NCERT Solutions for Class 6 English Honeysuckle Chapter 9 Desert Animals

NCERT Solutions for Class 6 English Honeysuckle Chapter 9 Desert Animals are part of NCERT Solutions for Class 6 English. Here we have given NCERT Solutions for Class 6 English Honeysuckle Chapter 9 Desert Animals.

Board CBSE
Textbook NCERT
Class Class 6
Subject English
Chapter Chapter 9
Chapter Name Desert Animals
Number of Questions Solved 7
Category NCERT Solutions

NCERT Solutions for Class 6 English Honeysuckle Chapter 9 Desert Animals

TEXTUAL QUESTIONS
(Page 117)
Working with the Text

A.
Question 1.
Talk to your partner and say whether the following statements are true or false.

  1. No animal can survive without water.
  2. Deserts are endless sand dunes.
  3. Most snakes are harmless.
  4. Snakes cannot hear, but they can feel vibrations through the ground.
  5. Camels store water in their humps.

Solution:

  1. True
  2. False
  3. True
  4. True
  5. False

Question 2.
Answer the following questions.

  1. How do desert animals survive without water ? (1)
  2. How do mongooses kill snakes ? (6)
  3. How does the hump of the camels help them to survive when there is no water ? (9)

Solution:

  1. The desert animals have to find different ways to survive without water. Some animals like gerbils spend the hottest part of the day in cool underground burrows. Some like darkling beetles catch drops of moisture on their legs. Some like camels get the necessary water from the desert plants they eat.
  2. The reactions of mongooses are so fast that they can dodge each time the snake strikes. They continually make a nuisance of themselves until after a while when the snake gets tired, they quickly dive in for a kill.
  3. The humps help the camels to survive by acting as storage containers. These humps are full of fat. The fat nourishes the camels in the absence of food and water.

Question :
B. Read the words/phrases in the box. With your partner find their meaning in the dictionary. Fill in the blanks in the following passage with the above words/ phrases.
NCERT Solutions for Class 6 English Honeysuckle Chapter 9 Desert Animals image 1
All animals in forests and deserts struggle to ____________ in ____________ Though most of the animals are ____________ , some are dangerous when . If an ____________ is noticed, they attack or bite to save themselves. They struggle ____________ for food and water. Some animals are called ____________ because they ____________ on other animals.

Solution:
Word                                 Meaning
harsh                                     hard
conditions                         situations
harmless                                safe
survive                             remain alive
intruder                          unwanted arrival
threatened                        feel unsafe
predators                             hunters
prey                                        hunt
continually                         all the time

All animals in forests and deserts struggle to survive in harsh conditions. Though most of the animals are harmless, some are dangerous when threatened. If an intruder is noticed, they attack or bite to save themselves. They struggle continually for food and water. Some animals are called predators because they prey on other animals.

Speaking

Question :
Look at these sentences.

  • Deserts are the driest places on earth.
  • Gerbils spend the hottest part of the day in cool underground burrows.

Now form pairs. Ask questions using a suitable form of the word in brackets. Try to answer the questions too.
Do you know

1. Which animal is the ____________ (tall) ?
2. Which animal runs the ____________ (fast) ?
3. Which place on earth is the _ (hot) or the ____________ (cold) ?
4. Which animal is the ____________ (large) ?
5. Which is the ____________ (tall) mountain in the world ?
6. Which is the ____________ (rainy) place on earth ?
7. Which is the ____________ (old) living animal ?
Can you add some questions of your own ?

Solution:

1. tallest
2. fastest
3. hottest … coldest
4. largest
5. tallest
6. most rainy
7. oldest

Students should try to add their own questions to this list.

Thinking about Language

Question :
A. Look at these sentences.

  • Most snakes are quite harmless, but a few are poisonous.
  • Most snakes lay eggs, but the rattlesnake gives birth to its young.

Now write five sentences like these using ‘most’ and the clues below.

  1. (90% of) people are honest (10%) are dishonest.
    __________________________________________________
  2. (Lots of) fruit have plenty of sugar, (some) citrus fruit are low in sugar.
    __________________________________________________
  3. (Every soft drink except this one) has lots of empty calories’.
    __________________________________________________
  4. (The majority of) films are romances, (a few) are on other topics.
    __________________________________________________
  5. (A majority of) people agree that he is a good leader, (just a few) disagree.
    __________________________________________________

Solution:

1. Most people are honest, but a few are dishonest.
2. Most fruit have plenty of sugar but citrus fruit are low in sugar.
3. Most soft drinks have lots of empty calories but this one is free from them.
4. Most films are romances but a few are on other topics.
5. Most of the people agree that he is a good leader but just a few disagree.

Question :
B. Look at these sentences.

  • Animals cannot survive for long without water.
  • So desert animals have to find different ways of coping.

The first sentence says what cannot happen or be done ; the second tells us what must, therefore, be done, what it is necessary to do. Complete these sentences using cannot and have to/has to.

1. You ____________ reach the island by land or air ; you go by boat.
2. We ____________ see bacteria with our eyes ; we, look at them through a microscope.
3. He ____________ have a new bicycle now ; he ____________ wait tili next year.
4. Old people often ____________ hear very well ; they ____________ use a hearing aid.
5. Road users ____________ do what they wish ; they ____________ follow the traffic rules.
6. She ____________ accept this decision ; she ____________ question it.
7. you ____________ believe everything you hear ; you ____________ use your own judgement.

Solution:

1. You cannot reach the island by land or air ; you have to go by boat.
2. We cannot see bacteria with our eyes ; we have to look at them through a microscope.
3. He cannot have a new bicycle now ; he has to wait till next year.
4. Old people often cannot hear very well ; they have to use a hearing aid.
5. Road users cannot do what they wish ; they have to follow the traffic rules.
6. She cannot accept this decision ; she has to question it.
7. You cannot believe everything you hear ; you have to use your own judgement.

Writing

Question :
Imagine you are journeying through a desert. Write a couple of paragraphs describing what you see and hear.

Solution:
Once I happened to pass through the deserts of Rajasthan. No vehicle like motor car could take me through the desert and so I had to accept the offer of a camel ride. It was my first chance to ride a camel. In the beginning, it was a little difficult but I got used to it.

We started in the evening because the camel-man told me that during the day the dust storm may become unbearable. He also asked me to fill my bottle with water because for the next six hours, I could not expect to get any water. So equipped with water and other necessary things, I rode the camel determined to cross the small part of the desert and reach the other side to a village. On the way, I saw many duns. Fortunately, the weather remained fine and the night seemed extremely beautiful. Nothing untoward happened on the way and I reached the village before it was dawn.

We hope the NCERT Solutions for Class 6 English Honeysuckle Chapter 9 Desert Animals help you. If you have any query regarding NCERT Solutions for Class 6 English Honeysuckle Chapter 9 Desert Animals, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Social Science History Chapter 6 Colonialism and the City

NCERT Solutions for Class 8 Social Science History Chapter 6 Colonialism and the City

These Solutions are part of NCERT Solutions for Class 8 Social Science.Here we have given. NCERT Solutions for Class 8 Social Science History Chapter 6 Colonialism and the City

Question 1.
State whether True or False:

  1. In the western world; modern cities grew with industrialization. True
  2. Surat and Machlipatnam developed in the nineteenth century. False
  3. In the twentieth century, the majority of Indians lived in cities. False
  4. After 1857 no worship was allowed in the Jama Masjid for five years. True 
  5. More money was spent on cleaning Old Delhi than New Delhi. False

Question 2.
Fill in the blanks:

  1. The first structure to successfully use the dome was called the Gol Gumbaz   
  2. The two architects who designed New Delhi and Shahjahanabad were
    Edward Lutyens and Herbert Baker
  3. The British saw overcrowded spaces as unhygienic  
  4. In 1888 an extension scheme called the Lahore gate improvement scheme was devised.

Question 3.
Identify three differences in the city design of New Delhi and Shahjahanabad.
Answer:
NCERT Solutions for Class 8 Social Science History Chapter 6 Colonialism and the City 1

Question 4.
Who lived in the “white” areas in cities such as Madras?
Answer:
The British lived in white areas of the cities such as Madras.

Question 5.
What is meant by de-urbanization?
Answer:
De-urbanisation is a process by which more and more people began to live in villages or rural areas.
In the late 18th century, Calcutta, Bombay and Madras emerged as Presidency cities. They became the centres of British power in different regions of India. At the same time, several smaller cities declined. Old trading centres and ports could not survive when the flow of trade shifted to new centres. Similarly, earlier centres of regional power collapsed with the defeat of local rurals by the British and new centres of administration grew. This process is described as de-urbanization.

Question 6.
Why did the British choose to hold a grand Durbar in Delhi although it was not the capital?
Answer:
During the Revolt of 1857, the British had realised that the Mughal emperor was still important to the people and they saw him as their leader. It was therefore important to celebrate British power with pomp and show in Delhi— the city the Mughal emperors had ruled earlier. The British thought that by doing this they would acknowledge people about their power and authority.

Question 7.
How did the Old City of Delhi change under British rule?
Answer:
The British changed the Old City of Delhi entirely. They wanted Delhi to forget its Mughal past. Hence, the area around the Fort was completely cleared of gardens, pavilions, and mosques. They either destroyed, the mosques or put them to other uses. For example, the Zinat-al-Masjid was converted into a bakery. No worship was allowed in the Jama Masjid for five years.
One-third of the city was demolished and its canals were filled up.

In the 1870s, the western walls of Shahjahanabad were broken to establish the railway and to allow the city to expand beyond the walls. The British began living in the sprawling Civil Lines area that came up in the north, away from the Indians in the Walled City. The Delhi College was turned into a school and shut down in 1877.

Question 8.
How did the Partition affect life in Delhi?
Answer:
1. Partition of India into India and Pakistan in. 1947 led to a massive transfer of populations on both sides of the new border.
2. Partition led to fierce rioting. Muslims left Delhi for Pakistan and Hindu and Sikh refugees came from Pakistan.

  • The population of Delhi swelled, jobs changed and culture became different.
  • Delhi became a city of refugees. Nearly 500,000 people were added to Delhi’s population and in 1951 this addition was little over 8,00,000.
  • Most of these migrants were from Punjab.
  • They stayed in camps, schools, military barracks, gardens etc.
  • New colonies like Lajpat Nagar and Tilak Nagar came up at this time.

