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NCERT Solutions for Class 12 History Chapter 8 Peasants, Zamindars and the State Agrarian Society and the Mughal Empire are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 8 Peasants, Zamindars and the State Agrarian Society and the Mughal Empire.

NCERT Solutions for Class 12 History Chapter 8 Peasants, Zamindars and the State Agrarian Society and the Mughal Empire

Question l.
What are the problems in using the Ain as a source for reconstructing agrarian history ? How do historians deal with this situation ?
Solution :
(a) Problems : The Ain was revised five times by the author so that it might become authentic. In all the quantitative sections, all numeric data were reproduced in words so as to minimise the chances of errors. But in spite this there are following problems in using Ain as a source for reconstructing agrarian history :

1. Numerous errors in totalling have been detected. These are ascribed to simple slips of arithmetic or of transcription by Abul Fazl’s assistants.
2. Another problem is about the somewhat skewed nature of the quantitative data. Data were not collected uniformly from all provinces. For example, detailed information about caste composition of the zamindars was collected from many subas. However, such information is not available for Bengal and Orissa.
3. Prices and wages that have been documented in the Ain pertains to areas in or around the imperial capital of Agra. It has, therefore, limited relevance for the rest of the country.
4. Ain was to present a vision of Akbar’s empire where social harmony was provided by a strong ruling class. There was no place for a successful revolt against the Mughal state. Thus, whatever we learn about the peasants from the Ain remains a view from the top.

(b)

1. The historians deal with the above problems by supplementing the descriptions contained in sources emanating from regions away from the Mughal capital, e.g., revenue records from Gujarat, Maharashtra and Rajasthan from the seventeenth and eighteenth centuries.
2. The extensive records of the East India Company too provide useful descriptions of agrarian relations in eastern India. These sources describe the instances of conflicts between peasants, zamindars and the state. Thus, they provide an insight into peasants’ perception of and their expectations of fairness from the state.

Question 2.
To what extent is it possible to characterise agricultural production in the sixteenth- seventeeth centuries as subsistence agriculture? Give reasons for your answer.
Solution :
(a) During Mughal, India was basically an agricultural country. In the Mughal state of India a different varieties of crops were produced. In Bengal two varieties of rices were produced. But the focus on the cultivation of basic crops does not mean that only subsistence agriculture existed in medieval India.
(b) The Mughal state encouraged peasants to cultivate varieties of crops which brought in revenue especially cotton and sugarcane.
(c) Cotton was mainly grown in vast area which was spread over central India and the deccan plateau, whereas in Bengal sugarcane was mainly produced.
(d) Many varieties of cash crops such as oilseeds including mustard and lentils.
(e) An average peasant of that time grew both commercial and subsistence crops.

Question 3.
Describe the role played by women in agricultural production.
Solution :
The role played by women in agricultural production was as mentioned below :

• Men and women worked shoulder to shoulder in the fields.
• Men tilled and ploughed, while women sowed, weeded, threshed and winnowed the harvest.
• During the medieval period, with the growth of nucleated villages and expansion in individuated peasant farming, the basis of production was the labour and resources of the entire household.
• Inspite of above, there were biases related to women’s biological functions. For example, menstruating women were not allowed to touch the plough or the potter’s wheel in western India, or enter the groves where betel-leaves (paan) were grown in Bengal.

Question 4.
Discuss, with examples, the significance of monetary transactions during the period under consideration.
Solution :
The significance of monetary transactions during the period (sixteenth and seventeenth centimes) was substantial because the Mughal Empire was among the greatest empires that had managed to consolidate its power and resources. There was stability in the Ming (China), Safavid (Iran) and Ottoman (Turkey) empires. This stability led to create vibrant networks of overland trade from China to the Mediterranean Sea. The discovery of New World resulted in massive expansion of trade of India with Europe.

With the expansion of trade the importance of monetary transactions increased. The expansion of trade brought huge amount of bullion and silver into India where there was no natural resource of silver. This led to remarkable stability in the availability of metal currency, particularly in silver rupya in India. This facilitated an unprecedented expansion of minting of coins and the circulation of money in the economy. At the same time, there was no exchange of goods or barter system during this period. The payments were made in gold or silver coins. According to Giovanni Careri, all the gold and silver which circulates throughout the world, ultimately, comes into India due to its overseas trade. Thus a large amounts of cash transactions took place in the sixteenth and seventeenth centuries.

Question 5.
Examine the evidence that suggests that land revenue was important for the Mughal fiscal system.
Solution :
(i) Agriculture was the mainstay of the economy. Land Revenue collected was used to pay salaries and to meet different kinds of administrative expenses. So it was considered important to establish an administrative apparatus to ensure control over agricultural production.

(ii) Thus, before fixing land revenue, Mughal state first acquired specific information about the extent of agricultural lands and their produce.

(iii) Land revenue collection arrangements was consisted of two stages of assessment. These were Jama and hasil. Cultivators were given the choice to pay land revenue either in cash or kind. The state preferred to collect land revenue as cash. Attempts were made to maximize profits from the land revenue collection.

(iv) Both cultivated and cultivable lands were measured in each province to fix land revenue. According to a decree of Akbar, it was the responsibility of malguzar to make cultivator pay land revenue in kind and it was also kept open. Thus, it is clear from the evidence that the monetary transactions were very important. To continue this policy efforts by subsequent emperors like Aurangzeb continued to measure land for collection of land revenue.

Question 6.
To what extent do you think caste was a factor in influencing social and economic relations in agrarian society ?
Solution :
Agricultural production involved the intensive participation and initiative of the peasantry. There were different social groups, on the basis of caste and other factors, that were involved in agricultural expansion. This affected their social and economic relations in the agrarian society in the following ways :

1. Deep inequalities on the basis of caste and other caste like distinctions made the cultivators a highly heterogeneous group.
2. Among those who tilled the land, there was a sizeable number who worked as menials or agricultural labourers (majur).
3. There was abundance of cultivable land but inspite of this certain caste groups were assigned menial tasks and were relegated to poverty. Such groups comprised a large section of the village population. They had the least resources and were constrained by their position in the caste hierarchy like the Dalits of modem India.
4. In Muslim communites too menials like halalkhoran (scavengers) were housed outside the boundaries of the village. The mallahzadas (sons of boatmen) in Bihar were comparable to slaves.

Thus, there was a direct correlation between caste, poverty and social status at the lower level of society. It was, however, not so at the intermediate levels. For example, in Marwar, Rajputs were considered peasants like Jats who were accorded a lower status in the caste hierarchy. The Gauravas, who cultivated land around Vrindavan sought Rajput status in the seventeenth century. Ahirs, Gujars and Malis rose in the hierarchy due to the profitability of cattle rearing and horticulture.

Question 7.
How were the lives of forest dwellers transformed in the sixteenth and seventeenth centuries ?
Solution :
The forest dwellers were people who earned their livelihood by gathering of forest produce, hunting and shifting agriculture. These activities changed according to seasons. For example, the Bhils collected forest produce in the spring. They did fishing in the summer, cultivated during the monsoon months. They did hunting in the winter and autumn. Such a division of activities presumed and perpetuated mobility which was a distinctive feature of the forest dwellers. However, their lives were transformed in the sixteenth and seventeenth centuries in the following ways :

1. In the sixteenth and seventeenth centuries, the state required elephants for the army. So, the peshkash – a form of tribute collected by the Mughal state – in the form of elephants too.
2. In the Mughal political ideology, the hunt symbolised the overwhelming concern of the state to ensure justice to all the subjects, rich and poor. Thus, during hunting expeditions, the emperor personally attended to the grievances of the people. Such hunting expeditions affected the lives of the forest dwellers.
3. Forests were cleaned for agricultural settlements. The spread of commercial agriculture impinged on the lives of forest dwellers, who lived on forest products like honey, beeswax and gum lac. Gum lac was exported from India. Elephants were captured and sold. There was [ exchange of commodities through barter too. All this changed due to commercial agriculture and agricultural settelments.
4. Social factors too transformed the lives of forest dwellers. For example, chieftains of many tribes had become zamindars, some even became kings. They built up their armies and demanded that their fraternity to provide military service. In Assam, the Ahom kings had their paiks. They were people who obliged to render military service in exchange for land. Not only this the capture of wild elephants was declared a royal monopoly by the Ahom kings.
5. With the establishment of tribal kingdoms in the north-east, war became a common feature.
6. The cultural influences as that of sufi saints encouraged the forest-dwellers particularly agricultural communities to accept Islam.

