NCERT Solutions for Class 4  पर्यावरण अध्ययन  Chapter 5 अनीता की मधुमक्खियाँ 

प्रश्न -अभ्यास

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NCERT Solutions for Class 4  पर्यावरण अध्ययन  Chapter 6 ओमना का सफ़र 

प्रश्न -अभ्यास

The post NCERT Solutions for Class 4 पर्यावरण अध्ययन Chapter 6 ओमना का सफ़र appeared first on Learn CBSE.

CBSE Class 12 Geography Sample Paper With Solutions Set 1

[Time Allowed : 3 hrs.]                                                                                                   [Max. Marks : 70]

Questions 1-7 (1 Mark), 8-13 (3 Marks), 14-20 (5 Marks), 21 and 22 (Map Question-5 Marks each)

Q.1.How is the actual growth of population calculated?
Ans. Actual growth = Births – Deaths + In migration – Out migration.

Q.2.Write down the name of any two Modern towns of India.
Ans. (i) Calcutta (Now Kolkata)
(ii)Bombay (Now Mumbai)
(iii)Madras (Now Chennai) (any two)

Q.3.Define manufacturing.
Ans. The transformation of raw material into finished goods, with the help of technology.

Q.4.What is Common property resource?
Ans. The common property resource is owned by the state and meant for the use of the community. Every member of the community has the right to access and use these resources with specified obligations.

Q.5.Which are the two main types of iron ore found in India?
Ans. Haematite and Magnetite are the two main types of iron ore found in India.

Q.6.Name the land locked sea-port of India. In which state is it located?
Ans. Vishakhapatnam (Vizag) is the land locked sea-port of India, located in Andhra Pradesh. Q.7. Which city is the main polluter of river Yamuna?
Ans. Delhi is the main polluter city of river Yamuna.

Q.8.Explain the concept of Possibilism with suitable examples.
Ans. Vidal-de-la Blache followed the approach of possibilism to study human geography. The possibilism school of thought offers many choices to men. In this approach, emphasis is on man rather than nature and man is seen choosing his needs according to his own culture and is also “The judge of their use”. The French School of Geographers is staunch supporters of possibilism. The main features of this thought are:
(i)Natural environment does not control human life
(ii)Environment offers some possibilities to man
(iii)Environment is inert and man is seen as an active force rather than passive one

Q.9.Define the term “Human Development”. State three indicators which form the basis of calculating the HDI. Name two countries forming highest and the lowest ranks each in HDI.
Ans. Human Development is defined as the process of widening people’s choice and the level of : well being they achieve. Since the quality of life and well being of people cannot be measured quantitatively, United Nations Development Programme (UNDP) developed a composite index known as HDI which is an integral organ of United Nation. Three indicators which form the basis of calculating HDI are as follows:
(i)Life expectancy as an index of longevity of life.
(ii)Literacy as an index of knowledge.
(iii)Per capita purchasing power as an index of decent standard of living.
Two countries having the highest HDI are Canada and Norway and the countries having the lowest HDI are Djibouti and Lao people’s Democratic Republic.

Q.10. Tourism is a travel undertaken for purposes of recreation rather than business. It has become the world’s single largest tertiary activities in total registered jobs (250 million) and total revenue (40 % of GDP). Besides, many local persons are employed to provide services like accommodation, meals, transport, entertainments and special shops serving tourist. Tourism fosters the growth of infrastructure and industries, retail, trading, and craft industries (souvenirs). In some regions, tourism is seasonal because the vacation period is dependent on favorable weather conditions, but many regions attract visitors all the year round.
Read the above paragraph and analyse the following:
(i)Define Tourism.
(ii)How does Tourism foster the economy of the local people?
Ans. (i) Tourism is a travel undertaken for purposes of recreation rather than business.
(ii)Tourism foster the economy of the local people in the following ways:
(a)Growth of infrastructure industries.
(b)Growth of the retail trading services.
(c)Growth of the handicraft industries.
Hence the above three factor promote ‘area development’ and attract tourist with a penchant for experiencing ethnic and local customs. It also enhances the quality of life with good moral values.

Q.11.Write six characteristics of the Panama canal.
Ans. The six characteristics of Panama Canal are as follows:
(i)It connects the Atlantic Ocean in the east and Pacific Ocean in the west.
(ii)It has been constructed^cross the Panama isthmus.
(iii)About 8km of area of the either side of the canal is owned by the U.S. government.
(iv)It has reduced the distance between eastern and western coast of North and South America.
(v)The canal has lock system. Ships cross different level of canal through three locks, prior entering to the Gulf of Panama.
(vi)It has reduced the distance between New York (east) and San Francisco, (west) by 13,000km.

Q.12.Explain any three ill effects of air pollution on human health in India.
Ans.The three effects of air pollution on human health in India are:
(i)Irritation to the eyes, nose and throat and respiratory infections like bronchitis and pneumonia.
(ii)Headaches, nausea and allergic reactions.
(iii)Chronic heart diseases, lung cancer, damage to brain, nerves, liver or kidneys.
(iv)It often affects the lungs of the growing children and aggravates or complicates medical conditions in the elderly.

Q.13. Explain any three characteristics of clustered rural settlements of India.
Ans. The three characteristics of clustered rural settlement of India are as follows:
(i)The houses in the villages are distinct and separated from the surrounding farms, barns and pastures.
(ii)Such rural settlements may be centered around some common point.
(iii)This settlement forms some recognizable pattern or shape like rectangular, radial or linear.

Q.14. What is the meaning of market gardening and horticulture ? Describe any four of this type of agriculture of the world.
Ans. The market gardening and horticulture is one of the most specialized forms of cultivation of vegetables, fruits and flowers, which has high value and have great demand in urban market. The four features of market gardening and horticulture are:
(i)It requires good and quick transportation system as fruits, vegetables and flowers are perishable in nature . Hence this farming is also stated as ” Truck farming”.
(ii)This type of farming is capital as well as labour intensive.
(iii)More stress is given on the optimum use of HYV seeds, pesticides, fertilizers, other means of irrigation and greenhouses for artificial heating in colder regions.
(iv)It is more common in the densely populated industrial regions of developed countries (North-west Europe, north eastern regions of the USA and Canada, Southern Europe, etc)

Q.15.”Globalisation along with free trade can adversely affect the economies of. the developing countries” . Support the statement with examples.
Ans. “Globalisation along with free trade can adversely affect the economies of the developing countries”
(i)Globalisation along with free trade has helped in widening the gap between the rich and the poor countries.
(ii)The rich and prosperous developed countries greatly influence world trade only for their own commercial interests.
(iii)Most of the developed countries use the markets of developing countries and restrict developing countries to enter their countries.
(iv)Even the local markets of the developing countries are completely under the influence of developed countries.
(v)Issues like worker’s right, child labour, health and environment concerns are completely ignored by the developed countries.

Q.16.What are the functions of rural settlements? Describes three main rural settlement patterns based on structures found in different parts of the world.
Ans. The arrangement of the streets, houses and other functions in rural settlements is
related to its form, environment and culture. As such the rural settlements present varied patterns. Following pattern of rural settlements are generally recognized:
(i)Linear Settlement: The linear settlement pattern is very common and develops parallelly to the roads, river banks and canals. In Kerala mainly linear settlement pattern is found along the either sides of roads and lagoons.
(ii)Circular or Square Pattern: Circular settlement develops in the flat level lands, around a pond , tank, crater, hill top and a cattle corral, for example in West Bengal and Bihar settlements around a village pond is a common feature . In Africa and Europe, circular villages may be seen.
(iii)Cross Shaped: Cross shaped settlements begin as a small hamlet at the intersection of roads. Gradually, it grows along the roads on all sides and appears as a cross or a star depending upon the number of roads joining at a junction.’-
(iv)Nebular Pattern: When the shape of a settlement resembles to nebula, it is stated as nebular settlements. The arrangement of roads is almost circular. Generally the size of nebular settlements is small and they develop around the house of landlord of village, the mosque, temple or church. This type of settlement is mainly found in several villages of Ganga-Yamuna doab.

Q.17. What is the meaning of population growth? Describe two main features each of the four phases of Indian demographic history.
Ans. Population Growth: The net change in the size of population between two points of time is said to be population growth. This change in the population growth is expressed in per cent and said to be the growth rate of population. The four phases of Indian demographic history are as follows:
(i)Stagnant Growth Phase (1901-1921): During this period the growth rate was irregular and there was sporadic increase in population. In this period both birth rate and death rate were very high. During this decade the growth rate of population declined by 7.70 lakh.
(ii)Steady Growth Phase (1921-1951): Due to increase in medical aid and sanitation, there was rapid decline in the death rate. Though the birth rate was very high, it influenced the high growth rate.
(iii)Rapid Growth Phase (1951-1981): During this period the growth rate doubled. The average annual growth rate of population during this period was 2.2 percent because of the developmental works undertaken on a large scale and a great improvement in the living conditions of people. So compared to birth rate, there was rapid decline in the death rate.
(iv)Declining Growth Phase (1981-2001): During this decade growth rate started declining due to success of family planning programme. The mass promotion of female literacy and family planning is responsible for the declining growth trend. The people started following small family norms.

Q.18. Explain any five measure necessary for the promotion of sustainable development in ‘Indira Gandhi Canal Command Area”.
Ans. The measures taken for the promotion or restoration of sustainanability in Indira Gandhi Canal Command Area are as follows:
(i)Implementation of the better water management, the canal project forms protective irrigation in stage -I and vast development of the crop and pastureland in stage- II.
(ii)Salinity occurred due to water logging should be properly checked, as it increases vast regime in the Thar Desert region.
(iii)Setting up proper infrastructure such as roads, markets, storage facilities and drinking water.
(iv)The construction of proper warabandi (the equal distribution of canal water in the command area of outlet) system for fair and efficient distribution of irrigation water.
(v)Selection and adoption of suitable agronomic practices .People of this region are encouraged to grow plantation crops such as citrus fruits, dates and palm.

Q.19.Describe the major challenges of agriculture in present India.
Ans. Agriculture is the main occupation of people of India. It is the source of living for about 70% of its working population. It is the base of Indian economy. Despite its dominant role, agriculture has not been able to provide the basic needs of the country. Our agriculture has failed to meet the food grains requirements of the country. Many factors that have hindered the development of agriculture in India are as follows:
(i)Pressure of population on land: Due to continuous increase in population, pressure of population on land is increasing. Due to overcrowding, the per capita cultivated land in India has been reduced to only 0.3 hectare. It has resulted in shortage of food grains. The growth rate in agriculture is also low.
(ii)Inadequate Irrigation Facilities: In India, agriculture depends on monsoonal rainfall. Due to uncertain and variable rainfall irrigation is necessary. Only 22% of cultivated land is irrigated. It is necessary to irrigate at least 50% of the cultivated land to make it a success. Moreover, irrigation is required to increase the yield, productivity and the intensity of cropping to check severe draught.
(iii)Low productivity: In India, the yield per hectare of food grains and other crop is low as compared to other countries of the world. High yielding varieties have been introduced. But only 16% of the cultivated land is under high yielding varieties.
(iv)Poor Techniques of production: Due to continuous agriculture over a long period, the fertility of soils is declining. To maintain its fertility, the use of chemical fertilisers is necessary. Indian farmers have been using old inefficient methods due to which productivity is low. Use of better quality seeds and pesticides can increase the productivity.
(v)Lack of Mechanised Farming: Agriculture is mostly labour intensive to get maximum output. Use of modern machines is limited due to low purchasing power ofTarmers. Farmers still use primitive tools and implements. Farmers need to be provided modern machines at subsidised prices so as to increase the yield.

Q.20.What geographical conditions are necessary for the development of Inland waterways? Write three advantages of inland waterways.
Ans. Rivers, canals and lakes are the main mediums of inland waterways.
Following conditions are necessary for the development on of Inland waterways:
1.Rivers should have sufficient water throughout the year. Seasonal rivers are not much suitable for inland waterways.
2.River must be free from rapids, gorges and waterfalls.
3.River meandering should be minimum so that boat and ships can have straight passage.
4.A good river mouth connects inland waterways to ocean routes.
5.Freezing of rivers in winter hinders inland navigation.
Advantages of inland waterways:
1.This is the cheapest means of transport and is useful for carrying heavy commodities.
2.There is neither any need of lying of track nor it has to be metalled.
3.They are the main source of transport in dense forests.

Q.21.On the given political outline map of the World, following five features are shown. Identify these features and write their correct names on the lines marked near each feature.
(i)A major area of commercial livestock rearing.
(ii)A major sea-port.
(iii)Megacity
(iv)Major airport
(v)Largest country of Europe by area.

Ans. (i) United States of America (ii) Valparaiso (iii) Moscow (iv) Darwin (v) France

Q.22.On the given political outline map of India, locate and label the following with appropriate symbols:
(i)Iron ore mine of Karnataka
(ii)A software Technology Park located in Himachal Pradesh
(iii)State with high HDI ranking
(iv)Telugu speaking state
(v)North Terminal of N-S Corridor
Ans.

The post CBSE Class 12 Geography Sample Paper With Solutions Set 1 appeared first on Learn CBSE.

NCERT Solutions for Class 4 Mathematics Unit-5

Unit 5: THE WAY THE WORLD LOOKS

NCERT Textbook Page 52
1. Draw how the fan looks from below.
Ans.

A ceiling fan looks like the given picture, when seen from below.

2. Can you think why Gappu could see the cheese on the jug but Chinky could not?
Ans. Cheese ‘was placed on the top of the jug. Gappu was hanging with a balloon which was stuck to the ceiling and he could easily saw the cheese, while Chinky was running on the floor, so she could not be able to see the cheese kept on the top of the jug.

3. When I ran around in my house, it looked so big! But from here, it looks small. How is that?
Ans. Objects are seen smaller from a distance. Gappu was seeing the house from higher in the air, so, that looked smaller.

NCERT Textbook Page 54
1. Imagine how your classroom looks from above. Try to draw it and mark the benches, blackboard, doors, window, etc.
Ans. The top view of a classroom is being given here.

NCERT Textbook Page 55
1. Look at these pictures and discuss why things look wide and big at this end but narrow and small at the other end.
Ans. The side which is nearer to us looks wider and which is farther to us looks narrow and smaller.

NCERT Textbook Pages 55-56
Match Two Views of the Same Pose
1. Only one of the photos below is the correct match of the same yoga pose. Mark it.
Ans.

2. These are two different views of the same bowls. In which photo are the bowls upside down?
.
Ans. In photo 2 the bowl has been kept upside down.

3. Draw lines to match the side view with the top view of—A pipe, A funnel.
Ans.

4. Try to draw pictures of a shoe from the side, top, front etc.
Ans. Different views of a shoe is given here

NCERT Textbook Page 57
The Park behind Gappu’s House
Here is a bigger picture of that part. Look at it carefully and answer the questions.

1. Mark the gate nearest to the sweet shop. A/B/C/D
Ans. Gate ‘A’ is nearest to the sweet shop.

2. Which gate is nearest to Gappu’s house?
Ans. The Gate ‘C’ is nearest to the Gappu’s house.

3. If you enter from gate B, the green bench will be to your-Left/Right/Front.
Ans. The green bench would be left to me, if I enter from gate B.

4. When Suhasinin entered the park, the flower bed was to her right. Which gate did she enter from?
Ans. Flower bed is right to the Gate ‘D’, so she entered from Gate ‘D’.

5. WThich of these is nearest to you if you enter from gate ‘C’?
(i) Basketball court (ii) Flower bed
(iii) Green bench (iv) See-saw
Ans. See-saw.

NCERT Textbook Page 58
Ismail’s Home
On the phone Ismail told Srijata the route to his house from her house. The route map is shown here.

This is what Ismail told to Srijata:
“From your house, reach the milk-booth and then take a left turn. From the second crossing take a right turn and go over the bridge. Go straight and then take the first right turn. After about 100 metres you will see a big park.
When you cross the park you will come to a side lane. My house is the first house in that lane”.
1. Did Ismail go wrong somewhere? Can you correct him?
Ans. Yes. Srijata has to take left going straight after crossing the bridge.

2. Show where Srijata will reach if she takes the route he told her.
Ans. Srijata would reach opposite to Ismail’s house.

3. Write the directions for going from Ismail’s house to Srijata’s house.
Ans. Start from Ismail’s house. Take first right turn. Go straight. After crossing the bridge take first left turn. Take the second right turn, which is from milk booth. Go straight, the red coloured house is Srijata’s house.

NCERT Textbook Page 59
Gibli and the Big Box
Well, one day Gibli saw a big box on her way. It looked like this.

Gibli moved across and turned left. Now she could see the other face 6f the big box.
Gibli was confused. What was this box? She climbed on a cup and tried to see from there. The box looked like this.
1. Can you guess what that box-like thing was?
Ans. The box was a dice.
(The numbers on the opposite faces of this bojt add up to 7)

2. Which number was on the opposite side of 5?
Ans. The number opposite of the 5 = 7 – 2 = 2.

3.In the picture, which number will be at the bottom? .
Ans. On the top the number is 1.
Hence, on the bottom the number = 7-1 = 6.

4.Which number will Gibli see if she again turns left from 5?
Ans. On turning left from 5, Gibli would see the number 1.

5.What will this box look like if you opened it up? Mark the correct picture.

Ans. The box looks like the figure given in ‘A’, if it would be opened.

