NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts

NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts.

Board CBSE
Textbook NCERT
Class Class 12
Subject History
Chapter Chapter 6
Chapter Name Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts
Number of Questions Solved 9
Category NCERT Solutions

NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts

Question 1.
Explain with examples what historians mean by the integration of cults.
Solution :
The integration of cults means that there was composition, compilation and preservation of Puranic texts in simple Sanskrit verse, explicitly meant to be accessible to women and Shudras, who were generally excluded from Vedic learning. The Brahmanas also accepted and reworked the beliefs and practices of these and other social categories. The example of this integration of cults is at Puri, Orissa where the principal deity was identified as Jagannatha (literally, the lord of the world), a form of Vishnu. Here the deity is represented in a very different way. In this case, a local deity, whose image was and continues to be made of wood by local tribal specialists, was recognised as a form of Vishnu. At the same time, Vishnu was visualised in a way, that was very different from that in other parts of the country. Such instances of integration were evident amongst goddess cults as well where the local deities were provided an identity as the wife of the principal male deities – Lakshmi, the wife of Vishnu or Parvati, the wife of Shiva.

Question 2.
To what extent do you think the architecture of mosques in the subcontinent reflects a combination of universal ideals and local traditions?
Solution :
With the arrival of Islam in the Medieval ages, the architecture of Islam also came to India. However, the Arab-cum-Islamic architecture got impacted by the local traditions and rites too. Hence, we see a fusion of the two. This can be further elaborated by the examples of architecture mainly the constructions of the mosques of those days.

Some features of the architecture of mosques are universal. All mosques have orientation towards Mecca. This is manifested in the placement of Mehrab and Minar within a mosque. But at the same time we have influences that can be described only as local influences. A 13th Century mosque in Kerala has a shikhar like roof unlike a normal mosque where it is dome. The Shah Hamdan Mosque in Kashmir is made of Kashmiri woods and its facade is like that of a temple. The Atia Mosque in Bangladesh is made of bricks, though its roof is round. Thus, we can see that the architecture of Mosques is that of fusion.

Question 3.
What were the similarities and differences between the be-shari‘a and ba- shari‘a sufi traditions ?
Solution :
(a) Similarities :

  • Both be-sharia and ba-shari‘a sufis turned to asceticism and mysticism and were against the growing materialism of the Caliphate as a religious and political institution.
  • They are critical of the dogmatic definitions and scholastic methods of interpreting the Quran and sunna adopted by the theologians.
  • Both laid emphasis on seeking salvation through intense devotion and love for God.
  • Both sought an interpretation of the Quran on the basis of their personal experience.

(b) Differences :

Be-Shari’s Ba-Shari’s
(i)   They scorned the Khanqah and took to mendicancy and observed Celibacy. (i) They organised communities around the khanqah controlled by a teaching master known as Shaikh, Pir or Murshid.
(ii)   They ignored rituals and observed extreme forms of asceticism. (ii) They observed special rituals of initiation. The initiates took an oath of allegiance, wore a patched garment and shaved their hair. Khanqah was the centre of social life.
(iii)   They were known by different names as Qalandars, Madaris, Malangs, and Haidaris. (iii) The Chishti silsila is the most important and influential tradition.
(iv)   They deliberately defied the shari‘a i.e., law governing the Muslim community. So they were called as be-shari‘a. (iv)     They complied with the sharia.

 Question 4.
Discuss the ways in which the Alvars, Nayanars and Virashaivas expressed critiques of the caste system.
Solution :
The Alvars, Nayanars and Virashaivas expressed critiques of the caste system in the following ways :
(a) Alvars and Nayanars :

  1. The Alvars and Nayanars protested against the caste system or at least attempted to reform the system. Their bhaktas hailed from diverse social backgrounds ranging from Brahmanas to artisans and cultivators and even from castes considered “untouchable”. Thus, people from all walks of society were welcomed by them.
  2. They stated that their compositions were as important as the Vedas. For example, one of the major anthologies of compositions by the Alvars, the Nalayira Divyaprabandham, was described as the Tamil Veda.
  3. Tondaradippodi, a Brahmana Alvar, opposed the cast system in the following way :
    “You (Vishnu) manifestly like those “servants” who express their love for your feet, though
    they may be bom outcastes, more than the Chaturvedins who are strangers and without allegiance to your service.”
  4. Another saint, a Nayanar, composed the following verse in protest of the caste system : “O rogues who quote the law books, Of what use are your gotra and kula, Just bow to Marperu’s lord (Shiva who resides in Marperu, in Thanjavur, Tamil Nadu) as your sole refuge.”

(b) Virashaiva:

  1. Virashaiva or Lingayats challenged the idea of caste and the “pollution” attributed to certain groups by Brahmanas. As a result of it, marginalised people within the Brahmanical order joined this tradition.
  2. They questioned the theory of rebirth.
  3. The Lingayats encouraged certain practices disapproved in the Dharmashastras, such as post-puberty marriage and remarriage of widows.

Question 5.
Describe the major teachings of either Kabir or Baba Guru Nanak, and the ways in which these have been transmitted.
Solution :

  1. The major teachings of Kabir and Baba Guru Nanak are as given below :
    • Teachings of Kabir :

      • He described the Ultimate Reality as Allah, Khuda, Hazrat and Pir. He also used terms drawn on Vedantic traditions, alakh (the unseen), nirakar (formless), Brahman and Atman. Terms such as shabda (sound) or shunya (emptiness) were drawn from yogic traditions.
      • There is only one God. He was against the distinction made between gods of different communities.
      • Kabir was against idol worship and Hindu polytheism.
      • He was in favour of the Hindu practice of nam-simaran (remembrance of God’s name).
    • Teachings of Guru Nanak : The message of Baba Guru Nanak is spelt out in his hymns and teachings which are as given below :
      • Baba Guru Nanak believed in nirguna bhakti.
      • He was against the external practices of religions.
      • He rejected sacrifices, ritual baths, image worship, austerities and the scriptures of both Hindus and Muslims.
      • For him, the Absolute or “arab” had no gender or form.
      • He favoured a simple way to connect to the Divine by remembering and repeating the Divine Name.
  2. The ways in which the major teachings of Kabir and Baba Guru Nanak were transmitted are as given below –
    • Compositions of the sants were sung by roadside musicians.
    • Baba Guru Nanak would sing his compositions in various ragas while his attendant Mardana played the rabab.
    • Baba Guru Nanak organised his followers into a community. He set up rules for congregational worship (sangat) involving collective recitation.

Question 6.
Discuss the major beliefs and practices that characterised Sufism.
Solution :
After the advent of Islam in the early, middle ages , it saw a new movement in later part. The movement has had great impact and reach in the Indian subcontinent. It is called Sufi movement. The Sufi saints were mystics. Their preachings included:
1. Sufi saints did not subscribe to the theological and rigid interpretations of religious scriptures of Islam. They believed that the interpretation have to be based on individual experiences. This way the theological interpretations became flexible. Further the control of the orthodox religious leaders got weakened. This was a people centric move.
2. They rejected the high sounding rituals. They also emphasised on simplicity in religious traditions and rites.
3. Sufi saints prescribed devotion to Almighty as path to salvation. They even approved of singing and dancing as part of devotion. It is notable that classical Islam has forbidden singing, dancing and any music.
4. The most important theme of Sufi philosophy was that serving people is the true religion. With the objective of serving the poor people they also held Langar. Today also one can go to Ajmer and can partake in the Langar organised on the tomb of Nijammudin Auliya, the great Sufi saint.
5. Sufi saints also emphasised on the equality among people and oneness among all.