Question 8.
Examine the role played by zamindars in Mughal India.
Solution :
Zamindars played a significant role in Mughal India as mentioned below :

1. The zamindars had landed properties and enjoyed certain social and economic privileges due to their superior status in the society and due performance of certain services (khidmat) for the state.
2. They had milkiyat lands which were cultivated for the private use of zamindars with the help of hired or servile labour. The zamindars could sell, bequeath or mortgage these lands at will.
3. The zamindars collected revenue on behalf of the state. They were compensated for this financially.
4. They had military resources such as armed contingent and fortresses (qilachas).
5. In the social hierarchy, the zamindars constituted its very narrow apex.
6. Contemporary documents give an impression that conquest may have been source of the origin of some zamindaris. A powerful military chieftain often dispossessed weaker people and expanded his zamindari.
7. Zamindars spearheaded the colonisation of agricultural land. They helped in settling cultivators by providing them with the means of cultivation, including cash loans. The buying and selling of zamindaris accelerated the process of monetisation in the countryside. Zamindars sold the produce from their milkiyat lands and established markets (haats) where peasants came to sell their produce too.
8. Although the zamindars were considered as an exploitative class but their relationship with the peasants were based on reciprocity, paternalism and patronage because the bhakti saints did not portray them as exploiters or oppressors of peasantry. Not only this, in a large ‘ number of agrarian uprisings in north India in the seventeenth century, zamindars often received the support of the peasantry in their struggle against the state.

Question 9.
Discuss the ways in which panchayats and village headmen regulated rural , society.
Solution :
Panchayats and village headmen regulated the rural society in the following ways :

1. The village panchayat was an assembly of elders. Its decisions were binding on the members.
2. The panchayat was headed by a headman known as muqaddam or mandal. The headman supervised the preparation of accounts, assisted by the accountant or patwari of the panchayat.
3. The panchayat ensured that the caste boundaries among the various communities inhabitating the village were upheld. In eastern India all marriages were held in the presence of the mandal or headman. Thus, the headman was to oversee the conduct of the members of the village community “chiefly to prevent any offence against their caste”.
4. Panchayats had the authority to levy fines and inflict serious punishments such as expulsion from the community. Such punishment was given as a deterrant to violation of caste norms.
5. There were Jati panchayats of each caste. In Rajasthan, Jati panchayats arbitrated civil disputes between members of different castes. They decided the disputes related to claims on lands and marriages. Generally, the state respected the decisions of Jati panchayats.
6. Sometimes petitions were presented to the panchayat complaining about extortionate taxation or the demand for unpaid labour (begar) imposed by the “superior” castes or officials of the state. These were submitted by the lower classes because they regarded the village panchayat as the court of appeal that would ensure that the state carried out its moral obligations and guaranteed justice. In such cases, the panchayats often suggested compromise and reconciliation. In case of failure of compromise, the peasants generally deserted the village because there was abundance of uncultivated land available in the villages.

We hope the NCERT Solutions for Class 12 History Chapter 8 Peasants, Zamindars and the State Agrarian Society and the Mughal Empire help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 8 Peasants, Zamindars and the State Agrarian Society and the Mughal Empire, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Physics Chapter 12 Atoms

NCERT Solutions for Class 12 Physics Chapter 12 Atoms are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 12 Atoms.

Question 1.
Choose the correct alternative from clues given at end of each statement:

1. The size of the atom in Thomson’s model is the atomic size in Rutherford’s model.(much greater than/no different from/much less than.)
2. In the ground state of…………… electrons are in stable equilibrium, while in ……………..  electrons always experience a net force. (Thomson’s model/Rutherford’s model.)
3. A classical atom based on…………. is doomed to collapse. (Thomson’s model/Rutherford’s model.)
4. An atom has a nearly continuous mass distribution in a………….. but has a highly non­uniform mass distribution in (Thomson’s model/Rutherford’s model.)
5. The positively charged part of the atom possesses most of the mass in…………. (Rutherford’s model/both the models.)

Answer:

1. no different from
2. Thomson’s model; Rutherford’s model.
3. Rutherford’s model.
4. Thomson’s model, Rutherford’s model.
5. both the models.

Question 2.
Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at a temperature 14 K). What results do you expect?
Answer:
The nucleus of a hydrogen atom is a proton. The mass of it is 1.67 x 10^-27 kg, whereas the mass of an incident α-particle is 6.64 x 10^-27 kg. Because the scattering particle is more massive than the target nuclei (proton). the α-particle won’t bounce back in even in a head-on collision. It is similar to a football colliding with a tennis ball at rest. Thus, there would be no large-angle scattering.

Question 3.
What is the shortest wavelength present in the Paschen series of spectral lines?
Answer:
The wavelength of the spectral lines forming the Paschen series is given by

Question 4.
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom transits from the upper level to the lower level?
Answer:

Question 5.
The ground state energy of the hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state?
Answer:
K.E. = -E (Total energy)
= -(-13.6) = 13.6 eV
P.E. = 2 X E = 2 X (-13.6)
= -27.2 eV

Question 6.
A hydrogen atom initially in the ground level absorbs a photon which excites it to the n = 4 level. Determine the wavelength and frequency of the photon.
Answer:
We know, the energy of an electron in the nth orbit of a hydrogen atom is given by

Question 7.
(a) Using Bohr’s model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels.
(b) Calculate the orbital period in each of these levels.
Answer:
(a) Speed of an electron in nth orbit of a hydrogen atom is given by

Question 8.
The radius of the innermost electron orbit of a hydrogen atom is 5.3 x 10^-11 m. What are the radii of the n = 2 and n = 3 orbits?
Answer:
We know, the radius of the nth orbit of a hydrogen atom is given by r_n = r₀n², where r₀ = 5.3 x 10^-u m is the radius of the innermost orbit of the hydrogen atom.
When n = 2, r₂ = 5.3 x 10^-u x 4
= 2.12 x 10-¹⁰m
When n = 3,
= 5.3 x 10-¹¹ x 9
= 4.77 x 10-¹⁰m.

Question 9.
A 12.75 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Answer:

Question 10.
In accordance with Bohr’s model, And the quantum number that characterizes the earth’s revolution around the sun in an orbit of radius 1.5 x 10¹¹ m with an orbital speed
3 x 10¹⁴ m s^-1. (Mass of earth = 6.0 x 10²⁴ kg.)
Answer:
According to Bohr’s postulate of quantization of angular momentum

Question 11.
Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.
(a) Is the average angle of deflection of a-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
(b) Is the probability of backward scattering (i.e., scattering of a-particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
(c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of a-particles scattered at moderate angles is proportional to What clue does this linear independence on t provide?
(d) In which model is it completely wrong to ignore multiple scattering for the calculation of the average angle of scattering of a-particles by a thin foil?
Answer:
(a) About the same
(b) Much less
(c) It suggests that the scattering is predominantly due to a single collision, because the chance of a single collision increases linearly with the number of target atoms, and hence linearly with thickness.
(d) In Thomson’s model, a single collision causes very little deflection. The observed average scattering angle can be explained only by considering multiple scattering. So it is wrong to ignore multiple scattering in the Thomson model. In Rutherford’s model, most of the scattering comes through a single collision and multiple scattering effects can be ignored as a first approximation.

Question 12.
The gravitational attraction between electron and proton in a hydrogen atom is weaker than the Coulomb attraction by a factor of about 10^-40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.
Answer:
If electron and proton were bound by gravitational attraction, then

It is astonishing this value of r is much greater than the size of the universe.

Question 13.
Obtain an expression for the frequency of radiation emitted when a hydrogen atom
de-excites from the level it to level (n – 1). For large it, shows that this frequency equals the classical frequency of revolution of the electron in the orbit.
Answer:
The energy of an electron in the nth orbit of a hydrogen atom is given by

i.e. frequencies are equal. This is called Bohr’s correspondence principle.