6.Can you use this box to play a game?
Ans. Yes. I can play ‘Ludo’ using this box.

The post NCERT Solutions for Class 4 Mathematics Unit-5 appeared first on Learn CBSE.

NCERT Solutions for Class 4 Mathematics Unit-6

Unit 6: THE JUNK SELLER

NCERT Textbook Page 60
1. What about you? Do you also find Maths difficult?
Ans. No. I do not find Maths difficult. It becomes easy if it is practised well regularly.

2. What is the most difficult thing in your Maths book?
Ans. Nothing, all topics look easy to me.

3. What do you think is the easiest lesson?
Ans. All lessons are of equal level. This depends upon a lot of practice of maths.

NCERT Textbook Page 61
Find out: How much for a cup of tea?
1. Ask people and find out the cost of a cup of tea –
• At a tea stall
Ans. At a tea stall a cup of tea costs Rs 3.
• At a hotel
Ans. At a hotel a cup of tea costs Rs 5.

2. If a person who runs a tea stall earns ? 30 a $ay, how much will he earn in 10 days?
Ans.
Earning in 1 day                            = Rs 30 x 1
Therefore, earning in 2 days      = Rs 30 x 2
Earning in 3 days                         = Rs 30 x 3
And so on_______
Earning in 10 days                       = Rs 30 x 10 = 300
•And in a month?
Ans.
A month has 30 days
If earning in 1 day                      = Rs 30
Therefore, earning in 30 days = Rs 30 X 30 = Rs 900.

FindOut: What is a Loan?
1. Have you ever heard of someone taking a loan? For what?
Ans.I heard a person took loan. He took loan to start his shop.

2. How much loan was taken?
Ans. He took Rs 10,000.

3. How much money was paid back?
Ans.He paid back Rs 11,000 after one year.

4. Who has to pay back more-Hariya or Babu?
Ans.Hariya has to pay Rs 51 every month for six months.
Total money he has to pay = 51 x 6 = Rs 306 Babu has to pay Rs 360
Hence, Babu will have to pay back more money.

NCERT Textbook Page 62
1. How much does Kiran earn from 9 rickshaws in a day?
Ans. From one rickshaw she earns          = Rs 20
From 2 rickshaws she will earn                 = Rs 20 x 2
From 3 rickshaws she will earn                 = Rs 20 x 3
Similarly from 9 rickshaws she will earn = Rs 20 x 9 = Rs 180
That is 9 times 20.

2. For 10 rickshaws she will get Rs 20 x 10 = Rs 200.
So, for 9 rickshaws, she will get Rs 200 – _____ = ______
Ans. For 9 rickshaws she will get Rs 200 – 20 = Rs 180.

NCERT Textbook Page 63
1. Ina week how much does Kiran earn from one rickshaw?
Ans. There are 7 days in a week.
In 1 day Kiran earns from one rickshaw               = Rs 20
So, in 7 days Kiran will earn from one rickshaw = Rs 20 x 7 = Rs 140.

2. Do it mentally and write the answers
2 x 6 = ________ 4 x 80 = ________
20 x 6 = ________ 4 x 81 = ________
2 x 60 = ________ 9 x 25 = ________
3 x 42 = ________ 31 x 9 = ________
Ans.
2 x 6      =      12              4 x 80   =       320
20 x 6   =       120           4 x 81    =       324
2 x 60   =       120           9 x 25    =       225
3 x 42   =       126            31 x 9    =       279

How Much to Pay for this Junk?
3. How much will Kiran pay for 31 kg newspaper?
Ans. The cost of 1 kg of newspaper = Rs 5
So, the cost of 31 kg of newspaper    = Rs 5 x 31
= Rs 155.

NCERT Textbook Page 64

1. How much will Kiran pay for 42 kg newspaper?
Ans. Kiran pays for 1 kg of newspaper  = Rs 5
So, she will pay for 42 kg of newspaper = Rs 15 x 42
= Rs 210.

2. Also find the cost of • 22 kg of plastic
Ans. Rate of 1 kg of plastic            = Rs 10
Therefore, cost of 22 kg of plastic = Rs 10 x 22
= Rs 220.
• 23 kg of waste paper
Ans. The cost of 1 kg of waste paper            = Rs 4
Therefore, the cost of 23 kg of waste paper = Rs 4 x 23
= Rs 92.
• 12 kg of iron
Ans. The cost of 1 kg of iron            = Rs 12
Therefore, the cost of 12 kg of iron = Rs 12 x 12
= Rs 144.

3. Guess the total money Kiran will pay to the junk collectors. Will it be: — More than 600?
— Less than 600?
Ans. The total money Kiran will pay to the junk collectors
The cost of 42 kg of newspaper       = Rs 210
The cost of 22 kg of plastic               = Rs 220
The cost of 23 kg of waste paper     = Rs 92
The cost of 12 kg of iron                    = Rs 144
Total cost of all junks                         = Rs 666
Hence Kiran will pay more than Rs 600.

NCERT Textbook Pages 64-66
Smart Kiran Sells the Junk

1. Kiran bought 1 kg plastic for Rs 10, but sold 1 kg plastic for Rs 12. How much
money does she earn on selling 1 kg plastic? Rs _______
Ans. Kiran bought 1 kg plastic for Rs 10 and sold 1 kg plastic for Rs 12 So, she earn Rs 12 – Rs 10 = Rs 2.
• So, how much money does she earn for 63 kg? Rs _______
Ans. For 1 kg plastic she earns Rs 2.
Therefore, for 63 kg plastic she earns = Rs 2 x 63
= Rs 126
It can be calculated as follows:

You will get 126 by adding the digits in last row.
B. Kiran sells 32 kg iron.

2. How much money will Dinu pay for 32 kg iron?
Ans. 1 kg of plastic is sold in Rs 14
Therefore, 32 kg of plastic = Rs 14 x 32 = Rs 448 This can be calculated in similar manner

You can get the desired numbers by adding the numbers in bold
300 + 20 + 120 + 8 = 448
So, Dinu will pay Rs 448 for 32 kg of plastic.

3. Kiran buys 1 kg iron for Rs 12, but sells it for Rs 14. How much does she earn when she sells 32 kg iron?
Ans. Kiran earns by selling 1 kg of iron  = Rs 14 – Rs 12
= Rs 2
So, she will earn by selling 32 kg of iron = Rs 2 x 32
= Rs 64
This can be calculated as follows also.

You will get the desired result by adding the numbers in last row
Which will be equal to 60 + 4 = 64
Hence, Kiran earns Rs 64 by selling 32 kg of iron.
C. What will Dinu pay for 152 kg newspaper?

4. I bought 1 kg newspaper for Rs 5, but sold it for Rs 6. How much money did I earn by selling 152 kg of newspaper?
Ans. She earn by selling 1 kg of newspaper
= Selling price of newspaper – Purchasing price of newspaper
=Rs 6-Rs 5 = Rs 1
Hence, she earned by selling 152 kg newspaper
= Rs 1 x 152 = Rs 152.
D. What does Dinu pay for brass?

5. How much money will Dinu pay for 4 kg brass?
Ans. The rate of 1 kg of brass according to Dinu’s price list = Rs 180
So, Dinu will pay for 4 kg of brass                                               = Rs 180 x 4
= Rs 720
This can be calculated as follows:

By adding the digits in last row you will get 400 + 320 = 720 Hence, Dinu will pay Rs 720 for 4 kg of brass.

NCERT Textbook Page 67
1. First, guess the answer and then calculate
(a) 37 x 18 =   (b) 45 x 24 =    (c) 69 x 52 =
(e) 142 x 5 =   (f) 382 x 3 =     (g) 2 x 175 =
(d) 77 x 55 =   (ft)4 x 206 =
Ans.
(a) Guess = 37 x 18 means 40 x 20 = about 800
Calculation

By adding the digits in third and fifth rows you will get
300 + 70 + 240 + 56 = 666
Hence, 37 x 18 = 666.
(b) Guess 45 x 24
Then will be near to 40 x 25 = 1000
By Calculation

By adding the numbers in third and fifth rows you will get
800 + 100 + 160 + 20 = 1080
So, 45 x 24 = 1080.
(c) On guessing 69 x 52, the numbers will be near to 70 x 50 = 3500
Calculation

By adding the numbers in third and fifth rows you will get
3000 + 450 + 120 + 18 = 3588
So, 69 x 52 = 3588.
(d) On guessing 77 x 55
The numbers will be near to 80 x 50 = 4000
Calculation:

By adding the numbers in third and fifth rows you will get
3500 + 350 + 350 + 35 = 4235
Hence, 77 x 55 = 4235.
(e) On guessing 142 x 5
The number will be near to 140 x 5 = 700
On calculation

By adding the numbers in last row you will get
500 + 200 + 10 = 710 .
Hence, 142 x 5 = 710.
(f) 382 x 3
On guessing this number will be near to 380 x 3 = 1140
On calculation:

By adding the numbers in last row you will get
900 + 240 + 6 = 1146
So,-382 x 3 = 1146.
(g) 2 x 175
This number will be near to 2 x 175 = 350
On calculation:

By adding numbers in last row you will get
200 + 140 + 10 = 350.
(h) 4 x 206
This number will be near to 5 x 200 = 1000
On calculation:

By adding the numbers in last row you will get
800 + 24 = 824
So, 4 x 206 = 824.

NCERT Textbook Page 67
Fill My Diary
1. Now you make a record in her diary.
Find out how much she earned this time.

The post NCERT Solutions for Class 4 Mathematics Unit-6 appeared first on Learn CBSE.

NCERT Solutions for Class 4 Mathematics Unit-7

Unit 7: JUGS AND MUGS

NCERT Textbook Pages 70-72
1. The donkey asked-500 millilitres of kheer? Isn’t that more than a litre?
Ans. No, 500 millilitre is not more than a litre, rather it is less than one litre. It is half litre.
The fox said-Come on, don’t behave like a donkey! One litre is 1000 millilitres, so 500 millilitres is half a litre.

2. The donkey looked confused and asked – Ten glasses of 100 ml each. How much
is that?
Ans. Ten glasses of 100 ml
= 10 x 100 ml .
= 1000 ml
= 1 litre.

3. The fox got another chance to show off! He said – Ah, That is simple! 10 times
hundred millilitres is_________ millilitres =_________litres.
Ans.
10 times hundred millilitre
= 10 x 100 ml
= 1000 ml
= 1 litre.

4. The cat said-Oh no, one thousand! We have to give kheer to 1000 suits!
After thinking the elephant said-No problem, I can manage. Each ant drinks 1 millilitre of kheer.
So, 1000 ants drink: 1000 x 1 mL =_________ mL
Ans. As each ant drinks 1 millilitre of kheer
So, 1000 ants drink = 1000 x 1 mL
= 1000 mL
= 1 litre.

NCERT Textbook Page 73
1. How much kheer can you have?
Ans. I can have only 200 ml of kheer.

2. Can you drink 1 L water at one time?
Ans. No. I cannot drink 1 L water at one time.

3. The donkey is trying to look for different ways to add up to 1 litre. Help him complete the chart.
Ans.

NCERT Textbook Page 74
Look Around
1. Look at these pictures. Now look for some other things we get in packets or bottles like these. Make your own list.
Ans.

My Litre Bottle
1. Have you seen a one-litre water bottle?
Ans. Yes. I have seen water bottle of one litre.

NCERT Textbook Page 75
1. Check if your guess is correct and fill the table.
Ans.

2. Look what Adithyan is saying. I poured two small bottles of water to fill this 1- litre bottle.
• How much water does his small bottle hold?
Ans. This small bottle holds 500 mL of water.
• Leela-to fill the 1 litre bottle I need to pour water 5 times from my small bottle. Then how much water does Leela’s bottle hold?
Ans. Leela’s bottle holds 200 mL of water. Because 200 mL x 5 = 1000 mL = 1 Litre

Ramu’s Measuring Bottle
1. Ramu got an empty 250 mL coconut oil bottle. Look
at the picture and discuss what he did to make his
big measuring bottle.
Ans.

(a) He took the 250 mL coconut bottle and fill it with water.
(b) He then pours one bottle of water in big bottle, and mark the water level as 250 mL.
(c) He again pours one 250 mL bottle of water in big bottle and mark the water level as 500 mL.
(d) By pouring third 250 mL bottle of water in big bottle, he mark the water level as 750 mL.
(e) After pouring four 250 mL bottle of water in big bottle, he mark the water level as 1000 mL or 1 L.

NCERT Textbook Page 76
My Measuring Bottle
1. Find your own way to make a bottle which can measure 200 mL, 400 mL, 600 mL, 800 mL and 1 litre. Discuss with your friends and teacher how you made this.
Ans. Take a large bottle and a 200 mL bottle.
Fill water in the 200 mL bottle and pour it in the big bottle. Mark the water level in big bottle as 200 mL.
Again fill water in the 200 mL bottle and pour it in the big bottle. Mark the water level in big bottle as 400 mL.
Again fill water in the 200 mL bottle and pour it in the big bottle. Mark the water level in big bottle as 600 mL.
Again fill water in the 200 mL bottle and pour it in the big bottle. Mark the water level in big bottle as 800 mL.
Again fill water in the 200 mL bottle and pour it in the big bottle. Mark the water level in big bottle as 1000 mL or 1 litre.
Doing this you can prepare your own measuring bottle.
Guess and Check .

2. Look at the buckets, mugs, glasses and other things in your house. Guess how much water each can hold. Check if your guess is right by using your measuring bottle.
Ans. My guess and measures

NCER.T Textbook Page 77
Neetu in Hospital
1. Neetu has to take 3 injections in a day for 5 days. One injection gives 5 mL of the medicine to your body.
(a) How much medicine will she need for one day?
Ans. One injection contains 5 mL of medicine
She requires 3 injections a day
So she needs medicine in one day = 5 mL x 3 = 15 mL of medicine.
(b) How much medicine in all for 5 days?
Ans. In one day she is getting 15 mL of medicine.
So, in 5 days medicine in all = 15 mL x 5 = 75 mL.

3. How much do we use at a time?
Eye drops               We use less than 1 mL at a time.
_________         __________________
Ans.
Eye drops             We use less than 1 mL at a time.
Cough syrup        We use 5 mL at a time.
Tea                        We use about 40 mL at a time
Milk                      We take about 200 mL at a time
Water                   We drink about 250 mL at a time.

4. List things we use more them one litre at a time.
Water for taking bath.
Ans.
1. Water for taking bath. 2. Water for washing cloths.
3. Water for flushing a toilet. 4. Petrol’to fill up the tank of a bike or a car
5. Water to wash kitchen utensils.

NCERT Textbook Page 78
Practice Time
1. Amina’s water bottle holds one litre of water. She drank 250 mL of water and her friend Govind drank 150 mL. How much water is left in her bottle?
Ans. Total water holds in bottle = 1 litre = 1000 mL
Total water they drank = Water drunk by Anima + Water drunk by Govind
= 250 mL + 150 mL = 400 mL
Water left in her bottle = 1000 mL – 400 mL
= 600 mL.
2. Yusuf runs a tea shop. For making a glass of tea he uses 20 mL of milk. Yesterday he made 100 glasses of tea. How much milk did he use?
Ans. For making 1 glass of tea milk is used
= 20 mL
Therefore, for making 100 glasses of teas, milk is used
= 20 mL x 100
= 2000 mL.

3. Radha’s grandma was ill. The doctor gave her a bottle with 200 mL of medicine. She has to take the medicine every morning for 10 days. How many millilitres of medicine does she have to take every morning?
Ans. In 10 days Radha’s gandma has to take 200 mL of medicine
So, in 1 day she has to take = 200 mL ÷ 10
= 20 mL.

NCERT Textbook Page 79
The table shows the water used in one day by a family of 5 people. They live in Goodallur village.

1. Total water used by them = ?
Ans. Total water used by them
= 30 L + 40 L + 20 L + 75 L = 165 L.

2. How many litres of water does your family use in a day? Guess and fill in this table.
Ans.

NCERT Textbook Pages 79-80
Total water used in a day = 20 L + 50 L + 10 L + 80 L + 10 L = 170 L
Drops and Drops Make an Ocean
1. Is there any tap in your school or your home which is leaking?
Ans. No, there is no tap in my school or in my home which is leaking.

2. How much water do you think we waste through a leaking tap?
Ans. I think about 50 litres of water is wasted through a leaking tap everyday.

3. Place your litre jar below the leaking tap so as to catch all the drops in the bottle. Note the time. After one hour check how much water is in the bottle.
Ans. In one hour water collected in bottle = 1 L.

4. Find out how much water is wasted in a day?
Ans. In one hour 1 L water is wasted
So, in 1 day or 24 hours water wasted = 1 L x 24 = 24 L.
• In a week?
Ans. In one day 24 L of water is wasted
So, in one week or 7 days = 24 L x 7 = 168 L.
• In a month?
Ans. In one day 24 L of water is wasted
Therefore, in a month which has 30 days, water wasted
= 24 L x 30 = 720 L.
• In a year?
Ans. In one month 720 L of water is wasted
Therefore in one year, which has 12 months water wasted
= 720 L x 12 = 8640 L.