Question 7.
Examine how and why rulers tried to establish connections with the traditions of the Nayanars and the sufis.
Solution :
The rulers tried to establish connections with the traditions of the Nayanars and the sufis to claim divine support and proclaim their own power and status.
(a) Rulars and Sufis :
(i) The Turks who had set up Delhi Sultanate also required legitimation from the sufis because they resisted the insistence of the ulama on imposing sharva as state law because they anticipated opposition from their non-Muslim subjects. The sultans, therefore, sought the support of the Sufis who derived their authority directly from God and did not depend on jurists to interpret the sharva.

(ii) The sultans wanted to have the blessings of Sufi saints because it was believed that the Auliya could intercede with God in order to improve the material and spiritual conditions of ordinary human beings. The Sultans of Delhi set up charitable trusts as endowments for hospices and granted tax-free land.

(iii) Akbar maintained connections to get their blessings for new conquests, fulfilment of vows and birth of sons. He used to visit the Dargah of Muinuddin Chishti. He went there 14 times and gave generous gifts, which were recorded in imperial documents. For example, in 1568, he offered a huge cauldron (degh) to facilitate cooking for pilgrims. He also had a mosque constructed within the compound of the dargah.

(b) Rulers and Nayanars :

(i) The Chola kings tried to establish connections with Nayanars and Alvars by building splendid temples that were adorned with stone and metal sculpture to recreate the visions of these popular saints who sang in the language of the people.

(iii) Those kings also introduced the singing of Tamil Shaiva hymns in the temples under royal patronage taking initiative to collect and organise them into a text.

(iii) An inscription suggests that Chola ruler Parantaka I had consecrated metal images of Appar, Sambandar and Sundarar in a Shiva temple. These were carried in processions during the festivals of these saints.

Question 8.
Analyse, with illustrations, why bhakti and sufi thinkers adopted a variety of languages in which to express their opinions.
Solution :
The bhakti and sufi thinkers adopted a variety of languages to express their opinions due to the following reasons :

  1. The local languages were used so that people might understand their verses easily.
  2. The language of the Vedas was Sanskrit. The Bhakti traditions were against the beliefs and practices of the Vedas. So, it was necessary to use local languages which people might understand and follow them in practice. That is why Kabir’s poems have survived in several languages. Similarly, Baba Guru Nanak expressed his ideas through hymns called “shabad” in Punjabi, the language of the region.
  3. The Chishtis too adopted local languages such as Hindavi because poetic compositions were recited in hospices, usually during ‘sama’. In Bijapur, sufi poetry was composed in Dakhani because these were probably sung by women while performing household chores like grinding grain and spinning.

Question 9.
Read any five of the sources included in this chapter and discuss the social and religious ideas that are expressed in them.
Solution :
The period of the Bhakti Movement and Sufi Movement also has many sources that contribute to the history of those days. Some of the major social and religious ideas expressed in the various sources of history are as follows:
1. The first is the architecture. The different types of stupas, temple, monasteries all symbolise different types of religious belief system and practices. Some of them exist as it is and enable us to look into the annals of history of those days. Some of them are in the form of ruins but they also throw light on the, religion and society of those days alike.
2. The next important source of history is the composition of the saints both Bhakti
and Sufi. In terms of content they are religious but they are not the divine textbooks of religion that are sacrosanct. The compilation throws light on the life of common men and village lifestyle. They also impact the music and art of those days.
3. Another very important source of the history of those days is the biographies of the Saints. The biographies include the description of the society and prevalent beliefs and practices. It is notable that such biographies may not be in the written form still they can give insight into the life of those days. It is the story prevalent that I when Kabirdas died, both Hindus and Muslims fought for his dead body later on his body turned into flowers. Some were taken by Muslims and others by Hindus.
This represents that there conflict and collaboration between both Hindus and Muslims of those days.
4. This was also the period of rise of religious leaders who were intermediaries between common men and God. Earlier it was only the Brahmins who got this role. Now many people from other background also joined in. To some extent it acted as the the force that idolised equality and fraternity.
5. The other source is the folklore. They are described in our art forms. It may be dance, paintings, and sculpture and so on. They all talk about the universal brotherhood of mankind and love for one and all.

We hope the NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 History Chapter 4 Thinkers, Beliefs and Buildings Cultural Developments

NCERT Solutions for Class 12 History Chapter 4 Thinkers, Beliefs and Buildings Cultural Developments are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 4 Thinkers, Beliefs and Buildings Cultural Developments.

Board CBSE
Textbook NCERT
Class Class 12
Subject History
Chapter Chapter 4
Chapter Name Thinkers, Beliefs and Buildings Cultural Developments
Number of Questions Solved 9
Category NCERT Solutions

NCERT Solutions for Class 12 History Chapter 4 Thinkers, Beliefs and Buildings Cultural Developments

Question 1.
Were the ideas of the Upanishadic thinkers different from those of the fatalists and materialists ? Give reasons for your answer.
Solution :
The ideas of the Upanishadic thinkers were different from those of the fatalists and materialists. Many ideas found in the Upanishadas show that people were curious about the meaning of life, the possibility of life sifter death and rebirth. There were debates on these issues. The thinkers were concerned with understanding and expressing the nature of the ultimate reality. The ideas such as the nature of the self and the true sacrifice were discussed in Upanishads.

On the other hand, the fatalists believed that everything was pre-determined. Pleasure and pain could not be altered, lessened or increased in the course of Samsara (transmigration). The materialists did not believe in alms or sacrifice or offerings. For them there was no such thing as this world or the next. A human being is made of four elements. When a human being dies the four elements; return to earth, water, fire and air.

Question 2.
Summarise the central teachings of Jainism.
Solution :
The main teachings of Jainism are as follows:
(i) The entire world is animated. Life exists even in rocks and stones normally considered non-living.
(ii) The principle of non-violence is practised in extreme form in Jainism. No harms should be caused to animals, plants and insects and any other living beings that may include rocks and stones too. This is notable that Jains are forbidden to eat late night lest they kill insects by mistake.
(iii) The cycle of birth and rebirth is shaped through Karma. If one is to escape this cycle of Karma, one must practise ascetism and penance. It is possible when one renounces the world. So one has to live in monastery to attain salvation.
(iv) Jain monks have to take vows to observe the following:
(a) Not to kill anyone
(b) Not to steal anything
(c) Not tell lies
(d) Not to possess property
(e) To observe celibacy.

Question 3.
Discuss the role of the Begums of Bhopal in preserving the stupa at Sanchi.
Solution :
The role of the Begums of Bhopal in preserving the stupa at Sanchi was as given below:

  • The French sought Shahjehan Begum’s permission to take away the eastern gateway for a museum in France. Englishmen too wanted to do the same. They were not allowed to do so. The French and the English were allowed to prepare plaster-cast copies only.
  • Shahjehan Begum and her successor Sultan Jehan Begum, provided money for the preservation of the ancient site.
  • Sultan Jehan funded the building of museum as well as guesthouse where persons like John Marshall lived and wrote his important volumes.
  • She funded the publication of the volumes.

