Category: CBSE Class 12

Here we are providing NCERT Solutions for Class 12 English Vistas Chapter 8 Memories of Childhood. Students can get Class 12 English Memories of Childhood NCERT Solutions, Questions and Answers designed by subject expert teachers.

Memories of Childhood NCERT Solutions for Class 12 English Vistas Chapter 8

Memories of Childhood NCERT Text Book Questions and Answers

Memories of Childhood Reading with insight

Question 1.
The two accounts that you read above are based on two distant cultures. What is the commonality of theme found in both of them?
Answer:
Both the autobiographical extracts, based upon two distant cultures, depict the lives of two women from marginalised communities who look back at their childhood and reflect on their relationship with mainstream culture. The first account is by an American Indian woman bom in the late 19th century. The account expressed the indignations suffered by the Native Americans at the hands of Christians.

She was ardently against the oppression of Native Americans in Western culture. Though she resented this mistreatment, she still aimed at bridging the wide gap between the dominant white and Native American cultures. She did not let herself get seduced into believing that her Native American traditions were folly or sin.

As a person of mixed blood, her life could be looked upon as an example of the beauty and accomplishments that are possible if the two cultures live in cooperation.The second account is by a contemporary Tamil Dalit writer. She voiced the discrimination she faced as a Dalit. She explored the impact of the discrimination, compounded by the poverty suffered by Dalit women.

The caste system had been so deeply ingrained in the Indian psyche that institutions that ought to promote egalitarianism became the means of perpetuating caste discrimination. Both these accounts are bound by the common theme of discrimination and indignity suffered by women in marginalised communities at the hands of the supposedly superior caste or culture.

Question 2.
It may take a long time for oppression to be removed, but the seeds of rebellion are sowed early in life. Do you agree that injustice in any form cannot escape being noticed even by children?
Answer:
Children are heavily influenced and conditioned by their upbringing. The social influences that they experience from childhood are often ingrained in their subconscious, and are manifested later. Children are also perceptive and keenly aware of positive and negative influences. Rebellious children manifest the ‘seed’ of rebellion planted at some point in their lives.

A sensitive perspective provides a means for understanding how the oppression of children occurs within multiple social contexts that interrelate to produce harmful outcomes for children. Children lack power and resources and are easy targets for adult oppression. Children are exposed to different levels and types of oppression that vary depending on their age, socio-economic class, race, and the beliefs of their parents.

According to the theory of differential oppression, oppression leads to adaptive reactions by children: passive acceptance, exercise of illegitimate coercive power, manipulation of one’s peers, and retaliation. Reducing the oppressive acts of adults and alleviating the damaging circumstances that characterise the social environment of children is critical to reducing the prevalence of juvenile delinquency and other kinds of problem behaviour. The reaction of Zitkala-Sa and Bama to the injustice they perceived as children ranged from defiance, anguish, resentment, and dejection to a fierce determination to excel.

Question 3.
Bama’s experience is that of a victim of the caste system. What kind of discrimination does Zitkala-Sa’s experience depict? What are their responses to their respective situations?
Answer:
Zitkala-Sa was ill-treated and discriminated against from the beginning of her journey. Native Americans, at that time, did not associate with white people and faced discrimination because of their appearance. She felt scared and extremely uncomfortable when unable to comprehend the ways of a foreign culture. Even though she hated the way she was treated, she still had to abide by rules and orders to avoid punishment. However much she suffered at the hands of cultural and racial discrimination, she managed to work her way through and never gave in to discrimination.

Similarly, Bama remained undeterred. She was convinced that she had a role to play. And it was this conviction that made her stand up for her beliefs. She championed the economic and social hardships faced by Dalits due to the discrimination by upper-caste people. Both the writers challenged accepted social practices and did not succumb to pressure. They made their voices heard, and brought to light the indignities faced by their races; they stood by what they believed to be true.

Memories of Childhood Extra Questions and Answers

Memories of Childhood Short Answer Questions

Question 1.
Who was Gertrude Simmons?
Answer:
Gertrude Simmons was an extraordinarily talented and educated Native American woman who struggled and triumphed at a time when severe prejudice prevailed against Native American culture and women. As a writer, she adopted the pen name, Zitkala-Sa. Her works criticized traditional dogma, and her life as a Native American woman was dedicated against the evils of oppression.

Question 2.
What were the first things that upset Zitkala-Sa at school?
Answer:
The severe cold weather and the alien surroundings upset Zitkala-Sa. She was neither familiar with the language nor the strict regime of the hostel. She was perturbed by the sound of the large bell with the annoying the clatter of shoes. Her spirit pined for its lost freedom.

Question 3.
How was the attire of the girls in school different from Zitkala-Sa?
Answer:
The girls at school wore stiff shoes and closely fitted dresses. The small girls wore sleeved aprons and shingled hair. Zitkala-Sa wore soft moccasins, and had wrapped a blanket on her shoulders. She found the tight-fitting clothes rather immodest.

Question 4.
Narrate Zitkala-Sa’s embarrassment at the breakfast table.
Answer:
At the breakfast table, a small bell was tapped, and each of the pupils drew a chair from under the table. Supposing this act meant that they were to be seated, Zitkala-Sa pulled out her chair and seated herself only to realize that the rest were standing. Just as she got up, a second bell was sounded. All were seated at that, and Zitkala-Sa had to crawl back into her chair again. She then heard a man’s voice and saw all had hung their heads over their plates. She saw a pale-faced woman stare at her and dropped her eyes. With the third bell, everyone picked up their knives and forks and began eating. Zitkala-Sa wept as she was confounded by the unfamiliarity of her surroundings and practices.

Question 5.
Why was cutting of her hair the greatest blow to Zitkala-Sa?
Answer:
Zitkala-Sa had been taught by her mother that, in Native American culture, unskilled warriors who were captured had their hair shingled by the enemy. Short hair was worn by mourners, and shingled hair by cowards according to their practices. She tried to resist being shorn of her cultural values.

Question 6.
What were the sights that fascinated little Bama as she walked home from school?
Answer:
Bama enjoyed the entertaining novelties and oddities on the streets, on her way home from school. The performing monkey, the antics of the snake-charmer, the cyclist at his cycle for three days, the Maariyaata temple and the huge bell, the pongal offerings being cooked, the dried fish stall, the sweet stall, the stall selling fried snacks, the street light changing colour, and other such sights fascinated her.

Question 7.
What was that one episode that left an indelible imprint on Bama’s mind?
Answer:
Bama noticed a man holding on to a food packet by its string. Without touching it, he handed it over to the landlord. The landlord opened the parcel and began to eat the vadais. At home, her elder brother explained to how lower cast people, such as the author herself, were prohibited to touch anything of use to the upper caste people. They were believed to sully the item they touched and therefore the package was carefully carried, by its string, for the landlord who belonged to higher caste.

Question 8.
What was the bitter truth of their life that Annan told Bama?
Answer:
Annan told Bama about the caste system that subjugated Dalits in the society. He added that being bom into a lower caste community, one could never get honour or dignity or respect. It was education alone that could help them earn respect.

Memories of Childhood Long Answer Questions

Question 1.
During her first meal in the missionary school, Simmons became painfully aware of the tension between tradition and acculturation and of the great lack of understanding people had about Native American culture. Explain.
Answer:
Before the meal, the girls were lined up. The Indian girls wore stiff shoes and closely clinging dresses, which seemed immodest to the author’s traditional taste. The small girls wore short hair. Simmons walked noiselessly in her soft leather shoes. She was embarrassed as her blanket had been taken off from her shoulders. Just as they entered, the boys came in from the opposite door. She was at a loss when the rest of them prayed, and ate with fork and knife, by the summons of bells.

She noticed the pale-faced woman staring at her and felt unnerved. The writer wept, she was at a loss over the strange practices, she lacked understanding of the western way of dressing, praying and eating. For a child, a sudden change of one’s faith, cultural and aesthetic conditioning is bewildering and often traumatic. The distance between the two cultures is pronounced by the example of the episode. What the west considered barbaric in Native American was the essence and beauty of their culture which was painfully and ignorantly stripped from their personality in an attempt to acclimatize them.

Question 2.
Narrate the trauma Simmons faced as a child during the hair-cutting episode.
Answer:
Judewin, who knew a few words of English, told Gertrude Simmons that they would cut their long, heavy hair. It was a blow to her as in their culture it signified humiliation or mourning. Judewin warned her about the futility of her mission to escape the hair-cutting ritual, but the writer was prepared to struggle. To hide from the situation, she crept up the stairs into a dim, large room, and crawled under the farthest bed in the dark comer. Every time footsteps were heard, she felt terrorized.

She knew that they were searching frantically for her and soon she was dragged out and tied to a chair. She wept loudly but her hair was clipped short. With her hair lost, she lost her spirit. She had suffered a great deal of disgrace earlier and now her hair was cut like that of a coward. She moaned for her mother, but no one came to comfort her. The brutality made her feel like a beast.

Question 3.
Bama like any other child enjoyed innocent pleasures. Justify.
Answer:
The writer walked home from school each day. It was a ten-minute walk but it took her thirty minutes to complete the distance. She would loiter along, watching all the fun and games and all the entertaining new and strange things that she saw on the way. She enjoyed everything, be it the performing monkey, or the snake charmer, or the man who cycled non-stop for days or the spinning wheels. She was fascinated by the Maariyaata temple, its huge bell and the pongal offerings.

She also stopped to hear the political activists, rallying, or to watch a street play, a puppet show, or a stunt performance.Even mundane activities such as waiters cooling coffee, or people chopping onions, or the almond tree with its fruit occasionally blown down held her attention for long. The people selling edibles were equally appealing to her.

Question 4.
Only an untouchable would know the pain of being one. Discuss with reference to Bama’s episode.
OR
How does “Memories of Childhood” bring out the plight of marginalized communities in India?
Answer:
Bama felt and experienced untouchability, early in life. As a child, she noticed an elderly person walking with a small packet, holding on to the packet by its string, without touching it. He brought it to the landlord, bowed low, and extended the packet towards him. The landlord opened the parcel and began to eat the vadais. Her elder brother explained that people believed that they were upper caste and therefore prohibited people from the lower caste to touch them, believing that the act would pollute them.

On hearing that, Bama felt sad and insulted. Her mind rebelled against fetching and carrying for people who considered people such as her untouchable. It made her very angry. She wondered why the upper-caste people behaved in such a manner just because they were richer. She strongly felt that her people must never run errands for those of the upper caste people.

NCERT Solutions for Class 12 History Chapter 10 Colonialism and the Countryside: Exploring Official Archives are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 10 Colonialism and the Countryside: Exploring Official Archives.

NCERT Solutions for Class 12 History Chapter 10 Colonialism and the Countryside: Exploring Official Archives

Question 1.
Why was thejotedar a powerful figure in many areas of rural Bengal ?
Solution :
The jotedar was a powerful figure in many areas of rural Bengal due to following factors :

• They had acquired vast areas of land.
• They controlled the local trade including moneylending.
• A large part of their land was cultivated through sharecroppers (adhiyars or bargadars). They brought their own ploughs and worked in the field. After the harvest they handed over half of the produce to the jotedars. They had become rich in the villages.
• Unlike zamindars, jotedars lived in the villages and exercised direct control over poor villagers.
• They resisted efforts by zamindars to increase the jama of the village and prevented zamindari officials from executing their duties.
• They mobilised ryots and deliberately delayed the payments of revenue to the zamindar.

Question 2.
How did Zamindars manage to retain control over their zamindaris?
Solution :
When zamindars were in bad times, they often resorted to various tactics to maintain control over their zamindari. These were in fact their survival tactics. Following are the important ones.

1. Zamindars created fictitious sales during auction. Their own men would make highest bid and later refused to pay up. After repeating this exercise for couple of occasions, the government would be tired and sell it back to zamindar at lesser rate.
2. A part of Estate was often transferred to female members of the family, and that part of property could not be taken by the government any more.
3. Zamindars put hurdle in purchase and occupation of the estate by others by use of sheer muscle power.
4. Sometimes even peasants under the influence of zamindars opposed auction of estate.

Question 3.
How did the Paharias respond to the coming of outsiders ?
Solution :

1. The Paharias were hunters, shifting cultivators, food gatherers, charcoal producers and silkworm rearers. Thus, their life depended on the forest produce. They lived in hutments and considered the entire region as their land. They considered it the basis of their identity as well as survival. They, therefore, resisted the intrusion of outsiders. Their chiefs maintained the unity of the group. They settled disputes and led the tribe in battles with other tribes and . plains’ people.
2. Paharias raided the plains. These raids were necessary for their survival. These were the ways of asserting power over settled communities. It was their means of negotiating political relations with outsiders. Thus to maintain peace, zamindars paid them regular tribute. Similarly, the traders gave them a small amount for permission to use the passes controlled by them.
3. When the British encouraged forests clearance for expansion of agriculture, the raids of Paharias on the settled villages increased. This led to conflict with the British. Ultimately, the Paharias withdrew deep into the mountains, insulating themselves from hostile forces, and carrying on a war with outsiders.
4. After the coming of the Santhals who cleared forests, ploughed land and grew rice and cotton in the areas of lower hills, the Paharias receded deeper into the Rajmahal hills. Thus, it is clear that the Paharias considered outsiders with suspicion and distrust. Every white man appeared to them to represent a power that was destroying their way of life and means of survival, snatching away their control over their forests and lands.

Question 4.
Why did the Santhals rebel against British rule ?
Solution :
The British failed in their attempt to transform the Paharias into settled agriculturists. They, therefore, persuaded the Santhals to settle in the foothills of Rajmahal. They demarcated a large area of land as Damin-i-Koh so that the Santhals might become settled peasants. The Santhals were expected to clear and cultivate one-tenth of the area within the first ten years. Santhals settlements expanded rapidly from 40 villages in 1838 to 1473 villages in 1851. Their population too increased during the same period from 3000 to over 82,000. As a result of it, the revenue for the company increased. The Santhals, who were in search of a place to settle, ultimately got a place and settled in the Damin-i-Koh on the peripheries of the Rajmahal hills.

But they soon found this was not an ideal world for them and they rebelled against the British rule due to the following factors :

• The state was levying heavy taxes on the land that they had cleared for cultivation;
• The moneylenders (dikus) were charging high rate of interest and taking over the land when debt remained unpaid;
• Zamindars were asserting control over Damin area.