5. Chelannur village has a milk society. Geeta and Ammini went there to buy 4 litres of milk. But the man could not find the one litre measure. He had only 3 litre and a 5 litre bottle with him. But he gave them exactly 4 litres of milk. Explain how he did this.
Ans. First of all he measure 2 time with 5 litre bottle in a container which becomes
= 5Lx2 = 10L
Then he pour out 3L of milk thrice, by this way he poured out total milk
=3Lx3=9L
The milk left in the container
= 10L-9L = 1 L
Using this process the milkman gave 1 L of milk using 3 L and 5 L bottles.

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NCERT Solutions for Class 4  पर्यावरण अध्ययन  Chapter 21 खाना -खिलाना 

प्रश्न – अभ्यास

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NCERT Solutions for Class 4  पर्यावरण अध्ययन  Chapter 22 दुनिया मेरे घर में 

प्रश्न – अभ्यास

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NCERT Solutions for Class 4  पर्यावरण अध्ययन  Chapter 23 पोचमपल्ली

प्रश्न – अभ्यास

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NCERT Solutions for Class 4  पर्यावरण अध्ययन  Chapter 24 दूर देश की बात 

प्रश्न – अभ्यास

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NCERT Solutions for Class 4  पर्यावरण अध्ययन  Chapter 25 चटपटी 

प्रश्न – अभ्यास

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NCERT Solutions for Class 4  पर्यावरण अध्ययन  Chapter 26 फ़ौजी वहीदा 

प्रश्न – अभ्यास

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NCERT Solutions for Class 4  पर्यावरण अध्ययन  Chapter 27 कोशिश हुई कामयाब

प्रश्न – अभ्यास

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NCERT Solutions For Class 12 History Chapter 13 Mahatma Gandhi and the Nationalist Movement Civil Disobedience and Beyond

                NCERT TEXTBOOK QUESTIONS SOLVED

1.How did Mahatma Gandhi seek to identify with the common people?
Ans. Mahatma Gandhi seeked to identify himself with the common people of India. For this action plan
(a)He began to live in a very simple life style. He wore simple clothes which a poor Indian would wear.
(b)He spoke the language of local people.
(c)Mahatma Gandhi opposed the caste system and attacked untouchability personally lived with the Harijan.
(d)Mahatma Gandhi attached dignity to labour and physical work. He worked on Charkha and cleaned toilets.
(e)He attacked the sentiment of the feeling of classifying people into low and high.

2.How was Mahatma Gandhi perceived by the peasants ?
Ans. India is a country of villagers and vast number of Indians are engaged in farming. Mahatma Gandhi knew that during freedom struggle his focus was to address the issues of farmers. He dressed like farmers. His involvement in Indian politics began in Champaran when he successfully resolved the issues of farmers. He stood for farmers against excesses of the British government like high taxes and oppressive tax collections.
Apart from all the above, mystery also surrounded the personality of Mahatma Gandhi. Many believed he was endowed with supernatural powers. Stories spread that those who spoke ill of Mahatma Gandhi suffered natural calamities.
Thus, farmers perceived Mahatma Gandhi as their saviour and still many believed he was bestowed with the power to perform miracles.

3.Why did the salt laws become an important issue of struggle?
Ans. Poorest of poor Indian consume food that has salt as one of its prime ingredient. British government brought tax on salt and making salt indigenously was forbidden. It was to become a big burden on the poor people of India. Some important points regarding salt law are as follows.
1.Salt law was to lead to monopoly of salt production and distribution. It was to fuel prices, and added to this was the tax levied by the government.
2.People were denied access to natural salt and tons of the same were destroyed.
3.Salt law was an attack on the local industry in the villages too.
Hence salt law was extremely unpopular and it became an important issue of the struggle.

4.Why are newspapers an important source for the study of national movement?
Ans. Contemporary newspapers are an important source of the study of national movement. Following points lay bare their importance as source of history with reference to Indian Freedom Movement.
(a)Many contemporary newspapers were published by those who were involved in the freedom struggle. For example, National Herald was issued by Motilal Nehru, further Mr Jinnah issued Dawn. These nespapers were mouthpieces and represented important voices of the movement. Hence, they made important source of information regarding the freedom movement.
(b)Newspapers do daily reporting, hence, their reporting is more detailed than perhaphs any other source can be. As they report on extremely recent events, the chances of misreporting is less. Reading different nespapers further makes our reading balanced and free from bias.
(c)Many newspapers were in local Indian languages, i.e. in vernacular languages and their circulation was limited. Hence, they published newspaper from local perspective which other sources of history may not have.
(d)They reflect the mood of the people too. These newspapers shaped what was published and the way events were reported. Accounts published in a London newspaper would be different from a report in an Indian nationalist paper.

5.Why was the charkha chosen as a symbol of nationalism?
Ans. Gandhiji used to work on charkha. He made it a symbol of our freedom movement. Following are the reasons for making it the symbol of our freedom struggle.
(a)Charkha symbolised manual labour.
(b)Gandhiji wanted to attach respect to manual labour. On charkha people worked with their own hand.
(c)Charkha was a low investment product hence anyone can afford it. It was a boost to the small scale industries.
(d)Charkha as it dignified manual labour. It also promoted the culture of doing one’s own work. It would also strike at the root of caste system.
(e)Charkha was used as tool to keep British imported clothes.Thus, Charkha became a symbol of Indian nationalism.

6.How was non-cooperation a form of protest?
Ans. Gandhiji believed that British empire in India could survive as long as the local people were cooperating with the foreign rule. Non-cooperating with the British government was to weaken it and also to protest against the same. Following points explain how it was a protest:
1.Non-coperation movement came along with the Khilaphate movement, The British has not seen Hindu Muslem unity of this level ever in history. The protest of the people was unified cutting across communities and at great scale.
2.People boycotted the pillars and symbols of British rule, courts, colleges and government offices. Lawyers stopped going to courts and students stayed away from colleges. At many places alternate arrangements were done to solve litigations out of court. Further many education institutions were established by the leaders of freedom struggle where students can study. One of them is Jamia Millia University in Delhi which exits today as one of the most reputed seats of higher education in India.
3.People boycotted tax collection also and they refused to pay taxes.
Thus, non-cooperation was a kind of protest too.

7.Why were the dialogues at the Round Table Conference inconclusive ?
Ans. The British Government has had the policy to review the progress of self-rule in India and bring reforms after the gap of ten years. This began in 1910 with Morley Minto Reform and was followed in 1920 with Montague Chemsford Report. Ten years later British government invited Round Table Conference in London for the way forward. The First Round Table Conference took place in November, 1930. The Conference failed as the most important stake holder of Indian Freedom Movement, the Indian National Congress was absent in the conference. The leaders of the Congress were behind bars due to civil disobedience movement.
The Second Table Conference took pace in February 1931. One month earlier Mahatma Gandhi was released from the jail. Hence, he participated in the conference. Gandhi Irwin pact was signed and the British government agreed to withdraw salt law partly. But the agreement came under criticism as it did not talk about complete independence of India.
Third and the most important Round Table Conference took place in the later part of 1931. The new constitutional developments were not agreed upon. The main reason was that the other participants of the conference described Congress as representative of small group of Indians and not the entire population. The major voice of dissent were, the Moslem League that claimed itself the sole representative of the Moslems in India, Dr B.R. Ambedkar claimed himself the sole representative of the low castes in India and the native rulers also claimed they would deal with the British independently and Congress could not have any say in that.
To conclude divisive politics of Moslem League, Dr Ambedkar and the attitude of the princely states are the main reasons for the failure of the round table conferences.

8.In what way did Mahatma Gandhi transform the nature of the national movement?
Ans.Gandhiji came to India back from South Africa in 1915. In 1917 he went to Champaran in Bihar to fight for the cause of farmers who were forced to grow indigo by the British government. The farmers movement proved successful as the British government accepted the demands of the farmers. Since that time to 1943 when he was assassinated, he occupied the central place in the politics of India. The fact is Mahatma Gandhi is the chief protagonist of the Indian Freedom Struggle.
Mahatma Gandhi changed the nature of freedom movement and this can be elaborated by the following points:
1.When Gandhiji joined Indian politics, the freedom movement was limited to the middle class. Everybody who participated in the political movements was educated and product of the English education. Gandhiji made it all pervasive, now people from villages, poor people, labours, workers, and students all became part of the freedom struggle. However, there are people who find fault with the act of Gandhiji. They point out that Mahatma Gandhi used religious symbols to popularise the freedom movement that in long term gave fillip to communal politics. It is notable that the Age of Gan-lhi is also the age of the Rise of Moslem League in Indian politics. Eminent author Nirad C Choudhary has also criticised Mahatma Gandhi for making the freedom movement a mass movement by short cuts.
2.Mahatma Gandhi has to be credited with emancipation of women and their participation in the public life at a scale not known in Indian history. Women were very prominent in picketing activities against shops selling foreign goods. The freedom movement gave some prominent woman leaders viz. Sarojini Naidu, Rajkumari Amrit Kaur and many more.
3.For Mahatma Gandhi freedom movement was also a platform for social reforms. He spoke in favour of place of dignity and respects for depressed classes. He made end to untouchability a fundamental objective of his political philosophy.
Thus Mahatma Gandhi made freedom movement a mass movement and a movement much beyond politics.

9.What do private letters and autobiographies tell us about an individual ? How are these sources different from official accounts ?
Ans. Private letters and autobiographies are important source of individual’s life and views. Many of our freedom struggle leaders wrote autobiographies and letters and today they are our great record about them and history too.
The autobiographies and letters tell us the following things about an individual.
1.Autobiographies and letters throw light on the interests of an individual. Let us take an example, Nehru wrote letters to his daughter Indira describing the events of world history, today it is known as the book, ” Glimpses of the World History”. These letters show that Nehru had great interest in history. These letters show also the views of the author. For example, Nehru talks highly of the socilaist government of USSR in his autobiography.
2.These autobiographies and letters are a good source of information of the social life of those days in India. Dr Rajendra Prasad has given vivid description of the village life that he saw as a child in his village.
3.Above all these autobiographies and letters are great source of history too. Nehru in his autobiography has explained in details about the obstinate approach of Moslem League towards solving the minority problem in India.
These sources were diffferent from the official accounts. This is manifested in the following points:
1.The official accounts are done by individuals but they work under the guidelines of the government. Thus, views that run against the government remain stifled. In addition, the author would not have the freedom of focused area. He would be required to write only on topics already defined. However, in autobiographies and letters one can choose anything of personal interest. Dr Rajendra Prasad gives a vivid description of his school and college days in his autobiography. This is not possible in any government account.
2.The autographic letters throw light on the personal life of individual leaders and show these events shaped the thought process of these leaders in future life. Mahatma Gandhi described how he was thrown out of the first class compartment of the train in South Africa because he was not a white man. He describes the struggle inside on how to protest and later how he took to non -violent means of protest.

10.Find out about the route of the Dandi March. On a map of Gujarat plot the line of the march and mark the major towns and villages that it passed along the route.
Ans. Dandi March was started from Sabarmati Ashram. This Ashram is in Ahmedabad (Gujarat). The route followed from Ahmedabad to Vadodara and from there to Surat. We have used triangle A, B and C to mark the Dandi expedition route.

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NCERT Solutions For Class 12 History Chapter 14   Understanding Partition Politics, Memories, Experiences

NCERT TEXTBOOK QUESTIONS SOLVED

1.What did the Muslim League demand through its resolution of 1940?
Ans. An important resolution was passed by the Muslim League on 23rd March, 1940. This resolution was drafted by Sikandar Hayat Khan, the leader of the Unionist Party and the Punjab Premier. Through this, the Muslim League demanded an autonomy for the Muslim -majority areas of the subcontinent. But in the resolution there was no mention either of the partition of the country or the creation of Pakistan.
Sikandar Hayat Khan was opposed to the idea of the formation of Pakistan. He opined of a loose federation with a lot of autonomy for the states.

2.Why did some people think of Partition as a very sudden development ?
Ans. Some people think that partition of India in 1947 was a sudden development. Many Muslim leaders were not serious in their demand for Pakistan as a separate nation. On many occasions, Jinnah used the idea of Pakistan to seek favours from the British and to block concessions into the Congress. Even the Muslims were confused about the idea of Pakistan. They could not think of their future in an independent country called Pakistan. Many people had migrated to the new country with the hope that they would soon come back to India as soon as the situation improved.
In fact, the partition was so sudden that nobody could imagine it.

3.How did ordinary people view Partition?(or)
Describe the harrowing experiences of ordinary people during the period of partition of India. 
Ans. For ordinary people, partition was full of challenges and brought sufferings. The division was not a territorial division for them. It was also not a party politics of Congress and the Muslim League for them. But for the ordinary people, partition was a challenge for them. It brought misery and troubles to them.
It meant death of their loved one, loss of property and wealth. Partition also uprooted them from their paternal land. People were forced to live in refugee camps. They were also forced to start their life once again from a new platform. So for ordinary people, partition was not a pleasant experience, but it was painful and full of sufferings.

4.What were Mahatma Gandhi’s arguments against Partition?
Ans.Mahatma Gandhi was in favour of unity among various communities of the country. He was a firm supporter of religious harmony. He never supported the idea of partition. He did not want the separation of the Muslims from the Hindus who had been living together for centuries.
In his view partition was wrong. He was ready to sacrifice his life for an undivided India. But he was not ready to accept the partition. In his view, Islam stood for unity and brotherhood of mankind and not for separation. So he said that the demand of Pakistan by the Muslim League was un-Islamic and sinful.In his view those who favoured the partition were enemies of both Islam and India.He opined the Hindu and the Muslims belonged to the same land. They were living in India together for centuries. They shared the same land, same food . They drank the same water. They speak the same language and they live in peace and harmony: So he appealed to the Muslim League not to demand for a separate nation.

5.Why is Partition viewed as an extremely significant marker in South Asian history ?
Ans.The following reasons can be put forward for the given view:
•The partition of India had a unique nature. This partition was based on religions. The partition took place in the name of the communities. History has never witnessed such type of partition.
•The partition marked a severe violence. Innumerable people were killed. People began to kill each other irrespective of their earlier relation. Earlier they lived with each other in harmony and peace but now started to kill each other. Government machinery failed to check this.
•People faced a lot of problems. Their life became miserable. Their near and dear ones were killed. Many people were abducted.
•People moved across the border. Most of the Muslims of India crossed over to Pakistan and almost all Hindus and Sikhs came to India from Pakistan. They were forced to start their life afresh.
•People lost all their movable and immovable property all of a sudden. They became homeless and forced to live in refugee camps.

6.Why was British India partitioned ?
Ans. Several factors can be attributed for the partition of British India. Some of them are discussed below:
Role of Communal Parties and Organisations: Several historians and scholars think that the main purpose of the foundation of the Muslim League was to serve the interests of the Muslims. In retaliation, the Hindu Mahasabha was founded. The Muslim League was demanding more and more political rights for the Muslims. In retaliation of this, some of the Hindus took steps and established the Hindu Mahasabha in the year 1915. The Hindu Mahasabha also demanded more political rights and representation of the Hindus in the different government organizations. Following in the footsteps, the Sikh League was founded. Akali Dal also put forward demand for their people. Directly or indirectly, these political parties helped separation. They created feeling of separation and isolation among different communities.
British Policy: In India , the British followed the policy of Divide and Rule. In India, before the arrival of the British, the Hindus and the Muslims lived happily. There was unity, mutual cooperation and brotherhood among them. But the British did not like this. They sowed the seeds of dissension and followed the policy of Divide and Rule. Most of the historians believe that this policy of Divide and Rule was the main reason of the partition.
The British historians, journalists and writers propagated through their writings that Muslim invaders made the Hindus enslaved and they had been exploited for centuries. Role of British Government: The British Government also encouraged partition. The British Government encouraged the Muslim League to demand for a separate state. They tried to disrupt the movement of independence by playing the game of imperialism.
Role of Leaders: Role of leaders was also responsible for the partition. Under the leadership of Jinnah , the Muslim League moved a resolution at Lahore demanding a measure of autonomy for the Muslim majority area and after that a new nation called Pakistan.The great poet Mohammad Iqbal also spoke about the need for a Muslim state in north west India as early as in 1930.

7.How did women experience Partition?
Ans.For women, partition was horrible. Women were raped , abducted and many times forced to live with strangers and start a new life. They were deeply traumatised and began to develop new family bonds in the changed circumstances.
Women became victims on both the sides of the border. They were forced to live in a strange circumstances. But the government officials of both the countries did not take any serious step to consult those women. Women were left on their fate.
They were even murdered by their own family members. When the men realized that the women of their family would fall into the hands of the enemy, they killed their women with their own hands. To escape from the hands of enemy, in a Sikh village, ninety women were said to have voluntarily jumped into a well.

8.How did the Congress come to change its view on Partition?
Ans. Initially the Indian National Congress was not in favour of the partition. But in March, 1947, the Congress high command agreed to Punjab into two parts. One part would consist of the Muslim -majority areas and the other part would consist of the areas having Hindu-Sikh majority. To most of the Sikh leaders and Congress leaders, partition of Punjab was a necessary evil. The Sikhs feared that their denial to the partition of Punjab may lead them to be overpowered by the Muslims. They would be under control of the Muslims.Situation was the same in Bengal. The Bhadralok Bengali Hindus of Bengal wanted to retain political power with them. They were also apprehensive of the Muslims. In Bengal, the Hindus were in minority. So they favoured the partition. They thought that partition would help them to retain political dominance. These reasons forced the Congress to change their view on Partition.