Question 14.
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, a thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learned in the text. To stimulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10-¹⁰m).
(a) Construct a quantity with the dimensions of length from the fundamental constants e, m_e, and c. Determine its numerical value.
(b) You will find that the length obtained in («) many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in a non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr discard c and look for something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognizing that h, m_e, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, m_g, and e and confirm that its numerical value has indeed the correct order of magnitude.
Answer:
(a) Here, the dimensional formula of e is A¹T¹, the dimensional formula of m_e is M¹, dimensional formula

Question 15.
The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?
Answer:
(a) K.E. = -E = – (-3.4 eV) = 3.4 eV
E. = 2E = 2 x (-3.4 eV)
= -6.8 eV
(b) Kinetic energy does not depend upon the choice of zero potential energy. Therefore, its value remains unchanged. However, the potential energy gets changed with the change in the zero levels of potential energy.

Question 16.
If Bohr’s quantisation postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantization of orbits of planets around the sun?
Answer:
Applying Bohr’s quantization postulate,

i.e., n is very large. Since n is very large, the difference between the two successive energy or angular momentum levels is very small and the levels may be considered continuous.

Question 17.
Obtain the first Bohr’s radius and the ground state energy of a ‘muonic hydrogen atom’ (i.e. an atom in which a negatively charged muon (μ^-1) of mass about 207 m_e orbits around a proton).
Answer:
Here the mass of the particle revolving around the proton is

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

NCERT Solutions for Class 12 History Chapter 15 Framing the Constitution The Beginning of a New Era are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 15 Framing the Constitution The Beginning of a New Era.

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

Question 1.
What were the ideals expressed in the Objectives Resolution ?
Solution :
The “Objectives Resolution” outlined the ideals of the Constitution of Independent India as mentioned below :

• Independent sovereign republic : India will be an independent state without any foreign or internal control. Its head will be elected by the people.
• Justice, equality and freedom : It will guarantee its citizens justice, equality and freedom.
• Safeguards for minorities : It assured that “adequate safeguards shall be provided for minorities, backward and tribal areas, and Depressed and Other Backward Classes”.

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

Question 3.
What were the arguments in favour of greater power to the provinces ?
Solution :
The following arguments were given in favour of greater power to the provinces :

1. K. Santhanam stated that by giving more powers to the Centre, we could not make it strong because in such a situation, the Centre would be overburdened with responsibilities. It would not be able to function effectively. On the other hand, if the Centre was relieved of some functions or powers it could function effectively and become stronger.
2. He argued that weak status would cripple them. Their financial position would be weaker and without finances they would not be able to take up any project for development. They would have to depend on the central aid for education, sanitation and other welfare work for the people. He added a strong centre might lead the states to “revolt against the Centre” in the future.
3. A member from Orissa warned that “the Centre is likely to break” since powers had been excessively centralised under the Constitution.

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

Question 5.
What historical forces shaped the vision of the Constitution?
Solution :
Historical forces that shaped the vision of the Constitution were as given below :

1. Historic efforts in the past : Nehru in his famous speech of 13 December 1946 referred to the American and French Revolutions. He thus linked the making of the Indian Constitution with the revolutionary moments in the past. But at the same time he emphasised not to copy the west but to learn from their experiments, achievements and failures.
2. The will of the people : Nehru stated that the source of the Constituent Assembly was its strength i.e., the will of the people. So, members always kept in mind that passions that lay in the hearts of the masses of the Indian people and tried to fulfil them. Thus, Constituent Assembly was expected to represent the people.
3. India, a large country with diversities : India is a large country with different religions, castes, communities, languages and groups. It was necessary to keep all united. So, the Constitution was prepared keeping in mind these diversities.
4. Protection of minorities : There were minorities and depressed classes. It was necessary to protect their interests. Gandhiji had already started movement for upliftment of the Harijan. Thus, there were debates in the Assembly and provisions were incorporated for their protection and upliftment.
5. Period of violence : There were riots and violence and communal frenzy. Under these circumstances it was necessary to have a strong Centre. There were arguments in favour of and against it. But ultimately more powers were given to the Centre.
6. Problem of princely states : There were more than five hundred princely states. To accommodate them, it was absolutely necessary to have a federal system of government.

Question 6.
Discuss the different arguments made in favour of protection of the oppressed groups.
Solution :
The different arguments made in favour of protection of the oppressed groups were as mentioned below :

1. It was argued that the problem of the “Untouchables” could not be resolved through protection and safeguards alone. Their disabilities were caused by the social norms and the moral values of caste society that had used their services and labour but kept them at a social distance.
2. J. Nagappa from Madras pointed out that the suffering of the Depressed Classes was i due to their systematic marginalisation and not due to their numerical insignificance. They had
   no access to education, or share in the administration.
3. K.J. Khanderkar from the Central Provinces argued that the Depressed Classes had been suppressed for thousands of years to such an extent that their bodies and minds were not able to march forward.

Question 7.
What connection did some of the members of the Constituent Assembly make between the political situation of the time and the need for a strong Centre ?
Solution :
The Constitution of India was framed between December 1946 and December 1949. It was a trouble-some time. There were riots and violence. There was the rising of the ratings of the Royal India Navy in Bombay and other cities in the spring of 1946. The violence culminated in the massacres that accompanied the transfer of populations when the Partition of India was announced. Some members of the Constituent Assembly made connection between the above political situation of the time and the need for a strong Centre as mentioned below :

1. Referring to riots and violence in the country, many members had repeatedly stated that the powers of the Centre had to be greatly strengthened to enable it to stop the communal frenzy.
2. Gopalaswami Ayyangar declared that “the Centre should be made as strong as possible”.
3. Balakrishna Sharma from the United Provinces reasoned at length that only a strong Centre could plan for the well-being of the country, mobilise the available economic resources and establish a proper administration.

Question 8.
How did the Constituent Assembly seek to resolve the language controversy ?
Solution :
There were two main views about the language of the nation as mentioned below :

1. A plea for Hindi : R.V. Dhulekar, from the United Provinces made a strong plea that Hindi be used as the language of making the constitution. He wanted Hindi to be declared a National Language.
2. The fear of domination : Shrimati Durgabai from Madras explained her worries. She informed the house that there was strong opposition against Hindi in the South. She stated
   that the erosion of inclusive and composite character of Hindustani was bound to create anxieties and fears amongst different language groups.

In view of the above differences, some members appealed for a spirit of accommodation and asked the members not to push the cause of Hindi aggressively. Thus, the language controversy was solved in the following way :

• Hindi in the Devanagari script would be the official language.
• Transition to Hindi would be gradual.
• For the first fifteen years, English would continue to be used for all official purposes.
• Each province was allowed to choose one of the regional languages for official work within the province.

Thus, referring to Hindi as the official rather that of the national language it was hoped that it would be acceptable to all.

We hope the NCERT Solutions for Class 12 History Chapter 15 Framing the Constitution The Beginning of a New Era help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 15 Framing the Constitution The Beginning of a New Era, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 History Chapter 12 Colonial Cities Urbanisation, Planning and Architecture are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 12 Colonial Cities Urbanisation, Planning and Architecture.

NCERT Solutions for Class 12 History Chapter 12 Colonial Cities Urbanisation, Planning and Architecture

Question 1.
To what extent are census data useful in reconstructing patterns of urbanisation in the colonial context ?
Solution :
The census data are useful in reconstructing patterns of urbanisation in the colonial context in the following ways :

1. The censuses reveal that after 1800 the urbanisation in India was slow. The proportion of the urban population to the total population in India was almost stagnant. Between 1900 and 1940 the urban population increased from about 10 per cent of the total population to about 13 per cent.
2. Smaller towns did not grow economically but Bombay, Madras and Calcutta grew rapidly.
3. The introduction of railways changed the centre of economic activity from traditional towns to towns connected with the railways.

The above facts provide us the patterns of urbanisation but the historians have found the census data misleading because the census operation was a means by which social data were converted into convenient statistics about the population. There were many shortcomings in it. For example, classification of different sections of population was arbitrary. There were overlapping identities of people. People w7ere too suspicious of census operations and did not cooperate with the officials. Thus, census data is invaluable but should be studied carefully in restructuring patterns of urbanisation in colonial India.