Thus, they found that the land was slipping away from their hands. Therefore, to create an ideal world for themselves where they would rule, they rebelled against zamindars, moneylenders and the British. After the revolt (1855-56), the British created the Santhal pargana in the hope to conciliate them.

Question 5.
What explains the anger of Deccan ryots against the moneylenders?
Solution :
The main reasons for the anger of Ryots against moneylenders are as follows:

1. In rural India it was traditional rule that the interest will always remain less than the principal amount. However, in many cases interest payable was more than the principal itself. In one case the interest was Rs 2000 against principal amount of Rs100.
2. No receipt was paid in case of payment of loan partly or fully. This opened the scope of manipulation by the moneylenders.
3. Ryots complained about forging of documents and other fraudulent activity by the moneylenders.
4. Ryots believed that moneylenders were insensitive to them and made an arrogant and exploitative lot.

Question 6.
Why were many zamindaris auctioned after the Permanent Settlement?
Solution :
Under the Permanent Settlement of Bengal 1793, the East India Company had fixed the land revenue that each zamindar had to pay. At the same time, it was stated that the estates of those who failed to pay would be auctioned to recover the revenue. The company fixed the total demand over the entire estate whose revenue the zamindar contracted to pay. The zamindar collected the rent from different villages, paid the revenue to the Company and retained the difference as his income. He was expected to pay the Company regularly. However in practice in the early decades after the Permanent Settlement, zamindars regularly failed to pay the revenue demand and unpaid balances accumulated due to the following reasons :

1. The initial demands were very high because it was felt that under Permanent Settlement, the Company would never be able to claim higher share in case of rise in prices and expansion of cultivation. So in anticipation of such loss high revenue was fixed. It was argued that the burden on zamindar would gradually decline as agricultural production expanded and prices rose.
2. At the time of the settlement, the prices of agricultural produce were low. It made it difficult for the ryots to pay their dues to the zamindar who in return could not pay revenue to the company.
3. The revenue, regardless of the harvest, had to be paid punctually under the Sunset Law.
4. The Permanent Settlement limited the power of the zamindar to collect rent from the ryot and manage his zamindari. The Company had disbanded their troops. Their courts (catcheries) were brought under the supervision of a collector.
5. Rent collection was a perennial problem due to bad harvests and low prices. Some times ryots deliberately delayed payments.

Under the above circumstances, the zamindars could not make payment punctually, and under the Sunset Law, if payments did not come in by sunset of the specified date, the zamindari was liable to be auctioned. This led to auction of many zamindaris after the Permanent Settlement in Bengal.

Question 7.
In what way was the livelihood of the Pah arias different from that of the Santhals ?
Solution :
The livelihood of the Paharias was different from that of the Santhals in the following way:

Question 8.
How did the American Civil War affect the lives of the ryots in India?
Solution :
American Civil War that began in 1860 had a huge impact on the ryots of Deccan region in India. Following events explains how the impact took shape:

1. Britain was the country where large cotton mills were operational. These cotton mills depended on cotton imported from North America.
2. When the USA was reeling under civil war, it was naturally very difficult to import cotton from there.
3. The cotton mills were forced to look for alternative suppliers of cotton apart from US. India made a good option.
4. The farmers in Deccan were encouraged to grow cotton. One way was the easy access of credit. The moneylenders would give credit of Rs 100 for every acre of land under cotton cultivation.
5. The farmers benefitted out of this demand for cotton. But the real beneficiary were the big farmers and traders.
6. However, things changed as normalcy returned to US. Now the demand of cotton in India declined and so declined the easy availability of credit. The ryots fell back to old days of penury and rose in rebellion in many places.

Question 9.
What are the problems of using official sources in writing about the history of peasants?
Solution :
In using official sources for writing the history of peasants, the following problems are faced:

1. The official sources reflect official concerns and interpretations of events. For example, the Deccan Riots Commission was specifically asked to judge whether the level of the government revenue demand was the cause of the revolt. The Commission reported that the government demand was not the cause of peasant anger. It was the moneylenders who were responsible for the riots. Generally, the colonial government never admitted that the discontent was due to government’s policy. In fact, the increase in revenue demand from 50 to 100 per cent was also responsible for the bad condition of the ryots.
2. Official sources explain certain events with exaggeration. For example, the evidence contained in the Fifth Report is invaluable. Researchers, however, indicate that, intent on criticising the maladministration of the Company, the Fifth Report exaggerated the collapse of traditional zamindari power. They also overestimated the scale on which zamindars were losing their land. In fact, zamindaris were auctioned but zamindars were not displaced in every case. In most of the cases, they retained their zamindaris.

Thus, the official sources are written from the point of view of the government. Even Buchanan wrote what he was asked to write by the East India Company. That is why the Paharias viewed him with suspicion and distrust. He was perceived as an agent of the sarkar.

We hope the NCERT Solutions for Class 12 History Chapter 10 Colonialism and the Countryside: Exploring Official Archives help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 10 Colonialism and the Countryside: Exploring Official Archives, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Biology Chapter 15 Biodiversity and Conservation

These Solutions are part of NCERT Solutions for Class 12 Biology. Here we have given NCERT Solutions for Class 12 Biology Chapter 15 Biodiversity and Conservation

Question 1.
Name the three important components of biodiversity
Solution:
The three important components of biodiversity are genetic diversity, species diversity and ecological diversity. These components are the basic building blocks of biodiversity. These are intimately linked and may have common elements.

Question 2.
How do ecologists estimate the total number of species present in the world?
Solution:
The diversity of living organisms present on the earth is very vast. According to an estimate by researchers, it is about seven million. The total number of species present in the world is calculated by ecologists by statistical comparison between species richness of a well-studied group of insects of temperate and tropical regions. Then, these ratios are extrapolated with other groups of plants and animals to calculate the total species richness present on the earth.

Question 3.
Give three hypotheses for explaining why tropics show the greatest levels of species richness.
Solution:
The three hypotheses for higher species richness in tropical areas are:

1. Prolong evolutionary time – Temperate areas have undergone frequent glaciation in the past. It killed most of the species. No such disturbance occurred in the tropics where species continued to flourish and evolve undisturbed for millions of years.
2. Favourable environment – There are no unfavourable seasons in the tropics. The continued favourable environment has helped tropical organisms to gain more niche specialisation and increased diversity.
3. More sunlight – More solar energy is available in the tropics. This promotes higher productivity and increased biodiversity.

Question 4.
What is the significance of the slope of regression in a species-area relationship?
Solution:
The slope of regression/regression co-efficient of species-area relationship indicates that species richness decreases with a decrease in area.

• The regression coefficient is between 0.1 – 0.2 regardless of taxonomic group or region, eg: Plants in Britain, Birds in California.
• But in large areas like continents value is eg:- Frugivorous birds, mammals is tropical forests.

Question 5.
What are the major causes of species losses in a geographical region?
Solution:
The major causes of species losses in a geographical area are:

1. Habitat loss and fragmentation
2. Overexploitation
3. Alien species invasion
4. Co-extinctions
5. Disturbance and degradation
6. Pollution
7. Intensive agriculture and forestry.

Question 6.
How is biodiversity important for ecosystem functioning?
Solution:
An ecosystem with high species diversity is much more stable than an ecosystem with low species diversity. Also, high biodiversity makes the ecosystem more stable in productivity and more resistant to disturbances such as alien species invasions and floods.

If an ecosystem is rich in biodiversity, then the ecological balance would not get affected. Various trophic levels are connected through food chains. If anyone organism or all organisms of any one trophic level is illed, then it will disrupt the entire food chain. For example, in a food chain, if all plants are killed, then all deer will die due to the lack of food.

If all deer are dead, soon the tigers will also die. Therefore, it can be concluded that if an ecosystem is rich in species, then there will be other food alternatives at each trophic level which would not allow any organism to die due to the absence of their food resource. Hence, biodiversity plays an important role in maintaining the health and ecological balance of an ecosystem.

Question 7.
What are sacred groves? What is their role in conservation?
Solution:
Sacred groves are forest patches around places of work. These are held in high esteem by tribal communities/state or central government. Tribals do not allow to cut even a single branch of trees in these sacred groves. Preserved over the course of many generations, sacred groves represent native vegetation in a natural or near-natural state & thus is rich in biodiversity & harbour many rare species of plants & animals. This is the reason why many endemic species flourish in these regions.

Question 8.
Among the ecosystem services are control of floods and soil erosion. How is this achieved by the biotic components of the ecosystem?
Solution:

• Control of soil erosion: Plant roots hold the soil particles tightly and do not allow the topsoil to be drifted away by winds or moving water. Plants increase the porosity and fertility of the soil.
• Control of floods: It is carried out by retaining water and preventing runoff rainwater. Litter and humus of plants function as sponges thus, retaining the water which percolates down and gets stored as underground water. Hence, the flood is controlled.

Question 9.
The species diversity of plants (22 percent) is much less than that of animals (72 percent). What could be the explanations for how animals achieved greater diversification?
Solution:
Scientists recorded 22% of plant species diversity including algae, fungi, bryophytes, pteridophytes, gymnosperms, and angiosperms. But they recorded 72% of animal species (including insects, mollusks, fishes, mammals, birds etc.) diversity. Plants have the less adaptive capacity as compared to animals. Animals show locomotory movements and can move from one place to another to suit the environment and also in search of food. On the contrary, plants are fixed. Moreover, animals have well organised body structure with various organs to help adjust to the environment.

Question 10.
Can you think of a situation where we deliberately want to make a species extinct? How would you justify it?
Solution:
Yes, there are various kinds of parasites and disease-causing microbes that we deliberately want to eradicate from the earth. Since these micro-organisms are harmful to human beings, scientists are working hard to fight against them. Scientists have been able to eliminate the smallpox virus from the world through the use of vaccinations. This shows that humans deliberately want to make these species extinct. Several other eradication programmes such as polio and hepatitis B vaccinations are aimed to eliminate these disease-causing microbes.

We hope the NCERT Solutions for Class 12 Biology Chapter 15 Biodiversity and Conservation help you. If you have any query regarding NCERT Solutions for Class 12 Biology Chapter 15 Biodiversity and Conservation, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Question 1.
Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Answer:

Question 2.
The work function of caesium metal is 2.14 eV. When light of frequency 6 x 10¹⁴ Hz is incident on the metal surface, photo emission of electrons occurs. What is the
(а)   maximum kinetic energy of the emitted electrons,
(b)   stopping potential, and
(c) maximum speed of the emitted photo electrons ?
Answer:

Question 3.
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Answer:
V₀ = 1.5 V
∴ Maximum kinetic energy of the emitted electrons,
\(\frac { 1 }{ 2 } m{ V }_{ max }^{ 2 }=e{ V }_{ 0 }=1.5j=15eV\)

Question 4.
Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam.
(b) How many photons per second, on average, arrive at a target irradiated by this beam? (Assume the beam to have a uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Answer:
Given, λ = 632.8 nm = 632.8 x 10^-9m
Power, P = 9.42 mW = 9.42 x 10^-3 W


Question 5.
The energy flux of sunlight reaching the surface of the earth is 1.388 x 10³ W/m². How many photons (nearly) per square meter are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Answer:

Question 6.
In an experiment on the photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 x 10^-15 V s. Calculate the value of Planck’s constant.
Answer:

Question 7.
A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the center of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm.
(a) What is the energy per photon associated with sodium light?
(b) At what rate are the photons delivered to the sphere?
Answer:
Power of sodium lamp,
P = 100 W

Question 8.
The threshold frequency for a certain metal is 3.3 x 10¹⁴ Hz. If the light of frequency 8.2 x 10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Answer:

Question 9.
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Answer:

Since the energy of incident radiation less than the work function (4.2 eV) of the metal, therefore, the photoelectric emission can not take place from the given metal. 11.10.

Question 10.
Light of frequency 7.21 x 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 x 10⁵ m/s are ejected from the surface. What is the threshold frequency for the photoemission of electrons?
Answer:
Using the relation

Question 11.
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the cathode, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the cathode is made.
Answer:

Question 12.
Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Answer:
Given V = 56 V.
(a) the momentum of the electron

Question 13.
What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with the kinetic energy of 120 eV?
Answer:
Here, E = 120 eV = 120 x 1.6 x 10^-19J
= 1.92 x 10-¹⁷ J

Question 14.
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron, and
(b) a neutron would have the same de Broglie wavelength.
Answer:

Question 15.
What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 x 10^-9 kg drifting with a speed of 2.2 m/s?
Answer:
(a) Here, m = 0.040 kg and υ = 1.0 km/s .
= 1000 m/s

Question 16.
An electron and a photon each have a wavelength of 100 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electrons.
Answer:

Question 17.
(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be
1.40 x 10-¹⁰ m?  (C.B.S.E. 2008)
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of \(\frac { 3 }{ 2 } \) kT at 300 K.
Answer:
(a) Here λ = 1.40 x 10^-10m
Also, h = 6.63 X 10^-34 Js
and m = 1.67 x 10^-27kg



Question 18.
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Answer:

Question 19.
What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Answer:
For nitrogen,


Question 20.
(a) Estimate the speed with which electrons emitted from a heated cathode of an evacuated tube impinge on the anode maintained at a potential difference of 500 V with respect to the cathode. Ignore the small initial speeds of the electrons. The ‘specific charge’ of the electron i.e., its elm is given to be 1.76 x 10¹¹ C kg^-1.
(b) Use the same formula you employ in
(a) to obtain electron speed for an anode potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?
Answer:




Question 21.
(a) A monoenergetic electron beam with an electron speed of 5.20 x 10⁶ ms^-1 is subject to a magnetic field of l.30 x 10^-4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given elm for electron equals 1.76 X 10¹¹ C kg^-1?
(b) Is the formula you employ in
(a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
Answer:
(a) υ  = 5.20 x 10⁶ms^_1
B = 1.30 x 10 ⁴T

(b) The formula employed in part (a) is not valid because with the increase in velocity, mass varies and in the above formula we have taken m as constant. Instead, m

Question 22.
An electron gun with its anode at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (~ 10^-2 mm of Hg). A magnetic field of 2.83 x 10 ⁴ T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data.
Answer:

Question 23.
(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at (45 Å. What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube.
Answer:

Question 24.
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as the annihilation of an electron-positron pair of total energy 10.2 BeV into two y-rays of equal energy. What is the wavelength associated with each γ-ray?   (1 BeV = 10⁹ eV)
Answer:
The energy carried by the pair of γ-rays = 10 .2 BeV
The energy of each γ ray is

Question 25.
Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a Mediumwave transmitter of 10 kW power, emitting radio waves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~ 10^_1° W m^-2). Take the area of the pupil to be about 0.4 cm², and the average frequency of white light to be about 6 x 10¹⁴
Answer:


This is quite a small number, but still large enough to be counted.
Comparison of cases (a) and (b) tells us that our eyes can not count the number of photons individually.