9.Examine the strengths and limitations of oral history. How have oral-history techniques furthered our understanding of Partition?
Ans. Oral history techniques help historians to write experiences of people during the time of partition. In fact, history of partition has been reconstructed with the help of oral narratives. It is not possible to extract such kind of information from government records. Government would not provide such information which paint them in bad colour. It will also not tell about the daily development of the events during the partition. Moreover, Government was involved in negotiation. Documents of government deal with policy matters and throw light on efforts of major political parties.
But the oral history tells the day to day account. It is told by the people who have actually gone through the trauma and pains of the partition.But the oral data is not free from limitations. Oral data lacks concrete details. It does not have the chronological order. Oral accounts are concerned with tangential issues and that small individual experiences are irrelevant to the unfolding of the larger canvas of history. In oral history people may not talk their personal aspects. They can hide even their fault or fault of their community as a whole. Many people may not remember all events. People tend to forget also. Accuracy of narration can also be questioned.

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NCERT Solutions For Class 12 History Chapter 15 Framing the Constitution The Beginning of a New Era

NCERT TEXTBOOK QUESTIONS SOLVED

l.What were the ideals expressed in the Objectives Resolution ?
Ans.Jawahar Lai Nehru presented the Objectives Resolution in the Constituent Assembly on 13 December, 1946 . It gave a brief account of the ideals and objectives of the Constitution. These are following:
•India was declared an independent sovereign republic .
•Justice, equality and fraternity were assured to all the citizens of India.
•Adequate safeguards were provided to minorities. It also referred to the well-being of the backward and depressed classes.
•It was made an objective that India would combine the liberal ideas of democracy with the socialist idea of economic justice.
•India would adopt that form of government which would be acceptable to its people. No imposition from the British would be acceptable by the people of India.
•India would work for peace and human welfare.

2.How was the term minority defined by different groups?
Ans.N.G. Ranga, a socialist who had been a leader of the peasant movement, urged that the term minorities be interpreted in economic terms. The real minorities were the poor and the downtrodden. Some considered that the real minorities were the masses of our country who were so depressed and oppressed that they were ot even able to take advantage of the ordinary civil rights. Singh spoke eloquently on the need to protect the tribes, and ensure conditions that could help them come up to the level of the general population.

3.What were the arguments in favour of greater power to the provinces ?
Ans.K.Santharam, a member from the Madras defended the rights of the states in the Constituent Assembly. He emphasised the need to strengthen the states.He was not in favour of vesting more powers with the Centre. He was of the opinion the Centre would not be able to perform its duties efficiently in case it is over-burdened. The Centre will become automatically strong if all states are made stronger. He advocated that the Centre should be given less powers and states should be given more powers.Proposed allocation of powers between the Centre and States was also a matter of concern for K. Santharam. He felt that such a distribution of power would cripple the states.

4.Why did Mahatma Gandhi think Hindustani should be the national language?
Ans.In view of Mahatma Gandhi Hindustani was a language that the common people could easily understand. Hindustani was a blend of Hindi and Urdu. It was also popular among a large section of the people. Moreover, it was a composite language enriched by the interaction of diverse cultures. Words and terms from many different languages got incorporated into this language over the years.It made this language easily understandable by people from various regions.
As per Mahatma Gandhi Hindustani would be the ideal language of communication between the communities. It would help to unify Hindus and Muslims and the people from north and south.Language came to be associated with the politics of religious identities from the end of the 19th century. But Mahatma Gandhi retained his faith in the composite character of Hindustani.

5.What historical forces shaped the vision of the Constitution ?
Ans. Following are some historical forces which shaped the vision of the Constitution. Certain basic values were accepted by all national leaders as a result of the Nehru Report and the Fundamental Rights Resolution passed the Karachi session of the Indian National Congress.Universal Adult Franchise, Right to Freedom and Equality and Protection of minority rights were these basic values.After the results of 1937 elections, the Congress and other political parties were able to form the governments in the provinces. This experience with legislative and political institutions helped in developing an agreement over institutional design.Many colonial laws were also the sources of the Indian Constitution. Government of India Act, 1935 was a major one. This wray, the Indian Constitution adopted many institutional details and procedures from the colonial laws.
The French Revolution also inspired the makers of the Constitution.The working of the Parliamentary democracy in Britain and the Bill of Rights in the USA also inspired the framers of the Constitution.

6.Discuss the different arguments made in favour of protection of the oppressed groups.
Ans. It was felt that oppressed classes like tribals and untouchables required special attention and safeguards to enable them to raise their status and come to the level of the general population.
Tribals were regarded backward. They were not accepted well in society. They were almost rejected. For their upliftment they were required to be assimilated in the society. They were also required to be brought into the mainstream of the society. So special protection and care were offered to them.
In society untouchables were treated as labourers. Society used their services but did not give them respectable position. They were treated as outcast and kept isolated. Their sufferings were due to their systematic marginalization.
Lands of the tribals have been confiscated and had been deprived of their forests and pastures. Tribals and untouchables had no access to education. They did not take part in administration. So some legislations were required to improve their conditions.

7.What connection did some of the members of the Constituent Assembly make between the political situation of the time and the need for a strong Centre?
Ans.On 15 of August 1947, India became independent from the British rule. It was declared an independent country. But this independence was painful also. India was divided and Pakistan came into existence. This partition was marred with communal violence. So many leaders like Jawaharlal Nehru and Ambedkar favoured a strong Central Government for India. For their view they referred riots and violence that were ripping the nation apart.
It was also felt that a strong centre was the need of the hour. Most of the members of the constituent Assembly also supported this view. Any deviation from this might jeopardize the interests of the nation. Peace, prosperity and political stability was not possible in case of a weak centre. It would fail to coordinate vital matters of common concern.
So Gopalaswami Ayyangar appealed to all the members of the Constituent Assembly that” the Centre should be made as strong as possible.”
It was also felt that only a strong and united centre could plan for the well-being of the country. Balakrishna Sharma also stated the similar view. It was also felt that it would mobilize all the resources , ensure strong defence against any aggressor and establish a proper administration.
Almost all the members of the Constituent Assembly supported a strong central government. They felt that it was necessary to check chaos, communal violence and to usher economic development of the country.

8.How did the Constituent Assembly seek to resolve the language controversy?
Ans.India is very big country. It has many different regions. Different varieties of people live here and speak different languages. So for a new nation like India it was necessary to give proper attention to the intricacies of different languages.
Hindustani: Hindustani was a choice for the Congress and Mahatma Gandhi. Congress had already decided to adopt Hindustani as the national language of the country. Mahatma Gandhi was also in favour of adopting Hindustani as the national language and supported strongly for this view. He argued that everyone should speak in a language which is understood by most of the common people. Hindustani was not a new language. It was a blend of Hindi and Urdu. It was enriched by the interaction of diverse cultures and spoken by most of the people of the country.
Hindi: R.V. Dhulekar pleaded in favour of Hindi for adopting it as the national language. He came from the United Province and a Congressman. He wanted that Hindi should be used as language of constitution-making . He even said that those who did not know Hindustani were not worthy to be the members of the Constituent Assembly.
Report of the Language Committee: The language Committee of the Constituent Assembly suggested a compromise formula in its report. It suggested that Hindi in Devnagri script should be the official language of the country and tried to resolve the issue. It also suggested that transition from English to Hindi should be gradual. It was also suggested that during first fifteen years since adoption of the Constitution, English would continue to serve as for official purposes. So it was clear that the Language Committee referred Hindi as the official language not the national language.
Threat to South: The members of the Constituent Assembly who belonged to the Southern India were apprehensive of the view. They felt that Hindi would be a threat to their provincial languages. Shankar Rao from Bombay . T.A. Ramalingam Chettiar and Mrs. G. Durgabai of Madras suggested that issue of language required utmost care and needed to be handled efficiently and dextrally. Hindi should not be thrust upon the people of South India.

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NCRT Exemplar Class 11 Biology Solutions Plant Kingdom

Multiple Choice Questions
1.Cyanobacteria are classified under (a) Protista (b) Plantae
(c) Monera (d) Algae.
Soln.(c): Cyanobacteria are classified under Kingdom Monera as they are prokaryotes. They are generally photosynthetic in nature and contain pigments, chlorophyll a, and carotenoids, etc. Nostoc and Oscillatoria are examples of this category.

2.Fusion of two motile gametes which are dissimilar in size is termed as
(a) oogamy
(b) isogamy
(c) anisogamy
(d) zoogamy.
Soln.(c): Anisogamy is fusion of two motile gametes dissimilar in size. It is observed in some species of Chlamydomonas. Oogamy is also fusion of two dissimilar sized gametes in which female gamete is larger but non-motile.

3.Holdfast, stipe and frond constitute the plant body in case of
(a) Rhodophyceae
(b) Chlorophyceae
(c) Phaeophyceae
(d) all of these.
Soln.(c): Phaeophyceae (Brown algae) are eukaryotic marine algae. The body consists of branched filamentous structure in lower forms (e.g. Ectocarpus) and parenchymatous structure in higher forms (e.g. Sargassum). The plant body is often differentiated into holdfast, stipe and lamina (frond). Lamina may be simple or divided variously and is photosynthetic.

4.A plant shows thallus level of organisation. It shows rhizoids and is haploid. It needs water to complete its life cycle because the male gametes are motile. It may belong to
(a) pteridophytes (b) gymnosperms (c) monocots (d) bryophytes.
Soln.(d): Bry ophytes are non-vascular terrestrial plants of moist habitat in which a multicellular diploid sporophyte lives as a parasite on an independent multicellular haploid gametophyte that develops multi-cellular jacketed sex organs. True roots are absent, instead rhizoids occur, which may be unicellular or multicellular. An external layer of water is essential for the swimming of male gametes to the archegonia.

5.A prothallus is
(a) a structure in pteridophytes formed before the thallus develops
(b) a sporophytic free living structure formed in pteridophytes
(c) a gametophyte free living structure formed in pteridophytes
(d) a primitive structure formed after fertilisation in pteridophytes.
Soln.(c): Prothallus is a small, flattened multicellular structure that represents the independent gametophyte generation in pteridophytes, e.g., club mosses, horsetails and ferns. In some of the pteridophytes a single prothallus bears both male and female sex organs. In others there are separate male and female prothalli.

6.Plants of this group are diploid and well adapted to extreme conditions. They grow bearing sporophylls in compact structures called cones. The group in reference is
(a) monocots
(b) dicots
(c) pteridophytes
(d) gymnosperms.
Soln.(d): Gymnosperms are those seed plants in which the seeds remain exposed over the surface of the megasporophylls because the latter are not folded to form pistils. Flowers are absent. Two types of sporophylls, microsporophylls and megasporophylls are usually aggregated to form distinct cones or strobili, pollen cones (male cones) and seed cones (female cones) respectively.

7.The embryo sac of an angiosperm is made up of
(a) 8 cells
(b) 7 cells and 8 nuclei
(c) 8 nuclei
(d) 7 cells and 7 nuclei.
Soln.(b) : Female gametophyte or embryo sac of angiosperms develops upto 8-nucleate, 7-celled state prior to fertilisation. There is a three celled apparatus (one egg cell or oosphere and two synergids), three antipodal cells and two polar nuclei. The two polar nuclei fuse to form a diploid secondary nucleus.

8.If the diploid number of a flowering plant is 36, what would be the chromosome number in its endosperm?
(a) 36 (b) 18
(c) 54 (d) 72
Soln.(C) : Endosperm of flowering plants is a triploid structure. As 2n = 36, then n = 18, therefore 3n = 54.

9.Protonema is
(a) haploid and is found in mosses
(b) diploid and is found in liverworts
(c) diploid and is found in pteridophytes
(d) haploid and is found in pteridophytes.
Soln.(a): The predominant stage in the life cycle of a moss (bryophyte) is the gametophyte which consists of two stages. The first stage is the protonema stage, which develops directly from a spore. It is a creeping, green, branched
and frequently filamentous stage. The second stage is the leafy stage, which develops from the secondary protonema as a lateral bud. It consists of upright, slender axes bearing spirally arranged leaves attached to the soil through multicellular and branched rhizoids. This stage bears the sex organs.

10.The giant Redwood tree (Sequoia sempervirens) is a/an
(a) angiosperm (b) free fern
(c) pteridophyte (d) gymnosperm.
Soln.(d): Sequoia sempervirens is a gymno-sperm. It is the sole living species of genus Sequoia. Its common names include coast red wood, California red wood. It is an evergreen, long living monoecious tree.

Short Answer Type Questions
1.Food is stored as floridean starch in Rhodophyceae. Mannitol is the reserve food material of which group of algae?
Soln. Mannitol is the reserve food material found in Class Phaeophyceae of Algae.

2.Give an example of plants with
(a) Haplontic life cycle
(b) Diplontic life cycle
(c) Haplo-diplontic life cycle
Soln. (a) Many algae such as Volvox, Spirogyra Chlamydomonas, Ulothrix, Chara, etc., represent haplontic life cycle pattern.
(b) All seed bearing plants (gymnosperms and angiosperms) such as Cycas, Pinus, Hibiscus rosq sinensis, Mangifera indica represent diplontic life cycle pattern.
(c) Haplodiplontic life cycle pattern is shown by bryophytes and pteridophytes, e.g., Marchantia, Marsilea, Sphagnum, Funaria, Selaginella, Equisetum.

3.The plant body in higher plants is well differentiated and well developed. Roots are the organs used for the purpose of absorption. What is the equivalent of roots in the less developed lower plants?
Soln. In less developed lower plants such as bryophytes, root-like structures called rhizoids are present instead of roots for the purpose of absorption.

4.Most algal genera show haplontic life style. Name an alga which is
(a) Haplo-diplontic (b) Diplontic
Soln.
(a): Algae such as Ectocarpus and Polysiphonia represent haplo-diplontic life cycle’pattern.
(b) Algae such as Fucus and Sargassum represent diplontic life cycle-pattern.

5.In bryophytes male and female sex organs are
called and .
Soln. In bryophytes male and female sex organs are called antheridia and archegonia.

Short Answer Type Questions
1.Why are bryophytes called the amphibians of the Plant Kingdom?
Soln. Bryophytes are non vascular terrestrial plants of moist habitats in which a multicellular diploid sporophyte lives as a parasite on an independent, multicellular, haploid gametophyte that develops multicellular jacketed sex organs. Bryophytes are called as amphibians of Plant Kingdom because they require an external layer of water on the soil surface for their existence. The external layer of water is required for :
(i) Dehiscence of antheridia and archegonia
(ii)Swimming of male gametes to archegonia
(iii)Protection from transpiration and hence dessication
(iv)Protection from transpiration and hence dessication as the plant body is not covered by cuticle
(v)Supply of water to all parts through capillary in the absence of vascular tissues.

2.The male and female reproductive organs of several pteridophytes and gymnosperms are comparable to floral structures of angiosperms. Make an attempt to compare the various reproductive parts of pteridophytes and gymnosperms with reproductive structures of angiosperms.
Soln. Comparison of various reproductive parts of pteridophytes and gymnosperms with reproductive structure of angiosperms is as follows:

3.Heterospory i.e., formation of two types of spores – microspores and megaspores is a characteristic feature in the life cycle of a few members of pteridophytes and all spermatophytes. Do you think heterospory has some evolutionary significance in PlantKingdom?
Soln. The occurrence of two kinds of spores in the same plant is called as heterospory. Among them the smaller spore is called microspore and the larger spore is called megaspore. Heterospory has evolutionary
significance in Plant Kingdom that has been described as follows:
(i) It expresses sex determining capability of the plant. In homosporous species, differentiation of sex takes place at the gametophytic stage whereas in heterosporous species the sex of the gametophyte can be predicted at the spore stage.
(ii) Biological significance of heterospory is that in heterosporous forms development of gametophyte is endosporic and the nutrition for the developing gametophyte is derived from the sporophyte. Hence, the development of gametophyte is not affected by ecological factors.
(iii) An important characteristic of seed habit is that the megaspore is retained by the parent even after fertilisation, which ensures nutrition for the developing embryo.

4.How far does Selaginella one of the few living members of Lycopodiales (Pteridophytes) fall short of seed habit.
Soln.Selaginella is the most common genus of the heterosporous pteridophytes. Most of the species of Selaginella are heterosporous and they have only one functional megaspore mother cell which gives rise to four megaspores after meiosis. Only a single functional megaspore in a sporangium is present in Selaginella rupestris, S. monospora and S. erythropus like spermatophytes. The development of female gametophyte, fertilisation and embryo development takes place within the megasporangium. Thus, evolution of seed habit took place in such species of Selaginella. However, the seeds developed in these species cannot be called true seeds. They fall short of seed habit due to:
(i)Absence of integument like covering around the megasporangium.
(ii)Occurrence of thick wall around the functional megaspore and absence of nucellus like tissue to provide nourish-ment to female gametophyte.
(iii)Absence of resting stage after embryo development.
(iv)Development of embryo accompanied with the development of shoot and rhizophore.
(v)Megasporangium containing the embryo is never shed as seed.