Question 4.
Read this short inscription and answer :
In the year 33 of the maharaja Huvishka (a Kushana ruler), in the first month of the hot season on the eighth day, a Bodhisatta was set up at Madhuvanaka by the hhikkhuni Dhanavati, the sister’s daughter of the bhikkhuni Buddhamita, who knows the Tipitaka, the female pupil of the hhikkhu Bala, who knows the Tipitaka, together with her father and mother.
(a) How did Dhanavati date her inscription ?
(b) Why do you think she installed an image of the Bodhisatta ?
(c) Who were the relatives she mentioned ?
(d) What Buddhist text did she know ?
(e) From whom did she learn this text ?
Solution :
(a) Dhanavati dated her inscription as in the year 33 of the maharaja Huvishka (a Kushan ruler), in the first month of the hot season on the eighth day.
(b) During the period of Kanishka, the Buddhism was divided into two branches i.e., Hinayana and the Mahayana. The worship of images of the Buddha and Bodhisattas had become an important part of this branch. So, she installed an image of the Bodhisatta in order to enable its followers to worship it.
(c) The relatives, she mentioned, were her father, mother, sister of her mother bhikkhuni Buddhamita.
(d) Tipitaka.
(e) She learnt this text from bhikkhu Bala.

Question 5.
Why do you think women and men joined the sangha?
Solution :
Women and men joined the sangha because within the sangha, all – workers, slaves, wealthy men, gahapatis and kings – were regarded as equal. It was an organisation of monks who could become teachers of dhamma. They shed their earlier social identities on becoming bhikkhus and bhikkhunis. Not only this, the internal functioning of the sangha was based on the traditions of ganas and sanghas where consensus was arrived at through discussions. If that failed, decisions were taken by a vote on the subject.

Question 6.
To what extent does knowledge of Buddhist literature help in understanding the sculpture at Sanchi?
Solution :
Buddhist literature help us upto some extent in understanding the sculpture at Sanchi. It is important that the sculptures at Sanchi depict the teachings of Buddha only. The teachings of Buddha are captured in the literature.

It is notable that Buddha used to roam around among people , preaching them on his teachings. However, he did not claim supernatural power. He told us that the world is ever changing. It is full of sorrows. Sorrow flows out of desire. Buddha asked the followers to take the middle path, not too much of penance, nor too much of indulgence. The literature of Buddhism is useful for the interpretation of the sculpture at Sanchi. People are shown in different moods and in sorrow. Different stages of life are depicted and so on. Hence, it can be stated that Buddhist literature throws valuable light on the sculptures of the Sanchi.

Question 7.
Figs A and B are two scenes from Sanchi. Describe what you see in each of them, focusing on the architecture, plants and animals, and the activities. Identify which one shows a rural scene and which an urban scene, giving reasons for your answer.
NCERT Solutions for Class 12 History Chapter 4 Thinkers, Beliefs and Buildings Cultural Developments 1NCERT Solutions for Class 12 History Chapter 4 Thinkers, Beliefs and Buildings Cultural Developments 2
Solution :

  1. In the figure A, depiction of animals, dead and alive, hunters with bow and arrow, trees have been made. These are one of the finest depictions. Several animal stories have been depicted at Sanchi, the above scene may be one of them. In the second figure B, the images are being worshipped.
  2. The scene in the figure A is a rural scene because it contains animals, hunters with bow and arrow and dead animals too. The second scene is an urban scene because it depicts worshipping of images.

Question 8.
Discuss how and why stupas were built. Describe the structure of stupa with example.
Solution :
About 200 years after the time of Buddha King Asoka erected a pillar at Lumbini. This was to announce the visit of Buddha to this place.Stupas were the mounds put on the bodily remains of the body of Lord Buddha or of any object that was used by him. At the place of stupas such objects were buried. These were places of great respect under the tradition of Buddhism, as they had the relics of Buddha. As per the description of Asokavadana winch a famous Buddhist book, Emperor Asoka gave Buddha’s relic to all major cities. Later on such places stupas were put. The most important stupas are at Sanchi, Bharhut and Saranath.

The structure of a stupa was like a dome and hemisphere. On the top of it, there would be a balcony called harmik. This balcony represented the abode of God. The harmik was covered with an umbrella. There used to be railings around the balcony.

The construction of the stupas was made possible by the contribution of many. On the forefront were the monarchs. The Satvahan Kings offered huge amount for the construction of the stupqs. Apart from the monarchs, merchants, artisans and common men and women also contributed to the construction of the stupas.

Question 9.
Discuss how and why stupas were built.
Solution :
(a) The stupas were built to bury the relics of Buddha such as his bodily remains or objects used by him. These were mounds known as stupas. Since they contained relics regarded as sacred, the entire stupa came to be venerated as an emblem of both the Buddha and Buddhism. According to Ashokavadana, Asoka distributed portions of the Buddha’s relics to every important town and ordered the construction of stupas over them.
(b)

  • Inscriptions found on the railing and pillars of stupas record donations made for building and decorating them. Some donations were made by kings such as the Satavahanas.
  • Some donations were made by guilds such as that of the ivory workers who financed part of one of the gateways at Sanchi.
  • Hundreds of donations were made by women and men who mention their names, sometimes adding the name of the place from where they came, as well as their occupations and names of their relatives.
  • Bhikkhus and bhikkhunis also contributed towards building the stupas.

We hope the NCERT Solutions for Class 12 History Chapter 4 Thinkers, Beliefs and Buildings Cultural Developments help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 4 Thinkers, Beliefs and Buildings Cultural Developments, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

NCERT Solutions for Class 12 Chemistry Chapter 6 contains concept wise details of the chapter. The solutions help the students prepare well for the examination. The diagrammatic representations and step wise solutions make it easy for the students to understand. Also the answers are accurate and are provided by the subject matter experts.

NCERT Solutions are provided for reference in various boards (CBSE, MP board, UP board, Gujarat board). The students appearing for these boards and competitive exams can score well by referring to the NCERT Solutions.

Board CBSE
Textbook NCERT
Class Class 12
Subject Chemistry
Chapter Chapter 6
Chapter Name General Principles and Processes of Isolation of Elements
Number of Questions Solved 31
Category NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements

Class 12 Chemistry chapter 6 explains the applications of various metals in our day-to-day life. Different terminologies such as refining, calcination, concentration, benefaction, roasting, etc. are very well explained here. The concepts of thermodynamics for the extraction of copper, aluminium and zinc are also mentioned here.

This chapter contains the details of various chemical processes associated with metallurgy. Being an important chapter, it will help the students score well during the examinations.

NCERT IN-TEXT QUESTIONS

Question 1.
Name some ores which can be concentrated by magnetic separation method.
Answer:
Only those ores can be concentrated by magnetic separation method in which either the ore particles or the impurities associated with it are of magnetic nature. For example, ores of iron such haematite (Fe2O3), magnetite (Fe3O4), siderite (FeCO3) are magnetic and can be concentrated by this method. Similarly, casseterite (SnO2) an ore of tin is non-magnetic while the impurities of tungstates of iron and chromium are of magnetic nature. Magnetic separation is effective in this case also.