3. Skills and professions of the refugees were different from those they replaced. Lives and occupations of people changed.
4. Social fabric of Delhi changed. Urban culture based on Urdu was overshadowed by new tastes in food, arts and dresses.

Question 9.
Find out the history of the town you live in or of any town nearby. Check when and how it grew, and how it has changed over the years. You could look” at the history of the bazaars, the buildings, cultural institutions, and settlements.
Answer:
Monu Nagar (An Imaginary Town):
1. Monu Nagar was a small village along G.T. Road.
2. People in the village lived a simple life, mostly of the agriculturist.
3. Slowly modern life style came.

  • People started shops along the road.
  • Some started repairs of vehicles, cycles, scooters and agriculture implements etc.

4. Agriculture as occupation declined.
5. Several schools, a college and health center developed over years.
6. Some air-conditioned restaurants were opened during last five years.
7. It has become a big town.

Question 10.
Make a list of at least ten occupations in the city, town or village to which you belong, and find out how long they have existed. What does this tell you about the changes within this area?
Answer:
NCERT Solutions for Class 8 Social Science History Chapter 6 Colonialism and the City 2

Some of these professions changed due to demographic changes. The changes were gradual. This showed that occupational changes bring a cultural and social change. Write to yourself about changes with the help of your teacher.

Objective Type Questions

1. Match the following:
NCERT Solutions for Class 8 Social Science History Chapter 6 Colonialism and the City 3
Answer:
(i)    c
(ii)   e
(iii)  f
(iv)  a
(v)   d
(vi)  b
(vii) g

2. State whether True or False:

  1. The British lived in white areas of the cities. True
  2. The British wanted Delhi to forget its Mughal past. True
  3. The Jama-Masjid was converted into a Bakery by the British. False
  4. In 1877, Queen Victoria was recognized as the Empress of India, True
  5. Lakpre gate improvement scheme was devised in the year 1905. False

3. Fill in the blanks:

  1. In the 1870s, the western walls of Shahjahanabad were broken to establish the railways.
  2. The Mughal aristocracy in the 17th and 18th century lived in Havelis 
  3. A haveli (is) housed by many families
  4. The central dome of the Viceroy’s Palace was copied from the Buddhist Stupa 
  5. The British exiled Bahadur Shah Zafar to Burma (now Myanmar)

 Multiple Choice Questions

Choose the correct answer:
1. Which of the following was a manufacturing town?

(a) Madurai
(b) Dacca
(c) Surat
(d) Agra

2. Which of the following city was Not developed as a Presidency city in colonial India?
(a) Agra  
(b) Bombay
(c) Madras
(d) Calcutta

3. How many Delhi Muslims migrated in the 1947 partition of India?
(a) Over two-third of the Delhi Muslims

(b) Over one-third of the Delhi Muslims
(c) Over three-fourth of the Delhi Muslims
(d) None of the above

4. Most of the migrants in Delhi were from
(a) Bengal
(b) Assam
(c) Punjab
(d) Rajasthan

5. Which was NOT the new British Port in the late eighteenth century?
(a) Bombay
(b) Machlipatnam
(c) Madras
(d) Calcutta

6. Which region was NOT de-urbanized in the 19th century?
(a) Machlipatnam
(b) Surat
(c) Seringapatam
(d) Bombay

7. Which was NOT the place of East India Company’s ‘factories’?
(a) Calcutta
(b) Surat
(c) Madras
(d) Delhi

8. When did the British gain control of Delhi?
(a) 1800
(b) 1803
(c) 1805
(d) 1810

9. When did Delhi become the capital of British India?
(a) 1900
(b) 1905
(c) 1911
(d) 1915

We hope the NCERT Solutions for Class 8 Social Science History Chapter 6 Colonialism and the City, help you. If you have any query regarding . NCERT Solutions for Class 8 Social Science History Chapter 6 Colonialism and the City, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Social Science History Chapter 8 Civilising the “Native”, Educating the Nation

NCERT Solutions for Class 8 Social Science History Chapter 8 Civilising the “Native”, Educating the Nation

These Solutions are part of NCERT Solutions for Class 8 Social Science.Here we have given. NCERT Solutions for Class 8 Social Science History Chapter 8 Civilising the “Native”, Educating the Nation

Question 1.
Match the following:
NCERT Solutions for Class 8 Social Science History Chapter 8 Civilising the Native, Educating the Nation 1
Answer:

NCERT Solutions for Class 8 Social Science History Chapter 8 Civilising the Native, Educating the Nation 2

Question 2.
State whether True or False:

  1. James Mill was a severe critic of the Orientalists. True
  2. The 1854 Despatch on education was in favour of English being introduced as a medium of higher education in India. True
  3. Mahatma Gandhi thought that promotion of literacy was the most important aim of education. False
  4. Rabindranath Tagore felt that children ought to be subjected to strict discipline. False

Question 3.
Why did William Jones feel the need to study Indian history, philosophy, and law?
Answer:

  1. In order to understand India, it was necessary to discover the sacred and legal texts that were produced in the ancient period.
  2. Only those texts could reveal the real ideas and laws of the Hindus and Muslims and only a new study of these texts could form the basis of future development in India.
  3. He believed that this project would not only help the British learn from Indian culture, but it would also help Indians rediscover their own heritage and understand the lost glories of their past.
  4. In this process, the British would become the guardians of Indian culture as well as its masters.

Question 4.
Why did James Mill and Thomas Macaulay think that European education was essential in India?
Answer:
Both James Mill and Thomas Macaulay saw India as an uncivilized country that needed to be civilized. And for this purpose, European education Was essential. They felt that knowledge of English would allow Indians to read some of the finest literature of the world, it would make them aware of the developments in Western science and philosophy. teaching English could thus be a way of civilizing people, changing their tastes, values, and culture.

Question 5.
Why did Mahatma Gandhi want to teach children handicrafts?
Answer:

  1. According to him, this would develop a person’s mind and soul.
  2. Simply, learning to read and write by itself does not count as education. So, people had to work with their hands, learn a craft, and know-how different things operated. This would develop their mind and their capacity to understand.

Question 6.
Why did Mahatma Gandhi think that English education had enslaved Indians?
Answer:
Mahatma Gandhi was dead against English education. He argued that this type of education had created a sense of inferiority in the minds of Indians. It had made them see Western civilization as superior and had destroyed the pride they had in their own culture. It had cast an evil spell on Indians. Education in English had crippled them, distanced them from their own surroundings, and made them strangers in their own lands. What is more, it had enslaved them.

Question 7.
Find out from your grandparents about what they studied in school.
Answer:

  • Urdu/Hindi language
  • Mathematics
  • The social study, Drawing.

Question 8.
Find out about the history of your school or any other school in the area you live.
Answer:
History of our school

  • Established as a middle school — Organised in tents.
  • No furniture.
  • Supplied furniture by Government.
  • Rooms got constructed.
  • Raised to secondary than to senior school.
  • After 10 years Pucca building got constructed.
  • All the amenities provided.
  • Now a full-fledged and flourishing Sarvodaya Bal Vidyalaya upto 12th standard.

Objectives Type Questions

1. Match the following:
NCERT Solutions for Class 8 Social Science History Chapter 8 Civilising the Native, Educating the Nation 3
NCERT Solutions for Class 8 Social Science History Chapter 8 Civilising the Native, Educating the Nation 4
Answer:
(i)  e
(ii)  f
(iii) a
(iv) b
(v)  d
(vi) c

2. State whether True or False:

  1. Mahatma Gandhi was the promotor of the English language. False
  2. William Jones had respect for ancient cultures. True
  3. Thomas Macaulay thought that European education was necessary for India. True
  4. William Carey had an appointment as a Supreme Court Judge. False
  5. Hindu College was set up at Banaras to encourage the study of ancient Hindi texts, False
  6. William Adam was a Scottish missionary True

3. Fill in the blanks:

  1. Mahatma Gandhi favoured Indian languages as a medium of instruction.
  2. William Jones had respect for Indian ancient cultures.
  3. Charles Wood emphasised the practical benefits of a system of European
  4. Rabindra Nath Tagore started the Santiniketan in 1901.
  5. According to Adam’s report, there were over 1 lakh Pathshalas in Bengal and Bihar.

Multiple Choice Questions

Choose the correct answer:

1. William Jones was a linguist because
(a) he had studied Greek and Latin
(b) he knew French and English
(c) he had learned Persian
(d) all of these

2. Who set up the Asiatic Society of Bengal?
(a) William Jones
(b) Henry Thomas Colebrooke
(c) Nathaniel Halhed
(d) All of these

3. Madrasa was set up in, Calcutta in the year
(a) 1750
(b) 1761
(c) 1771
(d) 1781

4. According to whom, “English education had enslaved Indians”?
(a) Rabindranath Tagore
(b) Mahatma Gandhi
(c) Subhas Chandra Bose
(d) Aacharya Vinoba Bhave

5. The Education Act was introduced in the year
(a) 1850
(b) 1835
(c) 1910
(d) 1900

6. Asiatick Researches (Journal) was NOT started by
(a) Henry Thomas Colebrooke
(b) Henry Thomas
(c) Nathaniel Halhed
(d) William Carey

7. Study of which of the following was NOT the purpose of setting up Madrasa in Calcutta in 1781?
(a) Arabic
(b) Sanskrit
(c) Persian
(d) Islamic laws

8. Who was Charles Wood?
(а) The President of the Board of Control of the Company

(b) Commissioner of the Board of Control of the Company
(c) An Educationist
(d) None of the above

9. The English Education Act was passed
(a) to materialize Macaulay’s thinking
(b) to make English the medium of instruction for higher education
(c) to stop the promotion of oriental institutions
(d) all of the above

10. What type of school did Tagore want to set up?
(a) Where the child was happy
(b) Where he/she could be free and creative
(c) He/she was able to explore her own thoughts and desire
(d) All of the above

11. Who said this “Education means all-round drawing out of the best in child and man-body, mind and spirit”?
(a) Rabindranath Tagore
(b) Mahatma Gandhi
(c) Swami Dayanand Saraswati
(d) None of these

We hope the NCERT Solutions for Class 8 Social Science History Chapter 8 Civilising the “Native”, Educating the Nation, help you. If you have any query regarding NCERT Solutions for Class 8 Social Science History Chapter 8 Civilising the “Native”, Educating the Nation, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter

NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter

NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter

Board CBSE
Textbook NCERT
Class Class 12
Subject Physics
Chapter Chapter 5
Chapter Name Magnetism and Matter
Number of Questions Solved 25
Category NCERT Solutions

Question 1.
Answer the following questions regarding earth’s magnetism :
(a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.
(b) The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain ?       (C.B.S.E. 1995)
(c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground ?
(d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole ? (C.B.S.E. 1995)
(e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of
magnetic moment 8 x 1022 JT_1 located at its center. Check the order of magnitude of this number in some way.
(f) Geologists claim that besides the main magnetic N­S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all ?
Answer:

(a) Magnetic elements

  • Declination
  • Dip and
  • Horizontal intensity

(b) Greater in Britain (it is about 70°), because Britain is closer to the magnetic north pole.

