Hostler, Mode of Study, IFSC Code, Account No., Admission Fees and Tuition Fees.However, corrections made by the Institute/State, if any, would be conveyed instantly to the student through SMS/email.
  • Candidates can fill up the online application in as many sittings as he/she wishes until he/she are satisfied that you have entered all desirable fields correctly. The software provides the facility to save their application at every stage.
  • Aadhaar No. is not Mandatory for the Students in order to Register and fills up the application form online. Students can apply for Scholarship without entering the Aadhaar no. but in that case, they have to enter Aadhaar Enrollment Id.
  • An Application ID (Permanent ID) will be provided to the candidate once his/her Registration is done. It will be conveyed to candidates through SMS and e-mail. Students should memorize their Application ID as it will be required while applying for Fresh/renewal scholarship.
  • Student cannot apply as a fresh if he/she is a Renewal candidate. Their application will be rejected in that case.
  • The student should immediately approach the institute to contact the nodal officer of the State where the institute is located. Student can also approach the Nodal Officer of that State directly through e-mail under intimation to the Ministry. If their institute is an eligible institution, the State Government concerned would enter it into the database and then they can apply.
  • The name and contact details of the Nodal Officer/State Department of all States/UTs are available in “Services->Know your State Nodal Officer” option.
  • Student can check the status of Online Application by submitting his/her Permanent id and Date Of Birth and open the link “Check your Status”.
  • Documents are required to be uploaded only for the Scholarship Amount more than ₹50000 per annum

FAQ’s

Question 1.
What is the last date for submitting applications online?

Answer:
Closure dates for acceptance of various scholarship applications are available in National Scholarships Portal.

Question 2.
How can a candidate apply online for scholarship?

Answer:
In order to apply online, please visit the website through URL www.scholarships.gov.in

Question 3.
Who are eligible to apply for Scholarship Schemes?

Answer:
Students fulfilling the Scheme guidelines of various Ministries are eligible to apply for these scholarships. These are available on the Home Page of the Portal.

Question 4.
Can an applicant, edit the information already saved and up-to what time?

Answer:
All the information can be edited till the closure of application form. After final submission, your application will be forwarded to the next level and application hereby cannot be edited.

The post Bihar Scholarship 2019 | List, Eligibility, Application, Timeline appeared first on Learn CBSE.

INSPIRE Scholarship 2019 | Important Dates, Eligibility, Application Procedure, Scholarship Rewards

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INSPIRE Scholarship: “Innovation in Science Pursuit for Inspired Research (INSPIRE)” is an innovative course sponsored and administered by the Department of Science & Technology(DST) for the pull of talent to Science. The basic goal of INSPIRE is to deliver to the youth of the country the activities of creative pursuit of science, pull talent to the art of science at an early age and thus create the required significant human resource pool for sustaining and expanding the Science & Technology system and R & D base.

A prominent feature of the program is that it does not consider in conducting competitive exams to identify talent at any level. It concludes in and relies on the ability of the existing educational edifice to identify the talent.

Government of India has passed the INSPIRE Scheme in November 2008 at a total expense of about Rs 1979.25 crores in the 11th Plan Period and Honourable Prime Minister originated the program on 13th December 2008. The plan is continuing in the 12th Plan period at a budgetary allocation of Rs. 2200 crores.

INSPIRE has three components

  1. Scheme for Early Attraction of Talent  (SEATS),
  2. Scholarship for Higher Education (SHE) and
  3. Assured Opportunity for Research Careers (AORC)

INSPIRE Scholarship Important Dates

The scholarship is advertised every year through the mid of October and ends at the end of the month of December. INSPIRE Scholarship for Higher Education 2019 is still to be declared. Find below the table showing the provisional dates of the scholarship.

Start Date of Online applicationOctober 2019
Closing Date of Online applicationDecember 2019
Release date of the list of shortlisted studentsApril-May 2020

INSPIRE Scholarship Eligibility

To apply for Scholarship for Higher Education (SHE), the candidates must have qualified class 12 and proceeding courses in Natural Sciences at the B.Sc. or Integrated M.Sc. level. Given below is the complete information about the eligibility guidelines of this scholarship.

  • The candidate must be in the age-group of 17-22 years.
  • The applicants must be among the top 1% of the successful learners in terms of overall marks in their 10+2 board exam.
  • The candidate should have registered in courses of Natural and Basic Sciences at the level of BSc, BS, and Integrated MSc/MS, in the same year of qualifying class 12th.
  • The candidates who are amongst the top 10000 rankers in Joint Entrance Examination (JEE)IIT, AIEEE (top 20000 rankers) and passed CBSE-Medical (AIPTM) and chosen to study Natural/Basic Sciences are eligible.
  • The applicants who got admission in Indian Institutes of Science Education and Research (IISER), National Institute of Science Education and Research (NISER), Department of Atomic Energy Centre for Basic Sciences (DAE-CBS) at the Mumbai University for pursuing programs in Natural/Basic Sciences commencing to B.Sc. and M.Sc. degree are also eligible.
  • The candidates who are KVPY (Kishore Vaigyanik Protsahan Yojna), National Talent Search Examination (NTSE), Olympiad Medallists and Jagadish Bose National Science Talent Search (JBNSTS) student and are seeking courses in Natural/Basic Sciences commencing to B.Sc. and M.Sc. degree are eligible.

INSPIRE Eligible Subjects

Candidate should be registered in a valid and regular degree level program in Natural/Basic sciences inside the scope of INSPIRE Scholarship at any UGC identified College or University or National Institute in India.

The subjects under Basic & Natural Sciences inside the scope of INSPIRE Scholarship for persevering B.Sc./BS/Int. MSc/Int. MS Course is:

Physics, Chemistry, Biology, Mathematics, Statistics, Geology, Astronomy, Electronics, Astrophysics, Botany, Zoology, Microbiology, Geophysics, Biochemistry, Anthropology, Atmospheric Sciences, Geochemistry, Ecology, Oceanic Sciences, BioPhysics, Marine Biology, and Genetics

INSPIRE Scholarship – Application Procedure

The applications for INSPIRE Scholarship for Higher Education (SHE) are received through online mode from INSPIRE portal. Given below is the entire guideline on how to apply for the scholarship.

  • The applicant has to register as a new user on the INSPIRE scholarship portal. The applicants must enter their name, gender, date of birth, email, password, eligibility criteria and verification code for successful registration.
  • The candidates will receive user ID and password on their registered email ID where they need to click on the activation link to begin his/her account.
  • The candidate should keep the scanned copies of all the demanded documents ready before filling the application form.
  • The candidate must log in to the INSPIRE scholarship portal using their credentials.
  • Now, the applicant must fill in all the necessary details like personal details, enrolment information at B.Sc. and Integrated M.Sc. level, senior secondary aggregates, competitive exam details, contact details and other mandatory details in the scholarship application form.
  • Also, the candidate has to upload all the needed documents in the online scholarship application.
  • The candidate has to review the listed details before the submission of the form by clicking on the ‘Print Preview’ button. They must make sure that all the details and uploaded documents are as per asked by the firm.
  • Now, the candidate has to click on the ‘Submit’ button to submit the application form.
  • The candidate must take the print out of the submitted application for their record.

INSPIRE Scholarship – Documents Required for Application

When candidates are applying for the Scholarship for Higher Education(SHE), they need to submit some necessary documents in support of their application, the absence of which sway lead to their refusal. Find the list of all the essential documents that a candidate requires to upload while filling their scholarship application.

  • Passport Size Photograph
  • Community/caste certificate in case of OBC/SC/ST category
  • Eligibility note or advisory note, provided by the State/Central Board
  • Class 12th or equivalent exam qualifying mark sheet
  • Class 10th or equivalent exam qualifying mark sheet or certificate(Proof as Date of Birth)
  • Certificate specifying rank/award in JEE (Main)/ JEE (Advanced)/ NEET/ KVPY /JBNSTS/ NTSE / International Olympic Medallists (if the candidate is eligible under this criterion)
  • Permission Form signed by the Principal of the College/Director of the Institute/Registrar of the University
  • SBI bank passbook details

INSPIRE Scholarship – Selection Procedure

All the applications sent by the candidates are judged on the basis of the defined eligibility criteria. On the basis of this, a list of shortlisted applicants is prepared. Here are the few details for the shortlisting of the scholars:

  • Shortlisted candidates list is presented on the online portal of INSPIRE and also on the website of the INSPIRE program.
  • The picked students have to download their provisional offer letter from the online portal of INSPIRE.
  • The shortlisted candidates also need to upload the important documents on their web portal within the mentioned timeline for further consideration.
  • The chosen candidates will receive their scholarship rewards directly into their State Bank of India (SBI) bank account through DBT (Direct Bank Transfer).

INSPIRE – Renewal of Scholarship

To continue SHE scholarship in the succeeding years, INSPIRE scholars have to fulfill the criteria specified below.

  • The scholars applying for renewal have to score at least 60% marks or 7.0 GPA on a 10.0-point scale in their annual exams each year.
  • If the applicant fails to achieve the required marks in their B.Sc/BS/Int. M.Sc/Int. MS exam, the scholarship for that academic year will be eliminated. The scholarship is renewed in the next academic year if the student’s academic accomplishment is satisfactory.

INSPIRE Scholarship – Rewards

Under the project of SHE, each elected candidate gets a learning amount of Rs.80,000 per annum and many other advantages for the study of Science and Technology and persevere research as a profession.

All the selected candidates need to begin summer research projects under the supervision of an active researcher in a verified research center of India. Find below the full details of the benefits granted under this scholarship.

SHE Scholarship RewardsRs. 80,000 to the elected student
An annual scholarship of Rs.60,000 distributed across the 12 months as Rs.5000 per monthRs.20,000 every year for offering the necessary summer research project

INSPIRE Scholarship – Rules To be Followed

There are a few rules that candidates need to follow in order to apply for this scholarship. Find below the list of terms and conditions that an applicant needs to keep in mind while applying for the SHE scholarship.

  • INSPIRE-SHE is granted to students for a period of five years, starting from the 1st year B.Sc., BS, Integrated M.Sc/MS or until the end of the course, whichever occurs earlier.
  • Applicants must provide completed application form else the form will be rejected.
  • Subjects such as Engineering, Medicine, Defence Studies, Agriculture, Military Science, Psychology, Home Science, Geography, Seed Technology, Anthropology, Economics, Education (including B.Sc.-B.Ed. dual degree course), Computer Applications, Bio-informatics, Biotechnology, Computer Science, Instrumentation, Information Technology, Physical Education, courses in Distance Education mode at the Open Universities and other professional courses are not carried under SHE scholarship.
  • The INSPIRE student have to open a regular Savings Bank Account in any branch of the State Bank of India (SBI) only. The account must be in the name of the candidate and it should not be a joint account.
  • The shortlisted candidates have to submit their year-wise performance report (signed by the Head of the Institution) and mark sheet of their BSc, BS, Int. MSc/MS course for the release of the scholarship. Head of the Institution refers to Vice-Chancellor/Registrar/Dean of Science in case of Universities and Principal/Vice Principal/Proctor/Officer-in-Charge in case of colleges.
  • The Head of the Institution should clearly state in the performance report if the scholarship should be extended or not in that academic year.
  • For the first time, the scholarship is published only after the approval of the performance report and mark sheet of 1st year B.Sc, BS, Int. M.Sc/MS. If the academic performance of the candidate is satisfying, two years of scholarship, i.e., Rs.1,20,000 is issued during the candidate’s 2nd year of B.Sc, BS, Int. M.Sc/MS.
  • Applicants availing any other scholarship from other sources are not eligible to win this scholarship.

About INSPIRE Programme

Generation and training of a human talent pool capable of using and developing first laws in science is both a pre-condition and integral part of such an innovation foundation. An India distinct model for pulling talent with an   for research and innovation, for a career in Basic & Natural sciences is needed. Department of Science & Technology (DST) has originated an innovative programme named Innovation in Science Pursuit for Inspired Research (INSPIRE) to pull talent to the enthusiasm and study of science at an early age, and to help the country build the demanded critical resource pool for strengthening and developing the S&T system and R&D base with a long term vision.

Let us discuss one by one the scholarship provided by INSPIRE:

Scheme for Early Attraction of Talent (SEATS) 

It strives to attract talented youth to study science by implementing INSPIRE Award of Rs 5000 to one million young students of the age group 10-15 years, ranging from Class VI to Class X standards, and also by providing summer camps for about 50,000 science students of Class XI with global leaders in science to feel the joy of innovations on an annual basis through INSPIRE Internship.

Scholarship for Higher Education (SHE) 

It tries to enhance rates of attachment of skilled youth to offer higher education in science-intensive programs, by granting scholarships and mentorship. The scheme offers 10,000 Scholarship every year at Rs 0.80 lakh per year for the skilled youth in the age group 17-22 years, for offering Bachelor and Masters level education in natural sciences. The main characteristic of the scheme is the mentorship assistance provided to every scholar.

Assured Opportunity for Research Careers (AORC) 

It endeavors to attract, attach, retain and nourish young scientific Human Resource for increasing the R&D foundation and base. It has two sub-components. In the first part i.e. INSPIRE Fellowship (age group of 22-27 years), it awards 1000 fellowships every year, for carrying out a doctoral degree in both basic and applied sciences including engineering and medicine. In the second component i.e. INSPIRE Faculty Scheme, it offers guaranteed opportunity every year for 1000 post-doctoral researchers in the age group of 27-32 years, through contractual and tenure track positions for 5 years in both basic and applied sciences area.

FAQ’s

Question 1.
How much is SHE valued for students?

Answer:
The scholarship is priced at Rs. 80,000/- per annum. Each candidate will receive an annual scholarship at Rs. 5000/- per month total value of Rs. 60,000/-. All the SHE scholars are required to undertake summer research projects under an active researcher in any recognized research centers across India. A summertime attachment fee of Rs. 20,000/- will be paid as Mentorship every year. The scholars have to submit the project report and Certificate from a mentor after the completion of the project in due course of time.

Question 2.
When can the student apply for INSPIRE Fellowship?

Answer:
Every year sometime during August/ September advertisement is announced in Websites. to seek fresh applications from eligible students in that year and applications are open for submission online through the portal: http://www.online-inspire.gov.in There is the provision of 1000 INSPIRE Fellowships every year to offer eligible students. Also, check our website http://www.inspire-dst.gov.in for regular updates.

Question 3.
Who can apply for INSPIRE Faculty Scheme? What are the eligibility criteria for availing INSPIRE Faculty Scheme?

Answer:
Indian citizens and people of Indian origin including NRI/PIO status with Ph.D. (in science, mathematics, engineering, pharmacy, medicine, and agriculture-related subjects) from any recognized university in the world. Those who have submitted their Ph.D. Theses and are awaiting award of the degree are also eligible. However, the award will be conveyed only after confirmation of the awarding the Ph.D. degree.

Question 4.
What courses are eligible for the offer of INSPIRE Scholarship?

Answer:
The following subjects under Basic and Natural Sciences are within the scope of INSPIRE Scholarship for pursuing BSc/BS/Int. MSc/Int. MS course: (1) Physics, (2) Chemistry, (3) Mathematics, (4) Biology, (5) Statistics, (6) Geology, (7) Astrophysics, (8) Astronomy, (9) Electronics, (10) Botany, (11) Zoology, (12) Bio-chemistry, (13) Anthropology, (14) Microbiology, (15) Geophysics, (16) Geochemistry, (17) Atmospheric Sciences & (18) Oceanic Sciences.

Courses other than these subjects such as Engineering, Medicine, Military Science, Defense Studies, Agriculture, Psychology, Seed Technology, Anthropology, Home Science, Geography, Economics, Education (including B.Sc.-B.Ed. dual degree course), Biotechnology, Computer Science, Computer Applications, Bioinformatics, Instrumentation, Information Technology, Physical Education, courses in Distance Education mode at the Open Universities and other professional courses are NOT supported.

The post INSPIRE Scholarship 2019 | Important Dates, Eligibility, Application Procedure, Scholarship Rewards appeared first on Learn CBSE.

CBSE Class 12 Sanskrit व्याकरणम् समास-प्रकरण

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CBSE Class 12 Sanskrit व्याकरणम् समास-प्रकरण

समास-प्रकरण :

दो या दो से अधिक समर्थ अर्थात् परस्पर सम्बद्ध अर्थवाले पदों का मिलकर एक पद बन जाता है। इस वृत्ति को समास वृत्ति कहते हैं। दो पदों का मिलकर एक समस्तपद होना ही समास है। समास वृत्ति से बने हुए पदों को समस्तपद कहते हैं तथा समस्तपद के पदों को अलग-अलग करने की प्रक्रिया विग्रह कहलाती है। समस्तपद बनाते समय पदों का परस्पर संबंध बताने वाली बीच की विभक्ति आदि का लोप हो जाता है तथा अंतिम पद में ही प्रकरण के अनुसार विभक्ति, वचन आदि का प्रयोग होता हैं समास निम्न प्रकार के होते हैं
(क) अव्ययीभाव (ख) तत्पुरुष (ग) कर्मधारय (घ) द्विगु (ङ) द्वन्द्व (च) बहुव्रीहि।

समास के भेद :

(क) अव्ययीभाव-समास – इस समास में पूर्व पद अव्यय होता है। पूर्वपद का अर्थ प्रधान होता है। समस्त पद नपुंसकलिङ्ग
एकवचन होता है तथा अव्यय होता है।
उदाहरण –
दयया सह = सदयम्

(ख) तत्पुरुष-समास – इसमें परवर्ती पद प्रधान होता है, बीच की विभक्ति का लोप होता है।
उदाहरण –
CBSE Class 12 Sanskrit व्याकरणम् समास-प्रकरण 1

तत्पुरुष के अन्य भेद – 1. उपपद तत्पुरुष, 2. नञ् तत्पुरुष।

1. उपपद तत्पुरुष – उत्तरपद क्रियापद होने पर यह समास होता है।
CBSE Class 12 Sanskrit व्याकरणम् समास-प्रकरण 2

2. नञ् तत्पुरुष – निषेध अर्थ को बताने के लिए ‘न’ समास होता है। इसमें उत्तरपद के पूर्ववर्ती वर्ण के व्यंजन होने पर ‘अ’ तथा स्वर होने पर ‘अन्’ जुड़ता है।
CBSE Class 12 Sanskrit व्याकरणम् समास-प्रकरण 3

(ग) कर्मधारय-समास – इसमें दोनों पद विशेषण-विशेष्य होते हैं। प्रायः दोनों में प्रथमा विभक्ति होती हैं। पूर्वपद विशेषण तथा उत्तरपद विशेष्य होता है।
उदाहरण –
CBSE Class 12 Sanskrit व्याकरणम् समास-प्रकरण 4

(घ) द्विगु समास – इसमें उत्तरपद संज्ञा तथा पूर्वपद संख्यावाचक होता है; जैसे
दशशतम् = दशानां शतानां समाहारः। , पञ्चानाम् पदानाम् समाहारः = पञ्चपदम् ।

(ङ) द्वन्द्व समास- इसमें दोनों पद प्रधान होते हैं।
दु:खं च सुखं च = दु:खसुखे
निघर्षणं च छेदनं च तापः च ताडनं च = निघर्षणछेदनतापताडनानि।

(च) बहुव्रीहि समास- इसमें अन्य पद प्रधान होता है और समस्तपद प्रायः अन्य पद का विशेषण होता है; जैसे –
CBSE Class 12 Sanskrit व्याकरणम् समास-प्रकरण 5

अभ्यासार्थ :

निम्नलिखित वाक्येषु रेखाकितानां पदानां समासविग्रहः समासो वा कृत्वा लिखत

1. सत्यम् अहिंसा च भारतस्य विशिष्टगुणद्वयम् अस्ति।
उत्तर:
विशिष्टं गुणद्वयम् (कर्मधारयः)

2. वयं भारतीयाः सर्वान् प्राणिनः आत्मवत् पश्यामः।
उत्तर:
सर्वप्राणिनः (कर्मधारयः)

3. वयं निरन्तरं मानवकल्याणाय उद्यताः भवेम।
उत्तर:
मानवानाम् कल्याणाय (षष्ठी तत्पुरुषः)

4. वयं अज्ञाननिद्रा परित्यज्य जीवन मूल्यानि जीवने धारयेम।
उत्तर:
अज्ञानस्य निद्राम् (षष्ठी तत्पुरुष:) /अज्ञानम् एव निद्रां (कर्मधारयः)

5. तृणानि भूमिरुदकं वाक् चतुर्थी च सुनृता।
उत्तर:
भूमिः च उदकं च तयोः समाहारः (द्वन्द्वः)

6. एतानि अपि सतां गेहे कदाचन न उच्छिद्यन्ते।
उत्तर:
सद्गृहे (षष्ठी तत्पुरुषः)

7. अहिंसया च भूतात्मा शुध्यति।
उत्तर:
न हिंसया (नञ् तत्पुरुषः)

8. सत्यमेव जयते न अनृतम्
उत्तर:
न ऋतम् (नञ् तत्पुरुष:)

9. देवयानः पन्थाः सत्येन विततः।
उत्तर:
देवयान पन्थाः (कर्मधारयः)

10. तत् हि सत्यस्य परमम् निधानम् अस्ति।
उत्तर:
परमनिधानम् (कर्मधारयः)

11. सर्वभूतेषु चात्मानं ततो न विजुगुप्सते।
उत्तर:
सर्वेषु भूतेषु (कर्मधारयः)

12. तत् दुर्गं पथः कवयो वदन्ति।
उत्तर:
दुर्गपथ: (कर्मधारयः)

13. सूर्यः एव प्रकृत्याधारः
उत्तर:
प्रकृते: आधारः (षष्ठी तत्पुरुषः)

14. अस्माकं सृष्टे: आधारः कः?
उत्तर:
सृष्ट्याधारः (षष्ठी तत्पुरुषः)

15. कथम् ऋतूणां दिनरात्र्योः च परिवर्तनं भवेत्?
उत्तर:
दिनस्य च रात्रेः च (द्वन्द्वः)

16. सर्वत्र एव सूर्यस्य महिमा वर्णिता।
उत्तर:
सूर्यमहिमा (षष्ठी तत्पुरुष:)

17. गुरुसेवने पटु बटुः उदेष्यन्तं भास्वन्तं प्रणमति।
उत्तर:
गुरुसेवन पटु (सप्तमी तत्पुरुषः)

18. अपितु भाषायाः सौन्दर्यस्य अपि उत्कृष्टम् उदाहरणम्।
उत्तर:
भाषासौन्दर्यस्य (षष्ठी तत्पुरुषः)

19. अरुण एष प्रकाशः पूर्वस्यां भगवतो मरीचिमालिनः
उत्तर:
मरीचयः एव मालाः यस्य तस्य (बहुव्रीहिः)

20. चक्रवर्ती खेचरचक्रस्य वर्तते।
उत्तर:
खेचराणाम् चक्रस्य (षष्ठी तत्पुरुषः)

21. आखण्डलदिशः कुण्डलम् वर्तते।
उत्तर:
आखण्डलस्य दिशः (षष्ठी तत्पुरुषः)

22. ब्रह्माण्डभाण्डस्य दीपक: सूर्य: वर्तते।
उत्तर:
ब्रह्माण्डम् एव भाण्डम्, तस्य (कर्मधारयः)

23. पुण्डरीकपटलस्य प्रेयान् वर्तते।
उत्तर:
पुण्डरीकाणाम् पटलस्य (षष्ठी तत्पुरुषः)

24. कोकलोकस्य शोकविमोकः वर्तते।
उत्तर:
कोकानाम् लोकस्य (षष्ठी तत्पुरुषः), शोकस्य विमोकः (षष्ठी तत्पुरुषः)

25. सूर्यः एव रोलम्बकम्बस्य अवलम्बः वर्तते।
उत्तर:
रोलम्बः कदम्बः, तस्य (कर्मधारयः)

26. सर्वव्यवहारस्य सूत्रधारः अस्ति।
उत्तर:
सर्व व्यवहारः, तस्य (कर्मधारयः)

27. अयमेव अहोरात्रम् जनयति।
उत्तर:
अहश्च रात्रिश्च (द्वन्द्वः)

28. अयम् एव वत्सरं द्वादशसु भागेषु विभनक्ति।
उत्तर:
द्वादशभागेषु (कर्मधारयः)

29. अयम् एव षण्णाम् ऋतूणाम् कारणम् अस्ति।
उत्तर:
षड़तूणाम् (कर्मधारयः)

30. एष एव दक्षिणम् अयनम् अङ्गीकरोति।
उत्तर:
दक्षिणायनम् (कर्मधारयः)

31. अनेन एव युगभेदाः सम्पादिताः।
उत्तर:
युगानाम् भेदाः (षष्ठी तत्पुरुषः)

32. अनेन एव कल्पभेदाः कृताः।
उत्तर:
कल्पानाम्/कल्पस्य भेदाः (षष्ठी तत्पुरुषः)

33. परमेष्ठिनः परार्द्धसङ्ख्या एनमेव अश्रित्य भवति।
उत्तर:
पराद्ध सङ्ख्या (कर्मधारयः)

34. धन्यः एष कुलमुलं श्रीरामचन्द्रस्य।
उत्तर:
कुलस्य मूलम् (षष्ठी तत्पुरुषः)

35. अन्यथा निराधारी पृथिवी कथम् आकाशे तिष्ठेत्।
उत्तर:
निराधारपृथिवी (कर्मधारयः)

36. निजपर्णकुटीरात् निर्गत्य बटुः सूर्यस्य महिमानं वर्णयति।
उत्तर:
निजपर्णानाम् कुटीरात् (षष्ठी तत्पुरुषः)

37. देशस्य कृते एव प्रजाधनस्य सदुपयोग: स्यात्।
उत्तर:
प्रजायाः धनम्, तस्य (षष्ठी तत्पुरुषः)

38. अमात्यः चाणक्यः सर्वदा देशहिताय एव प्रयत्नं करोति।
उत्तर:
देशस्य हिताय (षष्ठी तत्पुरुषः)

39. इति संस्कृत नाटकात् सङ्कलितः।
उत्तर:
संस्कृतस्य नाटकात् (षष्ठी तत्पुरुषः)

40. कुसुमपुरे चन्द्रगुप्तस्य प्रासादः
उत्तर:
चन्द्रगुप्तप्रासादः (षष्ठी तत्पुरुषः)

41. भोः भोः प्रासादाधिकृताः पुरुषाः
उत्तर:
प्रासादे अधिकृताः (सप्तमी तत्पुरुषः)

42. अत: सुगाङ्गप्रासादस्य उपरि स्थिता: प्रदेशाः संस्क्रियन्ताम्।
उत्तर:
सुगाङ्गः प्रासादः, तस्य (कर्मधारयः), स्थितप्रदेशा: (कर्मधारयः)

43. किं ब्रूथ? कौमुदी-महोत्सव: प्रतिषिद्धः?
उत्तर:
महान् उत्सवः (कर्मधारयः)

44. आः किम् एतेन? का प्राणहरेण कथा प्रसङ्गेन?
उत्तर:
कथायाः प्रसङ्गेन (षष्ठी तत्पुरुषः)

45. चन्दनवारिणी भूमि शीघ्र सिञ्चन्तु।
उत्तर:
चन्दनस्य वारिणा (षष्ठी तत्पुरुष:)

46. पुष्पमालाभिः स्तम्भान् अलङ्कुर्वन्तु।
उत्तर:
पुष्पाणाम् माला, ताभिः (षष्ठी तत्पुरुषः)

47. इदम् अनुष्ठीयते देवस्य शासनम् इति।
उत्तर:
देवशासनम् (षष्ठी तत्पुरुष:)

48. अयमागतः एव देव: चन्द्रगुप्त:
उत्तर:
देवचन्द्रगुप्तः (कर्मधारयः)

49. दुराराध्या हि राजलक्ष्मीः
उत्तर:
राज्ञः लक्ष्मी: (षष्ठी तत्पुरुषः)

50. आर्य वैहीनरे! सुगाङ्गमार्गम् आदेशय।
उत्तर:
सुगाङ्गस्य मार्गम् (षष्ठी तत्पुरुषः)

51. (सक्रोधम्) आः, केन?
उत्तर:
क्रोधेन सह/युगपत् (अव्ययी भाव:)

52. देव! कः अन्यः जीवितुकामो देवस्य शासनम् अतिवर्तेत?
उत्तर:
जीवितुम् कामयते येन सः (बहुव्रीहिः), देवशासनम् (षष्ठी तत्पुरुष:)

53. ततः प्रविशति स्वभवनगतः आसनस्थ: चाणक्यः।
उत्तर:
स्वभवनम् गतः (द्वितीया तत्पुरुषः)/ स्वम् भवनम् गतम् येन सः (बहुव्रीहिः), आसने स्थः यः सः (बहुव्रीहिः)

54. कथं स्पर्धते मयां सह दुरात्मा राक्षस:?
उत्तर:
दुष्टः आत्मा यस्य सः (बहुव्रीहिः)

55. अहो राजाधिराजमन्त्रिण: विभूतिः।
उत्तर:
राजसु अधिराजः (सप्तमी तत्पुरुषः) तस्य मन्त्रिणः (षष्ठी तत्पुरुषः)

56. तथाहि गोमयानाम् उपलभेदकम् एतत् प्रस्तरखण्डम्
उत्तर:
प्रस्तरस्य खण्डम् (षष्ठी तत्पुरुषः)

57. इत: शिष्यैः आनीतानां दर्भाणाम् स्तूपः।
उत्तर:
शिष्यानीत दर्भाणाम् (तृतीया तत्पुरुषः, कर्मधारयः)

58. अत्र जीर्णाः भित्तयः सन्ति।
उत्तर:
जीर्णभित्तयः (कर्मधारयः)

59. वैहीनरे! किम् आगमन-प्रयोजनम्?
उत्तर:
आगमनस्य प्रयोजनम् (षष्ठी तत्पुरुषः)

60. अतएव निस्पृहत्यागिभिः एतादृशैः जनैः राजा तृणवद् गण्यते।
उत्तर:
निस्पृहै: त्यागिभिः (कर्मधारयः), एतादृज्जनैः (कर्मधारयः)

61. यदि कार्य बाधा न स्यात्।
उत्तर:
कार्यबाधा (सप्तमी तत्पुरुषः)

62. आर्य! देवः चन्द्रगुप्तः आर्यं शिरसा प्रणम्य विज्ञापयति।
उत्तर:
देवचन्द्रगुप्तः (कर्मधारयः)

63. किं ज्ञात: कौमुदीमहोत्सव-प्रतिषेधः?
उत्तर:
कौमुदीमहोत्सवस्य प्रतिषेधः (षष्ठी तत्पुरुषः)

64. सः सुगाङ्गप्रासाद अस्ति।
उत्तर:
सुगाङ्गे प्रासादे (कर्मधारयः)

65. सुगाङ्गप्रासादस्य मार्गम् आदेशय।।
उत्तर:
सुगाङ्ग प्रासादमार्गम् (षष्ठी तत्पुरुष:)

66. अये सिंहासनम् अध्यास्ते वृषलः।
उत्तर:
सिंह इव आसनम् (कर्मधारयः)

67. आर्यप्रसादात् अनुभूयते एव सर्वम्।
उत्तर:
आर्यस्य प्रासादात् (षष्ठी तत्पुरुषः)

68. आर्यस्य दर्शनेन आत्मानम् अनुग्रहीतम्।
उत्तर:
आर्यदर्शनेन (षष्ठी तत्पुरुषः)

69. न निष्प्रयोजनं प्रभुभिः अधिकारिणः आहूयन्ते।
उत्तर:
प्रयोजनस्य अभावः (अव्ययीभाव:)

70. यदि एवं तर्हि शिष्येण गुरोः आज्ञा पालनीया।
उत्तर:
गुर्वाज्ञा (षष्ठी तत्पुरुषः)

71. किन्तु न कदाचित् आर्यस्य निष्प्रयोजना प्रवृत्तिः।
उत्तर:
निर्गतम् प्रयोजनम् यस्याः सा (बहुव्रीहिः)

72. नेपथ्ये वैतालिकौ काव्यपाठं कुरुतः।
उत्तर:
काव्यस्य पाठम् (षष्ठी तत्पुरुषः)

73. आभ्यां वैतालिकाभ्यां सुवर्णशतसहस्रं दापय।
उत्तर:
सुवर्णाणाम् शतम् च सहस्रम् च (षष्ठी तत्पुरुषः, द्वन्द्वः)

74. वृषल! किम् अस्थाने महान् प्रजाधनापव्ययः?
उत्तर:
न स्थाने (नञ् तत्पुरुषः), महाप्रजाधनापव्ययः (कर्मधारयः)

75. आर्येण एव सर्वत्र निरुद्धचेष्टस्य मे बन्धनम् इव राज्यम्।
उत्तर:
निरुद्धा चेष्टा यस्य तस्य (बहुव्रीहिः)

76. वृषल! स्वयम् अनभियुक्तानाम् राज्ञाम् एते दोषाः सम्भवन्ति।
उत्तर:
न अभियुक्तानाम् (नञ् तत्पुरुष:)

77. प्रथमं मम आज्ञायाः पालनम्।
उत्तर:
आज्ञापालनम् (षष्ठी तत्पुरुषः)

78. पर्वतकपुत्र: मलयकेतुः अस्मान् अभियोक्तुम् उद्यतः।
उत्तर:
पर्वतकस्य पुत्रः (षष्ठी तत्पुरुष:)

79. सोऽयं व्यायामकालो न उत्सवकालः इति।
उत्तर:
व्यायामस्य कालो (षष्ठी तत्पुरुषः)

80. अतः इदानीं दुर्गसंस्कारः प्रारब्धव्यः।
उत्तर:
दुर्गस्य संस्कारः (षष्ठी तत्पुरुष:)

81. राष्ट्रचिन्ता ननु गरीयसी।
उत्तर:
राष्ट्रस्य चिन्ता (षष्ठी तत्पुरुषः)

82. प्रथमं तु राष्ट्रसंरक्षम् ततः उत्सवाः इति।
उत्तर:
राष्ट्रस्य संरक्षम् (षष्ठी तत्पुरुषः)

83. दूरदृष्टि: फलप्रदा
उत्तर:
फलम् प्रददाति या सी (बहुव्रीहिः)

84. यः जनः दूरदर्शी भवति।
उत्तर:
दूरम् पश्यति इति (उपपद तत्पुरुषः)

85. पञ्चतन्त्रे विष्णुशर्मा मत्स्यानाम् उदाहरणेन
उत्तर:
पञ्चानाम् तन्त्राणाम् समाहारः, तस्मिन् (द्विगु:)

86. राजपुत्रान् एतदेव शिक्षायितुं प्रयत्नं करोति।
उत्तर:
राज्ञः पुत्रान् (षष्ठी तत्पुरुषः)

87. प्रत्युत्पन्नमति: शीघ्रमेव निर्णयं कृत्वा आत्मरक्षां करोति।
उत्तर:
प्रत्युत्पन्ना मति: यस्मिन् सः (बहुव्रीहिः), आत्मनः रक्षाम् (षष्ठी तत्पुरुषः)

88. कस्मिंश्चित् जलाशये त्रयः मत्स्याः प्रतिवसन्ति स्म।
उत्तर:
जलस्य आशयः, तस्मिन् (षष्ठी तत्पुरुषः)

89. जलाशयं दृष्ट्वा तत्र गच्छद्भिः मत्स्यजीविभिः उक्तम्।
उत्तर:
मत्स्यैः जीवन्ति इति, तैः (उपपद तत्पुरुष:)

90. अहो बहुमत्स्यः अयं हृदः।
उत्तर:
बहवः मत्स्याः यस्मिन् सः (बहुव्रीहिः)

91. अद्य तु आहारवृत्तिः सञ्जाता।
उत्तर:
आहारस्य वृत्तिः (षष्ठी तत्पुरुषः)

92. सन्ध्या समयः अपि संवृत्तः।
उत्तर:
सन्ध्यायाः समय: (षष्ठी तत्पुरुषः)

93. अशक्तैः बलिन:शत्रोः कर्तव्यम् प्रपलायनम्।
उत्तर:
न शक्तैः (नञ् तत्पुरुष:)

94. येषाम् अन्यत्र अपि विद्यमाना सुखावहा गतिः भवति।
उत्तर:
सुखम् आवहति या सा (बहुव्रीहिः)

95. ते विद्वांसः देहभङ्गम् कुलक्षयम् च न पश्यन्ति।
उत्तर:
देहस्य भङ्गम् (षष्ठी तत्पुरुष:), कुलस्य क्षयम् (षष्ठी तत्पुरुष:)

96. आयुः क्षयोऽस्ति चेत्।
उत्तर:
आयुषः क्षयः (षष्ठी तत्पुरुष:)

97. अरक्षितं दैवरक्षितं तिष्ठति।
उत्तर:
न रक्षितम् (नञ् तत्पुरुषः), दैवेन रक्षितम् (तृतीया तत्पुरुषः)

98. सुरक्षितं दैवहतं विनश्यति।
उत्तर:
सुष्ठ (शोभनम्) रक्षितम् (कर्मधारयः), दैवेन हतम् (तृतीया तत्पुरुष:)

99. अनाथः वने विसर्जितः अपि जीवति।
उत्तर:
न नाथः (नञ् तत्पुरुषः)

100. कृतप्रयत्नः गृहे अपि न जीवति।
उत्तर:
कृतम् प्रयत्नम् येन सः (बहुव्रीहिः)

101. छात्राः विद्यालयस्य पत्रिकायां चित्राणि पश्यन्ति।
उत्तर:
विद्यालयपत्रिकायाम् (षष्ठी तत्पुरुषः)

102. किम् एतानि चित्राणि पर्वतारोहणस्य सन्ति?
उत्तर:
पर्वते आरोहणम्, तस्य (सप्तमी तत्पुरुषः)

103. किं प्रक्षेपकमाध्यमेन तत् सर्वं यात्रावृत्तं द्रष्टुं वाञ्छथ?
उत्तर:
प्रक्षेपकस्य माध्यमेन (षष्ठी तत्पुरुषः), यात्रायाः वृत्तम् (षष्ठी तत्पुरुषः)

104. सर्वे कक्षायां यथास्थानम् उपविशन्ति।
उत्तर:
स्थानम् अनतिक्रम्य (अव्ययीभावः)

105. अहो! विपुला हिमराशिना धवला: एते लद्दाखप्रदेशीया गिरयः अतीव शोभन्ते।
उत्तर:
हिमस्य राशिः, तेन (षष्ठी तत्पुरुषः), लद्दाखप्रदेशीयगिरयः (कर्मधारयः)

106. शृणुत, उत्तुङ्गपर्वतानाम् उपत्यकाभूमिम् लद्दाख इति वदन्ति।
उत्तर:
उत्तुङ्गानाम् पर्वतानाम् (कर्मधारयः), उपत्यकायाः भूमिम् (षष्ठी तत्पुरुषः)

107. श्रूयते यत् लद्दाखमार्गेण एव तिब्बतक्षेत्रे बौद्धधर्मस्य प्रवेशः अभवत्।
उत्तर:
लद्दाखस्य मार्गेण (षष्ठी तत्पुरुषः), बौद्धधर्मप्रवेशः (षष्ठी तत्पुरुष:)

108. पश्यन्तु, कथं लद्दाखे आस्तृतः नीलाकाशः छत्रवत् प्रतीयते।
उत्तर:
नील: आकाशः (कर्मधारयः)

109. ग्रीष्मे नीलवर्णा भूमिः धूसरवर्णा जायते।
उत्तर:
नीलः वर्णः यस्य: सा (बहुव्रीहिः), धूसरः वर्ण: यस्याः सा (बहुव्रीहि:)

110. एषः ‘सेंग्ये’ नाम राजप्रासादः
उत्तर:
राज्ञः प्रासादः (षष्ठी तत्पुरुष:)

111. इदं ‘लेह’ इत्यभिधानेन प्रसिद्ध पर्यटनस्थलम्
उत्तर:
पर्यटनस्य स्थलम् (षष्ठी तत्पुरुषः)

112. एषः बौद्धधर्मस्य प्रसिद्धः प्राचीनश्च श्वेतस्तूपः।
उत्तर:
बौद्धनाम् धर्म: तस्य (षष्ठी तत्पुरुषः), श्वेतः स्तूपः (कर्मधारयः)

113. रात्रौ अयं दीपेषु प्रज्वलितेषु भव्यम् आलोकं वितरति।
उत्तर:
भव्यालोकम् (कर्मधारयः)

114. एते प्रख्याताः बौद्धमठाः सन्ति।
उत्तर:
बौद्धानाम् मठाः (षष्ठी तत्पुरुष:)

115. भगवतो बुद्धस्य विशालकाया मूर्तिः पर्यटकानाम् आकर्षणकेन्द्रम्
उत्तर:
विशालकाय मूर्तिः (कर्मधारयः)

116. एको विशालः ‘स्टाक पैलेस संग्रहालयः वर्तते।
उत्तर:
संग्रहस्य आलयः (षष्ठी तत्पुरुषः)

117. बौद्धानां सामाजिक जीवन कीदृशम् वर्तते?
उत्तर:
सामाजिक जीवनम् (कर्मधारयः)

118. किं तेऽपि उत्सवप्रिया:?
उत्तर:
उत्सव: प्रिय: यान् ते (बहुव्रीहि:)

119. बौद्धानां ‘गम्पा’ नाम वार्षिकोत्सवः शीते आयाति।
उत्तर:
वार्षिक उत्सवः (कर्मधारयः)

120. ‘ताहथोकदीनि ग्रीष्मपर्वाणि’ भगवन्तं बुद्ध प्रति भक्तिभावं दर्शयन्ति।
उत्तर:
भगवद्बुद्धं (कर्मधारयः), भक्तिः च भावः च तयोः समाहारः (द्वन्द्व:)

121. मठेषु उत्कीर्णा: लेखाः भित्तिलेखाःतिब्बतशैल्याः परिचायकाः।
उत्तर:
उत्कीर्णलेखा: (कर्मधारयः), भित्त्याः लेखा: (षष्ठी तत्पुरुषः), तिब्बतस्य शैल्याः (षष्ठी तत्पुरुषः)

122. लद्दाखस्य प्राकृतिक स्थलानां विषये किमपि भवती ब्रवीतु।
उत्तर:
प्राकृतिकानाम् स्थलानाम् (कर्मधारयः)

123. कारगिले आक्रमण कारिणाम् अपसारणाय भारतीयैः वीरैः यत् शौर्यं प्रदर्शितम्।
उत्तर:
भारतीयवीरैः (कर्मधारयः)

124. उपत्यकाभूमिषु शीत ऋतौ महान् हिमराशि: निपतति।
उत्तर:
शीत (कर्मधारयः), महाहिमराशिः (कर्मधारयः)

125. ग्रीष्मे सः द्रवीभूय कृषकाणां भूमिसेचने उपकरोति।
उत्तर:
भूमेः सेचने (षष्ठी तत्पुरुषः)

126. पर्वतारोहणाय ‘लिकिर’ ‘स्टाक’ नाम्नी स्थले उपयुक्ते स्त:।
उत्तर:
पर्वतस्य आरोहणाय (षष्ठी तत्पुरुष:)

127. घनीभूतं हिमं गिरिराजस्य शोभा सततं प्रवर्धयति।।
उत्तर:
घनीभूतहिम (कर्मधारयः), गिरीणाम् राजा, तस्य(षष्ठी तत्पुरुष:)

128. महाकवेः कालिदासस्य पद्यम् अस्य सौन्दर्यं सततं वर्णयति।
उत्तर:
महत: कवे:/महान् कविः, तस्य (कर्मधारयः)

129. अनन्तरत्नप्रभवस्य यस्य।
उत्तर:
अनन्तानि रत्नानि प्रभवन्ति यस्मात् सः, तस्य (बहुव्रीहिः)

130. एको हि दोषो गुणसन्निपाते
उत्तर:
गुणानाम् सन्निपाते (षष्ठी तत्पुरुष:)

131. सुधामुचः वाचः।
उत्तर:
सुधाम् मुञ्चन्ति याः ता: (बहुव्रीहि:)/सुधाम् मुञ्चन्ति इति (उपपद तत्पुरुषः)

132. सन्मार्ग च सूक्तयः सुभाषितानि का प्रदर्शयन्ति।
उत्तर:
सत् च तत् मार्गम् (कर्मधारयः)

133. विपदि निपतिताः मानवाः एतैः सुभाषितैः आश्वासनं प्राप्नुवन्ति।
उत्तर:
विपन्निपतिता: (सप्तमी तत्पुरुषः)

134. संस्कृतवाङ्मयं सहस्रशः सुमधुरवचनैः अलङ्कृतम् वर्तते।
उत्तर:
संस्कृतस्य वाङ्मयं (षष्ठी तत्पुरुषः), सुमधुरैः वचनैः (कर्मधारयः)

135. केषाञ्चिद् मधुरवचनानां सङ्कलनं अत्र प्रस्तूयते।।
उत्तर:
मधुरवचनसङ्कलनम् (षष्ठी तत्पुरुष:)

136. यूयम् अपि परियोजनारूपेण एतादृशमेव सङ्कलनं कर्तुं शक्नुथ।
उत्तर:
परियोजनाया: रूपेण (षष्ठी तत्पुरुषः)

137. वदनं प्रसादसदनं सदयं हृदयं सुधामुचः वाचः।
उत्तर:
प्रसादस्य सदनम् (षष्ठी तत्पुरुषः)

138. करणं परोपकरणं येषां केषां न ते वन्द्याः ।
उत्तर:
परेषाम् उपकरणम् (षष्ठी तत्पुरुष:)

139. कालपर्ययात् शिक्षा क्षयं गच्छति।
उत्तर:
कालस्य पर्ययात् (षष्ठी तत्पुरुषः)

140. सुबद्धमूला: पादपाः निपतन्ति।
उत्तर:
सुबद्धानि मूलानि येषाम् ते (बहुव्रीहिः), पादैः पिबन्ति ये ते (बहुव्रीहिः)

141. जलस्थानगतम् जलं च शुष्यति।
उत्तर:
जलस्य स्थानं गतम् (षष्ठी-द्वितीया तत्पुरुष:)