Question 26.
Ultraviolet light of wavelength 2271 A from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity
(~ 10^s W m^-2) red light of wavelength 6328 A produced by a He-Ne laser?
[C.B.S.E. 2005, UC, 13]
Answer:

Since the energy of a red photon is less than the work function for the metal, the photocell does not respond to red light.

Question 27.
Monochromatic radiation of wavelength 640.2 nm (1 nm = 10^-9 m) from a neon lamp irradiates photosensitive material made of caesium or tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Answer:

Question 28.
A mercury lamp is a convenient source for studying the frequency dependence of photoelectric emission since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used :
λ_x = 3650 Å, λ₂ = 4047 Å, λ₃ = 4358 Å, λ₄ = 5461 Å, λ₅ = 6907 Å.
The stopping voltages, respectively, were measured to be :
V₀₁ = 1.28 V, V₀₂ = 0.95 V, f₀₃ = 0.74 V, = 0.16 V,V₀₅ = 0 V
(a) Determine the value of Planck’s constant
(b) Estimate the threshold frequency and work function for the material.
Answer:
(a) From the Einstein photoelectric equation,




Question 29.
The work function for the following metals is given : Na : 2.75 eV ; K : 2.30 eV ; Mo : 417 eV ; Ni: 515 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away? (C.B.S.E. 2009)
Answer:
Here, λ = 3300 A = 3300 x 10^-10 m
Distance, r’ = 1 m and r’ = 50 cm = 0.5 m
Using the relation

Since the energy E of the incident photon of light is less than the work functions of Mo and Ni metals, so photoelectric emission will not occur in Mo and Ni. The distance of the source does not increase or decrease the energy of the photon of the light incident, therefore, the energy of electrons ejected will not change but the intensity of ejected electrons will increase (1 α 1/r² ) and become four times.

Question 30.
Light of intensity 10^-5 W m^-2 falls on a sodium photo­cell of surface area 2 cm². Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave- picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
Answer:
A = 2 cm² = 2 x 10^-4 m²
φ = 2eV = 2 x 1.6 x 10^-19 J
= 3.2 X 10^-19  J.
Taking the approximate radius of an atom as 10^-10 m. the effective area of sodium atom is ≈ r² = 10^-20 m.
.’. If there is one free electron per atom, then the number of electrons in five layers


The answer obtained implies that the time of emission of the electron is very large and is not in agreement with the observed time of emission, which is approximately 10 ⁹ s. Thus wave-picture of radiation is not applicable for photo-electric emission.

Question 31.
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å which is of the order of interatomic spacing in the lattice) (m_e = 9.11 x 10^­-31 kg).
Answer:

Clearly, the energy of photons is much greater than the energy of electrons.

Question 32.
(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 44, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable?
Explain. (m_n = 1.675 x 10^-27 kg).
(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27°C). Hence explain why a fast neutron beam needs to be thermalized with the environment before it can be used for neutron diffraction experiments.
Answer:

This wavelength is about hundred times smaller than the interatomic separation of crystals. Thus, neutrons are not suitable for diffraction experiments in case of crystals.

This wavelength is comparable to the interatomic spacing of crystals. Therefore, thermal electrons are able to interact with the crystal. Since

increasing the temperature, decreases their de Broglie wavelength and they become unsuitable for crystal diffraction. Thus, the fast beam of neutrons needs to be thermalised with the environment for neutron diffraction experiment.

Question 33.
An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow
light ?
Answer:



Question 34.
The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10^-15 m or less. This structure was first probed in early 1970’s using high-energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV).
Answer:

Thus, the energy of the proton ejected out of the linear accelerator is of the order of BeV.

Question 35.
Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.
Answer:
Here T = 27 °C = 27 + 273 = 300 K
P = 1 atm = 1.01 x 10⁵ Nm^-2
Also, the mass of helium atom,


Question 36.
Compute the typical de Broglie wavelength of an electron in metal at 27°C and compare it with the mean separation between two electrons in a metal which is given to be about 2 x 10^_10 m.
Answer:
Here, T = 27°C = 27 + 273 = 300 K


Question 37.
Answer the following questions:
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3) e; (- l/3)e]. Why do they not show up in *Millikan’s oil-drop experiment?
(b) What is so special about the combination elm? Why do we not simply talk of e and m separately?
(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?
(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?
(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations :

Answer:
(a) In case of the Millikan oil-drop experiment, the charge on the electron is measured. The electron revolves outside the nucleus and each has a charge e. Thus, we do not observe the fractional charges

(c) At ordinary pressure, molecules of gas keep on colliding with each other and the ions formed do not have a chance to reach the respective electrodes to constitute a current because of their recombination. At low pressure, however, ions do not collide frequently and are able to reach the respective electrodes to constitute a current.
(d) Work function in fact is the energy required to knock out the electron from the highest filled level of the conduction band of an emitter. In the conduction band, there are different energy levels which collectively form a continuous band of levels. Therefore, different amounts of energy are required to bring the electrons out of the different levels. Electrons emitted have different kinetic energies according to the energy supplied to the emitter.
(e) Since frequency for a given matter-wave remains constant for different layers of the matter but wavelength changes so X is more significant than v.
Similarly energy E = hv = \(\frac { 1 }{ 2 } \) m(λv)² is also constant for a given matter wave so phase λv is also not physically significant.

We hope the NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter, help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 History Chapter 2 Kings, Farmers and Towns Early States and Economies are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 2 Kings, Farmers and Towns Early States and Economies.

NCERT Solutions for Class 12 History Chapter 2 Kings, Farmers and Towns Early States and Economies

Question 1.
Discuss the evidence of craft production in Early Historic cities. In what ways is this different from the evidence from Harappan cities ?
Solution :
Various evidences of craft production in Early Historic cities have been found. These include fine pottery bowls and dishes, with a glossy finish, known as Northern Black Polished Ware, probably used by rich people and ornaments, tools, weapons, vessels, figurines, made of a wide range of materials – gold, silver, copper, bronze, ivory, glass, shell and terracotta. Iron was also used for making plough share, weapons and tools as well as to meet the growing demands in the cities.

On the other hand, the craft production in the Harappan cities included bead-making, shell¬cutting, metal-working, seal-making and weight-making. The material used was stones, jasper, crystal, quartz, copper, bronze, gold, shell, faience and terracotta.

The evidence of craft production in the Harappan civilisation have been found from excavations. The evidences for the Early Historic cities have been found from excavations as well as from inscriptions.
Another difference is that there were guilds in the early Historic cities. These were organisations of craft producers and merchants. These guilds or shrenis probably procured raw materials, regulated production and marketed the finished product.

Question 2.
Describe the salient features of Mahajanapadas.
Solution :
Mahajanapadas were states that existed between 6th and 4th BC centuries. Buddhist and Jain texts mention sixteen Mahajanapadas. The name of all these are not uniform in all texts but some names are common and uniform which means they were the powerful ones. These Mahajanapadas are Vajji, Magadha, Kaushal, Kuru, Panchal, and Gandhar.

The important features of the Mahajanapadas are as follows.

1. Most of the Mahajanapadas were ruled by powerful kings. However, there were some Mahajanapadas where rule was in the hands of people, we call them republics. In some states the king and the subject had collective control on the economic resources of the state.
2. Every Mahajanapadas had its own capital. The capital normally would be surrounded by fort. The fortification of the capital was needed for protection and economic resources.
3. It was around 6th Qentury BC, Brahmins began to compile scripture called “Dharmshastra” which states rules of morality including that of monarch. Herein it was mentioned that the king should be Kshatriya.
4. The main job of the king was collection of taxes from farmers, traders, craftsmen. They also accepted donations.
5. It was considered fair to plunder neighbouring countries for riches.
6. Gradually Mahajanapadas began to have full time army and officials. Soldiers were from the ranks of farmers.

Question 3.
How do historians reconstruct the lives of ordinary people?
Solution :
Ordinary people rarely left accounts of their thoughts and experiences. The historians reconstruct their lives by examining stories contained in anthologies such as the Jatakas and the Panchatantra. For example, one story known as the Gandatindu Jataka describes the plight of the subjects of a wicked king. The subjects included elderly women and men, cultivators, herders, village boys and even animals. When the king went in disguise to find out what his subjects thought about him, each one of them cursed him for their miseries, complaining that they were attacked by robbers at night and by tax collectors during the day. As a result of it, people abandoned their village and went to live in the forest.

Question 4.
Compare and contrast the list of things given to the Pandyan chief (Source 3) with those produced in the village of Danguna (Source 8). Do you notice any similarities or differences?
Solution :
(a) The defeated people gave the following things to the Pandya chief as a mark of respect to the victorious king : Ivory, fragrant wood, fans made of the hair of deer, honey, sandalwood, red ochre, antimony, turmeric, cardamom, pepper, coconuts, mangoes, medicinal plants, fruits, onions, sugarcane, flowers, areca*nut, bananas, baby tigers, lions, elephants, monkeys, bear, deer, musk deer, fox, peacocks, musk cat, wild cocks and speaking parrots.
(b) The village of Danguna produced the following things : Grass, animal hides, charcoal, fermenting liquors, salt, khadira trees, flowers and milk.
(c)

1. Similarities : Both the lists contain the things of daily use such as honey, turmeric, i cardamom, pepper, mangoes, fruits, onions, flowers (Source 3) and grass, salt, flowers and milk (Source 8).
2. Differences : The things given to the Pandya chief included precious things such as ivory, fragrant wood, sandalwood and wild animals like tigers, lions, elephants, wild cocks. These things and animals prove that the forest people were brave and their economic condition was good. On the other hand, the things of the Danguna village did not include precious things. It included things such as grass, animal hides, flowers and milk which prove that they were ordinary people and their economic condition was bad. That was probably the reason for granting them various exemptions by Prabhavati Gupta.

Question 5.
List some of the problems faced by the epigraphists.
Solution :
The specialists who study inscriptions are called Epigraphists. Some of the important problems they encounter when they try to decipher inscriptions are as follows:

1. Many of the inscriptions are not found in proper shape, they are partly damaged, hence deciphering them becomes a knotty problem.
2. The inscriptions are written from the point of view of those who have created it. Hence, in order to get an impartial understanding, we need to go beyond the written words, get into its interpretations.
3. Many of the inscriptions have descriptions in symbolic words. Hence deciphering them have become difficult.
4. Sometimes the inscriptions are engrafted in very light colors. Hence, deciphering them becomes difficult.

6. Question 6.
Solution :
Asokan inscriptions mention all the main features of the administration of the Mauryan Empire. Thus, the features of the administration are evident in the inscriptions of the Asokan age. The important features of the same are as follow:
1. The capital of the Mauryan Empire was Pataliputra. Apart from the capital there ‘ were four other centres of political power in the empire. They were Taxila, Ujjaini,
Tosali and Suvamagiri.
2. Committee and subcommittees were formed to run the administration and safety of boundaries. Megasthenes has mentioned that there were one committee and six sub-committees. The six subcommittees and their areas of activities are as follows:
(i) The first sub committee looked after navy.
(ii) The second sub committee looked after transport and communications.
(iii) The third sub committee looked after infantry.
(iv) The fourth sub committee had the responsibility of horses.
(v) The fifth had the responsibility of chariots.
(vi) The sixth had the responsibility of elephants.
3. Strong network of roads and communications were established. It is notable that no large empire can be maintained in the absence of the same.
4. Asoka made an attempt to keep the empire united by the philosophy of Dhamma. Dhamma are nothing but moral principles that actuated people towards good conduct. Special officers called Dhamma Mahamtras were appointed to propagate Dhamma. In fact Romila Thapar has made it the most important element of the Asokan state’s governing principle.

Question 7.
This is a statement made by one of the best-known epigraphists of the twentieth century. D.C. Sircar: “There is no aspect of life, culture and activities of the Indians that is not reflected in inscriptions.” Discuss.
Solution :
(a) The statement of D.C. Sircar that there is no aspect of life, culture and activities of the Indians that is not reflected in inscriptions does not seem to be correct because not everything that is politically or economically significant was necessarily recorded in inscriptions. Some examples are given below :

• Routine agricultural practices and the joys and sorrows of daily existence find no mention in inscriptions.
• The inscriptions generally focus on grand, unique events.
• The content of inscriptions almost invariably projects the perspective of the person who commissioned them. For example, in some inscriptions Asoka claims that earlier rulers had no arrangements to receive reports about the people. This does not seem to be correct.

(b) The inscriptions give us only the following information :

• Information about the administration particularly major political centres.
• Asoka’s Dhamma and its propagation by special officers known as the dhamma mahamatta.

Question 8.
Discuss the notions of kingship that developed in the post-Maury an period.
Solution :

1. The main notion that developed in the post-Mauryan period was that of divine kings. The kings identified themselves with a variety of deities to claim high status. This strategy was adopted by the Kushanas who ruled over a vast kingdom extending from Central Asia to northwest India.
2. Colossal statues of Kushana rulers were installed in a shrine at Mat near Mathura and in Afghanistan. This indicates that the Kushanas considered themselves godlike.
3. Many Kushana rulers adopted the title devaputra, or “son of god”. It was possibly inspired by Chinese rulers who called themselves sons of heaven.

Question 9.
To what extent were agricultural practices transformed in the period under consideration?
Solution :
The agricultural practices were transformed in the period under consideration i.e., 600 BCE – 6OO CE in the following ways :

• There was a shift to plough-agriculture in fertile alluvial river valleys such as those of Ganga and the Kaveri from c. sixth century BCE.
• The iron-tipped ploughshare was used to turn the alluvial soil in areas which had high rainfall.
• In some parts of the Ganga valley, production of paddy was dramatically increased by the introduction of transplantation.
• Those living in hilly tracts in the north-eastern and central parts of the subcontinent practised hoe agriculture, which was much better suited to the terrain.
• Irrigation was used to increase agricultural production. Wells, tanks, and canals were used for this purpose. Communities as well as individuals organised the construction of irrigation works.