Question 2.
What is the significance of leaching in the extraction of aluminium?
Answer:
Aluminium contains silica (SiO2), iron oxide (Fe2O3) and titanium oxide (TiO4) as impurities. These impurities can be removed by the process of leaching. During leaching, the powdered bauxite ore is heated with a concentrated (45%) solution of NaOH at 473-523 K, where alumina dissolves as sodium meta-aluminate and silica as sodium silicate leaving Fe2O3, TiO2 and other impurities behind:
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 1
The impurities are filtered off and solution of sodium meta-aluminate is neutralised by passing CO2 when hydrated alumina separates out while sodium silicate remains in solution. The hydrated aluminathus obtained is filtered, dried and heated to give back pure alumina.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 2
Thus, by leaching, pure alumina can be obtained from bauxite ore.

Question 3.
The reaction : Cr2O2(s) + 2Al(s) → Al2O3(s) + 2Cr(s) ; ∆G° = – 421 kJ is thermodynamically feasible as is apparent from the value of ∆G°. Why does not it take place at room temperature ?
Answer:
Though the reaction is feasible, it does not proceed at room temperature because all the reactants and products are solids. At elevated temperature, when chromium starts melting, the reaction becomes feasible.

Question 4
Is it true that under certain conditions, Mg can reduce Al2O3 and Al can reduce MgO ? What are those conditions ?
Answer:
If we look at the Ellingham diagram, it becomes evident that the plots for Al and Mg cross each other at 1350 °C (1623 K). Below this temperature, Mg can reduce Al2O3 and above this temperature, Al can reduce MgO.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 3

NCERT EXERCISE

Question 1.
Copper can be extracted by hydrometallurgy but not zinc. Explain.
Answer:
Copper is a comparatively less active metal as its reduction potential i.e. E° (Cu2+/Cu) is high (+0-34V). It can be displaced from a solution of Cu2+ ions by more active metals which have E° value lower than copper. For example, E° of zinc (Zn2+/Zn) is -0.76V and thus, zinc can displace copper from the solution of Cu2+ ions. In contrast, to displace zinc from a solution of Zn2+ ions a more reactive metal than zinc is required like, Na, K, Mg, Ca, etc. But, the more active metals readily react with water forming their corresponding ions and evolve hydrogen gas.
[2Na + 2H2O → 2NaOH + H2],
Thus, it is difficult to displace zinc from a solution of Zn2+ ions. Elance, copper can be extracted by hydrometallurgy but not zinc.

Question 2.
What is the role of the depressant in the froth floatation process?
Answer:
A depressant suppresses the formation of froth with a particular compound in the froth floatation process by reacting chemically with it. Thus, it helps in the separation of two metal sulphides present together in a particular ore.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 4
In actual process, the sulphide ore is finely powdered and is mixed with water to form a slurry in a tank as shown in Fig. 6.3. To this oily component of the sulphide ore particles by water. As a result, ore and oil constitute hydrophobic or water-repelling components while gangue and water form a lyophilic or water-attracting component.

Question 3.
Why is the extraction of copper from its sulphide ore difficult than that from its oxide through reduction?
Answer:
Sulphide ore of copper (Cu2S) cannot be directly reduced by either coke or hydrogen because ∆fG° of Cu2S is more than those of CS2 and H2S that will be formed as a result of the reaction.
These reactions are, therefore, not feasible. However, the ∆fG° of Cu2O is lower than that of CO2. Therefore, the sulphide ore is first roasted to Cu2O which is then reduced.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 5

Question 4.
Explain
(i) zone refining
(ii) column chromatography. (C.B.S.E. Sample Paper 2015)
Answer:
(i) Zone refining (Fractional Crystallisation). This method is used only if a metal in almost pure state is required. Metals like germanium and gallium which are used in semiconductors are purified by this method. The principle of zone refining is based on the fact that impurities are more soluble in molten metal than in solid metal.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 6

(ii) Chromatographic method. This method can also be employed on small scale for the purification of certain metals. Adsorption chromatography is normally used for this purpose. Different components present in a given sample are adsorbed to a different extent on the surface of the adsorbent.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 7

Question 5.
Out of C and CO, which is a better reducing agent at 673 K?
Answer:
When carbon reacts with dioxygen, two reactions are possible
C(s) + O2(g) → CO2(g) ……(i)
2C(s) + O2(g) → 2CO(g) …..(ii)
When CO is used as a reducing agent, it gets oxidized to CO2
2CO + O2 → 2CO2 …(iii)
It is clear from the Ellingham diagram that at 673K the ∆G° for the oxidation of CO to CO2 is more negative than the reaction (i) and reaction (ii). Therefore, CO is a better reducing agent than C. It is supported by the fact that the curve for the reaction (iii) lies below the curve for the reaction (i) and reaction (ii) at 673K. An element below in the Ellingham diagram reduces the oxide of other metal which lies above it.

Question 6.
Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?
Answer:
The common elements present in the anode mud are antimony, selenium, tellurium, silver, gold, and platinum. These elements settle down under anode as anode mud because they are less reactive and are not effected by CuSO4 – H2SO4solution.

Question 7.
Write the chemical reactions which take place in different zones in the blast furnace during the extraction of iron.
Answer:
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 8
The chemical reactions which take place in the blast furnace are briefly discussed.
Zone of combustion. At the bottom of the furnace, the blast of hot air causes the combustion of coke into carbon dioxide. The reaction is highly exothermic and a temperature of nearly 2170 K develops. It supplies most of the heat required for the process.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 9

Zone of heat absorption. As CO2 gas rises up, it combines with more coke to form carbon monoxide. Since the reaction is endothermic, the temperature in the middle of the furnace is nearly 1570 K. It further decreases as the reaction proceeds.

Zone of slag formation. In the middle of the furnace, the temperature is nearly 1123 K. Here limestone (CaCO3) decomposes to form CaO and CO2. The former (CaO) combines with silica (SiO2) which is an impurity in the haematite ore to form calcium silicate (CaSiO3) that is fusible. This implies that CaO has acted as a basic flux. It has combined with the acidic impurity of Si02 to form slag.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 10

Zone of reduction: This is the upper part of the furnace. The temperature ranges between 500K to 900K. Here haematite (Fe2O3) is reduced to FeO.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 11
Further reduction of FeO to Fe occurs at higher temperatures (1123 K) by CO gas.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 12
The direct reduction of iron ore left unreacted also occurs with carbon above 1123 K
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 13

Question 8.
Write the chemical reactions which take place in the extraction of zinc from zinc blende.
Answer:
Zinc blende is chemically zinc sulphide (ZnS). It undergoes the following reactions:
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 14
Zinc metal is in crude or impure form. It can be refined with the help of electro-refining. In this method, impure zinc is made anode while a plate of pure metal acts as the cathode. The electrolyte is aqueous zinc sulphate containing a small amount of dilute H2SO4. On passing electric current, Zn2+ ions from the electrolyte migrate towards the cathode and are reduced to zinc metal which gets deposited on the cathode.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 15

From anode, an equivalent amount of zinc gets oxidised to Zn2+ ions which migrate to the electrolyte
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 16

Question 9.
State the role of silica in the metallurgy of copper.
Answer:
During roasting, copper pyrites are converted into a mixture of FeO and Cu20. Thus, acidic flux silica is added during smelting to remove FeO (basic). FeO combines with SiO2to form famous silicate (FeSiO3) slag which floats over molten matte.