Question 2.
What do the terms “White” and “Black” Town signify?
Solution :
The British had white skin as they were often described ‘white’ and they considered themselves as superior to others. On the other hand, the blacks had brown or black skin. So they were known as the ‘black’. The White signified their superiority over the black due to the colour of their skin. The British symbolised the Black areas full of chaos and anarchy, filth and disease and on the other hand, the white areas stood for cleanliness and hygiene. In Black areas, epidemics like cholera and plague often broke out. So the British took stringent measures to ensure sanitation and public health to prevent diseases of the Black areas. They ensured underground piped water supply and introduced sewerage and drainage system in White areas. Thus, we can say, the White Towns were those parts of the colonial towns where the White people lived. These towns had wide roads, barracks, churches, paradeground, big bungalows and gardens, symbolised settled city life, whereas the Indian lived in Black Towns, were said to be unorganised and a source of filth and disease.

Question 3.
How did prominent Indian merchants establish themselves in the colonial city?
Solution :
The prominent Indian merchants established themselves in the colonial city in the following ways :

1. With the expansion of railways, the countryside from where raw materials and labour were obtained was linked to the cities like Bombay, Madras and Calcutta. This gave an opportunity to the Indian merchants to set up modern factories. Thus, after the 1850s, cotton mills were set up by Indian merchants and entrepreneurs in Bombay.
2. Kanpur specialised in leather, woollen, cotton textiles and Jamshedpur where steel factory was established by J. Tata, specialised in steel.
3. The American Civil War started in 1861 gave another opportunity to the Indian merchants for earning huge profits. Bombay was the most important city of India. By the late nineteenth century, Indian merchants in Bombay established cotton mills.

Question 4.
Examine how concerns of defence and health gave shape to Calcutta.
Solution :
Sirajudaula, the Nawab of Bengal in 1756, sacked the small fort from Britisher. In this fort the British traders had built to house their goods. Consequently, when Sirajudaula was defeated in the Battle of Plassey, the British built a new fort, Fort William which could not be easily attacked. Around this a vast open space was left. This open space ‘ was called the Maidan or garermath. This was done for security reasons, because there would be no obstructions to a straight time of fire from the Fort against an advancing enemy army. Soon the British began to move out of the Fort. They built residences along the periphery of the Maidan. This indicates that how the English Settlement in Calcutta began to take shape. The vast open space around the Fort William became the significant town planning measure in Calcutta (Now Kolkata).

Lord Wellesley was more concerned about the conditions that existed in the cities. Cities were overcrowded, and had no sanitation facilities. He issued an administrative order in 1803 on the need for town planning and set up various committees for this purpose open places in the city would make the city healthier. As a result of this, many bazaars, ghats, burial ground and tanneries were cleared or removed. After Wellesley’s departure, the Lottery Committee carried on with the work of town planning in Calcutta.

Question 5.
What are the different colonial architectural styles which can be seen in Bombay city ?
Solution :
The different colonial architectural styles which can be seen in Bombay city are as mentioned below :

1. European style : In mid nineteenth century, the buildings were constructed in the European style to create a familiar landscape in an alien country and to symbolise their superiority.
2. Indian style : As the Indians used European architecture, the British adopted Indian style that can be seen in the construction of bungalows in Bombay and all over India.
3. Neo-classical or new classical style : Its characteristics are construction of geometrical structures fronted with lofty pillars. Town Hall is an example of this style.
4. Neo-Gothic style : Its characteristics are high-pitched roofs, pointed arches and detailed decoration: Secretariat, University of Bombay and High Court were made in this style. The most spectacular example of the neo-Gothic style is the Victoria Terminus.
5. Indo-saracenic style : It was a hybrid architectural style that combined the Indian with the European. It was inspired by the medieval buildings in India with their domes, chhatris, jalis and arches. The Gateway of India and the Taj Mahal Hotel that was built by Jamsetji Tata belong to this style.

Question 6.
How were urban centres transformed during the eighteenth century?
Solution :
(i) The disintegration of the mughal empire after the death of Aurangzab paved the way of emergence of paverful regional powers. The capital cities of these regional kingdom likes Lucknow, Poona, Nagpur and Barda now become important. Taking the advantage of this opportunity many nobles and officials created new urban settlements such as the qasbah and ganj.

(ii) The European companies too had set up their bases in different parts of India during the sixteenth and seventeenth centuries. For example the Portugues (in Panaji in 1570) and the British in Madras in 1639. With the expansion in commercial activity, towns began to emerge as trading centres.

(iii) From the mid-eighteenth century trading. Centres like Surat and Dhakha which had grown in the seventeenth century now began to decline as trade shifted to other places. When the British acquired Bengal and the east Indian’s Company’s trade hereafter expanded the colonial port cities likes Madras and Calcutta. These new part cities began to emerge as the new economic capitals.

(iv) In these newly developed cities many new buildings were built and new occupations developed. People flocked to these cities in large numbers. By the nineteenth century, these newly developed cities become the biggest cities in India.

Question 7.
What were the new kinds of public places that emerged in the colonial city ? What functions did they serve ?
Solution :
New kinds of public places that emerged in the colonial city were as given below :

1. Fort St. George (Madras), Fort William (Calcutta), the Fort George (Bombay). These were the fortified areas of British settlement.
2. The Writers’ Building in Calcutta : It was the building where the servants of the East India Company in India stayed on arrival in the country. Later this building became a government office.
3. Clubs, racecourses and theatres were built for the ruling elites exclusively on racial grounds.
4. Cantonment places were developed. Here Indian troops under European command were stationed. These were considered safe enclaves for Europeans.
5. Simla, founded during the course of Gurkha war, and Darjeeling were hill stations that became strategic places for billeting troops, guarding frontiers and launching campaigns against enemy rulers.
6. Public places such as public parks, theatres, and cinema halls came into existence for providing new forms of entertainment and social interaction.
7. Government House Calcutta : It was built by Lord Wellesley for himself in Calcutta.

Question 8.
What were the concerns that influenced town planning in the nineteenth century?
Solution :

1. Two concerns which influenced the town planning in the nineteenth century were defence and health.
2. In many towns British built forts to protect their factories. Around the fort, a vast open space was left open. This vast space was known as the Maidan.
3. It was done so that there would be no obstructions to a straight line of fire from the Fort against an advancing enemy.
4. Attempts were also made to improve the sanitation and cleanliness by creating open spaces in the city.
5. For this purpose, in Calcutta many bazaars, ghats and burial grounds were cleared.

Question 9.
To what extent were social relations transformed in the new cities ?
Solution :
The social relations were transformed in the new cities in the following ways :

1. New transport facilities as horse-drawn carriages, trams and buses meant that people could live at a distance from the city centre. This led to separation of the place of work from the place of residence. People travelled from home to office or factory.
2. The sense of coherence and familiarity of the old towns disappeared. The public places such as parks, theatres and cinema halls provided new forms of entertainment and social interaction.
3. New social groups came into existence. The “middle classes” increased due to the coming of all types of people i.e., clerks, doctors, teachers, lawyers and others. With the spread of education, people could put forward their views in newspapers and journals. People started questioning old customs and practices.
4. Women entered new professions as factory workers, teachers, theatres and film actresses. However, their entry into public spaces remained the objects of social censure.
5. There was a dramatic contrast between extreme wealth and poverty. The new cities were bewildering places where life seemed always in a flux. Paupers from the villages came to cities in search of employment. The male migrants left their families in the villages because jobs were uncertain and food was expensive. But yet the villagers participated in religious festivals, tamashas and swangs which mocked the pretensions of their masters, Indian and European.

We hope the NCERT Solutions for Class 12 History Chapter 12 Colonial Cities Urbanisation, Planning and Architecture help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 12 Colonial Cities Urbanisation, Planning and Architecture, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

Question 1.
Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of
(a) reflected, and
(b) refracted light ? Refractive index of water is 1.33. (C.B.S.E. Sample Paper 1991)
Answer:
(a) For reflected light wavelength is unchanged i. e.
X = 589 x 10^-9 m = 589 nm
Also, speed of light in air c = 3 x 10⁸ m s ^-1

Question 2.
What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.
Answer:
(a) Spherical shape
(b) Plane wavefront
(c) Plane wavefront

Question 3.
(a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in a vacuum is 3.0 x 10⁸ m s^-1)
(b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism ?
Answer:
(a) Here ,n=105,c=3.0 x 10⁸ ms^-1

speed of light when passing through glass depends on colour of light. λ_r > λ_υ , therefore the speed of violet light is less than the red light.