(c) Field lines of B due to the earth’s magnetism would seem to come out of the ground.

(d) Compass needle can move only in the horizontal plane. Since the field is entirely vertical no direction is shown by the needle.
(e) Using the formula for magnetic field on the equatorial line of a magnetic dipole i.e.
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 1
This value tells the order of magnitude of magnetic field of earth.
(f) Geologists are correct to think so because it is an approximation to consider the magnetic field of earth to be a single dipole field. The magnetised mineral deposits can be treated as local dipoles on earth.

Question 2.
Answer the following questions :
(a) The earth’s magnetic field varies from point to point in space. Does it also change with time ? If so, on what time scale does it change appreciably ?
(b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why ?
(c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e. the source of energy) to sustain these currents ?
(d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past ?
(e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion ?(f) Interstellar space has an extremely weak magnetic field of the order of 1012 T. Can such a weak field be of any significant consequence ? Explain.
Answer:

(a) Yes, it changes with time. After a few hundred years, the earth’s magnetic field undergoes an appreciable change.

(b) The temperature inside the earth is so high that it is impossible for the iron to remain as a magnet and act as a source of the magnetic field. The magnetic field due to the earth is considered to be due to the circulating electric currents induced in the iron in the molten state and other conducting materials inside the earth.

(c) A possible explanation can be the phenomenon of radioactivity.

(d) Analysis of the rock magnetism /earth’s magnetic field gets recorded in certain rocks during solidification, (through weekly) provides clues to the geomagnetic history.

(e) At large distances, the earth’s magnetic field gets modified by the fields produced by the motion of ions in the earth’s ionosphere.

(f) At very-very large distances like interstellar distances the small fields can significantly affect the charged particles like that of cosmic rays. For small distances, the deflections are not noticeable for small fields but at very large distances the deflections are significant.
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 2NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 2
clearly small value of B gives a very large value of radius R

Question 3.
A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 X 10-2 J. What is the magnitude of magnetic moment of the magnet ?
Answer:
Using τ- MB sin θ, we get
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 3

Question 4.
A short bar magnet of magnetic moment m = 0.32 JT-1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium ? What is the potential energy of the magnet in each case ?
Answer:
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 4

Question 5.
A closely wound solenoid of 800 turns and area of cross section 2.5 x 10-4 m2 carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment ?
Answer:
When current is passed through the solenoid, the magnetic field is produced along with its axis. The magnetic field lines emanate from one end and enter the other just as in the case of a bar magnet. The two ends of the solenoid act as the two poles of a bar magnet.
Here, the number of turns in the solenoid = 800
I = 3A
A = 2.5 x 10-4m2
The magnetic moment of the solenoid,
M = (IA) x number of turns
= 3 x 2.5 x 10-4 x 800
= 0.6 Am2

Question 6.
If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of the torque on the solenoid when its axis makes an angle of 30° with the direction of the applied field ?
Answer:
Using x = MB sin θ, we get
x = 0.6 x 0.25 x sin 30
= 0.6 x 0.25 x \(\frac { 1 }{ 2 } \)
= 0.3 x 0.25 = 0.075 Nm
= 7.5 x 10-2 Nm.

Question 7.
A bar magnet of magnetic moment 1.5 JT-1 lies aligned with the direction of a uniform magnetic field of 0.22T.
(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment, (i) normal to the field direction, (ii) opposite to the field direction ?
(b) What is the torque on the magnet in cases (i) and (ii)?
Answer:
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 5

Question 8.
A closely wound solenoid of 2000 turns and area of cross-section 1.6 X 10-4 m2, carrying a current of 4.0 A, is suspended through its center allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid ?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 x 10-2 T is set up at an angle of 30° with the axis of the solenoid?
Answer:
N =2000, A= 1.6 x 10-4m2, I = 4.0 A
(a) m = ANI = 1.6 x 10-4 x 2000 x 4.0
= 1.28 Am2, along the axis

(b) B = 7.5 x 10-2T, θ = 30°
Net force = 0
τ = mB sin θ = 1.28 x 7.5 x 10-2 x sin 30
= 0.64 x 7.5 x 10-2
= 4.800 x 10-2 Nm
By the action of this, the solenoid can come to the direction of the external field.

Question 9.
A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 x 10-2 T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s_1. What is the moment of inertia of the coil about its axis of rotation ?
Answer:
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 6
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 7

Question 10.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.
Answer:
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 8
Question 11.
At a certain location in Africa, a compass points 12° west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60° above the horizontal. The horizontal component of the earth’s field is measured to be 0.16 G. Specify the direction and magnitude of the earth’s field at the location.
Answer:
Using BH = B cos δ, we get
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 9

Direction of B is 12° west of geographic meridian making upward angle of 60° with horizontal.

Question 12.
A short bar magnet has a magnetic moment of 0.48 JT-1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the center of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.
Answer:
On axial line
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 10

Question 13.
A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the center of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null-points (i.e., 14 cm) from the center of the magnet ? (At null points, Held due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)
Answer:
Magnetic field at the equatorial line of the magnet is given
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 11

Question 14.
If the bar magnet in Exercise 5.13 is turned around by 180°, where will the new null-points be located ?
Answer:
When magnet is turned around 180°, its south pole lies in the geographical south direction. Hence null point will lie on the equatorial line at a distance x from die center of the magnet.
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 12

Question 15.
A short bar magnet of magnetic moment 5.25 x 10-2 JT-1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the center of the magnet, the resultant field is inclined at 45° with the earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Answer:
Normal bisector
(a) Let resultant magnetic field of a magnet at point P makes an angle θ= 45° with the earth’s field. Therefore,
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 13
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 14
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 15
Question 16.
Answer the following questions:
(a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled? (C.B.S.E. 1991)
(b) Why is diamagnetism, in contrast, almost independent of temperature?
(c) If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?
(d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?
(e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point). Why?
(f) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet ?
Answer:
(a) When cooled, the tendency of the thermal agitation to disrupt the alignment of magnetic dipoles decreases in the case of paramagnetic materials. Hence they display greater magnetisation.

(b) The atoms of a diamagnetic do not have an intrinsic magnetic dipole moment. On placing a diamagnetic sample in a magnetic field, the magnetic moment of the sample is always opposite to the direction of the field. It is not affected by the thermal motion of the dipoles.

(c) Since bismuth is diamagnetic, the field in the core coil be sightly less than that when a core is empty.

(d) Permeability of a ferromagnetic material depends on applied magnetic field. Permeability is more for lower magnetic field.

(e) One of the reasons for this fact is that when a material has µr > > 1, the field lines meet the material nearly normally.

(f) Yes, a paramagnetic sample with saturated magnetisation will have the same .order of magnetisation as the magnetisation of a ferromagnetic substance. However, the saturated magnetisation will require magnetising field too high to achive. Further, there may be a minor difference in the strengths of the atomic dipoles of paramagnetic and ferromagnetic materials.

Question 17.
Answer the following questions:
(a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.
(b) The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetization, which piece will dissipate greater heat energy ?
(c) ‘A system displaying a hysteresis loop such as a ferromagnet is a device for storing memory.’ Explain the meaning of this statement.
(d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for budding ‘memory stores’ in a modern computer ?
(e) A certain region of space is to be shielded from magnetic fields. Suggest a method.
Answer:
A piece of carbon steel will dissipate a greater amount of heat energy as its hysteresis loop has a greater area.
The magnetisation produced in a ferromagnet does not have a unique value corresponding to the applied magnetizing field.

In addition, the magnetisation produced depends on the history of the magnetisation i.e. the number of cycles of magnetisation, it has been taken through. In other words, the value of magnetisation of a ferromagnet is a record or memory of its magnetisation. If information bits can be made corresponding to the cycles of magnetization, the system displaying the hysteresis loop of the ferromagnet can act as a device for storing the information.

  • ceramics are used for coating magnetic tapes in a cassette player or for building memory stores in a modem computer. Ceramics are specially treated barium iron oxides and are also called ferrates.
  • The shielding of the region can be done by surrounding it with soft iron rings. The magnetic field lines will be drawn into the rings and the enclosed region will become free of the magnetic field.

Question 18.
A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west to 10° north of east. The magnetic meridian of the place happens to be 10° west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (Ignore the thickness of the cable). (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)
Answer:
Let neutral point lies at a distance x from the cable. Now, at neutral point, magnetic field due to cable is equal in magnitude and opposite in direction of the earth’s magnetic field.
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 16

Question 19.
A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?
Answer:
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 17
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 18

Question 20.
A compass needle free to turn in a horizontal plane is placed at the center of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.
(a) Determine the horizontal component of the earth’s magnetic field at the location.
(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the place to be zero.
Answer:
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 19

Question 21.
A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 x 10-2 T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field ?
Answer:
Here B, = 1.2 X 10-2 T,θ1= 15°, θ2 = 45°.
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 20
The dipole will be in equilibrium, if torque acting on dipole due to B1 is equal and opposite to the torque acting on dipole due to B2.
That is, MBsin = MB2 sin θ2
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 21

Question 22.
A monoenergetic (18 keV) electron beam intially in the horizontal direction is subjected to a horizontal magnetic field of 0.40 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me = 9.11 x 10-19 Q.
[Note. Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]
Answer:
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 22
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 23

Question 23.
A sample of paramagnetic salt contains 2.0 x 1024 atomic dipoles each of dipole moment 1.5 x 10-23 JT-1. The sample is placed under a homogeneous magnetic field of 0.64 T and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K ? (Assume Curie’s law).
Answer:
Magnetic dipole moment of sample,
M = 15% of M (1.5 x 10-23) (2 x 1024) = 30 JT-1
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 24
Question 24.
A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetizing current of 1.2 A?
Answer:
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 25

Question 25.
The magnetic moment vectors μs and μl; associated with the intrinsic spin angular momentum S and orbital angular momentum Z, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by μs= -(e/m)S, μl= -(e/2m)l. Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
Answer:
NCERT Solutions for Class 12 Physics Chapter 5 Magnetism and Matter 26

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NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

NCERT Solutions for Class 12 Chemistry Chapter 3 provides detailed solutions for the questions asked in the textbook. The solutions are provided by subject experts and the students can refer to these to prepare well for the exams. NCERT Solutions are a guide to the students appearing in different boards like MP board, UP board, CBSE, Gujarat board, etc.