Question 10.
What is meant by the term chromatography?
Answer:
The term chromatography was originally derived from the Greek word chroma meaning colour and graphy meaning writing because the method was first used for the separation of coloured substances (plant pigments) into individual components. Now the term chromatography has lost its original meaning and the method is widely used for separation, purification and characterization of the components of a mixture whether coloured or colourless.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 17

Question 11.
What criterion is followed for the selection of the stationary phase in chromatography?
Answer:
The stationary phase is selected in such a way that it is capable of adsorbing the impurities more strongly than the elements to be purified. Under this condition, the impurities are retained by stationary phase i. e„ these cannot be eluted easily while the pure component, which is weakly adsorbed is easily eluted.

Question 12.
Describe a method for the refining of nickel.
Answer:
Nickel is refined by Mond’s process.
Mond’s process. Mond’s process is used for the refining of nickel metal. In the process, impure metal is heated in a current of carbon monoxide (CO) at 330 to 350 K to form nickel carbonyl which is of volatile nature. The vapours of the metal carbonyl escape leaving behind impurities. Upon heating to about 450 K, nickel carbonyl decomposes to give pure nickel.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 18
From the above discussion, we may conclude that the basic requirements for the refining of metal by Vand Alkel process and Mond’s process are:

  • The metal should form a volatile compound with the available reagent.
  • The volatile compound should be easily decomposable so that metal in pure form may be regenerated.

Question 13.
How can you separate alumina from silica in bauxite ore associated with silica? Give equations if any.
Answer:
The purification of bauxite containing silica as the main impurity is done by Serpeck’s process. The powdered ore is mixed with coke and heated to about 2073 K in an atmosphere of nitrogen. Silica (SiO2) is reduced to silicon which being volatile escapes. Alumina (Al2O3) is converted into aluminium nitride (AIN) by reacting with nitrogen. It is hydrolysed upon heating with water to get the precipitate of Al(OH)3. From the precipitate, Al2O3 is recovered upon strong heating.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 19

Question 14.
Giving examples, differentiate between calcination and roasting.
Answer:
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 20

Question 15.
How is ‘cast iron’ different from ‘pig iron’?
Answer:
Iron obtained from the blast furnace is called pig iron. It contains about 4% carbon and other impurities of S, P, Si, Mn etc. When pig iron is mixed with scrap iron and coke and then heated in a blast of hot air, some impurities are removed. Cast iron is obtained. It contains 3% carbon and some other impurities. It is hard and brittle.

Question 16.
Differentiate between mineral and ore.
Answer:
The naturally occurring chemical substances in form of which the metal occurs in the earth along with impurities are called minerals. The minerals from which the metal can be extracted conveniently and economically are called ore. Thus, all ores are minerals but all minerals are not ores. For example, iron is found in the earth’s crust oxides, carbonates and sulphides. Out of these minerals of iron, the oxides of iron are employed for the extraction of the metal. Therefore, oxides of iron are the ores of iron. Similarly, aluminium occurs in the earth’s crust in form of two minerals, i.e, bauxite (Al2O3. x H2O) and clay (Al2O3.2SiO. 2H2O). Out of these two minerals, Al can be conveniently and economically extracted from bauxite the ore of alluminium.

Question 17.
Why is copper matte put in the silicon-lined converter?
Answer:
Copper matte consists of a mixture of Cu2S and Cu2O. Along with that, it also contains a small amount of FeS and FeO. It is put in a silicon-lined converter known as a Bessemer converter. Some silica (SiO2) is also added and a blast of hot air is blown. As a result, a number of reactions take place.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 21

Question 18.
What is the role of cryolite in the metallurgy of aluminium?
Answer:
The role of cryolite is two-fold

  • It makes alumina a good conductor of electricity.
  • It lowers the fusion (melting point) temperature of both from 2323K to about 1140K.

Question 19.
How is leaching carried out in case of low-grade copper?
Answer:
Leaching in case of low-grade copper is carried out by reacting with an acid like H2SO4 in the presence of air when copper is oxidised to Cu2+ ions which pass into the solution. For example.
2 Cu(s) + 2 H2SO4 (aq) + O2 (g) → 2CuSO4 (aq) + 2 H2O(l)
or Cu + 2H+ (aq) + 1/2 O2(g) → Cu2+ (aq) + H2O(l)

Question 20.
Why is zinc not extracted from zinc oxide through reduction using CO?
Answer:
The standard free energy of formation (Δf G°) of CO2 from CO (fig 6.8) is higher than that of the formation-of ZnO from Zn. Therefore, CO cannot be used to reduce ZnO to Zn.

Question 21.
The value of ∆fG° for the formation of Cr2O3 is – 540 kJ mol-1 and that of Al2O3 is – 827 kJ mol-1. Is the reduction of Cr2O3 with Al possible? (Pb. Board2009)
Answer:
The two thermochemical equations may be written as follows :
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 22
Since ∆G° comes out to be negative, the reaction is feasible.

Question 22.
Out of C and CO, which is a better reducing agent for ZnO?
Answer:
The Δf G° of CO2 from CO is always a higher reduction of ZnO to Zn. In contrast, Δf G° of CO from C is lower at a temperature above 1180K while that of CO2 from C is lower at temperatures above 1270K than Δf G° of ZnO. Thus above 1270K, ZnO can be reduced to Zn by C. In actual practice, the reduction is usually carried out around 1673K. Thus out of C and CO, C is a better reducing agent than CO for ZnO.

Question 23.
The choice of a reducing agent in a particular case depends on the thermodynamic factors. How do you agree with this statement? Support your opinion with two examples.
Answer:
Thermodynamic factors have a major role in selecting the reducing agent for a particular reaction.
(i) Only that reagent will be preferred which will lead to decrease in free energy (∆G°) at a certain specific temperature.
(ii) A metal oxide placed lower in the Ellingham diagram cannot be reduced by the metal involved in the formation of the oxide placed higher in the diagram.
Examples:
(0 Al2O3 cannot be reduced by Cr present in Cr2O3 since the curve for Al2O3 is placed below that of Cr2O3 in the Ellingham diagram
(ii) CO cannot reduce ZnO because there is hardly any change in free energy (∆G°) as a result of the reaction.

Question 24.
Name the processes from which chlorine is obtained as the by-product. What will happen if an aqueous solution of NaCl is subjected to electrolysis?
Answer:
Chlorine is obtained as the by-product in the manufacture of sodium by Down’s process in which molten sodium chloride is subjected to electrolysis.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 23
Sodium obtained by this method is almost pure while chlorine is the by-product.
Chlorine can also be obtained by carrying out the electrolysis of an aqueous solution of sodium chloride. The process is carried in Nelson’s cell. The various reactions which take place are as follows :
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 24
At cathode. Both Na+ and H+ ions migrate towards the cathode but H+ ions are discharged in preference to Na+ ions since their discharge potential is less. Na+ ions remain in the solution.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 25
At anode: Both Cl and OH ions migrate towards the anode but Cl ions are discharged in preference to OH” since their discharge potential is less. OH ions remain in the solution.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 26
Thus, in the electrolysis of an aqueous NaCl solution, H2 gas is evolved at the cathode and chlorine at the anode. The solution contains NaOH and is, therefore, basic in nature.
It is always better to prepare chlorine by Down’s process.