142. यथा निघर्षणच्छेदनतापताडनै: चतुर्भिः कनकं परीक्ष्यते।
उत्तर:
निघर्षणेन च छेदनेन च तापेन च ताडनेन च (द्वन्द्वः)

143. तथा त्यागेन (च) शीलेन (च) गुणेन (च) कर्मणा (च) चतुर्भिः पुरुषः परीक्ष्यते।
उत्तर:
त्यागशीलगुणकर्मभिः (द्वन्द्वः)

144. ते मूढधियः पराभवं व्रजन्ति।
उत्तर:
मूढाः धियः येषां ते (बहुव्रीहि:)

145. इषवः तथाविधान् असंवृत्ताङ्गान् प्रविश्य तं ध्वन्ति।
उत्तर:
असंवृत्तानि अङ्गानि येषां तान् (बहुव्रीहिः)

146. यः अधिपं साधु ने शास्ति स किंसखा
उत्तर:
किम् च सः सखा (कर्मधारयः)

147. यः हितान् न संशृणुते स किंप्रभुः
उत्तर:
किम् च सः प्रभुः (कर्मधारयः)

148. नृपेषु अमात्येषु च अनुकूलेषु हि सर्वसम्पद: सदा रतिं कुर्वते।
उत्तर:
सर्वाः सम्पदः (कर्मधारयः)

149. लोभः चेत् (तदा) अगुणेन किम्?
उत्तर:
न गुणेन (नञ् तत्पुरुषः)

150. यदि सुमहिमा अस्ति तदा मण्डनैः किम्?
उत्तर:
शोभना महिमा (कर्मधारयः)

151. यदि सविद्या तदा धनैः किम्?
उत्तर:
सत् च सा विद्या (कर्मधारयः)

152. अयं पाठः ‘चारुदत्तं’ नाटकस्य प्रथमाङ्कात् सङ्कलितः।
उत्तर:
प्रथमः अङ्कः , तस्मात् (कर्मधारयः)

153. स उज्जयिनीवासी रूपवान् सङ्गीत् विद्यायाः प्रेमी, परोपकारपरायणः चासीत्।
उत्तर:
उज्जयिन्या:वासी (षष्ठी तत्पुरुष:), सङ्गीतविद्याप्रेमी (षष्ठी तत्पुरुषः), परोपकारे परायणः (सप्तमी तत्पुरुषः)

154. सः उदारतावशदानकारणात् शीघ्रं दरिद्रो जातः।
उत्तर:
उदारतायाः वशम् (षष्ठी तत्पुरुषः), तस्मात् दानकारणात् (पञ्चमी तत्पुरुषः)

155. दरिद्रावस्थायां मित्राणाम् उपेक्षायाः कारणात् कटुः अनुभवः भवति।
उत्तर:
मित्रोपेक्षाकारणात् (षष्ठी तत्पुरुष:)

156. मैत्रेयः विनोदप्रिय: विपत्तौ अपि चारुदत्तस्य विश्वासपात्रम् अस्ति।
उत्तर:
विनोदम् प्रियम् यस्मै सः (बहुव्रीहिः), विश्वासस्य पात्रम् (षष्ठी तत्पुरुषः)

157. ततः नान्द्यन्ते सूत्रधारः प्रविशति।
उत्तर:
नान्द्या: अन्ते (षष्ठी तत्पुरुष:), सूत्रम् धारयति इति (उपपद तत्पुरुष:)

158. किन्तु अद्य पुष्करपत्रपतितजलबिन्दू इव मेऽक्षिणी चञ्चलायेते।
उत्तर:
पुष्करपत्रे पतितजलबिन्दू (सप्तमी तत्पुरुषः)

159. किम् अस्त्यस्माकं गेहे कोऽपि प्रातराश:?
उत्तर:
प्रात: अश्यते इति (उपपद तत्पुरुषः)

160. एवं शोभनानां भोजनानाम् दात्री भव।
उत्तर:
सुभोजनानाम् (कर्मधारयः)

161. आः अनायें!
उत्तर:
न आयें (नञ् तत्पुरुषः)

162. यदि आर्यस्य अनुग्रह: स्यात्।
उत्तर:
आर्यानुग्रहः (षष्ठी तत्पुरुषः)

163. अहम् आर्यचारुदत्तस्य गेहे अहोरात्रम् आकण्ठमात्रम् अशित्वा दिवसान् अनयम्।
उत्तर:
अहश्च रात्रम् च तयो:समाहार: (द्वन्द्वः)

164. तदैव तत्र भवतः चारुदत्तस्य देवकार्य कारणात् गृहीतानि समनसः।
उत्तर:
देवानाम् कार्यम् (षष्ठी तत्पुरुषः) तस्य कारणात् (षष्ठी तत्पुरुषः)

165. एष चारुदत्तः यथाविभवं गृहदैवतानि अर्चयन् इत एवागच्छति।।
उत्तर:
विभवम् अनतिकम्य (अव्ययीभाव:), गृहस्य दैवतानि (षष्ठी तत्पुरुषः)

166. ततः प्रविशति चारुदत्तो, विदूषकः चङ्गेरिकाहस्ता चेटी च।।
उत्तर:
चङ्गेरिका हस्ते यस्याः सा (बहुव्रीहिः)

167. भो: दारिद्र्यं खलु नाम मनस्विनः पुरुषस्य, सोच्छ्वासं मरणम्।
उत्तर:
मनस्वीपुरुषस्य (कर्मधारयः), उच्छ्वासेन सह (अव्ययीभावः)

168. दानेन विपन्नविभवस्य, बहुलपक्षचन्द्रस्य ज्योत्स्ना परिक्षय इव।
उत्तर:
विपन्न:विभवः यस्य तस्य (बहुव्रीहिः), बहुलपक्षस्य चन्द्रस्य (कर्मधारयः)

169. भवत: रमणीयोऽयं दरिद्रभावः
उत्तर:
दरिद्रः भावः (कर्मधारयः)/दरिद्रस्य भावः (षष्ठी तत्पुरुषः)

170. न खलु अहम् नष्टां श्रियम् अनुशोचामि।।
उत्तर:
नष्टश्रियम् (कर्मधारयः)

171. गुणरसज्ञस्य तु पुरुषस्य व्यसनं दारुणतरं मां प्रतिभाति।
उत्तर:
गुणानां रसः (षष्ठी तत्पुरुषः), तम् ज्ञायते यः सः (बहुव्रीहिः)

172. यथान्धकारात् इव दीपदर्शनम्
उत्तर:
दीपस्य दर्शनम् (षष्ठी तत्पुरुषः)

173. किं भवान् अर्थविभवं चिन्तयति।
उत्तर:
अर्थः च विभवः च तयोः समाहारः (द्वन्द्वः)

174. सत्यं न मे धनविनाशगता विचिन्ता।
उत्तर:
धनविनाशम् गता (द्वितीया तत्पुरुषः)

175. भाग्यक्रमेण हि धनानि पुनर्भवन्ति।
उत्तर:
भाग्यस्य क्रमेण (षष्ठी तत्पुरुषः)

176. एतत्तु मेनष्टधनश्रियः मां दहति।
उत्तर:
नष्टे धनम् चे श्रीः च यस्य सः (बहुव्रीहिः)

177. तथैव धनविनाशदु:खस्य पुनः पुनः चिन्त्यमानस्य।
उत्तर:
धनविनाशस्य दु:खं तस्य (षष्ठी तत्पुरुषः)

178. नानाविद्याः चिन्ताङ्कुरा: प्रादुर्भवन्ति।
उत्तर:
चिन्तायाः अङ्कुराः (षष्ठी तत्पुरुषः)

179. विभवानुवशा भार्या समदु:खसुखो भवान्।
उत्तर:
विभवैः अनुवशा (तृतीया तत्पुरुषः), समम् दु:खे च सुखे च (सप्तमी तत्पुरुषः)

180. आश्चर्यमयं विज्ञानजगत्
उत्तर:
विज्ञानस्य जगत् (षष्ठी तत्पुरुषः)

181. विद्यालयस्य सूचनापट्टे विज्ञप्तिः।
उत्तर:
सूचनायाः पट्टे (षष्ठी तत्पुरुषः)

182. एकविंशतितारिकायां गुरुवासरे अर्धावकाशानन्तरं सभागारे एका सङ्गोष्ठी भविष्यति।
उत्तर:
अर्धावकाशात् अनन्तरम् (पञ्चमी तत्पुरुषः), सभायाः आगारे (षष्ठी तत्पुरुषः)

183. सर्वे छात्राः यथासमयं कृपया तत्र आगच्छन्तु।
उत्तर:
समयम् अनतिक्रम्य (अव्ययीभावः)

184. अस्माकं छात्रै: यद् विशिष्टाध्ययनं कृतं तस्य परिचयः अस्मभ्यम् लाभप्रदः भविष्यति।
उत्तर:
विशिष्टम् अध्ययनम् (कर्मधारयः)

185. अस्माकं मध्ये ग्रीष्मावकाशपरियोजनाकार्ये सर्वोत्तमान् अङ्कानु लब्धवन्तः छात्राः सन्ति।
उत्तर:
ग्रीष्मावकाशस्य परियोजना (षष्ठी तत्पुरुषः), तस्याः कार्ये (षष्ठी तत्पुरुषः), सर्वोत्तमाकान् (कर्मधारयः)

186. एते स्वाध्ययनस्य विशिष्टांशान् अत्र प्रस्तोष्यन्ति।
उत्तर:
स्वस्य अध्ययनम्, तस्य (षष्ठी तत्पुरुषः), विशिष्ट: अंशः, तान् (कर्मधारयः)

187. अस्त त्रिपुरस्य यथाक्रमम् भागत्रयं भवेत्।
उत्तर:
भागानाम् त्रयम् (षष्ठी तत्पुरुषः)

188. तेषु प्रथमभागस्य सञ्चारः पृथिवीतले
उत्तर:
पृथिव्याः तले (षष्ठी तत्पुरुषः)

189. त्रिपुरविमानस्य प्रथमः भागः, पृथिव्याः तले सञ्चरति।
उत्तर:
प्रथमभागः (कर्मधारयः), पृथिवीतले (षष्ठी तत्पुरुषः)

190. द्वितीयभाग: जलस्य अन्त: बहिः च विहरति।
उत्तर:
द्वितीयः भागः (कर्मधारयः), जलान्तः (षष्ठी तत्पुरुषः)

191. करतलध्वनिना अभिनवस्य उत्साहवर्धनं कुर्वन्तु।
उत्तर:
उत्साहस्य वर्धनम् (षष्ठी तत्पुरुषः)

192. करतलध्वनिना सभागार: गुञ्जति।।
उत्तर:
सभाया: आगार: (षष्ठी तत्पुरुषः)

193. इदानीं शालिनी अस्माकं ज्ञाने वृद्धिं करष्यति।
उत्तर:
ज्ञानवृद्धिः (सप्तमी तत्पुरुषः)

194. तर्हि सुश्रुतविरचिता सुश्रुतसंहिता अवश्यमेव पठनीया।
उत्तर:
सुश्रुतेन विरचिता (तृतीया तत्पुरुषः)

195. तत्र अष्टविधं शल्यकार्यं वर्णितम्।
उत्तर:
शलस्य कार्यम् (षष्ठी तत्पुरुषः)

196. सुश्रुतसंहितायां नासिका प्रत्यारोपणं विस्तरशः वर्णितम्।
उत्तर:
नासिकायाः प्रत्यारोपणम् (षष्ठी तत्पुरुषः)

197. सुश्रुतः प्रथमः त्वक्प्रत्यारोपकः आसीत्।
उत्तर:
त्वचः प्रत्यारोपणं करोति यः सः (बहुव्रीहिः)

198. प्रशंसनीया एव भवत्याः प्रस्तुतिः
उत्तर:
भवत्प्रस्तुतिः (षष्ठी तत्पुरुषः)

199. पुनः करतलध्वनिः भवति।
उत्तर:
करस्य तलम् (षष्ठी तत्पुरुष:) तस्य ध्वनिः (षष्ठी तत्पुरुष:)/करतलस्य ध्वनिः (षष्ठी तत्पुरुषः)

200. सस्वरं गायति।
उत्तर:
स्वरेण सहितम् (अव्ययीभाव:)

201. तस्माद् दक्षिणपार्वे सलिलं पुरुषद्वये स्वादु।
उत्तर:
दक्षिणे पार्वे (कर्मधारयः)

202. तालिकावादनेन सह सभागारं पूरयति।
उत्तर:
तालिकायाः वादनेन (षष्ठी तत्पुरुषः), सभायाः आगारम् (षष्ठी तत्पुरुष:)

203. जम्बूवृक्षस्य प्राग्वल्मीको यदि समीपस्थः भवेत्।
उत्तर:
समीपे स्थः (सप्तमी तत्पुरुष:)

204. यथा वृक्षायुर्वेद: वास्तुविज्ञानं, ज्योतिष, पशुविज्ञानम् इत्यादयः।
उत्तर:
वृक्षाणाम् आयुः (षष्ठी तत्पुरुष:) तस्य वेदः (षष्ठी तत्पुरुषः), पशूनाम् विज्ञानम् (षष्ठी तत्पुरुष:)

205. मिहिरेण बहुमूल्या सामग्री सङ्कलिता।
उत्तर:
बहुमूल्यसामग्री (कर्मधारयः)

206. परन्तु समयाभावात् तस्याः प्रस्तुतिः अत्र नः भविष्यति।
उत्तर:
समयस्य अभावात् (षष्ठी तत्पुरुष:)

207. इदानीं भारती आर्यभटीयम् इति ग्रन्थस्य वैशिष्ट्यं वर्णयिष्यति।
उत्तर:
ग्रन्थवैशिष्ट्यम् (षष्ठी तत्पुरुषः)

208. वयं वर्तमानकाले सङ्गणकस्य प्रयोगं कुर्मः।
उत्तर:
वर्तमाने काले (कर्मधारयः), सङ्गणक प्रयोगम् (षष्ठी तत्पुरुष:)

209. यदि आर्यभटेन शून्यस्य अविष्कार: न कृतः स्यात्।
उत्तर:
शून्याविष्कारः (षष्ठी तत्पुरुषः)

210. सूर्यं प्रति पूर्वाभिमुखा पृथिवी 365.25 वारं प्रतिवर्ष भ्रमति।
उत्तर:
पूर्वाभिमुख पृथिवी (कर्मधारयः), वर्षम् वर्षम् इति (अव्ययीभावः)

211. यथा नौकायां स्थितः मनुष्य: वृक्षादीन् पृष्ठं प्रति गच्छतः पश्यति।
उत्तर:
नौकास्थितः (सप्तमी तत्पुरुषः)

212. इदानीं नागार्जुन: कौटिल्यरचितात् अर्थशास्त्रात् रसायनशास्त्रम् अधिकृत्य।
उत्तर:
कौटिल्येन रचितम्, तस्मात् (तृतीया तत्पुरुषः), रसायनस्य शास्त्रम् (षष्ठी तत्पुरुषः)

213. अस्माकं ज्ञानवृद्धिं करिष्यति।
उत्तर:
ज्ञाने वृद्धिं (सप्तमी तत्पुरुषः)/ज्ञानस्य वृद्धिम् (षष्ठी तत्पुरुष:)

214. कौटिल्यस्य अर्थशास्त्रं तु अद्भुतः ग्रन्थः यत्र बहवः विषयाः वर्णिताः।
उत्तर:
अद्भुतग्रन्थः (कर्मधारयः), बहुविषयाः (कर्मधारयः)

215. धन्यवादः। प्रियबान्धवाः
उत्तर:
प्रिया: बान्धवाः (कर्मधारयः)

216. इयं सर्वे दिल्लीस्थ-लौहस्तम्भेन परिचिताः एव।
उत्तर:
दिल्लीस्थेन लौहस्तम्भेन (कर्मधारयः)

217. अन्ये विस्मयकरा: स्तम्भाः सन्ति।
उत्तर:
विस्मयम् कुर्वन्ति ये ते (बहुव्रीहिः)/विस्मयम् कुर्वन्ति इति (उपपद तत्पुरुषः)

218. सुचिन्दरं देवालये स्थिताः सङ्गीतमया: स्तम्भाः
उत्तर:
देवानाम् आलयः, तस्मिन् (षष्ठी तत्पुरुषः), सङ्गीतमयस्तम्भाः (कर्मधारयः)

219. फतेहपुरसीकरी मध्ये स्थिताः ध्वनिवाहकाः स्तम्भाः वैज्ञानिकानां गौरवगाथां वर्णयन्ति।
उत्तर:
ध्वनि वहन्ति इति (उपपद तत्पुरुषः)/ध्वनिम् वहन्ति ये ते (बहुव्रीहिः), गौरवस्य गाथाम् (षष्ठी तत्पुरुषः)

220. वयं सुषमाया: ज्ञानपूर्णेन सूचना प्रदानेन अनुगृहीताः।
उत्तर:
ज्ञानेन पूर्णम्, तेन (तृतीया तत्पुरुषः)

221. ये छात्राः भारतीयविज्ञानक्षेत्रे विशिष्टाः उपलब्धी: आधृत्य अध्ययनं करिष्यन्ति।
उत्तर:
भारतीयविज्ञानस्य क्षेत्रे (षष्ठी तत्पुरुषः), विशिष्टोपलब्धीः (कर्मधारयः)

222. अस्माकं प्राचार्यमहोदयः सम्मानपत्राणि प्रदास्यन्ति।
उत्तर:
प्राचार्य: महोदयः (कर्मधारयः)

223. पारितोषिकानि सम्मानपत्राणि च श्वः प्रार्थनासभायां प्रदास्यन्ते।
उत्तर:
प्रार्थनायाः सभायाम् (षष्ठी तत्पुरुषः)

224. सः सम्राजः चन्द्रगुप्तस्य आदेशस्य उल्लङ्घनम् अपि कर्तुम् उत्सहते स्म।
उत्तर:
सम्राट्चन्द्रगुप्तस्य (कर्मधारयः)

225. राज्यं हि नाम धर्मवृत्तिपरकस्य नृपस्य कृते महत् कष्टदायकम्
उत्तर:
धर्मवृत्तिपरकनृपस्य (कर्मधारयः), महाकष्टदायकम् (कर्मधारयः)

226. आर्य! अथ अस्मद्वचनात् आघोषित: कुसुमपुरे कौमुदीमहोत्सवः?
उत्तर:
अस्माकम् वचनम्, तस्मात् (षष्ठी तत्पुरुषः), कौमुद्या:महोत्सवः (षष्ठी तत्पुरुषः)

227. न खलु आर्यचाणक्येन अपहृतः प्रेक्षकाणाम् अतिशय रमणीयः चक्षुषो विषय:?
उत्तर:
आर्यः चाणक्यः, तेन (कर्मधारयः), चक्षुर्विषयः (षष्ठी तत्पुरुषः)

228. यद्येवं तर्हि कौमुदीमहोत्सव-प्रतिषेधस्य प्रयोजनं श्रोतुमिच्छामि।
उत्तर:
कौमुदी महोत्सवस्य प्रतिषेधः, तस्य (षष्ठी तत्पुरुषः)

229. पितृवधात् क्रुद्धः राक्षसोपदेशप्रवण: महीयसा म्लेच्छबलेन परिवृतः।
उत्तर:
पितुः वधात् (षष्ठी तत्पुरुषः), राक्षसस्य उपेदशे प्रवणः (षष्ठी तत्पुरुषः, सप्तमी तत्पुरुषः)

230. तत् नूनं प्रभातसमये मत्स्यजीविनः अत्र आगल मत्स्यसंक्षयं करिष्यन्ति।
उत्तर:
मत्स्यानाम् संक्षयम् (षष्ठी तत्पुरुषः)

231. विद्यालयस्य वार्षिकपत्रिकायां पर्वतारोहणं कुर्वतां छात्राणां चित्राणि प्रदर्शितानि।
उत्तर:
पर्वते (पर्वतस्य) आरोहणम् (सप्तमी (षष्ठी) तत्पुरुषः)

232. छात्राः शिक्षिकाम् उपेत्य तस्याः यात्रायाः वृत्तान्तं ज्ञातुम् अनुरोधं कुर्वन्ति।
उत्तर:
यात्रावृत्तान्तम् (षष्ठी तत्पुरुषः)

233. तेषाम् आग्रहं मत्त्वा शिक्षिका लेहलद्दाखयात्रायाः रोमाञ्चकारिणम् अनुभवं प्रस्तौति।
उत्तर:
लेहस्य लद्दाखस्य च यात्रायाः (षष्ठी तत्पुरुषः), रोमञ्चम् करोति, तम् (उपपद तत्पुरुषः)

NCERT Solutions for Class 12 Sanskrit

The post CBSE Class 12 Sanskrit व्याकरणम् समास-प्रकरण appeared first on Learn CBSE.


NTSE Manipur 2019 – 2020 for Class X | Application Form, Admit Card , Answer Key, Result

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NTSE Manipur 2020: The notification for NTSE Manipur is released by the education board for class 10 and the second stage exam is going to be conducted on June 2020. The candidates that are participating in the NTSE Manipur class 10 exam can find all the details in this article below.

The national-level talent search exam for students in class 10 is taken to nurture the talent. There are two stages in which this year’s NTSE Manipur exam is going to be held. For this, the 1st stage of the exam is going to be conducted at state or union territory level, while the stage 2 exam is going to be conducted by the NCERT. Below is the schedule for the exams going to be conducted under NTSE Manipur.

NTSE Manipur Important Dates

EventsImportant dates
NTSE Notification Release DateLast week of July 2019
Start of NTSE ApplicationsAugust 2019
Release of Stage 1 Admit CardThe second week of October 2019
Last date for NTSE application formAnnounced separately by each State/UT
NTSE Stage I 2020 Exam DateFirst Sunday of November 2019
Declaration of NTSE 2020 ResultJanuary-April 2020

NTSE Manipur Application Form

The application for the NTSE Manipur exam can be directly downloaded from the Manipur official website which is manipureducation.gov.in. It is advisable for the candidates to ensure that they are eligible for the exam. Also, all the conditions before filling the form should be fulfilled in the application form. For eligibility, the candidate should be currently studying in class 10 in recognized school to appear for the stage 1 exam.

This stage 1 exam is going to be conducted by the respective states or union territories where the schools are located. There is going to be no domicile restriction for the students. Thus, students that have registered under open distance learning are also eligible for this scholarship and it will be provided for students that are currently under 18 years of age. Also, that student should not be employed under any circumstances and he or she should be appearing for the first time in this exam.

NTSE Manipur Exam Pattern

For stage 1 and 2, the exam will be comprised of three papers which are:

PaperTest nameNo. of questionMarksDuration (mins)
Paper 1Mental Ability Test505045
Paper 2Language Test505045
Paper 3Scholastic Aptitude Tes10010090

For stage 2, there is going to be negative marking in each paper. For every incorrect answer, 1/3rd marks will be deducted.

NTSE Manipur Admit Card and Exam Center

NTSE Manipur admits card for the 2019-20 exam is going to be issued on their official website. Candidates can download the admit card from the official website or from the direct link given in the article above. The admit card can be downloaded online by entering the date of birth and application number of the student. It is compulsory to carry the admit card during the examination because no entry will be given to students without the admit card. NTSE Manipur test center details are

Exam center for Manipur NTSE stage 2 exam
Kendriya Vidyalaya no 1,
Lamphelpat, Imphal,
Manipur – 795004

NTSE Manipur Result and Answer Key

NTSE Manipur stage 1 result will be declared in one month from the date of the exam. The candidates that have cleared the stage 1 exam are eligible for the NTSE Manipur stage 2 exam. Candidates can also check the results for stage 1 and stage 2 from the NTSE Manipur official website. For answer key, the Manipur education board will release the answer key after the exam is over. A direct link will be activated from which candidates can check the answers.

NTSE Manipur Exam Important Information

  • Candidates can collect the admit card directly from the district liaison officer at least 1 week before the main exam.
  • Candidates should reach the exam centers at least half an hour before the exam commences and occupy their seats.
  • Calculators, physics or maths tables are not allowed inside the classroom.
  • Write the enrollment number as mentioned in the admit card on to the answer sheet along with the cover page of the question booklet in the space given.
  • Candidates should not write their names either on the answer sheet or the question booklet.
  • All the questions mentioned in the booklet will be objective type questions. For each question, there will be four possible answers.
  • Mark the answers using a blue or black pen only.

NTSE Manipur – Selection Procedure

The identification for NTSE Manipur talent is going to be a two-stage process. For the first stage, the individual state or union territories will conduct the exam. For the second stage, the selection is going to be held at the national level and will be carried out by the NCERT.

State Level Exam: State level will comprise of two stages for the students, part 1 is where mental ability test and in part 2 scholastic aptitude test is going to be conducted. This exam will determine the number of candidates required for the 2nd level of the tests to be conducted by the NCERT.

National Level Exam: Only the candidates that have given the state or union territory level exam and are qualified on the basis of the screening process will be eligible to appear for the 2nd stage of this exam. This second stage exam is going to be conducted by NCERT on the second Sunday of May every year.

The post NTSE Manipur 2019 – 2020 for Class X | Application Form, Admit Card , Answer Key, Result appeared first on Learn CBSE.

NTSE Haryana 2019-20 | Exam Dates (Released), Check Application, Eligibility

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NTSE Haryana 2019-20: SCERT Haryana has released the notification for the NTSE Haryana exam 2020. The process for application will begin from August 14 this year. Candidates applying needs to have a minimum of 55% marks to be eligible for this exam. The candidates that are eligible can fill the application form of NTSE Haryana exam till September 25, 2019. The application form should be filled by the candidates in online mode only. NTSE Haryana Stage 1 exam is to be conducted in November 1st week. The conducting body will release the admit card for NTSE Haryana 2020 exam online through the official website. NTSE Stage 1 exam is for shortlisting the candidates and deserving candidates for NTSE stage 2 exam. NTSE Stage 2 exam will be conducted on May 10, 2020, and these candidates will be awarded the NCERT scholarships.

NTSE exam is held individually for every state and union territory in India at two levels. The stage 1 exam is going to be conducted on the respective state-level and stage 2 will be held at the national level.

NTSE Haryana Exam Overview

Exam SpecificationExam Overview
Name of ExamNational Talent Search Examination, Haryana
Conducting BodyState Council of Educational Research and Training, Haryana
Examination LevelState Level
Mode of ExaminationOffline
Duration of NTSE Exam4 hours (2 hours per paper)
State/UT Liaison Officer & AddressSh. Sunil Vashishta
Lecturer Commerce, S.C.E.R.T., Sohna
Road, Opp. Panchayat Bhawan, Gurugram – 122001 (Haryana)
NTSE Official Websitehttps://scertharyana.gov.in/

NTSE Haryana Exam 2020 Important Dates

NTSE Haryana EventsDates
NTSE Haryana Application procedure startsAugust 14, 2019
Application closesSeptember 25, 2019
NTSE Haryana Stage I exam 2020November 3, 2019
NTSE Haryana 2020 Stage 1 cutoffFourth week of January 2020
NTSE Stage 2 Admit Card DateThird week of May 2020
NTSE Stage 2 examMay 10, 2020
Stage 2 resultsSeptember 2020

Eligibility Criteria for NTSE Haryana Exam

SCERT Haryana has out forward some eligibility criteria for the candidates that are appearing in the exam. Below are the eligibility criteria for the same:

  • The candidate that is appearing for NTSE Haryana 2020 exam should have cleared class 9 exams with a minimum of 55% marks. For candidates belonging to the reserved category, the criteria are different.
  • The candidate will be given the scholarship if he or she takes up studies only on a whole-time basis.
  • If the candidate gets the scholarship from other sources that this will lead to cancellation of the NTSE scholarship for that particular year.
  • Candidates that have undertaken any job with stipend, remuneration, or salary is not eligible for this scholarship.
  • The Indian candidates that are studying outside India in class 10 should be permitted to appear directly for the stage 2 exam based on norms and conditions by NCERT.

NTSE Haryana 2020 Exam Application Form

NTSE Haryana stage 1 application form is due to release on 14th August 2019. Candidates can fill the application form till September 25. The form will be available for everyone to download online on the SCERT official website. Also, the form will be available at respective schools of the candidates offline. Below is the step by step procedure to download the NTSE Haryana application form:

  • Go to the official website of NTSE Haryana to download the application form.
  • On the homepage, you will find a link under the news and events section. Click on this link to download the application form and print the form.
  • There will be a hall ticket or admit card attached to the application form.
  • Fill in the required details and it is important that the form is signed by the students, their school principal, and their parents. Keep the admit card or all hall ticket attached to the application form.
  • Photocopies of the caste certificate or any other special reservation should be attached if the candidate belongs to the reserved category.
  • Application form for the NTSE exam should be submitted to the state’s liaison officer. If not, then the candidates can also submit the form to their school authorities.

NTSE Haryana Exam Application Fee

CategoryWithout Late FeeWith Late Fee
GeneralRs. 200/-Rs. 250/-
SC/ST/PwDRs. 140/-Rs. 180/-

NTSE Haryana Exam Admit Card

The admit card will be available for everyone to download in the last week of October the NTSE Haryana 2020 admit card should include details such as candidate’s name, exam schedule, roll number, information related to the exam center, etc. It is mandatory for every student to carry the admit card for the exam. No candidate will be allowed to sit in the exam hall without having a valid admit card.

NTSE Haryana Exam Pattern

  • The question paper for the NTSE exam is divided into sections which are mental ability test or MAT and scholastic aptitude test or SAT.
  • Both these sections mentioned will have 100 questions each.
  • There is no negative marking for the NTSE Haryana stage 1 exam. Below table will give you complete information regarding the NTSE exam pattern.
SectionExam patternTimeQualifying scores
Mental Ability Test (MAT)100 questions120 minutesGeneral category: 40% of the maximum marks.
SC, ST, and PH: 32% of the maximum marks.
Scholastic Aptitude Test (SAT)100 questions120 minutesGeneral category: 40% of the maximum marks.
SC, ST, and PH: 32% of the maximum marks.

NTSE Haryana Exam Syllabus

There is no specific syllabus for NTSE Haryana. However, based on the previous exam trends and questions asked in the Haryana board there are some important topics that you can find below.

SectionsSubjectsImportant Topics
Mental Ability Test (MAT)Verbal Reasoning testArithmetic Reasoning Test, Blood Relations, Alphabet Test, Analogy, Classification, Mathematical Operations, Clocks, Puzzle test, Analytical Reasoning, Verification of truth of the statements, Series completion Test, Tests, etc.
Non-verbal ReasoningFolding paper cutting, Mirror images, Analogy, Analytical Reasoning, Transparent paper folding, Problems on cubes and dice, Water images, etc.
Scholastic Ability TestMathematicsAlgebra, basic geometry, Arithmetic progression, simple and compound interest, roots, statistics, coordinated geometry, surface area, and volume, etc.
ScienceCarbon and its compounds, physical and chemical changes, fibers and plastics, periodic classification of elements, magnetism, and electricity, source of energy, etc.
Social SciencesIntroduction of ancient Indian History, medieval architecture, and culture, Jainism, Mughal Empire, Vedic period, Buddhism, The Mauryas, etc.
UN and other International Agencies, Diversity and livelihood, Union government, Economic presence of the government, etc.
The atmosphere, industries, the motion of the earth, maps and globes, the internal structure of earth and rocks, solar system, natural vegetation, etc.

NTSE Haryana Answer Key and Result 2020

NTSE Haryana will release the official answer key on the SCERT official website after the results are declared. There are many coaching institutes that will provide the NTSE exam answer key. Students should check the official website for the NTSE exam frequently after the results are declared for the answer key. Students can use this answer key for comparing the answers with the correct answers and calculate their expected score.

For a result, NTSE Haryana will release the official result in the last of January 2020. The result will be available for everyone in the form of a merit list. Candidates can check the NTSE Haryana results online. NTSE Haryana results include candidate’s name, name of the school, roll number, overall score, section-wise marks, etc. Last year a total of 198 students managed to clear the NTSE stage 1 exam.

NTSE Haryana Cutoff

SCERT Haryana is going to release the cutoff for NTSE exam along with the results in March third week. The cutoff marks are the minimum marks that the candidates have to score to qualify the respective exam. Candidates will get their respective marks and will be able to appear before stage 2 NTSE exam during the third week of June 2020. Below is the previous year’s cutoff marks for NTSE Haryana:

CategoryTotal MAT and SAT scores out of 200Total MAT and SAT Cutoff
General181161
General-PH11484
OBC160149
OBC-PH158132
SC159128
SC-PHNilNil

NTSE Haryana Scholarship Amount 2020

NTSE program is solely for providing the scholarships to the deserving candidates. Under the program, 2000 scholarships are distributed for the candidates selected after the stage 2 process. Below is the category of wise scholarship data.

Education levelScholarship Amount
Classes 11-12Rs. 1,250/month
UndergraduateRs. 2,000/month
Post-graduateRs. 2,000/month
Ph.D.As per UGC norms.

The post NTSE Haryana 2019-20 | Exam Dates (Released), Check Application, Eligibility appeared first on Learn CBSE.

Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions

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Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions

Chemical Kinetics Class 12 Important Questions Very Short Answer Type

Question 1.
Define ‘rate of a reaction’. (Delhi 2010)
Answer:
Rate of a reaction: Either, The change in the concentration of any one of the reactants or products per unit time is called rate of a reaction. Or, The rate of a chemical reaction is the change in the molar concentration of the species taking part in a reaction per unit time.

Question 2.
Define ‘order of a reaction’. (All India 2011)
Answer:
The sum of powers of the concentration of the reactants in the rate law expression is called the order of reaction.

Question 3.
Define ‘activation energy’ of a reaction. (All India 2011)
Answer:
The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex is called activation energy.
The activation energy of the reaction decreases by the use of catalyst.

Question 4.
Express the rate of the following reaction in terms of the formation of ammonia :
N2(g) + 3H2(g) → 2NH3(g) (Comptt. All India 2013)
Answer:
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 1

Question 5.
If the rate constant of a reaction is k = 3 × 10-4 s-1, then identify the order of the reaction. (Comptt. All India 2013)
Answer:
S-1 is the unit for rate constant of first order reaction.

Question 6.
Write the unit of rate constant for a zero order reaction. (Comptt. All India 2013)
Answer:
Mol L-1 S-1 is unit of rate constant for a zero order reaction.

Question 7.
Define rate of reaction. (Comptt. Delhi 2016)
Answer:
The change in concentration of reactant or product per unit time is called rate of reaction.

Question 8.
Define rate constant (K). (Comptt. All India 2016)
Answer:
Rate constant. It is defined as the rate of reaction when the concentration of reaction is taken as unity.

Question 9.
For a reaction R → P, half-life (t1/2) is observed to be independent of the initial concentration of reactants. What is the order of reaction? (Delhi 2017)
Answer:
The t1/2 of a first order reaction is independent of initial concentration of reactants.

Chemical Kinetics Class 12 Important Questions Short Answer Type -I [SA-I]

Question 10.
A reaction is of second order with respect to a reactant. How will the rate of reaction be affected if the concentration of this reactant is
(i) doubled, (ii) reduced to half? (Delhi 2009)
Answer:
Since Rate = K[A]2
For second order reaction Let [A] = a then Rate = Ka2
(i) If [A] = 2a then Rate = K (2a)2 = 4 Ka2
∴Rate of reaction becomes 4 times

(ii) If [A] = \frac{a}{2} then Rate = K \left(\frac{a}{2}\right)^{2}=\frac{\mathrm{K} a^{2}}{4}
∴ Rate of reaction will be \frac{1}{4}^{\text { th }} .

Question 11.
Define the following :
(i) Elementary step in a reaction
(ii) Rate of a reaction (All India 2009)
Answer:
(i) Elementary step in a reaction: Those reactions which take place in one step are called elementary reactions.
Example : Reaction between H2, and I2 to form 2HI
H2 + I2 → 2HI
(ii) Rate of a reaction: The change in the concentration of any one of the reactants or products per unit time is called rate of reaction.

Question 12.
Define the following :
(i) Order of a reaction
(ii) Activation energy of a reaction (All India 2009)
Answer:
(i) Order of a reaction :

  • It is the sum of powers of molar concentrations of reacting species in the rate equation of the reaction.
  • It may be a whole number, zero, fractional, positive or negative.
  • It is experimentally determined.
  • It is meant for the reaction and not for its individual steps.

(ii) Activation energy of a reaction: The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex is called activation energy.

Question 13.
A reaction is of first order in reactant A and of second order in reactant B. How is the rate of this reaction affected when (i) the concentration of B alone is increased to three times (ii) the concentrations of A as well as B are doubled? (Delhi 2010)
Answer:
r = K[A]1 [B]2
(i) When concentration of B increases to 3 times, the rate of reaction becomes 9 times
r = KA(3B)2 ∴ r = 9KAB2 = 9 times
(ii) r = K(2A) (2B)2 ∴ r = 8KAB2 = 8 times

Question 14.
The rate constant for a reaction of zero order in A is 0.0030 mol L-1 s-1. How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M? (Delhi 2010)
Answer:
For a zero order reaction,
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 2

Question 15.
Distinguish between ‘rate expression’ and ‘rate constant’ of a reaction. (Delhi 2011)
Answer:
Rate expression: The expression which expresses the rate of reaction in terms of molar concentrations of the reactants with each term raised to their power, which may or may not be same as the stoichiometric coefficient of that reactant in the balanced chemical equation.
Rate constant: The rate of reaction when the molar concentration of each reactant is taken as unity.

Question 16.
What do you understand by the rate law and rate constant of a reaction? Identify the order of a reaction if the units of its rate constant are : (i) L-1 mol s-1 (ii) L mol-1 s-1 (All India 2011)
Answer:
The rate of reaction is found to depend on α concentration of term of reactant A and β concentration term of reactant B
Then Rate of reaction ∝ [A]α [B]β
or Rate = K [A]α [B]β
This expression is called Rate law.
‘K’ in this expression is called Rate constant. Rate constant’s unit :
(i) Unit = L-1 mol s-1 → Zero order reaction
(ii) Unit = L mol-1 s-1 → Second order reaction.

Question 17.
The thermal decomposition of HCO2H is a first order reaction with a rate constant of 2.4 × 10-3 s-1 at a certain temperature. Calculate how long will it take for three-fourths of initial quantity of HCO2 H to decompose. (log 0.25 = -0.6021) (All India 2011)
Answer:
Given : K = 2.4 × 10-3
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 3

Question 18.
A reaction is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is reduced to half? What is the unit of rate constant for such a reaction? (All India 2011)
Answer:
Rate = K [A]2 = Ka2
If [A] = \frac{1}{2}a Rate = K \left(\frac{a}{2}\right)^{2}=\frac{1}{4} Ka2
∴ Rate = 1/4th (one fourth of origina rate)
The unit of rate constant is L mol-1 s-1

Question 19.
What do you understand by the ‘order of a reaction’? Identify the reaction order from each of the following units of reaction rate constant:
(i) L-1 mol s-1 (ii) L mol-1 s-1 (Delhi 2012)
Answer:
Order of reaction: The sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction.
r = K[A]x[B]y Order = x + y
(i) Zero order
(ii) Second order

Question 20.
A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is (i) doubled (ii) reduced to half? (All India 2012)
Answer:
As Formula, r = K[R|2 …(Given)
(i) R’ = 2R ⇒ r = K[2R]-1 = 4KR-1
∴ Rate becomes 4 times than original rate
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 4

Question 21.
What is meant by rate of a reaction? Differentiate between average rate and instantaneous rate of a reaction. (Comptt. All India 2012)
Answer:
Rate of reaction: It is the change in concentration of the reactants or products in a unit time. Average rate : Average rate depends upon the change in concentration of reactants or products and the time taken for the change to occur. R → P
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 5
Instantaneous rate: It is defined as the rate of change in concentration of any one of the reactant or product at a particular moment of time.
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 6

Question 22.
(a) For a reaction A + B → P, the rate law is given by, r = k[A]1/2 [B]2.
What is the order of this reaction?
(b) A first order reaction is found to have a rate constant k = 5.5 × 10-14 s-1. Find the half life of the reaction. (All India 2013)
Answer:
(a) According to the formula : r = k[A]1/2 [B]2
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 7

Question 23.
Rate constant k for a first order reaction has been found to be 2.54 × 10-3 sec-1. Calculate its 3/4th life, (log 4 = 0.6020). (Comptt. India 2013)
Answer:
For first order reaction:
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 8

Question 24.
A first order gas phase reaction : A2B2(g) → 2A(g) + 2B(g) at the temperature 400°C has the rate constant k = 2.0 × 10-4 sec-1. What percentage of A2B2 is decomposed on heating for 900 seconds? (Antilog 0.0781 = 1.197) (Comptt. All India 2013)
Answer:
Since the reaction is of the first order
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 9

Question 25.
Write two differences between ‘order of reaction’ and ‘molecularity of reaction’. (Delhi 2014)
Answer:

Order of reactionMolecularity of reaction
(i) It is the sum of tire concentration terms on which die rate of reaction actually depends.It is the number of atoms, ions or molecules that must collide with one another simultaneously so as to result into a chemical reaction.
(ii) It can be fractional as well as zero.it is always a whole number.

Question 26.
Define the following terms :
(a) Pseudo first order reaction.
(b) Half life period of reaction (t1/2). (Delhi 2014)
Answer:
(a) Those reactions which are not truly of the first order but under certain conditions become first order reactions are called pseudo first order reaction.
(b) The time taken for half of the reaction to complete is called half life period.

Question 27.
Explain the following terms :
(i) Rate constant (k)
(ii) Half life period of a reaction (t1/2) (Delhi 2014)
Answer:
(i) Rate constant (k): It is a proportionality constant and is equal to the rate of reaction when the molar concentration of each of the reactants is unity.
(ii) Half life period of a reaction (t1/2): The time taken for half of the reaction to complete is called half life penod.(R)t

Question 28.
For a chemical reaction R → P, the variation in the f concentration (R) vs. time (f) plot is given as
(i) Predict the order of the reaction.
(ii) What is the slope of the curve? (All India 2014)
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 10
Answer:
(i) It is zero order reaction.
(ii) Slope of the curve = -K

Question 29.
(a) For a reaction, A + B → Product, the rate law is given by, Rate = k[A]1[B]2. What is the order of the reaction?
(b) Write the unit of rate constant ‘k’ for the first order reaction. (Comptt. Delhi 2014)
Answer:
(a) For a reaction, A + B
Rate = k [A]1 [B]2
This is the third order of reaction.
(b) Unit of rate constant for first order reaction is S-1

Question 30.
Define the following terms :
(i) Rate constant (k)
(ii) Activation energy (Ea) (Comptt. Delhi 2014)
Answer:
(i) Rate constant (k): It is a proportionality constant and is equal to the rate of reaction when the molar concentration of each of the reactants is unity.
(ii) Activation energy (Ea): The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex is called activation energy.

Question 31.
How does a change in temperature affect the rate, of a reaction? How can this effect on the rate constant of a reaction be represented quantitatively? (Comptt. All India 2014)
Answer:
The rate constant of a reaction increases with increase of temperature and becomes nearly double for every 10° rise in temperature.
The effect can be represented quantitatively by Arhenius equation K = Ae-Ea/RT
Where [Ea = Activation energy of the reaction; A = Frequency factor]

Question 32.
Define each of the following :
(i) Specific rate of a reaction
(ii) Energy of activation of a reaction (Comptt. All India 2014)
Answer:
(i) Specific rate of a reaction: Specific rate of reaction is the rate of reaction when the molar concentration of each of the reactants is unity.
(ii) Activation energy of a reaction: The minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to threshold value, is called activation energy.

Question 33.
A reaction is of second order with respect to its reactant. How will its reaction rate be affected if the concentration of the reactant is (i) doubled (ii) reduced to half? (Comptt. All India 2014)
Answer:
Since Rate = K[A]2
For second order reaction Let [A] = a then Rate = Ka2
(i) If [A] = 2a then Rate = K (2a)2 = 4 Ka2
∴Rate of reaction becomes 4 times
(ii) If [A] = \frac{a}{2} then Rate = K \left(\frac{a}{2}\right)^{2}=\frac{\mathrm{K} a^{2}}{4}
∴ Rate of reaction will be \frac{1}{4}^{\text { th }} .

Question 34.
Define the following terms :
(i) Half-life of a reaction (t1/2)
(ii) Rate constant (k) (Comptt. Delhi 2015)
Answer:
(i) Half-life of a reaction (t1/2): Half-life period (t1/2) is the time in which half of the substance has reacted and its concentration is reduced to one-half of its initial concentration.
(ii) Rate constant (k): Rate constant may be defined as the rate of reaction when the molar concentration of each reactant is taken as unity.

Question 35.
Write units of rate constants for zero order and for the second order reactions if the concentration is expressed in mol L-1 and time in second. (Comptt. All India 2015)
Answer:
Using formula of rate constant,
K = [mol L-1]1 – n s-1 (n = order of rxn)
Unit for zero order reaction,
K = [mol L-1]1 – 0 s-1
K = [mol L-1] s-1 = mol L-1 s-1
Unit for second order reaction,
K = [mol L-1]1 – 2 = [mol L-1]-1 s-1

Question 36.
For a reaction: 2NH2(g) \stackrel{\mathrm{Pt}}{\longrightarrow} N2(g) + 3H2(g)
Rate = k
(i) Write the order and molecularity of this reaction.
(ii) Write the unit of k. (Delhi 2016)
Answer:
2NH2(g) \stackrel{\mathrm{Pt}}{\longrightarrow} N2(g) + 3H2(g)
(i) It is a zero order reaction and its molecularity is two.
(ii) Unit of k is mol L-1 s-1.

Question 37.
For a reaction: H2 + Cl2 \stackrel{\mathrm{hv}}{\longrightarrow} 2HCl
Rate = k
(i) Write the order and molecularity of this reaction.
(ii) Write the unit of k. (All India 2016)
Answer:
H2 + Cl2 \stackrel{\mathrm{hv}}{\longrightarrow} 2HCl
This reaction is zero order reaction and molecularity is two.
(ii) Unit of k = mol L-1 s-1

Question 38.
For a chemical reaction R → P, variation in ln[R] vs time (f) plot is given below:
For this reaction:
(i) Predict the order of reaction
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 11
(ii) What is the unit of rate constant (k)? (Comptt. Delhi 2017)
Answer:
(i) It is zero order reaction.
(ii) The unit of rate constant (k) is mol L-1 S-1.