Chemistry Class 12 Chapter 3 Electrochemistry is very important from the examination point of view. All the analytical and conceptual details are provided in the NCERT Solutions Class 12 Chapter 3 that will help the students to score well.

Board CBSE
Textbook NCERT
Class Class 12
Subject Chemistry
Chapter Chapter 3
Chapter Name Electrochemistry
Number of Questions Solved 33
Category NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry

The production of electricity from chemical reactions and the use of electrical energy to bring non-spontaneous chemical transformations is known as electrochemistry. Both the theory and the practical portions are very important in this chapter.

Class 12 Chemistry chapter 3 Electrochemistry explains different types of cells and the differences between the two. Important concepts such as Gibb’s energy and equilibrium constant are also discussed here.

NCERT IN-TEXT QUESTIONS

Question 1.
How would you determine the standard electrode potential of the system ; Mg2+/Mg ?
Answer:
n order to determine E° value of Mg2+/Mg electrode, an electrochemical cell is set up in which a Mg electrode dipped in 1 M MgSO4 solution acts as one half cell (oxidation half cell) while the standard hydrogen electrode acts as the other half cell (reduction half cell). The deflection of voltmeter placed in the cell circuit is towards the Mg electrode indicating the flow of current. The cell may be represented as :
Mg/Mg2+ (1 M) || H+(l M)/H2(1 atm), Pt
The reading as given by voltmeter gives \({ E }_{ cell }^{ \circ }\)
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 1
The expected value of standard electrode potential (E°) = -2-36 V.

Question 2.
Can you store copper sulphate solution in a zinc pot ?
Answer:
No, it is not possible. The E° values of the copper and zinc electrodes are as follows :
Zn2+(aq) + 2e → Zn(s) ; E° = – 0·76 V
Cu2+(aq) + 2e → Cu(s) ; E° = + 0·34 V
This shows that zinc is a stronger reducing agent than copper. It will lose electrons to Cu2+ ions and redox reaction will immediately set in.
Zn(s) + Cu2+ (aq) → Zn2+(aq) + Cu(s)
Thus, copper sulphate solution cannot be stored in zinc pot.

Question 3.
Consultthe table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
Answer:
Oxidation of Fe2+ converts it to Fe3+, i.e.,Fe2+ –>Fe3+ +e ; E°ox= – 0.77 V Only those substances can oxidise Fe2+ to Fe3+ which are stronger oxidizing agents and have positive reduction potentials greater than 0.77 V, so that EMF of the cell reaction is positive. This is so for elements lying below Fe3+/Fe2+ in the series ex: Br2, Cl2 and F2.

Question 4.
Calculate the potential of hydrogen electrode in contact with a solution with pH equal to 10.
Answer:
For hydrogen electrode, H+ + e → 1/2H2
Applying Nernst equation,
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 2

Question 5.
Calculate e.m.f. of the cell in which the following reaction takes place
Ni(s) + 2Ag+(0·002M) → Ni2+(0·160M) + 2Ag(s) Given that \({ E }_{ cell }^{ \circ }\) = 1.05 V. (C.B.S.E. Outside Delhi 2015)
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 3

Question 6.
The cell in which the following reaction occurs
2Fe3+(aq) + 2I(aq) → 2Fe2+(aq) + I2(s) has \({ E }_{ cell }^{ \circ }\) = 0-236 V at 298 K.
Calculate standard Gibbs energy and equilibrium constant for the reaction.
Answer:
The two half reactions are :
2Fe3+ + 2e → 2Fe2+ and 2I → I2 + 2e
For the above reaction, n = 2
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 4

Question 7.
Why does the conductivity of a solution decrease with dilution?
Answer:
Conductivity of a solution is the conductance of ions present in a unit volume of the solutions. On dilution, no. of ions per unit volume decreases. Hence, the conductivity decreases.

Question 8.
Suggest a way to determine the \({ A }_{ m }^{ \circ }\) for water.
Answer:
Water (H2O) is a weak electrolyte. Its molar conductance at infinite dilution i.e., \({ A }_{ m }^{ \circ }\) can be determined in terms of \({ A }_{ m }^{ \circ }\) for strong electrolytes. This is in accordance with Kohlrausch’s Law.
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 5

Question 9.
The molar conductance of 0·025 mol L-1of methanoic acid is 46·15 cm2 mol-1. Calculate its degree of dissociation and dissociation constant. Given λ°(H+) = 349·6 S cm2 mol-1 and λ°(HCOO) = 54·6 S cm2 mol-1.
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 6

Question 10.
If a current of 0·5 ampere flows through a metallic wire for 2 hours, then how many electrons flow through the wire ?
Answer:
Quantity of charge (Q) passed = Current in amperes x Time in seconds = (0·5 A) X (2 x 60 x 60 s)
= 3600 As = 3600 C
No. of electrons flowing through the wire by passing a charge of one faraday (96500 C) = 6·022 x 1023
No. of electrons flowing through the wire by passing a charge of 3600 C
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 7

Question 11.
Suggest a list of metals that are extracted electrolytically.
Answer:
Na, Ca, Mg and Al

Question 12.
Consider the reaction :
Cr2\({ O }_{ 7 }^{ 2- }\) + 14H+ + 6e → 2Cr3+ + 7H2O.
What is the quantity of electricity in coulombs needed to reduce 1 mole of Cr2\({ O }_{ 7 }^{ 2- }\) ions ?
Answer:
The quantity of electricity in coulombs is 6 F or 6 x 96500 C = 5·76 x 105 C.

Question 13.
Write the chemistry of recharging the lead storage battery highlighting all the materials that are involved during recharging.
Answer:
Chemical reactions while recharging :
Chemical reactions while recharging :
2PbSO4 + 2H2O → PbO2 + Pb + 2H2SO4
Electricity is passed through the electrolyte PbSO4 which is converted into PbO2 and Pb.
Recharging is possible in this case because the PbSO4 formed during discharging is a solid and sticks to the electrodes. Therefore, it can either take up or give electrons during recharging.

Question 14.
Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Answer:
Methane and Methanol.

Question 15.
Explain how rusting of iron can be envisaged as the setting up of an electrochemical cell.
Answer:
Iron (Fe) is involved in the redox-reaction that is carried in the electrochemical cell which is set up. As a result, it slowly dissolves and the metal surface gets rusted or corroded.
The redox-reaction may be described as follows :
At anode: Fe (s) undergoes oxidation to release electrons
F2(s) → Fe2+(aq) + 2e–              ….(oxidation)
At cathode: The electrons which are released participate in the reduction reaction and combine with H+ ions released from carbonic acid (H2CO3) formed by the combination of CO2 and H2O present.
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 8

NCERT EXERCISE

Question 1.
Arrange the following metals in the order in which they displace each other from the solution of their salts: Al, Cu, Fe, Mg and Zn.
Answer:
Mg, Al, Zn, Fe, Cu, Ag.

Question 2.
Given the standard electrode potentials
K+/K = – 2·93 V, Ag+/Ag = 0·80 V
Hg2+/Hg = 0·79 V ; Mg2+/Mg = – 2·37V, Cr3+/Cr = – 0·74 V
Arrange these metals in increasing order of their reducing power.
Answer:
Less the electrode potential more will be the reducing power.
Ag < Hg < Cr < Mg < K.

Question 3.
Depict the galvanic cell in which the reaction
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place. Further show :
(i) which electrode is negatively charged ?
(ii) the carriers of the current in the cell.
(iii) individual reaction at each electrode. (C.B.S.E. Delhi 2008)
Answer:
The galvanic cell in which the given reaction takes place is depicted as:
Zn(s) | Zn2+ (aq) || Ag+ (aq) | Ag(s)
(i) Zn electrode (anode) is negatively charged
(ii) Tons are carriers of current in the cell and in the external circuit, current from silver to Zinc.
(iii) The reaction taking place at the anode is given by,
Zn(s) -H → Zn2+(aq) + 2e
The reaction taking place at the cathode is given
Ag++ e → Ag(s)

Question 4.
Calculate the standard cell potentials of galvanic cell in which the following reactions take place
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 9
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 10

Question 5.
Write the Nernst Equation and calculate e.m.f. of the following cells at 298 K :
(i) Mg(s) | Mg2+ (0·001 M) || Cu2+ (0·0001 M) | Cu(s) (C.B.S.E. Delhi 2008, 2013)
(ii) Fe(s) | Fe2+ (0·001 M) || H+ (1 M) | H2(g) (1 bar) | Pt(s)
(iii) Sn(s) | Sn2+ (0·050 M) || H+ (0·02 M) | H2(g) (1 bar) | Pt(s) (C.B.S.E. Outside Delhi 2013, 2015)
(iv) Pt(s) | Br2(l) | Br (0·010 M) || H+ (0·030 M) | H2(g) (1 bar) | Pt(s)
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 11
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 12

Question 6.
In the button cell widely used in watches and in other devices, the following reaction takes place:
Zn (s) + Ag2O (s) + H2O (l) → Zn2+ (aq) + 2Ag (s) + 2OH (aq)
Determine E° and ∆G° for the reaction. (C.B.S.B. Delhi 2005, Outside Delhi 2006 Supp., 2008, C.B.S.E. Sample Paper 2010)
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 13

Question 7.
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer:
The conductivity of a solution is defined as the conductance of a solution 1 cm in length and the area of cross-section cm2.1 is represented by K.

Conductivity always decreases with a decrease in concentration both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.

Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.

Molar conductivity increases with a decrease in concentration. This is because the total volume of the solution containing one mole of the electrolyte increases on dilution.