We hope the NCERT Solutions for Class 12 History Chapter 2 Kings, Farmers and Towns Early States and Economies help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 2 Kings, Farmers and Towns Early States and Economies, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity

NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity

Question 1.
The storage battery of a car has an e.m.f. of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery ?
Answer:

Question 2.
A battery of e.m.f. 10 V and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor ? What is the terminal voltage of the battery when the circuit is closed ?
Answer:

Question 3.
(a) Three resistors IΩ, 2 Ωand 3 Ω are combined in series. What is the total resistance of the combination?
(b) If the combination is connected to a battery of e.m.f. 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Answer:

Voltage drop across R₃ is given by, V₃ = IR₃ = 2 x 3 = 6V.

Question 4.
(a) Three resistors 2 Ω, 4 Ω, and 5 Ω, are combined in parallel. What is the total resistance of the combination?
(b) If the combination is connected to a battery of e.m.f. 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
Answer:

Question 5.
At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 x 10^-4 °C^-1.
Answer:

Question 6.
A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 x 10^-7 m², and its resistance is measured to be 50 Ω. What is the resistivity of the material at the temperature of the experiment ?
Answer:

Question 7.
A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.
Answer:

Question 8.
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C ? Temperature co­efficient of resistance of nichrome averages over the temperature range involved is 1.70 x 10^-4 °C^-1.
Answer:


Question 9.
Determine the current in each branch of the network shown in Figure

Answer:

Question 10.
(a) In a meter bridge (Figure), the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?

(b) Determine the balance point of the bridge above if X and Y are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current? (C.B.S.E. 2005)
Answer:


Question 11.
A storage battery of e.m.f. 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit ?
Answer:
During charging,
V = E + I(r + R)

Question 12.
In a potentiometer arrangement, a cell of e.m.f. 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the e.m.f. of the second cell ?
Answer:

Question 13.
The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 x 10²⁸ m³. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross­section of the wire is 2.0 x 10^-6 m² and it is carrying a current of 3.0 A.
Answer:

Question 14.
The earth’s surface has a negative surface charge density of 10^-9 cm^-2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric Held, how much time (roughly) would be required to neutralise the earth’s surface ? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe.)
(Radius of earth = 6.37 x 10⁶ m.)
Answer:

Question 15.
(a) Six lead-acid type of secondary cells each of e.m.f. 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage ?
(b) A secondary cell after long use has an e.m.f. of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell ? Could the cell drive the starting motor of a car ?
Answer:

It cannot be used for starting motor of a car because large current is needed to start the car.

Question 16.
Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (P_AL = 2.63 x 10⁸ Ω m, P_cu= 1.72 X 1(H Ω m. Relative density of A1 = 2.7, of Cu = 8.9.)
Answer:


Aluminium is lighter, so it is used for overhead power cables.

Question 17.
What conclusion can you draw from the following observations on a resistor made of alloy manganin?

Answer:
It indicates that Ohm’s law i.e.V α I is valid for a wide range.
Resistivity of Manganin remains nearly same with change in temperature.

Question 18.
Answer the following questions :
(a) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor : current, current density, electric field, drift speed?
(b) Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.
(c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?
(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?
Answer:
(a) Current
(b) No. Devices are and vacuum tubes, semiconductor diodes, transistors, thermisters, thyristors, etc.
(c) Lesser the value of ‘r’, the higher the current.

Clearly low r will ensure high current
(d) If internal resistance is not large, then the heavy current drawn during an accidental short circuit can damage the supply.

Question 19.
Choose the correct alternative:
(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.
(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy manganin is nearly independent of/increases rapidly with the increase of temperature.
(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (10²²/10)³).
Answer:
(a) Greater
(b) Lower
(c) Nearly independent
(d) 10²²

Question 20.
(a) Given n resistors each of resistance R, how will you combine them to get the
(i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?
(b) Given the resistances of 1 Q, 2 G, 3 Q, how will combine them to get an equivalent resistance of (i) (11/3)Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?
(c) Determine the equivalent resistance of networks shown in the figure. (N.C,E.R,T.)
Answer:

(i) Maximum resistance can be obtained by combining them in series with each other.
The maximum resistance R_max = R + R + R + ……………..n times = nR.
(ii) Minimum effective resistance can be obtained by combining them in parallel with each other. Minimum resistance R_max is found as

Question 21.
Determine the current drawn from a 12 V supply with internal resistance 0.5Ω by the infinite network shown in the figure. Each resistor has 1 Ω resistance. (N.C.E.R.T.)

Answer:
Let the total resistance of the circuit be Z and a set of three resistors of value R each is connected to it as shown in the figure.

Question 22.
The figure shows a potentiometer with a cell of 2.0 V and internal resistance of 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant e.m.f. of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown e.m.f. e and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(a) What is the value e?
(b) What purpose does the high resistance of 600 kΩ have?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an e.m.f. of 1.0 V instead of 2.0 V?
(f) Would the circuit work cell for determining an extremely small e.m.f., say of the order of a few mV (such as the typical e.m.f. of a Thermo ­couple not, how will you modify the circuit?
Answer:

(b) It allows only a small current to flow through the galvanometer when the circuit is not balanced.
(c) No
(d) No
(e) No
(f) No. The circuit will not work (E ∝ l).
The circuit can be modified by putting a suitable resistor ‘R’ in series with the wire AB.

Question 23.
The figure shows a potentiometer circuit for the comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε?
Answer:

(b) We fail to get the balance point with the given cell of emf E, if the potential difference across the wire AB. In order to obtain the balance point with the given cell E, either the emf of the auxiliary⁷ battery (between A and B) should be increased or a suitable resistance should be put in series with R and X (so as to decrease the potential drop across the wire AB).

Question 24.
The figure shows a 2.0 V potentiometer used for the determination of internal resistance

1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of
9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.
Answer:

We hope the NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity, help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Chemistry Chapter 15 provides answers to the questions provided in the textbook. The answers are accompanied by diagrammatic representations for better understanding. Also, the solutions are explained in such a language that the students find easy to understand.

NCERT Solutions are beneficial for the students appearing in UP board, MP board, CBSE, Gujarat board, etc. Also the students appearing for competitive exams such as NEET and JEE will find the NCERT Solutions for Class 12 Chemistry beneficial.

NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers

Polymers are very large molecules having high molecular mass. These are also known as macromolecules. This chapter explains the mechanical properties and applications of polymers. The classification of polymers is also described in detail.

The chapter gives an overview of different polymerization reactions. The chapter explains that the polymers are the backbone of four major industries, plastic, elastomers, fibres and paints.

NCERT INTEXT QUESTIONS

Question 1.
What are polymers?
Answer:
The word polymer has a Greek origin where poly means many and mers or meros stands for unit or part. Thus, the polymer may be defined as a substance of a high molecular mass (10³ – 10⁷ u) formed by the combination of a large number of simple
molecules called monomers by chemical bonds. The process by which monomers are converted into polymers is called polymerization.

Question 2.
How are polymers classified on the basis of structure?
Answer:
On the basis of structure, polymers are classified as:
(i) Linear polymers in which the monomers are joined together to form long straight chains of polymer molecules. Forex: HDPE, PVC, nylons, etc.
(ii) Branched-chain polymers in which the monomers not only join in a linear fashion but also form branches of different lengths along the main chain. For ex: LDPE, glycogen, etc.
(iii) Cross-linked polymers in which the initially formed linear polymer chains join together to form a 3D network structure. For ex: bakelite, Urea-formaldehyde resin, etc.

Question 3.
Write the names of the monomers of the following polymers:

Answer:
(i) The polymer is Nylon 66. Its monomer units are hexamethylenediamine and adipic acid.
(ii) The polymer is Nylon 6. Its monomer units are of caproilactum.
(iii) The polymer is Teflon. Its monomer units are of tetrafluoroethylene.

Question 4.
Classify the following as addition and condensation polymers:
(i) Terylene
(ii) Bakelite
(iii) Polyvinyl chloride
(iv) Polythene
Answer:
(i) Condensation polymer
(ii) Condensation polymer
(iii) Addition polymer
(iv) Addition polymer

Question 5.
Explain the difference between Buna-N and Buna-S.
Answer:
Both Buna-N and Buna-S are synthetic rubber and are co-polymers in nature. They differ in their constituents.
Buna-N: Constituents are : buta-1, 3-diene and acrylonitrile.
Buna-S: Constituents are : buta-1, 3-diene, and styrene. They condense in the presence of Na.

Buna – S: It is a co—polymer of 1. 3 – butadiene and styrene and is prepared by the polymerisation of these components in the
ratio of 3 : 1 in the presence of sodium.

Buna-N (Nitrile rubber):  h is a co-polymer of buta-1. 3-diene and acrylonitrile. It is formed as follows:

Question 6.
Arrange the following polymers in increasing order of their intermolecular forces :
(i) Nylon-66, Buna-S, Polythene
(ii) Nylon-6, Neoprene, Polyvinyl chloride
Answer:
We have studied the classification of polymers based upon intermolecular forces. These follow the order :
Elastomers < Plastics < Fibres.

The polymers listed may be arranged in increasing order of intermolecular forces as follows :
(i) Buna-S (Elastomer) < Polythene (Plastic) < Nylon-66 (Fibre)
(ii) Neoprene (Elastomer) < Polyvinyl chloride (Plastic) < Nylon-6 (Fibre).

NCERT Exercise

Question 1.
Explain the terms polymers and monomers.
Answer:
Polymers are a high molecular mass of macromolecules composed of repeating structural units derived from monomers. Polymers have a high molecular mass (10³-10⁷u). In a polymer, various monomer units are joined by strong covalent bonds. Polymers can be natural as well as synthetic. Polythene, rubber, and nylon 6,6 are examples of polymers.

Monomers are simple, reactive molecules that combine with each other in large numbers through covalent bonds to give rise to polymers. For example ethene, propene, styrene, vinyl chloride.

Question 2.
What are natural and synthetic polymers? Give two examples of each.
Answer:
Natural polymers:
Natural polymers are high molecular mass macromolecules and are found in nature mainly in plants and animals. For example, protein, nucleic acids, starch, cellulose, etc.

Synthetic polymer:
Synthetic polymers are man-made high molecular mass macromolecules. For example, plastics (Polyethene, P.V.C.), synthetic fibers (Polyesters, Nylon-6,6), synthetic rubber (Neoprene, Buna-S), etc.

Question 3.
Distinguish between the terms homopolymer and copolymer and give one example of each. (C.B.S.E. Outside Delhi 2008, Pb. Board 2008)
Answer:
A polymer in which the monomers are the same is called a homopolymer. For example, polystyrene is made from styrene only.
A polymer in which the monomers are different is known as a copolymer. For example, in Nylon-66 the monomers are adipic acid and hexamethylenediamine.

Question 4.
How do you explain the functionality of a monomer?
Answer:
The functionality of a monomer is the number of binding sites that is/are present in that monomer.
For example, the functionality of monomers such as ethene and propene is one and that of 1, 3-butadiene, and adipic acid is two.

Question 5.
Define the term polymerisation. (C.B.S.E. Delhi 2008)
Answer:
A very widely used polymer is polyethylene which is formed by the polymerisation of ethylene molecules (monomers) by heating under pressure in the presence of oxygen.

It may be noted that all the monomer units in a polymer may or may not be the same, In case, they are the same as in case of polyethylene, then the polymer is called homopolymer. On the other hand, if they happen to be different, then the polymer is known as copolymer or mixed polymer. For example, the monomer units in terylene are ethylene glycol and terephthalic acid. Similarly, hexamethylenediamine and adipic acid are the monomer units of nylon-66. Both are co-polymers or mixed polymers. We shall discuss these at a later stage in the present unit.

Question 6.

Answer:
It is a homopolymer in nature. It has only one type of monomer units i.e., NH₂ – CHR – COOH. These area-amino acids.

Question 7:
In which classes, are the polymers classified on the basis of molecular forces?
Answer:
Polymers are classified into four classes on the basis of molecular forces. These are:
elastomers, fibres, thermoplastic polymers, and thermosetting polymers.

1. Elastomers: In these polymers, the intermolecular forces are the weakest. As a result, they can be readily stretched by applying small stress and regain their original shape when the stress is removed. The elasticity can be further increased by introducing some cross-links in the polymer chains. Natural rubber is the most popular example of elastomers. A few more examples are of: buna-S, buna-N and neoprene.

2. Fibres: Fibres represent a class of polymers which are thread-like and can be woven into fabrics in a number of ways. These are widely used for making clothes, nets, ropes, gauzes etc. Fibres possess high tensile strength because the chains possess strong intermolecular forces such as hydrogen bonding. These forces are also responsible for close packing of the chains. As a result, the fibres are crystalline in nature and have also sharp melting points. A few common polymers belonging to this class are nylon – 66, terylene and polyacrylonitrile etc.

3. Thermoplastics: These are linear polymers and have weak van der Waals forces acting in the various chains and are intermediate of the forces present in the elastomers and in the fibres. When heated, they melt and form a fluid which sets into a hard mass on cooling, Thus, they can be cast into different shapes by using suitable moulds. A few common examples are polyethylene and polystyrene polyvinyls etc. These can be used for making toys, buckets, telephone apparatus, television cabinets etc.

4. Thermosetting plastics: These are normally semifluid substances with low molecular masses. When heated, they become hard and infusible due to the cross-linking between the polymer chains. As a result, they also become three-dimensional in nature. They do not melt when heated. A few common thermosetting polymers are bakelite, melamine-formaldehyde, urea-formaldehyde and polyurethane etc.

Question 8:
How can you distinguish between addition and condensation polymerisation?
Answer:
In addition polymerization: The monomers are unsaturated in nature i.e., they have atleast one double or triple bond in their molecules. They represent the functionality of the monomers which combine with each other at these sites. The addition polymerisation is generally chain growth polymerisation in nature. For example, polythene, polystyrene, PVC, Teflon etc. are all formed as a result of additional polymerisation.

In condensation polymerization: The monomer units have specific functional groups present which represent their functionality. The monomers combine through these functional groups and the polymerisation is step growth polymerisation in nature. For example, terylene, nylon-6, nylon-66, bakelite etc. are all formed as a result of condensation polymerisation.