Question 39.
(a) Explain why H2 and O2 do not react at room temperature.
(b) Write the rate equation for the reaction A2 + 3B2 → 2C, if the overall order of the reaction is zero. (Comptt. All India 2017)
Answer:
(i) H2 and O2 do not react at room temperature because they do not have enough activation energy to overcome the exceptionally high activation energy barrier.
(ii) A2 + 3B2 → 2C
Rate = \left(\frac{d x}{d t}\right) = K[A]0 [B]0 = K (rate constant)

Question 40.
Derive integrated rate equation for rate constant of a first order reaction. (Comptt. All India 2017)
Answer:
In a first order reaction, the rate of reaction, is directly proportional to the concentration of the reactant.
Let us consider the reaction,
A → Products
The instantaneous reaction rate can be expressed as:
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 12
If t = 0 and [A] = [A]0, where [A]0 is the initial concentration of the reactant.
Then equation (ii) becomes
-ln[A]0 = I ……………. (iii)
Substitute the value of I in equation (ii)
-ln[A] = Kt – ln[A]
ln[A]0 – ln[A] = Kt
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 13
This is called integrated rate equation for the first order reaction.

Question 41.
(i) What is the order of the reaction whose rate constant has same units as the rate of reaction?
(ii) For a reaction A + H2O → B; Rate ∝ [A],
What is the order of this reaction? (Comptt. All India 2017)
Answer:
(i) The reaction whose rate constant has same units as the rate of reaction, will have zero order of reaction.
(ii) The reaction A + H2O → B Rate ∝ [A]
The order of this reaction will be pseudo first order reaction as the rate of reaction depends only on concentration of A only.

Chemical Kinetics Class 12 Important Questions Short Answer Type – II [SA-II]

Question 42.
A first order reaction has a rate constant of 0.0051 min-1. If we begin with 0.10 M concentration of the reactant, what concentration of reactant will remain in solution after 3 hours? (Delhi & All India 2009)
Answer:
Given : [R]0 = 0.10 M, t = 3 hrs = 180 min
K = 0.0051 min-1 [R] = ?
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 14
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 15
∴ Concentration of reactant remains 0.040 M

Question 43.
For a decomposition reaction the values of rate constant k at two different temperatures are given below :
k1 = 2.15 × 10-8 L mol-1 s-1 at 650 K
k2 = 2.39 × 10-7 L mol-1 s-1 at 700 K
Calculate the value of activation energy for this reaction.
(R = 8.314 J K-1 mol-1) (All India 2009)
Answer:
Given: k1 = 2.15 × 10-8 L mol-1 s-1, T1 = 650 K
k2 = 2.39 × 10-7 L mol-1 s-1, T2 = 700 K
R = 8.314 J K-1 mol-1 Ea =?
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 16

Question 44.
Nitrogen pentoxide decomposes according to equation :
2N2O5(g) → 4 NO2(g) + O2(g).
This first order reaction was allowed to proceed at 40°C and the data below were collected :

[N2OJ (M)Time (min)
0.4000.00
0.28920.0
0.20940.0
0.15160.0
0.10980.0

(a) Calculate the rate constant. Include units with your answer.
(b) What will be the concentration of N2O5 after 100 minutes?
(c) Calculate the initial rate of reaction. (Delhi 2009)
Answer:
(a) K = \frac{2.303}{t} \log \frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}
Substituting the values, we get
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 17

Question 45.
For the reaction
2NO(g) + Cl2(g) → 2NOCl(g) the following data were collected. All the measurements were taken at 263 K :

Experiment No.Initial [NO](M)Initial

[Cl2](M)

Initial rate of tlisa/ipcarance of Cl2 (M/min)
10.150.150.60
20.150.301.20
30.300.152.40
40.250.25?

(a) Write the expression for rate law.
(b) Calculate the value of rate constant and specify its units.
(c) What is the initial rate of disappearance of Cl2 in exp. 4? (Delhi 2012)
Answer:
(a) Rate law = K[NO]2 [Cl2]
(b) 0.60 M min-1 = K[0.15]2 [0.15] M3
∴ K = 177.7 M-2 min-1
(c) Initial rate of disappearance of Cl2 in exp. 4
Formula : Rate = K[NO]2 [Cl2]
∴ Initial rate = 177.7M-2 min-1 × (0.25)2 × (0.25)M3
= 2.8 M min-1

Question 46.
(a) A reaction is first order in A and second order in B.
(i) Write differential rate equation.
(ii) How is rate affected when concentration of B is tripled?
(iii) How is rate affected when concentration of both A and B is doubled?
(b) What is molecularity of a reaction? (Comptt. All India 2009)
Answer:
(a) (i) Differential rate equation :
\frac{d x}{d t} = rate = K [A]1 [B]2
(ii) Rate, r1 = K [A]1[B]2 …………… (i)
When concentration of B is increased three times then
Rate, r2 = K [A]1 [3B]2 ………..(ii)
Dividing equation (ii) by (i) we get
r2 = 9r1 rate increases by n*ne times.
(iii) When concentration of both A and B are doubled, then
r3 = K [2A]1 [2B]2 ………….. (iii)
Dividing equation {Hi) by (t), we get
r3 = 8r1
Hence rate increases by eight times.

(b) The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction is called Molecularity of a reaction.

Question 47.
The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (Ea) of the reaction assuming that it does not change with temperature. [R = 8.314 JK-1 mol-1, log 4 = 0.6021] (All India 2013)
Answer:
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 18

Question 48.
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume :
SO2Cl2 (g) → SO2 (g) + Cl2(g)

ExperimentTime/s_1Total pressure/atm
100.4
21000.7

Calculate the rate constant.
(Given : log 4 = 0.6021, log 2 = 0.3010) (Delhi, All India 2014)
Answer:
SO2Cl2 (g) → SO2 (g) + Cl2(g)
Using formula, K = \frac{2.303}{t} \log \frac{\mathrm{P}_{0}}{2 \mathrm{P}_{0}-\mathrm{P}_{t}}
When t = 100 s
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 19

Question 49.
Hydrogen peroxide, H2O2 (aq) decomposes to H2O (l) and O2 (g) in a reaction that is first order in H2O2 and has a rate constant k = 1.06 × 10-3min-1.
(i) How long will it take for 15% of a sample of H2O2 to decompose?
(ii) How long will it take for 85% of the sample to decompose? (Comptt. Delhi 2014)
Answer:
(i) Given : k = 1.06 × 10-3min-1
For first order reaction
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 20

Question 50.
For a decomposition reaction, the values of k at two different temperatures are given below:
k1 = 2.15 × 10-8 L mol-1 s-1 at 650 K
k2 = 2.39 × 10-7 L mol-1 s-1 at 700 K
Calculate the value of activation energy for this reaction.
(Log 11.11 = 1.046) (R = 8.314 J K-1 mol-1) (Comptt. All India 2014)
Answer:
Given: k1 = 2.15 × 10-8 L mol-1 s-1, T1 = 650 K
k2 = 2.39 × 10-7 L mol-1 s-1, T2 = 700 K
R = 8.314 J K-1 mol-1 Ea =?
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 16

Question 51.
The rate constant of a reaction at 500 K and 700 K are 0.02 s-1 and 0.07 s-1 respectively. Calculate the value of activation energy, En (R = 8.314 J K-1 mol-1) (Comptt. Delhi 2015)
Answer:
Given : k2 = 0.07 s-1, k1, = 0.02 s-1, T1 = 500 K, T2 = 700 K, Ea = ?
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 21

Question 52.
The rate constant for a first order reaction is 60 s-1. How much time will it take to reduce the initial concentration of the reactant to its l/10th value? (Comptt. All India 2015)
Answer:
Given : k = 60 s-1, t = ?
If initial concentration is [A0]
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 22

Question 53.
The rate constant for the first order decomposition of H2O2 is given by the following equation:
log k = 14.2 – \frac{1.0 \times 10^{4}}{\mathbf{T}} \mathbf{K}
Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes.
(Given: R = 8.314 J K-1 mol-1) (Delhi 2016)
Answer:
Given: t1/2 = 200 min Ea = ?, T = ?
Using Arrhenius equation
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 23

Question 54.
For the first order thermal decomposition reaction, the following data were obtained:
C2H5Cl(g) → C2H4(g) + HCl(g)
Time/sec               Total pressure/atm
0                                        0.30
300                                   0.50
Calculate the rate constant (Given: log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021) (All India 2016)
Answer:
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 24

Question 55.
If the half-life period of a first order reaction in A is 2 minutes, how long will it take [A] to reach 25% of its initial concentration? (Comptt. Delhi 2016)
Answer:
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 25

Question 56.
The rates of most reactions double when their temperature is raised from 298 K to 308 K. Calculate their activation energy. [R = 8.314 JK-1 mol-1] (Comptt. All India 2016)
Answer:
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 26

Question 57.
Following data are obtained for the reaction:
N2O5 → 2NO2 + \frac{1}{2}O2

t/s0300600
[N205]/mol L-11.6 × 10-20.8 × 10-20.4 × 10-2

(a) Show that it follows first order reaction.
(b) Calculate the half-life.
(Given log 2 = 0.3010 log 4 = 0.6021) (Delhi 2016)
Answer:
(a) For first order reaction:
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 27

Question 58.
A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed.
(Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021) (All India 2016)
Answer:
Given:
t = 20 min, A0 = 100%, A = 100 – 25 = 75%, k = ?
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 28
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 29

Question 59.
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume :

ExperimentTime(s)Total pressure (atm)
100.4
21000.7

Calculate the rate constant (k).
[Given : log 2 = 0.3010; log 4 = 0.6021] (Comptt. Delhi 2016)
Answer:
SO2Cl2 (g) → SO2 (g) + Cl2(g)
Using formula, K = \frac{2.303}{t} \log \frac{\mathrm{P}_{0}}{2 \mathrm{P}_{0}-\mathrm{P}_{t}}
When t = 100 s
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 19

Question 60.
For the first order decomposition of azoisopropane to hexane and nitrogen at 543 K , the following data were obtained :
Calculate the rate constant. The equation for the reaction is :

ExperimentTime(s)Total (mmHg)
1035.0
272063.0

(CH3)2CHN = NCH(CH3)2 C6H14 (g) + N2 (g)
[Given : log 3 = 0.4771; log 5 = 0.6990] (Comptt. Delhi 2016)
Answer:
(CH3)2CHN = NCH(CH3)2 → N2 + C6H14

Initial pressureP0OO
After time tP0 – PPP

Total Pressure after time t(Pt) = (P0 – P) + P + P
P = Pt – P0
a ∝ P0
(a – x) ∝ P0 – P
Substituting the value of P
a – x ∝ P0 – (Pt – P0)
or, (a – x) ∝ P0 – Pt
As decomposition of azoisopropane is a first order reaction
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 30
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 31

Question 61.
For a first order reaction, show that time required for 99% completion is twice the time required for completion of 90% reaction. (Comptt. All India 2016)
Answer:
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 41
∴ Ea = 2.303 × 8.314 (JK-1 mol-1) × 4250 K
= 81.375 J mol-1 or 81.375 kJ mol-1

Question 62.
Half-life for a first order reaction 693 s. Calculate the time required for 90% completion of this reaction. (Comptt. All India 2016)
Answer:
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 32

Chemical Kinetics Class 12 Important Questions Long Answer Type (LA)

Question 63.
(a) Explain the following terms :
(i) Rate of a reaction
(ii) Activation energy of a reaction (b) The decomposition of phosphine, PH3, proceeds according to the following equation:
4 PH3 (g) → P4 (g) + 6 H2 (g)
It is found that the reaction follows the following rate equation :
Rate = K [PH3].
The half-life of PH3 is 37.9 s at 120° C.
(i) How much time is required for 3/4th of PH3 to decompose?
(it) What fraction of the original sample of PH3 remains behind after 1 minute? (All India 2010)
Answer:
(a) (i) Rate of a reaction: The change in the concentration of any one of the reactants or products per unit time is called rate of reaction.
(ii) The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex is called activation energy.
The activation energy of the reaction decreases by the use of catalyst.

(b) (i) According to the formula :
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 33
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 34

Question 64.
(a) Explain the following terms :
(i) Order of a reaction
(ii) Molecularity of a reaction
(b) The rate of a reaction increases four times when the temperature changes from 300 K to 320 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature. (R = 8.314 J K-1 mol-1) (All India 2016)
Answer:
(a) (i) Order of a reaction: It is the sum of powers of the molar concentrations of reacting species in the rate equation of the reaction.
(ii) Molecularity of a reaction :

  • It is the total number of reacting species (molecules, atoms or ions) which bring the chemical change.
  • It is always a whole number.
  • It is a theoretical concept.
  • It is meaningful only for simple reactions or individual steps of a complex reaction. It is meaningless for overall complex reaction.

(b) Given : T1 = 300 K T2 = 320 K
K1 = K (Consider)
K2 = 4 K R = 8.314 Ea = ?
Substituting these values in the formulae,
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 35
∴ Energy of activation, Ea = 55327.46 = 55.3 KJ mol-1

Question 65.
(a) With the help of a labelled diagram explain the role of activated complex in a reaction.
(b) A first order reaction is 15% completed in 20 minutes. How long will it take to complete 60% of the reaction ? (Comptt. Delhi 2012)
Answer:
(a) In order that the reactants may change into products, they have to cross an energy barrier as shown in the diagram
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 36
This diagram is obained by plotting potential energy vs. reaction coordinate. It is believed that when the reactant molecules absorb energy, their bonds are loosened and new bonds are formed between them. The intermediate complex thus formed is called activated complex. It is unstable and immediately dissociates to form the stable products.

(b) For the first order reaction
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 37

Question 66.
(a) What is the physical significance of energy of activation ? Explain with diagram.
(b) In general, it is observed that the rate of a chemical reaction doubles with every 10 degree rise in temperature. If the generalization holds good for the reaction in the temperature range of 295 K to 305 K, what would be the value of activation energy for this reaction ?
[R = 8.314 J mol-1 K-1] (Comptt. Delhi 2012)
Answer:
(a) The minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to threshold value is called activation energy. Less is the activation energy, faster is the reaction or greater is the activation energy, slower is the reaction
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 36
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 38
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 39

Question 67.
(a) A reaction is second order in A and first order in B.
(i) Write the differential rate equation,
(ii) How is the rate affected on increasing the concentration of A three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
(b) A first order reaction takes 40 minutes for 30% decomposition. Calculate t1/2 for this reaction. (Given log 1.428 = 0.1548) (Delhi 2013)
Answer:
(a) (i) Differential rate equation :
\frac{d x}{d t} = K [A]2[B]
(ii) When concentration of A is increased to three times, the rate of reaction becomes 9 times
r = K[3A]2B ∴ r = 9KA2B i.e. = 9 times
(iii) r = K[2A]2[2B] ∴ r = 8KA2B i.e. = 8 times

(b) Given : Time, t = 40 minutes, t =?
Let a = 100, ∴ x = 30% of 100 = 30
Using the formula :
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 40

Question 68.
(a) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. .
(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
log k = log A – \frac{\mathrm{E}_{a}}{2.303 \mathrm{R}}\left(\frac{1}{\mathrm{T}}\right)
where Ea is the activation energy. When a graph is plotted for log k vs. \frac{1}{\overline{\mathbf{T}}},a straight line with a slope of – 4250 K is obtained. Calculate ‘Ea‘ for the reaction. (R = 8.314 JK-1 mol-1) (Delhi 2013)
Answer:
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 41
∴ Ea = 2.303 × 8.314 (JK-1 mol-1) × 4250 K
= 81.375 J mol-1 or 81.375 kJ mol-1

Question 69.
(a) The decomposition of A into products has a value of K as 4.5 × 103 s-1 at 10°C and energy of activation 60 kj mol-1. At what temperature would K be 1.5 × 104 s-1?
(b) (i) If half life period of a first order reaction is x and 3/4,th life period of the same reaction is y, how are x and y related to each other?
(ii) In some cases it is found that a large number of colliding molecules have energy more than threshold energy, yet the reaction is slow. Why? (Comptt. Delhi 2013)
Answer:
(a) Given : K1 = 4.5 × 103 s-1,
T1 = 10K + 273K = 283K
K2 = 1.5 × 104 s-1, T2 = ?
Ea = 60 KJ mol-1
Using formula :
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 42
∴ Temperature, T2 will be = 297° – 273° = 24° C

(b) (i) t1/2 = \frac{0.693}{\mathrm{K}} (For first order reaction)
t3/4 = K ⇒ t3/4 = \frac{1.3864}{\mathrm{K}}
According to condition
(The value 1.3864 is double of 0.693)
From the above equation it is clear that
t3/4 = 2t1/2 ∴ y = 2X
(ii) It is due to improper orientation of the colliding molecules at the time of collision.

Question 70.
(a) A first order reaction takes 100 minutes for completion of 60% of the reaction. Find the time when 90% of the reaction will be completed.
(b) With the help of diagram explain the role of activated complex in a reaction. (Comptt. Delhi 2013)
Answer:
(a) For the first order reaction,
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 43
(b) In order that the reactants may change into products, they have to cross an energy barrier as shown in the diagram
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 36
This diagram is obained by plotting potential energy vs. reaction coordinate. It is believed that when the reactant molecules absorb energy, their bonds are loosened and new bonds are formed between them. The intermediate complex thus formed is called activated complex. It is unstable and immediately dissociates to form the stable products.

Question 71.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained :

t/s03060
[CH3COOCH3]/mol L-10.600.300.15

(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds. (Given log 2 = 0.3010, log 4 = 0.6021) (Delhi 2015)
Answer:
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 44
As k is constant in both the readings, hence it is a pseudo first order reaction.
(ii) Rate = – Δ[R] /Δt, Average rate between 30 to 60 seconds
= \frac{-(0.15-0.30)}{60-30}=\frac{0.15}{30}
= 0.5 × 10-2 mol L-1 sec-1

Question 72.
(a) For a reaction A + B → P, the rate is given by Rate = k[A] [B]2
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
(b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.
(log 2 = 0.3010) (Delhi 2015)
Answer:
(a) For the reaction A + B → P rate is given
by Rate = k[A]1[B]2
(i) r1 = k[A]1 [B]2
r2 = k[ A]1[2B]2 =
r2 = k[A]1 [2B]2=4k[A]1 [B]2
r1 = 4r2, rate will increase four times of actual rate.

(ii) When A is present in large amount, order w.r.t. A is zero.
Hence overall order = 0 + 2 = 2, second order reaction.
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 45
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 46

Question 73.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:

t/s01020
[CH3COOCH3]/mol L-10.100.050.025

(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 10 to 20 seconds. (Given : log 2 = 0.3010, log 4 = 0.6021) (All India 2015)
Answer:
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 47
As k1 and k2 are equal, hence pseudo rate constant is same.
It follows the pseudo first order reaction.
(ii) Average rate of reaction between 10 to 20 seconds
= \frac{-\Delta[\mathrm{R}]}{\Delta t}=\frac{-(0.025-0.05)}{(20-10)}=\frac{0.025}{10}
= 0.0025 mol lit-1 sec-1

Question 74.
(a) For a reaction A + B → P, the rate is given by Rate = k[A] [B]2
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
(b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction. (All India 2015)
Answer:
(a) For the reaction A + B → P
rate is given by Rate = k[A]1[B]2
(i) r1 = k[A]1 [B]2
r2= k[A]1 [2B]2 = 4k[A]1 [B]2
r1 = 4r2 (rate of reaction becomes 4 times)

(ii) When A is present in large amounts, order w.r.t. A is zero.
Hence overall order = 0 + 2 = 2
Important Questions for Class 12 Chemistry Chapter 4 Chemical Kinetics Class 12 Important Questions 48

Important Questions for Class 12 Chemistry

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Important Questions for Class 10 Maths Chapter 10 Circles

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Important Questions for Class 10 Maths Chapter 10 Circles

Circles Class 10 Important Questions Very Short Answer (1 Mark)

Question 1.
In the given figure, O is the centre of a circle, AB is a chord and AT is the tangent at A. If ∠AOB = 100°, then calculate ∠BAT. (2011D)
Important Questions for Class 10 Maths Chapter 10 Circles 1
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 2
∠1 = ∠2
∠1 + ∠2 + 100° = 180°
∠1 + ∠1 = 80°
⇒ 2∠1 = 80°
⇒ ∠1 = 40°
∠1 + ∠BAT = 90°
∠BAT = 90° – 40° = 50°

Question 2.
In the given figure, PA and PB are tangents to the circle with centre O. If ∠APB = 60°, then calculate ∠OAB, (2011D)
Important Questions for Class 10 Maths Chapter 10 Circles 3
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 4
∠1 = ∠2
∠1 + ∠2 + ∠APB = 180°
∠1 + ∠1 + 60° = 180°
2∠1 = 180° – 60° = 120°
∠1 = \frac{120^{\circ}}{2} = 60°
∠1 + ∠OAB = 90°
60° +∠OAB = 90°
∠OAB = 90° – 60° = 30°

Question 3.
In the given figure, O is the centre of a circle, PQ is a chord and PT is the tangent at P. If ∠POQ = 70°, then calculate ∠TPQuestion (2011OD)
Important Questions for Class 10 Maths Chapter 10 Circles 5
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 6
∠1 = ∠2
∠1 + ∠2 + 70° = 180°
∠1 + ∠1 = 180° – 70°
2∠1 = 110° ⇒ ∠1 = 55°
∠1 + ∠TPQ = 90°
55° + ∠TPQ = 90°
⇒ ∠TPQ = 90° – 55° = 35°

Question 4.
A chord of a circle of radius 10 cm subtends a right angle at its centre. Calculate the length of the chord (in cm). (2014OD)
Solution:
AB2 = OA2 + OB2 …[Pythagoras’ theorem
Important Questions for Class 10 Maths Chapter 10 Circles 7
AB2 = 102 + 102
AB2 = 2(10)2
AB = 10 \sqrt{2} cm

Question 5.
In the given figure, PQ R is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°. Find ∠PCA. (2016OD)
Important Questions for Class 10 Maths Chapter 10 Circles 8
Solution:
∠ACB = 90° …[Angle in the semi-circle
In ∆ABC,
∠CAB + ∠ACB + ∠CBA = 180°
30 + 90° + ∠CBA = 180°
∠CBA = 180° – 30° – 90° = 60°
∠PCA = ∠CBA …[Angle in the alternate segment
∴ ∠PCA = 60°

Question 6.
In the given figure, AB and AC are tangents to the circle with centre o such that ∠BAC = 40°. Then calculate ∠BOC. (2011OD)
Important Questions for Class 10 Maths Chapter 10 Circles 9
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 10
AB and AC are tangents
∴ ∠ABO = ∠ACO = 90°
In ABOC,
∠ABO + ∠ACO + ∠BAC + ∠BOC = 360°
90° + 90° + 40° + ∠BOC = 360°
∠BOC = 360 – 220° = 140°

Question 7.
In the given figure, a circle touches the side DF of AEDF at H and touches ED and EF produced at K&M respectively. If EK = 9 cm, calculate the perimeter of AEDF (in cm). (2012D)
Important Questions for Class 10 Maths Chapter 10 Circles 11
Solution:
Perimeter of ∆EDF
= 2(EK) = 2(9) = 18 cm

Question 8.
In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm, then calculate the length of AP (in cm). (2012OD)
Important Questions for Class 10 Maths Chapter 10 Circles 12
Solution:
2AP = Perimeter of ∆
2AP = 5 + 6 + 4 = 15 cm
AP = \frac{15}{2} = 7.5 cm

Question 9.
In the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA ⊥ PB, then find the length of each tangent. (2013D)
Important Questions for Class 10 Maths Chapter 10 Circles 13
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 14
Construction: Join AC and BC.
Proof: ∠1 = ∠2 = 90° ….[Tangent is I to the radius (through the point of contact
∴ APBC is a square.
Length of each tangent
= AP = PB = 4 cm
= AC = radius = 4 cm

Question 10.
In the given figure, PQ and PR are two tangents to a circle with centre O. If ∠QPR = 46°, then calculate ∠QOR. (2014D)
Important Questions for Class 10 Maths Chapter 10 Circles 15
Solution:
∠OQP = 900
∠ORP = 90°
∠OQP + ∠QPR + ∠ORP + ∠QOR = 360° …[Angle sum property of a quad.
90° + 46° + 90° + ∠QOR = 360°
∠QOR = 360° – 90° – 46° – 90° = 134°

Question 11.
In the given figure, PA and PB are tangents to the circle with centre O such that ∠APB = 50°. Write the measure of ∠OAB. (2015D)
Important Questions for Class 10 Maths Chapter 10 Circles 16
Solution:
PA = PB …[∵ Tangents drawn from external point are equal
∠OAP = ∠OBP = 90°
∠OAB = ∠OBA … [Angles opposite equal sides
∠OAP + ∠AOB + ∠OBP + ∠APB = 360° … [Quadratic rule
Important Questions for Class 10 Maths Chapter 10 Circles 17
90° + ∠AOB + 90° + 50° = 360°
∠AOB = 360° – 230°
= 130°
∠AOB + ∠OAB + ∠OBA = 180° … [∆ rule
130° + 2∠OAB = 180° … [From (i)
2∠OAB = 50°
⇒ ∠OAB = 25°

Question 12.
From an external point P, tangents PA and PB are drawn to a circle with centre 0. If ∠PAB = 50°, then find ∠AOB. (2016D)
Solution:
PA = PB …[∵ Tangents drawn from external point are equal
Important Questions for Class 10 Maths Chapter 10 Circles 18
∠PBA = ∠PAB = 50° …[Angles equal to opposite sides
In ∆ABP, ∠PBA + ∠PAB + ∠APB = 180° …[Angle-sum-property of a ∆
50° + 50° + ∠APB = 180°
∠APB = 180° – 50° – 50° = 80°
In cyclic quadrilateral OAPB
∠AOB + ∠APB = 180° ……[Sum of opposite angles of a cyclic (quadrilateral is 180°
∠AOB + 80o = 180°
∠AOB = 180° – 80° = 100°

Question 13.
In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If ∠QPT = 60°, find ∠PRQ. (2015OD)
Important Questions for Class 10 Maths Chapter 10 Circles 19
Solution:
PQ is the chord of the circle and PT is tangent.
∴ ∠OPT = 90° …[Tangent is I to the radius through the point of contact
Now ∠QPT = 60° … [Given
∠OPQ = ∠OPT – ∠QPT
⇒ ∠OPQ = 90° – 60° = 30°
In ∆OPQ, OP = OQ
∠OQP = ∠OPQ = 30° … [In a ∆, equal sides have equal ∠s opp. them
Now, ∠OQP + ∠OPQ + ∠POQ = 180°
∴ ∠POQ = 120° …[∠POQ = 180o – (30° + 30°)
⇒ Reflex ∠POQ = 360° – 120° = 240° …[We know that the angle subtended by an arc at the centre of a circle is twice the angle subtended by it at any point on the remaining part of the circle
∴ Reflex ∠POQ = 2∠PRO
⇒ 240° = 2∠PRQ
⇒ ∠PRQ = \frac{240^{\circ}}{2} = 120°

Question 14.
In the given figure, the sides AB, BC and CA of a triangle ABC touch a circle at P, Q and R respectively. If PA = 4 cm, BP = 3 cm and AC = 11 cm, find the length of BC (in cm). (2012D)
Important Questions for Class 10 Maths Chapter 10 Circles 20
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 21
AP = AR = 4 cm
RC = 11 – 4 = 7 cm
RC = QC = 7 cm
BQ = BP = 3 cm
BC = BQ + QC
= 3 + 7 = 10 cm

Question 15.
In a right triangle ABC, right-angled at B, BC = 12 cm and AB = 5 cm. Calculate the radius of the circle inscribed in the triangle (in cm). (2014OD)
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 22
AC2 = AB2 + BC2 …[Pythagoras’ theorem
= (5)2 + (12)2
AC2 = 25 + 144
AC = \sqrt{169} = 13 cm
Area of ∆ABC = Area of ∆AOB + ar. of ∆BOC + ar. of ∆AOC
Important Questions for Class 10 Maths Chapter 10 Circles 23
60 = r(AB + BC + AC)
60 = r(5 + 12 + 13)
60 = 30r ⇒ r = 2 cm

Question 16.
Find the perimeter (in cm) of a square circum scribing a circle of radius a cm. (2011OD)
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 24
Radius = R
AB = a + a = 2a
∴ Perimeter = 4(AB)
= 4(2a)
= 8a cm

Question 17.
In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides AB, BC, CD and AD at P, Q, R and S respectively. If the radius DA of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD ⊥ CD, then calculate the length of CD. (2013OD)
Important Questions for Class 10 Maths Chapter 10 Circles 25
Solution:
Const. Join OR
Proof. ∠1 = ∠2 = 90° … [Tangent is ⊥ to the radius through the point of contact
∠3 = 90° …[Given
Important Questions for Class 10 Maths Chapter 10 Circles 26
∴ ORDS is a square.
DR = OS = 10 cm …(i)
BP = BQ = 27 cm …[Tangents drawn from an external point
∴ CQ = 38 – 27 = 11 cm
RC = CO = 11 cm …[Tangents drawn from an external point
DC = DR + RC = 10 + 11 = 21 cm …[From (i) & (ii)

Circles Class 10 Important Questions Short Answer-I (2 Marks)

Question 18.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel. (2012OD)
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 27
Proof: ∠1 = 90° …(i)
∠2 = 90° …(ii)
∠1 = ∠2 … [From (i) & (ii)
But these are alternate interior angles
∴PQ || RS

Question 19.
In the figure, AB is the diameter of a circle with centre O and AT is a tangent. If ∠AOQ = 58°, find ∠ATQ.  (2015D)
Important Questions for Class 10 Maths Chapter 10 Circles 28
Solution:
∠ABQ = \frac{1}{2} ∠AOQ = \frac{58^{\circ}}{2} = 29°
∠BAT = 90° ….[Tangent is ⊥ to the radius through the point of contact
∠ATQ = 180° – (∠ABQ + ∠BAT)
= 180 – (29 + 90) = 180° – 119° = 61°

Question 20.
Two concentric circles are of radii 7 cm and r cm respectively, where r > 7. A chord of the larger circle, of length 48 cm, touches the smaller circle. Find the value of r. (2011D)
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 29
Given: OC = 7 cm, AB = 48 cm
To find: r = ?
∠OCA = 90° ..[Tangent is ⊥ to the radius through the point of contact
∴ OC ⊥ AB
AC = \frac{1}{2} (AB) … [⊥ from the centre bisects the chord
⇒ AC = \frac{1}{2} (48) = 24 cm
In rt. ∆OCA, OA2 = OC2 + AC2 … [Pythagoras’ theorem
r2 = (7)2 + (24)2
= 49 + 576 = 625
∴ r= \sqrt{625} = 25 cm

Question 21.
In the figure, the chord AB of the larger of the two concentric circles, with centre O, touches the smaller circle at C. Prove that AC = CB. (2012D)
Important Questions for Class 10 Maths Chapter 10 Circles 30
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 31
Const.: Join OC
Proof: AB is a tangent to smaller circle and OC is a radius.
∴ ∠OCB = 90° … above theorem
In the larger circle, AB is a chord and OC ⊥ AB.
∴ AC = CB … [⊥ from the centre bisects the chord

Question 22.
In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8 cm. Find the length of side AD. (2011OD)
Important Questions for Class 10 Maths Chapter 10 Circles 32
Solution:
AB + CD = AD + BC
6 + 8 = AD + 9
14 – 9 = AD ⇒ AD = 5 cm

Question 23.
Prove that the parallelogram circumscribing a circle is a rhombus. (2012D, 2013D)
Solution:
Given. ABCD is a ॥gm.
To prove. ABCD is a rhombus.
Proof. In ॥gm, opposite sides are equal
Important Questions for Class 10 Maths Chapter 10 Circles 33
AB = CD
and AD = BC ..(i)
AP = AS …[Tangents drawn from an external point are equal in length
PB = BQ
CR = CO
DR = DS
By adding these tangents,
(AP + PB) + (CR + DR) = AS + BQ + CQ + DS
AB + CD = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
AB + AB = BC + BC … [From (i)
2AB = 2 BC
AB = BC …(ii)
From (i) and (ii), AB = BC = CD = DA
∴ ॥gm ABCD is a rhombus.

Question 24.
In figure, a quadrilateral ABCD is drawn to circum- DA scribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, RA and S respectively. Prove that: AB + CD = BC + DA. (2013 OD, 2016 OD)
Important Questions for Class 10 Maths Chapter 10 Circles 34
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 35
AP = AS ……(i) (Tangents drawn from an external point are equal in length
BP = BO …(ii)
CR = CQ ….(iii)
DR = DS ..(iv)
By adding (i) to (iv)
(AP + BP) + (CR + DR) = AS + BQ + CQ + DS
AB + CD = (BQ + CQ) + (AS + DS)
∴ AB + CD = BC + AD (Hence proved)

Question 25.
In the given figure, an isosceles ∆ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC. (2012D)
Important Questions for Class 10 Maths Chapter 10 Circles 36
Solution:
Given: The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F F respectively.
Important Questions for Class 10 Maths Chapter 10 Circles 37
AB = AC
To prove: BD = CD
Proof: Since the lengths of tangents drawn from an external point to a circle are equal
∴ AF = AE … (i)
BF = BD …(ii)
CD = CE …(iii)
Adding (i), (ii) and (iii), we get
AF + BF + CD = AE + BD + CE
⇒ AB + CD = AC + BD
But AB = AC … [Given
∴ CD = BD

Question 26.
In Figure, a right triangle ABC, circumscribes a circle of radius r. If AB and BC are of lengths 8 cm and 6 cm respectively, find the value of r. (2012OD)
Important Questions for Class 10 Maths Chapter 10 Circles 38
Solution:
Const.: Join AO, OB, CO
Proof: area of ∆ABC
Important Questions for Class 10 Maths Chapter 10 Circles 39
From (i) and (ii), we get 12r = 24
∴ r = 2 cm

Question 27.
In the given figure, a circle inscribed in ∆ABC touches its sides AB, BC and AC at points D, E & F K respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF. (2013D)
Important Questions for Class 10 Maths Chapter 10 Circles 40
Solution:
Let AD = AF = x
BD = BE = y …[Two tangents drawn from and an external point are equal
CE = CF = z
Important Questions for Class 10 Maths Chapter 10 Circles 41
AB = 12 cm …[Given
∴ x + y = 12 cm …(i)
Similarly,
y + z = 8 cm …(ii)
and x + z = 10 cm …(iii)
By adding (i), (ii) & (iii)
2(x + y + z) = 30
x + y + z = 15 …[∵ x + y = 12
z = 15 – 12 = 3
Putting the value of z in (ii) & (iii),
y + 3 = 8
y = 8 – 3 = 5
x + 3 = 10
x = 10 – 3 = 7
∴ AD = 7 cm, BE = 5 cm, CF = 3 cm

Question 28.
The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC. (2014OD)
Solution:
Given: The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F respectively.
Important Questions for Class 10 Maths Chapter 10 Circles 42
AB = AC
To prove: BD = CD
Proof: AF = AE ..(i)
BF = BD …(ii)
CD = CE …(iii)
Adding (i), (ii) and (iii), we get
AF + BF + CD = AE + BD + CE
⇒ AB + CD = AC + BD
But AB = AC …[Given
∴ CD = BD

Question 29.
In the figure, a ∆ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively 6 cm 9 cm of lengths 6 cm and 9 cm. If the area of ∆ABC is 54 cm2, then find the lengths of sides AB and AC. (2011D, 2011OD, 2015 OD)
Important Questions for Class 10 Maths Chapter 10 Circles 43
Solution:
Given: OD = 3 cm; OE = 3 cm; OF = 3 cm ar(∆ABC) = 54 cm2
Important Questions for Class 10 Maths Chapter 10 Circles 44
Joint: OA, OF, OE, OB and OC
Let AF = AE = x
BD = BF = 6 cm
CD = CE = 9 cm
∴ AB = AF + BF = x + 6 …(i)
AC = AE + CE = x + 9 …(ii)
BC = DB + CD = 6 + 9 = 15 cm …(iii)
In ∆ABC,
Area of ∆ABC = 54 cm2 …[Given
ar(∆ABC) = ar(∆BOC) + ar(∆AOC) + ar(∆AOB)
Important Questions for Class 10 Maths Chapter 10 Circles 45

Question 30.
In the figure, a circle is inscribed in a ∆ABC, such that it touches the sides AB, BC and CA at points D, E and F respectively. If the lengths of sides AB, BC, and CA are 12 cm, 8 cm and 10 cm respectively, find the lengths of AD, BE and CF. (2016D)
Important Questions for Class 10 Maths Chapter 10 Circles 46
Solution:
AB = 12 cm, BC = 8 cm, CA = 10 cm
Important Questions for Class 10 Maths Chapter 10 Circles 47
As we know,
AF = AD
CF = CE
BD = BE
Let AD = AF = x cm
then, DB = AB – AD
= (12 – x) cm
∴ BE = (12 – x) cm ..[Tangents drawn from an external point are equal
Similarly,
CF = CE = AC – AF = (10 – x) cm
BC = 8 cm …[Given
⇒ BE + CE = 8 ⇒ 12 – x + 10 – x = 8
⇒ 22 – 8 = 2x ⇒ 2x = 14
∴ x = 7 ∴ AD = x = 7 cm
BE = 12 – x = 12 – 7 = 5 cm
CF = 10 – x = 10 – 7 = 3 cm

Question 31.
In Figure, common tangents AB and CD to the two circles with lo, centres O1 and O2 intersect at E. Prove that AB = CD. (2014OD)
Important Questions for Class 10 Maths Chapter 10 Circles 48
Solution:
EA = EC …(i) ….[Tangents drawn from an external point are equal
EB = ED …(ii)
EA + EB = EC + ED …[Adding (i) & (ii)
∴ AB = CD (Hence proved)

Question 32.
If from an external point P of a circle with centre O, two tangents PQ and PR are drawn such that ∠QPR = 120°, prove that 2PQ = PO. (2014D)
Solution:
∠OPQ = \frac{1}{2}(∠QPR) ..[Tangents drawn from an external point are equal
= \frac{1}{2}(120°) = 60° …[Tangent is ⊥ to the radius through the point of contact
∠OQP = 90°
In rt. ∆OQP, cos 60° = \frac{PQ}{PO}
\frac{1}{2}=\frac{P Q}{P O} ∴ 2PQ = PO

Question 33.
From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of line segment PQuestion (2015D)
Solution:
In ∆s’ TPC and TQC ….[Tangents drawn from an external point are equal
TP = TQ
TC = TC …[Common
∠1 = ∠2 …[TP and TQ are equally inclined to OT
∴ ∆TPC = ∆TQC … [SAS
∴ PC = QC …[CPCT
Important Questions for Class 10 Maths Chapter 10 Circles 49
∠3 = ∠4 …(i)
⇒ ∠3 + 24 = 180° … [Linear pair
⇒ ∠3 + ∠3 = 180°…[From (i)
⇒ 2∠3 = 180° ⇒ ∠3 = 90°
∴ ∠3 = ∠4 = 90°
∴ OT is the right bisector of PQuestion

Question 34.
In the figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120°, then prove that OR = PR + R. (2015OD)
Important Questions for Class 10 Maths Chapter 10 Circles 50
Solution:
Join OP and OO
∠OPR = 90°
PR = RQ … [Tangents drawn from an external point are equal
∠PRO = \frac{1}{2} ∠PRQ = \frac{1}{2} × 120° = 60°
Now, In ∆OPR,
Important Questions for Class 10 Maths Chapter 10 Circles 51
⇒ ∠OPR + ∠POR + ∠ORP = 180° …[∆ Rule
⇒ 90° + ∠POR + 60° = 180°
⇒ ∠POR + 150° = 180°
⇒ ∠POR = 30°
⇒ sin 30° = \frac{PR}{OR}\frac{1}{2}=\frac{P R}{O R}
⇒ OR = 2PR
⇒ OR = PR + QR (∵ PR = RQ) …(Hence proved)

Question 35.
In the figure, AP and BP are tangents to a circle with centre 0, such that AP = 5 cm and ∠APB = 60°. Find the length of chord AB. (2016D)
Important Questions for Class 10 Maths Chapter 10 Circles 52
Solution:
PA = PB …[Tangents drawn from an external point are equal
Given:
∠APB = 60°
∠PAB = ∠PBA … (i) …(Angles opposite to equal sides
In ∆PAB, ∠PAB + ∠PBA + ∠APB = 180° …[Angle-sum-property of a ∆
⇒ ∠PAB + ∠PAB + 60° = 180°
⇒ 2∠PAB = 180° – 60o = 120°
⇒ ∠PAB = 60°
⇒ ∠PAB = ∠PBA = ∠APB = 60°
∴ APAB is an equilateral triangle
Hence, AB = AP = 5 cm …[∵ All sides of an equilateral A are equal

Question 36.
In the figure, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP = 2r, show that ∠OTS = ∠OST = 30°. (2016OD)
Important Questions for Class 10 Maths Chapter 10 Circles 53
Solution:
Let ∠TOP = θ …[Tangent is ⊥ to the radius through the point of contact
∠OTP = 90°
OT = OS = r … [Given
In rt. ∆OTP, cos θ = \frac{\mathrm{OT}}{\mathrm{OP}}
⇒ cos θ = \frac{r}{2 r} ⇒ cos θ = \frac{1}{2}
⇒ cos θ = cos 60° ⇒ θ = 60°
∴ ∠TOS = 60° + 60° = 120°
In ATOS,
∠OTS = ∠OST …[Angles opposite to equal sides
In ∠TOS,
∠TOS + ∠OTS + ∠OST = 180° … [Angle-sum-property of a ∆
120° + ∠OTS + ∠OTS = 180° … [From (i)
2∠OTS = 180° – 120°
∠OTS = 60°/2 = 30°
∴ ∠OTS = ∠OST = 30°

Question 37.
In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Question If PA = 12 cm, QC = QD = 3 cm, then find PC + PD. (2017D)
Important Questions for Class 10 Maths Chapter 10 Circles 54
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 55
PA = PB = 12 cm …(i)
QC = AC = 3 cm …(ii)
QD = BD = 3 cm …(iii)
To find: PC + PD
= (PA – AC) + (PB – BD)
= (12 – 3) + (12 – 3) … [From (i), (ii) & (iii)
= 9 + 9 = 18 cm

Question 38.
In the figure, two circles touch each other at the point C. Prove that the common tangent to the circles at C, bisects the common tangent at P and Q. (2013 OD)
Important Questions for Class 10 Maths Chapter 10 Circles 56
Solution:
To prove: PR = RQ
Proof: PR = RC … (i)
QR = RC
From (i) and (ii), PR = QR (Hence proved)

Circles Class 10 Important Questions Short Answer-II (3 Marks)

Question 39.
In the figure, a circle is inscribed in a triangle PQR with PQ = 10 cm, QR = 8 cm and PR = 12 cm. Find the lengths of QM, RN and PI. (2012OD)
Important Questions for Class 10 Maths Chapter 10 Circles 57
Solution:
Let PL = PN = x cm
QL = QM = y cm
RN = MR = z cm
PQ = 10 cm = x + y = 10 …(i)
QR = 8 cm = y + z = 8 …(ii)
PR = 12 cm = x + z = 12 …(iii)
By adding (i), (ii) and (iii),
Important Questions for Class 10 Maths Chapter 10 Circles 58
We get,
⇒ 2x + 2y + 2z = 10 + 8 + 12
⇒ 2(x + y + z) = 30
⇒ x + y + z = 15
⇒ 10 + z = 15 … [From (i)
∴ z = 15 – 10 = 5 cm
From (ii)
y + 5 = 8
y = 8 – 5
y = 3 cm
From (iii)
x + 5 = 12
x = 12 – 5
x = 7 cm
∴ QM = 3 cm, RN = 5 cm, PL = 7 cm

Question 40.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle. (2012D)
Solution:
1st method:
To prove. (i) ∠AOD + ∠BOC = 180°
(ii) ∠AOB + ∠COD = 180°
Proof. In ∆BPO and ∆BQO …[Tangents drawn from an external point are equal
Important Questions for Class 10 Maths Chapter 10 Circles 59
PO = 20 … [radii
BO = BO … [Common
∆BPO = ∆BQO … [SSS Congruency rule
∠8 = ∠1 …(i) (c.p.c.t.)
Similarly,
∠2 = ∠3, ∠4 = ∠5 and ∠6 = ∠7
∠1 + ∠2 + 23+ 24 + 25 + 26+ 27 + ∠8 = 360° …(Complete angles
∠1 + ∠2 + 22+ 25 + 25 + 26+ ∠6+ ∠1 = 360°
2(∠1 + ∠2 + 25 + 26) = 360°
∠BOC + ∠AOD = 180°…(i) [Proved part I
∠AOB + ∠BOC + ∠COD + ∠DOA = 360° …(Complete angles
∠AOB + ∠COD + 180o = 360° … [From (i)
∴ ∠AOB + ∠COD = 360° – 180o = 180° …(proved)

2nd method:
To prove:
(i) ∠6 + ∠8 = 180°
(ii) ∠5 + ∠7 = 180°
Proof. As AS and AP are tangents to the circle from a point A
∴ O lies on the bisector of ∠SAP
∴ ∠1 = \frac{1}{2} ∠BAD …(i)
Similarly BO, CO and DO are the bisectors of
∠ABC, ∠BCD and ∠ADC respectively. …(ii)
Important Questions for Class 10 Maths Chapter 10 Circles 60
∴ ∠1 + ∠4 + ∠3 + ∠2 =180°…(iii) ..[From (1) & (ii)
In ∆AOD, ∠1 + ∠2 + 26 = 180° …[Angle-sum-Prop. of a ∆
In ∆BOC, ∠3 + ∠4 + ∠8 = 180° …(v)
Adding (iv) and (v)
(∠1 + ∠2 + 23 + 24) + 26 + 28 = 180° + 180°
180° + 26 + 28 = 180° + 180° … [From (iii)
∴∠6 + 28 = 180°
Now ∠5 + ∠6 + ∠7 + ∠8 = 360° … (Complete angles
(∠5 + ∠7) + (∠6 + ∠8) = 360°
(∠5 + ∠7) + 180° = 360°
∠5 + ∠7 = 360° – 180° = 180°
∠5 + ∠7 = 180°