Question 25.
What is the role of a graphite rod in the electrometallurgy of aluminium?
Answer:
In the electrometallurgy of aluminium, oxygen gas is evolved at the anode. It reacts with graphite or carbon (graphite electrodes) to form carbon monoxide and carbon dioxide. In case some other metal electrodes act as an anode, then oxygen will react with aluminium formed during the process to form aluminium oxide (Al2O3) which will pass into the reaction mixture. Since graphite is cheaper than aluminium, its wastage or consumption can be tolerated.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 27

Question 26.
Outline the principles of refining of metals by the following methods:
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour phase refining
Answer:
(i) 1. This method is based on the principle that the impurities are more soluble in the molten state of metal (the melt) than in the solid-state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal. As the heater moves the molten zone of the rod also moves with it. As a result, pure metal crystallizes out of the melt and the impurities pass onto the adjacent molten zone. This process is repeated several times, which leads to the segregation of impurities at one end of the rod. Then the end with impurities is cut off. Silicon, boron, gallium, indium, etc can be purified by this process.

2. Column chromatography is a technique used to separate components of a mixture where components are in minute quantities. In chromatography, there are two phases: the mobile phase and the stationary phase. The stationary phase is immobile and immiscible. Al2O3 column is usually used as the stationary phase in column chromatography. The mobile phase may be a gas, liquid, or supercritical fluid in which the sample extracts dissolve. Then the mobile phase is forced to wave through the stationary phase. The component that is more strongly absorbed on the column takes a long time to travel than the component weakly, absorbed.

(ii) Electrolytic refining is the process of refining impure metal by using electricity. In this process, impure metal is made the anode and a strip of pure metal is made the cathode. A solution of a soluble salt of the same metal is taken as the electrolyte. When an electric current is passed, metal ions from the electrolyte are deposited at the cathode as pure metal and the impure metal from the anode dissolves into the electrolyte in the form options. The impurities present in the impure metal gets collected below the anode. This is called anode mud.

Anode : M → Mn++ ne
Cathode : Mn+ + ne → M.

(iii) Vapour phase refining is the process of refining metal by converting it into its volatile compound and then, decomposing is to obtain a pure metal to carry out this process.
(a) The metal should form a volatile compound with available reagent, and
(b) The volatile compound should be easily decomposable so that the metal can be easily recovered.
Nickel, Zirconium, and titanium are refined using this method.

Question 27.
Predict the conditions under which aluminium can be expected to reduce magnesium oxide.
Answer:
The equations for the formation of the two oxides are :
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 228
If we look at the plots for the formation of the two oxides on the Ellingham diagram, we find that they intersect at a certain point. The corresponding value of ∆G° becomes zero for the reduction of MgO by Al metal.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 29
This means that the reduction of MgO by A1 metal can occur below this temperature.
Aluminium (Al) metal can reduce MgO to Mg above this temperature because ∆fG for Al2O3 is less as compared to that of MgO.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 30NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 30
We hope the NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements, drop a comment below and we will get back to you at the earliest.

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

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

Board CBSE
Textbook NCERT
Class Class 12
Subject History
Chapter Chapter 10
Chapter Name Colonialism and the Countryside Exploring Official Archives
Number of Questions Solved 9
Category NCERT Solutions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Paharias Santhals
 (i) They lived around the Rajmahal hills. (i) They came into Bengal around the 1780s
(ii) They subsisted on forest produce and practised shifting cultivation. They did not cut forests. They grew a variety of pulses and millets for consumption. (ii) They cleared forest, cut down timber and ploughed the land for cultivation. They grew rice and cotton.
(iii) They scratched the ground lightly with hoes. (iii) They used plough for cultivation.
(iv) The Paharias were hunters, shifting cul­tivators, food gatherers, charcoal producers, silkworm rearers. They were intimately connected to the forest. They resisted the intrusion of the outsiders. They did not take to plough agriculture. (iv) The Santhals gave up their earlier life of mobility and settled down, cultivating a range of commercial crops for the market, and dealing with traders and moneylenders.

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

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

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

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

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

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

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

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

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

Board CBSE
Textbook NCERT
Class Class 12
Subject Physics
Chapter Chapter 11
Chapter Name Dual Nature of Radiation and Matter
Number of Questions Solved 37
Category NCERT Solutions

Question 1.
Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 1
Question 2.
The work function of caesium metal is 2.14 eV. When light of frequency 6 x 1014 Hz is incident on the metal surface, photo emission of electrons occurs. What is the
(а)   maximum kinetic energy of the emitted electrons,
(b)   stopping potential, and
(c) maximum speed of the emitted photo electrons ?
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 2
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 3

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

Question 4.
Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam.
(b) How many photons per second, on average, arrive at a target irradiated by this beam? (Assume the beam to have a uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Answer:
Given, λ = 632.8 nm = 632.8 x 10-9m
Power, P = 9.42 mW = 9.42 x 10-3 W
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 4
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 5
Question 5.
The energy flux of sunlight reaching the surface of the earth is 1.388 x 103 W/m2. How many photons (nearly) per square meter are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 6

Question 6.
In an experiment on the photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 x 10-15 V s. Calculate the value of Planck’s constant.
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 7
Question 7.
A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the center of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm.
(a) What is the energy per photon associated with sodium light?
(b) At what rate are the photons delivered to the sphere?
Answer:
Power of sodium lamp,
P = 100 W
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 8
Question 8.
The threshold frequency for a certain metal is 3.3 x 1014 Hz. If the light of frequency 8.2 x 1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 9


Question 9.

The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 10
Since the energy of incident radiation less than the work function (4.2 eV) of the metal, therefore, the photoelectric emission can not take place from the given metal. 11.10.

Question 10.
Light of frequency 7.21 x 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 x 105 m/s are ejected from the surface. What is the threshold frequency for the photoemission of electrons?
Answer:
Using the relation
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 11
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 12

Question 11.
Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the cathode, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the cathode is made.
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 13
Question 12.
Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Answer:
Given V = 56 V.
(a) the momentum of the electron
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 14
Question 13.
What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with the kinetic energy of 120 eV?
Answer:
Here, E = 120 eV = 120 x 1.6 x 10-19J
= 1.92 x 10-17 J

NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 15

Question 14.
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
(a) an electron, and
(b) a neutron would have the same de Broglie wavelength.
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 16

Question 15.
What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 x 10-9 kg drifting with a speed of 2.2 m/s?
Answer:
(a) Here, m = 0.040 kg and υ = 1.0 km/s .
= 1000 m/s
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 17

NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 18
Question 16.
An electron and a photon each have a wavelength of 100 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electrons.
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 19


Question 17.