Question 4.
In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 Determine the wavelength of light used in the experiment.
Answer:

Question 5.
In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ is K units. What is the intensity of light at a point where path difference is λ/2 ?
Answer:

Question 6.
A beam of light consisting of two wavelengths 650 nm and 520 nm is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide ?
The distance between two slits is 2 mm and distance between the plane of the slits and the screen is 1.2 m.
Answer:
(a) λ = 650 nm = 650 x 10^-9 m,
d = 2 mm = 2 x 10^-3 m,
D = 1.2 m
Distance of mth bright fringe from the central maximum is given by

Question 7.
In a double-slit experiment, the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experiment apparatus is immersed in water ? Take
refractive index of water to be \(\frac { 4 }{ 3 } \)
Answer:


Question 8.
What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)
Answer:
µ = 1.5
According to Brewster’s law,
µ = tan p
tan p = 1.5
⇒ p = 56.31º

Question 9.
Light of wavelength 5000 A falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light ? For what angle of incidence is the reflected ray normal to the incident ray ?
Answer:
Here X = 5000 A = 5000 X 10^-10 m,
c =3 x 10⁸ m s^-1
Wave length of reflected light
= Wavelength of incident light = 5000 Å


Question 10.
Estimate the distance for which ray optics is good approximation  for an aperture  of
4 mm and wavelength 400 nm.
Answer:
Here X = 400 nm = 400 x 10^-9 m, Aperture, a = 4 mm = 4 x 10^-3 m
.’. Distance for which  ray optics is a good approximation is Fresnel’s distance


Question 11.
The 6563 Å Hα line emitted by hydrogen in a star is found to be red-shifted by 15Å. Estimate the speed with which the star is receding from the Earth.
Answer:

Negative sign shows that the star is receding away from the earth.

Question 12.
Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water ? If not, which alternative picture of light is consistent with experiment ?
Answer:
According to Corpuscular theory, when light in the form of particles enters into denser medium from a rarer medium, a force of attraction comes into play on the particles normal to the surface. Thus, the component of velocity normal to the surface of water increases whereas the component of velocity parallel to surface does not change. Therefore,

velocity in water is greater than velocity of light in air. However, in actual case, c > υ. Huygen’s wave theory of light is consistent with the experiment.

Question 13.
You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.
Answer:
Let there be a point object A at a distance y from a plane mirror. Treating this point to be a point source of light, we can assume spherical wavefronts progressing from A of radius y. Let there be no mirror then after time t, the wavefront will reach A’ as wavefront I. If a mirror is placed as shown in the figure then image will be formed at A’ represented by II.

It is seen that OA’ = OA i. e. virtual image is formed at a distance equal to the distance of object from the mirror.

Question 14.
Let us list some of the factors, which could possibly influence the speed of wave propagation:
(1) nature of the source,
(2) direction of propagation.
(3) the motion of the source and/or observer.
(4) wavelength
(5) intensity of the wave.
On which of these factors, if any, does
(a) the speed of light in vacuum,
(b) the speed of light in a medium (say, glass or water), depend ?
Answer:
(a) Speed of light in a vacuum is an absolute constant (universal constant). It is independent of any factor. It is independent of the relative motion between source and observer even.
(b)

1. Speed of light in a medium depends upon wavelength.
2. It is independent of the nature of the source and motion of the source relative to the medium.
3. It depends upon the properties of the medium of propagation and motion of the observer relative to the medium,
4. It is independent of the direction of propagation for isotropic medium,
5. It is independent of the intensity of the wave.

Question 15.
For sound waves, the Doppler formula for frequency shift differs slightly between the two situations :
(1) a source at rest; observer moving, and
(2) source moving ; observer at rest. The exact Doppler formulae for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulae to be strictly identical for the two situations in case of light travelling in a medium ?
Answer:
Sound requires material medium for propagation. Though situations
(1) and (2) may correspond to the same relative motion, yet they are not identical physically as the motion of observer relative to medium may be different in both situations. Hence, Doppler effect for sound cannot be same in both situations. Light when passing through material medium is also governed by different Doppler formulae for
(1) a source at rest; observer moving and
(2) source moving; observer at rest.
But when light passes through vacuum the formulae become exactly same for the two different situations because speed of light and frequency/wavelength of light remain unchanged in vacuum.

Question 16.
In a double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between two slits?
Answer:

Question 17.
Answer the following questions :
(a) In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction
band ?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment? (C.B.S.E. 2013, 2013 )
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the center of the shadow of the obstacle.
Explain why? (C. B. S. E. 2013 )
(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily ?   (C.B.S.E. 1990 )
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?   (C.B.S.E. 1990)
Answer:
(a) The width of central maxima = 2λD/d.
When the width (d) of slit is doubled, then the width of central diffraction maxima reduces to half and the intensity of the central band increases four times as amplitude of light wave is doubled.
(b) Intensity of fringes produced in the double-slit experiment is changed due to diffraction pattern superposing due to each slit.
(c) Light waves diffract at the edges of the circular obstacle. These diffracted waves interfere constructively and give rise to the bright spot at the center of the geometrical shadow.
(d) Diffraction is observed when the wavelength of the wave is of the order of the size of the obstacle. The wavelength of sound wave (≈ 0.33 m) is larger than the light wave (≈10^-7 m) and is also comparable to wall, so diffraction of sound waves takes place and hence the students can converse easily. On the other hand, the wavelength of light is very small as compared to the obstacle e. 1 m high wall so the diffraction of light waves does not take place.
(e) In optical instruments, size of apertures are much larger than the wavelength of light. So diffraction of light is negligible. Hence, the assumption that light can travel in straight line is used in optical instruments.

Question 18.
Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects ?
Answer:
If A and B are two hills and C is the hill peak midway

Question 19.
A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2-5 mm from the center of the screen. Find the width of the slit.( C.B.S.E. 2013)
Answer:

Question 20.
Answer the following questions :
(a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TY screen. Suggest a possible explanation.
(b) As you have learned in the text, the principle of linear superposition of wave displacement is basic to understanding distributions in diffraction and interference patterns. What is the justification of this principle?
Answer:
(a) When a low flying aircraft passes overhead, the metallic body of the aircraft reflects the TV signal. A slight shaking of the picture on the TV screen takes place due to the interference of the reflected signal from the aircraft and the direct signal received by the antenna.
(b) The linear combination of wave equations is also a wave equation. This is the very basis of the superposition principle.

Question 21.
In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angle of nλ/α. Justify this by suitably dividing the slit to bring out the cancellation.
Answer:
Let us suppose that we have n slits each of width

Therefore, each of the n slits of width d’ each sends zero intensity in the direction 9. As a result, the net resultant of intensity due to n such slits is zero.

We hope the We hope the NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics, help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

Question 1.
A small candle 2.5 cm in size is placed 27 cm in front of a concave mirror of a radius of curvature of 36 cm. At what distance from the mirror should a screen be placed in order to receive a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Answer:
Here, h = 2.5 cm, u = – 27 cm, R = – 36 cm.

Nature of image : Real, inverted and magnified. When the position of the object (i.e. candle) is moved closer to the concave mirror, the distance of the image moves away from the screen till the distance of the candle from the concave mirror is less than 18 cm. Hence, the screen has to be moved away from the concave mirror. When the distance of the candle is less than 18 cm from the concave mirror, a virtual and magnified image of the candle is formed behind the mirror. This image is not obtained on the screen.

Question 2.
A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and magnification. Describe what happens as the needle is moved farther from the mirror.
Answer:
Here, h₁ = 4.5 cm, μ = – 12 cm, f= 15 cm

Thus, the image is virtual, erect and diminished. As we move the needle away from the mirror, the image goes on decreasing in size and moves towards the principal focus on the other side.