Question 41.
Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that ∠PTQ = 2∠OPQ. (2017D)
Solution:
We are given a circle with centre O, an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact (see Figure).
Important Questions for Class 10 Maths Chapter 10 Circles 61
We need to prove that:
∠PTQ = 2∠OPQ
Let ∠PTQ = θ
Now, TP = TQuestion ….[∵ Lengths of tangents drawn from an external pt. to a circle are equal
So, TPQ is an isosceles triangle.
Important Questions for Class 10 Maths Chapter 10 Circles 62

Circles Class 10 Important Questions Long Answer (4 Marks)

Question 42.
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. (2011OD, 2012OD, 2013D, 2014OD, 2015D)
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 63
Given: XY is a tangent at point P to the circle with centre O.
To prove: OP ⊥ XY
Const.: Take a point Q on XY other than P and join to OQuestion
Proof: If point Q lies inside the circle, then XY will become a secant and not a tangent to the circle.
∴ OQ > OP
This happens with every point on the line XY except the point P.
OP is the shortest of all the distances of the point O to the points of XY
∴ OP ⊥ XY … [Shortest side is ⊥

Question 43.
Prove that the lengths of tangents drawn from an external point to a circle are equal. (2011D, 2012OD, 2013OD, 2014, 2015D & OD
2016D & OD, 2017D)
Solution:
Given: PT and PS are tangents from an external point P to the circle with centre O.
Important Questions for Class 10 Maths Chapter 10 Circles 64
To prove: PT = PS
Const.: Join O to P,
T & S
Proof: In ∆OTP and
∆OSP,
OT = OS …[radii of same circle
OP = OP …[circle
∠OTP – ∠OSP …[Each 90°
∴ AOTP = AOSP …[R.H.S
PT = PS …[c.p.c.t

Question 44.
Important Questions for Class 10 Maths Chapter 10 Circles 65
In the above figure, PQ is a chord of length 16 cm, of a circle of radius 10 cm. The tangents at P and Q intersect at a point T. Find the length of TP. (2014OD)
Solution:
TP = TQuestion .. [Tangents drawn from an external point
∆TPQ is an isosceles ∆ and TO is the bisector of ∠PTQ ,
OT ⊥ PQ …[Tangent is ⊥ to the radius through the point of contact
∴ OT bisects PQ
∴ PR = RQ = 16 = 8 cm …[Given
In rt. ∆PRO,
PR2 + RO2 = PO2 … [Pythagoras’ theorem
82 + RO2 = (10)2
RO2 = 100 – 64 = 36
∴ RO = 6 cm
Let TP = x cm and TR = y cm
Then OT = (y + 6) cm
In rt. ∆PRT, x2 = y2 + 82 …(i) …[Pythagoras’ theorem
In rt. ∆OPT,
OT2 = TP2 + PO2 …(Pythagoras’ theorem
(y + 6)2 = x2 + 102
y2 + 12y + 36 = y2 +64 + 100 …[From (i)
12y = 164 – 36 = 128 ⇒ y = \frac{128}{12}=\frac{32}{3}
Putting the value of y in (i),
Important Questions for Class 10 Maths Chapter 10 Circles 66

Question 45.
In the figure, tangents PQ and PR are drawn from an external point P to a circle with centre O, such that ∠RPQ = 30°. A chord RS is drawn parallel to the tangent PQuestion Find ∠RQS. (2015D)
Important Questions for Class 10 Maths Chapter 10 Circles 67
Solution:
PR = PO …[∵ Tangents drawn from an external point are equal
⇒ ∠PRQ = ∠PQR …[∵ Angles opposite equal sides are equal
In ∆PQR,
Important Questions for Class 10 Maths Chapter 10 Circles 68
⇒ ∠PRQ + ∠RPQ + ∠POR = 180°…[∆ Rule
⇒ 30° + 2∠PQR = 180°
\angle \mathrm{PQR}=\frac{(180-30)^{\circ}}{2}
= 75°
⇒ SR || QP and QR is a transversal
∵ ∠SRQ = ∠PQR … [Alternate interior angle
∴ ∠SRO = 75° …..[Tangent is I to the radius through the point of contact
⇒ ∠ORP = 90°
∴ ∠ORP = ∠ORQ + ∠QRP
90° = ∠ORQ + 75°
∠ORQ = 90° – 75o = 150
Similarly, ∠RQO = 15°
In ∆QOR,
∠QOR + ∠QRO + ∠OQR = 180° …[∆ Rule
∴∠QOR + 15° + 15° = 180°
∠QOR = 180° – 30° = 150°
⇒ ∠QSR = \frac{1}{2}∠QOR
⇒ ∠QSR = \frac{150^{\circ}}{2} = 750 … [Used ∠SRQ = 75° as solved above
In ARSQ, ∠RSQ + ∠QRS + ∠RQS = 180° … [∆ Rule
∴ 75° + 75° + ∠RQS = 180°
∠RQS = 180° – 150o = 30°

Question 46.
Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc. (2015OD)
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 69
B is the mid point of arc (ABC)
OA = OC …[Radius
OF = OF …[Common
∴ ∠1 = ∠2 …[Equal angles opposite equal sides
∴ ∆OAF = ∆OCF (SAS)
∴ ∠AFO = ∠CFO = 90° …[c.p.c.t
⇒ ∠AFO = ∠DBO = 90° …[Tangent is ⊥to the radius through the point of contact
But these are corresponding angles,
∴ AC || DE

Question 47.
In the figure, O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects circle at E. If AB is a tangent to the circle at E, find the length of AB, where TP and TQ are two tangents to the circle. (2016D)
Important Questions for Class 10 Maths Chapter 10 Circles 70
Solution:
∠OPT = 90° …[Tangent is ⊥ to the radius through the point of contact
We have, OP = 5 cm, OT = 13 cm
In rt. ∆OPT,
OP2 + PT2 = OT? …[Pythagoras’ theorem
⇒ (5)2 + PT2 = (13)2
⇒ PT2 = 169 – 25 = 144 cm
⇒ PT = \sqrt{144}
= 12 cm
OP = OQ = OE = 5 cm … [Radius of the circle
ET = OT – OE
= 13 – 5 = 8 cm
Let, PA = x cm, then AT = (12 – x) cm
PA = AE = x cm …[Tangent drawn from an external point
In rt. ∆AET,
AE2 + ET2 = AT2 …(Pythagoras’ theorem
⇒ x2 + (8)2 = (12 – x)2
⇒ x2 + 64 = 144 + x2 – 24x
⇒ 24x = 144 – 64
x = \frac{80}{24}=\frac{10}{3} cm
AB = AE + EB = AE + AE = 2AE = 2x :
∴ AB = 2\left(\frac{10}{3}\right)=\frac{20}{3} \mathrm{cm}=6 \frac{2}{3} cm
or 6.67 cm or 6.6 cm

Question 48.
In the figure, two equal circles, with centres 0 and O’, touch each other A at X. OO’ produced meets the circle with centre O’ at A. AC is tangent to the circle with centre O, at the point C. O’D is perpendicular to AC. Find the value of \frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}. (2016OD)
Important Questions for Class 10 Maths Chapter 10 Circles 71
Solution:
Given: Two equal circles, with centres O and O’, touch each other at point X. OO’ is produced to meet the circle with centre (at A. AC is tangent to the circle with centre O, at the point C. OʻD is perpendicular to AC.
To find: = \frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}
Proof: ∠ACO = 90° … [Tangent is ⊥ to the radius through the point of contact
In ∆AO’D and ∆AOC
∠O’AD = ∠OAC …(Common
∴ ∠ADO = ∠ACO …[Each 90°
∴ ∆AO’D ~ ∴AOC …(AA similarity
\frac{\mathrm{AO}^{\prime}}{\mathrm{AO}}=\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}} … [In ~ As corresponding sides are proportional
\frac{r}{3 r}=\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}} …[Let AO’ = O’X = OX = r ⇒ AO = r +r+ r = 3r
\frac{\mathrm{DO}^{\prime}}{\mathrm{CO}}=\frac{1}{3}

Question 49.
In the figure, l and m are two parallel tangents to a circle with centre O, touching the circle at A and B respectively. Another tangent at C intersects the line I at D and m at E. Prove that ∠DOE = 90°. (2013D)
Important Questions for Class 10 Maths Chapter 10 Circles 72
Solution:
Proof: Let I be XY and m be XY’
∠XDE + ∠X’ED = 180° … [Consecutive interior angles
Important Questions for Class 10 Maths Chapter 10 Circles 73
\frac{1}{2}XDE + \frac{1}{2}∠X’ED =
= \frac{1}{2} (180°)
= ∠1 + ∠2 = 90° …[OD is equally inclined to the tangents
In ∆DOE, ∠1 + ∠2 + 23 = 180° …[Angle-sum-property of a ∆
90° + 23 = 180°
⇒ ∠3 = 180° – 90o = 90°
∴ ∠DOE = 90° …(proved)

Question 50.
In the figure, the sides AB, BC and CA of triangle ABC touch a circle with centre o and radius r at P, Q and R. respectively. (2013OD)
Prove that:
(i) AB + CQ = AC + BQ
(ii) Area (AABC) = \frac{1}{2} (Perimeter of ∆ABC ) × r
Important Questions for Class 10 Maths Chapter 10 Circles 74
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 75
Part I:
Proof: AP = AR …(i)
BP = BQ … (ii)
CQ = CR … (iii)
Adding (i), (ii) & (iii)
AP + BP + CQ
= AR + BQ + CR
AB + CQ = AC + BQ
Part II: Join OP, OR, OQ, OA, OB and OC
Proof: OQ ⊥ BC; OR ⊥ AC; OP ⊥ AB
ar(∆ABC) = ar(∆AOB) + ar(∆BOC) + ar (∆AOC)
Area of (∆ABC)
Important Questions for Class 10 Maths Chapter 10 Circles 76

Question 51.
In the figure, a triangle ABC is drawn to circumscribe a circle of radius 4 cm, such that the segments BD and DC are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC. (2014OD)
Important Questions for Class 10 Maths Chapter 10 Circles 77
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 78
Let AE = x ∴ AF = x
BC = 8 + 6 = 14 cm
AB = (x + 8) cm
AC = (x + 6) cm
∠1 = ∠2 = 23 = 90° …[ Tangent is ⊥ to the radius B [through the point of contact
Important Questions for Class 10 Maths Chapter 10 Circles 79
4 \sqrt{3 x(x+14)} = 2(2x + 28)
4 \sqrt{3 x(x+14)} = 2.2(x + 14)
3x(x + 14) = (x + 14)2 … [Squaring both sides
3x(x + 14) – (x + 14)2 = 0
(x + 14) [3x – (x + 14)] = 0
(x + 14) (2x – 14) = 0
x = -14 or x = 7
∴ x = 7 … [As side of ∆ cannot be -ve
∴ AB = x + 8 = 15 cm
and AC = x + 6 = 13 cm

Question 52.
Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre. (2014 D)
Solution:
Important Questions for Class 10 Maths Chapter 10 Circles 80
Given: CD and EF are two C parallel tangents at points A and B of a circle with centre O.
To prove: AB passes through centre O or AOB is diameter of the circle.
Const.: Join OA and OB. Draw OM || CD.
Proof: ∠1 = 90° … (i)
…[∵ Tangent is I to the radius through the point of contact
OM || CD
∴ ∠1 + ∠2 = 180° …(Co-interior angles
90° + ∠2 = 180° …[From (i)
∠2 = 180° – 90o = 90°
Similarly, ∠3 = 90°
∠2 + ∠3 = 90° + 90° = 180°
∴ AOB is a straight line.
Hence AOB is a diameter of the circle with centre O.
∴ AB passes through centre 0.

Important Questions for Class 10 Maths

The post Important Questions for Class 10 Maths Chapter 10 Circles appeared first on Learn CBSE.

Important Questions for Class 10 Maths Chapter 11 Constructions

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Important Questions for Class 10 Maths Chapter 11 Constructions

Constructions Class 10 Important Questions Short Answer-I (2 Marks)

Question 1.
Draw a line segment of length 6 cm. Using compasses and ruler, find a point P on it which divides it in the ratio 3 : 4. (2011D)
Solution:
Important Questions for Class 10 Maths Chapter 11 Constructions 1
Hence, PA : PB = 3 : 4

Question 2.
Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB such that \frac{\mathbf{A} \mathbf{P}}{\mathbf{A B}}=\frac{3}{5}. (2011OD)
Solution:
AB = 7 cm, AB = \frac{\mathbf{A} \mathbf{P}}{\mathbf{A B}}=\frac{3}{5} … [Given
∴ AP : PB = 3 : 2
Important Questions for Class 10 Maths Chapter 11 Constructions 2
Hence, AP : AB = 3 : 5 or \frac{\mathbf{A} \mathbf{P}}{\mathbf{A B}}=\frac{3}{5}

Constructions Class 10 Important Questions Short Answer-II (3 Marks)

Question 3.
Draw a triangle ABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°. Then construct a triangle whose sides are \frac{5}{7} times the corresponding sides of ∆ABC. (2011D)
Solution:
In ∆ABC
AB = 5 cm
BC = 6 cm
∠ABC = 60°
Important Questions for Class 10 Maths Chapter 11 Constructions 3
Hence, ∆A’BC’ is the required ∆.

Question 4.
Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle, whose sides are \frac{3}{5} times the corresponding sides of the given triangle. (20140D)
Solution:
Important Questions for Class 10 Maths Chapter 11 Constructions 4
∴ ∆AB’C’ is the required ∆.

Question 5.
Construct a right triangle in which the sides, (other than the hypotenuse) are of length 6 cm and 8 cm. Then construct another triangle, whose sides are \frac{3}{5} times the corresponding sides of the given triangle. (2012D)
Solution:
Here AB = 8 cm, BC = 6 cm and
Ratio = \frac{3}{5} of corresponding sides
Important Questions for Class 10 Maths Chapter 11 Constructions 5
∴ ∆AB’C’ is the required triangle.

Question 6.
Draw a triangle PQR such that PQ = 5 cm, ∠P = 120° and PR = 6 cm. Construct another triangle whose sides are \frac{3}{4} times the corresponding sides of ∆PQR. (2011D)
Solution:
In ∆PQR,
PQ = 5 cm, PR = 6 cm, ∠P = 120°
Important Questions for Class 10 Maths Chapter 11 Constructions 6
∴ ∆POʻR’ is the required ∆.

Question 7.
Draw a triangle ABC with BC = 7 cm, ∠B = 45° and ∠C = 60°. Then construct another triangle, whose sides are \frac{3}{5} times the corresponding sides of ∆ABC. (2012OD)
Solution:
Here, BC = 7 cm, ∠B = 45°, ∠C = 60° and ratio is \frac{3}{5} times of corresponding sides
Important Questions for Class 10 Maths Chapter 11 Constructions 7
∴ ∆A’BC’ is the required triangle.

Question 8.
Construct a triangle with sides 5 cm, 4 cm and 6 cm. Then construct another triangle whose sides are \frac{2}{3} times the corresponding sides of first triangle. (2013D)
Solution:
Important Questions for Class 10 Maths Chapter 11 Constructions 8
Steps of Construction:

  • Draw ∆ABC with AC = 6 cm, AB = 5 cm, BC = 4 cm.
  • Draw ray AX making an acute angle with AÇ.
  • Locate 3 equal points A1, A2, A3 on AX.
  • Join CA3.
  • Join A2C’ || CA3.
  • From point C’ draw B’C’ || BC.

∴ ∆AB’C’ is the required triangle.

Question 9.
Draw a pair of tangents to a circle of radius 3 cm, which are inclined to each other at an angle of 60°. (2011OD)
Solution:
Important Questions for Class 10 Maths Chapter 11 Constructions 9
∴ PA & PB are the required tangents.

Question 10.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm. (2013D)
Solution:
Important Questions for Class 10 Maths Chapter 11 Constructions 10
Steps of Construction:
Draw two circles with radius OA = 4 cm and OP = 6 cm with O as centre. Draw ⊥ bisector of OP at M. Taking M as centre and OM as radius draw another circle intersecting the smaller circle at A and B and touching the bigger circle at P. Join PA and PB. PA and PB are the required tangents.
Verification:
In rt. ∆OAP,
OA2 + AP2 = OP2 … [Pythagoras’ theorem
(4)2 + (AP)2 = (6)2
AP2 = 36 – 16 = 20
AP = + \sqrt{20}=\sqrt{4 \times 5}
= 2 \sqrt{5} = 2(2.236) = 4.472 = 4.5 cm
By measurement, ∴ PA = PB = 4.5 cm

Question 11.
Draw a pair of tangents to a circle of radius 4.5 cm, which are inclined to each other at an angle of 45°. (2013OD)
Solution:
Important Questions for Class 10 Maths Chapter 11 Constructions 11
Draw ∠AOB = 135°, ∠OAP = 90°, ∠OBP = 90°
∴ PA and PB are the required tangents.

Question 12.
Draw two tangents to a circle of radius 3.5 cm, from a point P at a distance of 6.2 cm from its cehtre. (2013OD)
Solution:
OP = OC + CP = 3.5 + 2.7 = 6.2 cm
Important Questions for Class 10 Maths Chapter 11 Constructions 12
Hence AP & PB are the required tangents.

Question 13.
Draw a right triangle ABC in which AB = 6 cm, BC = 8 cm and ∠B = 90°. Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle. (2014D, 2015OD)
Solution:
Important Questions for Class 10 Maths Chapter 11 Constructions 13
Steps of Construction:

  • Draw BC = 8 cm.
  • From B draw an angle of 90°.
  • Draw an arc BA = 6 cm cutting the angle at A.
  • Join AC. ∴ ∆ABC is the required ∆.
  • Draw ⊥ bisector of BC cutting BC at M.
  • Take Mas centre and BM as radius, draw a circle.
  • Take A as centre and AB as radius draw an arc cutting the circle at E. Join AE.

AB and AE are the required tangents.
Justification: ∠ABC = 90° …[Given
Since, OB is a radius of the circle.
∴ AB is a tangent to the circle.
Also, AE is a tangent to the circle.

Question 14.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. (2014OD)
Solution:
Important Questions for Class 10 Maths Chapter 11 Constructions 14
Draw two circles on A and B as asked.
Z is the mid-point of AB.
From Z, draw a circle taking ZA = ZB as radius,
so that the circle intersects the bigger circle at M and N and smaller circle at X and Y.
Join AX and AY, BM and BN.
BM, BN are the reqd. tangents from external point B.
AX, AY are the reqd. tangents from external point A.
Justification:
∠AMB = 90° …[Angle in a semi-circle
Since, AM is a radius of the given circle.
∴ BM is a tangent to the circle
Similarly, BN, AX and AY are also tangents.

Constructions Class 10 Important Questions Long Answer (4 Marks).

Question 15.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45° and ∆A = 105°. Then construct a triangle whose sides are \frac{3}{5}times the corresponding sides of ∆ABC. (2011D)
Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180° … [angle sum property of a ∆
105° + 45° + C = 180°
∠C = 180° – 105° – 45o = 30°
BC = 7 cm
Important Questions for Class 10 Maths Chapter 11 Constructions 15
∴ ∆A’BC’ is the required ∆.

Question 16.
Draw a triangle ABC with side BC = 6 cm, ∠C = 30° and ∠A = 105°. Then construct another triangle whose sides are \frac{2}{3} times the corresponding sides of ∆ABC. (2012D)
Solution:
Here, BC = 6 cm, ∠A = 105o and ∠C = 30°
Important Questions for Class 10 Maths Chapter 11 Constructions 16
In ∆ABC,
∠A + ∠B + ∠C = 180° …[Angle-sum-property of a ∆
105° + ∠B + 30o = 180°
∠B = 180° – 105° – 30o = 45°
∴ ∆A’BC’ is the required ∆.

Question 17.
Draw a triangle with sides 5 cm, 6 cm and 7 cm. Then construct another triangle whose sides are \frac{2}{3} times the corresponding sides of the first triangle. (2012OD)
Solution:
Here, AB = 5 cm, BC = 7 cm, AC = 6 cm and ratio is \frac{2}{3} times of corresponding sides.
Important Questions for Class 10 Maths Chapter 11 Constructions 17
∴ ∆A’BC’ is the required triangle.

Question 18.
Construct an isosceles triangle whose base is 6 cm and altitude 4 cm. Then construct another triangle whose sides are \frac{3}{4} times the corresponding sides of the isosceles triangle. (2015D)
Solution:
Important Questions for Class 10 Maths Chapter 11 Constructions 18
∴ ∆A’BC’ is the required triangle.

Question 19.
Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle. (2015D)
Solution:
Important Questions for Class 10 Maths Chapter 11 Constructions 19
Draw two circles on A and B as asked.
Z is the mid-point of AB.
From Z, draw a circle taking ZA = ZB as radius,
so that the circle intersects the bigger circle at M and N and smaller circle at X and Y.
Join AX and AY, BM and BN.
BM, BN are the required tangents from external point B.
AX, AY are the required tangents from external point A.
Justification:
∠AMB = 90° …[Angle in a semi-circle
Since, AM is a radius of the given circle.
∴BM is a tangent to the circle
Similarly, BN, AX and AY are also tangents.

Question 20.
Construct a ∆ABC in which AB = 6 cm, ∠A = 30° and ∆B = 60°. Construct another ∆AB’C’ similar to ∆ABC with base AB’ = 8 cm. (2015OD)
Solution:
Important Questions for Class 10 Maths Chapter 11 Constructions 20
Steps of construction:

  • Draw a ∆ABC with side AB = 6 cm, ∠A = 30° and ∠B = 60°.
  • Draw a ray AX making an acute angle with AB on the opposite side of point C.
  • Locate points A1, A2, A3 and A4 on AX.
  • Join A3B. Draw a line through A4 parallel to A3B intersecting extended AB at B’.
  • Draw a line parallel to BC intersecting ray AY at C’.

Hence, ∆AB’C’ is the required triangle.

Question 21.
Construct a triangle ABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°. Now construct another triangle whose sides are \frac{5}{7} times the corresponding sides of ∆ABC. (2015OD)
Solution:
In ∆ABC, AB = 5 cm; BC = 6 cm; ∠ABC = 60°
Important Questions for Class 10 Maths Chapter 11 Constructions 21
∴ ∆A’BC’ is the required ∆.

Question 22.
Construct a triangle ABC in which BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct another triangle whose sides are \frac{3}{4} times the corresponding sides of ∆ABC. (2016D)
Solution:
Important Questions for Class 10 Maths Chapter 11 Constructions 22
Steps of Construction:

  • Draw ∆ABC with the given data.
  • Draw a ray BX downwards making an acute angle with BC.
  • Locate 4 points B1, B2, B3, B4, on BX, such that BB1 = B1B2 = B2B3 = B3B4.
  • Join CB4.
  • From B3 draw a line C’B3 || CB4 intersecting BC at C’.
  • From C’ draw A’C’ || AC intersecting AB at B’.

Then ∆AB’C’ in the required triangle.
Justification:
Important Questions for Class 10 Maths Chapter 11 Constructions 23

Question 23.
Draw a triangle ABC with BC = 7 cm, ∠B = 45° and ∠A = 105°. Then construct a triangle whose sides are \frac{4}{5} times the corresponding sides of ∆ABC. (2016D)
Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180° ..[Angle-sum-property of a ∆
105° + 45° + ∠C = 180°
∠C = 180° – 105° – 45o = 30°
Important Questions for Class 10 Maths Chapter 11 Constructions 24
∴ A’BC’ is the required triangle.

Question 24.
Draw an isosceles ∆ABC in which BC = 5.5 cm and altitude AL = 3 cm. Then construct another triangle whose sides are \frac{3}{4} of the corresponding sides of ∆ABC. (2016OD)
Solution:
Important Questions for Class 10 Maths Chapter 11 Constructions 25
∴ ∆A’BC’ is the required ∆.

Question 25.
Draw a triangle with sides 5 cm, 6 cm and 7 cm. Then draw another triangle whose sides are \frac{4}{5} of the corresponding sides of first triangle. (2016OD)
Solution:
Important Questions for Class 10 Maths Chapter 11 Constructions 26
∴ ∆A’BC’ is the required ∆.

Question 26.
Draw two concentric cirlces of radii 3 cm and 5 cm. Construct a tangent to smaller circle from a point on the larger circle. Also measure its length. (2016D)
Solution:
We have, OD = 3 cm and OP = 5 cm
Important Questions for Class 10 Maths Chapter 11 Constructions 27
PA and PB are the required tangents
By measurement PA = PB = 4 cm.

Question 27.
Draw a circle of radius 4 cm. Draw two tangents to the circle inclined at an angle of 60° to each other. (2013, 2016OD)
Solution:
Draw a circle with O as centre and radius 4 cm.
Draw any ∠AOB = 120°. From A and B draw ∠PAO = ∠PBO = 90° which meet at P.
Important Questions for Class 10 Maths Chapter 11 Constructions 28
∴ PA and PB are the required tangents.

Important Questions for Class 10 Maths

The post Important Questions for Class 10 Maths Chapter 11 Constructions appeared first on Learn CBSE.

Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions

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Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions

Alternating Current Class 12 Important Questions Very Short Answer Type

Question 1.
The instantaneous current and voltage of an a.c. circuit are given by i = 10 sin 300 t A and V = 200 sin 300 t V. What is the power dissipation in the circuit? (All India 2008)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 1

Question 2.
The instantaneous current and voltage of an a.c. circuit are given by
i = 10 sin 314 t A and v = 50 sin 314 t V. What is the power dissipation in the circuit? (All India 2008)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 2

Question 3.
The instantaneous current and voltage of an a.c. circuit are given by i = 10 sin 314 tA and v = 50 sin \left(314 t+\frac{\pi}{2}\right)V. (All India 2008)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 3
Phase difference between current voltage
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 4

Question 4.
Define the term ‘wattless current’. (Delhi 2011)
Answer:
Wattless current is that component of the circuit current due to which the power consumed in the circuit is zero.

Question 5.
Mention the two characteristic properties of the material suitable for making core of a transformer. (All India 2012)
Answer:
Characteristic properties of material suitable for core of a transformer :

  • It should have high permeability
  • It should have low hysteresis loss.
  • It should have low coercivity/retentivity.
  • It should have high resistivity. (Any two)

Question 6.
When an ac source is connected across an ideal inductor, show on a graph the nature of variation of the voltage and the current over one complete cycle. (Comptt. Delhi 2012)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 5

Question 7.
A heating element is marked 210 V, 630 W. What is the value of the current drawn by the element when connected to a 210 V dc source? (Delhi 2013)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 6

Question 8.
A heating element is marked 210 V, 630 W. Find the resistance of the element when connected to a 210 V dc source.
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 7

Question 9.
Why is the core of a transformer laminated? (Comptt. Delhi 2013)
Answer:
The core of a transformer is laminated to minimize eddy currents in the iron core.

Question 10.
Why is the use of a.c. voltage preferred over d.c. voltage? Give two reasons. (All India 2013)
Answer:
a.c. voltage is preferred over d.c. voltage because of following reasons :

  1. it can be stepped-up or stepped-down by a transformer.
  2. carrying losses are much less.

Question 11.
Define capacitor reactance. Write its S.I. units. (Delhi 2015)
Answer:
‘Capacitor reactance’ is defined as the opposition to the flow of current in ac circuits offered by a capacitor.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 8
S.I. Unit : Ohm.

Question 12.
A variable frequency AC source is connected to a capacitor. Will the displacement current change if the frequency of the AC source is decreased? (Comptt. All India 2015)
Answer:
On decreasing the frequency of AC source, reactance, x_{C}=\frac{1}{\omega C} will increase, which will lead to decrease in conduction current. In this case
ID = IC
Hence, displacement current will decrease.

Question 13.
Plot a graph showing variation of capacitive reactance with the change in the frequency of the AC source. (Comptt. All India 2015)
Answer:
Graph showing a variation of xc capacitive reactance with the change in frequency of AC source.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 9

Question 14.
Define ‘quality factor’ of resonance in series LCR circuit. What is its SI unit? (Delhi 2016)
Answer:
Quality factor (Q) is defined as, Q = \frac{\omega_{0} L}{R}
It gives the sharpness of the resonance circuit. It has no SI unit.

Question 15.
For an ideal inductor, connected across a sinusoidal ac voltage source, state which one of the following quantity is zero :
(i) Instantaneous power
(ii) Average power over full cycle of the ac voltage source (Comptt. All India 2016)
Answer:
Average power over full cycle of the ac voltage source is zero, when connected with an ideal inductor.

Alternating Current Class 12 Important Questions Short Answer Type SA-I

Question 16.
Prove that an ideal capacitor in an a.c. circuit does not dissipate power. (Delhi 2008)
Answer:
Average power associated with a capacitor :
When an a.c. is applied to a capacitor, the current leads the voltage in phase by \frac{\pi}{2}radian. So we write the expressions for instantaneous voltage and current as follows :
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 10
Work done in the circuit in small time dt will be
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 11
The average power dissipated per cycle in the capacitor is,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 12
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 13
Thus the average power dissipated per cycle in a capacitor is zero.

Question 17.
Prove that an ideal inductor does not dissipate power in an a.c. circuit. (Delhi 2016)
Answer:
Average power associated with an inductor.
When a.c. is applied to an ideal inductor,current lags behind the voltage in phase by \frac{\pi}{2} radian. So we can write the instantaneous values of voltage and current as follows :
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 14
The average power dissipated per cycle in the inductor is
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 15
Thus, the average power dissipated per cycle in an inductor is zero.

Question 18.
Derive an expression for the impedance of an a.c. circuit consisting of an inductor and a resistor. (Delhi 2008)
Answer:
From the phasor diagram, we get R
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 16
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 17
Thus the average power dissipated per cycle in a capacitor is zero.

Question 19.
The circuit arrangement as shown in the diagram shows that when an a.c. passes through the coil A, the current starts flowing in the coil B.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 18
(i) State the underlying principle involved.
(ii) Mention two factors on which the current produced in the coil B depends.(All India 2008)
Answer:
(i) It is based on the principle of “mutual induction”.
(ii) Two factors are:

  • distance between the coils.
  • orientation of the coils.
  • Number of turns in the coil, (any two)

Question 20.
The figure given shows an arrangement by which current flows through the bulb (X) connected with coil B, when a.c. is passed through coil A.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 19
(i) Name the phenomenon involved.
(ii) If a copper sheet is inserted in the gap between the coils, explain, how the brightness of the bulb would change. (All India 2008)
Answer:
(i) The phenomenon involved is mutual induction.
(ii) When the copper sheet is inserted, eddy currents are set up in it which opposes the passage of magnetic flux. The induced emf in coil B decreases. This decreases the brightness of the bulb.

Question 21.
A 15.0 µF capacitor is connected to 220 V, 50 Hz source. Find the capacitive reactance and the rms current. (All India 2009)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 20

Question 22.
An electric lamp having coil of negligible inductance connected in series with a capacitor and an a.c. source is glowing with certain brightness. How does the brightness of the lamp change on reducing the
(i) capacitance, and
(ii) the frequency? Justify your Answer. (Delhi 2009)
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 21
Answer:
Brightness of lamp \propto I0,
Assuming zero resistance and zero inductance of lamp
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 22
On reducing C or v; It would decrease
∴ Brightness of the lamp will decrease.

Question 23.
State the principle of working of a transformer. Can a transformer be used to step up or step down a d.c. voltage? Justify your Answer. (All India 2009)
Answer:
Transformer works on the principle of mutual induction, i.e., when a changing current is passed through one of the two inductively coupled coils, an induced emf is set up in the other coil.

No, transformer cannot be used to step up or step down a d.c. voltage because d.c. voltage cannot produce a change in magnetic flux.

Question 24.
Mention various energy losses in a transformer. (All India 2009)
Answer:
(i) A transformer is an electrical device for converting an alternating current at low voltages into that at high voltage or vice versa.
If it increases the input voltage, it is called step- up-transformer.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 83
Principle : It works on the principle of mutual induction i.e., “when a changing current is passed through one of the two inductively coupled coils, an induced emf is set up in the other coil.”

Working : As the alternating current flows through the primary, it generates an alternating magnetic flux in the core which also passes through the secondary. This changing flux sets up an induced emf in the secondary, also a self- induced emf in the primary. If there is no leakage of magnetic flux, then flux linked with each turn
of the primary will be equal to that linked with each turn of the secondary.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 84
…where [Np and Ns are number of turns in the primary and secondary respectively,
Vp and Vs are their respective voltages]
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 85
This ratio \frac{\mathrm{N}_{\mathrm{S}}}{\mathrm{N}_{\mathrm{P}}} is called the turns ratio.
Assuming the transformer to be ideal one, so that there are no energy losses, then
Input power = output power
Vplp = VSIS
…where [IP and IS are the current in the primary and secondary respectively
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 86

In a step up transformer, Ns > Np i.e., the turns ratio is greater than 1 and therefore Vs > Vp.
The output voltage is greater than the input voltage.

Main assumptions :

  1. The primary resistance and current are small.
  2. The same flux links both with the primary and secondary windings as the flux leakage from due core is negligible (small).
  3. The terminals of the secondary are open or the current taken from it, is small, (any two)

For long distance transmission, the voltage output of the generator is stepped-up (so that current is reduced and consequently, IR loss is reduced). It is transmitted over long distance and is stepped- down at distributing substations at consumers’ end.

(ii) Two sources of energy loss in a transformer:
1. Copper loss”: Some energy is lost due to heating of copper wires used in the primary and secondary windings. This power loss (= I2R) can be minimised by using thick copper wires of low resistance.

2. Eddy current loss : The alternating magnetic flux induces eddy currents in the iron core which leads to some energy loss in the form of heat. This loss can be reduced by using laminated iron core.

(iii) No, a step up transformer does not violate law of conservation of energy because whatever is gained in voltage ratio is lost in the current ratio and vice-versa. It steps up the voltage while it steps down the current.

Question 25.
State the underlying principle of a transformer.
How is the large scale transmission of electric energy over long distances done with the use of transformers? (All India 2012)
Answer:
A transformer is an electrical device for converting an alternating current at low voltages into that at high voltage or vice versa.
If it increases the input voltage, it is called step- up-transformer.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 83
Principle : It works on the principle of mutual induction i.e., “when a changing current is passed through one of the two inductively coupled coils, an induced emf is set up in the other coil.”

Working : As the alternating current flows through the primary, it generates an alternating magnetic flux in the core which also passes through the secondary. This changing flux sets up an induced emf in the secondary, also a self- induced emf in the primary. If there is no leakage of magnetic flux, then flux linked with each turn
of the primary will be equal to that linked with each turn of the secondary.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 84
…where [Np and Ns are number of turns in the primary and secondary respectively,
Vp and Vs are their respective voltages]
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 85
This ratio \frac{\mathrm{N}_{\mathrm{S}}}{\mathrm{N}_{\mathrm{P}}} is called the turns ratio.
Assuming the transformer to be ideal one, so that there are no energy losses, then
Input power = output power
Vplp = VSIS
…where [IP and IS are the current in the primary and secondary respectively
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 86

In a step up transformer, Ns > Np i.e., the turns ratio is greater than 1 and therefore Vs > Vp.
The output voltage is greater than the input voltage.

Main assumptions :

  1. The primary resistance and current are small.
  2. The same flux links both with the primary and secondary windings as the flux leakage from due core is negligible (small).
  3. The terminals of the secondary are open or the current taken from it, is small, (any two)

For long distance transmission, the voltage output of the generator is stepped-up (so that current is_ reduced and consequently, IR loss is reduced). It is transmitted over long distance and is stepped- down at distributing substations at consumers’ end.

Question 26.
A light bulb is rated 100 W for 220 V ac supply of 50 Hz. Calculate
(i) the resistance of the bulb;
(ii) the rms current through the bulb. (All India 2012)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 23

Question 27.
A light bulb is rated 200 W for 220 V ac supply of 50 Hz. Calculate
(i) the resistance of the bulb;
(ii) the rms current through the bulb. (All India 2012)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 23
Hint: (i) 242Ω
(ii) Irms = 0.90 atmosphere

Question 28.
A light bulb is rated 150 W for 220 V ac supply of 60 Hz. Calculate
(i) the resistance of the bulb; ,
(ii) the rms current through the bulb. (All India 2012)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 23
Hint :
(i) P = 322.67 Ω
(ii) Irms = 0.68 ampere

Question 29.
An alternating voltage given by V = 140 sin 314 t is connected across a pure resistor of 50 Ω. Find
(i) the frequency of the source.
(ii) the rms current through the resistor. (All India 2012)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 24

Question 30.
An alternating voltage given by V = 280 sin 50πt is connected across a pure resistor of 40 Ω. Find
(i) the frequency of the source.
(ii) the rms current through the resistor. (All India 2012)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 24
Hint:
(i) v = 25 Hz
(ii) Irms = 4.95 A

Question 31.
An alternating voltage given by V = 70 sin 100πt is connected across a pure resistor of 25 Ω . Find
(i) the frequency of the source.
(ii) the rms current through the resistor. (All India 2012)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 24
(i) v = 50Hz
(ii) Irms = 1.98 Ampere.

Question 32.
A lamp is connected in series with a capacitor. Predict your observation when this combination is connected in turn across
(i) ac source and
(ii) a ‘dc’ battery. What change would you notice in each case if the capacitance of the capacitor is increased? (Comptt. Delhi 2012)
Answer:
When dc source is connected, the condenser is charged but no current flows in the circuit, therefore, the lamp does not glow. No change occurs even when capacitance of capacitor is increased.

When ac source is connected, the capacitor offers capacitive reactance X_{c}=\frac{1}{\omega C}. The current flows in the circuit and the lamp glows. On increasing capacitance, Xc decreases. Therefore, glow pf the bulb increases.

Question 33.
A capacitor ‘C’, a variable resistor ‘R’ and a bulb ‘B’ are connected in series to the ac mains in a circuit as shown. The bulb glows with some brightness. How will the glow of the bulb change if
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 25
(i) a dielectric slab is introduced between the plates of the capacitor, keeping resistance R to be the same;
(ii) the resistance R is increased keeping the same capacitance? (Delhi 2012)
Answer:
(i) Brightness will increase due to increase in capacitance on introducing dielectric slab.
(ii) Brightness will decrease, as the resistance (R) is increased, the potential drop across the bulb will decrease (since both are connected in series).

Question 34.
The figure shows a series LCR circuit connected to a variable frequency 200 V source with L = 50 mH, C = 80 µF and R = 40 Ω.
Determine
(i) the source frequency which derives the circuit in resonance;
(ii) the quality factor (Q) of the circuit. (Comptt. All India 2014)
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 26
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 27

Question 35.
The figure shows a series LCR circuit connected to a variable frequency 250 V source with L = 40 mH, C = 100 µF and R = 50 Ω.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 119
Determine :
(i) the source frequency which derives the circuit in resonance;
(ii) The quality factor (Q) of the circuit. (Comptt. All India 2014)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 27
[Hint]:
(i) 80 Hz
(ii) Q = 0.4.

Alternating Current Class 12 Important Questions Short Answer Type SA-II

Question 36.
An inductor of unknown value, a capacitor of 100 μF and a resistor of 10 Ω are connected in series to a 200 V. 50 Hz a.c. source. It is found that the power factor of the circuit is unity. Calculate the inductance of the inductor and the current amplitude. (Delhi 2008)
Answer:
As power factor of circuit is unity :
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 28

Question 37.
Two heating elements of resistances R1 and R2 when operated at a constant supply of voltage, V, consume powers P1 and P2 respectively. Deduce the expressions for the power of their combination when they are, in turn, connected in
(i) series and
(ii) parallel across the same voltage supply. (All India 2008)
Answer:
When two resistances R1 and R2 are operated at a constant voltage supply V, their consumed power will be P1 and P2
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 29
When they are connected in series, Power will be
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 30

Question 38.
The figure shows a series LCR circuit with L = 5.0 H, C = 80 μF, R = 40 Ω connected to a variable frequency 240 V source. Calculate
(i) The angular frequency of the source which drives the circuit at resonance.
(ii) The current at the resonating frequency.
(iii) The rms potential drop across the capacitor at resonance. (Delhi 2008)
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 120
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 31

Question 39.
A series LCR circuit with L = 4.0 H,C = 100 μF and R = 60 Ω. is connected to a variable frequency 240 V source as shown in the figure.
Calculate :
(i) the angular frequency of the source which derives the circuit at resonance;
(ii) the current at the resonating freqency;
(iii) the rms potential drop across the inductor at (Delhi 2008)
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 121
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 32

Question 40.
The figure shows a series LCR circuit with L = 10.0 H, C = 40 μF, R = 60 Ω connected to a variable frequency 240 V source.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 122
Calculate:
(i) The angular frequency of the source which drives the circuit at resonance.
(ii) The current at the resonating frequency.
(iii) The rms potential drop across the inductor at resonance. (Delhi 2008)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 32
(i) Angular frequency, to = 50 ‘radiance/sec.
(ii) Irms = 4A
(iii) Vrms = IrmsXc = 2000 V

Question 41.
A series LCR circuit is connected to an ac source. Using the phasor diagram, derive the expression for the impedance of the circuit. Plot a graph to show the variation of current with frequency of the source, explaining the nature of its variation. (All India 2008)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 33

Question 42.
(a) The graphs
(i) and
(ii) shown in the figure represent variation of opposition offered by the circuit elements, X and Y, respectively to the flow of alternating current vs. the frequency of the applied emf. Identify the elements X and Y.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 34
(b) Write the expression for the impedance offered by the series combination of these two elements connected to an ac source of voltage V = V0 sin ωt.
Show on a graph the variation of the voltage and the current with ‘out’ in the circuit. (Comptt. All India 2008)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 35

Question 43.
Draw a sketch showing the basic elements of an a.c. generator. State its principle and explain briefly its working. (Comptt. All India 2008)
Answer:
(a) Principle of A.C. generator : The working of an a.c. generator is based on the principle of electromagnetic induction. When a closed coil is rotated in a uniform magnetic field with its axis perpendicular to the magnetic field, the magnetic flux linked with the coil changes and an induced emf and hence a current is set up in it.

(b) Let N = number of turns in the coil
A = Area of face of each turn
B = magnitude of the magnetic field
θ = angle which normal to the coil makes with field B at any instant
ω = the angular velocity with which coil rotates
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 68
The magnetic flux linked with the coil at any instant f will be,
ϕ = NAB cos θ = NAB cos ωt
By Faraday’s flux rule, the induced emf is given by,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 69
When a load of resistance R is connected across the terminals, a current I flows in the external circuit.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 70

Question 44.
In a series LCR circuit connected to an ac source of variable frequency and voltage v = Vm sin ωt, draw a plot showing the variation of current (I) with angular frequency (ω) for two different values of resistance R1 and R2 (R1 > R2). Write the condition under which the phenomenon of resonance occurs. For which value of the resistance out of the two curves, a sharper resonance is produced? Define Q-factor of the circuit and give its significance. (Delhi 2013)
Answer:
(a) Condition for resonance to occur is XL = XC, and Z = R.
(b) Sharper resonance is produced for R2
(c) The Q-factor (quality factor) of series resonant circuit is defined as the ratio of the voltage developed across the inductance of capacitance at resonance to the impressed voltage, which is the voltage applied across the R.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 36
Significance : Higher the value of Q, the narrower and sharper is the resonance and therefore circuit will be more selective

Question 45.
(i) For a given a.c., i = im sin ωt, show that the average power dissipated in a resistor R over a complete cycle is \frac{1}{2} i_{m}^{2} \mathbf{R}.
(ii) A light bulb is rated at 100 W for a 220 V a.c. supply. Calculate the resistance of the bulb. (All India 2013)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 37

Question 46.
(a) For a given a.c., i = im sin ωt, show that the average power dissipated in a resistor R over complete cycle is \frac{1}{2} i_{m}^{2} \mathbf{R}.
(b) A light bulb is rated at 125 W for 250 V a.c. supply. Calculate the resistance of the bulb. (All India 2013)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 37

(ii) P = 125 W, V = 250 V
Where P = Power and V = Voltage
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 38

Question 47.
(a) When an a.c. source is connected to an ideal capacitor show that the average power supplied by the source over a complete cycle is zero.
(b) A lamp is connected in series with a capacitor. Predict your observations when the system is connected first across a d.c. and then an a.c. source. What happens in each case if the capacitance of the capacitor is reduced? (Comptt. Delhi 2013)
Answer:
(a) Average power associated with a capacitor :
When an a.c. is applied to a capacitor, the current leads the voltage in phase by \frac{\pi}{2}radian. So we write the expressions for instantaneous voltage and current as follows :
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 10
Work done in the circuit in small time dt will be
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 11
The average power dissipated per cycle in the capacitor is,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 12
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 13
Thus the average power dissipated per cycle in a capacitor is zero.

(b) (i) In this case, the bulb will glow initially for a very short Bulb duration depending upon its time constant during the charging of capacitor. Once the capacitor is fully charged, it will not allow current to pass, hence the bulb ceases to glow.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 39

(ii) In this case, when connected to a.c. source, the bulb will glow with the same brightness.
When the capacity of capacitor is reduced, it will have no appreciable effect when connected to d.c. source.
However, in case when connected to a.c. source, capacitance is reduced, hence
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 40
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 41
\chi_{\mathrm{C}} capactive reactance will increase and thus the brightness of bulb will reduce.