5.Each plant or group of plants has some phylogenetic significance in relation to evolution: Cycas, one of the few living members of gymnosperms is called as the ‘relic of past’. Can you establish a phylogenetic relationship of Cycas with any other group of plants that justifies the above statement?
Soln.Cycas is one of the few living members of gymnosperms and is called as the ‘relic of past’. Cycas has phylogenetic relationship with pteridophytes which justifies the above statement. These are discussed as follows:
(i)Stem when young is underground and subterranean.
(ii)Leaf base scars are like the tree ferns.
(iii)Young leaves exhibit circinate vernation.
(iv)Sporophylls are leaf like structure.
(v)Microsporophylls are aggregated to form male strobilus : Microsporangia are aggregated in definite groups known as sori on the lower or adaxial surface of the microsporophylls.
(vi)Sperms are multiciliated and motile structures.
(vii)Megasporophylls are not organised into cones instead they occur in close spirals in acropetal succession around the stem apex of the female plant.

6.The heterosporous pteridophytes show certain characteristics, which are precursor to the seed habit in gymnosperms. Explain.
Soln. The heterosporous pteridophytes show certain characteristics which are precursor to the seed habit in gymnosperms. These are discussed as follows:
(i) Formation of two types of spores, microspores and megaspores and hence two types of gametophytes, male and female.
(ii)Gametophytes are nutritionally dependent on parent sporophyte.
(iii)Gametophytes show precocious development.
(iv)In some species, a single megaspore mother cell is functional and gives rise to four megaspores. Out of these, one functional megaspore develops into a megasporangium, e.g., S. monospora.
(v)In some species, the megaspore is not shed but develops completely inside partially opened megasporangia, e.g., S. apus, S. rupestris. Microspores reach there and form male gametophyte. Fertilisation and development of embryo also occur there. However, seed formation does not occur.

7.Comment on the life cycle and nature of a fern prothallus.
Soln. In fern the diploid sporophyte is represented by a dominant, independent, photosynthetic, vascular plant body. It alternates with multicellular, saprophytic/ autotropic, independent but short-lived haploid gametophyte. Such a pattern is known as haplo-diplontic life cycle. Prothallus in fem is green thalloid, nonvascular, free living, inconspicuous,small,multicellular,independent gametophyte which is monoecious, i.e., bears both the types of sex organs, male antheridia and female archegonia.

8.How are the male and female gametophytes of pteridophytes and gymnosperms different from each other?
Soln. Male and female gametophytes of pteridophytes and gymnosperms are different from each other, in the following manner.
A.Differences between pteridophytes and gymnosperms with respect to male gameto-phyte are as follows:
(i)In pteridophytes, male gametophytes may not be present whereas in gymnosperms a distinct male gemetophyte is always present.
(ii)Pteridophytes contain an antheridium whereas in gymnosperms antheridium is not formed.
(iii)Male gametes in pteridophytes are flagellate whereas in gymnosperms male gametes may or may not be flagellated.
(iv)In pteridophytes, male gametes reach the female gamete by swimming in a thin film of water whereas in gymnosperms, male gametes reach the female gamete through a pollen tube, water is not required.
B.Differences between pteridophytes and gymnosperms with respect to female gametophyte are as follows:
(i)In pteridophytes, a distinct female gametophyte may or may not be present whereas in gymnosperms, a distinct female gametophyte is always present.
(ii)In pteridophytes, female gametophyte is largely independent whereas in gymnosperms, female gametophyte does not leave the parent plant.
(iii)In pteridophytes, female gametophyte is not enclosed in an ovule whereas in gymnosperms, female gametophyte is enclosed inside an ovule.

9.In which plant will you look for mycorrhiza and corolloid roots? Also explain what these terms mean.
Soln.Coralloid roots are peculiar feature of Cycas plant. These are specialised apogeotropic roots which grow on surface of soil. They are irregular, dichotomously branched coral like roots which do not possess root hairs and root caps. Coralloid roots have a symbiotic association with blue green algae like Nostoc
and Anabaena species. The cortex of these roots possess specific algal zone where colonies
• of these blue-green algae are present. The coralloid roots possess lenticels which help in respiration.
Mycorrhiza is mutually beneficial or symbiotic association of a fungus with the root of higher plant. Mycorrhizal roots often show a wooly covering of fungal hyphae. The shape is different from normal roots. Root cap and root hairs are absent. Mycorrhizal association are present in plants like Pinus, Cedrus, Abies, and Picea.

Long Answer type Questions
1.Gametophyte is a dominant phase in the life cycle of a bryophyte. Explain.
Soln. Bryophytes are a group of the simplest and primitive plants belonging to embryophyta gametophytic and sporophytic phases are present in the life cycle of bryophytes and both these phases are morphologically distinct (heteromorphic). Gametophytic phase in bryophytes is more conspicuous, long lived, independent, green and freely branched whereas, the sporophytic phase is short lived and dependent upon the gametophyte. The main plant body of the bryophyte is haploid and bears sex organs i.e., antheridium and archegonium.
Antheridium produces a number of flagellate male gametes called sperms or antherozoids and archegonium is flask shaped with tubular neck and swollen venter. The gametophytic plant body of bryophytes is thalloid in liverworts whereas foliose in mosses. In liverworts, the thallus is differentiated into a dorsal photosynthetic and ventral storage region. Sex organs antheridia and archegonia are either distributed on the dorsal surface of the thallus or are on distinct receptacles. In mosses, the gametophyte has two growth stages (i) Protonema stage- It is the juvenile stage represented by prostrate, creeping, green and branched filamentous structure; it develops from the spore and is only a transitory vegetative stage and (ii) leafy stage or gametophore – an erect cylindrical shoot with persistent leaves and sex organs.

2.With the help of a schematic diagram describe the haplo-diplontic life cycle pattern of a plant group.
Soln. Haplo-diplontic life cycle pattern, involves sequential recurrence of two well developed somatic phases, gametophytic phase and sporophytic phase.
The sporophyte possesses diploid chromo-some number (2n). Meiosis takes place in it at the time of formation of meiospores. The haploid meiospores germinate to produce haploid gametophytes. The gametophytes produce gametes. The fusion product of gametes is a diploid zygote which develops into the sporophytic thallus of the progeny. There is a clear alternation of generations between a haploid gamete producing gametophyte and a diploid spore producing sporophyte.

In bryophytes, a dominant, independent, photosynthetic, thalloid or erect phase is represented by a haploid gametophyte and it alternates with the short-lived multioelluler sporophyte totally or partially dependent on the gametophyte for its anchorage and nutrition.
In pteridophytes, the diploid sporophyte is represented by a dominant, independent, photosynthetic, vascular plant body. It alternates with multicellular, usually autotrophic, independent but short-lived haploid gametophyte.
While most algal genera are haplontic, some of them such as Ectocarpus, Polysiphonia, kelps are haplo-diplontic.

3.Lichen is usually cited as an example of ‘symbiosis’ in plants where an algal and a fungal species live together for their mutual benefit. Which of the following will happen if algal and fungal partners are separated from each other?
(a).Both will survive and grow normally and independent from each other.
(b).Both will die
(c).Algal component will survive while the fungal component will die.
(d).Fungal component will survive while algal partner will die.
Based on your answer how do you justify this association as symbiosis.
Soln.(a) Lichens are small group of plants which represent symbotic association between fungi and algae. The algal component is known as phycobiont and the fungal component is known as mycobiont. Lichens have a composite thallus of a specific structure which neither resembles algae nor fungi. The algal components of lichens forms carbohydrates by photosynthesis whereas the fungal component is responsible for absorption and retention of water. Lichens cite good example of mutualism, where algae and fungi can survive in adverse conditions, viz high temperature and dry conditions for long time. Both fungal and algal species found in lichens are capable of living independently, although habitat requirements may be greatly different from those of the lichen.
In nature, algal and fungal components of a lichen remain together sharing an obligatory relationship. Algal and fungal partners can be separated only under laboratory conditions. In this case, both will survive and grow normally if provided with suitable growth conditions. However, on a barren rock or dry environment, algal and fungal Components of lichen cannot grow independently, if separated. Therefore they get mutually associated to form lichens for survival and perpetuation in such unfavourable conditions. ,

4.Explain why sexual reproduction in angiosperms is said to take place through double fertilisation and triple fusion. Also draw a labelled diagram of embryo sac to explain the phenomena.
Soln.In sexual reproduction of angiosperms, one male gamete fuses with oosphere or egg cell to form zygote. It is called generative fertilisation or syngamy. The second male gamete fuses with diploid, secondary nucleus to form triploid primary endosperm nucleus (PEN). It represents vegetative fertilisation on triple fusion. Since two types of fusions, syngamy and triple fusion take place in an embryo sac of angiosperms therefore, this phenomenon is termed as double fertilisation. Thus, in angiosperms sexual reproduction is said to take place through double fertilisation.

5.Draw labelled diagrams of
(a) Female and male thallus of a liverwort.
(b) Gametophyte and sporophyte of Funaria.
(c) Alternation of generation in Angiosperm.
Soln.(a): Labelled diagrams of female and male thallus of liverwort are as follow

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NCRT Exemplar class 11 Biology Solutions Animal Kingdom

Multiple Choice Questions
1.In some animal groups, the body is found divided into compartments with serial repetition of at least some organs. This characteristic feature is called
(a) segmentation (b) metamerism (c) metagenesis (d) metamorphosis.
Soln.(b) : Metamerism is true segmentation where external divisions correspond to internal divisions. The body is divided both externally and internally into number of segments (100-120) called metameres. E.g., In earthworm, genital papillae are present in 17lh and 19th segments and nephridipores are scattered all over body surface except first two. Metagenesis is alternation of two life forms in an organism. Metamorphosis is phenomenon of passing through different juvenile stages before attaining adult form. Segmentation is division of- body into different segments and in some organisms, only external divisions are seen as head, thorax and abdomen.

2.Given below are types of cells present in some animals. Which of the following cells can differentiate to perform different functions?
(a) Choanocytes
(b) Interstitial cells
(c) Gastrodermal cells
(d) Nematocytes
Soln.(b) : Interstitial cells are reserve cells and are called totipotent cells, which can
differentiate into any type of cells. Nemato- cvtes are stinging cells used for offence and defence. Gastrodermal cells line the gastrodermis and intracellular digestion takes place inside these cells. All these cells are found in cnidarians. Choanocytes or collar cells are found in sponges; they are specialised flagellated cells that line spongocoel and canals.

3.Which one of the following sets of animals share a four chambered heart?
(a) Amphibians, Reptiles, Birds
(b) Crocodiles, Birds, Mammals
(c) Crocodiles, Lizards, Turtles
(d) Lizards, Mammals, Birds
Soln.(b): Amphibians have three-chambered heart, while reptiles have incomplete four-chambered heart (except crocodiles). Crocodiles, birds and mammals have complete, four-chambered heart.

4. Which of the following pairs of animals has non- glandular skin?
(a) Snake and Frog
(b) Chameleon and Turtle
(c) Frog and Pigeon
(d) Crocodile and Tiger
Soln.(b) Amphibians and mammals have glandular skin, i.e., they have glands on their skin to keep it moist, which include sweat’gland and sebaceous glands, present in mammals. While aves have dry skin and does not contain any glands, except, preen/
oil gland at the base of the tail. Reptiles are devoid of glands. Therefore, Chameleon and turtle have non-glandular skin.

5.Birds and mammals share one of the following characteristics as a common feature.
(a) Pigmented skin (b) Pneumatic bones
(c) Viviparity (d) Warm blooded
Soln.(d) : Both birds and mammals are homoiotherms (warm blooded) i.e., they have ability to maintain constant body temperature inspite of changes in ambient conditions.

6.Which one of the following sets of animals belong to a single taxonomic group?
(a) Cuttlefish, Jellyfish, SHverfish, Dogfish, Starfish
(b) Bat, Pigeon, Butterfly
(c) Monkey, Chimpanzee, Man
(d) Silkworm, Tapeworm, Earthworm
Soln. (c): Monkey, chimpanzee and man are primates belonging to Class Mammalia, Phylum Chordata.

7.Which one of the following statements is incorrect?
(a) Mesoglea is present in between ectoderm and endoderm in Obelia.
(b) Asterias exhibits radial symmetry.
(c) Fasciola is a pseudocoelomate animal.
(d) Taenia is a triploblastic animal.
Soln. (c) : Fasciola (a platyhelminth) is an acoelomate animal.

8.Which one of the following statements is incorrect?
(a) In cockroaches and prawns excretion of waste material occurs through Malpighian tubules.
(b) In ctenophores, locomotion is mediated by comb plates.
(c) In Fasciola, flame cells take part in excretion.
(d) Earthworms are hermaphrodites and yet cross fertilisation take place among them.
Soln. (a): In cockroaches excretion takes place by Malpighian tubules while in prawns it takes place by green glands.

9.Which one of the following is oviparous?
(a) Platypus (b) Flying fox (Bat)
(c) Elephant (d) Whale
Soln.(a): Platypus, bat, elephant and whale, all belong to Class Mammalia of Sub-phylum
Vertebrata. Out of the four, only platypus is oviparous (egg laying), and rest all are viviparous (give birth to young ones).

10.Which one of the following is not a poisonous snake?
(a) Cobra (b) Viper
(c) Python (d) Krait
Soln. (c): Python is a large and non-poisonous snake. It kills its prey by coiling around and crushing and swallowing the prey. ,

11.Match the following list of animals with their
level of organisation.
Division of Labour                       Animal
A.Organ level                              i.Pheretima
B.Cellular aggregate level     ii.Fasciola
C.Tissue level                            iii.Spongilla
D.Organ system level              iv. Obelia
Choose the correct match showing division of labour with animal example.
(a) i-B, ii-C, iii-D, and iv-A
(b) i-B, ii-D, iii-C, and iv-A
(c) i-D, ii-A, iii-B, and iv-C
(d) i-A, ii-D, iii-C, and iv-B
Soln.(c)

12.Body cavity is the cavity present between body wall and gut wall. In some animals the body cavity is not lined by mesoderm. Such animals are called
(a) acoelomate (b) pseudocoelomate
(c) coelomate (d) haemocoelomate.
Soln.(b) : Acoelomates are characterised by absence of coelom or body cavity while coelomates are those organisms in which body cavity arises as a cavity in embryonic mesoderm and mesoderm lines the coelom. In pseudocoelomates body cavity is derived from blastocoel of embryo but is not lined with mesoderm.

13.Match the column A with column B and choose
the correct option.
Column A                               Column B
A.Porifera                              i.Canal system
B.Aschelminthes                ii.Water-vascular system
C.Annelida                           iii.Muscular Pharynx
D.Arthropoda                      iv.Jointed appendages
E.Echinodermata                v. Metameres
(a) A-ii, B-iii, C-v, D-iv, E-i
(b) A-ii, B-v, C-iii, D-iv, E-i
(c) A-i, B-iii, C-v, D-iv, E-ii
(d) A-i, B-v, C-iii, D-iv, E-ii
Soln.(c)

Very Short Answer Type Questions
1.Identify the phylum in which adults exhibit
radial symmetry and larva exhibit bilateral
symmetry.
Soln.Phylum Echinodermata.

2.What is the importance of pneumatic bones and air sacs in Aves?
Soln.Pneumatic bones and air sacs in birds (Aves) help them in flying. Pneumatic bones, i.e., hollow bones filled with air cavities reduce weight which help in flight and air sacs serve as reservoirs of air. Air sacs also aid as cooling devices and regulate body temperature.

3.What is metagenesis? Mention an example which exhibits this phenomenon.
Soln.In some cnidarians, polyps reproduce medusae asexually and medusae form the polyps sexually. Such alternation of asexual and sexual phases in the life cycle of cnidarians is called metagenesis e.g., Obelici.

4.What is the role of feathers?
Soln.Feathers are flight adaptations in birds, they provide passage for air and reduce friction to minimum. They help to maintain constant body temperature by preventing heat loss.

5.Which group of chordates possess sucking and circular mouth without jaws?
Soln.In Cyclostomata, mouth is circular and jawless. It feeds by sucking blood of host fishes. E.g:, Lamprey.

6.Give one example each for an animal possessing placoid scales and that with cycloid scales.
Soln.Placoid scales – Scoliodon (Class Chon- drichthyes)
Cycloid scales – Exocoetus (Class Osteichthyes)

7.Mention two modifications in reptiles required for terrestrial mode of life.
Soln.Two modifications in reptiles required for terrestrial mode of life are:
(i) Respiration by lungs.
(ii)Internal fertilisation.

8.Mention one example each for animals with
chitinous exoskeleton and those covered by a
calcareous shell.
Soln. Chitinous exoskeleton – Cockroach
(Arthropoda)
Calcareous shell – Pila (Mollusca)

9.What is the role of radula in molluscs?
Soln. Radnla is a rasping organ having rows
of horny teeth and used in feeding especially
for scraping and cutting.

10. Name the animal, which exhibits the
phenomenon of bioluminescence.
Mention the phylum to which it belongs.
Soln. Pleurobrachia exhibits the phenomenon of bioluminescence i.e., phenomenon of emission of light by living organisms. It belongs to the Phylum Ctenophora.

11 .Write one example each of the following in the space provided.
(a) Cold blooded animal_______
(b) Warm blooded animal_______
(c) Animal possessing dry and cornified skin_______
(d) Dioecious animal_______
Soln. (a) Rana (Frog) (b) Corvus (Crow)
(c) Crocodilus (Crocodile) (d) Nereis

12. Differentiate between a diploblastic and a triploblastic animal.
Soln. Animals, which have two germinal layers, an external ectoderm and an internal endoderm, are called diploblastic animals e.g., animals of Phylum Coelenterata and Phylum Porifera. Animals which have three layers, an external ectoderm, middle mesoderm and an internal endoderm, are called triploblastic animals e.g., animals from Phylum Platyhelminthes to Phylum Chordata.