Question 3.
A tank is filled with water to a height of 12.5 cm. The apparent depth of the needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water ? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again ? (C.B.S.E. 2009 )
Answer:

Question 4.
The following figures (a) and (b) show refraction of an incident ray in air at 60° with the normal to glass-air and water-air interface, respectively. Predict the angle of refraction of an incident ray in water at 45° with the normal to a water-glass interface [Fig. (c)]

Answer:

Question 5.
A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out ? Refractive index of water is 1.33 (consider the bulb to be a point source).
Answer:

The light rays from the bulb B, which fall on the surface of water at an angle equal to critical angle (θ_C), grazes on the surface of water and the rays of light which fall on the surface of water at an angle greater than θ_C are totally internally reflected back into the water. The rays of light images emerges out of water through a circular patch of radius r.

Question 6.
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. By rotating the prism, the angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism ? If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light. The refracting angle of the prism is 60°.
Answer:

Question 7.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm ?
Answer:
Here, n = 1.55, R₁ = R and R₂ = – R, f= 20 cm Using lens maker formula, we get

Question 8.
A beam of light converges to a point P. A lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm?
Answer:
(a) When a convex lens is placed in the path of light converging at P, the beam converges at P_t. Thus, point P acts as virtual object for the convex lens.
Now, u = 12 cm, f= 20 cm.

Question 9.
An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved farther from the lens?
Answer:

Thus, the image is virtual, erect, diminished and is formed on the same side of the lens at a distance of 8.4 cm from the lens. If the object is moved away from the lens, the image moves towards the principal focus and goes on decreasing in size.

Question 10.
What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
Answer:
Here, f₁ = 30 cm and f₂ = -20 cm
For the combination of two thin lenses, the focal length of the combination is given by

Since the focal length of the system of lens is negative, therefore, the combination behaves as a diverging lens.

Question 11.
A compound microscope consists of an objective lens of focal length 2-0 cm and an eye-piece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct   vision (25 cm), (b) infinity ? What is the magnifying power of the microscope in each case? (C.B.S.E. 2008)
Answer:
Here, f_n =2.0 cm, f = 6.25 cm,
Distance between object lens and eye piece = 15 cm (a) For the formation of image at the least distance of distinct vision,

Question 12.
A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eye-piece of focal length 2.5 cm can bring an object placed 9.0 mm from the objective in sharp focus. What is the separation between the two lenses ? How much is the magnifying power of the microscope?
Answer:

Question 13.
A small telescope has an objective lens of focal length 144 cm and an eye-piece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eye-piece?
Answer:
Here, focal length of objective lens, f₀ = 144 cm Focal

Question 14.
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eye-piece of focal length 1.0 cm is used, what is the angular magnification of the telescope ?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens ? The diameter of the moon is 3.48 x 10⁶m and the radius of the lunar orbit is 3.8 x 10⁸m. (C.B.S.E. 2008, 2011)
Answer:

Question 15.
Use the mirror equation to deduce that
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. (C.B.S.E. 2011)
Answer:

Question 16.
A small pin fixed on the table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table?
Refractive index of glass = 1.5. Does the answer depend upon the location of the slab?
Answer:
Here, t = 15 cm, n = 1.5
The lateral displacement

For small angles of incidence, the answer does not depend upon the location of the slab.

Question 17.
(a) Following figure shows a cross-section of a ‘light pipe’ made of glass fibre of refractive index 1.68. The outer covering of the pipe is made of material of refractive index 1.44. What is the range of the angles of incident rays with the axes of the pipe for which the total internal reflection inside the pipe take place as shown in the figure ?
(b) What is the answer if there is no outer covering of pipe ?

Answer:


Therefore all incident rays in the range 0 to 90°suffer total internal refletion.

Question 18.
Answer the following questions:
(a) You have learned that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances ? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?
(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake, would the fisherman look taller or shorter to the diver than what he actually is ?
(d) Does the apparent depth of a tank of water change if viewed obliquely ? If so, does the apparent depth increase or decreases
(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter ?
Answer:
(a) Yes. They can produce real images if the object is a virtual object as shown in figure.

(b) here is no contradiction in this case. The virtual image of the object acts as an object for the convex lens of our eye and the lens of our eye make a real image of this object on the ratina.
(c) Let AB be the fisherman standing on the bank of the lake. The rays of light from the head of the fisherman bends towards the normal on refraction at the interface separating water and air. The refracted rays appear to come from point B’ instead of point B for the fish. Thus, for a diver the height of the fisherman is AB’ which is greater than his actual height AB.

(d) The apparent depth of a pond of water decreases when viewed obliquely. This is due to the refraction of light from the surface of water.

Question 19.
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose ?
Answer:
The minimum distance between a real object and its real image formed by a convex lens of focal length/is given by L = 4f

Question 20.
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
Answer:
Let O be the position of object and I is the position of image when lens is at L₁ and then at L₂


Question 21.
(a) Determine the effective focal length of the combination of the two lenses in the question 910 if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side a beam of parallel light is incident ? Is the notion of effective focal length of this system useful at all ?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the above arrangements. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.
Answer:

Question 22.
At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face ? The refractive index of the material of the prism is 1.524.
Answer:
Let the ray of light be incident on the face AB at angle i so that it is totally internally reflected at face AC.

Question 23.
You are given prism made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will (a) deviate a pencil of white light without much dispersion (b) disperse (and displace) a pencil of white light without much deviation.
Answer:
(a) Angular dispersion produced by two prisms i.e. crown glass and flint glass should be zero in this case


In the combination of prisms, flint glass prism of greater angle may be tried but in any case still this angle will be smaller than the angle of the crown glass prism in opposite order as shown in figure.

Question 24.
For a normal eye, the far point is at infinity and the near point of distinct vision is about 2.5 cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation {i.e., the range of converging power of the eye-lens) of a normal eye.
Answer:
When the object is placed at infinity, the eye makes use of the least converging power, Therefore, total converging power of cornea and the eye lens = 40 + 20 = 60 dioptre.


Question 25.
Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation ? If not, what might cause these defects of vision ?
Answer:
No. Myopia may arise due to the elongation of the eye ball and hypermetropia may arise due to the decrease in the size of the eye ball even when the eye has the normal ability of accommodation. There is another defect in the eye called presbyopia similar to hypermetropia. However, the causes of presbyopia and hypermetropia are different. Presbyopia arises in elderly persons and is corrected by using a bi-focal lens.

Question 26.
A myopic person has been using spectacles of power – 1.0 dioptres for distant vision. During old age he also needs to use separate reading glass of power +2 dioptres. Explain what may have happened.
Answer:
For -1.0 dioptre, the far point for eyes is 1 m i.e. 100 cm. The near point is 25 cm. The objects lying at infinity are brought at 100 cm from his eyes using the concave lens and the objects lying in between 25 cm and 100 cm are brought to focus using the ability of accommodation of the eye lens. In the old age, this ability of accommodation is reduced and the near point reaches 50 cm from his eyes.


Question 27.
A person looking at a mesh of crossed wires is able to see the vertical wires more distinctly than the horizontal wires. What is this defect due to ? How is such a defect of vision corrected ?
Answer:
This is due to the defect of lenses called astigmatism. The defect arises because of the fact that curvature of the eye-lens and the cornea is not same in different planes. This defect is removed by using cylindrical lens with vertical axis.

Question 28.
A man with normal near point (25 cm) reads a book with small print using a magnifing glass: a thin convex lens of focal length 5 cm.
(a) What is the closest and the farthest distance at which he can read the book when viewing through the magnifying glass ?
(b) What is the maximum and minimum angular magnification (magnifying power) possible using the above simple microscope ?
Answer:
(a) To see the object at a closest distance, the image of object should be formed at the least distance of distinct vision.


Question 29.
A card sheet divided into squares each of size 1 mm² is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to eye.
(a) What is the magnification (image size/object size) produced by the lens ? How much is the area of each square in the virtual object ?
(b) What is the angular magnification (magnifying powers) of the lens ?
(c) Is the magnification in
(1) equal to magnifying power in
(2) ? Explain.
Answer:

(c) Clearly magnification and power magnification are not equal to each other unless the image is located near the least distance of distinct vision, e. v = D.