Question 48.
A voltage V = V0 sin est is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle.
Under what condition is
(i) no power dissipated even though the current flows through the circuit,
(ii) maximum power dissipated in the circuit? (All India 2014)
Answer:
Average power in LCR circuit :
Let the alternating emf applied to an LCR circuit,
V = V0 sin ωt …(i)
If alternating current developed lags behind the applied emf by a phase angle ϕ
then, I = I0 sin(ωt – ϕ ) …(ii)
Total work done over a complete cycle is,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 42
∴ Average power in LCR circuit over complete cycle is,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 43
(i) No power is dissipated in purely inductive or purely capacitive circuit, because phase difference between voltage and current is \frac{\pi}{2} and cos ϕ = 0. It is known as wattless current.

(ii) Maximum power is dissipated in a LCR circuit at Resonance, because XC – XL = 0 and ϕ = 0, cos ϕ = 1
Power = I2Z = I2R

Question 49.
An inductor L of inductance XL is connected in series with a bulb B and an ac source. How would brightness of the bulb change when
(i) number of turns in the inductor is reduced,
(ii) an iron rod is inserted in the inductor and
(iii) a capacitor of reactance XC = XL is inserted in series in the circuit. Justify your Answer in each case. (Delhi 2015)
Answer:
(i) Increases. XL = ωL
As number of turns decrease, L decreases, hence current through the bulb increases. Also voltage across bulb increases.
(ii) Decreases : Iron rod increases the inductance which increases XL, hence current through the bulb decreases./Voltage across the bulb decreases.
(iii) Increases. Under this condition (XC = XL) the current through the bulb will become maximum.

Question 50.
(a) Determine the value of phase difference between the current and the voltage in the given series LCR circuit.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 44
(b) Calculate the value of the additional capacitor which may be joined suitably to the capacitor C that would make the power factor of the circuit unity. (All India 2014)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 45

(b) Let the additional capacitor be C’ which is to be connected in parallel with C, to increase the value of combined capacitances; hence resulting into ‘capacitive reactance’ reduced. In parallel Cnet = C + C’
When the power factor is unity
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 46

Question 51.
A circuit containing an 80 mH inductor and a 250 µF capacitor in series connected to a 240 V, 100 rad/s supply. The resistance of the circuit is negligible.
(i) Obtain rms value of current. ‘
(ii) What is the total average power consumed by the circuit? (Comptt. Delhi 2014)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 47
(ii) Average power consumed = 0 (Zero)
(As there is no ohmic resistance in the current)

Question 52.
A source of ac voltage V = V0 sin cat is connected to a series combination of a resistor ‘R’ and a capacitor ‘C’. Draw the phasor diagram and use it to obtain the expression for
(i) impedance of the circuit and
(ii) phase angle. (Comptt. All India 2014)
Answer:
Phasor diagram and circuit diagram for the given circuit are,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 48
Expression for impedance and phase angle : A resistor and a capacitor are connected in series to a source of alternating current, V = V0 sin ωt
Let ‘I’ be the instantaneous value of current in this circuit.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 49
Which is the effective resistance of L – C circuit and is called its ‘impedance’
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 50

Question 53.
(i) When an AC source is connected to an ideal inductor show that the average power supplied by the source over a complete cycle is zero.
(ii) A lamp is connected in series with an inductor and an AC source. What happens to the brightness of the lamp when the key is plugged in and an iron rod is inserted inside the inductor? Explain.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 123
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 51
(ii) Brightness decreases because as iron rod is inserted its value of inductance increases. Thus, current decreases and also brightness decreases.

Question 54.
Derive the expression for the average power dissipated in a series LCR circuit for an ac source of a voltage, v = vm sin ωt, carrying a current, i = im sin(ωt + ϕ ).
Hence define the term “Wattless current”. State under what condition it can be realized in a circuit. (Comptt. Delhi 2014)
Answer:
Average power in LCR circuit :
Let the alternating emf applied to an LCR circuit,
V = V0 sin ωt …(i)
If alternating current developed lags behind the applied emf by a phase angle ϕ
then, I = I0 sin(ωt – ϕ ) …(ii)
Total work done over a complete cycle is,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 42
∴ Average power in LCR circuit over complete cycle is,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 43
(i) No power is dissipated in purely inductive or purely capacitive circuit, because phase difference between voltage and current is \frac{\pi}{2} and cos ϕ = 0. It is known as wattless current.
(ii) Maximum power is dissipated in a LCR circuit at Resonance, because XC – XL = 0 and ϕ = 0, cos ϕ = 1
Power = I2Z = I2R

Question 55.
Obtain the expression for the magnetic energy stored in an ideal inductor of self inductance L when a current I passes through it.
Hence obtain the expression for the energy density of magnetic field B produced in the inductor.
Answer:
Expression for Magnetic Energy density in an ideal inductor :
Instantaneous induced emf in an inductor when current changes through it
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 52
Hence instantaneous applied voltage
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 53
Putting the values of (LI) and (I) from equations (ii) and (iii) in equation (i), we have
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 54

Question 56.
The current, in the LCR circuit shown in the figure is observed to lead the voltage in phase. Without making any other change in the circuit, a capacitor, of capacitance C0, is (appropriately) joined to the capacitor C. This results in making the current, in the ‘modified’ circuit, flow in phase with the applied voltage.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 55
Draw a diagram of the ‘modified’ circuit and obtain an expression for C0 in terms of ω, L and C. (Comptt. All India)
Answer:
Since the current leads the voltage in phase, hence, XC > XL
For resonance, we must have
New value of X’C = XL
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 56
This requires an increase in the value of C. Hence, capacitor C0 should be connected in parallel across C.
The diagram of the modified circuit is shown. For resonance, we then have
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 57

Question 57.
A 200 mH (pure) inductor, and a 5µF (pure) capacitor, are connected, one by one, across a sinusoidal ac voltage source V = [70.7 sin (1000 t)] voltage. Obtain the expressions for the current in each case. (Comptt. All India 2016)
Answer:
Given: For the applied voltage V = 70.7 sin(1000 t),
we have V0 = 70.7 volts, ω = 1000s-1
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 58
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 59

Question 58.
(i) Find the value of the phase difference between the current and the voltage in the series LCR circuit shown here. Which one leads in phase: current or voltage?
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 60
(ii) Without making any other change, find the value of the additional capacitor Cv to be connected in parallel with the capacitor C, in order to make the power factor of the circuit unity. (Delhi 2017)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 61

(ii) To make power factor unity, ϕ = 0°, hence we need to adjust C to a new value C’, the condition is :
XC = XL = 100 Ω
Thus, phase angle is 45° with the current leading the voltage.

To make power factor unity, we need to have XC also equal to 100 Ω. For this, C needs to have a value of 10µ.

We, therefore, need to put an additional capacitor of (10 – 2), i.e., 8 µF in parallel with the given capacitor.

Alternating Current Class 12 Important Questions Long Answer Type

Question 59.
An a.c. source generating a voltage v = vm sin ω t is connected to a capacitor of capacitance C. Find the expression for the current, i, flowing through it. Plot a graph of v and i versus tat to show that the current is π/2 ahead of the voltage. A resistor of 200Ω and a capacitor of 15.0 µF are connected in series to a 220 V, 50 Hz a.c. source. Calculate the current in the circuit and the rms voltage across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox. (All India 2008)
Answer:
(a) Voltage applied to the capacitor, v = vm sin ωt
Let instantaneous capacitor = v
we have, v = \frac{q}{C}
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 62
According to Kirchoff’s loop rule, the voltage across the source and the capacitor are equal at any instant of time.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 63
From equations (i) and (ii) we conclude that current leads the voltage by a phase angle of π/2
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 64
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 65
(ii) As the current is same throughout the series circuit, we have
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 66
The algebraic sum of the two voltages, VR and VC is 311.3 V which is more than the source voltage of 220 V. These two voltages are 90° out of phase. These cannot be added like ordinary numbers. The voltage is obtained by using Pythagoras theorem,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 67
Thus if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.

Question 60.
Explain briefly, with the help of a labelled diagram, the basic principle of the working of an a.c. generator.
In an a.c. generator, coil of N turns and area A is rotated at v revolutions per second in a uniform magnetic field B. Write the expression for the emf produced.
A 100-turn coil of area 0.1 m2 rotates at half a revolution per second. It is placed in a magnetic field 0.01 T perpendicular to the axis of rotation of the coil. Calculate the maximum voltage generated in the coil. (All India 2008)
Answer:
(a) Principle of A.C. generator : The working of an a.c. generator is based on the principle of electromagnetic induction. When a closed coil is rotated in a uniform magnetic field with its axis perpendicular to the magnetic field, the magnetic flux linked with the coil changes and an induced emf and hence a current is set up in it.

(b) Let N = number of turns in the coil
A = Area of face of each turn
B = magnitude of the magnetic field
θ = angle which normal to the coil makes with field B at any instant
ω = the angular velocity with which coil rotates
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 68
The magnetic flux linked with the coil at any instant f will be,
ϕ = NAB cos θ = NAB cos ωt
By Faraday’s flux rule, the induced emf is given by,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 69
When a load of resistance R is connected across the terminals, a current I flows in the external circuit.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 70

Question 61.
(a) Derive an expression for the average power consumed in a series LCR circuit connected to a.c. source in which the phase difference between the voltage and the current in the circuit is 0.
(b) Define the quality factor in an a.c. circuit. Why should the quality factor have high value in receiving circuits? Name the factors on which it depends. (Delhi 2009)
Answer:
(a) Let an alternating current I = Im sin cot be passing through a network of L, C and R creating a potential difference of V = Vm sin (ωt ± ϕ ) where ϕ is the phase difference. Then the power consumed is
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 71

(b) Quality factor should be high to have the current corresponding to a particular frequency to be more and to avoid the other unwanted frequencies. Q-factor depends on f, L, R and C.
Sharpness of resonance is determined by quality factor (Q) of the circuit i.e.,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 72
Larger the value of Q,
Sharper is the resonance i.e. sharper peak in the current.

Question 62.
(a) Derive the relationship between the peak and the rms value of current in an a.c. circuit.
(b) Describe briefly, with the help of a labelled diagram, working of a step-up transformer. A step-up transformer converts a low voltage into high voltage. Does it not violate the principle of conservation of energy? Explain. (Delhi 2009)
Answer:
(a) R.M.S. value of current say I = Im sin ωt is given by
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 73
where [Im is the peak value of current]

(b) The supply of ac to the primary will bring a varying flux in the secondary causing emf.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 74
in the secondary flux will be more than the primary as the condition NS > NP is satisfied. Production of high voltage does not violate the law of conservation of energy as the current will be reduced in the process.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 75
As VS > VP with NS > NP, IS will become less than IP.

Question 63.
Describe briefly, with the help of a labelled diagram, the basic elements of an a.c. generator. State its underlying principle. Show diagrammatically how an alternating emf is generated by a loop of wire rotating in a magnetic field. Write the expression for the instantaneous value of the emf induced in the rotating loop. (Delhi 2010)
Answer:
(a) Principle of A.C. generator : The working of an a.c. generator is based on the principle of electromagnetic induction. When a closed coil is rotated in a uniform magnetic field with its axis perpendicular to the magnetic field, the magnetic flux linked with the coil changes and an induced emf and hence a current is set up in it.

(b) Let N = number of turns in the coil
A = Area of face of each turn
B = magnitude of the magnetic field
θ = angle which normal to the coil makes with field B at any instant
ω = the angular velocity with which coil rotates
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 68
The magnetic flux linked with the coil at any instant f will be,
ϕ = NAB cos θ = NAB cos ωt
By Faraday’s flux rule, the induced emf is given by,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 69
When a load of resistance R is connected across the terminals, a current I flows in the external circuit.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 70

Diagram of how an alternating emf is generated by a loop of wire
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 76

Question 64.
A series LCR circuit is connected to an a.c. source having voltage v = vm sin ωt.
Derive the expression for the instantaneous current I and its phase relationship to the applied voltage.
Obtain the condition for resonance to occur. Define ‘power factor’. State the conditions under which it is
(i) maximum and
(ii) minimum. (Delhi 2010)
Answer:
(a) Suppose a resistance R, an inductance L and capacitance C are connected in series to a source of alternating emf ε given by
ε = ε0 sin ωt
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 77
Let I be the instantaneous value of current in the series circuit. Then voltage across the three components are
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 78
resultant is equal to the applied emf \vec{\varepsilon}, as given by the diagonal of the parallelogram.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 79
Clearly \frac{\varepsilon}{i} is the effective resistance of the series LCR circuit and is called its impedance (Z)
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 80
When XL = XC, the voltage and current are in the same phase. In such a situation, the circuit is known as non-inductive circuit.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 81
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 82
(i) Power factor is maximum when the circuit contains only resistance R. In that case ϕ = 0, cos ϕ = 1.
(ii) Power factor is minimum when the circuit contains purely capacitive or inductive circuit. In this case cos ϕ = 0 and no power is dissipated even though a current is flowing in the circuit.

Question 65.
Draw a schematic diagram of a step-up transformer. Explain its working principle. Deduce the expression for the secondary to primary voltage in terms of the number of turns in the two coils. In an ideal transformer, how is this ratio related to the currents in the two coils? How is the transformer used in large scale transmission and distribution of electrical energy over long distances? (All India 2010)
Answer:
A transformer is an electrical device for converting an alternating current at low voltages into that at high voltage or vice versa.
If it increases the input voltage, it is called step- up-transformer.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 83
Principle : It works on the principle of mutual induction i.e., “when a changing current is passed through one of the two inductively coupled coils, an induced emf is set up in the other coil.”

Working : As the alternating current flows through the primary, it generates an alternating magnetic flux in the core which also passes through the secondary. This changing flux sets up an induced emf in the secondary, also a self- induced emf in the primary. If there is no leakage of magnetic flux, then flux linked with each turn
of the primary will be equal to that linked with each turn of the secondary.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 84
…where [Np and Ns are number of turns in the primary and secondary respectively,
Vp and Vs are their respective voltages]
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 85
This ratio \frac{\mathrm{N}_{\mathrm{S}}}{\mathrm{N}_{\mathrm{P}}} is called the turns ratio.
Assuming the transformer to be ideal one, so that there are no energy losses, then
Input power = output power
Vplp = VSIS
…where [IP and IS are the current in the primary and secondary respectively
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 86

In a step up transformer, Ns > Np i.e., the turns ratio is greater than 1 and therefore Vs > Vp.
The output voltage is greater than the input voltage.

Main assumptions :

  1. The primary resistance and current are small.
  2. The same flux links both with the primary and secondary windings as the flux leakage from due core is negligible (small).
  3. The terminals of the secondary are open or the current taken from it, is small, (any two)

For long distance transmission, the voltage output of the generator is stepped-up (so that current is reduced and consequently, IR loss is reduced). It is transmitted over long distance and is stepped- down at distributing substations at consumers’ end.

Question 66.
(i) With the help of a labelled diagram, describe briefly the underlying principle and working of a step-up transformer.
(ii) Write any two sources of energy loss in a transformer.
(iii) A step up transformer converts a low input voltage into a high output voltage. Does it violate law of conservation of energy? Explain. (Delhi 2011)
Answer:
(i) A transformer is an electrical device for converting an alternating current at low voltages into that at high voltage or vice versa.
If it increases the input voltage, it is called step- up-transformer.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 83
Principle : It works on the principle of mutual induction i.e., “when a changing current is passed through one of the two inductively coupled coils, an induced emf is set up in the other coil.”

Working : As the alternating current flows through the primary, it generates an alternating magnetic flux in the core which also passes through the secondary. This changing flux sets up an induced emf in the secondary, also a self- induced emf in the primary. If there is no leakage of magnetic flux, then flux linked with each turn
of the primary will be equal to that linked with each turn of the secondary.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 84
…where [Np and Ns are number of turns in the primary and secondary respectively,
Vp and Vs are their respective voltages]
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 85
This ratio \frac{\mathrm{N}_{\mathrm{S}}}{\mathrm{N}_{\mathrm{P}}} is called the turns ratio.
Assuming the transformer to be ideal one, so that there are no energy losses, then
Input power = output power
Vplp = VSIS
…where [IP and IS are the current in the primary and secondary respectively
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 86

In a step up transformer, Ns > Np i.e., the turns ratio is greater than 1 and therefore Vs > Vp.
The output voltage is greater than the input voltage.

Main assumptions :

  1. The primary resistance and current are small.
  2. The same flux links both with the primary and secondary windings as the flux leakage from due core is negligible (small).
  3. The terminals of the secondary are open or the current taken from it, is small, (any two)

For long distance transmission, the voltage output of the generator is stepped-up (so that current is reduced and consequently, IR loss is reduced). It is transmitted over long distance and is stepped- down at distributing substations at consumers’ end.

(ii) Two sources of energy loss in a transformer:
1. Copper loss”: Some energy is lost due to heating of copper wires used in the primary and secondary windings. This power loss (= I2R) can be minimised by using thick copper wires of low resistance.

2. Eddy current loss : The alternating magnetic flux induces eddy currents in the iron core which leads to some energy loss in the form of heat. This loss can be reduced by using laminated iron core.

(iii) No, a step up transformer does not violate law of conservation of energy because whatever is gained in voltage ratio is lost in the current ratio and vice-versa. It steps up the voltage while it steps down the current.

Question 67.
Derive an expression for the impedance of a series LCR circuit connected to an AC supply of variable frequency.
Plot a graph showing variation of current with the frequency of the applied voltage.
Explain briefly how the phenomenon of resonance in the circuit can be used in the tuning mechanism of a radio or a TV set. (Delhi 2011)
Answer:
Expression for impedance:
(a) Let an alternating current I = Im sin cot be passing through a network of L, C and R creating a potential difference of V = Vm sin (ωt ± ϕ ) where ϕ is the phase difference. Then the power consumed is
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 71

(b) Quality factor should be high to have the current corresponding to a particular frequency to be more and to avoid the other unwanted frequencies. Q-factor depends on f, L, R and C.
Sharpness of resonance is determined by quality factor (Q) of the circuit i.e.,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 72
Larger the value of Q,
Sharper is the resonance i.e. sharper peak in the current.
Resonant circuit can be used in the tuning mechanism of a radio or a TV set.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 87
The antenna of a radio accepts signals from many broadcasting stations. The signals picked up in the antenna acts as a source in the tuning circuit of the radio, so the circuit can be driven at many frequencies. But to hear one particular radio station, we tune the radio. In tuning, we vary the capacitance of a capacitor in the tuning circuit such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal received. When this happens, the amplitude of the current with the frequency of the signal of the . particular radio station in the circuit is maximum.

Question 68.
State the working of a.c. generator with the help of a labelled diagram.
The coil of an a.c. generator having N turns, each of area A, is rotated with a constant angular velocity to. Deduce the expression for the alternating emf generated in the coil.
What is the source of energy generation in this device? (All India 2011)
Answer:
AC generator: A dynamo or generator is a device which converts mechanical energy into electrical energy. It is based on the principal of electromagnetic energy into electrical energy. It is based on the principle of electromagnetic induction.
Construction : It consists of the four main parts :
(i) Field Magnet : It produces the magnetic field. In the case of a low power dynamo, the magnetic field is generated by a permanent magnet, while in the case of large power dynamo, the magnetic field is produced by an electromagnet. ;

(ii) Armature : It consists of a large number of turns of insulated wire in the soft iron drum or ring. It can revolve round an axle between the two poles of the field magnet. The drum or ring serves the two purposes :

  1. It serves as support to coils and
  2. It increases the magnetic field due to air core being replaced by an iron core.

(iii) Slip Rings : The slip rings R1 and R2 are the two metal rings to which the ends of armature coil are connected. These rings are fixed to the shaft which rotates the armature coil so that the rings also rotate along with the armature.

(iv) Brushes : These are two flexible metal plates or carbon rods (B1 and B2) which are fixed and constantly touch the revolving rings. The output current in external load RL is taken through these brushes.
Working : When the armature coil is rotated in the strong magnetic field, the magnetic flux linked with the coil changes and the current is induced in the coil, its direction being given by Fleming’s right hand rule. Considering the armature to be in vertical position and as it rotates in anticlockwise direction, the wire ab moves upward and cd downward, so that the direction of induced current is shown in fig. In the external circuit, the current flows along B1RLB2. The direction of current remains unchanged during the first half turn of armature. During the second half revolution, the wire ab moves downward and cd upward, so the direction of current is reversed and in external circuit it flows along B2RLB1. Thus the direction of induced emf and current changes in the external circuit after each half revolution.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 88

Numerical :
N = number of turns in the coil
A = area enclosed by each turn of coil
\overrightarrow{\mathrm{B}} = strength of magnetic field
θ = angle, which is normal to the coil, makes with \overrightarrow{\mathrm{B}} at any instant t
∴ Magnetic flux linked with the coil in this position
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 89
If the coil rotates with an angular velocity ω and turns through an angle θ in time t
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 90
Source of energy : This induced emf is the source of energy generations in this device.

Question 69.
(a) A voltage V = V0 sin ωt applied to a series LCR circuit drives a current i = i0 sin ωt in the circuit. Deduce the expression for the average power dissipated in the circuit.
(b) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain.
(c) Define the term ‘wattless current’. (Comptt. Delhi 2012)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 91
If the instantaneous power remains constant for a small time dt
then small amount of work done in maintaining the current for a small time dt is,
dW = V0i0 sin2 ωtdt
Total work done or energy spent in maintaining current over one full cycle,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 92
(b) A transformer is an electrical device for converting an alternating current at low voltages into that at high voltage or vice versa.
If it increases the input voltage, it is called step- up-transformer.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 83
Principle : It works on the principle of mutual induction i.e., “when a changing current is passed through one of the two inductively coupled coils, an induced emf is set up in the other coil.”

Working : As the alternating current flows through the primary, it generates an alternating magnetic flux in the core which also passes through the secondary. This changing flux sets up an induced emf in the secondary, also a self- induced emf in the primary. If there is no leakage of magnetic flux, then flux linked with each turn
of the primary will be equal to that linked with each turn of the secondary.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 84
…where [Np and Ns are number of turns in the primary and secondary respectively,
Vp and Vs are their respective voltages]
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 85
This ratio \frac{\mathrm{N}_{\mathrm{S}}}{\mathrm{N}_{\mathrm{P}}} is called the turns ratio.
Assuming the transformer to be ideal one, so that there are no energy losses, then
Input power = output power
Vplp = VSIS
…where [IP and IS are the current in the primary and secondary respectively
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 86

In a step up transformer, Ns > Np i.e., the turns ratio is greater than 1 and therefore Vs > Vp.
The output voltage is greater than the input voltage.

Main assumptions :

  1. The primary resistance and current are small.
  2. The same flux links both with the primary and secondary windings as the flux leakage from due core is negligible (small).
  3. The terminals of the secondary are open or the current taken from it, is small, (any two)

For long distance transmission, the voltage output of the generator is stepped-up (so that current is_ reduced and consequently, IR loss is reduced). It is transmitted over long distance and is stepped- down at distributing substations at consumers’ end.

(c) Wattless current. The current which consumes no power for its maintenance in the circuit is called wattless current or idle current.

Question 70.
(a) An ac source of voltage v = v0 sin ωt is connected across a series combination of an inductor, a capacitor and a resistor. Use the phasor diagram to obtain the expression for
(i) impedance of the circuit and
(ii) phase angle between the voltage and the current.
(b) A capacitor of unknown capacitance, a resistor of 100 Ω and an inductor of self-inductance L = (4/π2) henry are in series connected to an ac source of 200 V and 50 Hz. Calculate the value of the capacitance and the current that flows in the circuit when the current is in phase with the voltage. (Comptt. All India 2012)
Answer:
(a) (i) Impedance of circuit : The effective resistance offered by a series LCR circuit is called its impedance.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 93
Suppose an inductance L, capacitance C and resistance R are connected in series to a source of alternating emf, V = V0 sin ωt.
Let I be the instantaneous value of current in the series circuit.
Then voltages across the three components are : ‘
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 94
These voltages are shown in the phasor in Figure 2.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 95
Clearly V/I is the effective resistance of the series LCR circuit and is called its impedance (Z).
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 96
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 97

Question 71.
(a) Explain with the help of a labelled diagram, the principle and working of a transformer. Deduce the expression for its working formula.
(b) Name any four causes of energy loss in an actual transformer. (Comptt. All India 2012)
Answer:
(a) Principle. It is a device which converts high voltage a.c. into low voltage a.c. and vice versa. It is based upon the principle of mutual induction. When alternating current passes through a coil, an induced emf is set up in the neighbouring coil.
Construction. A transformer consists of two coils of many turns of insulated copper wire wound on a closed laminated iron core. One of the coils known as Primary ‘P’ is connected to A.C. supply. The other coil known as Secondary ‘S’ is connected to the load. Working. When an alternating current passes through the primary, the magnetic flux through the iron core changes which does two things. It produces emf in the primary and an induced emf is also set up in the secondary. If we assume that the resistance of primary is negligible, the back emf will be equal to the voltage applied to the primary.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 98
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 99

(b) Four causes of energy loss :

  1. Magnetic flux loss
  2. Hystersis loss
  3. Iron loss
  4. Losses due to the resistance of primary and secondary coils.

Question 72.
(a) Draw a schematic sketch of an ac generator describing its basic elements. State briefly its working principle. Show a plot of variation of
(i) Magnetic flux and
(ii) Alternating emf versus time generated by a loop of wire rotating in a magnetic field.
(b) Why is choke coil needed in the use of fluorescent tubes with ac mains? (Delhi 2014)
Answer:
(a) AC generator.
(a) Principle of A.C. generator : The working of an a.c. generator is based on the principle of electromagnetic induction. When a closed coil is rotated in a uniform magnetic field with its axis perpendicular to the magnetic field, the magnetic flux linked with the coil changes and an induced emf and hence a current is set up in it.
(b) Let N = number of turns in the coil
A = Area of face of each turn
B = magnitude of the magnetic field
θ = angle which normal to the coil makes with field B at any instant
ω = the angular velocity with which coil rotates
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 68
The magnetic flux linked with the coil at any instant f will be,
ϕ = NAB cos θ = NAB cos ωt
By Faraday’s flux rule, the induced emf is given by,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 69
When a load of resistance R is connected across the terminals, a current I flows in the external circuit.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 70

(i) Graph between magnetic flux and time, according to equation (i), shown below in Graph (i)

(ii) As the coil rotates, angle θ changes. Therefore, magnetic flux ϕ linked with the coil changes and an emf is induced in the coil. At this instant f, if e is the emf , induced in the coil, then
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 100

(iii) The graph between alternating emf ‘ versus time is shown below in Graph (ii).
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 101
(b) A choke coil is an electrical appliance used for controlling current in a a.c. circuit. Therefore, if we use a resistance R for the same purpose, a lot of energy would be wasted in the form of heat etc.

Question 73.
(a) A series LCR circuit is connected to an a.c. source of variable frequency. Draw a suitable phasor diagram to deduce the expressions for the amplitude of the current and phase angle.
(b) Obtain the condition at resonance. Draw a plot showing the variation of current with the frequency of a.c. source for two resistances R1 and (R1 > R2). Hence define the quality factor, Q and write its role in the tuning of the circuit. (Comptt. Delhi 2014)
Answer:
(a) Suppose a resistance R, an inductance L and capacitance C are connected in series to a source of alternating emf ε given by
ε = ε0 sin ωt
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 77
Let I be the instantaneous value of current in the series circuit. Then voltage across the three components are
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 78
resultant is equal to the applied emf \vec{\varepsilon}, as given by the diagonal of the parallelogram.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 79
Clearly \frac{\varepsilon}{i} is the effective resistance of the series LCR circuit and is called its impedance (Z)
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 80
When XL = XC, the voltage and current are in the same phase. In such a situation, the circuit is known as non-inductive circuit.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 81
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 82
(i) Power factor is maximum when the circuit contains only resistance R. In that case ϕ = 0, cos ϕ = 1.
(ii) Power factor is minimum when the circuit contains purely capacitive or inductive circuit. In this case cos ϕ = 0 and no power is dissipated even though a current is flowing in the circuit.
(b)
(a) Condition for resonance to occur is XL = Xc, and Z = R.
(b) Sharper resonance is produced for Rr
(c) The Q-factor (quality factor) of series resonant circuit is defined as the ratio of the voltage developed across the inductance of capacitance at resonance to the impressed voltage, which is the voltage applied across the R.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 36
Significance : Higher the value of Q, the narrower and sharper is the resonance and therefore circuit will be more selective

Question 74.
(a) Draw a labelled diagram of a.c. generator and state its working principle.
(b) How is magnetic flux linked with the armature coil changed in a generator?
(c) Derive the expression for maximum value of the induced emf and state the rule that gives the direction of the induced emf.
(d) Show the variation of the emf generated
versus time as the armature is rotated with respect to the direction of the magnetic field. (Comptt. Delhi 2014)
Answer:
(a) Principle of A.C. generator : The working of an a.c. generator is based on the principle of electromagnetic induction. When a closed coil is rotated in a uniform magnetic field with its axis perpendicular to the magnetic field, the magnetic flux linked with the coil changes and an induced emf and hence a current is set up in it.

(b) Let N = number of turns in the coil
A = Area of face of each turn
B = magnitude of the magnetic field
θ = angle which normal to the coil makes with field B at any instant
ω = the angular velocity with which coil rotates
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 68
The magnetic flux linked with the coil at any instant f will be,
ϕ = NAB cos θ = NAB cos ωt
By Faraday’s flux rule, the induced emf is given by,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 69
When a load of resistance R is connected across the terminals, a current I flows in the external circuit.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 70
(c) Direction of induced emf can be determined by using Fleming’s Left hand rule.
(d)
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 102

Question 75.
(a) Draw a schematic arrangement for winding of primary and secondary coil in a transformer when the two coils are wound on top of each other.
(b) State the underlying principle of a transformer and obtain the expression for the ratio of secondary to primary voltage in terms of the
(i) number of secondary and primary windings and
(ii) primary and secondary currents.
(c) Write the main assumption involved in deriving the above relations.
(d) Write any two reasons due to which energy losses may occur in actual transformers. (Comptt. All India 2014)
Answer:
A transformer is an electrical device for converting an alternating current at low voltages into that at high voltage or vice versa.
If it increases the input voltage, it is called step- up-transformer.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 83
Principle : It works on the principle of mutual induction i.e., “when a changing current is passed through one of the two inductively coupled coils, an induced emf is set up in the other coil.”

Working : As the alternating current flows through the primary, it generates an alternating magnetic flux in the core which also passes through the secondary. This changing flux sets up an induced emf in the secondary, also a self- induced emf in the primary. If there is no leakage of magnetic flux, then flux linked with each turn
of the primary will be equal to that linked with each turn of the secondary.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 84
…where [Np and Ns are number of turns in the primary and secondary respectively,
Vp and Vs are their respective voltages]
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 85
This ratio \frac{\mathrm{N}_{\mathrm{S}}}{\mathrm{N}_{\mathrm{P}}} is called the turns ratio.
Assuming the transformer to be ideal one, so that there are no energy losses, then
Input power = output power
Vplp = VSIS
…where [IP and IS are the current in the primary and secondary respectively
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 86

In a step up transformer, Ns > Np i.e., the turns ratio is greater than 1 and therefore Vs > Vp.
The output voltage is greater than the input voltage.

Main assumptions :

  1. The primary resistance and current are small.
  2. The same flux links both with the primary and secondary windings as the flux leakage from due core is negligible (small).
  3. The terminals of the secondary are open or the current taken from it, is small, (any two)

For long distance transmission, the voltage output of the generator is stepped-up (so that current is_ reduced and consequently, IR loss is reduced). It is transmitted over long distance and is stepped- down at distributing substations at consumers’ end.

(d) Reasons due to which energy losses may occur :

  1. Flux leakage
  2. Losses due to the resistance of primary and secondary coils
  3. eddy currents
  4. hystersis

Question 76.
(i) An a.c. source of voltage V = V0 sin ωt is connected to a series combination of L, C and R. Use the phasor diagram to obtain expressions for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called?
(ii) In a series LR circuit XL = R and power factor of the circuit is P1. When capacitor with capacitance C such that XL = XC is put in series, the power factor becomes P2. Calculate P1/P2.
Answer:
(a) (i) Impedance of circuit : The effective resistance offered by a series LCR circuit is called its impedance.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 93
Suppose an inductance L, capacitance C and resistance R are connected in series to a source of alternating emf, V = V0 sin ωt.
Let I be the instantaneous value of current in the series circuit.
Then voltages across the three components are : ‘
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 94
These voltages are shown in the phasor in Figure 2.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 95
Clearly V/I is the effective resistance of the series LCR circuit and is called its impedance (Z).
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 96
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 97
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 103

Question 77.
(i) Write the function of a transformer. State its principle of working with the help of a diagram. Mention various energy losses in this device.
(ii) The primary coil of an ideal step up transformer has 100 turns and transformation ratio is also 100. The input voltage and power are respectively 220 V and 1100 W. Calculate
(a) number of turns in secondary
(b) current in primary
(c) voltage across secondary
(d) current in secondary
(e) power in secondary (Delhi 2016)
Answer:
(i) (a) Principle. It is a device which converts high voltage a.c. into low voltage a.c. and vice versa. It is based upon the principle of mutual induction. When alternating current passes through a coil, an induced emf is set up in the neighbouring coil.

Construction. A transformer consists of two coils of many turns of insulated copper wire wound on a closed laminated iron core. One of the coils known as Primary ‘P’ is connected to A.C. supply. The other coil known as Secondary ‘S’ is connected to the load. Working. When an alternating current passes through the primary, the magnetic flux through the iron core changes which does two things. It produces emf in the primary and an induced emf is also set up in the secondary. If we assume that the resistance of primary is negligible, the back emf will be equal to the voltage applied to the primary.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 98
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 99
(b) Four causes of energy loss :

  1. Magnetic flux loss
  2. Hystersis loss
  3. Iron loss
  4. Losses due to the resistance of primary and secondary coils.

Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 104
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 105

Question 78.
(i) Draw a labelled diagram of a step-down transformer. State the priciple of its working.
(ii) Express the turn ratio in terms of voltages.
(iii) Find the ratio of primary and secondary currents in terms of turn ratio in an ideal transformer
(iv) How much current is drawn by the primary of a transformer connected to 220 V supply when it delivers power to a 110V – 550 W refrigerator? (All India 2016)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 106

Labelled diagram of a step-down transformer :

(b) Mutual Induction, which means “whenever an alternative voltage is applied in the primary windings /coil, an emf is induced in the secondary windings (Coil).”
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 107

Question 79.
Discuss how Faraday’s law of e.m. induction is applied in an ac-generator for converting mechanical energy into electrical energy.
Obtain an expression for the instantaneous value of the induced emf in an ac generator. Draw graphs to show the ‘phase relationship’ between the instantaneous
(i) magnetic flux (ϕ) linked with the coil and
(ii) induced emf (ε) in the coil. ‘ (Comptt. Delhi 2016)
Answer:
Faradays law of e.m. induction and expression for instantaneous value of induced emf
(a) Principle of A.C. generator : The working of an a.c. generator is based on the principle of electromagnetic induction. When a closed coil is rotated in a uniform magnetic field with its axis perpendicular to the magnetic field, the magnetic flux linked with the coil changes and an induced emf and hence a current is set up in it.

(b) Let N = number of turns in the coil
A = Area of face of each turn
B = magnitude of the magnetic field
θ = angle which normal to the coil makes with field B at any instant
ω = the angular velocity with which coil rotates
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 68
The magnetic flux linked with the coil at any instant f will be,
ϕ = NAB cos θ = NAB cos ωt
By Faraday’s flux rule, the induced emf is given by,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 69
When a load of resistance R is connected across the terminals, a current I flows in the external circuit.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 70
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 108

Question 80.
Draw an arrangement for winding of primary and secondary coils in a transformer with two coils on a separate limb of the core.
State the underlying principle of a transformer. Deduce the expression for the ratio of secondary voltage to the primary voltage in terms of the ratio of the number of turns of primary and secondary winding. For an ideal transformer, obtain the ratio of primary and secondary currents in terms of the ratio of the voltages in the secondary and primary voltages.
Write any two reasons for the energy losses which occur in actual transformers. (Comptt. Delhi 2016)
Answer:
(a) Principle. It is a device which converts high voltage a.c. into low voltage a.c. and vice versa. It is based upon the principle of mutual induction. When alternating current passes through a coil, an induced emf is set up in the neighbouring coil.

Construction. A transformer consists of two coils of many turns of insulated copper wire wound on a closed laminated iron core. One of the coils known as Primary ‘P’ is connected to A.C. supply. The other coil known as Secondary ‘S’ is connected to the load. Working. When an alternating current passes through the primary, the magnetic flux through the iron core changes which does two things. It produces emf in the primary and an induced emf is also set up in the secondary. If we assume that the resistance of primary is negligible, the back emf will be equal to the voltage applied to the primary.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 98
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 99
(b) Four causes of energy loss :

  1. Magnetic flux loss
  2. Hystersis loss
  3. Iron loss
  4. Losses due to the resistance of primary and secondary coils.

Question 81.
(a) Draw a labelled diagram of AC generator. Derive the expression for the instantaneous value of the emf induced in the coil.
(b) A circular coil of cross-sectional area 200 cm2 and 20 turns is rotated about the vertical diameter with angular speed of 50 rad s-1 in a uniform magnetic field of magnitude 3.0 × 10-2 T. Calculate the maximum value of the current in the coil. (Delhi 2017)
Answer:
(a) Principle of A.C. generator : The working of an a.c. generator is based on the principle of electromagnetic induction. When a closed coil is rotated in a uniform magnetic field with its axis perpendicular to the magnetic field, the magnetic flux linked with the coil changes and an induced emf and hence a current is set up in it.

(b) Let N = number of turns in the coil
A = Area of face of each turn
B = magnitude of the magnetic field
θ = angle which normal to the coil makes with field B at any instant
ω = the angular velocity with which coil rotates
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 68
The magnetic flux linked with the coil at any instant f will be,
ϕ = NAB cos θ = NAB cos ωt
By Faraday’s flux rule, the induced emf is given by,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 69
When a load of resistance R is connected across the terminals, a current I flows in the external circuit.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 70
(b) Given : A = 200 cm2 = 200 × 10-4 m2
N = 20 turns, ω = 50 rad s-1,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 109
Maximum value of i0 = mA
[ ∵ Numerical value of R is not given]

Question 82.
(a) Draw a labelled diagram of a step-up transformer. Obtain the ratio of secondary to primary voltage in terms of number of turns and currents in the two coils.
(b) A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 3000 turns. Find the number of turns in the secondary to get the power output at 220 V. (Delhi 2017)
Answer:
(a) A transformer is an electrical device for converting an alternating current at low voltages into that at high voltage or vice versa.
If it increases the input voltage, it is called step- up-transformer.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 83
Principle : It works on the principle of mutual induction i.e., “when a changing current is passed through one of the two inductively coupled coils, an induced emf is set up in the other coil.”

Working : As the alternating current flows through the primary, it generates an alternating magnetic flux in the core which also passes through the secondary. This changing flux sets up an induced emf in the secondary, also a self- induced emf in the primary. If there is no leakage of magnetic flux, then flux linked with each turn
of the primary will be equal to that linked with each turn of the secondary.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 84
…where [Np and Ns are number of turns in the primary and secondary respectively,
Vp and Vs are their respective voltages]
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 85
This ratio \frac{\mathrm{N}_{\mathrm{S}}}{\mathrm{N}_{\mathrm{P}}} is called the turns ratio.
Assuming the transformer to be ideal one, so that there are no energy losses, then
Input power = output power
Vplp = VSIS
…where [IP and IS are the current in the primary and secondary respectively
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 86

In a step up transformer, Ns > Np i.e., the turns ratio is greater than 1 and therefore Vs > Vp.
The output voltage is greater than the input voltage.

Main assumptions :

  1. The primary resistance and current are small.
  2. The same flux links both with the primary and secondary windings as the flux leakage from due core is negligible (small).
  3. The terminals of the secondary are open or the current taken from it, is small, (any two)

For long distance transmission, the voltage output of the generator is stepped-up (so that current is reduced and consequently, IR loss is reduced). It is transmitted over long distance and is stepped- down at distributing substations at consumers’ end.

(b) Given : Vp = 2200 V, Np = 3000 turns,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 110

Question 83.
A device ‘X’ is connected to an ac source V = V0 sin cot. The variation of voltage, current and power in one cycle is shown in the following graph:
(a) Identify the device ‘X’.
(b) Which of the curves A, B and C represent the voltage, current and the power consumed in the circuit? Justify your
answer.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 111
(c) How does it impedance vary with frequency of the ac source ? Show graphically.
(d) Obtain an expression for the current in the circuit and its phase relation with ac voltage. (All India 2017)
Answer:
(a) The device X is a capacitor.
(b) Curve B ➝ voltage
Curve C ➝ current
Curve A ➝ power consumption over a full cycle.
Reason: The current leads the voltage in phase, by a plane angle of \frac{\pi}{2}, for a capacitor.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 112
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 113
Current leads the voltage, in phase, by \frac{\pi}{2}

Question 84.
(a) Draw a labelled diagram of an ac generator.
Obtain the expression for the emf induced in the rotating coil of N turns each of cross-sectional area A, in the presence of a magnetic field \overrightarrow{\mathbf{B}}.
(b) A horizontal conducting rod 10 m long extending from east to west is falling with a speed 5.0 ms-1 at right angles to the horizontal component of the Earth’s magnetic field, 0.3 × 10-4 Wb m-2. Find the instantaneous value of the emf induced in the rod. (All India 2017)
Answer:
(a) Principle of A.C. generator : The working of an a.c. generator is based on the principle of electromagnetic induction. When a closed coil is rotated in a uniform magnetic field with its axis perpendicular to the magnetic field, the magnetic flux linked with the coil changes and an induced emf and hence a current is set up in it.

(b) Let N = number of turns in the coil
A = Area of face of each turn
B = magnitude of the magnetic field
θ = angle which normal to the coil makes with field B at any instant
ω = the angular velocity with which coil rotates
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 68
The magnetic flux linked with the coil at any instant f will be,
ϕ = NAB cos θ = NAB cos ωt
By Faraday’s flux rule, the induced emf is given by,
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 69
When a load of resistance R is connected across the terminals, a current I flows in the external circuit.
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 70

(b) Given : l = 10 m, v = 5.0 ms-1,
B = 0.3 × 10-4 wb m-2
Induced emf = B/V
∴ E = (0.3 × 10-4) × (10) × (5) volt
E = 1.5 × 10-3 V = 1.5 mV

Question 85.
In the given circuit, calculate
(a) the capacitance of the capacitor, if the power factor of the circuit is unity,
(b) the Q-factor of this circuit. What is the significance of the Q-factor in a.c. circuit? Given the angular frequency of the a.c. source to be 100/s. Calculate the average power dissipated in the circuit. (Comptt. Delhi 2017)
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 114
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 115

Question 86.
(a) Prove that the current flowing through an ideal inductor connected across a.c. source, lags the voltage in phase by \frac{\pi}{2}.
(b) An inductor of self inductance 100 mH, and a bulb are connected in series with a.c. source of rms voltage 10 V, 50 Hz. It is found that effective voltage of the circuit \frac{\pi}{4}. Calculate the inductance of the inductor used and average power dissipated in the circuit, if a current of 1 A flows in the circuit. (Comptt. Delhi)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 116
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 117

Question 87.
(a) Prove that an ideal capacitor in an ac circuit does not dissipate power.
(b) An inductor of 200 mH, capacitor of 400 f and a resistor of 10 Q are connected in series to ac source of 50 V of variable frequency. Calculate the
(i) angular frequency at which maximum power dissipation occurs in the circuit and the corresponding value of the effective current, and
(ii) value of Q-factor in the circuit. (Comptt. All India 2017)
Answer:
Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions 118

Important Questions for Class 12 Physics

The post Important Questions for Class 12 Physics Chapter 7 Alternating Current Class 12 Important Questions appeared first on Learn CBSE.

UGC Scholarship 2019-20 | Get Complete List, Eligibility, Rewards, Application Procedure

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UGC Scholarship 2019-20: University Grants Commission (UGC), a famous sanctioned body established beneath the Ministry of Human Resource Development by the government of India, proposes a number of scholarships for the learners across the country to help them continue their education aa t the higher level. This scholarship focuses on providing financial support for college and university level studies. Both meritorious and underserved students can avail these scholarships.

We will discuss here the importance of each UGC scholarship including the application procedure, eligibility criteria, rewards, etc. Candidates can check here and seek their dream academic profession with these UP accomplishments.

List of UGC Scholarship

The University Grants Commission provides both scholarships and fellowships, catering the needs of over 20,000 students every year. Whether you are pursuing your studies at graduation level, postgraduation level, doctorate level, or post-doctorate level, you can find a UGC scholarship for your studies. The table given below contains a comprehensive list of all UGC scholarships and fellowships.

Name of the Scholarship Application DurationNo. of Scholarships Offered
Post Graduate Indira Gandhi Scholarship Scheme for Single Girl ChildJuly-October3000
Post Graduate Merit Scholarship for University Rank HolderJuly-October3000
Post Graduate Scholarships for Professional Courses for SC/ST CandidatesJuly-October1000
Special Scholarship Scheme Ishan Uday for NERJuly-October10000
Swami Vivekananda Single Girl Child Scholarship for Research in Social SciencesAvailable Throughout the yearNA
Emeritus FellowshipAvailable Throughout the year200
Dr. S. Radhakrishnan Post-Doctoral Fellowship in Humanities and Social SciencesNA200
Dr. D. S. Kothari Postdoctoral Fellowship SchemeAvailable throughout the yearNA
Post-Doctoral Fellowship to Women CandidatesNA100
Post-Doctoral Fellowship to SC/ST CandidatesNA100
National Fellowship for Scheduled Caste StudentsNA2000
National Fellowship for Higher Education of ST StudentsNA750
National Fellowship for Persons with DisabilitiesOctober-November200
Maulana Azad National Fellowship for Minority StudentsNA756
National Fellowship for OBC CandidateNA300

Note: The dates mentioned here are tentative and may change on the discretion of the scholarship provider.