13. Give an example of the following
(a) Round worm
(b) Fish possessing poison sting
(c) A limbless reptile/ amphibian
(d) An oviparous mammal
Soln. (a) Ascaris (b) Trygon (c) Limbless reptile – Naja naja, Limbless amphibian- lchthyophis (d) Ornithorhynchus

14. Provide appropriate technical term in the space provided.
(a) Blood-filled cavity in arthropods_______
(b) Free-floating form of cnidaria_______
(c) Stinging organ of jelly fishes_______
(d) Lateral appendages in aquatic annelids_______
Soln. (a) Blood-filled cavity in arthropods haemocoel.
(b) Free-floating form of cnidaria medusa.
(c) Stinging organ of jelly fishes nematocvst.
(d) Lateral appendages in aquatic annelids parapodia.

15 .Match the following.
Animals Locomotory Organ
(a) Octopus (i) Limbs
(b) Crocodile (ii) Comb plates
(c) Catla (iii) Tentacles
(d) Ctenoplana (iv) Fins
Soln. (a) – (iii), (b) – (i), (c) – (iv), (d) – (ii)

Short Answer Type Questions
1.Differentiate between:
a.Open circulatory system and closed circulatory system
b.Oviparous and viviparous characteristic
c.Direct development and Indirect development
Soln.(a) The differences between open circulatory system and closed circulatory system are as follows:

(b) The differences between oviparous and viviparous characteristic i.e., ovipary and vivipary are as follows:

(c)Thedifferencesbetweendirectdevelopment and indirect development are as follows:

2.Sort out the animals on the basis of their symmetry (radial or bilateral) coelenterates, ctenophores, annelids, arthropods, and echinoderms.
Soln. Radial symmetry: Coelenterates,
Ctenophores, Echinoderms (larva has bilateral symmetry).
Bilateral symmetry : Annelids, Arthropods.

3.There has been an increase in the number of chambers in heart during evolution of vertebrates. Give the names of the class of vertebrates having two, three or four-cham-bered heart.
Soln. Two-chambered heart: Pisces Three chambered heart: Amphibia, Reptilia Four chambered heart: Aves, Mammalia.

4. Fill up the blank spaces appropriately

Soln.

5.Match the following
(a) Amphibia                  (i)  Air bladder
(b) Mammals                 (ii) Cartilaginous
                                                    notochord
(c) Chondrichthyes    (iii) Mammary glands
(d) Osteichthyes           (iv) Pneumatic bones
(e) Cyclostomata           (v) Dual habitat
(f) Aves                             (vi) Sucking and circular
mouth without jaws.
Soln.(a)-(v), (b)-(iii), (c)-(ii), (d)-(i), (e)-(vi), (f)-(iv)

6.Endoparasites are found inside the host body. Mention the special structure, possessed by these and which enables them to survive in those conditions.
Soln. Most Platyhelminthes are endoparasites. Special structures in parasitic Platyhelminthes include presence of thick integument which is resistant to host’s digestive enzymes and anti¬toxins and they possess adhesive organs like
suckers in flukes and hooks and suckers in tapeworms for firm grip in host’s body.

7.Match the following and write correct choice in space provided
Animal                               Characteristics
(a) Pila                               (i)  Jointed appendages
(b) Cockroach                 (ii) Perching
(c) Asterias                      (iii) Water vascular system
(d) Torpedo                     (iv) Electric organ
(e) Parrot                           (v) Presence of shell
(f) Dogfish                        (vi) Placoidscales
(a)______, (b)______, (c)______, (d)______,
(e)______, (f)______
Soln. (a)-(v), (b)-(i), (c)-(iii), (d)-(iv), (e)-(ii), .
(f) -(vi)
8.Differentiate between:
(a) Open and closed circulatory system
(b) Oviparity and viviparity
(c) Direct and indirect development
(d) Acoelomate and pseudocoelomate
(e) Notochord and nerve cord
(f) Polyp and medusa
Soln.
(a) Refer answer 1 (a).
(b) Refer answer 1 (b).
(c) Refer ansiver 1 (c).
(d) The differences between acoelomate and pseudocoelomate are as follows:

(e) The differences between notochord and
nerve cord are as follows:

(f)The differences between polyp and medusa are as follows:

9.Givethecharacteristicfeaturesofthefollowing, citing one example of each:
(a) Chondrichthyes and osteichthyes
(b) Urochordata and cephalochordata
Soln. (a) Characteristic features of chondri-chthyes are:
(i) They are mostly marine called poikilothermic animals (i.e., they have the capacity to regulate their body temperature.
(ii)TKey have cartilaginous endoskeleton. Notochord is persistent throughout life.
(iii)Mouth is ventral in position, skin is tough containing minute placoid scales and teeth are modified placoid scales. They have strong jaw and are predaceous by nature.
(iv)Gill slits are generally five pairs and gill cover (operculum) are absent.
(v) Heart is two chambered with one auricle and one ventricle.
(vi)Some of them possess electric organs (e.g., Torpedo) and some possess poison sting (e.g., Trygon).
(vii)They are called ureotelic animals as they secrete urea.
(viii)Sexes are separate and fertilisation is internal.
(ix)Males usually have claspers on pelvic fins, which help in copulation.
(x)Many of them are viviparous.
E.g.r Scoliodon (shark or dog fish), Pristis (saw fish), Trygon (stingray).
Characteristic features of osteichthyes are as follows:
(i) They are found in marine water as well as in freshwater.
(ii) The body is streamlined to facilitate easy movement through water.
(iii) Exoskeleton contains dermal scales like cycloid or ctenoid scales.
(iv) Gills are covered with operculum (gill cover) and are generally four pairs.
(v) Heart is two-chambered with one auricle and one ventricle.
(vi) Air bladders are present. They are the hydrostatic organs.
(vii) These are ammonotelic animals that excrete ammonia.
(viii) Sexes are separate. Fertilisation is usually external, mostly viviparous and development is direct.
E.g., Labeo (rohu), Mrigal (carps), Catla (catla), Hippocampus (sea horse), Exocoetus (flying fish), Remora (sucker fish), Anabas (climbing percfi), Protopterus (African lung fish).
(b) Characteristic features of Urochordata
are:
(i) Urochordates are commonly called Tunicates because adult body is enclosed
in test or tunic.
(ii)The notochord is present only in the tail of the larva and disappears in the adult.
(iii)Adult is sedentary, while larva is motile.
(iv)Nerve cord is found only in the larva. It is replaced by a dorsal ganglion in the adult.
(v)The larva undergoes retrogressive metamorphosis.
Example: Herdmania.
Characteristic features of Cephalochordata
are:
(i) The notochord is present throughout life and extends from the anterior end to the posterior end of the body.
(ii)Both larva and adult are motile.
(iii)Nerve cord is present in both larva and adult.
(iv)The larva undergoes progressive meta-morphosis.
Example: Branchiostoma.

10 .Mention two similarities between
(a) Aves and mammals
(b) A frog and crocodile
(c) A turtle and Pila
Soln. (a) Both have 4-chambered heart and are warm-blooded.
(b) Both are cold-blooded and oviparous.
(c) Both are aquatic and their body is covered by shell.

11. Name
(a) A limbless animal
(b) A cold blooded animal
(c) A warm blooded animal
(d) An animal possessing dry and cornified skin
(e) An animal having canal system and spicules
(f) An animal with cnidoblasts
Soln. (a) Ichthyophis
(b) Chelone (Turtle) (c) Columba (Pigeon)
(d) Crocodilus (e) Sycon
(f) Adamsia

12. Give an example for each of the following
(a) A viviparous animal
(b) A fish possessing a poison sting
(c) A fish possessing an electric organ
(d) An organ, which regulates buoyancy
(e) Animal, which exhibits alternation of generation
(f) Oviparous animal with mammary gland
Soln. (a) Macropus (Kangaroo)
(b) Trygon (Sting ray) (c) Torpedo (Electric ray)
(d) Air bladder (e) Obelia (Sea fur)
(f) Ornithorhynchus (Duck-billed Platypus)

13.Excretory organs of different animals are given below. Choose correctly and write in the space provided.
Animal Excretory Organ/Unit
(a) Balanoglossus (i) Metanephridia
(b) Leech                 (ii) Nephridia
(c) Locust               (iii) Flame cells
(d) Liver fluke       (iv) Absent
(e) Sea urchin        (v) Malpighian tubule
(f) Pila                      (vi) Proboscis gland
(a)______, (b)______,(c)______,(d)______,
(e)______,(f)______
Soln. (a) – (vi), (b) – (ii), (c) – (v), (d) – (iii),
(e) – (iv), (f) – (i)

Long Answer Type Questions
1. Give three major differences between chordates and non-chordates and draw a schematic
sketch of a chordate showing those features.
Soln.Differences between chordates and non- chordates are as follows:

2.What is the relationship between germinal layers and the formation of body cavity in case of coelomate, acoelomates and pseudocoelomates?
Soln. There are three germinal layers: ectoderm, endoderm and mesoderm. Presence or absence of a cavity between the body wall and gut wall is very important in classification. The body cavity lined by mesoderm is called coelom. The animals having coelom are called coelomates. In coelomates, body cavity arises as a cavity in embryonic mesoderm. In this, mesoderm of embryo provides cellular lining to the cavity and coelom is filled with coelomic fluid, E.g., annelids, echinoderms, and chordates. In Pseudocoelomata, body cavity is derived from blastocoel of the embryo and the coelom is not lined by mesoderm. Instead, mesoderm is present as scattered pouches in between the ectoderm and mesoderm. Such a coelom is called pseudocoelom. Aschelminthes are pseudocoelomates. Acoelomates are those animals in which the body cavity is absent i.e., they do not have coelom. E.g., sponges, cnidarians, ctenophores, flatworms.

3.Comment upon the habitats and external features of animals belonging to class, amphibia and reptilia.
Soln. Class – Amphibia
Habitat : Amphibians are first cold blooded
vertebrates which can Jive on land as well
as in water. They are mostly found in warm
countries.
External features:
(i) Body is compressed and cylindrical and differentiated into head and trunk.
(ii)Nostrils are connected to the buccal cavity eyes have eyelids.
(iii)Skin is mostly smooth, moist, highly vascular and rich in gland. Scales, are generally absent. It helps in cutaneous respiration.
(iv)Amphibians mostly have two pairs of pentadactyl limbs. They are used for locomotion.
(v)A pair of external nares for olfaction, a pair of eyes with movable eyelids are present.Tympanum represents the ear. E.g., Ram (frog), Bufo (toad), Elyla (tree frog), Rhacophorus (flying frog), Salamandra (salamander) etc.
Class – Reptilia
Habitat : Reptiles are mostly terrestrial
animals, mainly found in warmer parts of the
world. Only few of them live in water such as
crocodiles, turtles.
External features :
(i) The body may be long, cylindrical or short and broad. It is divisible as head,
neck, trunk and tail.
(ii) They have dry, rough and non-glandular
skin. It is provided with horny, epidermal
scales or scutes.
(iii) Appendages are of two pairs of
pentadactyl limbs with powerful horny
claws. Limbs are locomotory organs.
(iv) Sense organs like eyes, ears and nose are
well developed. ….
(v) Eyelids and nictitating membrane are present in lizards but absent in snakes.
E.g., Calotes (garden lizard), Draco (flying
lizard), Naja (cobra), Hemidactyhis (wall lizard)

4. Mammals are most adapted among the
vertebrates. Elaborate.
Soln. Mammals are most adapted among the
vertebrates. This can be elaborated by the
following characteristics of mammals.
(i) They are found in a variety of habitats
like polar ice caps, mountains, deserts, forests and oceans.
(ii)Most of them are terrestrial but some of them are adapted to fly (Bat) or live in water (Whale).
(iii)They have 2 pairs of limbs, adapted for
walking, running, climbing, burrowing,
swimming or flying.
(iv)They have mammary glands and are the
only animals which nourish their young . ones with milk.
(v) They are capable of learning due to
presence of developed brain and thus
are dominant animals.
(vi)Presence of muscular diaphragm is a
characteristic feature of mammals which
help in breathing.
(vii)Four-chambered heart pumps only oxygenated blood.
(viii)A well developed placenta is present except in egg laying mammals.
(ix) They are warm blooded, capable of
maintaining constant body temperature.
E.g., Macropus (Kangaroo), Equus
(Horse), Macaca (Monkey) etc.

The post NCRT Exemplar class 11 Biology Solutions Animal Kingdom appeared first on Learn CBSE.

NCRT Exemplar class 11 Biology Solutions  Morphology of Flowering Plants

Multiple Choice Questions
1. Rearrange the following zones as seen in the root in vertical section and choose the correct option.
A. Root hair zone
B. Zone of meristems
C. Rootcapzone D. Zone of maturation
E. Zone of elongation
(a) C, B, E, A, D (b) A,B,C,D,E
(c) D, E, A, C, B (d) E, D, C, B, A
Soln. (a): A typical root possesses five parts
or regions:
(i) Root cap covers the root meristem. The cells of the root cap secrete mucilage which lubricates the passage of root through the soil. Cells of root cap possess starch grains which are believed to take part in graviperception. Function cap is the protection of root meristem from soil particles.
(ii) Growing point or meristematic zone produces new cells for the root cap and basal region of the root. Therefore, it is essential for the growth of the root.
(iii) Cells of zone of elongation are newly formed cells which lose the power of division. They elongate rapidly which increases length of the root. The external cells possess the power of absorption of water and mineral salts from the soil.
(iv) Root hair zone also represents the zone of differentiation or maturation. Most of the water absorption occurs in this region. Some of the outer cells of this zone give rise tq lateral tubular outgrowths called root hairs. The root hairs increase the exposed surface of the root for absorption.
(v) Region or zone of mature cells forms the bulk of the root without undergoing any – further change. The outermost layer of this region cannot help the root in water absorption because they have thick walled or impermeable cells. Its only function is to anchor the plant firmly in the soil. Lateral roots also arise from the interior of this region.

2. In an inflorescence where flowers are borne laterally in an acropetal succession, the position of the youngest floral bud shall be
(a) proximal (b) distal
(c) intercalary (d) anywhere.
Soln.(b): In acropetal succession, the youngest floral bud is towards growing point and oldest is towards the base.

3. The mature seeds of plants such as gram and peas, possess no endosperm, because
(a) these plants are not angiosperms
(b) there is no double fertilisation in them
(c) endosperm is not formed in them
(d) endosperm gets used up by the developing embryo during seed development.
Soln. (d): Majority of dicot seeds (e.g., pea, gram, bean, mustard, groundnut) and a few monocot seeds (e.g., orchids, Sagittaria), are . called nonendospermic or exalbuminous seeds because the endosperm gets consumed during seed development and the food is stored in cotyledons and other regions.

4. Roots developed from parts of the plant other than radicle are called
(a) taproots
(b) fibrous roots
(c) adventitious roots
(d) nodular roots.
Soln.(c): Adventitious roots are those which develop from any part of the plant other than radicle.

5. Venation is a term used to describe the pattern of arrangement of
(a) floral organs
(b) flower in inflorescence
(c) veins and veinlets in a lamina
(d) all of them.
Soln.(c): The arrangement of veins and veinlets on the lamina of a leaf is called
venation. Venation is of three main types reticulate, parallel and furcate.
Reticulate venation : The veinlets form a reticulum or network. Reticulate venation is found in most dicots.
Parallel venation : Veinlets are inconspicuous. Reticulations are absent. The veins run parallel to one another. Parallel venation is characteristic of most monocots.
Furcate venation : The veins branch
dichotomously. The finer branches do not form a reticulum. It is common in ferns (e.g., Adiantum). Among higher plants furcate venation is found in Circeaster.

6. Endosperm, a product of double fertilisation in angiosperms is absent in the seeds of
(a) coconut (b) orchids
(c) maize (d) castor.
Soln.(b): Refer answer 3.

7. Many pulses of daily use belong to one of the families below (tick the correct answer).
(a) Solanaceae (b) Fabaceae
(c) Liliaceae (d) Poceae
Soln.(b) : A number of legumes or pulses are obtained from Fabaceae- broad bean (Vicia faba), soyabean (Glycine max), kidney bean (Phaseolus vulgaris), cowpea (Vigna unguiculata = V. sinensis), pea (Pisum sativum), gram (Cicer arietinum), green gram (Vigna radiata = Phaseolus radiatus = P. aureus, mung), pigeon pea (Cajanus cajan, arhar), etc.

8. The placenta is attached to the developing seed near the
(a) testa (b) hilum
(c) micropyle (d) chalaza.
Soln.(d): Seed contains an embryo or miniature plant in suspended animation, adequate reserve food for future development of the embryo and a covering for protection against mechanical injury, loss of water, pathogens, etc. A seed may have one or two coverings called seed coats. The outer or the only seed coat (if one is present) is called testa while the inner one is named as tegmen. Surface of the seed possesses a fine pore at one end called micropyle. Hilum is a place where funiculus or stalk of seed is borne. Some seeds also show chalaza (place of origin of seed coats) and raphe (part of funiculus fused with seed wall).