Question 8.
The conductivity of 0·20 M solution of KCl at 298 K is 0·0248 S cm-1. Calculate its molar conductivity.
(C.B.S.E. Delhi 2008, 2013)
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 14

Question 9.
The resistance of a conductivity cell containing 0·001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if the conductivity of 0·001M KCl solution at 298 K is 0·146 x 10-3 S cm-2? (C.B.S.E. Outside Delhi 2007, 2008, 2013)
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 18

Question 10.
The conductivity of sodium chloride solution at 298 K has been determined at different concentrations and results are given below :
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 19
Calculate molar conductivity for all the concentrations and draw a plot between \({ A }_{ m }^{ c }\) and \(\sqrt { c } \). Find the value \({ A }_{ m }^{ \circ }\) from the graph.
Answer:
\(\frac { 1S{ cm }^{ -1 } }{ 100S{ m }^{ -1 } } =1\) (unit conversion factor)
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 20
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 21
\({ A }_{ m }^{ \circ }\) can be obtained on extrapolation to zero concentration along Y-axis. It is 124·0Scm2mol-1.

Question 11.
The conductivity of 0·00241 M acetic acid is 7·896 x 10-5 S cm-1. Calculate the molar conductivity. If A° for acetic acid is 390·5 S cm2 mol-1, what is its dissociation constant? (C.B.S.E. Delhi 2008, C.B.S.E. Outside Delhi 2016)
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 22

Question 12.
How much charge is required for the reduction of :
(i) 1 mol of Al3+ to Al
(ii) 1 mol of Cu2+ to Cu
(iii) 1 mol of Mn\({ O }_{ 4 }^{ – }\) to Mn2+.
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 23

Question 13.
How much electricity in terms of Faraday is required to produce.
(i) 20.0 g of Ca from molten CaCl2?
(ii) 40.0 g of Al from molten Al203?
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 24

Question 14.
How much electricity is required in coulomb for the oxidation of
(i) 1 mol of H2O to O2
(ii) 1 mol of FeO to Fe2O3.
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 25
For the oxidation of two moles of FeO, charge required = 2 F
For the oxidation of one mole of FeO, charge required = 1 F = 96500 C.

Question 15.
A solution of Ni(NO3)2 is electrolyzed between platinum electrodes using a current of 5·0 ampere for 20 minutes. What weight of Ni will be produced at the cathode? (Atomic mass of Ni = 58·7). (Jharkhand Board 2009)
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 26

Question 16.
Three electrolytic cells A, B, and C containing electrolytes zinc sulphate, silver nitrate, and copper sulphate respectively, were connected in series. A steady current of 1·50 ampere was passed through them until 1·45 g of silver was deposited at the cathode of cell B. How long did the current flow? What weight of copper and of zinc were deposited? (Atomic mass of Cu = 63·5 ; Zn = 65·3; Ag = 108) (C.B.S.E. Outside Delhi 2008, Jharkhand Board 2010)
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 27
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 28

Question 17.
Predict if the reaction between the following is feasible:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 29
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 30
Answer:
The reaction is feasible if the EMF of the cell reaction is positive.
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 31

Question 18.
Predict the products of electrolysis of each of the following :
(i) An aqueous solution of AgNO3 using silver electrodes.
(ii) An aqueous solution of AgNO3 using platinum electrodes.
(iii) A dilute solution of H2SO4 using platinum electrodes. (C.B.S.E. Outside Delhi 2007)
(iv) An aqueous solution of CuCl2 using platinum electrodes. (C. B. S. E. Sample Paper 2010)
Answer:
(i) An aqueous solution of AgNO3 using silver electrodes :
Both AgNO3 and water will ionise in aqueous solution
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 32
At cathode: Ag+ ions with less discharge potential are reduced in preference to H+ ions which will remain in solution. As a result, silver will be deposited at cathode.
Ag+ (aq) + e → Ag (deposited)
At anode: An equivalent amount of silver will be oxidised to Ag+ ions by releasing electrons.
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 33
(ii) An aqueous solution of AgNO3 using platinum electrodes:

In this case, the platinum electrodes are the non-attackable electrodes. On passing current the following changes will occur at the electrodes.
At cathode: Ag+ ions will be reduced to Ag which will get deposited at the cathode.
At anode: Both \({ NO }_{ 3 }^{ – }\) and OH ions will migrate. But OH ions with less discharge potential will be oxidised in preference to \({ NO }_{ 3 }^{ – }\) ions which will remain in solution.
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 34
Thus, as a result of electrolysis, silver is deposited on the cathode while O2 is evolved at the anode. The solution will be acidic due to the presence of HNO3.
(iii) A dilute solution of H2SO4 using platinum electrodes:
On passing current, both acid and water will ionise as follows:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 35
At cathode:
H+ (aq) ions will migrate to the cathode and will be reduced to H2.
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 36
Thus, H2 (g) will be evolved at the cathode.
At anode: OH ions will be released in preference to \({ SO }_{ 4 }^{ 2- }\) ions because their discharge potential is less. They will be oxidized as follows:
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 37
Thus, O2 (g) will be evolved at the anode. The solution will be acidic and will contain H2SO4.
(iv) An aqueous solution of CuCl2 using platinum electrodes :
The electrolysis proceeds in the same manner as discussed in the case of AgNO3 solution. Both CuCl2 and H2O will ionise as follows :
NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry 38
At cathode:
Cu2+ ions will be reduced in preference to H+ ions and copper will be deposited at the cathode

Cu2+ (aq) + 2e → Cu (deposited)

At anode: C ions will be discharged in preference to OH ions which will remain in solution.

Cl → Cl + e; Cl + Cl → Cl2 (g)

Thus, Cl2 will be evolved at the anode.

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

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

NCERT Solutions for Class 12 History Chapter 14 Understanding Partition Politics, Memories, Experiences are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 14 Understanding Partition Politics, Memories, Experiences.

Board CBSE
Textbook NCERT
Class Class 12
Subject History
Chapter Chapter 14
Chapter Name Understanding Partition Politics, Memories, Experiences
Number of Questions Solved 9
Category NCERT Solutions

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

Question 1.
What did the Muslim League demand through its resolution of 1940?
Solution :
The resolution of 23 March 1940, demanded a measure of autonomy for the Muslim- majority areas of the subcontinent. It never mentioned partition or Pakistan. Sikandar Hayat Khan, who had drafted the resolution was the Punjab Premier and leader of the Unionist Party. He declared in the Punjab Assembly on 1 March, 1941 that he was opposed to a Pakistan that would mean “Muslim Raj here and Hindu Raj elsewhere… “If Pakistan means unalloyed Muslim Raj in the Punjab then I will have nothing to do with it”. He reiterated his plea for a loose (united) confederation with considerable autonomy for confederating units.

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

Question 3.
How did ordinary people view Partition?
Solution :
Ordinary people did not know what the Partition was and it would affect their lives in the future. They even did not know about the different areas of the subcontinent. Migrants thought they would return to their original place as soon as peace prevailed again.

Question 4.
What were Mahatma Gandhi’s arguments against Partition?
Solution :

  1. Mahatma Gandhi opposed the Partition by arguing that both Hindus and Muslims were bom of same soil and they had the same blood, ate the same food, drank the same water and spoke the same language. So, they were similar to each other.
  2. He stated that the demand for Pakistan put forward by the Muslim League was un- Islamic and sinful because Islam stands for the unity and brotherhood of mankind and not for disrupting the oneness of the human family. So, those who wanted Partition were enemies alike of Islam and India.

Question 5.
Why is Partition viewed as an extremely significant marker in South Asian history?
Solution :
The following reasons can be put forward for the given view:

  • The partition of India had a unique nature. This partition was based on religions. The partition took place in the name of the communities. History has never witnessed such type of partition.
  • The partition marked a severe violence. Innumerable people were killed. People began to kill each other irrespective of their earlier relation. Earlier they lived with each other in harmony and peace but now started to kill each other. Government machinery failed to check this.
  • People faced a lot of problems. Their life became miserable. Their near and dear ones were killed. Many people were abducted.
  • People moved across the border. Most of the Muslims of India crossed over to Pakistan and almost all Hindus and Sikhs came to India from Pakistan. They were forced to start their life afresh.
  • People lost all their movable and immovable property all of a sudden. They became homeless and forced to live in refugee camps.

Question 6.
Why was British India partitioned?
Solution :
Partition of India was not a sudden event because even in its resolution of March 1940, the Muslim League had only demanded a measure of autonomy for the Muslim majority areas of the subcontinent. It was a culmination of events such as communal politics that started developing in the opening decades of the twentieth century as mentioned below :

  1. Government of India Act 1909 and 1919 – The British Government granted separate electorate for Muslims in 1909. These were expanded in 1919. Separate electorates implied that Muslims could elect their own representatives in designated constituencies. Thus, religious identities were encouraged. Community identities no longer indicated simple difference in faith and belief: but they led to active opposition and hostilities between communities.
  2. Events during 1920s and 1930s – During the 1920s and 1930s, Muslims were agitated by the activities of the Hindus such as “music-before-mosque”, cow protection movement, and shuddhi movement of Arya Samaj. Similarly, Hindus were angered by the rapid spread of tabligh (propaganda) and tanzim (organisation). These activities led to riots at different places and deepened differences between two communities.
  3. The provincial elections of 1937 and the Congress ministries – In the elections of 1937, Congress did well but Muslim League failed poorly in the constituencies reserved for Muslims. The Muslim League wanted to form a joint government with the Congress in United Provinces where Congress had won an absolute majority. The Congress had, therefore, rejected the offer. This led to drifting away of the Muslim League but thereafter Muslim League doubled its efforts at expanding its social support.
  4. Policies of the Congress ministries – The Congress ministry in UP wanted to abolish landlordism which was supported by the Muslim League. The Congress also could gain much in its mass contact programme in UP. But its policies alarmed the conservative Muslims.
  5. Rise of Hindu Mahasabha and Rashtriya Swayamsevak Sangh – The rise of Hindu Mahasabha and Rashtriya Swayamsevak Sangh which had over 100,000 trained and highly disciplined cadres pledged to an ideology of Hindu nationalism, convinced Muslims that India was a land of the Hindus.