UGC Scholarship Eligibility

Each UGC scholarship has its defined set of eligibility criteria to be fulfilled by the students that majorly centers around the age of the students, educational qualification and the annual income of the family. Also, there are some category-specific scholarships too, that are being offered by UGC for students belonging to SC, ST or OBC categories. What are the eligibility conditions for each scholarship that you need to fulfill? Know in detail in the table given below. Depending on your current class of study, you can select one of these scholarships and apply for them to stay advantaged.

Post Graduate Indira Gandhi Scholarship Scheme for Single Girl Child (All Categories)

  • Girl students up to the age of 30 years at the time of admission in PG courses are eligible.
  • The candidate should be the only girl child of the family.
  • The scheme is applicable to such a single girl child who has taken admission in regular, full-time 1st year Masters Degree course in any approved university or a post-graduate college.

Post Graduate Merit Scholarship for University Rank Holder (All Categories)

  • The scholars who have completed their graduation in the following subjects can apply:
    • Life Sciences
    • Chemical Sciences
    • Physical Sciences
    • Social Sciences
    • Languages
    • Commerce
    • Mathematical Sciences
    • Earth Sciences
  • Candidate should be the first or second rank holder at the UG level and enlisted in a full-time, regular master’s degree course from a recognized university/institute.
  • The age of the candidate should not be more than 30 years while taking admission in the PG course.
  • The minimum percentage required by the student in graduation is 60%.

Post Graduate Scholarships for Professional Courses for SC/ST Candidates

  • The scholarship is available for students who are seeking the first year of the professional program at the PG level during the current year from a known university/institution.
  • The programs approved from guiding regulatory organizations like MCI, NCTE, Bar Council of India, PCI, AICTE, Development Authority of India, NCTIS, RCI, DCI, Forensic Regulatory, INC, and ICAR are regarded as professional courses for which the scholarship is relevant.

Special Scholarship Scheme Ishan Uday for NER

  • The scholarship is available for students who are a domicile of North Eastern Region (NER) state and who have passed their class 12th or equivalent examination.
  • The student must have acquired admission in a technical, professional or general degree course from an identified institute/college/university.
  • The annual income of the family should not be more than Rs.4.5 Lakh.

Swami Vivekananda Single Girl Child Scholarship for Research in Social Sciences (All categories)

  • The scholarship is open for only girl child of a family.
  • The candidate must be pursuing a regular, full-time Ph.D. program in Social Sciences from a  known Indian University/ College.
  • The age limit for general category candidates is 40 years and 45 years for SC/ST/OBC and PwD.
  • Candidates who are pursuing their Ph.D. through distance learning are not eligible to apply for this scholarship.
  • The scholarship is also available for transgender candidates.

Emeritus Fellowship (All categories)

  • The fellowship is available to highly qualified, antiquated and trained teachers of recognised institutions/colleges/ universities.
  • The candidate must have given their service career with quality research and distributed work.
  • The fellowship gives the rewarded candidate, to serve with a well-defined time-bound action plan until the age of 70 years.

Dr.S.Radharkrishnan Post-Doctoral Fellowship in Humanities and Social Sciences (All categories)

  • The fellowship is available for unemployed young researchers, aged below 35 years before the deadline online application submission. (Relaxation of 5 years is given for SC, ST, OBC, PwD and Women candidates).
  • The minimum percentage marks to be secured by general category scholars at UG level is 55% marks and at PG level is 60% marks. (For reserved classes including SC, ST, OBC, and PwD, a relaxation of 5% is available).
  • The fellowship is also available for transgender candidates.

Dr.D.S.Kothari Postdoctoral Fellowship Scheme (All categories)

  • The fellowship is available for applicants below 35 years of age who hold a Ph.D. degree under science faculty. (Note: A relaxation of up to 3, 5 and 10 years will be given to SC, ST, OBC, Women and PwD candidates).
  • Teachers, who are working on a confirmed post and are below 35 years of age can also apply.
  • The previous applicants of the fellowship can also apply if they were declared unsuccessful.

Post-Doctoral Fellowship to Women Candidates (All categories)

  • The fellowship is applicable for women candidates only who are unemployed and hold a Ph.D. degree.
  • The age of the applicant should not be more than 55 years (Note: A relaxation of 5 years in age is given to SC, ST, OBC, PwD candidates).
  • The minimum percentage of marks obtained by general category applicants at UG level is 55% and at PG level is 60% (Note: A relaxation of 5% in the minimum percentage of marks is given to SC, ST, OBC, PwD candidates).
  • Transgender candidates can also apply for this scheme.

Post-Doctoral Fellowship to SC/ST Candidates

  • The fellowship is open for unemployed candidates who have a Ph.D. degree in the relevant subject.
  • The maximum age limit for male applicants is 50 years while for female applicants is 55 years.
  • The applicants should have scored 50% marks at the UG level and 55% marks at PG level.
  • Transgender candidates can also apply.

National Fellowship for Scheduled Caste (SC) Students

  • The students who have passed their postgraduation examination and have registered for an M.Phil./PhD course can apply for this fellowship.
  • The fellowship is open for transgender candidates.

National Fellowship for Higher Education of ST Students

  • The students having a postgraduate degree are eligible for the fellowship.
  • The applicant must be registered in a full-time regular M.Phil./PhD course at a university/college/institution recognized by UGC.

National Fellowship for Persons with Disabilities

  • The fellowship is open for students with disabilities who have been enrolled in a Ph.D./MPhil program at a University or academic institution.

Maulana Azad National Fellowship for Minority Students

  • The fellowship is applicable to students belonging to minority communities i.e. Muslims, Christians, Sikh, Buddhists, Parsi, and Jain.
  • The student must be enrolled in a full-time and regular M.Phil./PhD program.
  • The annual income of the family should not be more than Rs.2.5 Lakh from all sources.
  • The minimum percentage of marks to be obtained by the candidates in postgraduation in 50%.
  • The transgender applicants can also apply for the fellowship.

National Fellowship for OBC Candidate

  • The students who have passed their postgraduation and have registered for M.Phil./PhD courses can apply for the fellowship.
  • The annual income of the family should not be more than Rs.6 Lakh from all sources.
  • Transgender candidates can also apply for the fellowship.

UGC Scholarship Application Procedure

Candidates can know here, where and when to apply for UGC scholarship. Students are provided with online mode to apply or to complete the application process. There is no offline mode available. While the application process for most of the scholarships is done through National Scholarship Portal, you can apply for the rest of the scholarships through the official website of UGC. The below-given table contains details about how and where can you apply for the UGC scholarship.

Name of the Scholarship Application Procedure
Post Graduate Indira Gandhi Scholarship Scheme for Single Girl ChildCandidates have to apply online through the National Scholarship Portal (NSP).
Post Graduate Merit Scholarship for University Rank HolderCandidates have to apply online through the National Scholarship Portal.
Post Graduate Scholarships for Professional Courses for SC/ST CandidatesCandidates have to apply online through the National Scholarship Portal.
Special Scholarship Scheme Ishan Uday for NERCandidates have to apply online through the National Scholarship Portal.
Swami Vivekananda Single Girl Child Scholarship for Research in Social SciencesCandidates have to apply online through the official website of UGC.
Emeritus FellowshipCandidates have to apply online through the official website of UGC.
Dr. S. Radhakrishnan Post-Doctoral Fellowship in Humanities and Social SciencesCandidates have to apply online through the official website of UGC.
Dr. D. S. Kothari Postdoctoral Fellowship SchemeCandidates have to apply online through the official website of the fellowship.
Post-Doctoral Fellowship to Women CandidatesCandidates have to apply online through the official website of UGC.
Post-Doctoral Fellowship to SC/ST CandidatesCandidates have to apply online through the official website of UGC.
National Fellowship for Scheduled Caste StudentsCandidates have to apply online through the official website of UGC.
National Fellowship for Higher Education of ST StudentsCandidates have to apply online through the official website of UGC.
National Fellowship for Persons with DisabilitiesCandidates have to apply online via the official website of UGC.
Maulana Azad National Fellowship for Minority StudentsCandidates have to apply online via the official website of UGC.
National Fellowship for OBC CandidateCandidates have to apply online through the official website of UGC.

UGC Scholarship – Rewards

The details of the scholarship amount that a student receives through a UGC scholarship is provided here. The amount may vary for different scholarships, the students get the awards in the form of monetary benefits only. Find out what each UGC scholarship holds for you in the table below. It comprises a piece of detailed information about scholarship rewards.

Scholarship NameMonetary Rewards
Post Graduate Indira Gandhi Scholarship Scheme for Single Girl ChildRs.36,200 per annum for a period of 2 years
Post Graduate Merit Scholarship for University Rank HolderRs.3,100 per month for a period of 2 years
Post Graduate Scholarships for Professional Courses for SC/ST CandidatesRs. 7,800 per month for a period of 2 years for ME/MTech students.
Rs. 4,500 per month for a period of 2 years for other professional courses.
Special Scholarship Scheme Ishan Uday for NERStudents pursuing general degree courses will receive Rs. 5,400 per month.
Students pursuing technical/ professional/ medical courses will receive Rs. 7,800 per month.
Swami Vivekananda Single Girl Child Scholarship for Research in Social SciencesFellowship of Rs. 25,000 per month for the initial 2 years and Rs. 28,000 per month for remaining tenure.
A contingency grant of Rs.10,000 per annum for initial 2 years and Rs. 20,500 per annum for remaining tenure.
Escort and reader assistance of Rs.2,000 per month (for persons with disabilities).
Emeritus FellowshipAn honorarium of Rs.31,000 per month for 2 years.
A contingency of Rs.50,000 per annum
Dr. S. Radhakrishnan Post-Doctoral Fellowship in Humanities and Social SciencesFellowship of Rs.38,800 for the first year, Rs.40,300 for the second year and Rs.41,900 for the third year.
A contingency of Rs.50,000 per annum for 3 years.
Escorts and readers assistance of Rs.2,000 per month for PwD (Persons with Disabilities) candidates.
Dr. D. S. Kothari Postdoctoral Fellowship SchemeNormal Fellowship of Rs.43,400, Rs. 45,000 and Rs. 46,500 for first, second and third year respectively.
Higher fellowship of Rs. 46,500 for the first, second and third year.
Contingency grant of Rs. 1 Lakh per annum.
Post-Doctoral Fellowship to Women CandidatesFellowship of Rs. 38,800 per month for the first 2 years and Rs. 46,500 per month after two years.
A contingency of Rs. 50,000 per annum for 5 years.
Escorts and reader allowance of Rs. 2,000 per month (for PwD applicants).
Post-Doctoral Fellowship to SC/ST CandidatesFellowship of Rs. 38,800 per month for the first 2 years and Rs. 46,500 per month after two years.
A contingency of Rs. 50,000 per annum for 5 years.
Escorts and reader allowance of Rs. 2,000 per month (for PwD applicants).
National Fellowship for Scheduled Caste StudentsFellowship in Social Sciences, Humanities and Sciences include JRF of Rs. 25,000 per month for initial two years and SRF of Rs.28,000 per month for remaining tenure.
Contingency for Humanities and Social Sciences is Rs.10,000 per annum for initial 2 years and Rs.20,500 per annum for remaining tenure.
Contingency for Science, Engineering & Technology is Rs.12,000 per annum for initial 2 years and Rs.25,000 per annum for remaining tenure.
Escorts and reader assistance of Rs.2,000 per month (for blind and physically handicapped students).
National Fellowship for Higher Education of ST StudentsFellowship in Social Sciences, Humanities and Sciences include JRF of Rs.25,000 per month for initial two years and SRF of Rs.28,000 per month for remaining tenure.
Contingency for Humanities and Social Sciences is Rs.10,000 per annum for initial 2 years and Rs.20,500 per annum for remaining tenure.
Contingency for Science, Engineering & Technology is Rs.12,000 per annum for initial 2 years and Rs.25,000 per annum for remaining tenure.
Escorts and reader assistance of Rs.2,000 per month (for blind and physically handicapped students).
National Fellowship for Persons with DisabilitiesThe fellowship amount includes:
Monthly stipend of Rs.16,000 for initial 2 years and for remaining tenure Rs.18,000 per month.
Contingency allowance for Humanities, Commerce and Social Science is Rs.10,000 per annum for initial 2 years and Rs.20,500 for remaining tenure.
Contingency allowance for Science and Engineering & technology is Rs.12,000 and Rs.25,000 for remaining tenure.
Departmental assistance of Rs.3,000 per annum
HRA as per the rules of Institution/University
Escorts/reader assistance of Rs.2,000 per month (for blind and physically disabled candidates)
Maulana Azad National Fellowship for Minority StudentsFellowship of Rs.25,000 per month (JRF) for initial two years and Rs.28,000 per month (SRF) for remaining tenure.
Contingency for Humanities, Social Sciences and commerce is Rs.10,000 per annum for initial 2 years and Rs.20,500 per annum for remaining tenure.
Contingency for Science is Rs.12,000 per annum for initial 2 years and Rs.25,000 per annum for remaining tenure.
Escorts and reader assistance of Rs.2,000 per month (for blind and physically handicapped students).
National Fellowship for OBC CandidateFellowship in Social Sciences, Humanities and Sciences include JRF of Rs.25,000 per month for initial two years and SRF of Rs.28,000 per month for remaining tenure.
Contingency for Humanities and Social Sciences is Rs.10,000 per annum for initial 2 years and Rs.20,500 per annum for remaining tenure.
Contingency for Science, Engineering & Technology is Rs.12,000 per annum for initial 2 years and Rs.25,000 per annum for remaining tenure.
Escorts and reader assistance of Rs.2,000 per month (for blind and physically handicapped students).

Note: The scholarship award is valid for one year which can be renewed each year for the degree course subject to your good conduct and maintenance of prescribed attendance.

FAQ’s

Question 1.
What are the functions of UGC?

Answer:
The mandate of the UGC is to take all steps as it may think fit for the promotion and co-ordination of University education and for the determination and maintenance of standards of teaching, examination, and research in Universities

Question 2.
Where can I get the answer-keys of UGC-NET?

Answer:
The answer keys of the all NET subjects of UGC-NET examinations from June 2012 to June 2014 are available on the UGC website www.ugcnetonline.in. For UGC-NET conducted from December 2014 onwards, they are available on CBSE website www.cbsenet.nic.in.

Question 3.
How is the scholarship amount paid to the students?

Answer:
The UGC pays the amount of scholarship directly through Direct Benefit Transfer (DBT). They transfer the amount into the bank accounts of the beneficiaries. Thus, it is necessary for the applicants to have their own bank account at any Indian bank when they are applying for the scholarships or fellowships.

Question 4.
Where is the list of selected candidates displayed?

Answer:
All the results are declared on the official website of UGC, www.ugc.ac.in.

Question 5.
What is the method of UGC Scholarship Renewal?

Answer:
The renewal of scholarship demands the student to submit a yearly progress report to their respective bank’s branch duly signed by the HoD and Registrar of the University of Principal of the College or Head of the Institution. If the student fails to maintain satisfactory progress during his/her study, the scholarship may get canceled.

The post UGC Scholarship 2019-20 | Get Complete List, Eligibility, Rewards, Application Procedure appeared first on Learn CBSE.


NTSE Kerala 2019-20 for Class X | Application Form, Admit Card, Answer Key, Question Paper

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NTSE Kerala 2019-20: The state council for education and training are also known as SCERT has released the notification for the NTSE Kerala stage 1 exam 2019. The exam for this year will happen in the third week of November. Only the students of class 10 can apply tor the NTSE Kerala exam.

NTSE Kerala 2019 Important Dates

All the important dates for NTSE Kerala 2019 exam have been released. Below is the table that provides the details regarding the exam:

EventsDates
Start of  online registration03rd week of Sep 2019
Last date for fee application01st week of Oct 2019
Last date for online application01st week of Oct 2019
Last date of document verification04th week of Oct 2019
Issuing the admit card02nd week of Nov 2019
Date of Stage I Kerala NTSE exam03rd week of Nov 2019
Date of Stage I Kerala NTSE 2 result02nd week of Feb 2020
Date of Stage II Kerala NTSE exam02nd week of Jun 2020

NTSE Kerala Application Form

The application form for NTSE 2019 exam will be available online on NTSE Kerala official website. The application fee to be collected for the exam needs to submit through SBI collect which is an online payment getaway. For general category students, the application fee is Rs. 250. While for students belonging to SC/ST/PWD category, the application fee is Rs. 100.

Once the application fee is remitted and application form is submitted correctly, an application number, as well as application ID, will be given to the candidates. To access the candidate application form, students can log in by submitting an application number and ID.

How To Fill NTSE Kerala Application Form?

  • Collect the application form from the state liaison officer for the NTSE Kerala exam or from the school authorities.
  • Fill the required details like preferred exam center, category, etc.
  • Now fill the admit card form also that is attached with the application form and paste the photograph for the given column.
  • Do not separate the admit card from the application form.
  • Submit the application form after filling all the required details to respective school authorities by the August last week
  • Collect the admit card from state liaison’s office at least 10 days before the NTSE Kerala stage 1 exam

Documents To Be Uploaded Along With NTSE Kerala Exam Application Form

The candidates that belong to respective categories need to submit the following documents

  • For physically challenged candidates, there needs to an attested copy of the medical certificate from the concerned authority to ensure that the candidate is more than 40% physically handicapped.
  • For students of SC, ST, and BPL categories, they should submit an attested copy of cast certificate or BPL certificate as proof.

NTSE Kerala Exam – Eligibility Criteria

Candidates should fulfill the eligibility criteria for the NTSE exam before applying. Below are the eligibility criteria for NTSE Kerala exam:

  • Only the students that are currently studying in class 10 in government, government-aided, recognized schools can apply for this exam.
  • The candidates that are applying need to have a minimum of 55% marks in all non-language subjects in class 9.
  • For candidates applying through open distance learning are eligible to apply. But the maximum age of these should be less than 18 years.

NTSE Kerala – Exam Scheme

The stage 1 and stage 2 exam at state as well as union territories level will have 3 papers.  At stage 2, there will be negative marking for any of the paper. For every wrong answer, the 1/3rd mark will be deducted in the stage 2 exam.

The exam pattern for NTSE Kerala is given below

PaperTest NameNo. of questionMarksDuration (mins)
Paper 1Mental Ability Test505045
Paper 2Language Test505045
Paper 3Scholastic Aptitude Tes10010090

NTSE Kerala Admit Card

The admit card for the candidates for NTSE Kerala will be issued through the conducting body which is a state of the council for educational research and training after the application process is successfully completed. The admit card for the candidates will be released online on the NTSE Kerala official website which is scert.kerala.gov.in. Last year, the candidates were required to download the admit card and also had to produce a hard copy during the exam. If the candidates do not produce the admit card during the exam hall, the candidate will not be allowed to enter the exam hall.

Details for the test of NTSE Kerala is here

Kendriya Vidyalaya,
Willingdon island,
Kochi,
Ernakulam, Kerala – 682003

NTSE Kerala Answer Key and Result

The answer key for the NTSE exam will be uploaded after the successful completion of the exam.  Candidates that have appeared for the exam can estimate the scores using the answer key provided by the official website. Results for the NTSE exam can be checked by candidates online on the official NTSE website. As a result, the overall marks of the candidates along with the marks in both the sections are going to be uploaded.

NTSE Kerala Exam – Selection Procedure

The identification for NTSE Kerala talent is going to be a two-stage process. For the first stage, the individual state or union territories will conduct the exam. For the second stage, the selection is going to be held at the national level and will be carried out by the NCERT.

State-level Exam: State level will comprise of two stages for the students, part 1 is where mental ability test and in part 2 scholastic aptitude test is going to be conducted. This exam will determine the number of candidates required for the 2nd level of the tests to be conducted by the NCERT.

National Level Exam: Only the candidates that have given the state or union territory level exam and are qualified on the basis of the screening process will be eligible to appear for the 2nd stage of this exam. This second stage exam is going to be conducted by NCERT on the second Sunday of May every year.

The post NTSE Kerala 2019-20 for Class X | Application Form, Admit Card, Answer Key, Question Paper appeared first on Learn CBSE.

NTSE Jharkhand 2019-20 for Class 10 | Application Form, Admit Card, Answer Key, Question Paper

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NTSE Jharkhand 2019-20: NTSE Jharkhand is held for choosing 1000 toppers and giving them the schola for their future studies. The NTSE Jharkhand exam is going to be conducted in 2 stages where one will be conducted at the state level while the other exam will be conducted at the national level. The stage 2 exam will be for those candidates that are able to clear the stage 1 exam. The NTSE Jharkhand result for the 2018 exam is now available on the official website.

The exam will be conducted by the Jharkhand academic council on 17th November 2019. The NTSE Jharkhand stage 1 application for class 10 students starts from September last week. The candidates that are eligible through the stage 1 exam can submit the online application form till the second week of October 2019. Candidates will also be able to download the admit card 10 days before the stage 1 exam. NTSE stage 1 exam for Jharkhand will be divided into 2 stages which are a mental ability test (MAT) and a scholastic aptitude test (SAT). Also, candidates that have qualified the stage 1 exam are also eligible for appearing into the stage 2 NTSE Jharkhand exam.

NTSE Jharkhand Overview

Name of the examinationNational Talent Search Examination, Jharkhand
Level of the examState/UT
Conducted byJharkhand Academic Council (JAC), Ranchi
State/UT Liaison Officer &
Address
Dr. Arbind Pd. Singh
Chairman
Jharkhand Academic Council Ranchi,
Gyandeep Campus, Bargawan Namkum, Ranchi-834010.
Mode of examinationPen and paper
Official websitejac.jharkhand.gov.in
Medium of examinationEnglish

NTSE Jharkhand Important Dates

EventsDates
Application beginsLast week of September 2019
Application concludesThe second week of October 2019
NTSE Jharkhand 2020 examNovember 17, 2019
NTSE Jharkhand 2020 ResultsThe second week of April 2020
Stage 2 Admit Card DateApril 2020
NTSE 2020 Stage 2 examMay 10, 2020
NTSE 2020 Stage 2 ResultSeptember 2020

NTSE Jharkhand Eligibility Criteria

Students that wish to apply for NTSE Jharkhand exam should satisfy the following eligibility criteria:

  • Every student should have passed the class 9 exam with a minimum of 55% marks.
  • Jharkhand education board does not provide scholarships to students that are working along with their studies.
  • The student should be pursuing a class 10 exam from a reputable school or government-aided school.
  • For PH, OBC, SC, and ST candidates, one should at least score 40% in class 9 to be eligible for the exam.
  • Students that have NMMS scholarship can also appear for the NTSE exam.

NTSE Jharkhand Exam Application Form

Students can download the application form online through SCERT from the last week of September. The process for downloading the NTSE application form will end by the second week of October 2019.

How To Apply for NTSE Jharkhand Exam?

  1. Go the official website of Jharkhand board which is jac.jharkhand.gov.in.
  2. Click on the NTSE registration link given on the website on the left side.
  3. Enter the required credentials in the space given and click on the print button.
  4. Put your signature on the printed copy along with parents’ sign on the designated columns.
  5. Students should submit this application form to the state’s liaison officer before the deadline with the required documents.

The documents that are required by the students along with the application form are:

  • Copy of 9th class marksheet
  • Two recent passport size photographs of the candidate
  • Copy of the reservation certificates for the candidates that belong to reserve categories
  • Disability status certificate for the physically handicapped students from a government recognized a medical practitioner

NTSE Jharkhand Admit Card

Candidates that are aspiring for the exam can download their admit card or hall ticket from the main website of JAC. For downloading the admit card one must enter the password and registration number. Candidates should carry the NTSE exam admit card to the exam because, without it, candidates won’t be allowed to sit in the exam hall.

NTSE Jharkhand Exam Pattern

The NTSE Jharkhand exam for stage 1 will be conducted in a very specified manner. NTSE exam pattern will have 2 sections which are mental ability test and scholastic ability test. For more details related to the exam pattern and marking scheme, refer the table below:

SectionExam patternTime DurationQualifying scores
MAT100 questions – 1 mark each.120 minutesFor General: 40%
For SC/ ST/ PH: 32%
SAT100 questions – 1 mark each.120 minutesFor General: 40%
For SC/ ST/ PH: 32

NTSE Jharkhand Exam Syllabus

There is no prescribed syllabus mentioned by NCERT or JAC for NTSE exams. But students can refer to Jharkhand class 9, 10 syllabi as well as CBSE class 10 syllabus for preparation. MAT  is conducted for evaluating the candidate’s logical reasoning skills as well as problem-solving skills. While the SAT is conducted for testing the conceptual knowledge in core subjects like science, mathematics, science, and social science. Below are some of the important topics for NTSE Jharkhand exam

SectionsSubjectsImportant Topics
Mental Ability Test (MAT)Verbal Reasoning testArithmetic Reasoning Test, Blood Relations, Alphabet Test, Analogy, Classification, Mathematical Operations, Clocks, Puzzle test, Analytical Reasoning, Verification of truth of the statements, Series completion Test, Tests, etc.
Non-verbal ReasoningFolding paper cutting, Mirror images, Analogy, Analytical Reasoning, Transparent paper folding, Problems on cubes and dice, Water images, etc.
Scholastic Ability Test (SAT)MathematicsAlgebra, basic geometry, Arithmetic progression, simple and compound interest, roots, statistics, coordinated geometry, surface area, and volume, etc.
ScienceCarbon and its compounds, physical and chemical changes, fibers and plastics, periodic classification of elements, magnetism, and electricity, source of energy, etc.
Social SciencesIntroduction of ancient Indian History, medieval architecture, and culture, Jainism, Mughal Empire, Vedic period, Buddhism, The Mauryas, etc.
UN and other International Agencies, Diversity and livelihood, Union government, Economic presence of the government, etc.
The atmosphere, the industries, motion of the earth, maps and globes, the internal structure of earth and rocks, solar system, natural vegetation, etc.

How To Prepare for NTSE Jharkhand Exam?

  • Students should focus on conceptual clarity rather than just learning the topics. Thus, practical knowledge will increase the chances of the probability of getting good scores for students in the NTSE exam.
  • Students should go for the best NTSE books for preparing and making notes and also revising these notes on a regular basis.
  • One key preparation tips are practicing the sample papers to get a good grip on the subject and understanding the type of questions asked.
  • For more practice, candidates can refer to the previous year’s papers of NTSE Jharkhand exam.

NTSE Jharkhand Answer Key and Result

NTSE Jharkhand will release the official answer key on the SCERT official website by the second week of November. There are many coaching institutes that will provide the NTSE exam answer key. Students should check the official website for the NTSE exam frequently after the results are declared for the answer key. Students can use this answer key for comparing the answers with the correct answers and calculate their expected score.

As a result, NTSE Jharkhand will release the official result in the second week of April 2020. The result will be available for everyone in the form of a merit list. Candidates can check the NTSE Jharkhand results online. NTSE Jharkhand results include the candidate’s name, name of the school, roll number, overall score, section-wise marks, etc.

NTSE Jharkhand 2020 Exam Cutoff

Students that are appearing for NTSE 2020 exam cab expect the cutoff to be released soon after the results are released. Below is the table for previous year’s cutoff:

The NTSE program is solely for providing scholarships to the deserving candidates. Under the program, 2000 scholarships are distributed for the candidates selected after the stage 2 process. Below is the category of wise scholarship data.

CategoryCut off
General124
SC107
ST87
PH67

NTSE Jharkhand Exam Scholarship

The NTSE program is solely for providing scholarships to the deserving candidates. Under the program, 2000 scholarships are distributed for the candidates selected after the stage 2 process. Below is the category of wise scholarship data.

Education levelScholarship Amount
Classes 11-12Rs. 1,250/ month
UndergraduateRs. 2,000/month
Post-graduateRs. 2,000/month
Ph.D.As per UGC norms.

The post NTSE Jharkhand 2019-20 for Class 10 | Application Form, Admit Card, Answer Key, Question Paper appeared first on Learn CBSE.

Reading Comprehension Class 12 Passages, Exercises, Worksheets

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Reading Comprehension CBSE Class 12 Passages, Exercises, Worksheets

Step 1: Skim once as rapidly as possible to determine the main idea before you look at the questions. Do not worry about words you do not know at this stage.
Step 2: Underline the words that you do not understand to facilitate a complete understanding of the passage. This will enable you to solve the vocabulary questions quicker.
Step 3: Look through the words carefully. You are advised to maintain the order in which the questions appear in the test paper. Read intensively the portion relevant to the answer.
Step 4; Concentrate on the vocabulary items and puzzle out from the context the meanings of those words you do not know.

♦ Ideas To Follow While Skimming

  1. Read the title of the passage/ poem very carefully, if given. Determine what clues it gives you about the passage/poem.
  2. Watch for keywords like causes, results, effects etc. Do not overlook signal words such as those suggesting controversy (e.g., versus, pros and cons), which indicate that the author is intending to present both sides of an argument.
  3. Concentrate on the main ideas and ignore details. Most passages require at least two readings. Before writing the answer, check the questions again to be sure you have understood them.

♦ Guidelines To Attempt Comprehension Passage

  1. First of all, read the passage quickly study the questions given at the end of your passage.
  2. Start your second reading of the passage. This reading should be thorough. Underline key sentences or words related to the given questions.
  3. An alternate method could be to go through the questions first, which gives a rough idea about the content or subject of the passage. It becomes easier to underline the keywords while going through the passage and will help to reach the answers faster.
  4. While answering the questions, try not to give vague or general answers; be specific; sometimes students use one general description when four or five points have to be made. Avoid general answers.
  5. Write in short, simple sentences unless required to do otherwise.
  6. Do not repeat yourself. This is a waste of time. Avoid using slang. Do not use vague words when a precise one will do.
  7. Make sure that you use your own words as far as possible. This means that you must summarise and’ interpret information; never copy whole ‘chunks’ from the passage.
  8. When answering factual questions, i.e., questions that involve words like ‘what7, ‘when’, ‘how’ and ‘why’, do not include information not given in the passage. While answering the ‘why’ question, you may begin your answer with ‘This is because of a similar phrase.
  9. While answering vocabulary questions, determine the part of speech of the word. Your answer should have the same part of speech. ,

♦ Previous Years’ CBSE Examination Questions

Q.l. Read the passage given below and answer the questions that follow: (Delhi, All India 2010)

1. Today’s woman is a highly self-directed person, alive to the sense of her dignity and the importance of her functions in the private domestic domain and the public domain of the world of work. Women are rational in approach, careful in handling situations and want to do things as best as possible. The Fourth World Conference of Women held in Beijing in September 1995 had emphasized that no enduring solution of society’s most threatening social, economic and political problems could be found without the participation and empowerment of the women. The 1995 World Summit for Social Development had also emphasised the pivotal role of women in eradicating poverty and mending the social fabric.

2. The Constitution of India had conferred on women equal rights and opportunities political, social, educational and of employment with men. Because of oppressive traditions, superstitions, exploitation and corruption, a majority of women are not allowed to enjoy the rights and opportunities, bestowed on them. One of the major reasons for this state of affairs is the lack of literacy and awareness among women. Education is the main instrument through which we can narrow down the prevailing inequality and accelerate the process of economic and political change in the status of women.

3. The role of women in a society is very important. Women’s education is the key to a better life in the future. A recent World Bank study says that educating girls is not a charity, it is good economics and if developing nations are to eradicate poverty, they must educate the girls. The report says that the economic and social returns on investment in education of the girls considerably affect the human development index of the nation. Society would progress only if the status of women is respected and the presence of an educated woman in the family would ensure education of the family itself. Education and empowerment of women are closely related.

4. Women’s education has not received due care and attention from the planners and policymakers. The National Commission for Women has rightly pointed out that even after 50 years of independence, women continue to be treated as the single largest group of backward citizens of India. The role of women in overall development has not been fully understood nor has it been given its full weight in the struggle to eliminate poverty, hunger, injustice and inequality at the national level. Even when we are at the threshold of the 21st century, our society still discriminates against women in matters of their rights and privileges and prevents them from participating in the process of national and societal progress.

Various Committees and Commissions have been constituted before and after the independence to evaluate the progress in women’s education and to suggest ways and means to enhance the status of women. The female literacy rate has gone up in the 20th century from 0.6 per cent in 1901 to 39.29 per cent in 1991 but India still possesses the largest number of illiterate women in the world. The female literacy index for the year 1991 shows that there are eight States which fall below the national average. The most populous States of the country, UP, MP, Bihar and Rajasthan fall in the category of most backward States as far as female literacy is concerned.

5. The prevailing cultural norms of gender behaviour and the perceived domestic and reproductive roles of women tend to affect the education of girls. Negative attitude towards sending girls to schools, restrictions on their mobility, early marriage, poverty and illiteracy of parents affect the girl’s participation in education.

6. Women’s political empowerment got a big boost with the Panchayati Raj Act of 1993 which gave them 30 per cent reservation in Village Panchayats, Block Samities and Zila Parishads throughout the country. The National Commission for Women was also set up in 1992 to act as a lobby for women’s issues.

7. The educational system is the only institution which can counteract the deep foundations of inequality of sexes that are built in the minds of people through the socialization process. Education is the most important instrument of human resource development. Educational system should be used to revolutionize the traditional attitudes and inculcate new values of equality.

(a)
(i) Mention any two attributes of a modern woman. 2
(ii) Why are women’s participation and empowerment considered necessary? 2
(iii) Which factors adversely affect the education of girls? 2
(iv) What benefits did the women get with the enactment of the Panchayati Raj Act of 1993? 2
(v) By what process can we remove the sense of inequality of sexes from the minds of the people? 1
Answer:
(i) The modem woman of today is a highly self-dependent person alive to the sense of her dignity. She is rational in her approach and handles situations carefully.
(ii) At the Fourth World Conference of Women held in Beijing, it was emphasized that without women’s participation and empowerment no enduring solution to society’s most threatening social, economic and political problems could be found. Also women’s empowerment is extremely important in eradicating poverty and mending the social fabric.
(iii) The prevailing cultural norms of gender behaviour and the perceived domestic and reproductive roles of women tend to adversely affect the education of girls. Restrictions on the mobility of girls, early marriage of girls, poverty and illiteracy of parents affect the girl’s participation in education.
(iv) Women’s political empowerment got a major boost with the Panchayati Raj Act of 1993 which gave them 30 per cent reservation in Village Panchayats, Block Samities and Zila Parishads throughout the country.
(v) The sense of inequality of sexes can be removed from the minds of the people only through the process of education.

(b) Pick out words from the passage which mean the same as each of the following: 1×3=3
(i) cruel and unfair (para 2)
(ii) remove (para 3)
(iii) full of people (para 4)
Answer:
(i) oppressive
(ii) eradicate
(iii) populous

Q.2. Read the passage given below and answer the questions that follow: (Delhi, All India 2011)
1. For many years now the Governments have been promising the eradication of child labour in hazardous industries in India. But the truth is that despite all the rhetoric no Government so far has succeeded in eradicating this evil, nor has any been able to ensure compulsory primary education for every Indian child. Between 60 and 100 million children are still at work instead of going to school, and around 10 million are working in hazardous industries. India has the biggest child population of 380 million in the world; plus the largest number of children who are forced to earn a living.

2. We have many laws that ban child labour in hazardous industries. According to the Child Labour (Prohibition and Regulation) Act 1986, the employment of children below the age of 14 in hazardous occupations has been strictly banned. But each state has different rules regarding the minimum age of employment. This makes the implementation of these laws difficult.

3. Also, there is no ban on child labour in nonhazardous occupations. The act applies to the organised or factory sector and not the unorganized or informal sector where most children find employment as cleaners, servants, porters, waiters, among other forms of unskilled work. Thus, child labour continues because the implementation of the existing law is lax.

4. There are industries, which have a special demand for child labour because of their nimble fingers, high level of concentration and capacity to work hard at abysmally low wages. The carpet industry in U.P. and Kashmir employs children to make hand-knotted carpets. There are 80,000 child workers in Jammu & Kashmir alone. In Kashmir because of the political unrest, children are forced to work while many schools are shut. Industries like gem cutting and polishing, pottery and glass want to remain competitive by employing children.

5. The truth is that it is poverty which is pushing children into the brutish labour market. We have 260 million people below the poverty line in India, a large number of them are women. Poor and especially woman-headed families have no option but to push their little ones in this hard life in hostile conditions, with no human or labour rights.

6. There is a lobby which argues that there is nothing wrong with children working as long as the environment for work is conducive to learning new skills but studies have shown that the children are made to do boring, repetitive and tedious jobs and are not taught new skills as they grow older. In these hellholes like the sweet shops of the old, there is no hope.

7. Children working in hazardous industries are prone to debilitating diseases which can cripple them for life. By sitting in cramped, damp and unhygienic spaces, their limbs become deformed for life. Inside matchstick, fireworks and glass industries they are victims of bronchial diseases and T.B. Their mental and physical development is permanently impaired by long hours of work. Once trapped, they can’t get out of this vicious circle of poverty. They remain uneducated and powerless. Finally, in later years, they too are compelled to send their own children to work. Child labour perpetuates its own nightmare.

8. If at all the Government was serious about granting children their rights, an intensive effort ought to have been made to implement the Supreme Court’s Directive of 1997 which laid down punitive action against employers of child labour. Only compulsory primary education can eliminate child labour.

9. Surely, if 380 million children are given a better life and elementary education, India’s human capital would be greatly enhanced. But that needs, as former President Abdul Kalam says, “a Second

(a)
(i) On which two counts has the Government not succeeded so far in respect of children? 2
(ii) What makes the implementation of child labour law difficult? 2
(iii) Why do industries prefer child labour?
(iv) What are the adverse effects of hazardous industries on children? Given any two. 2
(v) What does the Supreme Court’s Directive of 1997 provide? 1
Answer:
(i) In respect to children the Government has not yet succeeded in eradication of child labour in hazardous industries and ensuring compulsory primary education.
(ii) Implementation of child labour becomes difficult because each state has different rules regarding the minimum age of employment and there is no ban on child labour in the nonhazardous occupations.
(iii) Industries prefer child labour because children have a capacity to work hard, a high level of concentration and can be employed at low wages.
(iv) Children working in hazardous industries are prone to debilitating diseases which can cripple them for life. By sitting in cramped, damp and unhygienic spaces their limbs too become deformed for life. In matchstick, fireworks and glass industries, children become victims of bronchial diseases and T.B.
(v) The Supreme Court’s Directive of 1997 provides punitive action against employers of child labour.

(b) Find words from the passage which mean the same as the following:
(i) risky/dangerous (para 1)
(ii) very unfriendly (para 5)
(iii) intended as punishment (para 8)
Answer:
(i) hazardous
(ii) hostile
(iii) punitive

Q.3. Read the passage given below and answer the questions that follow: (Delhi, All India 2012)
1. While there is no denying that the world loves a winner, it is important that you recognize the signs, of stress in your behaviour and be healthy enough to enjoy your success. Stress can strike anytime, in a fashion that may leave you unaware of its presence in your life. While a certain amount of pressure is necessary for performance, it is important to be able to recognize your individual limit. For instance, there are some individuals who accept competition in a healthy fashion. There are others who collapse into weeping wrecks before an exam or on comparing marks sheets and finding that their friend has scored better.

2. Stress is a body reaction to any demands or changes in its internal and external environment. Whenever there is a change in the external environment such as temperature, pollutants, humidity and working conditions, it leads to stress. In these days of competition when a person makes up his mind to surpass what has been achieved by others, leading to an imbalance between demands and resources, it causes psychosocial stress. It is a part and parcel of everyday life.

3. Stress has a different meaning, depending on the stage of life you are in. The loss of a toy or a reprimand from the parents might create a stress shock in a child. An adolescent who fails an examination may feel as if everything has been lost and life has no further meaning. In an adult the loss of his or her companion, job or professional failure may appear as if there is nothing more to be achieved.

4. Such signs appear in the attitude and behaviour of the individual, as muscle tension in various parts of the body, palpitation and high blood pressure, indigestion and hyperacidity. Ultimately the result is self-destructive behaviour such as eating and drinking too much, smoking excessively, relying on tranquilisers. There are other signs of stress such as trembling, shaking, nervous blinking, dryness of throat and mouth and difficulty in swallowing.

5. The professional under stress behaves as if he is a perfectionist. It leads to depression, lethargy and weakness. Periodic mood shifts also indicate the stress status of the students, executives and professionals.

6. In a study sponsored by World Health Organization and carried out by Harvard School of Public Health, the global burden of diseases and injury indicated that stress diseases and accidents are going to be the major killers in 2020.

7. The heart disease and depression both stress diseases are going to rank first and second in 2020. Road traffic accidents are going to be the third-largest killers. These accidents are also an indicator of psychosocial stress in a fast-moving society. Other stress diseases like ulcers, hypertension and sleeplessness have assumed epidemic proportions in modern societies.

8. A person under stress reacts in different ways and the common ones are flight, fight and flee depending upon the nature of the stress and capabilities of the person. The three responses can be elegantly chosen to cope with the stress so that stress does not damage the system and become distressed.

9. When stress crosses the limit, peculiar to an individual, it lowers his performance capacity. Frequent crossings of the limit may result in chronic fatigue in which a person feels lethargic, disinterested and is not easily motivated to achieve anything. This may make the person mentally undecided, confused and accident-prone as well. Sudden exposure of unnerving stress may also result in a loss of memory. Diet, massage, food supplements, herbal medicines, hobbies, relaxation techniques and dance movements are excellent stress busters.

(a)
(i) What is stress? What factors lead to stress? 2
(ii) What are the signs by which a person can know that he is under stress? 2
(iii) What are the different diseases a person gets due to stress? 2
(iv) Give any two examples of stress busters. 1
(v) How does a person react under stress? 2
Answer:
(i) Stress is a body reaction to any demands or changes in its external and internal environment. A change in the external environments such as temperature, pollutants, humidity and working conditions lead to stress.
(ii) Certain signs appear in the attitude and behaviour of an individual under stress. These include muscle tension in various body parts, palpitation, high blood pressure, indigestion and hyperacidity. Other stress-related signs are trembling, shaking, nervous blinking, dryness of throat and mouth and difficulty in swallowing.
(iii) Heart disease and depression are the two major stress-related diseases. Other stress diseases include ulcers, hypertension and sleeplessness.
(iv) Herbal medicines and relaxation techniques are two examples of stress busters.
(v) A person under stress reacts in different ways, the most common ones being flight, fight and flee depending upon the nature of the stress and capabilities of a person.

(b) Which words in the above passage mean the same as the following?
(i) Fall down (para 1)
(ii) rebuke (para 3)
(iii) inactive (para 9)
Answer:
(i) collapse
(ii) reprimand
(iii) lethargic

Q.4. Read the passage given below and answer the questions that follow: (Delhi, All India 2013)
1. Air pollution is an issue which concerns us all alike. One can willingly choose or reject a food, a drink or a life comfort, but unfortunately there is little choice for the air we breathe. All, what is there in the air is inhaled by one and all living in those surroundings.

2. Air pollutant is defined as a substance which is present while normally it is not there or present in an amount exceeding the normal concentrations. It could either be gaseous or a particulate matter. The important and harmful polluting gases are carbon monoxide, carbon dioxide, ozone and oxides of sulphur and nitrogen. The common particulate pollutants are the dusts of various inorganic or organic origins. Although we often talk of the outdoor air pollution caused by industrial and vehicular exhausts, the indoor pollution may prove to be as or a more important cause of health problems.

3. Recognition of air pollution is relatively recent. It is not uncommon to experience a feeling of ‘suffocation’ in a closed environment. It is often ascribed to the lack of oxygen. Fortunately, however, the composition of air is remarkably constant all over the world. There is about 79 per cent nitrogen and 21 per cent oxygen in the air the other gases forming a very small fraction. It is true that carbon dioxide exhaled out of lungs may accumulate in a closed and overcrowded place. But such an increase is usually small and temporary unless the room is really airtight. Exposure to poisonous gases such as carbon monoxide may occur in a closed room, heated by burning cctal inside. This may also prove to be fatal.

4. What is more common in a poorly ventilated home is a vague constellation of symptoms described as the sickbuilding syndrome. It is characterized by a general feeling of malaise, headache, diiiiness and irritation of mucous membranes. It may also be accompanied by nausea, itching, aches, pains and depression. Sick building syndrome is getting commoner in big cities with the small houses, which are generally overfurnished. Some of the important pollutants whose indoor concentrations exceed those of the outdoors include gases such as carbon monoxide, carbon dioxide, oxides of nitrogen and organic substances like spores, formaldehydes, hydrocarbon aerosols and allergens. The sources are attributed to a variety of construction materials, insulations, furnishings, adhesives, cosmetics, house dusts, fungi and other indoor products.

5. Byproducts of fuel combustion are important in houses with indoor kitchens. It is not only the burning of dried dung and fuel wood which is responsible, but also kerosene and liquid petroleum gas. Oxides of both nitrogen and sulphur are released from their combustion.

6. Smoking of tobacco in the closed environment is an important source of indoor pollution. It may not be high quaniiiatively, but signiiicantly hazardous for health. It is because of the fact that there are over 3,000 chemical consiiiuents in tobacco smoke, which have been identiiied. These are harmful for human health.

7. Microorganisms and allergens are of special signiiicance in the causation and spread of diseases. Most of the infective illnesses may involve more persons of a family living in common indoor environment. These include viral and bacterial diseases like tuberculosis.

8. Besides infections, allergic and hypersensitivity disorders are spreading fast. Although asthma is the most common form of respiratory allergic disorders, pneumonias are not uncommon, but more persistent and serious. These are attributed to exposures to allergens from various fungi, moulds, hay and other organic materials. Indoor air ventilation systems, coolers, airconditioners, dampness, decay, pet animals, production or handling of the causative items are responsible for these hypersensitivitydiseases.