Question 9:
Explain the term co-polymerization and give two examples.
Answer:
Co-polymerisation is a process in which a mixture of more than one different monomeric units are allowed to polymerise. The copolymer thus formed contains multiple units of each monomer in the chain. The formation of buna-S copolymer in which the monomers are : buta-1, 3-diene, styrene and sodium is an example of co­polymerisation.
Another Example of co-polymer: Buna-S is a co-polymer of 1 : 3 – butadiene and styrene in the presence of sodium metal.

Thus, as a result of addition polymerisation, the double bonds change to single bonds. A large number of monomer units can be linked in this way.
The addition polymerisation is normally a chain reaction in which the chain is built up by the successive addition of the monomer units. It involves a number of intermediates such as free radicals, carbocations or carbanions. Let us illustrate the chain growth polymerisation based upon different mechanisms.

Question 10:
Write the free radical mechanism for the polymerisation of ethene.
Answer:
The polymerisation of ethene to polythene consists of heating or exposing to light a mixture of ethene with a small amount of benzoyl peroxide initiator The process starts with the addition of phenyl free radical formed by the peroxide to the ethene double bond thus regenerating a new and larger free radical. This step is called chain initiating step. As this radical reacts with another molecule of ethene, another bigger sized radical is formed. The repetition of this sequence with new and bigger radicals carries the reaction forward and the step is termed as chain propagating step. Ultimately; at some stage the product radical thus formed reacts with another radical to form the polymerised product. This step is called the chain terminating step. The sequence of steps may be depicted as follows :
Chain initiating steps:

Chain terminating step:
For termination of the long-chain, these radicals can combine in different ways to form polythene. One mode of termination of the chain is shown as under:

Question 11:
Define thermosetting and thermoplastic polymers with two examples of each.
Answer:
Thermoplastics are polymers which can be easily softened repeatedly on heating and hardened on cooling. Therefore, it can be used again and again. For example, polyethylene and polyvinyl chloride. Thermosetting polymers are those which undergo permanent change on heating. They become hard and infusible on heating and cannot be softened again. For example, Bakelite, and Melamine formaldehyde.

Question 12:
Write the monomers used for getting the following polymers :

1. Polyvinyl chloride
2. Teflon
3. Bakelite

Answer:

1. vinyl chloride
2. tetrafluoroethylene
3. phenol and formaldehyde.

Question 13:
Write the name and structure of one of the common initiators used in free radical addition polymerisation.
Answer:

Question 14:
How does the presence of double bonds in rubber molecules influence their structure and reactivity?
Answer:
Chemically natural rubber is polyisoprene in which monomer units are of isoprene i.e., 2-methyl-i, 3-butadiene. It is in fact, 1, 4 polymer in which monomers are linked through CH₂ groups located at 1.4 positions. The residual double bonds are located at C₂ and C₃ positions in the isoprene units. All the double bonds have cis configurations. Thus, natural rubber is cis polyisoprene.

The high elasticity of natural rubber is due to the absence of polar groups and the cis-configurations about double bonds do I not allow the polymer chains to come closer. Therefore, only weak van der Waals’ forces are present. Since the chains are not linear, they can be stretched just like springs and exhibit elastic properties.

Question 15:
Discuss the main purpose of vulcanisation of rubber.
Answer:
The main purpose of vulcanization of rubber is to improve the following draw-back of natural rubber:

• At high temperature (T >335K) natural rubber becomes soft.
• At low temperature (T< 283K) natural rubber becomes brittle.
• Natural rubber is soluble in non-polar solvents.
• It is non-resistant to attack by oxidizing agents.

Question 16:
What are the monomers repeating units of Nylon 6 and Nylon 66?
Answer:
The monomeric repeating unit of nylon 6 is [NH – (CH₂)₅ – CO], which is derived from caprolactam.
The monomeric repeating unit of nylon 6, 6 is [NH – (CH₂)₆ – NH – CO – (CH₂)₄ – CO], which is derived from hexamethylene diamine and adipic acid.

Question 17:
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
(ii) Buna-N
(iii) Dacron
(iv) Neoprene (C.B.S.E. Delhi 2008)
Answer:

Question 18:
Identify the monomers in the following polymeric structures:

Answer:
(i) Decanedioic acid: HOOC(CH₂)₈COOH
Octamethylenediamine: H₂N(CH₂)₈NH₂

(ii)

Question 19:
How is dacron obtained from ethylene glycol and terephthalic acid?
Answer:

Question 20:
What is a biodegradable polymer? Give an example of an aliphatic biodegradable polyester.
Answer:
A polymer that can be decomposed by bacteria is called a biodegradable polymer. Poly – β- hydroxybutyrate – Co -β – hydroxy valerate (PHBV) is a biodegradable aliphatic polyester.

We hope the NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 15 Polymers, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Biology Chapter 4 Reproductive Health

These Solutions are part of NCERT Solutions for Class 12 Biology. Here we have given NCERT Solutions for Class 12 Biology Chapter 4 Reproductive Health

Question 1.
What do you think is the significance of reproductive health in a society?
Solution:
Significance of reproductive health in society are:

• Control over the transmission of STDs.
• Less death due to reproduction-related diseases like-AIDS, cancer of the reproductive tract.
• Control in a population explosion.
• Not only the reproductive health of men and women affects the health of the next generation.

Question 2.
Suggest the aspects of reproductive health which need to be given special attention in the present scenario.
Solution:
Providing medical facilities and care to the problems like menstrual irregularities, pregnancy related aspects, delivery, medical termination of pregnancy, STDs, birth control, infertility. Post-natal child maternal management is another important aspect of the reproductive and child health care programme.

Question 3.
Is sex education necessary in schools? Why?
Solution:
Yes, sex education is necessary for schools because:

• It will provide proper information about reproductive organs, adolescence, safe, hygienic sexual practices, and Sexually Transmitted Diseases (STDs).
• It will provide the right information to avoid myths and misconceptions about sex-related queries.

Question 4.
Do you think that reproductive health in our country has improved in the past 50 years? If yes, mention some such areas of improvement.
Solution:
The reproductive health in our country has improved in the last 50 years. Some areas of improvement are :

• Massive child immunization.
• Increasing use of contraceptives.
• Better awareness about sex related matters.
• Increased number of medically assisted deliveries and better post-natal care leading to decreased maternal and infant mortality rates.
• Increased number of couples with small families.
• Better detection and cure of STDs and overall increased medical facilities for all sex related problems.

Question 5.
What are the suggested reasons for the population explosion?
Solution:

• Improved medical facilities
• Decline in death rate, IMR, MMR
• Slower decline in birth rate.
• Longer life span.
• Lack of 100% family planning and education among the village.

Question 6.
Is the use of contraceptives justified? Give reasons.
Solution:
Yes, the use of contraceptives is justified: To overcome the population growth rate, contraceptive methods are used. It will help in bringing birth rate down & subsequently curb population growth. With the rapid spread of HIV/ AIDS in the country, there is now a growing realization about the need to know about contraception & condoms.

Question 7.
Removal of gonads cannot be considered as a contraceptive option. Why?
Solution:
Removal of gonads not only stops the production of gametes but will also stop the secretions of various important hormones, which are important for bodily functions. This method is irreversible and thus, can not be considered as a contraceptive method.

Question 8.
Amniocentesis for sex determination is banned in our country. Is this ban necessary? Comment.
Solution:
Amniocentesis is a prenatal diagnostic technique to find out the genetic disorders and metabolic disorders of the foetus. Unfortunately, the useful technique of amnio-centesis had been misused to kill the normal female foetuses as it could help detect the sex of foetus also. Hence, this technique is now banned in our country. This ban is necessary as this technique was promoting female foeticide in our country.

Question 9.
Suggest some methods to assist infertile couples to have children.
Solution:
If the couples are enabled birth to the children and corrections are not possible, the couples could be assisted to have children through certain special techniques, commonly known as Assisted Reproductive Technologies (ART). Some methods are given as:

1. In Vitro Fertilization (IVF): In this method, ova from the female and the sperm from the male are collected and induced to form a zygote under simulated conditions in the laboratory. This process is called In Vitro Fertilization (IVF). Some method is given as follows:

• Zygote Intrafallopian Transfer (ZIFT): The zygote or early embryo with up to 8 blastomeres is transferred into the fallopian tube.
• Intra-Uterine Transfer (IUT): Embryo with more than 8 blastomeres is transferred into the uterus in females who cannot conceive embryos formed by the fusion of gametes in another female are transferred.
• Test tube baby: In this method, ova from the donor (female) and sperm from the donor (male) are collected and are induced to form a zygote under simulated conditions in the laboratory. The zygote could then be transferred into the fallopian tube and embryos transferred into the uterus, to complete its further development. The child born from this method is called a test-tube baby.

2. Gamete Intra Fallopian Transfer (GIFT): It is the transfer of an ovum collected from a donor into the fallopian tube 8 another female who cannot produce one, but can provide a suitable environment for fertilization and further development of the embryo.

3. Intra Cytoplasmic Sperm Injection (ICSI) : It is a procedure to form an embryo HI* the laboratory by directly injecting the sperm into an ovum.

4. Artificial Insemination (AI): In this method, the semen collected either from the husband or a healthy donor is artificially introduced into the vegina or into the uterus (Intra Uterine Insemination, IUI). This technique is used in cases where the male is unable to inseminate sperms in the female reproductive tract or due to very low sperm counts in the ejaculation.

5. Host Mothering: In this process, the embryo is transferred from the biological mother to a surrogate mother. The embryo then develops till it is fully developed or partially developed. It is then transferred to the biological mother or into any other. This technique is useful for females in which embryo forms but is not able to develop.

Question 10.
What are the measures one has to take to prevent contracting STDs?
Solution:
Diseases or infections which are transmitted through sexual intercourse are collectively called sexually transmitted diseases (STDs) or reproductive tract infections (RT), e.g., gonorrhea, syphilis, genital herpes, AIDS, etc. The measures that one has to take to prevent from contracting STDs are:

• Avoid sex with unknown partners/multiple partners.
• use condoms during coitus.
• In case of doubt, go to a qualified doctor for early detection and get complete treatment if diagnosed with the disease.

Question 11.
State True/False with an explanation.

1. Abortions could happen spontaneously too.
2. Infertility is defined as the inability to produce viable offspring and is always due to abnormalities/defects in the female partner.
3. Complete lactation could help as a natural method of contraception.
4. Creating awareness about sex related aspects is an effective method to improve the reproductive health of people.

Solution:

1. True: One-third of all pregnancies abort spontaneously (called miscarriage) within four weeks of conception and abortion passes unrecognized with menses.
2. False: Infertility is defined as the inability of the couple to produce viable offspring. It is due to abnormalities/defects in either male or female or both.
3. True: Complete lactation is a natural method of contraception as during this period ovulation does not occur, but this is limited to a period of 6 months after parturition.
4. True: Creating awareness in people about sex-related aspects like right information about reproductive organs, accessory organs of reproduction, safe and hygienic sexual practices, birth control methods, care of pregnant women, post-natal care of mother and child, etc., can help in improving the reproductive health of people.

Question 12.
State True/False with an explanation.
(a) Abortions could happen spontaneously too. (True/False)
Answer:
False, Abortion does not happen under normal conditions. It happens accidentally or under the will of Parents.

(b) Infertility is defined as the inability to produce a viable offspring and is always due to abnormalities/defects in the female partner. (True/False)
Answer:
False, Sterility always does not occur due to females sometimes. Males are also responsible for this.

(c) Complete lactation could help as a natural method of contraception. (True/False)
Answer:
True, the Menstrual cycle does not occur after parturition which can act as natural
contraception but this method is functional for a period of six months from parturition.

(d) Creating awareness about sex related aspects is an effective method to improve the reproductive health of dead people. (True/False)
Answer:
True, this creates better reproductive health among people.

We hope the NCERT Solutions for Class 12 Biology Chapter 4 Reproductive Health help you. If you have any query regarding NCERT Solutions for Class 12 Biology Chapter 4 Reproductive Health, drop a comment below and we will get back to you at the earliest.

NCERT Class 12 Chemistry Solutions for Chapter 12 provides answers for the questions provided in the textbook. The answers are accurate to the best of our knowledge and are provided by subject experts. The students can refer to these or sure shot success.

The students appearing for various boards and competitive exams can find these solutions helpful for practice. Most of the questions asked in UP board, MP board, Gujarat board, CBSE, etc. are asked from these. To score well in the examinations, the students should go through these solutions atleast once after finishing the entire syllabus.

NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones and Carboxylic Acids

This chapter explains the structure, physical and chemical properties and applications of aldehydes, ketones and carboxylic acid. It also explains a correlation between the three. It also explains the mechanism of different reactions of aldehydes, ketones and carboxylic acids. The factors affecting the acidity of carboxylic acids and their reactions help in understanding the advanced concepts related to this chapter.

NCERT Solutions for Class 12 Chapter 12 will help you revise the chapter during the examination. It also helps in clearing doubts if any.

NCERT INTEXT QUESTIONS

Question 1:
Write the structures of the following compoimds :
(i) α -Methoxypropionaldehyde
(ii) 3-Hydroxybutanal
(iii) 2-Hydroxycyclopentanecarbaldehyde
(iv) 4-Oxopentanal
(v) Di-sec butylketone
(vi) 4-Fluoroacetophenone
Answer:

Question 2.
Write the structures of products of following reactions:

Answer:

Question 3.
Arrange the following compounds in increasing order of their boiling points :
CH₃CHO,  CH₃CH₂OH,  CH₃OCH₃, CH₃CH₂CH₃
Answer:
The increasing order of boiling points of all these compounds of comparable molecular masses is :

Explanation: We know that the boiling points of liquids are directly related to the magnitude of the intermolecular forces of attraction.

1. Hydrocarbons (alkanes) are completely non-polar. The only attractive forces in their molecules are Van der Waals forces which are quite weak. That is why propane (CH₃CH₂CH₃) has the least boiling point. It is a gas at room temperature.
2. Ethers have bent structures and are also polar. However, there is no hydrogen bonding in their molecules. The only attractive forces are dipolar forces. Therefore, boiling point of dimethyl ether (CH₃OCH₃) is higher than that of propane. However, it is also a gas at room temperature.
3. Aldehydes contain polar carbonyl group and have strong dipolar interactions in their molecules. It is more than in ethers. Therefore, the boiling point of acetaldehyde (CH₃CHO) is more than that of dimethyl ether.
   