The above factors created differences between two communities but inspite of this fact remains that the Cabinet Mission (1946) plan that recommended a loose three-tier confederation was accepted by all the major parties. It was due to later developments such as ‘Direct Action da/ (16 August, 1946), riots and violence, fear of Sikh leaders and Congressmen in the Punjab and a section of bhadralok. Bengali Hindus in Bengal which compelled the Congress to accept the partition of the country.

Question 7.
How did women experience Partition?
Solution :
For women, partition was horrible. Women were raped, abducted and many times forced to live with strangers and start a new life. They were deeply traumatised and began to develop new family bonds in the changed circumstances.

Women became victims on both the sides of the border. They were forced to live in a strange circumstances. But the government officials of both the countries did not take any serious step to consult those women. Women were left on their fate.

They were even murdered by their own family members. When the men realized that the women of their family would fall into the hands of the enemy, they killed their women with their own hands. To escape from the hands of enemy, in a Sikh village, ninety women were said to have voluntarily jumped into a well.

Question 8.
How did the Congress come to change its views on Partition?
Solution :
Initially, the proposals of the Cabinet Mission were accepted by all the major political parties but due to differences over interpretation of the plan, neither the Congress, nor the League agreed to the Cabinet Mission’s proposal. Thereafter, following developments took place:

  • The Muslim League announced 16 August 1946 as “Direct Action Day” for winning its Pakistan demand.
  • “Direct Action Day” led to riots at Calcutta and other places.
  • At that time, many Sikh leaders and Congressmen in the Punjab were convinced that Partition was a necessary evil, otherwise they would be swamped by Muslim majority and Muslim leaders would dictate their terms to them.
  • Similarly, a section of bhadralok, Bengali Hindus, who wanted political power to remain with them, began to fear the “permanent tutelage of Muslims”. They were in a numerical minority so only a division of the province could ensure their political dominance.

Thus, under these circumstances, the Congress had no option except to agree to the Partition.

Question 9.
Examine the strengths and limitations of oral-history. How have oral-history techniques furthered our understanding of Partition ?
Solution :
The strengths and limitations of oral-history are as mentioned below :
(a) Strengths :

  • It helps us grasp experiences and memories in detail. It enables historians to write richly textured, vivid accounts of what happened to people during Partition.
  • Oral-history enables historians to broaden the scope of their discipline by writing experiences of the poor and the powerless who have been generally ignored in mainstream history.

(b) Limitations :

  • The oral-history lacks concreteness. Its chronology is not precise.
  • The uniqueness of personal experience makes generalisation difficult because a large picture cannot be built from micro-evidence and one witness is no witness.
  • Oral accounts deal with tangible issues. Small individual experiences are not relevant to unfold larger processes of history.

But inspite of above shortcomings the oral-history is important because it can be corroborated by other sources. The experiences of the people during Partition are significant and should be used to check other sources and vice-versa.

We hope the NCERT Solutions for Class 12 History Chapter 14 Understanding Partition Politics, Memories, Experiences help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 14 Understanding Partition Politics, Memories, Experiences, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital Vijayanagara

NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara.

Board CBSE
Textbook NCERT
Class Class 12
Subject History
Chapter Chapter 7
Chapter Name An Imperial Capital: Vijayanagara
Number of Questions Solved 9
Category NCERT Solutions

NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara

Question 1.
What have been the methods used to study the ruins of Hampi over the last two centuries ? In what way do you think they would have complemented the information provided by the priests of the Virupaksha temple ?
Solution :
(a) The methods used to study the ruins of Hampi over the last two centuries were as given below:

  • Colonel Colin Mackenzie, an engineer, antiquarian and an employee of the English East India Company discovered the ruins at Hampi in 1800. He prepared the first survey map of site.
  • From 1856, photographers began to record the monuments which enabled scholars to study them.
  • Since 1836, epigraphists began collecting several dozen inscriptions found at the temples at Hampi.
  • Thereafter the historians collated the information from above sources with accounts of foreign travellers and other literature written in Telugu, Kannada, Tamil and Sanskrit.

(b) The above methods would have complemented the information provided by the priests of the Virupaksha temple because that was based on the memories of the priests. The earlier information was corroborated by the inscriptions, photographs, maps and accounts of foreign travellers and other material.

Question 2.
How were the water requirements of Vijayanagara met ?
Solution :
The water requirements of Vijayanagara were met in the following ways :

  1. Its location is the natural basin formed by the river Tungabhadra which flows in a north-easterly direction. The stunning granite hills form a girdle around the city. A number of streams flow down to the river from these rocky outcrops.
  2. The embankments were built along these streams to create reservoirs of varying
    sizes.
  3. Tanks were built to store rainwater and conduct it to the city. The most important tank built is now called Kamalapuram tank. Water from this tank irrigated fields nearby as well as was also conducted through a channel to the “royal centre”.
  4. The Hiriya canal drew water from a dam across the Tungabhadra and irrigated the cultivated valley that separated the “sacred centre” from the “urban core”.

Question 3.
What do you think were the advantages and disadvantages of enclosing agricultural land within the fortified area of the city?
Solution :
Advantages of enclosing agriculture land within fortified area:
(i) It had an elaborate canal system which drew water from the Tungabhadra to provide irrigation facilities.
(ii) It enclosed agricultural tracts, cultivated fields, gardens, and forests.
(iii) This enclosure saved crops from being eaten by wild animals.
(iv) In the medieval period, sieges were laid to starve the defending armies into submission. These sieges lasted for many months or many years. So the rulers of Vijayanagara adopted and elaborated a strategy to protect the agricultural belt and built large granaries.

Disadvantages
(i) This system was very expensive.
(ii) During adverse, circumstances this system proved inconvenient to the farmers.
(iii) The farmers had to seek the permission of gate-keeper to reach their field.
(iv) If enemy encircled the field the farmer could not look after their field.

Question 4.
Figure given below is an illustration of another pillar from the Virupaksha temple. Do you notice any floral motifs? What are the animals shown? Why do you think they are depicted? Describe the human figures shown.
NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital Vijayanagara
Solution :
(a) There are many floral motifs.
(b) Horses and elephants have been shown in the pillar.
(c) The Vijayanagara kings competed with contemporary rulers including the Sultans of the Deccan and the Gajapati rulers of Orissa – for control of the fertile river valleys and resources generated by lucrative overseas trade. The kingdom remained in constant state of military preparedness. So, the kings paid attention to improve harbours and encouraged its commerce so that horses, elephants, precious gems, etc. are freely imported. Thus, due to the importance of horses and elephants in the warfare, these animals had been depicted on the pillars.
(d) The images of gods have been shown in the pillar. One devotee has also been shown before a Shiva linga.

Question 5.
What do you think was the significance of the rituals associated with the mahanavami dibba?
Solution :
The mahanavami dibba was one of the most impressive platforms in the “king’s palace”. It was located on one of the highest points in the city. Rituals associated with the structure probably coincided with mahanavami (literally the great ninth day) of the ten day Hindu festival during the autumn months of September and October, known variously as Dusehra (northern India), Durga Puja (in Bengal) and Navaratri or Mahanavami (in peninsular India). The Vijayanagara kings displayed their prestige, power and suzerainty on this occasion.

The ceremonies such as worship of the image, worship of the state horse, the sacrifice of buffaloes and other animals were performed on this occasion. Dances, wrestling matches, processions of caparisoned horses, elephants, chariots, soldiers and ritual presentations were held before the kings, guests, the chief nayakas were held. On the last day, the king inspected his army and the armies of the nayakas who brought rich gifts for the king as well as the stipulated tribute. Thus, there was great significance of the rituals associated with the mahanavami dibba.

Question 6.
Discuss whether the term “royal centre” is an appropriate description for the part of the city for which it is used.
Solution :
The term “royal centre” is an appropriate description for the part of the city for which it is used because the Royal center had more than 60 temples. Most of these temples were constructed by the ruler of Vijayanagara Empire to express their supremacy. The royal centre had 30 palaces. These were made of perishable material. A brief description of the building of Royal centre are as given below:
(i) One of the most beautiful buildings in the royal centre is the Lotus Mahal. It was named by British travellers in the nineteenth century. While the name is certainly romantic, historians are not quite sure what the building was used for. One suggestion, found in a map drawn by Mackenzie, is that it may have been a council chamber, a place where the king met his advisers.
(ii) Most temples were located in the sacred centre. One of the most spectacular of these is the Hazara Rama Temple. This was probably meant to be used only by the king and his family.

Question 7.
What does the architecture of buildings like the Lotus Mahal and elephant stables tell us about the rulers who commissioned them ?
Solution :
The Lotus Mahal had nine towers – a high central one, and eight along the sides. Although it is not clear for what the building was used for but according to Mackenzie, it may have been a council chamber, place where the king met his advisers. Elephant stables were located close to the Lotus Mahal.

The architecture of Lotus Mahal tells us that the rulers used to consult their advisers on various issues and problems and meetings were held in the council chamber i.e., Lotus Mahal. The construction of “elephant stables” shows that the rulers took interest in the trade of elephants as well as in keeping them properly because elephants were very important factor in the warfare. It is perhaps one of reasons that elephants and horses have been depicted on the panels of the Hazara Rama temple.

Question 8.
What are the architectural traditions that inspired the architects of Vijayanagara ? How did they transform these traditions ?
Solution :
(a) The architectural traditions that inspired the architects of Vijayanagara were as given below :

  1. Prior to Vijayanagara, Cholas in Tamil Nadu and the Hoysalas in Karnataka had extended patronage to elaborate temples such as the Brihadishvara temple, Thanjavur and the
    Chennakeshava temple at Belur. The rulers of Vijayanagara built on these traditions and carried them literally to new heights.
  2. Like Indo-Islamic architecture, there was an arch on the gateway leading into fortified settlement as well as dome over the gate.
  3. The architecture of tombs and mosques located in the urban core resembles that of the mandapas found in the temples of Hampi.
  4. The Pallavas, Chalukyas, Hoysalas and Cholas encouraged temple building as a means of associating themselves with the divine – the deity was identified with the king. The choice of the site Vijayanagara was perhaps inspired by the existence of the shrines of Virupaksha and Pampadevi.
  5. The arches in the Lotus Mahal were inspired by Indo-Islamic technique.