9. Obviously, the spectrum of pollution is very wide and our options are limited. Indoor pollution may be handled relatively easily by an individual. Moreover, the good work must start from one’s own house. (Extracted from The Tribune)

(a)
(i) What is an air pollutant? 1
(ii) In what forms are the air pollutants present? 2
(iii) Why do we feel suffocated in a closed environment? 1
(iv) What is sick building syndrome? How is it increasing? 2
(v) How is indoor smoking very hazardous? 1
(vi) How can one overcome the dangers of indoor air pollution? 2
Answer:
(i) An air pollutant is a substance which is present while normally it is not there in an amount exceeding the normal , concentrations.
(ii) Air pollutants are present as gaseous or particulate matter. The harmful polluting gases are carbon monoxide, carbon dioxide, ozone and oxides of sulphur and nitrogen. The common particulate pollutants are the dusts of various inorganic or organic origins.
(iii) We often feel suffocated in a closed environment due to the lack of oxygen.
(iv) Sick building syndrome is a vague constellation of symptoms in a poorly ventilated room. Sick building syndrome is characterised by a general feeling of malaise, headache, diiiiness and irritation of mucous membranes. It may also be accompanied by nausea, itching, aches, pains and depression. This syndrome is increasing in big cities which have an increasing number of small houses which are generally overfurnished.
(v) Indoor smoking is very hazardous because over 3,000 chemical consiiiuents are present in tobacco smoke and these are harmful for human health.
(vi) The dangers of indoor pollution can be avoided through well ventilated houses and improving greenery around houses. Also, it can be avoided by not smoking tobacco inside houses and by not burning coal inside closed rooms.

(b) Find the words from the above passage which mean the same as the following: 3
(i) giddiness (para 4)
(ii) constant (para 8)
(iii) humidity (para 8)
Answer:
(i) diiiiness
(ii) persistent
(iii) dampness

Q.5. Read the following passage and answer the questions that follow: (Delhi, All India 2014)

1. Too many parents these days can’t say no. As a result, they find themselves raising ‘children’ who respond greedily to the advertisements aimed right at them. Even getting what they want doesn’t satisfy some kids; they only want more. Now, a growing number of psychologists, educators and parents think it’s time to stop the madness and start teaching kids about what’s really important: values like hard work, contentment, honesty and compassion. The struggle to set limits has never been tougherand the stakes have never been higher. One recent study of adults who were overindulged as children, paints a discouraging picture of their future: when given too much too soon, they grow up to be adults who have difficulty coping with life’s disappointments. They also have distorted sense of eniiilement that gets in the way of success in the work place and in relationships.

2. Psychologists say that parents who overindulge their kids, set them up to be more vulnerable to future anxiety and depression. Today’s parents themselves raised on values of thrift and selfsacriiice, grew up in a culture where no was a household word. Today’s kids want much more, partly because there is so much more to want. The oldest members of this generation were bom in the late 1980s, just as PCs and video games were making their assault on the family room. They think of MP3 players and flat screen TV as essential utilities, and they have developed strategies to get them. One survey of teenagers found that when they crave for something new, most expect to ask nine times before their parents give in. By every measure, parents are shelling out record amounts. In the heat of this buying blitz, even parents who desperately need to say no find themselves reaching for their credit cards.

3. Today’s parents aren’t equipped to deal with the problem. Many of them, raised in the 1960s and ’70s, swore they’d act differently from their parents and have closer relationships with their own children. Many even wear the same designer clothes as their kids and listen to the same music. And they work more hours; at the end of a long week, it’s tempting to buy peace with ‘yes’ and not mar precious family time with conflict. Anxiety about future is another factor. How do well intentioned parents say no to all the sports gear and arts and language lessons they believe will help their kids thrive in an increasingly compeiiiive world? Experts agree : too much love won’t spoil a child. Too few limits will.

4. What parents need to find, is a balance between the advantages of an affluent society and the critical life lessons that come from waiting, saving and working hard to achieve goals. That search for balance has to start early. Children need limits on their behaviour because they feel better and more secure when they live within a secured structure. Older children learn selfcontrol by watching how others, especially parents act. Learning how to overcome challenges is essential to becoming a successful adult. Few parents ask kids to do chores. They think their kids are already overburdened by social and academic pressures. Every individual can be of service to others, and life has meaning beyond one’s own immediate happiness. That means parents eager to teach values have to take a long, hard look at their own.

(a) Answer the following:
(i) What values do parents and teachers want children to learn? 2
(ii) What are the results of giving the children too much too soon? 2
(iii) Why do today’s children want more? 1
(iv) What is the balance which the parents need to have in today’s world? 2
(v) What is the necessity to set limits for children? 2
Answer:
(i) Parents and teachers want children to learn values like hard work, contentment, honesty and compassion.
(ii) When children are given too much too soon, they grow up to be adults who have difficulty coping with life’s disappointments. They also have distorted sense of eniiilement that gets in the way of success in the work place and in relationships.
(iii) Today’s children want more, partly because there is so much more to want.
(iv) The balance that parents need to have in today’s world is between the advantages of an affluent society and the critical life lessons that come from waiting, saving and working hard to achieve goals.
(v) It is necessary to set limits on the behaviour of children because they feel better and more secure when they live within a secured structure.

(b) Pick out words from the passage that mean the same as the following: 3
(i) a feeling of satisfaction (para 1)
(ii) valuable (para 3)
(iii) important (para 4)
Answer:
(i) a feeling of satisfaction contentment
(ii) valuable precious
(iii) important essential/critical

Q.6. Read the passage carefully: (Delhi, Comptt. Delhi, Comptt. All India 2015)

1. For four days, I walked through the narrow lanes of the old city, enjoying the romance of being in a city where history still livesin its cobblestone streets and in its people riding asses, carrying vine leaves and palm as they once did during the time of Christ.

2. This is Jerusalem, home to the sacred sites of Christianity, Islam and Judaism. This is the place that houses the church of the Holy Sepulchre, the place where Jesus was finally laid to rest. This is also the site of Christ’s cruciiixion, burial and resurrection.

3. Built by the Roman Emperor Constantine at the site of an earlier temple to Aphrodite, it is the most venerated Christian shrine in the world. And justiiiably so. Here, within the church, are the last five stations of the cross, the 10th station where Jesus was stripped of his clothes, the 11th where he was nailed to the cross, the 12th where he died on the cross, the 13th where the body was removed from the cross, and the 14th, his tomb.

4. For all this weighty tradition, the approach and entrance to the church is nondescript. You have to ask for directions. Even to the devout Christian pilgrims walking along the Via Dolorosathe Way of Sorrowsfirst nine stations look clueless. Then a courtyard appears, hemmed in by other buildings and a doorway to one side. This leads to a vast area of huge stone architecture.

5. Immediately inside the entrance is your first stop. It’s the stone of anointing: this is the place, according to Greek tradition, where Christ was removed from the cross. The Roman Catholics, however, believe it to be the spot where Jesus’ body was prepared for burial by Joseph.

6. What happened next? Jesus was buried. He was taken to a place outside the city of Jerusalem where other graves existed and there, he was buried in a cave. However, all that is long gone, destroyed by
continued attacks and rebuilding; what remains is the massiveand impressiveRotunda (a round building with a dome) that Emperor Constantine built. Under this, and right in the centre of the Rotunda, is the structure that contains the Holy Sepulchre.

7. “How do you know that this is Jesus’ tomb?” I asked one of the pilgrims standing next to me. He was clueless, more interested, like the rest of them, in the novelty of it all and in photographing it, than in its history or tradition.

8. At the start of the first century, the place was a disused quarry outside the city walls. According to the gospels, Jesus’ cruciiixion occurred ‘at a place outside the city walls with graves nearby. Archaeologists have discovered tombs from that era, so the site is compatible with the biblical period.

9. The structure at the site is a marble tomb built over the original burial chamber. It has two rooms, and you enter four at a time into the first of these, the Chapel of the Angel. Here the angel is supposed to have sat on a stone to recount Christ’s resurrection. A low door made of white marble, partly worn away by pilgrims’ hands, leads to a smaller chamber inside. This is the ‘room of the tomb’, the place where Jesus was buried.

10. We entered in a single file. On my right was a large marble slab that covered the original rock bench on which the body of Jesus was laid. A woman knelt and prayed. Her eyes were wet with tears. She pressed her face against the slab to hide them, but it only made it worse.

On the basis of your understanding of this passage answer the following questions with the help of the given options: 1×4=4
(a) How does Jerusalem still retain the charm of the ancient era?
(i) There are narrow lanes.
(ii) Roads are paved with cobblestones,
(iii) People can be seen riding asses.
(iv) All of the above.

(b) Holy Sepulchre is sacred to
(i) Christianity
(ii) Islam
(iii) Judaism
(iv) Both (i) and (iii)

(c) Why does one have to constantly ask for directions to the church?
(i) Its lanes are narrow.
(ii) Entrance to the church is nondescript.
(iii) People are not tourist-friendly.
(iv) Everyone is lost in enjoying the romance of the place.

(d) Where was Jesus buried?
(i) In a cave
(ii) At a place outside the city
(iii) In the Holy Sepulchre
(iv) Both (i) and (ii)

Answer the following questions briefly: 1 x 6 = 6
(e) What is the Greek belief about the ‘stone of anointing’?
(f) Why did Emperor Constantine build the Rotunda?
(g) What is the general attitude of the pilgrims?
(h) How is the site compatible with the biblical period?
(i) Why did the pilgrims enter the ‘room of the tomb’ in a single file?
(j) Why did ‘a woman’ try to hide her tears?
(k) Find words from the passage which mean the same as 1 x 2 = 2
(i) A large grave (para 3)
(ii) Having no interesting features/dull (para 4)
Answer:
(a) (iv) All of the above
(b) (i) Christianity
(c) (ii) Entrance to the church is non-descript.
(d) (iv) Both (z) and (ii)
(e) According to Greek belief the ‘stone of anointing’ is the place where Christ was removed from the cross
(f) Emperor Constantine built the Rotunda to venerate the place of burial of Jesus. He built this structure to protect the Holy Sepulchre.
(g) The pilgrims are not interested in the history or tradition of the place and the tomb where Jesus was buried. They are more interested in the novelty of it all and in photographing it.
(h) Archaeologists have discovered tombs from that era. So this is compatible with the biblical period according to which Jesus crucifixion occurred ‘at a place outside the city walls with graves nearby.
(i) The pilgrims enter the room of the tomb in a single file (line) because it has a narrow passage and a low door which leads to a smaller chamber.
(j) ‘A woman’ tried to hide her tears because she did not want anyone to see her crying. Like a true Christian, she was overwhelmed as Jesus was buried there, while others seemed unconcerned.
(k) (i) tomb (ii) nondescript

Q.7. Read the passage given below: (Delhi)

1. We often make all things around us the way we want them. Even during our pilgrimages we have begun to look for whatever makes our heart happy, gives comfort to our body and peace to the mind. It is as if external solutions will fulfil our needs, and we do not want to make any special efforts even in our spiritual search. Our mind is resourceful works to find shortcuts in simple and easy ways.

2. Even pilgrimages have been converted into tourism opportunities. Instead, we must, awaken our conscience and souls and understand the truth. Let us not tamper with either our own nature or that of the Supreme.

3. All our cleverness is rendered ineffective when nature does a dance of destruction. Its fury can and will wash away all imperfections. Indian culture, based on Vedic treatises; assists in human evolution, but we are now using our entire energy in distorting these traditions according to our convenience instead of making efforts to make ourselves worthy of them.

4. The irony is that humans are not even aware of the complacent attitude they have allowed themselves to sink to. Nature is everyone’s Amma and her fierce blows will sooner or later comer and force us to understand this truth. Earlier, pilgrimages to places of spiritual significance were rituals that were undertaken when people became free from their worldly duties. Even now some seekers take up this pious religious journey as a path to peace and knowledge. Anyone travelling with this attitude feels and travels with only a few essential items that his body can carry. Pilgrims traditionally travelled light, on foot, eating light, dried chickpeas and fruits, or whatever was available. Pilgrims of olden days did not feel the need to stay in special AC bedrooms, or travel by luxury cars or indulge themselves with delicious food and savouries.

5. Pilgrims traditionally moved ahead, creating a feeling of belonging towards all, conveying a message of brotherhood among all they came across whether in small caves, ashrams or local settlements. They received the blessings and congregations of yogis and mahatmas in return while conducting the dharma of their pilgrimage. A pilgrimage is like penance or sadhana to stay near nature and to experience a feeling of oneness with it, to keep the body healthy and fulfilled with the amount of food, while seeking freedom from attachments and yet remaining happy while staying away from relatives and associates.

6. This is how a pilgrimage should be rather than making it like a picnic by taking a large group along and living in comfort, packing in entertainment, and tampering with environment. What is worse is giving a boost to the ego of having had a special darshan. Now alms are distributed, charity done while they brag about their spiritual experiences!

7. We must embark on our spiritual journey by first understanding the grace and significance of a pilgrimage and following it up with the prescribed rules and rituals this is what translates into the ultimate and beautiful medium of spiritual evolution. There is no justification for tampering with nature.

8. A pilgrimage is symbolic of contemplation and meditation and acceptance and is a metaphor for the constant growth or movement and love for nature that we should hold in our hearts.

9. This is the truth! On the basis of your understanding of the above passage answer the questions that follow with the help of the given options: 1 x 2 = 2
(a) How can a pilgrim keep his body healthy?
(i) By travelling light
(ii) By eating a small amount of food
(iii) By keeping free from attachments
(iv) Both (i) and (ii)

(b) How do we satisfy our ego?
(i) By having a special darshan
(ii) By distributing alms
(iii) By treating it like a picnic
(iv) Both (i) and (ii)

Answer the following as briefly as possible: 1 x 6 = 6
(c) What change has taken place in our attitude towards pilgrimages?
(d) What happens when pilgrimages are turned into picnics?
(e) Why are we complacent in our spiritual efforts?
(f) How does nature respond when we try to be clever with it?
(g) In olden days with what attitude did people go on a pilgrimage?
(h) What message does the passage convey to the pilgrims?
(i) Find words from the passage which mean the same as the following: 1 x 2 = 2
(i) made/turned (para 3)
(ii) very satisfied (para 4)
Answer:
(a) (iv) Both (i) and (it)
(b) (i) By having a special darshan
(c) During our pilgrimages we have begun to look for whatever makes our heart happy, gives comfort to our body and peace to the mind.
(d) When pilgrimages are turned into picnics, we travel with a large group consisting of our relatives, friends and associates. We live in comfort, pack in entertainment and tamper with the environment.
(e) We have become complacent in our spiritual efforts. We feel external solutions will fulfil our needs, and we do not want to make any special efforts in our spiritual search. We often make all things around us the way we want them.
(f) When we try to be clever with nature it does a dance of destruction and we have to face the fierce blows which will sooner or later comer us and wash away all imperfections.
(g) In olden days when people went on a pilgrimage, they created a feeling of belonging towards all, conveying a message of brotherhood among all they came across.
(h) The passage conveys the message that a pilgrimage symbolizes contemplation, meditation, acceptance growth and love for nature. The message the passage conveys to pilgrims is that we must embark on our spiritual journey by first understanding the grace and significance of a pilgrimage and following it up with prescribed rules and rituals.
(k) (i) rendered (ii) complacent

Q.8. Read the passage given below: (Delhi, All India 2016)

1. Maharana Pratap ruled over Mewar only for 25 years. However, he accomplished so much grandeur during his reign that his glory surpassed the boundaries of countries and time turning him into an immortal personality. He along with his kingdom became a synonym for valour, sacrifice and patriotism. Mewar had been a leading Rajput Kingdom even before Maharana Pratap occupied the throne. Kings of Mewar, with the cooperation of their nobles and subjects, had established such traditions in the kingdom, as augmented their magnificence despite the hurdles of having a smaller area under their command and less population. There did come a few thorny occasions when the flag of the kingdom seemed sliding down. Their flag once again heaved high in the sky thanks to the gallantry and brilliance of the people of Mewar.

2. The destiny of Mewar was good in the sense that barring a few kings, most of the rulers were competent and patriotic. This glorious tradition of the kingdom almost continued for 1500 years since its establishment, right from the region of Bappa Rawal. In fact only 60 years before Maharana Pratap, Rana Sanga drove the kingdom to the pinnacle of fame. His reputation went beyond Rajasthan and reached Delhi. Two generations before him, Rana Kumbha had given a new stature to the kingdom through victories and developmental work. During his reign, literature and art also progressed extraordinarily. Rana himself was inclined towards writing and his works are read with reverence even today. The ambience of his kingdom was conducive to the creation of high quality work of art and literature. These accomplishments were the outcome of a longstanding tradition sustained by several generations.

3. The life of the people of Mewar must have been peaceful and prosperous during the long span of time; otherwise such extraordinary accomplishment in these fields would not have been possible. This is reflected in their art and literature as well as their loving nature. They compensate for lack of admirable physique by their firm but pleasant nature. The ambience of Mewar remains lovely thanks to the cheerful and liberal character of its people.

4. One may observe astonishing pieces of workmanship not only in the forts and palaces of Mewar but also in public utility buildings. Ruins of many structures which are still standing tall in their grandeur are testimony to the fact that Mewar was not only the land of the brave but also a seat of art and culture. Amidst aggression and bloodshed, literature and art flourished and creative pursuits of literature and artists did not suffer. Imagine, how glorious the period must have been when the Vijaya Stambha which is the sample of our great ancient architecture even today, was constructed. In the same fort, Kirti Stambha is standing high, reflecting how liberal the then administration was which allowed people from other communities and kingdoms to come and carry out construction work. It is useless to indulge in the debate whether the Vijaya Stambha was constructed first or the Kirti Stambha. The fact is that both the capitals are standing side by side and reveal the proximity between the king and the subjects of Mewar.

5. The cycle of time does not remain the same. Whereas the reign of Rana Sanga was crucial in raising the kingdom to the acme of glory, it also proved to be his nemesis. History took a turn. The fortune of Mewarthe land of the brave, started waning. Rana tried to save the day with his acumen which was running against the stream and the glorious traditions for sometime.

On the basis of your understanding of the above passage answer each of the questions given below with the help of options that follow: 1 x 4 = 4
(a) Maharana Pratap became immortal because:
(i) he ruled Mewar for 25 years.
(ii) he added a lot of grandeur to Mewar.
(iii) of his valour, sacrifice & patriotism.
(iv) both (ii) and (iii).

(b) Difficulties in the way of Mewar were:
(i) lack of cooperation of the nobility.
(ii) ancient traditions of the kingdom.
(iii) its small area and small population.
(iv) the poverty of the subjects.

(c) During thorny occasions:
(i) the flag of Mewar seemed to be lowered.
(ii) the flag of Mewar was hoisted high.
(iii) the people of Mewar showed gallantry.
(iv) most of the rulers heaved a sigh of relief.

(d) Mewar was lucky because:
(i) all of its rulers were competent,
(ii) most of its people were competent.
(iii) most of its rulers were competent.
(iv) only a few of its people were incompetent.

Answer the following questions briefly: 1 x 6 = 6
(e) Who is the earliest king of Mewar mentioned in the passage?
(f) What was Rana Kumbha’s contribution to the glory of Mewar?
(g) What does the writer find worth admiration in the people of Mewar?
(h) How could art and literature flourish in Mewar?
(i) How did the rulers show that they cared for their subjects?
(j) What does the erection of Vijaya Stambha and Kirti Stambha in the same fort signify?

(k) Find words from the passage which mean the same as each of the following:
(i) surprising (para 4)
(ii) evidence (para 4) 1 x 2 = 2
Answer:
(a) (iv) both (ii) & (iii)
(b) (iii) its small area and small population.
(c) (i) the flag of Mewar seemed to be lowered.
(d) (ii) most of its rulers were competent.
(e) The earliest king of Mewar mentioned in the passage is Bappa Rawal.
(f) Rana Kumbha gave new stature to the kingdom through victories and developmental work. During his reign, literature and art progressed extraordinarily.
(g) The writer finds the cheerful and liberal character of the people of Mewar and their loving and pleasant nature worth admiration.
(h) Art and literature flourished in Mewar as the ambience of Rana Sanga’s Kingdom was conducive to the creation of high-quality work of art and literature. Also, the people of Mewar led peaceful and prosperous lives for a long period of time which helped art flourish. The rulers were inherently inclined towards art and culture.
(i) The rulers of Mewar created an atmosphere where cooperation existed between the nobles and subjects. The people of Mewar lived peacefully and had prosperous lives. Not just the palaces but public utility buildings built by the rulers had astonishing workmanship.
(j) The erection of Vijaya Stambha and Kirti Stambha in the same fort signifies how liberal the then administration of Mewar was which allowed people from other communities and kingdoms to come and carry out construction work. It also depicts the proximity between the king and the subjects of Mewar.
(k) (i) astonishing (ii) testimony

Q.9. Read the passage given below: (Delhi)

1. To ensure its perpetuity, the ground is well held by the panther both in space and in time. It enjoys a much wider distribution over the globe than its bigger cousins, and procreates sufficiently profusely to ensure its continuity for all time to come.

2. There seems to be no particular breeding season of the panther, although its sawing and caterwauling is more frequently heard during winter and summer. The gestation period is about ninety to hundred days (Whipsnade, ninety-two days). The litter normally consists of four cubs, rarely five. Of these, generally two survive and not more than one reaches maturity. I have never come across more than two cubs at the heels of the mother. Likewise, graziers in the forest have generally found only two cubs hidden away among rocks, hollows of trees, and other impossible places.

3. Panther cubs are generally in evidence in March. They are born blind. This is a provision of Nature against their drifting away from the place of safety in which they are lodged by their mother, and exposing themselves to the danger of their being devoured by hyenas, jackals, and other predators. They generally open their eyes in about three to four weeks.

4. The mother alone rears its cubs in seclusion. It keeps them out of the reach of the impulsive and impatient male. As a matter of fact the mother separates from the male soon after mating and forgets all about their tumultuous union. The story that the male often looks in to find out how the mother is progressing with her cubs has no foundation except in what we wish it should do at least.

5. The mother carries its cubs about by holding them by the scruff of their neck in its mouth. It trains them to stalk and teaches them how to deliver the bite of death to the prey. The cubs learn to treat all and sundry with suspicion at their mother’s heels. Instinctively the cubs seek seclusion, keep to cover and protect their flanks by walking along the edge of the forest.

6. I have never had an opportunity to watch mother panther train its cubs. But in Pilibhit forests, I once saw a tigress giving some lessons to its little ones. I was sitting over its kill at Mala. As the sun set, the tigress materialized in the twilight behind my machan. For about an hour, it scanned and surveyed the entire area looking and listening with the gravest concern. It even went to the road where my elephant was awaiting my signal. The mahout spotted it from a distance and drove the elephant away.

7. When darkness descended upon the scene and all was well and safe, the tigress called its cubs by emitting a low halogen. The cubs, two in number and bigger than a full-grown cat, soon responded. They came trotting up to their mother and hurried straight to the kill in indecent haste. The mother spat at them so furiously that they doubled back to its heels immediately. Thereafter, the mother and its cubs sat undercover about 50 feet (15 m) away from the kill to watch, wait, look, and listen. After about half an hour’s patient and fidget less vigil the mother seemed to say ‘paid for’. At this signal, the cubs cautiously advanced, covering their flanks, towards the kill. No longer did they make a beeline for it, as they had done before.

8. The mother sat watching its cubs eat, and mounted guard on them. She did not partake of the meal.

On the basis of your understanding of the above passage complete the statements given below with the help of options that follow: 1 x 2 = 2

(a) To protect its cubs the mother panther hides them:
(i) among rocks
(ii) in the branches of the trees
(iii) behind the tree trunks
(iv) at its heels

(b) The male panther.
(i) is protective of its cubs
(ii) trains its cubs
(iii) watches the progress of the mother
(iv) is impulsive and impatient

(c) How many cubs does the mother panther rarely deliver?
(d) What may happen if the panther cubs are not born blind?
(e) Why did the mahout drive his elephant away?
(f) Why did the tigress spit at its cubs?
(g) From the narrator’s observation, what do we learn about the nature of the tigress?
(h) Why does the panther not face the risk of extinction?

(i) Find words from the passage which mean the same as each of the following:
(i) moving aimlessly (para 3)
(ii) came down/fell (para 7) 1 x 2 = 2
Answer:
(a) (i) among rock
(b) (iv) is impulsive and impatient
(c) The mother panther rarely delivers five cubs.
(d) If the panther cubs are not born blind they may drift away from the place of safety in which they are lodged by their mother and expose themselves to the danger of being devoured by hyenas, jackals and other predators.
(e) The mahout did not want to disturb the tigress. Thus, on spotting the tigress the mahout drove his elephant away because he knew the presence of his elephant there would deter the tigress from summoning her cubs to devour the kill.
(f) The mother was furious with its cubs and so she spit at them to discipline and train them to come back to her heels. The mother spat at them as they hurried straight to the kill in the indecent haste without watching, waiting, looking and listening.
(g) From the narrator’s observation, we get to know the tigress was extremely concerned about the safety and wellbeing of her cubs. She was protective and caring and yet remained on guard and was vigilant.
(h) The panther does not face the risk of extinction because it enjoys a much wider distribution over the globe and procreates sufficiently profusely to ensure its continuity for all times to come.
(i)
(i) drifting
(ii) descended

Q.10. Read the passage carefully: (Comput. Delhi)

1. Can you imagine a college without walls, professors or classrooms? Educator Bunker Roy can. More than 40 years ago, Roy, now 69, founded the Barefoot College in Tilonia, Rajasthan. His school admits rural women, often grandmothers and teaches them the basics of solar engineering and freshwater technology. His efforts have yielded enormous benefits. When the women return to their homes, they are trained enough to provide their communities, some of the world’s most lonely places, with electricity and clean water. They also gain something important: a newfound self-confidence. The Barefoot model has already been used to empower women throughout Asia, Africa, and Latin America. Last year, former President, Bill Clinton presented Roy with a Clinton Global Citizen Award, which honours leaders who are solving the world’s problems in effective ways.

2. If you go all over the world, to very remote villages, you will often find only very old people and very young people. The men have already left. So two ideas were put into practice in order to make the Barefoot Model work. First it was declared that men are untrainable, restless, always ready to move, ambitious, and they all want a certificate to show for their efforts. And the moment you give one of them a certificate, he leaves the village looking for a job in the city. That is how, the simple, practical solution of training grandmothers came up. They are sympathetic, tolerant, willing to learn, and patient. All the qualities you need are there. And the second idea was not to give out certificates. Because the moment a certificate is given, a woman, like a man, will see it as a passport for leaving rural areas and going to urban areas to find a job.

3. Barefoot College follows the lifestyle of Mahatma Gandhi: Students eat, sleep, and work on the floor. They can work for 20 years or they can go home the next day. As of today, 604 women solar engineers from 1083 villages in 63 countries have been trained. The engineers have given solar power to 45,000 houses. These were done by women who had never left their homes before. They hate the idea of leaving their families and getting on a plane. When they reach India, sometimes after 19 hours of travel, they are faced with strange food, strange people, and a strange language. All the training is done in sign language. Yet in six months, they will know more about solar engineering than most university graduates. Some women face problems at home for attending college. Most of the husbands do not like their wives going to these colleges and tell them not to come back if they do so. But, on her return when she is able to help provide her village with solar electricity, her husband wants her to get back home. The respect she now has is enormous and she considers herself no less than solar engineers. Bunker Roy dreams of providing the world’s 47 least developed countries with Barefoot College trained grandmothers and solar electrify more than 1,00,000 houses.

On the basis of your understanding of the above passage answer the questions that follow with the help of the given options: 1 x 2 = 2

(a) Why did the promoter of Barefoot Model decide to train grandmothers?
(i) Men do not want to be trained.
(ii) Grandmothers were patient, willing to learn and tolerant.
(iii) Men are lazy, want to make money.
(iv) Men and women are not skilled.

(b) The attitude of the husbands to their wives on their return from training is different because
(i) of the respect they gain from the villagers
(ii) they were away for a long time
(iii) they will again be looked after
(iv) they will not go back

Answer the following questions briefly:lx6=6
(c) How is Barefoot College different from other colleges?
(d) What did the women gain from the college apart from technology?
(e) Why were certificates not given out after training?
(f) What are the difficulties the women have to face during their travel and their life in Tilonia, Rajasthan?
(g) How do the women consider themselves professionally, after their training?
(h) What is the narrator’s dream about solar electrification?

Find words from the passage which mean the same as the following: 1 x 2 = 2
(i) many/great in size (para 1)
(ii) far off (para 2)
Answer:
(a) (ii) Grandmothers were patient, willing to learn and tolerant.
(b) (i) of the respect they gain from the villagers.
(c) Barefoot college is a college without walls, professors or classrooms. It admits rural women, often grandmothers. Students eat, sleep and work on the floor. They can work for 20 years or they can go home the next day.
(d) Apart from technology the women gain a newfound self-confidence and are trained enough to provide their communities, some of the world’s most lonely places, with electricity and clean water.
(e) Certificates were not given out after training because the moment a certificate is given a person sees it as a passport for leaving rural areas and going to urban areas to find a job.
(f) The women hate the idea of leaving their families and getting on a plane. When they reach India, sometimes after 19 hours of travel they are faced with strange food, strange people and a strange language. Some women also face problems at home for attending the college as their husbands do not like their wives going to these colleges.
(g) After their training, the women professionally consider themselves no less than solar engineers. They are able to help provide their village with solar electricity.
(h) The narrator’s dream about solar electrification is to provide the world’s 47 least developed countries with Barefoot college-trained grandmothers and solar electrify more than 1,00,000 houses.
(i) many/great in size enormous
(j) far off remote

Q.11. Read the passage given below and answer the questions that follow: (Delhi, All India 2017)

1. We sit in the last row, bumped about but free of stares. The bus rolls out of the dull crossroads of the city, and we are soon in open countryside, with fields of sunflowers as far as the eye can see, their heads all facing us. Where there is no water, the land reverts to desert. While still on level ground, we see in the distance the tall range of the Mount Bogda, abrupt like a shining prism laid horizontally on the desert surface. It is over 5,000 metres high, and the peaks are under permanent snow, in powerful contrast to the flat desert all around. Heaven Lake lies part of the way up this range, about 2,000 metres above sea level, at the foot of one of the higher snow-peaks.

2. As the bus climbs, the sky, brilliant before, grows overcast. I have brought nothing warm to wear: it is all down at the hotel in Urumqi. Rain begins to fall. The man behind me is eating overpoweringly smelly goat’s cheese. The bus window leaks inhospitably but reveals a beautiful view. We have passed quickly from desert through arable land to pasture, and the ground is now green with grass, the slopes dark with pine. A few cattle drink at a clear stream flowing past moss-covered stones; it is a Constable landscape. The stream changes into a white torrent, and as we climb higher I wish more and more that I had brought with me something warmer than the pair of shorts that have served me so well in the desert. The stream (which, we are told, rises in Heaven Lake) disappears, and we continue our slow ascent. About noon, we arrive at Heaven Lake and look for a place to stay at the foot, which is the resort area. We get a room in a small cottage, and I am happy to note that there are thick quilts on the beds.

3. Standing outside the cottage we survey our surroundings. Heaven Lake is long, sardine shaped and fed by snowmelt from a stream at its head. The lake is an intense blue, surrounded on all sides by green mountain walls, dotted with distant sheep. At the head of the lake, beyond the delta of the inflowing stream, is a massive snowcapped peak which dominates the vista; it is part of a series of peaks that culminate, a little out of view, in Mount Bogda itself.

4. For those who live in the resort, there is a small mess hall by the shore. We eat here sometimes, and sometimes buy food from the vendors outside, who sell kabab and naan until the last buses leave. The kababs, cooked on skewers over charcoal braziers, are particularly good; highly spiced and well done. Horse’s milk is available too from the local Kazakh herdsmen, but I decline this. I am so affected by the cold that Mr. Cao, the relaxed young man who runs the mess, lends me a spare pair of trousers, several sizes too large but more than comfortable. Once I am warm again, I feel a predinner spurt of energy dinner will be long in coming and I ask him whether the lake is good for swimming in.

5. “Swimming ?” Mr. Cao says. “You aren’t thinking of swimming, are you?”

6. ” I thought I might,” I confess. “What’s the water like?”

7. He doesn’t answer me immediately, turning instead to examine some receipts with exaggerated interest. Mr. Cao, with great offhandedness, addresses the air. “People are often drowned here,” he says. After a pause, he continues. “When was the last one ?” This question is directed at the cook, who is preparing a tray of mantou (squat white steamed bread rolls), and who now appears, wiping his doughy hand across his forehead. “Was it the Beijing athlete ?” asks Mr. Cao.

On the basis of your understanding of the above passage, complete the statements given below with the help of the options that follow: 1 x 4 = 4
(a) One benefit of sitting in the last row of the bus was that:
(i) the narrator enjoyed the bumps.
(ii) no one stared at him
(iii) he could see the sunflowers.
(iv) he avoided the dullness of the city.

(b) The narrator was travelling to:
(i) Mount Bogda
(ii) Heaven Lake
(iii) a 2,000 m high snow peak
(iv) Urumqi

(c) On reaching the destination the narrator felt relieved because:
(i) he had got away from the desert
(ii) a difficult journey had come to an end
(iii) he could watch the snow peak
(iv) there were thick quilts on the beds

(d) Mount Bogda is compared to:
(i) a horizontal desert surface
(ii) a shining prism
(iii) a Constable landscape
(iv) the overcast sky

Answer the following questions briefly: 1 x 6 = 6
(e) Which two things in the bus made the narrator feel uncomfortable?
(f) What made the scene look like a Constable landscape?
(g) What did he regret as the bus climbed higher?
(h) Why did the narrator like to buy food from outside?
(i) What is ironic about the pair of trousers lent by Mr. Cao?
(j) Why did Mr. Cao not like the narrator to swim in the lake?
(k) Find words from the passage which mean the same as each of the following:
(i) sellers (para 4)
(ii) increased (para 7) 1 x 2 = 2
Answer:
(a) (ii) no one started at him.
(b) (ii) Heaven lake.
(c) (iv) there were thick quilts on the bed.
(d) (ii) a shining prism.
(e) The two things that made the narrator feel uncomfortable in the bus were that the man behind him was eating overpoweringly smelly goat’s cheese and the bus window leaked inhospitably. Moreover, he had to endure a bumpy ride.
(f) The pasture green with grass, the slopes dark with pine and the sight of a few cattle drinking at a clear stream flowing past moss-covered stones made the scene look like a Constable landscape.
(g) The stream changed into a white torrent as the bus climbed higher and the narrator regretted that he had not brought something warmer than a pair of shorts with him.
(h) The narrator likes to buy kababs and naan from outside as the kababs, cooked on skewer over charcoal braziers, are rather good, highly spiced and well done.
(i) Though the pair of trousers lent by Mr. Cao was several sizes too large for the narrator but they were more than comfortable for him.
(j) Mr. Cao did not like the narrator to swim in the lake as he says many people often drowned in it.
(k)
(i) sellers vendors
(ii) increased exaggeratedly

Q.12. Read the passage given below: (Delhi)

1. Thackeray reached Kittur along with a small British army force and a few of his officers. He thought that the very presence of the British on the outskirts of Kittur would terrorise the rulers and people of Kittur, and that they would lay down their arms. He was quite confident that he would be able to crush the revolt in no time. He ordered that tents be erected on the eastern side for the fighting forces, and a little away on the western slopes tents be put up for the family members of the officers who had accompanied them. During the afternoon and evening of 20th October, the British soldiers were busy making arrangements for these camps.

2. On the 21st morning, Thackeray sent his political assistants to Kittur fort to obtain a written assurance from all the important officers of Kittur rendering them answerable for the security of the treasury of Kittur. They, accordingly, met Sardar Gurusiddappa and other officers of Kittur and asked them to comply with the orders of Thackeray. They did not know that the people were in a defiant mood. The commanders of Kittur dismissed the agent’s orders as no documents could be signed without sanction from Rani Chennamma.

3. Thackeray was enraged and sent for the commander of the Horse Artillery, which was about 100 strong, and ordered him to rush his artillery into the fort and capture the commanders of the Desai’s army. When the Horse Artillery stormed into the fort, Sardar Gurusiddappa, who had kept his men on full alert, promptly commanded his men to repel and chase them away. The Kittur forces made a bold front and overpowered the British soldiers.

4. In the meanwhile, the Desai’s guards had shut the gates of the fort and the British Horse Artillerymen, being completely overrun and routed, had to get out through the escape window. Rani’s soldiers chased them out of the fort, killing a few of them until they retreated to their camps on the outskirts.

5. A few of the British had found refuge in some private residences, while some were hiding in their tents. The Kittur soldiers captured about forty persons and brought them to the palace. These included twelve children and a few women from the British officers’ camp. When they were brought in the presence of the Rani, she ordered the soldiers to be imprisoned. For the women and children she had only gentleness and admonished her soldiers for taking them into custody. At her orders, these women and children were taken inside the palace and given food and shelter. Rani came down from her throne, patted the children lovingly and told them that no harm would come to them.

6. She, then, sent word through a messenger to Thackeray that the British women and children were safe and could be taken back any time. Seeing this noble gesture of the Rani, he was moved. He wanted to meet this gracious lady and talk to her. He even thought of trying to persuade her to enter into an agreement with the British to stop all hostilities in lieu of an inam (prize) of eleven villages. His offer was dismissed with a gesture of contempt. She had no wish to meet Thackeray. That night she called Sardar Gurusiddappa and other leading Sardars, and after discussing all the issues came to the conclusion that there was no point in meeting Thackeray who had come with an army to threaten Kittur into submission to British sovereignty.

On the basis of your understanding of the above passage, complete the statements given below with the help of options that follow:

(a) Thackeray was a/an
(i) British tourist
(ii) army officer
(iii) advisor to the Rani of Kittur
(iv) treasury officer

(b) British women and children came to Kittur to
(ii) enjoy life in tents
(iii) stay in the palace
(iv) give company to officers 1 x 2 = 2

Answer the following questions briefly: 1 x 6 = 6

(c) Why did Thackeray come to Kittur?
(d) Why did the Kittur officials refuse to give the desired assurance to Thackeray?
(e) What happened to the Horse Artillery?
(f) How do we know that the Rani was a noble queen?
(g) How, in your opinion, would the British women have felt after meeting the Rani?
(h) Why did the Rani refuse to meet Thackeray?

(i) Find words from the passage which mean the same as the following:
(i) entered forcibly (para 3)
(ii) aggressiv^refusing to obey (para 2)
Answer:
(a) (ii) army officer
(b) (iv) give company to officers
(c) Thackeray had come with an army to threaten the rulers and people of Kittur into submission to British sovereignty and to crush the revolt.
(d) The Kittur officials were in a defiant mood and declared that no document could be signed without sanction from Rani Chennamma so they refused to give the desired assurance to Thackeray.
(e) The Horse Artillery was repelled and chased away by Sardar Gurusiddappa’s men who had been kept on high alert by him. The Desai’s guards shut the gates of the fort and the British Horse Artillerymen were thus completely overrun and routed and had to get out through the escape window.
(f) The Rani was indeed a noble queen. When the forty captured persons were brought in her presence, she ordered only the soldiers to be imprisoned. For the women and children, she had only gentleness and admonished her soldiers for taking them into custody. She then ordered that they be given food and shelter. Moreover, she patted the children lovingly and told them that no harm would come to them.
(g) The British women must have felt very relieved and happy to meet the Rani who was a kind and gentle queen. They must have been overwhelmed by her noble and gracious gesture.
(h) The Rani felt there was no point in meeting Thackeray who had come with an army to threaten Kittur into submission to British sovereignty.

(i)
(i) entered forcibly stormed
(ii) aggressive/refusing to obeydefiant

Q.13. Read the passage given below and answer the questions that follow: (Comput. Delhi, Comput. All India)

The Art Of Living
1. The art of living is learnt easily by those who are positive and optimistic. From humble and simple people to great leaders in history, science or literature, we can learn a lot about the art of living, by having a peep into their lives. The daily routines of these great men not only reveal their different, maybe unique lifestyles but also help us learn certain habits and practices they followed. Here are some; read, enjoy and follow in their footsteps as it suits you.

2. A private workplace always helps. Jane Austen asked that a certain squeaky hinge should never be oiled so that she always had a warning whenever someone was approaching the room where she wrote. Willliam Faulkner, lacking a lock on his study door, detached the doorknob and brought it into the room with him. Mark Twain’s family knew better than to breach his study door they would blow a horn to draw him out. Graham Green went even further, renting a secret office; only his wife knew the address and the telephone number. After all, every one of us needs a workplace where we can work on our creation uninterruptedly. Equally, we need our private space too!

3. A daily walk has always been a source of inspiration. For many artists, a regular stroll was essentially a creative inspiration. Charles Dickens famously took three hour walks every afternoon, and what he observed on them fed directly into his writing. Tchaikovsky made do with a two-hour jaunt but wouldn’t return a moment early; convinced that doing so would make him ill. Ludwig van Beethoven took lengthy strolls after lunch, carrying a pencil and paper with him in case inspiration struck.

Nineteenth-century composer Erik Satie did the same on his long hikes from Paris to the working-class suburb where he lived, stopping under streetlamps to jot down ideas that came on his journey; it’s rumoured that when those lamps were turned off during the war years, his music declined too. Many great people had a limited social life too. One of Simone de Beauvoir’s close friends puts it this way. “There were no receptions, parties. It was an uncluttered kind of life, a simplicity deliberately constructed so that she could do her work.” To Pablo, the idea of Sunday was an “at home day”.

4. The routines of these thinkers are difficult. Perhaps it is because they are so unattainable. The very idea that you can organize your time as you like is out of reach for most of us, so I’ll close with a toast to all those who worked with difficulties. Like Francine Prose, who began writing when the school bus picked up her children and stopped when it brought them back; or T.S. Eliot, who found it much easier to write once he had a day job in a bank than he had as a starving poet and even F. Scott Fitzgerald, whose early books were written in his strict schedule as a young military officer. Those days were not as interesting as the nights in Paris that came later, but they were much more productive and no doubt easier on his liver.

5. Being forced to follow someone else’s routine may irritate, but it makes it easier to stay on the path. Whenever we break that trail ourselves or take an easy path of least resistance, perhaps what’s most important is that we keep walking.

On the basis of your understanding of the above passage, complete each of the statements given below with help of the options that follow: 1 x 4 = 4

(a) The passage is about:
(i) how to practise walking
(ii) walking everyday
(iii) the life of a genius
(iv) what we can learn from the routines of geniuses

(b) The writers in the past:
(i) followed a perfect daily routine
(ii) enjoyed the difficulties of life
(iii) can teach us a lot
(iv) wrote a lot in books

(c) In their daily routines:
(i) they had unique lifestyles
(ii) they read books and enjoyed them
(iii) they did not get any privacy
(iv) they did not mind visitors

(d) Some artists resorted to walking as it was:
(i) an exercise
(ii) a creative inspiration
(iii) essential for improving their health
(iv) helpful in interaction with others

On the basis of your understanding of the above passage, answer the following questions: 1×2=2
(e) What did Jane Austen like?
(f) Why do you think Graham Green hired a secret office?
(g) What was the rumour about Erik Satie’s productivity?
(h) How did her limited social life affect Simone de Beauvoir?
(i) In what way did T.S. Eliot’s day job help him to write?
(j) What makes it easier for one to stay on the path?

Find words from the passage which mean the same as each of the following:
(i) glance/look (para 1)
(ii) noisy (para 2) 1×2=2
(a) (iv) what we can learn from the routines of geniuses.
(b) (i) followed a perfect daily routine.
(c) (i) they had unique lifestyles.
(d) (ii) a creative inspiration.
(e) Jane Austen liked a private workplace. She did not want a certain squeaky hinge to be oiled so that it warned her whenever someone was approaching the room where she wrote.
(j) Graham Green hired a secret office because he needed a workplace where he could work on his creation uninterruptedly.
(g) Erik Satie used to stop under streetlamps on his long hikes from Paris to the place where he lived in the working-class suburb, and jot down ideas that came to his mind during the walk. It is rumoured that when those street lights were turned off during the war years, his music too declined.
(h) Simone de Beauvoir led a limited social life. There were no receptions or parties. It was an uncluttered kind of life, a deliberately constructed simplicity so that she could concentrate on her work.
(i) T.S. Eliot found it much easier to write once he had a day job in a bank than he could as a starving poet.
(j) Following a routine and leading an organized life may be irritating but it makes it easier for one to stay on the path.
(i) glance/look: peep
(ii) noisy: squeaky

Q.14. Read the passage given below: (Comput. Delhi)

1. Ammon means “fragrant spice plant” in Arabic and Hebraic and in Italian, canella means “little tube”. These are a few of the many terms given to the popular spice known as cinnamon. Dating back as far as 2800 B.C., Chinese writings describe cinnamon as an important part of the culture, so much so that over the years this spice was traded right up there with silver. Nowadays we find it in sweetened cereals, baked goods and sprinkled on various foods such as yoghurt. Yet, many do not consider its wealth of healing capabilities including the potential as a weight loss remedy.

2. Cinnamon is derived from the inner bark of the cinnamon tree grown and harvested mostly in Sri Lanka but also found in Brazil, Indonesia, Vietnam, China and Burma. After a cinnamon tree grows for about six to eight years it is cut down leaving a stump to allow it to grow again making it a very sustainable practice. It is then stripped from the bark, dried and packaged as sticks for export.

3. Several studies have been published regarding the weight loss properties of cinnamon which include its unique ability to be used for type 2 diabetes which is a disease often resulting from obesity. When eaten, the spice seems to slow down glucose absorption within the intestines while stimulating insulin production. This normalizes blood glucose levels which in turn can indirectly decrease weight gain.

4. “The results of study demonstrate that intake of 1, 3 or 6 g of cinnamon per day reduces serum glucose, triglyceride, LDL cholesterol and total cholesterol in people with type 2 diabetes and suggest that the inclusion of cinnamon in the diet of people with type 2 diabetes will reduce risk factors associated with diabetes and cardiovascular diseases”.

5. A study from the Department of Family and Consumer Sciences, called “Effect of ground cinnamon on after-meal blood glucose level in normal-weight and obese adults” found that cinnamon may be effective in moderating post-meal glucose level in normal weight and obese adults.

6. Columbia University nutritionist Tara Ostrowe comments to Reader’s Digest on the benefits of this spice: ” Cinnamon really is the new skinny food ……………………………. Scientists already credit cinnamon with helping lower blood sugar concentration and improving insulin sensitivity. When less sugar is stored as fat, this translates into more help for your body when it comes to weight loss”.