(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be
1.40 x 10-10 m?  (C.B.S.E. 2008)
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of \(\frac { 3 }{ 2 } \) kT at 300 K.
Answer:
(a) Here λ = 1.40 x 10-10m
Also, h = 6.63 X 10-34 Js
and m = 1.67 x 10-27kg
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 20
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 21

Question 18.

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

NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 22
Question 19.

What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Answer:
For nitrogen,
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 23

Question 20.

(a) Estimate the speed with which electrons emitted from a heated cathode of an evacuated tube impinge on the anode maintained at a potential difference of 500 V with respect to the cathode. Ignore the small initial speeds of the electrons. The ‘specific charge’ of the electron i.e., its elm is given to be 1.76 x 1011 C kg-1.
(b) Use the same formula you employ in
(a) to obtain electron speed for an anode potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 24
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 25
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 26

Question 21.

(a) A monoenergetic electron beam with an electron speed of 5.20 x 106 ms-1 is subject to a magnetic field of l.30 x 10-4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given elm for electron equals 1.76 X 1011 C kg-1?
(b) Is the formula you employ in
(a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?
Answer:
(a) υ  = 5.20 x 106ms_1
B = 1.30 x 10 4T
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 27
(b) The formula employed in part (a) is not valid because with the increase in velocity, mass varies and in the above formula we have taken m as constant. Instead, m
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 28
Question 22.
An electron gun with its anode at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (~ 10-2 mm of Hg). A magnetic field of 2.83 x 10 4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data.
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 29

Question 23.
(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at (45 Å. What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube.
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 30

Question 24.
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as the annihilation of an electron-positron pair of total energy 10.2 BeV into two y-rays of equal energy. What is the wavelength associated with each γ-ray?   (1 BeV = 109 eV)
Answer:
The energy carried by the pair of γ-rays = 10 .2 BeV
The energy of each γ ray is
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 31
Question 25.
Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a Mediumwave transmitter of 10 kW power, emitting radio waves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~ 10_1° W m-2). Take the area of the pupil to be about 0.4 cm2, and the average frequency of white light to be about 6 x 1014
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 32
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 33
This is quite a small number, but still large enough to be counted.
Comparison of cases (a) and (b) tells us that our eyes can not count the number of photons individually.

Question 26.
Ultraviolet light of wavelength 2271 A from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity
(~ 10s W m-2) red light of wavelength 6328 A produced by a He-Ne laser?
[C.B.S.E. 2005, UC, 13]
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 34
Since the energy of a red photon is less than the work function for the metal, the photocell does not respond to red light.

Question 27.
Monochromatic radiation of wavelength 640.2 nm (1 nm = 10-9 m) from a neon lamp irradiates photosensitive material made of caesium or tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 35

Question 28.
A mercury lamp is a convenient source for studying the frequency dependence of photoelectric emission since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used :
λx = 3650 Å, λ2 = 4047 Å, λ3 = 4358 Å, λ4 = 5461 Å, λ5 = 6907 Å.
The stopping voltages, respectively, were measured to be :
V01 = 1.28 V, V02 = 0.95 V, f03 = 0.74 V, = 0.16 V,V05 = 0 V
(a) Determine the value of Planck’s constant
(b) Estimate the threshold frequency and work function for the material.
Answer:
(a) From the Einstein photoelectric equation,
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 36
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 37
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 38
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 39
Question 29.

The work function for the following metals is given : Na : 2.75 eV ; K : 2.30 eV ; Mo : 417 eV ; Ni: 515 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away? (C.B.S.E. 2009)
Answer:
Here, λ = 3300 A = 3300 x 10-10 m
Distance, r’ = 1 m and r’ = 50 cm = 0.5 m
Using the relation
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 40
Since the energy E of the incident photon of light is less than the work functions of Mo and Ni metals, so photoelectric emission will not occur in Mo and Ni. The distance of the source does not increase or decrease the energy of the photon of the light incident, therefore, the energy of electrons ejected will not change but the intensity of ejected electrons will increase (1 α 1/r2 ) and become four times.

Question 30.
Light of intensity 10-5 W m-2 falls on a sodium photo­cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave- picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
Answer:
A = 2 cm2 = 2 x 10-4 m2
φ = 2eV = 2 x 1.6 x 10-19 J
= 3.2 X 10-19  J.
Taking the approximate radius of an atom as 10-10 m. the effective area of sodium atom is ≈ r2 = 10-20 m.
.’. If there is one free electron per atom, then the number of electrons in five layers
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 41
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 42
The answer obtained implies that the time of emission of the electron is very large and is not in agreement with the observed time of emission, which is approximately 10 9 s. Thus wave-picture of radiation is not applicable for photo-electric emission.

Question 31.
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å which is of the order of interatomic spacing in the lattice) (me = 9.11 x 10­-31 kg).
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 43
Clearly, the energy of photons is much greater than the energy of electrons.

Question 32.
(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 44, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable?
Explain. (mn = 1.675 x 10-27 kg).
(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27°C). Hence explain why a fast neutron beam needs to be thermalized with the environment before it can be used for neutron diffraction experiments.
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 44
This wavelength is about hundred times smaller than the interatomic separation of crystals. Thus, neutrons are not suitable for diffraction experiments in case of crystals.
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 45
This wavelength is comparable to the interatomic spacing of crystals. Therefore, thermal electrons are able to interact with the crystal. Since
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 46
increasing the temperature, decreases their de Broglie wavelength and they become unsuitable for crystal diffraction. Thus, the fast beam of neutrons needs to be thermalised with the environment for neutron diffraction experiment.

Question 33.
An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow
light ?
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 47
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 48

Question 34.

The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10-15 m or less. This structure was first probed in early 1970’s using high-energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV).
Answer:
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 49
Thus, the energy of the proton ejected out of the linear accelerator is of the order of BeV.

Question 35.
Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.
Answer:
Here T = 27 °C = 27 + 273 = 300 K
P = 1 atm = 1.01 x 105 Nm-2
Also, the mass of helium atom,
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 50

Question 36.

Compute the typical de Broglie wavelength of an electron in metal at 27°C and compare it with the mean separation between two electrons in a metal which is given to be about 2 x 10_10 m.
Answer:
Here, T = 27°C = 27 + 273 = 300 K
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 51

Question 37.

Answer the following questions:
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3) e; (- l/3)e]. Why do they not show up in *Millikan’s oil-drop experiment?
(b) What is so special about the combination elm? Why do we not simply talk of e and m separately?
(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?
(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?
(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations :
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 52
Answer:
(a) In case of the Millikan oil-drop experiment, the charge on the electron is measured. The electron revolves outside the nucleus and each has a charge e. Thus, we do not observe the fractional charges
NCERT Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter 53
(c) At ordinary pressure, molecules of gas keep on colliding with each other and the ions formed do not have a chance to reach the respective electrodes to constitute a current because of their recombination. At low pressure, however, ions do not collide frequently and are able to reach the respective electrodes to constitute a current.
(d) Work function in fact is the energy required to knock out the electron from the highest filled level of the conduction band of an emitter. In the conduction band, there are different energy levels which collectively form a continuous band of levels. Therefore, different amounts of energy are required to bring the electrons out of the different levels. Electrons emitted have different kinetic energies according to the energy supplied to the emitter.
(e) Since frequency for a given matter-wave remains constant for different layers of the matter but wavelength changes so X is more significant than v.
Similarly energy E = hv = \(\frac { 1 }{ 2 } \) m(λv)2 is also constant for a given matter wave so phase λv is also not physically significant.