9. Which of the following plants is used to extract the blue dye?
(a) Trifolium (b) Indigofera
(c) Lupin (d) Cassia
Soln.(b): Indigo (blue dye) is obtained from the leaves of Indigofera tinctoria and
I.sujfruticosa. The leaves contain a colourless chemical which on exposure to air turns bluish.

10. Match the followings and choose correct option
Group A Group B
A. Aleurone layer ( i)without
fertilisation
B.Parthenocarpic fruit (ii)Nutrition
C.vule (iii)Double
fertilisation
D.Endosperm (iv)Seed
Options
(a) A-(i), B-(ii), C-(iii),D-(iv)
(b) A-(ii), B-(i), C-(lv), D-(iii)
(c) A-(iv), B-(ii), C-(i), D-(iii)
(d) A-(ii), B-(iv), C-(i), D-(iii)
Soln.(b)

Short Answer Type Questions
1. Roots obtain oxygen from air in the soil for respiration. In the absence or deficiency of 02, root growth is restricted or com pletely stopped. How do the plants growing in marshlands or swamps obtain their 02 required for root respiration?
Soln. Plants growing in marshlands or swamps obtain their 02 through modified negatively geotropic tap roots called pneumatophores. Pneumatophores or respiratory roots come out of water and pick up oxygen for root respiration.

2. Write floral formula for a flower which, is bisexual; actinomorphic; sepals five, twisted aestivation, petals five; valvate aestivation; stamens six; ovary tricarpellary, syncarpous, superior, trilocular with axile placentation.
Soln. Floral formula of a flower which, is bisexual; afctinomorphic; sepals five, twisted aestivation, petals five; valvate aestivation; stamens six; ovary tricarpellary, syncarpous, superior, trilocular with axile placentation is

3. In Opuntia the stem is modified into a flattened green structure to perform the function of leaves (/’.e., photosynthesis). Cite some other examples of modifications of plant parts for the purpose of photosynthesis.
Soln. In some plants, roots are become assimilatory or photosynthetic. Adventitious roots are modified into assimilatory or photosynthetic roots in Trapa, Tinospora. Cladode is an aerial modification of stem in Asparagus to perform photosynthesis.

4. In swampy areas like the Sunderbans in West Bengal, plants bear special kind of roots called _______
Soln. In swampy areas like the Sunderbans in West Bengal, plants bear special kind of roots called pneumatophores.

5. In aquatic plants like Pistia and Eichhornia, leaves and roots are found near_______
Soln.ln aquatic plants like Pistia and £ ichhorn ia, leaves and roots are found near surface of water

6. Reticulate and parallel venation are
characteristic of_______and_______
Soln. respectively.
Reticulate and parallel venation are
characteristic of dicots. and monocots
respectively.

7. Which parts in ginger and onion are edible?
Soln. In ginger, edible part is rhizome which is modified shoot that stores food materials. The edible part of onion is fleshy scale leaves.

8. In epigynous flower, ovary is situated below
the _______
Soln.In epigynous flower, ovary is situated below other floral organs viz sepals, petals and stamens.

9. Add the missing floral organs of the given floral
formula of Fabaceae.
Soln. Floral formula of Fabaceae is

10. Namedhe body part modified for food storage in the following.
(a) Carrot________________
(b) Colocasia_______________
(c) Sweet potato_______________
(d) Asparagus_______________
(e) Radish_______________
(f) Potato_______________
(g) Dahlia_______________
(h) Turmeric_______________
(i) Gladiolus_______________
(j) Ginger________________
(k) Portulaca_______________
Soln.(a) Carrot – modified fleshy conical
tap root
(b) Colocasia-highly condeused and specialised underground stem, corm
(c) Sweet potato – tuberous fleshy
adventitious root
(d) Asparagus – fasciculated fleshy
adventitious root
(e) Radish – fusiform fleshy tap root.
(f) Potato – modified underground stem, tuber
(g) Dahlia – fasciculated fleshy adventitious root
(h) Turmeric – modified underground stem, rhizome
(i) G/fld/o/iis-highly condensed and specialised underground stem, corm
(j) Ginger – modified underground stem, rhizome
(k) Portulaca – moniliform or beaded adventi¬tious root.

Short Answer Type Questions
1. Give two examples of roots that develop from . different parts of the angiospermic plant other
than the radicle.
Soln. Two examples of roots that develop from different parts of the angiospermic plant other than the radicle are as follows:
(i) Stilt roots develop from the basal node of the main stem, examples: maize and sugarcane.
(ii) Prop roots develop from the upper part of the stem, especially the horizontal branches. Example: Banyan.

2. The essential functions of roots are anchorage and absorption of water and minerals in the terrestrial plant. What functions are associated
with the roots of aquatic plants. How are roots of aquatic plants and terrestrial plants different?
Soln. The root system in hydrophytes is feebly developed and root, root hairs and cap arq absent. In some floating plants such as Utricularia, Ceratophyllum, etc., no roots are developed, and in submerged plants such as Vallisneria, Hydrilla etc., water dissolved mineral salts and gases are absorbed by their whole surface. In plants like Pistia, Eichhornia, Lemna, etc., no root cap develops, but root pocket is formed instead. An aquatic plant is, in reality, submerged in or floating up on a nutrient solution. In hydrophytes the root system is functioning chiefly as holdfast or anchors, and a large part of the absorption takes place through the leaves and stems.

3. Draw diagrams of a typical monocot and dicot leaves to show their venation pattern.
Soln. Labelled diagrams of monocot leaves showing parallel venation are as follows:

Labelled diagrams of monocot leaves showing parallel venation are as follows
.

4.A typical angiosperm flower consists of four
floral parts. Give the names of the floral parts . and their arrangements sequentially.
Soln. A typical angiosperm flower consists of
following four parts:
(i)Calyx : This is the outermost whorl of the flower and first whorl of the non-essential organs. Calyx consists of sepals. Sepals are usually small, green and protect the other floral parts in the bud condition.
(ii)Corolla : Corolla is the next whorl of non¬essential organs. It is composed of petals. Petals are usually brightly coloured to attract insects for pollination.
(iii)Androecium : Androecium forms the third whorl of the flower. Androecium consists of stamens which are the male reproductive organs. Stamen is morphologically equivalent to microsporophyll.
(iv)Gynoecium : The gynoecium or pistil is composed of one or more carpels. It is innermost whorl.

5. Given below are a few floral formulae of some well known plants. Draw floral diagrams from these formula.

Soln.(i) Floral diagram for floral formula,

(ii)Floral diagram for floral formula,


(iii)Floral diagram for floral formula,

6. Reticulate venation is found in dicot leaves while in monocot leaves venation is of parallel type. Biology being a ‘Science of exceptions’, find out any exception to this generalisation.
Soln. Exceptionally, reticulate venation is present in monocot leaves such as Smilax Dioscorea etc., and parallel venation is present in dicot leaves such as Calophyllum, Eryngium etc.

7. You have heard about several insectivorous plants that feed on insects. Nepenthes or the pitcher plant is one such example, which usually grows in shallow water or in marsh lands. What part of the plant is modified into a ‘pitcher’? How does this modification help the plant for food even though it can photosynthesise like any other green plant?
Soln. Insectivorous plants are those plants which grow in marsh lands or soils deficient in nitrogen. These plants contain chlorophyll and are capable of manufacturing carbohydrates by the process of photosynthesis but in order to fulfil their nitrogen requirement they capture and digest insects. In insectivorous plant Nepenthes, the leaf lamina gets modified into pitcher like structure and the apex of lamina changes into lid of pitcher. Pitcher contains digestive enzymes which digest the trapped insects. This type of modification supplements nitrogen nutrition in the nitrogen deficient substratum

8. Mango and coconut are ‘drupe’ type of fruits. In mango fleshy mesocarp is edible. What is the edible part of coconut? What does milk of tender coconut represent?
Soln. Mango and coconut are drupe type of fruits. In mango fleshy mesocarp is edible whereas in coconut, endosperm is edible part. Coconut meal represents cellular endoperm whereas milk of tender coconut represents the nuclear endosperm in liquid form.

9. How can you differentiate between free central and axile placentation?
Soln. Differences between free central and axile placentation are as follows:
(i) In axile placentation, ovary is partitioned into chambers whereas in free central placentation ovary is single chambered.
(ii)The ovule bearing column occurs in the central region where septa meet in axile placentation whereas in free central placentation ovule bearing column lies free in the centre of ovary and septa are absent.

10. Tendrils are found in the following plants. Identify whether they are stem tendrils or leaf tendrils.
(a) Cucumber
(b) Peas
(c) Pumpkins
(d) Grapevine
(e) Watermelons
Soln.
(a) Cucumber: Stem tendril.
(b) Peas: Leaf tendril.
(c) Pumpkins: Stem tendril.
(d) Grapevine: Stem tendril.
(e) Watermelon: Stem tend i ii.

11 .Why is maize grain usually called as a fruit and not a seed?
Soln. Maize grain is caryopsis type of fruit which develops from monocarpellary pistil
with superior unilocular uniovuled ovary. The thin pericarp is inseparable and completely fused with the seed coat. It is most advanced kind of dry fruit.

12. Tendrils of grapevines are homologous to the tendril of pumpkins but are analogous to that of pea. Justify the above statement.
Soln. Tendrils of grapevine and tendrils of pumpkin have the same origin i.e., they are modification of stem, but different functions such as tendrils of grapevine help in climbing whereas tendrils of pumpkin help in creeping. So, tendrils of grapevine and tendrils of pumpkin are homologous because of same origin but different function. On the other hand, tendrils of pea are modification of leaves. Both tendrils of grapevine and tendrils of pea help in climbing. So, they are analogous because they have different origin but same function.

13. Rhizome of ginger is like the roots of other plants that grows underground. Despite this fact ginger is a stem and not a root. Justify.
Soln. Rhizome of ginger is like the roots of other plants that grows underground but rhizome of ginger is a modification of stem is not a root because it bears nodes, internodes, terminal buds, axillary buds and scale leaves, and it doesn’t bear root hair and root cap.

14 .Differentiate between:
(a) Bract and bracteole
(b) Pulvinus and petiole
(c) Pedicel and peduncle
(d) Spike and spadix
(e) Stamen and staminode
(f) Pollen and pollenium
Soln.
(a) Differences between bract and bracteole are as follows:

(b) Differences between petiole and pulvinus are as follows:

(c) Differences between pedicel and peduncle are as follows:

(d) Differences between spike and spadix are as follows:


(e) Differences between stamen and stami-node are as follows:

(f) Differences between pollen and pollinium . are as follows:

Long Answer Type Questions
1. Distinguish between families Fabaceae, Solanaceae, Liliaceae on the basis of gynoecium characteristics (with figures). Also write economic importance of any one of the above families.
Soln. In the Family Fabaceae, gynoecium is monocarpellary, with unilocular superior ovary. Ovary has marginal placentation, and many ovules are arranged in two alternate rows. Style is bent, and stigma is simple or capitate.

In Family Solanaceae, gynoecium is bicarpellary, syncarpous, ovary is superior, carpels are placed obliquely, generally bilocular, placentation is axile. Ovules are many in each locule, placentae are swollen. A nectariferous disc or lobe may be present, stigma is capitate or lobed.

In Family Liliaceae gynoecium is tricarpellary, syncarpous. Ovary is superior, trilocular with two to many ovules in each locule. Placentation is axile, rarely parietal, styles are united or separate, stigma is free or fused and trilobed.

Economic importance of Family Fabaceae is as follows:
(i) Food : A number of legumes or pulses are obtained from Fabaceae. E.g., Pisum sativum (pea), Cicer arietinum (gram), Lens culinaris (Masur), Phaseolus mungo (urd),
Phaseolus aureus (mung) Cajanus cajan (arhar), Glycine max (soybean), etc.
(ii)Fodder : Medicago sativa (Alfalfa),
Medicago denticulata (toothed, bur clover), Trifolium alexandrium (Egyptian clover), etc’are common herbs used as fodder.
(iii)Oils : They are extracted from seeds of Arachis hypogaea (groundnut or peanut) and Glycine max (soybean).
(iv)Soil fertility: Nodule bearing papilionace¬ous plants increase nitrogen content of the soil. E.g., Crotolaria, Sesbania, Cyamopsis.
(v) Fibres : Obtained from the stems of Crotolaria juncea (sunn hemp) and Sesbania species, used in making cordage, sacks, nets, tissue paper, etc.
(vi)Dyes : Indigo is obtained from the leaves of Indigofera tinctoria.
(vii)Gum : Cyamopsis tetragonoloba yields guar gum which is employed in pharmaceuticals, textiles, paper, plastics and photography.
(viii)Meditines : The roots of Glycyrrhiza glabra (malatthi) are used as demulcent, expectorant and in treating gastric ulcers. Fresh leaf juice of Abrus precatorius (Jeweller’s weights, ratti) is useful in treating leucoderma.
Juice of flowers of Sesbania grandiflora improves eye sight.
(ix)Timber : Dalbergia sissoo (Shisham) and Dalbergia latifolia (Indian rosewood) provide wood for making furniture.
(x)Ornamentals : Lathyrus odoratus, Clitoria ternatea, Lupinus. Erythrena indica are ornamental plants.

2. Describe various stem modifications associated
with food storage, climbing and protection.
Soln.Stem modifications associated with food
storage are as follows:
(i) Rhizome: Rhizome is a thickened,
underground, dorsiventral stem that grows horizontally at a particular depth within the soil. They store food materials and appear tuberous. E.g., Zingiber officinale (ginger), Curcuma longa (turmeric), Canna indica.
(ii) Corms: Corms is an underground, modified main stem. It grows vertically at a particular depth in the soil. It stores food materials. E.g., Colocasia.
(iii) Tubers: Tuber is the tuberous tip of an underground branch. The axillary branches (stolons) that are produced near the soil surface grow into the soil and their tips become swollen due to the accumulation of starch and proteins. E.g., Solanum tuberosum (potato).
Stem modifications associated with climbing are as follows:
(i) Tendrils: Tendrils are thin, wiry, leafless and spirally coiled branches. The terminal part of a tendril is sensitive. It holds the support by coiling around it. The tendrils help the weak stems to climb the support. In some weak stemmed plants, the axillary bud or terminal bud may be modified to form tendrils which are specially called stem tendrils, e.g., Passiflora, Cissus.
(ii) Hooks : Hook is hard, woody, persistent curved, climbing organs found in strangers. E.g., Hugonia, Artabotrys. The axillary buds in Hugonia are modified into hooks.
Stem modifications associated with protection are as follows:
Thoms: Thom is a hard, woody, pointed structure. Thom is formed by the modification of axillary bud or terminal bud which has lost the apical growth. Thoms are endogenous in origin and differ from spines and prickles. E.g., Bougainvillea, Carissa, Duranta. The terminal bud in Carissa and the axillary buds in Duranta and Bougainvillea are modified into thorns. Thoms protect the plant against attack by herbivorous animals.

3.Stolon, offset and rhizome are different forms of stem modifications. How can these modified forms of stem be distinguished from each other?
Soln. Stolon, offset and rhizome are different forms of stem modifications. Differences
between these modified stems are as follows: Stolon is modification of subaerial stem. It is a thin lateral stem branch which when comes in contact with the soil produces roots. Usually shrubs produce some weak thin branches (stolons) at ground level. If these branches happen to touch the soil, roots are produced at the point of contact. When these branches are separated from the mother plant, they grow into individual plants.

Offset: Offset is modification of subaerial stem. It is a weak, elongated, horizontal branch of one internode that arises in the axil of a leaf. At the trip, it produces cluster of leaves above and tuft of roots below. The offset may break off from the parent plant and act as an individual plant. Offset are found usually in aquatic plants and rarely in terrestrial plants. They are helpful for vegetative propagation e.g., Eichhornia, Agave, Pistia.

Rhizome: Rhizome is modification of
underground stem. It is a thickened, underground, dorsiventral stem that grows horizontally at a particular depth within the soil. It stores food materials and appears tuberous. E.g., Zingiber officinale (ginger), Curcuma longa (turmeric), Canna indica

4. The mode of arrangements of sepals or petals in a floral bud is known as aestivation. Draw the various types of aestivation possible for a typical pentamerous flower.
Soln. The various types of aestivation possible for a typical pentamerous flower are as follows:
(i) Valvate aestiva tion: In valvate aestivation, the margins of petals just touch each other without any overlapping, e.g., Brassica.
(ii) Twisted aestivation : In twisted or contorted type, one margin of each petal overlaps the margin of an adjacent petal and the other margin being overlapped by margin of another adjacent petal, e.g., china rose.
(iii) Imbricate aestivation : Imbricate is aestivation of five parts, where one is exterior, one is interior and rest three are having one margin exterior’ and other interior.
(iv) Quincuncial aestivation : Quincuncial is aestivation of five parts, where two are exterior, two interior and the fifth is having one margin exterior and the other interior.
(v) Vexillary aestivation: Vexillary aestivation when the standard petal is large and overlaps the two wing petals which in turn
overlap the keel petals. It is technically known as vexillary aestivation. E.g. pea,
bean, Irtdigofera, Tephrosia, etc.
overlap the keel petals. It is technically known as vexillary aestivation. E.g. pea,
bean, Irtdigofera, Tephrosia, etc.