Question 30.
(a) At what distance should the lens be held from the figure in the above exercise in order to view the squares distinctly with maximum possible magnifying power ?
(b) What is the magnification (image size/object size) in this case ?
(c) Is the magnification equal to magnifying power in this case ? Explain.
Answer:
(a) The magnifying power is maximum if the image is formed at the least distance of distinct point from the eye, i.e., if υ = -25 cm ; Also, f = 10 cm

=> the linear magnification and magnifying power is equal in this case.

Question 31.
What should be the distance between the object in the previous exercise and the magnifying glass if the virtual image of each square in the figure is to have an area 6.25 mm²? Would you be able to see the squares distinctly with your eyes very close to the magnifier?
Answer:


Question 32.
Answer the following questions :
(а)  The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virutal image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eye-piece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eye-piece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
Answer:
(a) When magnifying glass is not used, object to be seen clearly is to be placed at 25 cm. However, while using magnifying glass, object can be placed closer to eye than at 25 cm. The closer object has large angular size than the same object placed at 25 cm. It is in this sense that magnifying glass provides angular magnification.
(b) Yes. The angular magnification decreases slightly because angle subtended at eye is somewhat less than the angle subtended at the lens.
(c) The aberrations like spherical and chromatic aberrations start croping up if the convex lens of smaller and smaller focal length is made.
(d) Angular magnification of eye piece is given by (1+D/f_e)  and angular magnification of objective is approximately given by υ/f₀. Clearly for better magnification focal length of eye piece f_e and focal length of objective f_e should be small.
(e) If we position our eye very close to the eyepiece, the whole light will not fall on our eye and the field of view will decrease. So we place our eye a short distance away from the eye-piece to collect the large amount of light refracted through the eyepiece to increase the field of view.

Question 33.
An angular magnification (magnifying power) of 30 X is desired using an objective of focal length 1.25 cm and eye piece of focal length 5 cm. How will you set up the compound microscope ?
Answer:
For the image formed at the least distance of distinct vision, the magnifying power is given by



Question 34.
A small telescope has an objective lens of focal length 140 cm and an eye piece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e.,when the final image is formed at infinity) ?
(b) the final image is formed at the least distance of distinct vision (25 cm) ?
Answer:

Question 35.
For the telescope described in the last exercise, in 9.34
(a) what is the separation between the objective lens and the eye-piece ?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of image of the tower formed by objective lens ?
(c) What is the height of final image of the tower if it is formed at 25 cm ?
Answer:
(a) Since the final image is formed at infinity, the distance between the object lens and the eye-piece is f₀ + f_e = 140 + 5 = 145 cm


Question 36.
A cassegrain telescope uses two mirrors as shown in figure. Such a telescope is built with mirrors 20 mm apart. If the radius of curvature of large mirror is 220 mm and the small mirror is 140 mm, where will be the final image of an object at infinity be ?

Answer:

Question 37.
The adjoining figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of liquid ?

Answer:
In the presence of the liquid, the distance of the needle from the lens is equal to the focal length f of the combination of the convex lens and the piano concave lens formed by the liquid below it i.e. f = 45 cm. Also n = 1.5
In the absence of the liquid, the distance of the needle and the lens is equal to the focal length of the convex lens only i.e. f = 30 cm
.’.  If f₂ is the focal length of plane concave lens formed

We hope the NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments, help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

Question 1.
What is the force between two small charged spheres having charges of 2 x 10^-7 C and 3 x 10^-7 C placed 30 cm apart in the air?
Answer:

Question 2.
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge -0.8 μC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?
Answer:
(a) Force on charge 1 due to charge 2 is given by the relation

Question 3.
Check that the ratio ke²/G m_em_p. is dimensionless. Lookup a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Answer:

Thus, the given ratio is a number and is dimensionless. This ratio signifies that electrostatic force between electron and proton is very-very large as compared to the gravitational force between them.

Question 4.
(a) Explain the meaning of the statement ‘electric charge of a body is quantised.’
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large-scale charges?
Answer:
(a) Charge occurs in nature as a discrete entity. One packet of charge (least quantity) is called quantum charge. It is represented by ‘e‘. Generally, charge of an electron is represented by ‘e’ and charge of a proton is represented by ‘+ e’. Therefore, the charge possessed by any charged body will be an integral multiple of ± e, ie., ne where n = 1, 2, 3,
∴ q = ± ne
The fact that electric charge collections are integral multiples of the fundamental electronic charge was proved experimentally by Millikan.

(b) While dealing with the large-scale electrical phenomenon, we ignore the quantization of charge because the magnitude of charge of proton and electron is so small. For continuous charge distribution, charge can be accounted in terms of charge density such as linear charge density λ etc. We need not go for individual charges.

Question 5.
When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
Answer:
Before rubbing, both the glass rod and the silk cloth are electrically neutral. In other words, the net charge on the glass rod and the silk cloth is zero. When the glass rod is rubbed with silk cloth, a few electrons from the glass rod get transferred to the silk cloth. As a result, the glass rod becomes positively charged and the silk cloth negatively charged. Since the magnitude of positive charge on the glass rod is the same as that of negative charge on the silk, the net charge on the system is zero. Thus the appearances of charge on the glass rod and the silk cloth is in accordance with the law of conservation of charges.

Question 6.
Four-point charges q_A = 2 μC, q_B = -5 μC, or q_c = -2 μC and q_D = -5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1μC placed at the center of the Square?
Answer:
The symmetry of the figure clearly indicates that 1μC charge will experience equal and opposite forces due to equal charges of 2μC placed at A and C. Similarly, 1μC charge will experience equal and opposite forces due to -5 μC charges placed at D and B.

Thus, net force = zero.

Question 7.
(a) An electrostatic field line is a continuous curve.
That is, a field line cannot have sudden breaks. Why not?
(b) Explain why two field lines never cross each other at any point?
Answer:
(a) Electric lines of force exist throughout the region of an electric field. The electric field of a charge decreases gradually with increasing distance from it and becomes zero at infinity (i)e., electric field can’t vanish abruptly. So a line of force can’t have sudden breaks, it must be a continuous curve.

(b) If two lines of force intersect, then there would be two tangents and hence two directions of electric fields at the point of intersection, which is not possible.

Question 8.
Two-point charges q_A = 3μC and q_B = -3 μC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 x 10^-9 C is placed at this point, what is the force experienced by the test charge? (C.B.S.E. 2003)
Answer:
(a) Electric field at the midpoint of the separation between two equal and opposite charges is given by

(b) Force experienced by test charge = q₀^E
= (1.5 x 10^-9) (5.4 x 10⁶)
= 8.1 x 10^-3 N along BA

Question 9.
A system has two charges q_A = 2.5 x 10^-7 C and q_B = -2.5 x 10^-7 C located at points
A : (0,0, -15 cm) and B : (0, 0, + 15 cm), respectively. What are the total charge and electric dipole moments of the system?
Answer:
Clearly, the given points are lying on the z-axis.
Distance between charges, 21

= 15 + 15 = 30 cm = 0.3 m
Total charge = (2.5 x 10-⁷) – (2.5 x 10-⁷) = 0
Dipole moment
= q x 21 = 2.5 x 10-⁷ x 0.3
= 7.5 x 10^-8 C m along negative z-axis.

Question 10.
An electric dipole with dipole moment 4 x 10^-9 Cm is aligned at 30° with the direction of a uniform electric field of magnitude 5 x 10⁴ NC^-1. Calculate the magnitude of the torque acting on the dipole.
Answer:
Using τ = pE sin 0, we get
= (4 x 10-⁹) (5x 10-⁴ sin 30°)
= 2 x 10-⁴ x \(\frac { 1 }{ 2 } \) = 10-⁴ N m 2

Question 11.
A polythene piece rubbed with wool is found to have a negative charge of 3.2 x 10-⁴ C.
(a) (Estimate the number of electrons transferred (from which to which ?)
(b) Is there a transfer of mass from wool to polythene?
Answer:
(a) Using q = ne, we get

(b) Yes, but of negligible amount because mass of an electron is very-very small (mass transferred = m_e x n = 91 x 10-³¹ x 2 x 10¹² = 1.82 x 10^-18 kg).