7. Talk to your doctor about adding cinnamon daily into your healthy diet and exercise program. Add it to your tea, oatmeal, fruit, toast or anything else you can think of, as a small amount will go a long way and potentially assist in your weight loss mission.

On the basis of your understanding of the above passage, complete each of the statements given below with the help of options that follow: 1 x 2 = 2
(a) Cinnamon is called _______ in Hebraic.
(i) little tube
(ii) canella
(iii) Ammon
(iv) a fragrant spice plant

(b) In ‘Yet, many do not consider its wealth of healing capabilities ‘ The writer refers to the word ‘wealth’ to:
(i) the payment in silver
(ii) the cost of cinnamon
(iii) health of people
(iv) the healing power of cinnamon

On the basis of your understanding of the above passage, answer the following: 1 x 6 = 6
(c) Which country produces most of the cinnamon in the world?
(d) Pick out the phrase from the passage (para 1) which shows that cinnamon was much in demand in China.
(e) From what is cinnamon derived?
(f) How is it used today?
(g) In what way does cinnamon help people suffering from type2 diabetes?
(h) How is cinnamon helpful in weight loss?

Find words from the passage which mean the same as the following:
(i) sweet-smelling (para 1)
(ii) reaped/cultivated (para 2)
Answer:
(a) (ii) Ammon
(b) (iv) the healing power of cinnamon.
(c) Sri Lanka produces most of the cinnamon in the world.
(d) Chinese writings describe cinnamon as an important part of the culture, so much so that over the years this spice was ‘traded right up there with silver’.
(e) Cinnamon is derived from the inner bark of the Cinnamon tree from which it is stripped, dried and packaged as sticks.
(f) Nowadays we find cinnamon in sweetened cereals, baked goods and it is sprinkled on various foods such as yoghurt.
(g) Cinnamon helps people suffering from type 2 diabetes. It seems to slow down the glucose absorption within the intestines while stimulating insulin production. This normalizes blood glucose levels which indirectly decreases weight gain.
(h) Scientists credit cinnamon with helping lower blood sugar concentration and improving insulin sensitivity. When less sugar is stored as fat, it helps the body in terms of weight loss.
(i) sweet-smelling fragrant
(ii) reaped/cultivated harvested

Important Questions for Class 12 English

The post Reading Comprehension Class 12 Passages, Exercises, Worksheets appeared first on Learn CBSE.

CBSE Class 12 Sanskrit व्याकरणम् उपपदविभक्ति प्रयोग

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CBSE Class 12 Sanskrit व्याकरणम् उपपदविभक्ति प्रयोग

CBSE Class 12 Sanskrit व्याकरणम् उपपदविभक्ति प्रयोग 1
CBSE Class 12 Sanskrit व्याकरणम् उपपदविभक्ति प्रयोग 2

अभ्यासार्थ

1. निम्नवाक्येषु मजूषायां प्रदत्तानां शब्दानाम् सहायतया रिक्तस्थानपूर्तिः कर्तव्यः

1. (i) तृणानि भूमिः उदकं चतुर्थी च ………………….. वाक्; सताम् गेहे एते कदाचन न उच्छिद्यन्ते।
(ii) …………… ऋषयः येन अत्र आक्रमन्ति तत् सत्यस्य परमं निधानं भवति।
(iii) …………… पन्थाः सत्येन एव विततोऽस्ति।
(iv) यः तु ………………… भूतानि आत्मनि एव अनुपश्यति।
(v) कवयः तत् पथः ………………… वदन्ति।
मञ्जूषा – देवयानः, सुनृता, सर्वाणि, दुर्गम्, आप्तकामाः।
उत्तर:
(i) सूनृता
(ii) आप्तकामाः
(iii) देवयान:
(iv) सर्वाणि
(v) दुर्गम्

2. (i) ………………… पृथिवी अन्यथा कथम् आकाशे तिष्ठेत्?
(ii) शिवराजविजयस्य प्रारम्भेऽपि सूर्योदयस्य रम्यम् अद्भुतं ………………… वर्णनम् उपलभ्यते।
(iii) निजपर्णकुटीरात् निर्गत्य ………………… बटुः सूर्यस्य महिमानं वर्णयति।
(iv) इदम् भाषायाः सौन्दर्यस्य अपि ………….. उदाहरणम् अस्ति।
(v) अरुण एष पूर्वस्यां ………………… मरीचिमालिनः प्रकाशोऽस्ति।
मञ्जूषा – विज्ञानमयं, भगवतः, निराधारा, गुरुसेवनपटुः, उत्कृष्टम्।
उत्तर:
(i) निराधारा
(ii) उत्कृष्टम्
(iii) गुरुसेवनपटुः
(iv) विज्ञानमयं
(v) भगवतः

3. (i) अयमेव भगवान् ………………… ऋतूणाम् कारणाम् अस्ति।
(ii) वेदाः एतस्य एव ………………… सन्ति ।
(iii) विश्वेषाम् एषः सूर्यः एव ………………… अस्ति।
(iv) एषः एव ………………… दक्षिणं च अयनम् अङ्गीकरोति।
(v) वनस्पतीनां सूर्यं विना अस्तित्वं ………………… भविष्यति।
मञ्जूषा – उत्तरम्, वन्दिनः, षण्णाम्, समाप्तं, प्रणम्यः।
उत्तर:
(i) षण्णाम्
(ii) वन्दिनः
(iii) प्रणम्यः
(iv) उत्तरम्
(v) समाप्तं

4. (i) चन्द्रगुप्तस्य राज्ये ………………… चाणक्य: देशहिताय एव प्रयत्नं करोति।
(ii) चाणक्यः ………………… चन्द्रगुप्तस्य आदेशस्य उल्लङ्घनं कर्तुमपि उत्सहते स्म।
(iii) सदैव राष्ट्रचिन्ता ………………… भवति।
(iv) भोः भोः प्रासादाधिकृताः पुरुषा ………………… चन्द्रगुप्तः वः विज्ञापयति।
(v) अतः सुगाङ्गप्रासादस्य उपरि ………………… प्रदेशाः संस्क्रियन्ताम्।
मजूषा – देवः, स्थिताः, अमात्यः, गरीयसी, सम्राजः।
उत्तर:
(i) अमात्यः
(ii) सम्राजः
(iii) गरीयसी
(iv) देवः
(v) स्थिताः

5. (i) आः किम् एतेन व ………………… कथाप्रसङ्गेन?
(ii) आर्य! देवस्य ………………… शासनम् अनुष्ठीयते।
(iii) अहो! राज्यं हि नाम ………………… नृपस्य कृते महत् कष्टदायकम्।
(iv) संसारे ………………… हि राजलक्ष्मीः भवति।
(v) आर्य! अथ अस्मद् वचनात् ………………… कुसुमपुरे कौमुदीमहोत्सव:?
मञ्जूषा – दुराराध्या, धर्मवृत्तिपरकस्य, प्राणहरेण, आघोषितः, इदम्।
उत्तर:
(i) प्राणहरेण
(ii) आघोषितः
(iii) धर्मवृत्तिपरकस्य
(iv) दुराराध्यो
(v) इदम्

6. (i) खलु आर्यचाणक्येन प्रेक्षकाणाम् ………………… चक्षुषो विषयः न अपहृतः?
(ii) देव! कः ………………… जीवितुकामो देवस्य शासनम् अतिवर्तेत?
(iii) ततः ………………… स्वभवनगतः चिन्तां नाटयन् चाणक्यः प्रविशति।
(iv) इत: शिष्यैः ………………… दर्भाणां स्तूपः दृश्यते।
(v) अत्र ………………… समिद्भिः अतिनमित: छदिप्रान्तः वर्तते।
मञ्जूषा – शुष्यमाणैः, अन्यः, अतिशयरमणीयः, आनीतानाम्, आसनस्थः।
उत्तर:
(i) अतिशयरमणीयः
(ii) अन्यः
(iii) आसनस्थः
(iv) आनीतानाम्
(v) शुष्यमाणैः

7.(i) अत: ………………… एतादृशैः जनैः राजा तृणवद् गण्यते।
(ii) आः ज्ञातम्। भवद्भिः एव प्रोत्साह्य ………………… वृषलः।
(iii) अलम् ………………… विनयेन।
(iv) किन्तु न कदाचित् आर्यस्य ………………… प्रवृत्तिः।
(v) वृषल! किम् अस्थाने ………………… प्रजा धनापव्ययः?
मजूषा – निस्पृहत्यागिभिः, अनेन, महान्, निष्प्रयोजना, कोपितः।
उत्तर:
(i) निस्पृहत्यागिभिः
(ii) कोपितः
(iii) अनेन
(iv) निष्प्रयोजना
(v) महान्

8. (i) आर्येण एव सर्वत्र ………………… मे बन्धनम् इव राज्यं, न राज्यम् इव।
(ii) वृषल! स्वयम् ………………… राज्ञाम् तु एते दोषाः सम्भवन्ति।
(iii) अथ ………………… अपि प्रयोजनं श्रोतुमिच्छसि तदपि कथयामि।
(iv) पितृवधात् क्रुद्धः ………………… महीयसा म्लेच्छबलेन परिवृत्त: मलयकेतुः अस्मान् अभियोक्तुम् उद्यतः।
(v) ………………… समये कि कौमुदी-महोत्सवेन; इति प्रतिषिद्धः।
मञ्जूषा – अस्मिन्, निरुद्धचेष्टस्य, राक्षसोपदेशप्रवणः, अनभियुक्तानां, अपरम्।
उत्तर:
(i) निरुद्धचेष्टस्य
(ii) अनभियुक्तानां
(iii) अपरम्
(iv) राक्षसोपदेशप्रवणः
(v) अस्मिन्

9. (i) प्रायः वयं ………………… सुखमेव पश्यामः।
(ii) येन ………………… कष्टानि आपतन्ति।
(iii) दूरदर्शी जन: ………………… पक्षान् विपक्षान् च गणयित्वा निर्णयं करोति।
(iv) अत्र अनागतविधाता ………………… मत्स्यः विपत्तेः निराकरणस्य उपायं करोति।
(v) ………………… जलाशये त्रयो मत्स्याः प्रतिवसन्ति स्म।
मञ्जूषा – अनेकानि, नामकः, कस्मिंश्चित्, तात्कालिकं, सर्वान्।
उत्तर:
(i) तात्कालिकं
(ii) अनेकानि
(iii) सर्वान्
(iv) नामकः
(v) कस्मिंश्चित्

10. (i) अथ कदाचित् तं जलाशयं दृष्ट्वा ………………… मत्स्यजीविभि: उक्तम्।
(ii) अहो, बहुमत्स्य: ………………… हृदः, कदापि न अस्माभिः अन्वेषित:।
(iii) तेषां तत् ………………… वचः समाकर्त्य अनागतविधाता सर्वान् मत्स्यान् अवदत्।
(iv) तद् रात्रौ अपि किञ्चित् ………………… सर: गम्यताम्।
(v) अशक्तै: ………………… शत्रोः प्रपलायनं कर्तव्यम्।
मञ्जूषा – कुलिशपातोपमं, बलिनः, अयं, गच्छद्भिः , निकटम्।
उत्तर:
(i) गच्छद्भिः
(ii) अयं
(iii) कुलिशपातोपमं
(iv) निकटम्
(v) बलिनः

11. (i) तेषाम् ………………… गतिः न भवेत्।
(ii) येषाम् अन्यत्र अपि विद्यमाना ………………… गतिः भवति। वे विद्वांसः कुलक्षयं न पश्यन्ति।
(iii) किं वाङमात्रेणापि ………………… एतत् सर: त्यक्तुं युज्यते।
(iv) मम अपि ………………… अभीष्टम् अस्ति ।
(v) वने ………………… अनाथः अपि जीवति।
मजूषा – पितृपैतामहादिकं, अन्या, एतत्, विसर्जितः, सुखावहा।
उत्तर:
(i) अन्या
(ii) सुखावहा
(iii) पितृपैतामहादिकं
(iv) एतत्
(v) विसर्जितः

12. (i) अहो! कीदृशी ………………… हिमानी राजते।
(ii) विद्यालयस्य वार्षिक पत्रिकायां पर्वतारोहणं ………………… छात्राणां चित्राणि प्रदर्शितानि।
(iii) शिक्षिका तेषाम् आग्रहं मत्वा लेहलद्दाख यात्राया: ………………… अनुभवं प्रस्तौति।
(iv) इदम् अभियानम् अतीव रोचकम् ………………… चासीत्।
(v) विपुल हिमराशिना धवला: एते ………………… गिरयः अतीव शोभन्ते।
मजूषा – साहासिकं, लद्दाख प्रदेशीयाः, इयं, रोमाञ्चकारिणम्, कुर्वतां।
उत्तर:
(i) इयं
(ii) कुर्वतां
(iii) रोमाञ्चकारिणम्
(iv) साहासिकं
(v) लद्दाख प्रदेशीयाः

13. (i) किं लद्दाख शब्दस्य कश्चिद् ………………… अर्थः भवति?
(ii) पश्यन्तु, कथं लद्दाखे ………………… नीलाकाश: छत्रवत् प्रतीयते।
(iii) ग्रीष्मे ………………… भूमि: धूसरवर्णा जायते।
(iv) अयं बौद्धधर्मस्य स्तूपः रात्रौ ………………… दीपेषु भव्यम् आलोकं वितरति।
(v) मान्ये! ………………… एतानि स्थानानि किं मठाः सन्ति?
मजूषा – नीलवर्णा, आस्तृतः, विशिष्टः, दीर्घ-दीर्घाणि, प्रज्वलितेषु।
उत्तर:
(i) विशिष्टः
(ii) आस्तृतः
(iii) नीलवर्णा
(iv) प्रज्वलितेषु
(v) दीर्घ-दीर्घाणि

14. (i) सिन्धुनद्याः पूर्वत: लेहनगरस्य दक्षिणपूर्व भागे एते ………………… बौद्धमठा: सन्ति।
(ii) भगवत: बुद्धस्य सप्तदशशताब्या: ………………… मूर्ति: पर्यटकानाम् आकर्षण-केन्द्रम् अस्ति।
(iii) लद्दाख स्थित राजप्रासादस्य ………………… भागे एकः विशाल: संग्रहालयो वर्तते।
(iv) मठेषु ………………… लेखा: भित्तिलेखाः च तिब्बत शैल्या: परिचायकाः सन्ति।
(v) ………………… हिमं गरिराजस्य शोभा सततं प्रवर्धयति।
मञ्जूषा – उत्कीर्णाः, विशालकाया, प्रख्याताः, आन्तरिके, घनीभूतं।
उत्तर:
(i) प्रख्याता:
(ii) विशालकाया
(iii) आन्तरिके
(iv) उत्कीर्णाः
(v) घनीभूतं

15. (i) विपदि ………………… मानवाः एतैः सुभाषितैः आश्वासनं प्राप्नुवन्ति।
(ii) संस्कृत वाङ्मयं सहस्रशः सुमधुरवचनैः सम्यग् ………………… वर्तते।
(iii) जीवनस्य प्रत्येक क्षेत्रे ………………… एतानि सुभाषितानि साहित्ये सुलभानि सन्ति।
(iv) अस्मिन् पाठे केषाञ्चित् ………………… मधुरवचनानां सङ्कलनम् अत्र प्रस्तूयते।
(v) येषां ………………… वदनम्, सदयम् हृदयं, सुधामुचः वाचः; ते केषां न वन्द्याः सन्ति।
मञ्जूषा – अमृतवर्षिणां, अलङ्कृतं, प्रसादसदनम्, प्रेरणाप्रदानानि, निपतिताः।
उत्तर:
(i) निपतिताः
(ii) अलङ्कृतं
(iii) प्रेरणाप्रदानानि
(iv) अमृतवर्षिणां
(v) प्रसादसदनम्

16. (i) ………………… पादपाः निपतन्ति।
(ii) ………………… जलम् च शुष्यति।
(iii) शठा: हि ………………… इषवः इव तथाविधान् असंवृत्ताङ्गान् प्रविश्य घ्नन्ति।
(iv) ………………… हि नृपेषु अमात्येषु च सर्वसम्पद: सदा रतिं कुर्वते।
(v) यदि ………………… मनः अस्ति तीर्थेन किम्?
मञ्जूषा – जलस्थानगतं, निशिताः, सुबद्धमूलाः, शुचि, अनुकूलेषु।
उत्तर:
(i) सुबद्धमूलाः
(ii) जलस्थानगतं
(iii) निशिताः।
(iv) अनुकूलेषु
(v) शुचि

17. (i) दारिद्र्ये ………………… सत्त्वम् भवति।
(ii) दरिद्रावस्थायां मित्राणाम् उपेक्षायाः कारणात् ………………… अनुभवः भवति।
(iii) दैन्ये दयितस्य मन: ………………… न भवति।
(iv) तस्य मित्रम् मैत्रेयः ………………… विपत्तौ चापि तस्य विश्वासपात्रम्, भवति।
(v) पूर्व ………………… चारुदत्तः पश्चात् उदारतावश दान कारणात् शीघ्रं दरिद्रो जातः।
मञ्जूषा- भ्रष्टं, धनवान्, विनोदप्रियः, कटुः, दुर्लभं।
उत्तर:
(i) दुर्लभं
(ii) कटुः
(iii) भ्रष्टं
(iv) विनोदप्रियः
(v) धनवान्

18. (i) अद्य खलु प्रत्यूष एव ………………… मे अक्षिणी बुभुक्षया पुष्कर-पत्रपतित-जलबिन्दु इव चञ्चलायेते।
(ii) किम् अस्त्यस्माकं गेहे ………………… प्रातराशः।
(iii) चिरंजीव, एवं ………………… भोजनानां दात्री भव।
(iv) तर्हि अस्मादृश योग्यं ………………… जनं निमन्त्रयितुम् इच्छामि।
(v) कुत्र नु खलु दरिद्र ………………… जनं लभेय।
मञ्जूषा – कोऽपि, शोभनानां, योग्यं, गेहान्निष्क्रान्तस्य, कञ्चिद्।
उत्तर:
(i) गेहान्निष्क्रान्तस्य
(ii) कोऽपि
(iii) शोभनानां
(iv) कञ्चिद्
(v) योग्य

19. (i) एषः आर्यचारुदत्तस्य ………………… आर्यमैत्रेयः इत: एवागच्छति।
(ii) ………………… अशनम् अशितव्यम् भविष्यति इति।
(iii) स एव इदानीम् अहम् ………………… चारुदत्तस्य दरिद्रतया अन्यत्र भुक्त्वा तस्यावासमेव गच्छामि।
(iv) तत: चारुदत्त: विदूषकः ………………… चेटी च प्रविशन्ति।
(v) भोः दारिद्र्यं खलु नाम ………………… पुरुषस्य सोच्छ्वासं मरणम्।
मञ्जूषा – चङ्गरिकाहस्ता, तत्रभवतः, मनस्विनः, वयस्यः, सम्पन्नम्।
उत्तर:
(i) वयस्यः
(ii) सम्पन्नम्
(iii) तत्रभवतः
(iv) चङ्गरिकाहस्ता
(v) मनस्विनः

20. (i) दानेन विपन्नविभवस्य, बहुलपक्षस्य चन्द्रस्य ज्योत्स्ना परिक्षय इव भवतः ………………… अयं दरिद्रभावः।
(ii) अहं खलु ………………… श्रियं न अनुशोचामि।
(iii) ………………… पुरुषस्य तु व्यसनं दारुणतरं मां प्रतिभाति।
(iv) ‘दानं श्रेयस्करं’ इति प्रत्ययादेव ममार्था: ………………… जाता:।
(v) धनविनाश दु:खस्य पुनः पुनः चिन्त्यमानस्य ………………… चिन्ताङकुरा: प्रादुर्भवन्ति।
मजूषा – नानाविधाः, रमणीयः, गुणरसज्ञस्य, क्षीणाः, नष्टाम्।
उत्तर:
(i) रमणीयः
(ii) नष्टाम्
(iii) गुणरसज्ञस्य
(iv) क्षीणाः
(v) नानाविधा:

21. (i) इदं विज्ञानजगत् ………………… अस्ति।
(i) गुरुवासरे सभागारे प्राचीन भारतीय विज्ञानम् अधिकृत्य ………………… सङ्गोष्ठी भविष्यति।
(iii) ………………… छात्राः यथासमयं कृपया तत्र आगच्छन्तु।
(iv) अस्माकं मध्ये परियोजना कार्ये ………………… अङ्कान् लब्धवन्तः छात्राः समुपस्थिताः।
(v) त्रिपुरविमानस्य ………………… भागः पृथिव्याः तले सञ्चरित।
मजूषा – सर्वे, आश्चर्यमयम्, प्रथमः, एका, सर्वोत्तमान्।
उत्तर:
(i) आश्चर्यमयम्
(ii) एका
(iii) सर्वे
(iv) सर्वोत्तमान्
(v) प्रथमः

22. (i) ………………… भागः तु स्वतः एवं आकाशे सञ्चरति।
(ii) विमानशास्त्रे ………………… लेपः अपि वर्णितः।
(iii) शालिनी भारतस्य ………………… सुश्रुतस्य शल्यक्रियामधिकृत्य अस्माकं ज्ञानवृद्धिं करिष्यति।
(iv) तर्हि ………………… सुश्रुत संहिता अवश्यमेव पठनीया।
(v) तत्र ………………… शल्यकार्यं वर्णितम्।
मञ्जूषा – सुश्रुतविरचिता, एतादृशः, तृतीयः, अष्टविधं, प्रमुखचिकित्सकस्य। |
उत्तर:
(i) तृतीयः
(ii) एतादृशः
(iii) प्रमुखचिकित्सकस्य
(iv) सुश्रुतविरचिता
(v) अष्ट्रविधं

23. (i) ऋषि सुश्रुतः प्रथमः ………………… आसीत्।।
(ii) खलु ………………… सुश्रुतस्य शल्यक्रिया बहुप्रसिद्धा अस्ति।
(iii) तर्हि वृक्षस्य दक्षिणे दिशि पुरुषद्वये ………………… सलिलं भविष्यति।
(iv) वस्तुतः वराहमिहिरेण स्वग्रन्थे ………………… विषयाः वर्णिताः।
(v) मिहिरेण ………………… सामग्री सङ्कलिता।
मञ्जूषा – भगवतः, बहुमूल्या, विविधाः, त्वक्प्रत्यारोपकः, स्वादु।
उत्तर:
(i) त्वक्प्रत्यारोपकः
(ii) भगवतः
(iii) स्वादु
(iv) विविधाः
(v) बहुमूल्या

24. (i) सङगणके तु ………………… शून्यञ्च द्वे एव संख्ये महत्त्वपूर्णे।
(ii) सूर्य प्रति ………………… पृथिवी 365.25 वारं प्रतिवर्ष भ्रमति।
(iii) नागार्जुनः ………………… अर्थशास्त्रात् अस्माकं ज्ञानवृद्धिं करिष्यति।
(iv) कौटिल्यस्य अर्थशास्त्रं तु ………………… ग्रन्थः अस्ति।
(v) अशुद्धं स्वर्णं ………………… सीसेन शोधयते।
मञ्जूषा – चतुर्गुणेन, कौटिल्यरचितात्, अद्भुतः, एकं, पूर्वाभिमुखा।
उत्तर:
(i) एकं
(ii) पूर्वाभिमुखा
(iii) कौटिल्यरचितात्
(iv) अद्भुतः
(v) चतुर्गुणेन

25. (i) इदानीं सुषमा भारते ………………… विचित्र शिल्पकलाकृतीनां परिचयं प्रदास्यति।
(ii) अहमदनगरे ………………… स्तम्भाः अस्माकं भारतस्य वैज्ञानिकानां गौरवगाथा वर्णयन्ति।
(iii) वयं ………………… सुषमाया: ज्ञानपूर्णेन सूचनाप्रदानेन अनुगृहीताः।
(iv) ये छात्रा: भारतीय विज्ञानक्षेत्रे ………………… उपलब्धी: आधृत्य अध्ययनं करिष्यति।
(v) सुचिन्दरं देवालये स्थिता: ………………… स्तम्भा: भारतीयवैज्ञानिकानां वर्णनं कुर्वन्ति।
मञ्जूषा – भवत्याः, विशिष्टा:, कम्पमानाः, सङ्गीतमयाः, उपलब्धानां।
उत्तर:
(i) उपलब्धानां
(ii) कम्पमानाः
(iii) भवत्याः
(iv) विशिष्टाः
(v) सङ्गीतमयाः

2. कारक एवम् उपपद विभक्तीनाम् उचितं प्रयोगं कृत्वा कोष्ठगत पदैः रिक्तस्थानपूर्तिः कर्तव्या

(i) बुद्धिः ………………… शुध्यति। (ज्ञान)
उत्तर:
ज्ञानेन

(ii) ………………… परितः एषा पृथिवी नित्यं परिभ्रमति। (किम्)
उत्तर:
कम्

(iii) ओषधीनाम् अपि ………………… विना अस्तित्वं समाप्तं भविष्यति। (सूर्य)
उत्तर:
सूर्यम्

(iv) ………………… निर्गत्य गुरुसेवनपटुः बटुः उदेष्यन्तं भास्वन्तं प्रणमति। (निजपर्णकुटीर)
उत्तर:
निजपर्णकुटीरात्

(v) ………………… एव कल्पभेदाः कृताः। (इदम्)
उत्तर:
अनेन

(vi) ………………… कृते एव प्रजाधनस्य सदुपयोग: स्यात्। (देश)
उत्तर:
देशस्य

(vii) एषः अंश: ‘मुद्राराक्षसम्’ इति ………………… सङ्कलितः। (संस्कृतनाटक)
उत्तर:
संस्कृतनाटकात्

(viii) अत: ………………… उपरि स्थिता: प्रदेशाः संस्क्रियन्ताम्। (सुगाङ्गप्रासाद)
उत्तर:
सुगाङ्गप्रासादम्

(ix) आः किम् एतेन; कः प्राणहरेण …………………? (कथाप्रसङ्ग)
उत्तर:
कथाप्रसङ्गेन

(x) राज्यं हि नाम ………………… कृते महत् कष्टदायकम्। (नृप)
उत्तर:
नृपस्य

(xi) कथं स्पर्धते ………………… सह दुरात्मा राक्षस:? (अस्मद्)
उत्तर:
अस्माभिः

(xii) विरम विरम ………………… दुर्व्यसनात्। (इदम्)
उत्तर:
अस्मात्

(xiii) आर्य! देव: चन्द्रगुप्त: आयं ………………… प्रणम्य विज्ञापयति। (शिरस्)
उत्तर:
शिरसा

(xiv) ………………… एव प्रोत्साह्य कोपित: वृषलः। (भवत्)
उत्तर:
भवद्भिः

(xv) अये ………………… अध्यास्ते वृषलः। (सिंहासन)
उत्तर:
सिंहासनम्

(xvi) अलम् अनेन ………………… । (विनय)
उत्तर:
विनयेन

(xvii) न ………………… अन्तरा चाणक्य: स्वप्नेऽपि चेष्टते। (प्रयोजन)
उत्तर:
प्रयोजनम्।

(xviii) आर्य वैहीनरे! ………………… सुवर्णशतसहस्रं दापय। (वैतालिक-द्विवचन)
उत्तर:
वैतालिकाभ्याम्

(xix) ………………… परिवृत: पर्वतक-पुत्र: मलयकेतुः अस्मान् अभियोक्तुम् उद्यत:। (म्लेच्छबल)
उत्तर:
म्लेच्छबलेन

(xx) जीवने ………………… विना सफलता न लभ्यते। (दूरदृष्टि)
उत्तर:
दूरदृष्टि

(xxi) अनागतविधाता प्रत्युत्पन्न मतिश्च ………………… सह निष्क्रान्तौ। (परिजन)
उत्तर:
परिजनैः

(xxii) तैः तं जलाशयम् ………………… सह निर्मत्स्यतां नीतम्। (यद्भविष्य)
उत्तर:
यद्भविष्येण

(xxiii) सर्वे छात्रा: ………………… उपेत्य यात्रायाः वृत्तान्तं ज्ञातुम् अनुरोधं कुर्वन्ति। (शिक्षिका)
उत्तर:
शिक्षिकाम्

(xxiv) ‘ताहथोक’ आदीनि ग्रीष्मपर्वाणि ………………… प्रति भक्तिभावं दर्शयन्ति। (बुद्ध)
उत्तर:
बुद्ध

(xxv) विपदि निपतिताः मानवाः एतैः ………………… आश्वासनं प्राप्नुवन्ति। (सुभाषित)
उत्तर:
सुभाषितैः

(xxvi) त्यागेन शीलेन गुणेन ………………… (इति) चतुर्भिः पुरुषः परीक्ष्यते। (कर्मन्)
उत्तर:
कर्मणा

(xxvii) अपयशः यदि अस्ति कि ………………… (मृत्यु)
उत्तर:
मृत्युना

(xxviii) अहं ………………… दूरम् आरोप्य पातितोऽस्मि। (पर्वत)
उत्तर:
पर्वतात्

(xxix) इदानीम् अहं चारुदत्तस्य दरिद्रतया ………………… समम् अन्यत्र भुक्त्वा तस्यावासमेव गच्छामि। (पारवत)
उत्तर:
पारावतेन/पारावतैः

(xxx) तद् अलं भवत: ………………… । (सन्ताप)
उत्तर:
सन्तापेन

(xxxi) नमः सभाभ्य: ………………… च। (सभापति)
उत्तर:
सभापतिभ्यः

(xxxii) द्वितीय भागस्य सञ्चारः जलस्यान्तर्बहिः ………………… (क्रम)
उत्तर:
क्रमति/क्रम्यते

(xxxiii) ………………… अभिनवस्य उत्साहवर्धनं कुर्वन्तु। (करतलध्वनि)
उत्तर:
करतलध्वनिना

(xxxiv) शालिनी ………………… अधिकृत्य अस्माकं ज्ञाने वृद्धिं करिष्यति। (शल्यक्रिया)
उत्तर:
शल्यक्रियाम्

(xxxv) सम्प्रति मिहिर: बृहत्संहितात: ………………… अधः कुत्र-कुत्र जलं भवति, इति सूचयिष्यति । (भूमि)
उत्तर:
भूमेः

(xxxvi) यदि ………………… पूर्वदिशि वल्मीक: भवेत् तर्हि तस्य दक्षिणदिशि जलं भविष्यति। (जम्बूवृक्ष)
उत्तर:
जम्बूक वृक्षस्य

(xxxvii) अपि च ………………… प्रति पूर्वाभिमुखा पृथिवी 365.25 वारं प्रतिवर्ष भ्रमति। (सूर्य)
उत्तर:
सूर्यं

(xxxviii) मनुष्यः वृक्षादीन् ………………… प्रति गच्छतः पश्यति। (पृष्ठ)
उत्तर:
पृष्ठं

(xxxix) नक्षत्रादयः भूमध्यरेखा स्थितस्य ………………… कृते पश्चिमं प्रति धावन्तः प्रतीयन्ते। (नर)
उत्तर:
नरस्य

(xl) ………………… नमो नमः। (सर्व)
उत्तर:
सर्वेभ्यः

(xli) सः स्तम्भ: ………………… विना तथैव तिष्ठति। (विकृति)
उत्तर:
विकृति

(xlii) अद्य यैः प्रस्तुतिः कृता, ………………… श्वः प्राचार्य महोदयः सम्मानपत्राणि प्रदास्यन्ति। (तत्)
उत्तर:
तेभ्यः

3. निम्न वाक्येषु उपपद विभक्तिं आधृत्य रिक्त स्थान पूर्ति कुरुत

1. बुद्धि: ………………… शुध्यति। (ज्ञान)
उत्तर:
ज्ञानेन

2. देवयानः पन्था ………………… विततः। (सूर्य)
उत्तर:
सत्येन

3. आः किम् ………………… वः प्राणहरेण कथा प्रसंगेन? (एतत्)
उत्तर:
एतेन

4. राज्यं हि नाम धर्मवृत्तिपरकस्य ………………… कृते महत् कष्टदायकम्। (नृप)
उत्तर:
नृपस्य

5. न खलु ………………… अपहृतः प्रेक्षकाणाम् चक्षुषो विषयः? (आर्यचाणक्य)
उत्तर:
आर्यचाणक्येन

6. कथं स्पर्धते ………………… सह दुरात्मा राक्षस? (अस्मद्)
उत्तर:
मया

7. विरम विरम अस्माद् ………………… (दुर्व्यसन)
उत्तर:
दुर्व्यसनात्

8. अये ………………… अध्यास्ते वृषलः। (सिंहासन)
उत्तर:
सिंहासनम्

9. अलम् अनेन ………………… । (विनय)
उत्तर:
विनयेन ।

10. न निष्प्रयोजनं ………………… अधिकारिणः आहूयन्ते। (प्रभु)
उत्तर:
प्रभुम्

11. न ………………… अन्तरा चाणक्य: स्वप्नेऽपि चेष्टते। (प्रयोजन)
उत्तर:
प्रयोजनम्

12. आर्य वैहीनरे। न वैतालिकाभ्यां सुवर्णशत सहस्त्रं दापय। (इदम्)
उत्तर:
आभ्यां

13. ………………… समये किं कौमुदी-महोत्सवेन इति प्रतिषिद्धः। (इदम्)
उत्तर:
अस्मिन्

14. श्रुतम् ………………… यत् मत्स्यजीविभि: अभिहितम्। (भवत्)
उत्तर:
भवद्भिः

15. अनागतविधाता प्रत्युत्पन्नमतिश्च ………………… सह निष्क्रान्तौ। (परिजन)
उत्तर:
परिजनैः

16. तैः मत्स्यजीविभिः ………………… सह तत् सरो निर्मत्स्य तां नीतम्। (यद्भविष्य)
उत्तर:
यद्भविष्येण

17. अहम् ………………… दर्शयामि। (प्रक्षेपक)
उत्तर:
प्रक्षेपकेण

18. बौद्धमठेषु किं-किम् ………………… अवलोकितम्। (भवती)
उत्तर:
भवत्या

19. ग्रीष्मपर्वाणि भगवन्तं ………………… प्रति भक्ति भावं दर्शयन्ति। (बुद्ध)
उत्तर:
बुद्धम्

20. शुचिमनो यद्यस्ति ………………… किम् (तीर्थ)
उत्तर:
तीर्थन

21. ………………… सति गुणेन प्रयोजनं न भवति। (लोभ)
उत्तर:
लोभे

22. ………………… सत्याम् धनस्य आवश्यकता नास्ति। (सविद्या)
उत्तर:
सविद्यायाम्

23. परनिन्दायाः भाव: यदि जीवने आगतः तदा ………………… प्रयोजनं न भवति। (पातक)
उत्तर:
पातकेन

24. ये ………………… सरलतया व्यवहरन्ति ते सफलता न लभन्ते। (धूर्तजन)
उत्तर:
धूर्तजनेषु

25. योग्यः नृपः ………………… ध्यानेन हितं आकर्णयति। (आप्तजन)
उत्तर:
आप्तजनानां

26. अपयश: यदि जीवने, कि ………………. ? (मृत्यु)
उत्तर:
मृत्युना

27. आर्य ………………… खलु आगतोऽसि। (दिष्टि)
उत्तर:
दिष्ट्या

28. अहं ………………… दूरमारोप्य पतितोऽस्मि । (पर्वत)
उत्तर:
पर्वतात्

29. तदलं भवतः …………………। (सन्ताप)
उत्तर:
सन्तापेन

30. अहं ………………… समम् यत्र-तत्र गच्छामि। (पारावत)
उत्तर:
पारावतेन

31. सत्त्वं च न परिभ्रष्टं यद् ………………… दुर्लभम्। (द्रारिद्य)
उत्तर:
दारिद्यम्

32. ………………… , ………………… नमः । (सभा, सभापति)
उत्तर:
सभायै, सभापतये

33. मिहिर: वृहत्संहितात: ………………… अधः कुत्र जलं भवति इति विषये सूचनां प्रदास्यति। (भूमि)
उत्तर:
भूमेः

34. अतिशोभनम् इति ध्वनिः ………………… सह सभागारं पूरयति। (तालिकावादन)
उत्तर:
तालिकावादनेन ।

35. अपि च ………………… प्रति पूर्वाभिमुखा पृथिवी 365.25 वारं प्रतिवर्ष भ्रमति। (सूर्य)
उत्तर:
सूर्यम्

36. यथा नौकायां स्थितः मनुष्यः वृक्षादीन् ………………… प्रतिगच्छत: पश्यति। (पृष्ठ)
उत्तर:
पृष्ठ

37. तथैव नक्षत्रादयः भूमध्यरेखास्थितस्य ………………… कृते ………………… प्रति धावन्तः प्रतीयन्ते। (नर, पश्चिम)
उत्तर:
नरस्य, पश्चिम

38. ………………… नमो नमः। (सर्व)
उत्तर:
सर्वेभ्यः

39. कथं सः स्तम्भ: ………………… विना तथैव तिष्ठति। (विकृति)
उत्तर:
विकृतिं

40. अद्य यैः प्रस्तुतिः कृता ………………… श्वः सम्मान पत्राणि प्रदास्यन्ति। (तत्)
उत्तर:
तेभ्यः

NCERT Solutions for Class 12 Sanskrit

The post CBSE Class 12 Sanskrit व्याकरणम् उपपदविभक्ति प्रयोग appeared first on Learn CBSE.

NTSE Madhya Pradesh 2019-20 for Class X | Exam Dates, Application Process and Syllabus

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NTSE Madhya Pradesh 2019-20: Madhya Pradesh NTSE Stage 1 is scheduled to be held on 3rd November 2019. Rajya Shiksha Kendra will conduct NTSE Madhya Pradesh Stage 1 for students who are currently studying in class X . Shortlisted students will be able to submit the online application form from the last week of August till the last week of September 2019. Students who want to appear for the NTSE Madhya Pradesh can download their admit card in the third week of October 2019. Shortlisted students of Madhya Pradesh has to first clear the NTSE Madhya Pradesh Stage 1 exam which is the first level of NTSE. After that only qualified candidates in the stage1 will be invited to appear for NTSE stage 2, going to be held on May 10, 2020.

NCERT organizes NTSE every year to recognize and encourage young talent in India. NCERT will award scholarships to the NTSE stage 2 qualified candidates till the doctorate level of their academics. Candidates should read the complete article to know more information about NTSE Madhya Pradesh 2019-20. Information such as exam dates, admit card, eligibility, exam pattern, and syllabus, etc.

NTSE Madhya Pradesh Exam Dates

NTSE EventsDates
Releasing of Application formLast week of August 2019
Closing of ApplicationLast week of September 2019
Admit Card for Stage 1Third week of October 2019
NTSE Madhya Pradesh Stage 1November 3, 2019
NTSE Madhya Pradesh Answer KeyNovember 2019
NTSE Madhya Pradesh Stage 1 Result & Cut-off ScoresFirst Week of March 2020
NTSE Admit Card for Stage 2April 2020
NTSE Stage 2May 10, 2020
NTSE Stage 2 ResultSeptember 2020

NTSE Madhya Pradesh Eligibility Criteria

Candidates must know the following eligibility criteria to appear for NTSE Madhya Pradesh 2019-20.

  • Age limit for NTSE Stage 1 exam is below 18 years as on 1st July 2019.
  • All class X students of Madhya Pradesh in the academic year of 2019-20 are eligible.
  • Reserved category candidates must secure at least 32% marks in Class IX to be eligible for the exam.
  • General category candidates must secure at least 40% marks in class IX to be eligible for the exam.
  • Candidates who are studying in Madhya Pradesh Government, CBSE, ICSE or Top schools are eligible.
  • Candidates from open schools or distance learners who have not crossed the age limit of 18 are also eligible.

NTSE Madhya Pradesh Application Form

NTSE Madhya Pradesh application form will be released online in the last week of August 2019. The last date of NTSE Madhya Pradesh online application form submission is the last week of September 2019. Candidates can check the below-mentioned information to apply for NTSE Madhya Pradesh.

  • Candidates have to visit the official website mponline.gov.in.
  • After filling the print out of the application form, make sure to verify it by the Principal of the respective school.
  • After verifying the application form, candidates have to take it to the Kiosk Centers of Madhya Pradesh to arrange for online entry.
  • After online entry, candidates will get two receipts. One they can keep for themselves and the other one will be attached to the application form.
  • Submit the application form to the Principal on or before the due date.
  • It is mandatory for reserved candidates to attach the caste certificates with their application form.
  • Candidates who submit application form through postal or manual would not be considered.
  • Candidates are advised to contact liaison officer to avoid misperception in the application. Candidates can also contact the assistant manager of Rajya Siksha Kendra to avoid misperception.

NTSE Madhya Pradesh Application Fee

  • Candidates no need to pay any application fee for NTSE Madhya Pradesh Stage 1 exam.

NTSE Madhya Pradesh Admit Card

NTSE Madhya Pradesh Stage 1 admit card will be released in the third week of October 2019. The following points contain information about collecting the admit card.

  • Candidates should have their application number and date of birth.
  • Candidates should contact the authorized person of the school to collect the admit card.
  • It is important that candidates must carry their admit card into the examination hall.
  • Without the admit card candidate will not be allowed to enter into the examination hall.
  • The NTSE Stage 1 admit card contains the candidate’s name, roll number, date, time and exam center details.

NTSE Madhya Pradesh Question Pattern

The NTSE Madhya Pradesh Stage 1 exam is divided into 2 parts. These are the Scholastic Ability Test (SAT) and the Mental Ability Test (MAT). Only objective type questions will be asked in the NTSE Madhya Pradesh stage 1. The Stage 1 exam will be held in pen and paper mode. The exam will be organized in English and Hindi medium. The following table shows the complete details on exam pattern for NTSE Madhya Pradesh.

PapersNumber of questionsQuestion TypesTime Allotted
Scholastic Ability Test (SAT)100objective type120 minutes
Mental Ability Test (MAT)100objective type120 minutes
  • Each correct answer carries 1 mark. There is no negative marking for incorrect answers.
  • SAT evaluates the candidate’s interpretative, reading ability, analytical and general awareness.
  •  The total number of questions asked in the SAT is 100 with the time duration of 120 minutes.
  • The questions asked in the exam will be equivalent to the syllabus of class IX and X.
  • MAT evaluates the candidates reasoning, imagining and judging, distinguish the thinking ability.
  • MAT questions are based on series, analogies, pattern perception, logical reasoning, problem-solving, coding-decoding, hidden figures, etc.
  • The total number of questions asked in the MAT is 100 with the time duration of 120 minutes.

NTSE Madhya Pradesh Answer Key

The answer key for the NTSE Madhya Pradesh will be released by the Rajya Shiksha Kendra. The official answer key will be published after some days of the exam. The answer key contains the correct answers to all questions asked in the MAT and SAT papers. Candidates can evaluate their expected scores with the help of answer keys and guess the chances of qualifying for NTSE Stage 2. The following points are given to check the NTSE Madhya Pradesh answer key:

  1. Candidate has to check the official website mponline.gov.in. Then, click on the answer key link.
  2. Download and save the answer key.
  3. It will help the candidate to guess the chances of qualifying for the NTSE stage 2 exam.
  4. The accurate marking approach should be followed. Candidates answer key can be reviewed based on some significant proof.
  5. The final reviewed answer key will be published.

NTSE Madhya Pradesh Exam Result

NTSE Madhya Pradesh exam result will be announced by Rajya Shisha Kendra on its official website on March 2020. The exam result contains the marks secured in SAT, MAT & the total marks of each candidate. The exam result will be released in the form of merit list of qualified candidates for NTSE stage 2 exam which will be held on May 10, 2020. Candidates can check the following steps to know the NTSE Madhya Pradesh exam results:

  • Candidates can visit the official website mponline.gov.in to check their results.
  • The result link can be seen on the homepage of the portal, click on it.
  • Candidates can enter their roll number and date of birth and download the result list.
  • Candidates can keep the printed copies of their result list for future reference.

Details Mentioned on NTSE Madhya Pradesh Exam Result

Candidates can check their NTSE Madhya Pradesh merit list on the official website of Rajya Shiksha Kendra. The merit list contains the following information of the qualified candidates.

  • Name, Roll Number and Address of the candidate
  • Marks obtained in MAT & SAT
  • Candidates caste category, gender, date of birth and disability status

NTSE Madhya Pradesh Stage 1 Cut-Off

Cut-off scores are the minimum scores candidates must obtain to get selected for the NTSE stage 2 exam. NTSE Madhya Pradesh qualifying scores will be declared along with the result by Rajya Siksha Kendra on March 2020. Candidates qualifying scores depends on various factors as given below:

  • Number of candidates in each category
  • Maximum marks obtained in the exam
  • Minimum marks obtained in the exam

The following table shows the previous year’s cut-off marks for NTSE Madhya Pradesh Stage 1 category candidates

CategoryCut-Off 2018-19
General173
SC149
ST132
PH125
OBC164

The following table shows the previous year’s qualifying marks for MAT & SAT for NTSE Madhya Pradesh Stage 1 category candidates

CategoriesMATSATLT
General448340
SC346646
ST355134
PH374637

NTSE Madhya Pradesh Reservation Criteria

NTSE scholarships are circulated as per the Government rules for reservation. Candidates who qualify both the stages will be awarded a total of 2000 scholarships. The following table contains the details about reservation criteria.

CategoriesReservation Criteria
SC15%
ST7.5%
PH3%
OBC27%

NTSE Madhya Pradesh Stage 1 Syllabus

There is no specific syllabus for the NTSE Madhya Pradesh Stage 1 exam. So, candidates must cover their CBSE class IX and X syllabus in order to get selected in this exam.

NTSE Scholarship Amount

Candidates who selected in both the stages with minimum cut-off marks will receive the scholarship amount every month.

Education LevelScholarship Amount
Class XI to XIIRs. 1250/-
UndergraduateRs. 2000/-
PostgraduateRs. 2000/-
PhDAccording to the UGC norms

Hope this article will help you to get more information about NTSE Madhya Pradesh 2019-20.

The post NTSE Madhya Pradesh 2019-20 for Class X | Exam Dates, Application Process and Syllabus appeared first on Learn CBSE.

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