5. The arrangements of ovules within the ovaryis known as placentation. What does the term placenta refer to? Name and draw varioustypes of placentations in the flower as seen inT.S.orV.S.
Soln. Placenta is a parenchymatous cushion
present inside the ovary where ovules are borne. The arrangements of ovules within the ovary is known as placentation. __
(i) Marginal placentation : In marginal
placentation the placenta forms a ridge
along the ventral suture of the ovary and
the ovules are borne on this ridge forming
two rows, as in pea.
(ii) Axile placentation : When the placenta is axial and the ovules are attached to it in
a multilocular ovary, the placentation is
said to be axile, as in china rose, tomato
and lemon.
(iii) Parietal placentation: In parietal
placentation, the ovules develop on the
inner wall of the ovary or on peripheral
part. Ovary is one-chambered but it
becomes two chambered due to the formation of the false septum, e.g.,
mustard and Argemone. ‘
(iv) Free central placentation : When the
ovules are borne on central axis and septa
are absent, as in Dianthus and Primrose the placentation is called free central.
(v) Basal placentation : In basal placentation,
the placenta develops at the base of ovary ,
an’d a single ovule is attached to it, as in sunflower, marigold.

6.Sunflower is not a flower. Explain.
Soln. Sunflower is not a single flower rather it represents a complete inflorescence, called capitulum or racemose head. Small sessile flowers called florets develop compactly over discoid peduncle. They open in centripetal order. They are surrounded and protected by an involucre of bracts. The florets are of two types namely, ray florets and disc florets. Ray florets are usually female and zygomorphic and disc florets are bisexual and actionomorphic.
In Helianthus, the ray florets are towards the periphery and disc florets are at the centre of the inflorescence.

7. How do you distinguish between hypogeal germination and epigeal germination? What is the role of cotyledon (s) and the endosperm in the germination of seeds?
Soln. In hypogeal germination, epicotyl grows first and only the plumule is pushed out of the soil, while cotyledons and all other parts remain under the soil. It is shown by some dicotyledons e.g., pea, gram, broad bean (Vicia faba), mango, etc. and by most of the monocotyledons, e.g., rice, maize, coconut, etc.
In epigeal germination, hypocotyl grows first and it pushes the cotyledons and other parts of the seed out of the soil. It takes place in cucurbits, mustard, Tatnarindus, French bean (Phaseolus vulgaris), Castor, Onion, Alusma plantago, etc.

Food is necessary for embryo during seed germinations which comes from cotyledons or endosperm. So, they have nutritive function.

8. Seeds of some plants germinate immediately after shedding from the plants while in other plants they require a period of rest before germination. The later phenomena is called as dormancy. Give the reasons for seed dormancy and some methods to break it.
Soln. The reasons for seed dormancy are as follows:
(i) Impermeability of seed coats to oxygen, E.g., Xanthium.
(ii)Impermeability of seed coats to water, e.g., many plants of Leguminosae.
(iii)Hard seed coat which does not allow proper growth of developing embryo, etg., mustard.
(iv)Some seeds contain an imperfectly developed immature embryo.
(v) Embryos require after-ripening in dry storage. These embryos although developed fully, do not germinate unless kept in storage in a dry place for sometime after harvest.
(vi) Some plants produce such chemical compounds (germination inhibitors) that inhibit the germination of their own seeds, e.g., tomato. The pulp of the fruit produces an inhibitor.
Various methods have been employed for breaking dormancy of seeds. Some of which are discussed as follows:
(i) Mechanical scarification : Weakening of hard seed coat with anything of sharp edge, e.g., pieces of glass.
0983
(ii)Chemical scarification: Treating the seeds with dilute acids, fat solvents etc.
(iii)Chilling treatment : Low temperature treatment, alternating low and high temperature treatment.
(iv)High temperature treatment: Permeability of seed coat in Alfaalfa seed increases when they are kept in water at the temperature of 85-90°C for sometime.
(v) Neutralising the effect of inhibitors: Effect of germination inhibitors in the seeds can be counteracted by giving low and high temperature treatment to seeds or by treating the seeds with KNO3, thiourea, gibberellin, ethylene chlorohydrin, etc.

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NCERT Exemplar problems for Class 10 Science Chapter 1 Chemical Reactions And Equation

Short Answer Type Questions

1.Write the balanced chemical equations for the following reactions and identify the type of reaction in each case.
(a) Thermit reaction, iron (III) oxide reacts with aluminium and gives molten iron and aluminium oxide.
(b) Magnesium ribbon is burnt in an atmosphere of nitrogen gas to form solid magnesium nitride.
(c) Chlorine gas is passed in an aqueous potassium iodide solution to form potassium chloride solution and solid iodine.
(d) Ethanol is burnt in air to form carbon dioxide, water and releases heat.
Ans.

2.A solution of potassium chloride when mixed with silver nitrate solution, an insoluble white substance is formed. Write the chemical reaction involved and also mention the type of the chemical reaction.
Ans.

It is a double displacement reaction. It is also a precipitation reaction as AgCl is a white precipitate.

3.Write the balanced chemical equations for the following reactions.
(a)Sodium carbonate on reaction with hydrochloric acid in equal molar concentrations gives sodium chloride and sodium hydrogen-carbonate.
(b)Sodium hydrogencarbonate on reaction with hydrochloric acid gives sodium chloride, water and liberates carbon dioxide.
(c)Copper sulphate on treatment with potassium iodide precipitates cuprous iodide \((Cu_{ 2 }{ I }_{ 2 })\), liberates iodine gas and also forms potassium sulphate.
Ans.

4.Why do fireflies glow at night?
Ans. It is because protein present in fireflies undergoes oxidation in presence of air and an enzyme. This chemical reaction involves emission of visible light. Therefore, fireflies glow at night.

5.Grapes hanging on the plant do not ferment but after being plucked from the plant can be fermented. Under what conditions do these grapes ferment? Is it a chemical or a physical change?
Ans. Grapes when attached to plants are living and therefore they have their immune system due to which they cannot get fermented. When microbes attack plucked grapes in absence of air, they undergo fermentation to form alcohol. This is a chemical process.

6.During the reaction of some metals with dilute hydrochloric acid, following observations were made.
(a) Silver metal does not show any change.
(b) The temperature of the reaction mixture rises when aluminium (Al) is added.
(c) The reaction of sodium metal is found to be highly explosive.
(d) Some bubbles of a gas are seen when lead (Pb) is reacted with the acid. Explain these observations giving suitable reasons.
Ans. (a) It is because silver is less reactive than hydrogen. It cannot displace hydrogen from dilute acid.
(b) It is because the reaction is exothermic.
(c) It is because sodium is highly reactive and forms hydrogen gas in the presence of moisture \((H_{ 2 }O)\) which catches fire as the reaction is highly exothermic and \(H_{ 2 }\) is highly inflammable.
(d) It is due to formation of hydrogen gas. The reaction becomes slow after some time as \(PbCl_{ 2 }(s)\) covers Pb metal.

7.A substance X, which is an oxide of a group 2 element, is used intensively in the cement industry. This element is present in bones also. On treatment with water, it forms a solution which turns red litmus blue. Identify X and also write the chemical reactions involved.
Ans.The substance ‘X’ is calcium oxide (CaO), element is calcium. Calcium oxide is used in cement industry. Calcium is present in bones in form of calcium phosphate.
Calcium oxide dissolves in water forming alkali which turns red litmus blue. CaO(s)+

8.Why do we store silver chloride in dark coloured bottles?
Ans.It is because silver chloride decomposes to silver and chlorine gas in presence of sunlight.

9.A magnesium ribbon is burnt in oxygen to give a white compound X accompanied by emission of light. If the burning ribbon is now placed in an atmosphere of nitrogen, it continues to burn and forms a compound Y.
(а)Write the chemical formulae of X and Y.
(b)Write a balanced chemical equation, when X is dissolved in water.
Ans.

10.A silver article generally turns black when kept in the open for a few days. The article when rubbed with toothpaste again starts shining.
(a) Why do silver articles turn black when kept in the open for a few days? Name the phenomenon involved.
(b) Name the black substance formed and give its chemical formula.
Ans.(a) Silver reacts with \({ H }_{ 2 }S\) gas present in atmosphere to form a black compound \({ Ag }_{ 2 }S\) (silver sulphide) on its surface. This phenomenon is called corrosion.
(b) \({ Ag }_{ 2 }S\)(silver sulphide) is a black coloured solid.

11.Write the balanced chemical equation for the following equations for the following reaction and identify the type of reaction in each case.
(a)Nitrogen gas is treated with hydrogen gas in the presence of a catalyst at 773 K to form ammonia gas.
(b)Sodium hydroxide solution is treated with acetic acid to form sodium acetate and water.
(c)Ethanol is warmed with ethanoic acid to form ethyl acetate in presence of cone. \({ H }_{ 2 }SO_{ 4 } \).
(d)Ethene is burnt in presence of oxygen to form carbon dioxide, water and releases heat and light.
Ans.

12.Complete the missing components/variables given as x and y in the following reactions :

Ans.

13.Which among the following changes are exothermic or endothermic in nature?
Ans.(a)Decomposition of ferrous sulphate
(b)Dilution of sulphuric acid
(c)Dissolution of sodium hydroxide in water
(d)Dissolution of ammonium chloride in water Ans. (a) It is endothermic reaction.
(b)It is exothermic process.
(c)It is exothermic process.
(d)It is endothermic process.

14. Identify the reducing agent in the following reactions :

Ans.(a)\(HN_{ 3 }\) is reducing agent.
(b)\(F_{ 2 }\) is reducing agent.
(c)CO (Carbon monoxide) is reducing agent.
(d)\(H_{ 2 }\) is reducing agent.

15.Identify the oxidising agent (oxidant) in the following reactions :

Ans.(a)\({ Pb }_{ 3 }O_{ 4 }\) (Red lead). It is also called Sindur used by married ladies. It is an oxidant (oxidising agent).
(b)\(O_{ 2 }\) is oxidising agent.
(c)\(CuSO_{ 4 }\) is oxidising agent.
(d)\({ V }_{ 2 }O_{ 5 }\) is oxidising agent.
(e)\(H_{ 2 }O\) is oxidising agent.
(f)CuO  is oxidising agent.

16.Ferrous sulphate decomposes with the evolution of gas having a characteristic order of burning sulphur. Write the chemical reaction involved and identify the type of reaction.
Ans.

It is decomposition reaction.

17.Which among the following are physical or chemical changes
(a)Evaporation of petrol.
(b)Burning of liquefied petroleum gas (LPG).
(c)Heating of an iron rod to red hot.
(d)Curdling of milk.
(e)Sublimation of solid ammonium chloride
Ans.(a)Physical change
(b)Chemical change
(c)Physical change
(d)Chemical change
(e)Physical change

18.Write a balanced chemical equation for each of the following reactions and also classify them.
(a)Lead acetate solution is treated with dilute hydrochloric acid to form lead chloride and acetic acid solution.
(b)A piece of sodium metal is added to absolute ethanol to form sodium ethoxide and hydrogen gas.
(c)Iron (III) oxide on heating with carbon monoxide reacts to form solid iron and liberates carbon dioxide gas.
(d)Hydrogen sulphide gas reacts with oxygen gas to form solid sulphur and liquid water.
Ans.

19.Balance the following chemical equations arid identify the type of chemical reaction:

Ans.

20.Zinc liberates hydrogen gas when reacted with dilute hydrochloric acid, whereas copper does not. Explain why?
Ans.Zinc is more reactive than hydrogen whereas copper is less reactive than hydrogen. Therefore, Zn displaces hydrogen from dil.HCl whereas copper does not

21.On heating blue coloured powder of copper (II) nitrate in boiling tube, copper oxide (black), oxygen gas and a brown gas ‘X’ is formed.
(a)Write a balanced chemical equation of the reaction.
(b)Identify the brown gas ‘X’ evolved.
(c)Identify the type of reaction.
(d)What could be the pH range of aqueous solution of the gas ‘X’?
Ans.

(b) The brown gas ‘X,’ is nitrogen dioxide \((NO_{ 2 })\).
(c) The reaction is a thermal decomposition reaction.
(d) The gas ‘X’ is acidic in nature because it is a non-metallic oxide. Its aqueous solution has pH less than 7.

22.Give the characteristic tests for the following gases.
(a) \(CO_{ 2 }\) (b) \(SO_{ 2 }\) (c) \(O_{ 2 } \)(d) \(H_{ 2 }\)
Ans.(a)Pass the gas through limewater. Limewater turns milky which confirms the presence of \(CO_{ 2 }\).

Limewater Carbon dioxide Calcium carbonate Water
(b) Pass the gas through acidified potassium dichromate solution (orange in colour). The solution will become green in colour due to formation of\( Cr_{ 2 }(SO_{ 4 })_{ 3 }\). It is a redox reaction. Acidified \( { K }_{ 2 }Cr_{ 2 }O_{ 7 }\) is oxidising agent whereas SO_{ 2 } is reducing agent.

Alternative:
Pass the gas through acidified potassium permanganate solution (pink/purple in r colour). It will become colourless because \(SO_{ 2 }\) reduces \(KMnO_{ 4 }\) to\( Mn_{ 2 }+ \)ion which is colourless. It is a redox reaction. Acidified\( KMnO_{ 4 } \)is an oxidising agent.

(c) Bring a candle near oxygen gas. The intensity of candle flame is increased, it shows the presence of oxygen gas which is a supporter of combustion.
(d) Bring a burning matchstick near hydrogen gas. The gas will burn explosively ‘ with a pop sound. It confirms the presence of hydrogen.

23.What happens when a piece of
(a)zinc metal is added to copper sulphate solution?
(b)aluminium metal is added to dilute hydrochloric acid?
(c)silver metal is added to copper sulphate solution? Also, write the balanced chemical equation if the reaction occurs.
Ans.(a)The solution will become colourless due to formation of zinc sulphate and reddish brown

(b)aluminium chloride solution will be formed.

(c) No reaction will take place because silver is less reactive than copper, it cannot displace copper from copper sulphate.

24.What happens when zinc granules are treated with dilute solutions of \(H_{ 2 }SO_{ 4 }\), HCl, \(HNO_{ 3 }\), NaCl and NaOH? Also write the chemical equations if reaction, occurs.
Ans.

25.On adding a drop of barium chloride solution to an aqueous solution of sodium sulphite, white precipitate is obtained.
(a)Write a balanced chemical equation of the reaction involved.
(b)What other name can be given to this precipitation reaction?
(c)On adding dilute hydrochloric acid to the reaction mixture, white precipitate disappears. Why?
Ans.

(b)It is also called double displacement reaction.

It is because barium sulphite reacts with HC1 to form barium chloride which is soluble in water and liberates \(SO_{ 2 }(g)\).

26.You are provided with two containers made up of (A) copper and (B) aluminium. You are also provided with solutions of (a) dilute HCl (b) dilute \(HNO_{ 3 }\)(c) ZnCl2 and (d) H20. In which of the above containers these solutions can be kept?
Ans.A. When solutions are kept in copper container-
(a)Dilute HC1: No reaction takes place as copper is less reactive than hydrogen. Therefore, dil. HC1 can be kept in copper container.
(b)Dilute HNOs: Copper will react with dil. HNOs as it is an oxidising agent and will form copper nitrate and nitrogen monoxide gas.

Therefore, \(HNO_{ 3 }\)(dil.) cannot be kept in copper container.
(c) \(ZnCl_{ 2 }\): No reaction takes place because copper is less reactive than Zn, therefore \(ZnCl_{ 2 }\) can be kept in copper container.
(d)\( H_{ 2 }O\) : NO reaction takes place because copper is less reactive than hydrogen. Therefore, water can be kept in copper container.
Thus, dil. HCl, \(ZnCl_{ 2 }\) and \( H_{ 2 }O\)  can be kept in copper container whereas dil. \(HNO_{ 3 }\) cannot be kept in it.
B.When solutions are kept in aluminium container-
(a)Dilute HCl: Aluminium reacts with dil. HCl to form aluminium chloride . and hydrogen gas is formed.
\(2Al(s)+6Hcl(dil.)\longrightarrow 2AlCl_{ 3 }(aq.)+3H_{ 2 }(g)\)
So, dil. HCl cannot be kept in aluminium container.
(b)Dilute \(HNO_{ 3 }\): Aluminium gets oxidised by \(HNO_{ 3 }\) to form a layer of aluminium oxide which does not react further therefore dil. \(HNO_{ 3 }\)can be kept in aluminium container.
(c) \(ZnCl_{ 2 }\): It reacts with ‘AT to form aluminium chloride and zinc metal will get deposited.
\(3ZnCl_{ 2 }+2Al\longrightarrow 2AlCl_{ 3 }+3Zn\)
Therefore, Zinc chloride cannot be stored in aluminium container.
(d)\( H_{ 2 }O{ 2 }\) : It can he stored in aluminium container because aluminium does not react with cold as well as hot water.
A1 reacts with steam to form aluminium oxide and hydrogen
\(2Al(s)+3H_{ 2 }O(g)\longrightarrow Al_{ 2 }O_{ 3 }(s)+3H_{ 2 }( g)\)
Thus.dil. \(HNO_{ 3 }\) and \( H_{ 2 }O\)  can be kept in aluminium container whereas dil. HCl and \(ZnCl_{ 2 }\) solution cannot be stored in it.

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