Question 12.
(a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 x 10–⁷ C ? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
Answer:

Question 13.
Suppose the spheres A and B in Q 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?
Answer:
Charge on sphere A on contact with the third sphere (say C) having no charge is given by

When third sphere, how having charge 3.25 10^-7 C is brought in contact with sphere B, the charge left on sphere B is given by,

Question 14.
Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Answer:
Unlike charges attract each other, therefore particle 1 and 2 are negatively charged whereas particle 3 has a positive charge. Particle 3 gets maximum deflection so it has the highest charge (e) to mass (m) ratio because deflection, y α e/m

Question 15.
Consider a uniform electric field \(\overrightarrow { E } \)
= 3 x 10³ \(\hat { i } \)N/C.
(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?
(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?
Answer:
(a) Electric flux through the square,

Question 16.
What is the net flux of the uniform electric field of Problem 1.15 through a cube of the side 20 cm oriented so that its faces are parallel to the co­ordinate planes?
Answer:
Zero, because the number of field lines entering the cube is equal to the number of field lines coming out of the cube.

Question 17.
Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 x 10³ Nm2/C.
Answer:
(a) What is the net charge inside the box?
(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?
Ans.
(a) Using φ = φ /ε₀ we get q =φ ε₀
= (8 x 10³) (8.854 x 10^-12)
= 70.8 x 10-⁹ C = 0.07 μC
(b) No, it cannot be said so because there may be an equal number of positive and negative elementary changes inside the box. It can only be said that the net charge inside the box is zero.

Question 18.
A point charge + 10 μC is a distance 5 cm directly above the center of a square of side 10 cm, as shown in the given figure. What is the magnitude of the electric flux through the square? [Hint. Think of the square as one fact of a cube with edge 10 cm.]

Answer:
The charge can be assumed to be placed as shown in the figure.

Question 19.
A point charge of 2.0 μC is at the center of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Answer:

Question 20.
A point charge causes an electric flux of -1.0 x 10³ Nm²/C to pass through a spherical Gaussian surface of a 10.0 cm radius centered on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?
Answer:
(a) The electric flux depends only on the charge enclosed by the Gaussian surface and independent of the size of the Gaussian surface. The electric flux through new Gaussian surface remains same
i.e. -1 x 10³ Nm² C-¹ because the charge enclosed remains same in this case also.
(b) Using φ = q/ε₀, we get q = ε₀ φ = (8.85 x 10-¹²) (-1 x 10³)
i.e. q = -8.85 x 10^-9 C.

Question 21.
a conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the center of the sphere is 1.5 x 10³ N/C and points radially inward, what is the net charge on the sphere?
Answer:

Question 22.
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.4 μC /m².
(a) Find the charge on the sphere.
(b) What is the total electric flux leaving the surface of the sphere? (B.S.E. 2009 C)
Answer:
(a) Charge on the sphere is given by

Question 23.
An infinite line charge produces a field of 9 X 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density
Answer:

Question 24.
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 x 10^-22 C/m². What is E?

(a) in the outer region of the first plate,
(b) in the outer region of the second plate, and
(c) between the plates?
Answer:

Question 25.
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 x 10⁴ NC^-1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm^-3. Estimate the radius of the drop, (g = 9.81 ms^-2 ; e = 1.60 x 10^-19 C.)
Answer:
Charge on the drop,

Question 26.
Which among the curves shown in the figure cannot possibly represent electrostatic field lines?



Answer:
(a) Incorrect, because the field lines should be normal to the surface of a conductor.
(b) Incorrect, because the field lines cannot start from a negative charge.
(c) Correct
(c) Incorrect, because electric field lines cannot intersect with each other.
(d) Incorrect, because electrostatic field lines cannot form a closed loop.

Question 27.
In a certain region of space, the electric field is along the z-direction throughout. The magnitude of the electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 10⁵ N C^-1 per meter. What are the force and torque experienced by a system having a total dipole moment equal to 10^-7 C m in the negative z-direction?
Answer:
Suppose the dipole is along the z-axis.


Question 28.
(a) A conductor A with a cavity as shown in figure (a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.
(b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Figure (b)].
(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Answer:
Select a Gaussian surface lying wholly inside the conductor but very near to the surface of the conductor.
(a) There is no electric field inside the conductor so electric flux through Gaussian surface is zero or in other words, net charge inside the Gaussian surface is zero. Then it can be said that the charge lies outside the Gaussian surface e. on the outer surface of the conductor.
(b) Charge q inside the cavity will induce a charge -q on the inner side of the cavity and thus +q will appear on the outer surface. Thus total charge will be (q + Q).
(c) The instrument should be enclosed in a metallic shell so that the effect of the electrostatic field is cancelled out.

Question 29.
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/2ε₀) \(\hat { n } \), where \(\hat { n } \) is the unit vector in the outward normal direction, and a is the surface charge density near the hole.
Answer:
Let the tiny hole of the conductor be considered as filled up. The field inside the conductor is zero, whereas outside it is given by

This field is infact due to
(1) field (E₁) due to plugged hole and
(2) field E₂ due to rest of the charged conductor. Inside the conductor, these fields are equal but opposite, whereas outside they are exactly same. i.e.

Question 30.
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density X without using Gauss’s law.
[Hint. Use Coulomb’s law directly and evaluate the necessary integral.]
Answer:
Consider a long thin wire of uniform linear charge density X placed along the X-axis. Let P be a point lying on the y-axis






Question 31.
It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves build-out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge + (2/3)e, and the ‘down’ quark (denoted by d) of charge
(-l/3)e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.)
Answer:

Question 32.
(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
Answer:
(a) To prove the result, let us assume that the test charge placed at the null point is in stable equilibrium. If it is so, then on being displaced slightly away from the null point, the test charge should return to its position., It implies that if a closed surface is drawn around the test charge, there will be a net inward flux of the electric field through its surface. According to Gauss law, there cannot be any electric flux through its surface as it does not enclose any charge. Hence our assumption is wrong and the test

(b) For the configuration of the two charges of the same magnitude and sign, the null point is the midpoint of the line joining the two charges. If the test charge is displaced slightly from the null point along the line, it will return back due to the restoring force that comes into the day. But if the charge is displaced slightly from the null – point along normal to the line it will not return. This is because the resultant force due to the configuration of two charges will take it away from the null point. For the test charge to be in stable equilibrium restoring force must come into play, when it is displaced in any direction. Hence the test charge cannot be in stable equilibrium.

Question 33.
A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed σ_x, (like particle 1 in figure.) The length of plate is L and a uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL²/(2m υ_x²).
Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10. of Class XI Textbook of Physics.
Answer:
Consider a uniform electric field \(\overrightarrow { E } \) set up between two oppositely charged parallel plates (Figure). Let a positively charged particle having charge +q and mass m enters the region of electric field E at O with velocity E along X-direction.
Step 1.

Force acting on the charge +q due to electric field E is given by
\(\overrightarrow { F } \) =Q \(\overrightarrow { E } \)
The direction of the force is along the direction of
\(\overrightarrow { E } \)and hence the charged particle is deflected accordingly.
Acceleration produced in the charged particle is given by

Step 2.
The charged particle will accelerate in the direction of E . As soon as the particle leaves the region of electric field, it travels due to inertia of motion and hits the screen at point P. Let t be the time taken by the charged particle to traverse the region of electric field of length L. Let y be the distance travelled by the particle along y-direction (i.e. direction of electric field). Using a standard equation of motion,
S = ut + \(\frac { 1 }{ 2 } \) at².
For horizontal motion. S = L, u = υ_x and a = 0.
(∴ no force acts on the particle along x-direction)
From equation (ii), we have

Equation (i) is the equation of a parabola.
Hence a charged particle moving in a uniform electric field follows a parabolic path.

Question 34.
Suppose that the particle in Q 1.33 is an electron projected with velocity
υ_x = 2.0 x 10⁶ ms^-1. If E between the plates separated by 0.5 cm is 9.1 x 10² N/C, where
will the electron strike the upper plate ? ( |e| = 1.6 x 10^-19 C, m_e = 9.1 x 10-³¹ kg.)
Answer:

We hope the NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields, help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields, drop a comment below and we will get back to you at the earliest.