(b) They transformed these traditions in the followings ways :

  1. In the fortifications, according to Abdur Razzaq, no mortar or cementing agent was employed. The stone blocks were wedge shaped, which held them in place, and the inner portion of the walls was of earth packed with rubble.
  2. Royal portrait sculpture was innovated and developed. It was displayed in temples and the king’s visits to temples were treated as important state occasions.
  3. In temple architecture, new features were structures of immense scale best exemplified by the Raya gopurams or royal gateways. They often dwarfed the towers on the central shrines and signalled the presence of the temple from a great distance. Other features include mandapas or pavilions and long, pillared corridors.

Question 9.
What impressions of the lives of the ordinary people of Vij ay an agar a can you cull from the various descriptions in the chapter ?
Solution :
The various descriptions in this chapter give the following impression of the lives of the ordinary people of Vijayanagara :

  1. Horses were imported from Arabia and Central Asia. This trade was done by the traders and local communities of merchants i.e., kudirai chettis or horse merchants.
  2. There were markets dealing in spices, textiles and precious stones.
  3. Vijayanagara boasted of a wealthy population that demanded high-value exotic goods.
  4. Portuguese traveller Barbosa described the houses of ordinary people, which have not survived as “the other houses of the people are thatched, but nonetheless well-built and arranged according to occupations, in long streets with many open places”. Besides this there is little archaeological evidence of the houses of ordinary people.
  5. There were numerous shrines and small temples which implies that there were variety of cults, supported by different communities.
  6. There were wells, rainwater tanks, temple tanks which may have served as sources of water to the ordinary town dwellers.
  7. Paes gives us a vivid description of a bazaar. He states that the provisions, such as rice, wheat, barley, etc. were available cheaply and abundantly. This means that the life of the ordinary people was good and they did not suffer for want of essential things.

We hope the NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 7 An Imperial Capital: Vijayanagara, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds

NCERT Class 12 Chemistry Solutions for Chapter 9 Coordination Compounds provides in-depth knowledge of the concepts given in the chapter. It helps the students prepare well for boards as well as competitive exams. It also helps the students strengthen their basics for advanced concepts.

The students appearing for UP board, MP board, CBSE, Gujarat board, Maharashtra board, etc. can refer to these NCERT Solutions and score well in the examination.

Board CBSE
Textbook NCERT
Class Class 12
Subject Chemistry
Chapter Chapter 9
Chapter Name Coordination Compounds
Number of Questions Solved 43
Category NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds

Complex compounds are known as coordination compounds. It is an important chapter from the examination perspective. The students need to be thorough with the concepts such as properties, examples and applications of coordination compounds.

A thorough knowledge of the basic concepts helps the students while studying the advanced concepts. This not only helps them during boards and home exams but also during competitive exams.

NCERT IN-TEXT QUESTIONS

Question 1.
Write the formulas of the following coordination compounds:
(a) tetraamminediaquacobalt (III) chloride
(b) potassium tetracyanonickelate (II)
(c) tris (ethane-1, 2-diamine) chromium (III) chloride
(d) amminebromidochloridonitrito-N-platinate (II)
(e) dichlorobis (ethane-1, 2-diamine) platinum (TV) nitrate
(f) iron (III) hexacyanoferrate (II).
Answer:
(a) [Co(NH3)4(H2O)2]Cl3
(b) K2[Ni(CN)4]
(c) [Cr(en)3]Cl3
(d) [Pt(NH3)BrCl(NO2)]
(e) [PtCl2(en)2](NO3)2
(f) Fe4[Fe(CN)6]3.

Question 2.
Write the IUPAC names of the following coordination compounds:

  1. [Co(NH3)6]Cl3
  2. [Co(NH3)5Cl]Cl2
  3. K3[Fe(CN)6l
  4. K3lFe(C2O4)3]
  5. K2[PdCl4]
  6. [Pt(NH3)2Cl(NH2CH3)]Cl

Answer:

  1. Hexaamminecobalt (III) chloride
  2. Pentaamminechloridecobalt (III) chloride
  3. Potassium hexacyanoferrate (III)
  4. Potassiumtrioxalatoferrate(III)
  5. Potassium tetrachloridopalladate (II)
  6. Diamminechloride (methylamine) platinum (II) chloride

Question 3.
Give the types of isomerism exhibited by the following complexes and draw the structures of these isomers:
(a) K[Cr(H2O)2(C2O4)2]
(b) [Co(en)3]Cl3
(c) [CO(NH3)5(NO2)] (NO3)2
(d) [Pt(NH3) (H2O)Cl2] (C.B.S.E. Outside Delhi 2013)
Answer:
(a) K[Cr(H2O)2(C2O4)2] or K[Cr(H2O)2(OX)2]
(i) It exists as geometrical isomers :
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 1
(ii) The cis isomer can also exist as pair of optical isomers.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 2
(b) The complex can exist as optical isomers. For structure, consult section 9.7.
(c) The complex can exist as pair of ionisation isomers as well as linkage isomers.
Ionisation isomers: [Co(NH3)5(NO2)] (NO3)2 and [Co(NH3)5(NO3)](NO2)(NO3)
Linkage isomers: [Co(NH3)5(NO2)](NO3)2 and [Co(NH3)5ONO](NO3)2
(d) The complex can exist as pair of geometrical isomers.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 3

Question 4.
Give evidence to show that [Co(NH3)5Cl]SO4 and [CO(NH3)5SO4]Cl exist as ionisation isomers.
Answer:
Dissolve both the complexes separately in water. First add a few drops of BaCl2 solution to both these complexes. Only one of these will give white precipitate with the solution indicating that SO4 is not a part of complex entity. If exists as an anion.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 4
Now again add a few drops of AgN03 solution to both these complexes. Only one of these will give white precipitate with the solution indicating that in this case Cl is not a part of complex entity. It exists as anion.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 5
This shows that the complexes exist as pair of ionisation isomers.

Question 5.
Explain on the basis of valence bond theory that [Ni(CN)4]2- ion with square planar structure is diamagnetic and the[Ni(CN)4]2- ion with tetrahedral geometry is paramagnetic.
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 6
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 7

Question 6.
[MCl4]2- is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why ? (C.B.S.E. Outside Delhi2012)
Answer:
In the complex [NiCl4]2-, Ni is in + 2 oxidation state and has the configuration 3d840. The CL ion being a weak ligand cannot pair the two unpaired electrons present in 3d orbitals. This means that 3d orbitals are not involved in hybridisation. The complex is sp3 hybridised (tetrahedral) and is paramagnetic in nature. In the other complex [Ni(CO)4], the oxidation state of Ni is zero and electronic configuration is 3d84s2. In the presence of the ligand CO, the 4s electrons shift to the two half filled 3d orbitals and make all the electrons paired. The valence 4s and 3p orbitals are involved in hybridisation. The complex is tetrahedral but diamagnetic. For more details, consult text part.

Question 7.
[Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3- is weakly paramagnetic. Explain.
Answer:
Outer electronic configuration of iron (Z = 26) in-ground state is 3d64s2. Iron in this complex is in +3 oxidation state. Iron achieves +3 oxidation state by the loss of two 4s electrons and one 3d electron. The resulting Fe3+ ion has outer electronic configuration of 3d5. Since H2O is not a strong field ligand, it is unable to cause electron pairing.

Outer electronic configuration of iron (Z=26) in ground state is 3d64s2. Iron in this complex is in a +3 oxidation state. Iron achieves +3 oxidation state by the loss of two 4s electrons and one 3d electron. The resulting Fe3+ ion has an outer electronic configuration as 3d5. CN ion is a strong field ligand

Question 8.
Explain [Co(NH3)6]3+ is an inner orbital complex while [Ni(NH3)6]2+ is an outer orbital complex.
Answer:
In the complex [Co(NH3)6]3+, the oxidation state of cobalt is +3 and has 3d6 configuration. In the presence of NH3 molecules (ligands), two 3d electrons pair up and two 3d orbitals remain empty. Since six ligands are to be accommodated the hybridisation of the metal ion is d2sp3.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 8
As inner d-electrons are involved, the complex is inner orbital complex and is diamagnetic in nature.
In the complex [Ni(NH3)6]3 + , the oxidation state of Ni is +2 and has 3d8 configuration. Since six NH3 molecules (ligands) are to be accomodated, the hybridisation of metal ion is sp3d2. This means that 4d orbitals are involved in the hybridisation.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 9
The complex is paramagnetic as well as outer orbital complex since outer (4d) electrons are involved in the hybridisation.

Question 9.
Predict the number of unpaired electrons in the square planar [Pt(CN)4]2- ion.
Answer:
The element Pt(Z = 78) is present in group 10 with electronic configuration 5d96s1. The divalent cation Pt2+ has 5d8 configuration.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 10
For square planar complex, Pt (II) is in dsp2 hybridisation state. To achieve this, the two unpaired electrons present in 5d orbitals get paired. The complex has, therefore, no unpaired electrons.

Question 10.
The hexaquomanganese(II) ion contains five unpaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 11
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 12

Question 11.
Calculate the overall complex dissociation equilibrium constant for [Cu(NH3)4]2+ ion, given that p4 for the complex is 2·1 x 1013.
Answer:
The dissociation constant is the reciprocal of overall stability constant (β4)
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 13

NCERT EXERCISE

Question 1.
Explain bonding in co-ordination compounds in terms of Werner’s postulates.
Answer:
Werner’s coordination theory: Alfred Werner gave his co-ordination theory in 1893. The important postulates of this theory are:
(i) All metals in atomic or ionic form exhibit two types of valencies in coordination compounds :
(a) Primary or principal or ionic valency (—–),
(b) Secondary or auxiliary or nonionic valency (—).
The primary valency is ionizable and it is shown by dotted lines. The secondary valency is non-ionizable and is shown by a continuous line.
(ii) Primary valency represents oxidation states of a metal atom or ion and secondary valency represents the co-ordination number of metal ion which is fixed for a particular atom.
(iii) The primary valencies are satisfied by negative ions whereas the secondary valencies may be satisfied either by negative ions (e.g., Cl, Br, CN etc.) or neutral molecules (ag., H2O).
(iv) Secondary valencies are directed towards a fixed position in space.
(v) Every element tends to satisfy both its primary and secondary valencies. For this purpose a negative ion may often act a dual behaviour i.e., it may satisfy primary as well as secondary valency (—–,-).
Example : (i) Luteo cobaltic chloride CoCl3.6NH3 or [Co(NH3)6]Cl3.
(ii) Purpureo cobaltic chloride COC13.5NH3 or [Co(NH3)5Cl]Cl2

Question 2.
FeSO4 solution mixed with (NH4)2SO4solution in 1:1 molar ratio gives the test of, Fe2+ion but CuSO4solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu2+ ion. Explain why.
Answer:
FeSO4 solution mixed with (NH4),SO4 solution in 1 : 1 molar ratio forms a double salt, FeS04 (NH4)2SO4-6H2O (Mohr’s salt) which ionizes in the solution to give Fe2+ions. Hence it gives the tests of Fe2+ ions. CuSO4 solution mixed with aqueous ammonia in 1:4 molar ratio forms a complex salt, with the formula [CU(NH3)4]SO4. The complex ion [Cu(NH3)4]2+ does not ionize to give Cu2+ ions. Hence, it does not give the tests of Cu2+ ion.