4. Out of all the families listed, alcohols have maximum intermolecular forces in the form of hydrogen bonding
   
   As a result, ethyl alcohol (C₂H₅OH) has the maximum boiling point.

Question 4.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions
(i) Ehtanal, propanaL, propanone, butanone
(ii) Benzaldehyde,p-Tolualdehyde, p-Nitrobenzaldehyde, acetophenone.
Ans: (i) Butanone < Propanone < Propanal < Ethanal .This is because as the no. of alkyl groups attached to carbonyl carbon increases, +I-effect increases. As a result, e^– density

Question 5.
Predict the products of the following reactions :

Answer:

Question 6.
Give the IUPAC names of the following compounds:

Answer:

Question 7.
Show how each of the following compounds can be converted into benzoic acid?
(i) Ethylbenzene (C.B.S.E. Outside Delhi 2017)   
(ii) Acetophenone  (C.B.S.E. Outside Delhi 2017)
(iii) Bromobenzene
(iv) Phenylethene (Styrene)
Answer:

Question 8.
Which acid of each pair would you expect to be stronger?
(i) CH₃CO₂H or FCH₂CO₂H
(ii) FCH₂CO₂H or ClCH₂CO₂H
(iii) FCH₂CH2CH₂CO₂H or CH₃CH(F)CH₂CO₂H

Answer:

NCERT EXERCISE

 Question 1.
What is meant by the following terms? Give an example in each case.
(a) Cyanohydrin
(b) Semicarbazone
(c) Acetal
(d) Oxime
(e) Cyanohydrin
(f) Ketal
(g) Aldol
(h) Schiff’s base
(i) 2, 4-D.N.P.
(ii) Imine
Answer:

Question 2.
Name the following compounds according to IUPAC system of nomenclature.
(i) CH₃CH(CH₃)CH₂CH₂CHO
(ii) CH₃CH=CHCHO
(iii) CH₃CH(CH₃)CH₂C(CH₃)₂COCH₃
(iv) OHCC₆H₄CHO(-p)
(v) CH₃CH₂COCH(C₂H₅)CH₂CH₂Cl
(vi) CH₃COCH₂COCH₃
(vii) (CH₃)₃CCH₂COOH
(viii) (CH₃)₂CHCH(CH₃)COCl
Answer:
(i) 4-Methylpentanal
(ii) But-2-enal
(iii) 3 3 5-Trimethylhexan-2-one
(iv) Benzene- 1, 4-dicarbaldehyde
(v) 6-Chloro-4-ethylhexan-3-one
(vi) Pentane-2, 4-dione
(vii) 3, 3-Dimethylbutanoic acid
(viii) 2, 3-Dimethylbutanoyl chloride

Question 3.
Draw the structures of the following compounds:
(i) 3-Methylbutanal
(ii) p-Nitropropiophenone
(iii) p-Methylbenzaldehyde
(iv) 4-Methylpent-3-en-2-one
(v) 4-Chloropentan-2-one
(vi) 3-Bromo-4-phenylpentanoic acid
(vii) pp’-Dihydroxybenzophenone
(viii) Hex-2-en-4-ynoic acid
Answer:

Question 4.
Write the IUPAC names of the following aldehydes and ketones. Also give the common names wherever possible.

Answer:

Answer:

Question 7.
Which of the following will undergo Aldol condensation, which Cannizzaro’s reaction, and which neither of these? Write the structures of the expected products in each case
(i) Methanal
(ii) 2-Methylpentanal
(iii) Benzaldehyde
(iv) Benzophenone
(v) Cyclohexanone
(vi) 1-Phenylpropanone
(vii) Phenylacetaldehyde
(viii) Butan-1-ol
(ix) 2, 2-Dimethyl butanal
Answer:
(i) Methanal (HCHO): It will give Cannizzaro’s reaction since the α-hydrogen atom is absent.

(ii) 2-Methylpentanal [CH₃CH₂CH₂CH (CH₃)CHO]: It will give Aldol condensation since the α-hydrogen atom is present.

(iii) Benzaldehyde (CgH₅CHO): It will give Cannizzaro’s reaction since a-hydrogen is not present.

(iv) Benzophenone (C₆H₅COC₆H₅): It will not give any of the two reactions. Being ketone, does not take part in Cannizzaro’s reaction. Without a-hydrogen, it fails to participate in Aldol condensation.

(vi) 1-Phenylpropanone (C₆H₅COCH₂CH₃): It will undergo Aldol condensation since the α-hydrogen atom is present.

(vii) Phenylacetaldehyde (C₆H₅CH₂CHO): It will give Aldol condensation since the α-hydrogen atom is present.

(viii) Butan-1-ol: It will not give any of the reactions.

Question 8.
How will you convert ethanal to the following compounds?
(i) Butane-1, 3-diol
(ii) But-2-enal
(iii) But-2-enoic acid.
Answer:
(i) Ethanal to butane -1, 3-diol

(ii) Ethanal to but-2-enal

(iii) Ethanal into but-2-enoic acid

Question 9.
Write the structural formulae and names of four possible aldol condensation products from propanal and butanal. In each case, indicate which aldehyde serves as a nucleophile and which as an electrophile.
Answer:
Both propanal and butanal have a-hydrogen atoms present. These can undergo self aldol condensation as well as cross aldol condensation to give four compounds as follows:
(i) Condensation involving propanal: It is a case of a self aldol condensation.

(ii) Condensation involving butanal: It is self aldol condensation.

(iii) Condensation involving butanal (electrophile) and propanal (nucleophile):  It is cross-aldol condensation.

(iv) Condensation involving propanal (electrophile) and butanal (nucleophile): It is cross-aldol condensation.

Question 10.
An organic compound with molecular formula CgHjoO forms 2, 4-DNP derivative, reduces Tollen’s reagent, and undergoes Cannizzaro’s reaction. On vigorous oxidation, it gives 1, 2 benzene dicarboxylic acid. Identify the compound.      (C.B.S.E. Outside Delhi 2012; Haryana Board 2013)
Answer:
Since the compound forms 2, 4-DNP derivative on reacting with 2, 4-DNP, it is a carbonyl compound. As the compound reduces Tollen’s reagent and undergoes Cannizzaro’s reaction, it is an aldehyde and not a ketone. The data further reveals that the compound on vigorous oxidation gives 1, 2-benzene dicarboxylic acid. This clearly shows that in the compound which is of aromatic nature, CHO group is present at position-1 and C₂H₅ side chain at position-2. The given compound is 2-ethyl benzaldehyde. The reactions involved are given below :

Question 11.
An organic compound [A] with molecular formula C₈H₁₆O₂ was hydrolysed with dilute sulphuric acid to give a carboxylic acid [B] and alcohol [C]. Oxidation of [C] with chromic acid produced [B]. The alcohol [C] on dehydration gave but-1-ene. Write equations for the reactions involved. (C.B.S.E. 2008 Supp., C.B.S.E. 2009)
Answer:
(i) The available data shows that the compound [A] upon hydrolysis gave a carboxylic acid [B] and alcohol [C]. It must be an ester.
(ii) Since the alcohol [C] upon oxidation with chromic acid gave back the carboxylic acid [B], both the acid and alcohol must have the same number of carbon atoms (four each).
(iii) The alcohol [C] upon dehydration gave an alkene.
The equations for the reactions are given:

Question 12.
Arrange the following compounds in increasing order of their property as indicated:
(i) Acetaldehyde, Acetone, Di-tert butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN)
(ii) CH₃CH₂CH(Br)COOH, CH₃CH(Br)CH₂COOH, (CH₃)₂CH COOH, CH₃CH₂CH₂COOH (acid strength).
(iii) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic add, 4-Methoxybenzok acid (acid strength).
Answer:
(i) The reactivity of aldehydes and ketones towards HCN addition decreases as the +1 – effect of the alkyl groups increases. Secondly, it decreases with increase in steric hindrance to the nucleophilic attack by CN^– at the carbonyl carbon. Thus the decreasing order of reactivity towards HCN is,

(ii) We know that the + I-effect decreases while -I-effect increases the acidic strength of carboxylic acids. Since + I-effect of isopropyl group is more than that of propyl group, therefore, (CH₃)₂CHCOOH is a weaker acid than CH₃CH₂CH₂COOH. Further since -I-effect decreases with distance, therefore CH₃CH₂CHBrCOOH is a stronger acid than CH₃CHBrCH₂COOH. Thus, the overall acid strength increases in the order:

(iii) Since electron-donating groups decrease the acidic strength, therefore, 4-methoxy benzoic acid is a weaker acid than benzoic acid. Further, since electron-withdrawing groups increase the acidic strength, therefore, both 4-nitrobenzoic acid and 3,4-dinitrobenzoic acid are stronger acids than benzoic acid. Further due to the presence of an additional -NO₂ group at /w-position with respect to -COOH group, 3,4-dinitrobenzoic acid is a stronger acid than 4-nitrobenzoic acid. Thus, the overall acidic strength increases in the order:4-methoxy benzoic acid < benzoic acid < 4-nitrobenzoic acid < 3,4-dinitrobenzoic acid.

Question 13.
Give chemical tests to distinguish between the following pairs of compounds :
(i) Propanal and propanone  (C.B.S.E. Delhi 2011, 2012)
(ii) Phenol and benzoic acid
(iii) Acetophenone and benzophenone
(iv) Benzoic acid and ethyl benzoate  (C.B.S.E. Outside Delhi 2009, 2011)
(v) Pentan-2-one and pentane-3-one
(vi) Benzaldehyde and acetophenone (C.B.S.E. Outside Delhi 2015)
(vii) Ethanal and propanal (C.B.S.E. Outside Delhi 2009, 2011, 2012)
Answer:
(i) Propanal and propanone:  Propanal will give a silver mirror upon heating with Tollen’s reagent but propanone will not respond.
(ii) Phenol and benzoic acid: Benzoic acid will give brisk effervescence with sodium hydrogen carbonate (NaHC0₃) but phenol will not respond.
(iii) Acetophenone and benzophenone: Acetophenone is a methyl ketone. It will give a yellow precipitate upon heating with I₂ and NaOH. Benzophenone will not respond.
(iv) Benzoic acid and ethyl benzoate: Benzoic acid will give brisk effervescence with sodium hydrogen carbonate (NaHC0₃) but ethyl benzoate (ester) will not respond.
(v) Pentan-2-one and pentan-3-one: Pentan-3-one is a methyl ketone and will give a yellow precipitate upon heating with I₂ and NaOH. Pentan-3-one will not respond.
(vi) Benzaldehyde and acetophenone: The distinction can also be made by iodoform test. Acetophenone will give yellow precipitate while benzaldehyde will not react.
(vii) Ethanal and propanal: Ethanal will respond to iodoform test and give yellow precipitate. Propanal will not react.

Question 14.
How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom.
(i) Methylbenzoate
(ii) m-Nitrobenzoic acid
(iii) p-Nitrobenzoic acid
(iv) Phenylacetic acid
(v) p-nitrobenzaldehyde
Answer:
(i) Benzene to methylbenzoate

(ii) Benzene to m-nitrobenzoic acid


(iii) Benzene to p-nitrobenzoic acid


(vi) Benzene to phenylacetic acid


(v) Benzene to p-nitrobenzaldehyde

Question 15.
How will you bring about the following conversions in not more than two steps?
(i) Propanone to propene   (C.B.S.E. Delhi 2009, Uttarakhand Board 2009)
(ii) Propanal to butanone
(iii) Ethanol to 3-hydroxybutanal (C.B.S.E. Outside Delhi 2012)
(iv) Benzaldehyde to benzophenone (C.B.S.E. Outside Delhi 2012)
(v) Benzaldehyde to 3-PhenyIpropan-1-ol
(vi) Benzaldehyde to a-Hydroxyphenylacetic acid
(vii) Benzoic acid to benzaldehyde (C.B.S.E. Delhi 2009, Outside Delhi 2017)
(viii) Benzene to m-nitroacetophenone
(ix) Benzoic acid to /n-nitrobenzyl alcohol.      (C.B.S.E. Delhi 2012)
Answer:
(i) Propanone to propene

(ii) Propanal to butanone

(iii) Ethanol to 3-hydroxybutanal

(iv) Benzaldehyde to benzophenone

(v) Benzaldehyde to 3-phenylpropan-1-ol

(vi) Benzaldehyde to a-Hydroxyphenylacetic acid

(vii) Benzoic acid to benzaldehyde

(viii) Benzene to m-nitroacetophenone

(ix) Benzoic acid to m-nitrobenzyl alcohol

Question 16:
Describe the following:
(i) Acylation
(ii) Cross-aldol condensation
(iii) Cannizzaro’s reaction
(iv) Decarboxylation.
Answer:

Question 17:
Complete each synthesis by giving missing starting material, reagent, or products.


Answer:

Question 18:
Give a plausible explanation for each of the following:
(i) Cyclohexanone forms cyanohydrin in good yield but 2, 4, 6-trimethylcyclohexanone does not.
(ii) There are two – NH₂ groups in semicarbazide. However, only one is involved in the formation of semicarbazone.
(iii) During the preparation of esters from carboxylic acid and alcohol in the presence of an acid catalyst, the water or the ester should be removed as fast as it is formed.
Answer:

In cyclohexanone, the attack of CN“ ion (nucleophile) can easily take place at the carbonyl carbon atom. However, in 2, 4, 6-trimethylcyclohexanone, the three CH₃ groups being electron releasing in nature (+ I effect) will considerably increase the electron density on the carbonyl carbon atom and the nucleophile attack does not seem to be feasible. Moreover, the two —CH₃ substituents at the ortho positions will also hinder the attack of nucleophile CN^– ion on the carbonyl group.

(ii) The structural formula of semi-carbazide is NH₂NHCONH₂. Although both the amino groups have lone electron pairs, one of these is in conjugation with the electron-withdrawing carbonyl group and acquires a positive charge. Therefore, it is not in a position to act as the nucleophile, and only one -NH₂ group is involved in the formation of semicarbazone.

(iii) The esterification carried in the presence of acid is of reversible nature and the reverse reaction is called ester hydrolysis.

In order that the reaction may proceed in the forward direction, ester or water formed in the reaction must be removed. Sulphuric acid added in esterification helps in removing molecules of H₂0 as it is a dehydrating agent.