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

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

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

Board CBSE
Textbook NCERT
Class Class 12
Subject History
Chapter Chapter 2
Chapter Name Kings, Farmers and Towns Early States and Economies
Number of Questions Solved 9
Category NCERT Solutions

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

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

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

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

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

The important features of the Mahajanapadas are as follows.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity

NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity

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

Board CBSE
Textbook NCERT
Class Class 12
Subject Physics
Chapter Chapter 3
Chapter Name Current Electricity
Number of Questions Solved 24
Category NCERT Solutions

Question 1.
The storage battery of a car has an e.m.f. of 12 V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery ?
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 1
Question 2.
A battery of e.m.f. 10 V and internal resistance 3Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor ? What is the terminal voltage of the battery when the circuit is closed ?
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 2

Question 3.
(a) Three resistors IΩ, 2 Ωand 3 Ω are combined in series. What is the total resistance of the combination?
(b) If the combination is connected to a battery of e.m.f. 12 V and negligible internal resistance, obtain the potential drop across each resistor.
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 3
Voltage drop across R3 is given by, V3 = IR3 = 2 x 3 = 6V.

Question 4.
(a) Three resistors 2 Ω, 4 Ω, and 5 Ω, are combined in parallel. What is the total resistance of the combination?
(b) If the combination is connected to a battery of e.m.f. 20 V and negligible internal resistance, determine the current through each resistor, and the total current drawn from the battery.
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 4

Question 5.
At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 x 10-4 °C-1.
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 5

Question 6.
A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0 x 10-7 m2, and its resistance is measured to be 50 Ω. What is the resistivity of the material at the temperature of the experiment ?
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 6

Question 7.
A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C. Determine the temperature coefficient of resistivity of silver.
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 7
Question 8.
A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of 2.8 A. What is the steady temperature of the heating element if the room temperature is 27.0 °C ? Temperature co­efficient of resistance of nichrome averages over the temperature range involved is 1.70 x 10-4 °C-1.
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 8
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 9
Question 9.
Determine the current in each branch of the network shown in Figure
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 10
Answer:

NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 11
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 12
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 13
Question 10.

(a) In a meter bridge (Figure), the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips?
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 14
(b) Determine the balance point of the bridge above if X and Y are interchanged.
(c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current? (C.B.S.E. 2005)
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 15

Question 11.

A storage battery of e.m.f. 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit ?
Answer:
During charging,
V = E + I(r + R)
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 16

Question 12.
In a potentiometer arrangement, a cell of e.m.f. 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the e.m.f. of the second cell ?
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 17
Question 13.
The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 x 1028 m3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross­section of the wire is 2.0 x 10-6 m2 and it is carrying a current of 3.0 A.
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 18

Question 14.
The earth’s surface has a negative surface charge density of 10-9 cm-2. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe. If there were no mechanism of sustaining atmospheric electric Held, how much time (roughly) 
would be required to neutralise the earth’s surface ? (This never happens in practice because there is a mechanism to replenish electric charges, namely the continual thunderstorms and lightning in different parts of the globe.)
(Radius of earth = 6.37 x 106 m.)
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 19
Question 15.
(a) Six lead-acid type of secondary cells each of e.m.f. 2.0 V and internal resistance 0.015 Ω are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current drawn from the supply and its terminal voltage ?
(b) A secondary cell after long use has an e.m.f. of 1.9 V and a large internal resistance of 380 Ω. What maximum current can be drawn from the cell ? Could the cell drive the starting motor of a car ?
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 20
It cannot be used for starting motor of a car because large current is needed to start the car.

Question 16.
Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (PAL = 2.63 x 108 Ω m, Pcu= 1.72 X 1(H Ω m. Relative density of A1 = 2.7, of Cu = 8.9.)
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 21
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 22
Aluminium is lighter, so it is used for overhead power cables.

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

Current Voltage Current Voltage
A V A V
0.2 3.94 3.0 59.2
0.4 78.7 4.0 78.8
0.6 11.8 5.0 98.6
0.8 15.7 6.0 118.5
1.0 19.7 7.0 138.2
2.0 39.4 8.0 158.0

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

Question 18.
Answer the following questions :
(a) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor : current, current density, electric field, drift speed?
(b) Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law.
(c) A low voltage supply from which one needs high currents must have very low internal resistance. Why?
(d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?
Answer:
(a) Current
(b) No. Devices are and vacuum tubes, semiconductor diodes, transistors, thermisters, thyristors, etc.
(c) Lesser the value of ‘r’, the higher the current.
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 23
Clearly low r will ensure high current
(d) If internal resistance is not large, then the heavy current drawn during an accidental short circuit can damage the supply.

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

Question 20.
(a) Given n resistors each of resistance R, how will you combine them to get the
(i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?
(b) Given the resistances of 1 Q, 2 G, 3 Q, how will combine them to get an equivalent resistance of (i) (11/3)Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?
(c) Determine the equivalent resistance of networks shown in the figure. (N.C,E.R,T.)
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 24
(i)
Maximum resistance can be obtained by combining them in series with each other.
The maximum resistance Rmax = R + R + R + ……………..n times = nR.
(ii) Minimum effective resistance can be obtained by combining them in parallel with each other. Minimum resistance Rmax is found as
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 25
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 26
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 27

Question 21.
Determine the current drawn from a 12 V supply with internal resistance 0.5Ω by the infinite network shown in the figure. Each resistor has 1 Ω resistance. (N.C.E.R.T.)
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 28
Answer:
Let the total resistance of the circuit be Z and a set of three resistors of value R each is connected to it as shown in the figure.
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 29
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 30

Question 22.
The figure shows a potentiometer with a cell of 2.0 V and internal resistance of 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant e.m.f. of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown e.m.f. e and the balance point found similarly, turns out to be at 82.3 cm length of the wire.
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 31
(a) What is the value e?
(b) What purpose does the high resistance of 600 kΩ have?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an e.m.f. of 1.0 V instead of 2.0 V?
(f) Would the circuit work cell for determining an extremely small e.m.f., say of the order of a few mV (such as the typical e.m.f. of a Thermo ­couple not, how will you modify the circuit?
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 32
(b) It allows only a small current to flow through the galvanometer when the circuit is not balanced.
(c) No
(d) No
(e) No
(f) No. The circuit will not work (E ∝ l).
The circuit can be modified by putting a suitable resistor ‘R’ in series with the wire AB.

Question 23.
The figure shows a potentiometer circuit for the comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε?
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 33
(b) We fail to get the balance point with the given cell of emf E, if the potential difference across the wire AB. In order to obtain the balance point with the given cell E, either the emf of the auxiliary7 battery (between A and B) should be increased or a suitable resistance should be put in series with R and X (so as to decrease the potential drop across the wire AB).

Question 24.
The figure shows a 2.0 V potentiometer used for the determination of internal resistance
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 34
1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of
9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.
Answer:
NCERT Solutions for Class 12 Physics Chapter 3 Current Electricity 35

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