Question 19:
An organic compound contains 69-77% carbon, 11-63% hydrogen and the rest is oxygen. The molecular mass of the compound is 86. It does not reduce Tollen’s reagent but forms an addition compound with sodium hydrogen sulphite and gives a positive iodoform test. On vigorous oxidation, it gives ethanoic acid and propanoic acid. Write the possible structure of the compound. (C.B.S.E. Delhi 2008, 2009, Uttarakhand Board 2015)
Answer:
Step I: Calculation of molecular formula of the compound
Percentage of oxygen = 100 – (% C + % H) = 100 – (69.77 + 11.63) = 18.6%

Step II. Predicting the structure of the compound

• Since the compound forms an addition compound with NaHS0₃, it must be a carbonyl compound.
• As the compound does not reduce Tollen’s reagent but gives a positive iodoform test, it must contain in it a methyl
  

Keeping in view these characteristics, the compound is CH₃CH₂CH₂COCH₃ (Pentan-2-one).
All the reactions in which pentan-2-one participates, are given for the benefit of the students.

Question 20.
Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why?
Answer:

We hope the NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones, and Carboxylic Acids help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 12 Aldehydes, Ketones, and Carboxylic Acids, drop a comment below and we will get back to you at the earliest.

Here we are providing NCERT Solutions for Class 12 English Flamingo Chapter 2 Lost Spring. Students can get Class 12 English Lost Spring NCERT Solutions, Questions and Answers designed by subject expert teachers.

Lost Spring NCERT Solutions for Class 12 English Flamingo Chapter 2

Lost Spring NCERT Text Book Questions and Answers

Lost Spring Think as you read 

Question 1.
What is Saheb looking for in the garbage dumps? Where is he and where has he come from?
Answer:
Unlike his parents who sifted through the garbage dumps for their survival, Saheb took it to be a treasure trove, wondrous and magical. He sometimes chanced upon a coin and hoped of finding more. He lived in Seemapuri. His family had arrived from Bangladesh in 1971.

Question 2.
What explanations does the author offer for the children not wearing footwear?
Answer:
The author disagreed with the usual explanation that is offered for the children going barefoot as a part of tradition in India. She felt it was only an excuse for the lack of money. They could ill-afford shoes as they lived in “a perpetual state of poverty”.

Question 3.
Is Saheb happy working at the tea-stall? Explain.
Answer:
Saheb was secure working at a tea-stall where he received daily wages and was given regular meals. However, it can be guessed that he was unhappy as he does not answer the writer when asked if he was happy. The writer also noticed that his face no longer carried the carefree look. He looked burdened with responsibilities.

Question 4.
What makes the city of Firozabad famous?
Answer:
Firozabad was famous for its bangles. Most families in Firozabad were engaged in making bangles. It engaged most of the families in its central industry. They worked around furnaces, welding glass and making bangles.

Question 5.
Mention the hazards of working in the glass bangles industry?
Answer:
Bangles were manufactured in glass furnaces with high temperatures, in dingy cells without air or light. As a consequence, the children, who slogged away in cloistered rooms close to the hot furnaces, often lost the brightness of their eyes, even their vision.

Question 6.
How is Mukesh’s attitude to his situation different from that of his family?
Answer:
Mukesh’s father worked as a tailor and as a bangle maker. He passed on his bangle-making skill to his family.
However, Mukesh dreamt of becoming a car mechanic, and wanted to break away from the occupation his family had been involved in for generations.

Lost Spring Understanding the text

Question 1.
What could be some of the reasons for the migration of people from villages to cities?
Answer:
Although it is difficult for the people to relocate from villages to cities, migration of such a nature continues unabated. People migrate for various reasons. These could be:

• de-fragmentation of land holdings
• lack of job opportunities
• lack of physical infrastructure no factories or other forms of employment, no medical support, no educational institutions, etc.
• lack of public health amenities such as sewage, drainage, etc.
• inability to deal with environmental hazards such as rains, storms, etc.
• glamour of the city life lures youngsters
• limited opportunities for progress
• aspirations for a better lifestyle

Question 2.
Would you agree that promises made to the poor children are rarely kept? Why do you think this happens in the incidents narrated in the text?
Answer:
Promises made to the poor children are rarely kept. Promises are made both at the national and international levels to provide healthy lives, quality education, protection against abuse, exploitation and violence, and combating HIV or AIDS. Yet, it is estimated that 246 million children are engaged in child labour. Of those, almost 171 million work in hazardous conditions, such as in mines, with chemicals and pesticides in agriculture or with dangerous machinery.

In some sectors their presence is kept under wraps where they toil away as domestic servants in homes, and labouring behind the walls of workshops. Millions of girls work as domestic servants and as unpaid household help and are especially vulnerable to exploitation and abuse. As the world looks ahead to prosperity, many children barely have a future, trapped in the conditions of poverty, conflict, and a degraded environment.

Just like Saheb and Savita, their future continues to be bleak. The children become disillusioned and often turn cynical. This happens because of utter poverty and the failure of the government to provide social security to the people.

Question 3.
What forces conspire to keep the workers in the bangle industry of Firozabad in poverty?
Answer:
Despite a government ban on child labour, 20,000 children in Firozabad work in horrific conditions to support their poor families. The workers are exposed to hazards such as blindness, tuberculosis, bronchitis, etc. In spite of working in such hazardous conditions the children are underpaid.They are forced to lead this life of poverty, as bangle-making continues to be their only means of livelihood. They can barely afford two square meals a day. They lack money and enterprise to do anything except carry on the business of

making bangles. The young men follow in the path of their elders as the profession is carried from one generation to the next. Years of mind-numbing toil kills their drive and their ability to dream. They lack the education and awareness to organise themselves into a cooperative and escape the vicious circle created by middlemen. The fear of the police and lack of leadership keep them back They remain caught in a web of poverty, burdened by the stigma of caste. The bureaucrats and the politicians exploit them further.

Lost Spring Talking about the text

Question 1.
How, in your opinion, can Mukesh realise his dream?
Answer:
Mukesh, one of the many children in Firozabad, aspired to be a motor mechanic. His dreams were unlike those of his peers, who worked in bangle manufacturing units amidst appalling conditions. Most of the people there, caught in the vicious circle, were bom and died in the same miserable plight as their forefathers. Mukesh, however, dared to dream.

He was determined to go to a garage and learn how to become a garage mechanic. He realized that the garage was a long way from his home, yet he was resolute and decided to walk all his way there. He dreamt of driving cars that he saw hurtling down the streets of his town. His passion and perseverance would help him achieve his goal. Mukesh was able to dream of breaking away from tradition, and that was the first step towards the realisation of his dreams.

Question 2.
Mention the hazards of working in the glass bangle industry.
Answer:
Child labour gives rise to a situation where the children are forced to work in dangerous and unhealthy conditions that scar them physically, emotionally, and mentally for the rest of their lives. The glass and glassware industry in India is concentrated in Firozabad. These factories produce a number of glass items, such as bangles, chandeliers, wine glasses, beads, crockery, bulbs, and cut glass items. The industry employs about 8,000 to 50,000 children, some as young as eight years old. The factory floor is like an inferno, due to the intense heat, poor ventilation, broken glass, dangling electric wires, and lack of protective equipment.

Often, glass splinters injure the workers, and pieces of glass cut into the bare feet of children. The children bump into each other and may scorch their bodies. Children are seen walking barefoot over glass littered floors, some with scarred eyes and burnt scalps. Child workers in the glass factories in Firozabad suffer from mental regression, asthma, bronchitis, eye problems, liver ailments, skin bums, chronic anaemia, and tuberculosis. Studies conducted at the Maulana Azad Medical College, in New Delhi, show genetic damage in the body cells of the labourers who have worked close to furnace heat for three years or more.Children, working in factories, often suffer from emotional, mental, and psychological scars.

Question 3.
Why should child labour be eliminated and how?
Answer:
There are various harmful effects of child labour. These include:
(a) Economic exploitation: Children may only receive one quarter of adult wages.
(b) Long working hours: Some children are expected to work for excessive hours, often up to 12-16 hours per day.
(c) Loss of educational opportunities: Children, who work, either give up their school education or find that their educational performance declines because of their work.
(d) Physical harm: Working children experience physical harm in a number of ways in terms of:require concerted efforts from all sections of the society to make a dent in the problem. It can be eliminated by:

• Increased risk of accidents
• Risk of physical violence from people in authority
• Theft is a risk faced by children who work as street vendors
• Risk of illness from poor hygiene and exposure to bad weather
• Harmful effects of chemicals

(e) Abuse and exploitation: It is essentially a socio-economic problem inextricably linked to poverty and illiteracy.
It will

• Legislative action plan
• Focusing general developmental programmes for benefitting child labour
• Subsidising education
• Providing basic necessities

Lost Spring Thinking about language

Question.
Carefully read the following phrases and sentences taken from the text and name the figures of speech used.
Answer:

• Saheb-e-Alam which means the lord of the universe is directly in contrast to what Saheb was in reality. The figure of speech used – irony
• “Drowned in an air of desolation”. The figure of speech used – hyperbole
• “Seemapuri, a place on the periphery of Delhi yet miles away from it, metaphorically.” The figure of speech
  used –  metaphor/irony
• “For the children it is wrapped in wonder; for the elders it is a means of survival.” The figure of speech
  used – contrast
• “As her hands move mechanically like the tongs of a machine, I wonder if she knows the sanctity of the bangles she helps make.” The figure of speech used – simile
• “She still has bangles on her wrist, but no light in her eyes.” The figure of speech used – paradox/contrast
• “Few airplanes fly over Firozabad.” The figure of speech used – metaphor/sarcasm/contrast
• “Web of poverty”, the figure of speech used – metaphor
• “Scrounging for gold”, the figure of speech used – metonymy/hyperbole
• “And survival in Seemapuri means rag-picking. Through the years, it has acquired the proportions of a fine art.” The figure of speech used – hyperbole/sarcasm
• “The steel canister seems heavier than the plastic bag he would carry so lightly over his shoulders.” The figure of speech used – metaphor

Lost Spring Extra Questions and Answers

Lost Spring Short Answer Questions

Question 1.
Who was Saheb? Where was he and where had he come from? What did he look for in the garbage dumps?
Answer:
Saheb was a child who had been forced by circumstances to become a ragpicker. His family had migrated from the green fields of Dhaka, Bangladesh in 1971. They had been forced to move out because storms had swept away all they had. They had shifted to Delhi to make a living. They lived in the slums of Seemapuri. Each day the child went looking for money in garbage heaps.

Question 2.
Saheb was a victim of circumstances. Justify.
Answer:
Saheb had once lived in the green fields of Dhaka but the storms swept away their fields and homes. Consequently, he ended up in Delhi as a ragpicker. There was an inherent desire in him to attend school and study. This could not be fulfilled because of poverty. When Anees suggested that he go to school, he was excited and a few days later asked her if her school was ready.

Question 3.
Bring out the irony in Saheb’s name.
Answer:
Saheb’s name was “Saheb-e-Alam” Ironically, it meant, lord of the universe. But that was something he would never know. Even if he did, he would have found it hard to believe. He roamed the streets barefoot scrounging the garbage heaps, but hardly managed to get one full meal.

Question 4.
Explain: “Seemapuri, a place on the periphery of Delhi, yet miles away from it, metaphorically”.
Answer:
Geographically, Seemapuri is a place on the outskirts of Delhi. It housed migrants from Bangladesh, who earned their living as ragpickers. A run-down place that lacked amenities of sewage, drainage, or running water, it was unlike the life of glitter and glamour in Delhi. People in Delhi lived a luxurious life in contrast to the poverty prevailing in Seemapuri.

Question 5.
What does garbage mean to adults and the children in the slum?
Answer:
Garbage meant different things to the adults and to the children in the slum. To the adults in Seemapuri, rag¬picking meant survival. It had assumed proportions of fine art. On the other hand, to the children garbage was like a mysterious package. They scrounged through it to discover unknown valuables.

Question 6.
Saheb is resigned to his fate and does not covet for what he considers is beyond his means. Justify.
Answer:
Saheb, a poor ragpicker, had resigned himself to his fate. He knew the areas that were out of bounds for him. He used to stand by the fenced gate of the club and was content watching others play tennis. He ventured into the club, to swing when no one was around. He had accepted his place in the society where he had to subsist on the items discarded by the privileged—tennis shoes, shirt and shorts. He gladly accepted work at a tea stall although it robbed him of his freedom.

Question 7.
How was Mukesh different from Saheb?
Answer:
Saheb was more resigned to his fate and had given up the freedom he enjoyed as a ragpicker for a salaried job at a tea stall. On the other hand, Mukesh insisted on being his own master. He was determined to be a motor mechanic. He was not prepared to compromise his dreams and give in like Saheb. He had even chalked out a path to achieve his dreams.

Question 8.
What did most slum dwellers do for a living in Firozabad?
Answer:
Firozabad is known for its bangles; it is the centre of India’s glass-blowing industry. Many people are employed in this industry. Families have spent generations working around furnaces, welding glass and making bangles. Since child labour is cheap, this place has around twenty thousand children who work in the hot furnaces. These children often lose vision before they become adults because of the environment they work in.

Question 9.
Describe the scene in Mukesh’s house as viewed by the narrator.
Answer:
Mukesh was another young child who had been forced by poverty into child labour. He lived in a dilapidated shanty with garbage strewn around. His house was a half-built shack, thatched with dead grass and a wobbly iron door. When the narrator visited the place, she noticed a firewood stove with a large vessel of spinach leaves. She also noticed a frail very young woman cooking. Later she realized she was wife of Mukesh’s elder brother. The narrator also saw Mukesh’s grandmother and his father, who were weak having spent their . lives making bangles.

Question 10.
How do you know that everyone in Mukesh’s family had resigned to their fate?
Answer:
Like most people in Firozabad, Mukesh was bom in the caste of bangle makers. Mukesh’s father was a poor bangle maker who had worked hard, first as a tailor. But despite slogging all his life, he had not been able to renovate a house or educate his two sons. Hence he had no option but to pass on the art of making bangles to his sons. Mukesh’s grandmother was an old woman who had watched her husband go blind but she did not complain. She accepted it as her fate of her “god-given lineage”. It was only Mukesh who dreamt of breaking out and being a mechanic.

Question 11.
The future of the slum dwellers in Firozabad is as bleak as their present. Why does the writer feel so?
Answer:
The future of the slum dwellers in Firozabad was as bleak as their present. The families were trapped in poverty, a curse that continued for generations. They also faced the stigma of having been bom in a lower caste and were victimized by the inhuman sahukars, the middlemen, the policemen, the keepers of law, the bureaucrats and the politicians. With all forces working against them, they were unable to defy the norms.

Question 12.
Why didn’t the people in Firozabad organize themselves into a cooperative to fight the system?
Answer:
Despite being exploited, the people in Firozabad were unable to organize themselves into a cooperative to escape from being manipulated and fight the system. Had they organized themselves into cooperatives, they ran the risk of running into trouble with the authorities. Moreover, there was no leader among them who could lead them. They were trapped in the vicious cycle of poverty, indifference, greed and injustice.

Question 13.
Mukesh dreams of a different future. What does he dream of? How does he want to accomplish his dreams?
Answer:
Mukesh was bom in a family of poor slum dwellers. He has been taught to work in the glass factory but Mukesh did not want to follow the traditional profession. He dreamt of being a motor mechanic. He wanted to learn about cars and was determined to leam the skill in order to ensure for himself a better life.

Question 14.
Explain: Few airplanes fly over Firozabad.
Answer:
Mukesh wanted to be a motor mechanic and he was prepared to walk to the garage to leam. He never dreamt of flying a plane as to the slum dwellers in Firozabad, planes were a far-fetched reality. Because of the limited exposure in the slums of Firozabad, Mukesh dreamt within his means.

Lost Spring Long Answer Questions

Question 1.
Describe the miserable plight of the people in Seemapuri.
Answer:
Seemapuri was a locality on the outskirts of Delhi which housed unlawful residents who came from Bangladesh back in 1971. This area was a place where thousands of ragpickers lived. The people lived in structures of mud, with roofs of tin and tarpaulin. There was no sewage system, no drainage or running water. The migrant poor lived there for more than thirty years without an identity, without permits but with ration . cards that got their names on voters’ lists. The children in these slums grew up to become partners in survival as ragpickers. To them, garbage heaps were like gold mine that would ensure their daily bread and a roof over their heads.

Question 2.
“But promises like mine abound in every comer of his bleak world.” What does the writer mean?
Answer:
The narrator, Anees, met Saheb every morning looking for money in the garbage. Saheb confessed to the narrator that he scrounged the rubbish heaps as he had nothing better to do. He longed to go to school but there was not one in the neighbourhood. When Anees asked him half joking that if she started a school, would he join, he consented very gladly. In fact he was so enamoured with the idea that a few days later he asked her if her school was ready. Anees was embarrassed at having made a promise that she was not serious about. She felt that most people made promises to children like him that were never fulfilled.

Question 3.
What do people have to say about people walking barefoot? What is the writer’s opinion?
Answer:
The writer narrates an experience when she asked a child why they were barefoot. One replied that his mother did not bring them down from the shelf, while the other boy felt that he would throw them off anyway. Yet another boy expressed his desire for shoes. The writer recalled having seen children walking barefoot. She had been informed that people walked barefoot not because of lack of money but due to the tradition in India. She wondered if this was an excuse to explain away the perpetual poverty. She had also heard of a boy from Udipi, who prayed every morning for a pair of shoes.

Question 4.
Explain: The steel canister seems heavier than the plastic bag.
Answer:
The writer, one morning, saw Saheb on his way to the milk booth carrying a steel canister. He had relinquished his job as a ragpicker and had taken up employment in a tea stall for eight hundred rupees.His face had lost the carefree look when he roamed the streets like a vagabond. As a ragpicker he seemed carefree, constantly looking for things—“wrapped in wonder”.

At the tea stall he was now burdened with the responsibility of a job. He was literally and metaphorically fraught with the weight of the steel canister. The canister was heavier than the plastic bag that he carried “lightly over his shoulder”. The plastic bag was light because the bag gave him the freedom to lead a life that was not governed by adults. He was no longer the master of his own free will after he was burdened by the job at the tea stall.

Question 5.
Describe the poverty of living conditions of the people in Firozabad.
Answer:
Firozabad is a city known for bangles and glass industry. However the people working in the industry led a pitiable life. They lived in dilapidated, dingy houses in cloistered lanes that were foul smelling and clogged with garbage. Their homes were hovels with crumbling walls, shaky doors, no windows, and crowded with families of humans and animals living together.

Mukesh’s house was no better. It was a half-built shack. One part of the house was thatched with dead grass and it had a wobbly iron door. Most of the houses were similar dark huts. The children worked under flickering .oil lamps with their parents, welding pieces of coloured glass into circles of bangles. Their eyes were more adjusted to the dark than to the light outside. Hence, they often ended up losing their eyesight before they became adults. The people were exploited by sahukars, the middlemen, the policemen, the keepers of law, the bureaucrats and the politicians.

Question 6.
Describe the writer’s visit to Mukesh’s place.
Answer:
Anees visited Mukesh’s house in Firozabad, a place known for its glass industry. Mukesh was bom in the caste of bangle makers. Anees noticed a weak young woman, the wife of Mukesh’s elder brother, cooking . the evening meal for the whole family. She was very young but as the daughter-in-law of the house, was in charge of three men—her husband, Mukesh and their father.

Mukesh’s father was an impoverished bangle maker. He had worked hard, first as a tailor, then a bangle maker. Despite years of relentless labour, he had neither been able to renovate a house, nor send his two sons to school.Mukesh’s grandmother had watched her husband go blind with the dust from polishing the glass of bangles. But she had accepted it as her fate. She felt that “god-given lineage” could never be broken.

Question 7.
Mukesh is not like the others. His “dream(s) loom like a mirage amidst the dust of streets that fill his town Firozabad.” Justify.
Answer:
Mukesh was born in a family riddled with poverty. He and his family were leading a difficult life that was not uncommon to the people of that socio-economic strata. But that had not deterred his desire to lead a different life. Mukesh was determined to be his own master. He had decided to become a motor mechanic and was . determined to leam to drive a car.

When Anees heard of that, she felt that Mukesh’s dreams were like a mirage—unattainable because it was difficult for him to break out of the generations of bangle-making tradition. She was convinced it would be difficult for Mukesh to achieve his unconventional dream.

Question 8.
In your opinion, can Mukesh realize his dream?
Answer:
Yes, Mukesh could certainly achieve his dream as he dared to dream in the first place. He was unlike most people who spent their lives doing what their families had done for generations. Mukesh had a tangible plan in action to realize his dreams. He was determined to go to a garage and leam more about cars. Despite the fact that the garage was at a distance, he was resolute. “I will walk,” he said. His passion and perseverance would certainly help him achieve his goals.
OR
No, Mukesh will not be able to realize his dream because there were thousands of families trapped in poverty ‘ who face the stigma of caste system. To cap it all, they live with insensitive people who exploit the situation. The inhuman sahukars, the middlemen, policemen, the keepers of law, the bureaucrats and the politicians— all work against them. These poor people are unable to come out of their misery because they lack education or leadership. Mukesh’s dream will too die a death like many others of his station.

Lost Spring Value Based Question

Question 1.
What societal evils are depicted in the “Stories of Stolen Childhood”?
Answer:
Anees Jung voices her concern over the exploitation of children in hazardous jobs such as bangle making and rag-picking. Grinding poverty and thoughtless traditions result in the loss of childhood innocence and education. They are denied a life of dignity, having been born into and conditioned by a life of poverty.

The miserable plight of Saheb-e-Alam and Mukesh brings out the grinding poverty and traditions which condemn children to a life of exploitation. It also spells out the callousness of the society towards the underprivileged.

Saheb, a ragpicker, was a young boy who had been denied education and was engaged in ragpicking as a profession. Mukesh was a bangle maker, born into the bangle-making legacy of his poor family. He, however, nurtures dreams of becoming a motor mechanic someday. Through examples of these slum children, constricted by the narrow bounds of poverty and child labour, the author voices the relevant concerns of societal evils.

NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance

These Solutions are part of NCERT Solutions for Class 12 Biology. Here we have given NCERT Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance

Question 1.
Group the following as nitrogenous bases and nucleosides:
Adenine, cytidine, thymine, guanosine, uracil, and cytosine.
Solution:
Adenine, Guanosine, Thymine, Uracil, and Cytosine are nitrogenous bases. (Adenine and Guanosine → Purine, Thymine, Uracil and Cytosine → Pyrimidine) Cytidine is a nucleoside.

Question 2.
If a double-stranded DNA has 20 percent of cytosine, calculate the percent of adenine in the DNA.
Solution:
According to Chargaff’s rule, in a double-stranded DNA, the total number of cytosine molecules will be equal to the number of guanine molecules and the number of adenine molecules will be equal to the number of thymine molecules. Therefore, if a double-stranded DNA has 20 percent of cytosine then the guanine will also be 20 per cent. The remaining 60% will consist of adenine and thymine in equal amount. Thus adenine will be 30%.

Question 3.
If the sequence of one strand of DNA is written as follows:
5′-ATGCATGCATGCATGCATGCA
TGCATGC-3′
Write down the sequence of complementary strand in 5′ -> 3′ direction.
Solution:
5′-GCATGCATGCATGCATGCAT G C ATG CAT-3′.

Question 4.
If the sequence of the coding strand in a transcription unit is written as follows: 5′-ATGCATGCATGCATGCATGCATGCATGC-3′ Write down the sequence of mRNA.
Solution:
If the sequence of coding strand is :
5′ – ATGCATGCATGCATGCATGCATGCATGC – 3′
Then template strand is :
3′ – TACGTACGTACGTACGTACGTACGTACG – 5′
The mRNA will be formed on the template strand in 5′ —> 3’ direction. Thus mRNA sequence will be:
5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′
Thymine in DNA is substituted by uracil in RNA.

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesise a semi-conservative mode of DNA replication? Explain.
Solution:
The two strands of DNA show complementary base pairing. This property of DNA led Watson and Crick to suggest a semi-conservative mechanism of DNA replication in which one strand of a parent is conserved while the other complementary strand formed is new.

Question 6.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases
Solution:
DNA dependent DNA polymerases and DNA dependent RNA polymerases.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic
material?
Solution:
They raised 2 types of bacteriophages

• On radioactive phosphorous (³²P)
• On radioactive sulphur (³⁵S).

³⁵S gets into protein and ³²P into DNA When both bacteriophages infected bacteria differently and by shaking them, the viral protein coat was separated

After raising these bacteria it was found that those infected with ³²P bacteriophage → radioactivity were found. But with ³⁵S → no radioactivity was found.

Question 8.
Differentiate between the following:

1. Repetitive DNA and Satellite DNA
2. Template strand and Coding strand
3. mRNA and tRNA

Solution:

1. Differences between repetitive DNA and satellite DNA are as follows:
   
   
2. Differences between template strand and coding strand are as follows:
   
   
3. Differences between mRNA and tRNA are as follows:
   
   
   
   

Question 9.
List two essential roles of ribosome during translation
Solution:
Two essential roles of the ribosome during translation are:

1. One of the RNA acts as a peptidyl transferase ribozyme for the formation of peptide bonds.
2. The ribosome provides sites for attachment of mRNA and charged tRNA for polypeptide synthesis.

Question 10.
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?
Solution:
The lac operon is regulated by the amount of lactose in the medium where the bacteria are grown. When the amount of lactose is exhausted in the medium, the lac operon shuts down.

Question 11.
Explain (in one or two lines) the function of the followings:

1. Promoter
2. tRNA
3. Exons

Solution:

1. Promoter: It is located at the 5′ end of the transcription unit and provides site for attachment of transcription factors (TATA Box) and RNA polymerase.
2. tRNA: It takes part in the transfer of activated amino acids from cellular pool to ribosome so that they can take part in protein formation.
3. Exons: In eukaryotes, DNA is mosaic of exons and introns. Exons are coding sequences of DNA which are both transcribed and translated.

Question 12.
Why is the Human Genome Project called a mega project?
Solution:
The human genome was a megaproject that aimed to sequence every base in the human genome. The estimated cost of the project would be a billion (1 billion = 100 crores) US dollars.

Question 13.
What is DNA fingerprinting? Mention its application.
Solution:
DNA fingerprinting is the identification of differences in specific regions of DNA sequences based on DNA polymorphism, repetitive DNA, and satellite DNA.
Application of DNA fingerprinting: Settling, paternity disputes and identity of criminal by different DNA profiles in forensic laboratories.

Question 14.
Briefly describe the following:

1. Transcription
2. Polymorphism
3. Translation
4. Bioinformatics

Solution:
1. Transcription – It is the process of copying genetic information from the anti-sense or template strand of the DNA into RNA. It is meant for taking the coded information from DNA in nucleus to the site where it is required for protein synthesis. Principle of complementarity is used even in transcription. The exception is that uracil is incorporated instead of thymine opposite adenine of template. The segment of DNA that takes part in transcription is called transcription unit. It has three components

• • a promoter,
  • the structural gene and
  • a terminator.

2. Polymorphism – It is the variation at genetic level, arisen due to mutations. Such variations are unique at particular site of
DNA. They occur approximately once in every 500 nucleotides or about 107 times per genome. These are due to deletions, insertions, and single-base substitutions. These alterations in healthy people, occur in non-coding regions of DNA and do not code for any protein but are heritable. The polymorphism in DNA sequences is the basis of genetic mapping of human genome as well as DNA fingerprinting.

3. Translation – It is the mechanism by which the triplet base sequence of mRNA guides the linking of a specific sequence of amino acids to form a polypeptide chain (protein) on ribosomes in the cell cytoplasm. All the protein that a cell needs are synthesised by the cell within itself.
The raw materials required in protein synthesis are ribosomes, amino acids, mRNA, tRNAs and amino acyl tRNA synthetase. Mechanism of protein synthesis involves following steps:

• • Activation of amino acids
  • Charging or aminoacylation of tRNA
  • Initiation
  • Elongation (Polypeptide chain formation)
  • Termination

The ribosomes move along the mRNA ‘reading’ each codon in turn. Molecules of transfer RNA (tRNA), each bearing a particular amino acid, are brought to their correct positions along the mRNA, molecule base pairing occurs between the bases of the codons and the complementary base triplets of tRNA. In this way, amino acids are assembled in the correct sequence to form the polypeptide chain.

4. Bioinformatics – Bioinformatics is the combination of biology, information technology and computer science. Basically, bioinformatics is a recently developed science which uses information technology to understand biological phenomenon. It broadly involves the computational tools and methods used to manage, analyse and manipulate volumes of biological data.

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