Question 3.
Explain with two examples each of the following:
coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heterolepic.
Answer:
(a) Co-ordination Entity: A coordination complex or entity is the species enclosed in square bracket. It is also called co-ordinate sphere. It contains in it a certain metal atom or ion to which a fixed number of neutral molecules or ions capable of donating electron pairs are linked with co-ordinate bonds. e.g. [COCl3(NH3)3]

(b) Ligand: Ligands are the electron donor molecules or ions which may be either neutral, or anionic (sometimes cationic as well) and are linked to the central metal atom or ion by co-ordinate bonds also called dative bonds. In fact, the electrons needed for the bond are provided by the ligands. e.g. H2NCH2CH2NH2 or N(CH2CH2NH2)3

(c) Co-ordination Number: The coordination number [C.N.] of a metal ion in a complex can be defined as the number of ligand or donor atoms to which the metal is directly bonded. For example, in the complex ions, [PtCl6]2- and [Ni(NH3)4]2+, the coordination number of Pt and Ni are 6 and 4 respectively. Similarly, in the complex ions, [Fe(C2O4)3]3- and [CO(en)3]3+,the coordination number of both, Fe and Co, is 6 because C2O42-and en (ethane-1,2-diamine) are bidentate ligands.

(d) Coordination polyhedron : The spatial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron about the central atom. The most common coordination polyhedra are octahedral, square planar and tetrahedral. For example, [CO(NH3)6]3+ is octahedral, [Ni(CO)4] is tetrahedral and [PtCl4]2- is square planar.

(e) Homoleptic and heteroleptic complexe : Complexes in which a metal is bound to only one kind of donor groups, e.g., [CO(NH3)6]3+, are known as homoleptic. Complexes in which a metal is bound to more than one kind of donor group, e.g., [CO(NH3)4Cl2]+, are known as heteroleptic.

Question 4.
What is meant by unidentate, didentate and ambidentate ligands ? Give two examples for each.
Answer:
Unidentate ligands are those which bind to the metal ion through a single donor atom, e.g., Cl , H2O.

Bidentate ligands are those which bind to the metal ion through two donor atoms. e.g., ethane-1,2-diamine (H2NCH2CH2NH2), oxalate (C2O42-) ion.

Ambidentate ligands are those which can bind to metal ion through two different donor atoms, e.g., NO2 and SCN ion.

Question 5.
Specify the oxidation numbers of metals in the following co-ordination entities :
(a) [Co(H2O)(CN)(en2)]2+
(b) [PtCl4]2-
(c) [Cr(NH3)3Cl3]
(d) [CoBr2(en)2]+
(e) K3[Fe(CN)6].
Answer:
(a) O.N. of Co : x + 0 + (-1) + 2(0)= + 2 or x = + 2+ 1 = + 3
(b) O.N. of Pt : x + 4 (-1) =-2 or x =-2 + 4 = + 2
(c) O.N. of Cr : x + 3(0) + 3(- 1) = 0 or x = + 3
(d) O.N. of Co : x + 2(- 1) + 2(0) = + 1 or x = + 1 + 2 = + 3
(e) O.N. of Fe : x + 6 (- 1) = – 3 or x = – 3 + 6 = + 3

Question 6.
Using IUPAC norms write the formulas for the following:
(i)Tetrahydroxozincate(Il)
(ii)Potassium tetrachloridopalladate (II)
(iii)Diamminedichlorido platinum (II)
(iv)Potassium tetracyanonickelate (II)
(v)Pentaamminenitrito-O-cobalt(III)
(vi)Ilexaamminccobalt (III) sulphate
(vii)Potassium tri(oxalato) chromate (III)
(yiii)Hexaammineplatinum (IV)
(ix)Tetrabromidocuprate(II)
(x) Pentaamminenitrito-N-cobalt (III)
Answer:
(i)[Zn(OH)4]2- (ii)K2[PdCl4]
(iii)[Pt(NH3)2Cl2]
(iv)K2[Ni(CN)4]
(v)[Co(NH3)5(ONO)]2+
(vi)[Co(NH3)6]2(SO4)3
(vii)K3[Cr(C2O4)3]
(viii)[Pt(NH3)6]4+
(ix)[CuBr4]2-
(x)[Co(NH3)5(N02)]2+

Question 7.
Using IUPAC norms write the systematic names of the following :
(a) [CO(NH3)6]Cl3
(b) [CO(NH3)4Cl(NO2)]Cl
(c) [Ni(NH3)6]Cl2
(d) [Pt(NH3)2Cl(NH2CH3)]Cl
(e) [Mn(H2O)6]2+
(f) [NiCl4]2-
(g) [Co(en)3]3+
(h) [Ti(H2O)6]3+
(i) [Ni(CO)4]. (Jharkhand Board 2015)
Answer:
(a) hexamminecobalt(III) chloride
(b) tetramminechloriodonitrito-N-cobalt(III) chloride
(c) hexaamminenickel(II) chloride
(d) diamminechlorido (methaneamine) platinum(II) chloride
(e) hexaaquamanganese(II) ion
(f) tetrachloriodonickelate(II) ion
(g) tris(ethane-l, 2-diammine) cobalt(III) ion
(h) hexaaquatitanium(III) ion
(i) tetracarbonylnickel (0).

Question 8.
List various types of isomerism possible for coordination compounds, giving an example of each.
Answer:
(i) Geometrical isomerism: The isomer in which similar ligands occupy adjacent positions is referred to as cis isomer and the isomer in which similar ligands occupy opposite positions is referred to as trans isomer. Therefore, this type of isomerism is also known as cis-trans isomerism.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 14

(ii) Optical isomerism: The isomer which rotates the plane of polarised light to the right is called dextro rotatory and designated as d- and the one which rotates the plane of polarised light to the left is called laevo rotatory and designated as l. These optical isomers have identical physical and chemical properties except their behaviour towards the plane polarised light.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 15

(iii) Linkage isomerism: Linkage isomerism arises in a coordination compound containing ambidentate ligand. A simple example is provided by complexes containing the thiocyanate ligand, NCS, which may bind through the nitrogen to give M-N CS or through sulphur to give M-SCN. This behaviour was seen in the complex [CO(NH3)5(NO2)]Cl2, which is obtained as the red form, in which the nitrite ligand is bound through oxygen ( ONO), and as the yellow form, in which the nitrite ligand is bound through nitrogen (-NO2).

(iv) Coordination isomerism : This type of isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex. An example is provided by [Co(NH3)6] [Cr(CN)6], in which the NH3 ligands are bound to CO3+ and the CN ligands to Cr3+. In its coordination isomer [Cr(NH3)6][CO(CN)6], the NH3 ligands are bound to Cr3+ and the CN ligands to CO3+.

(v) Ionisation isomerism : This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion. An example is provided by the ionisation isomers [CO(NH3)5SO4]Br and [CO(NH3)5Br]SO4.

(vi) Solvate isomerism : This form of isomerism is known as ‘hydrate isomerism in case where water is involved as a solvent. This is similar to ionisation isomerism. Solvate isomers differ by whether or not a solvent molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice. An example is provided by the aqua complex [Cr(H2O)6]Cl3 (violet) and its solvate isomer [Cr(H2O)5Cl]Cl2 H2O (grey green).

Question 9.
How many geometrical isomers are possible in the following coordination entities ?
(a) [Cr(C2O4)3]3-
(b) [Co(NH3)3Cl3]
Answer:
(a) [Cr(C2O4)2]3- : No geometrical isomerism is possible.
(b) [CO(NH3)3Cl3] : Two geometrical isomers : fac and mer
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 16

Question 10.
Draw the structures of optical isomers of :
(a) [Cr(C2O4)3]3-
(b) [PtCl2(en)2]2+
(c) [Cr(NH3)2Cl2(en)]+ (C.B.S.E. Foreign 2015)
Answer:
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 17

Question 11.
Draw all the isomers (geometrical and optical) of :
(a) [CoCl2(en)2]+
(b) [CoNH3Cl(en)2]2+
(c) [Co(NH3)2Cl2(en)]+
Answer:
(a)
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 18
(b)
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 19
(c)
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 20

Question 12.
Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)]. How many of these will exhibit optical isomerism ?
Answer:
There are three geometrical isomers.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 21
Optical isomerism is generally not shown by the square planar complexes with CN = 4.

Question 13.
Aqueous copper sulphate solution (blue in colour) gives (a) green precipitate with aqueous potassium fluoride and (b) a bright green solution with aqueous potassium chloride solutions. Explain these experimental results.
Answer:
Aqueous solution of copper sulphate which is blue in colour exists as [Cu(H2O)4]SO4 and gives [Cu(H2O)4]2+ in solution. It is a labile complex entity in which the ligands H2O get easily replaced by F ions of KF and by Cl ions of KCl.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 22

Question 14.
What is the coordination entity formed when excess aqueous KCN is added to an aqueous solution of copper sulphate? Why is that no precipitate of copper sulphide is obtained when H2S(g) is passed through the solution?
Answer:
When an excess aqueous KCN is added to an aqueous solution of CuSO4, Potassiumtetra-cyanocuprate (II) is formed. When H2S(g) is passed through the above solution, no precipitate of copper sulphide is obtained because CN ions are strong ligands so the complex [Cu(CN)4]2- is very stable. As Cu2+ ions are not available so CuS precipitate is not formed.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 23

Question 15.
Discuss the nature of bonding in the following co-ordination complexes on the basis of valence bond theory :
(a) [Fe(CN)6]4-
(b) [FeF6]3-
(c) [Co(C2O4)3]3-
(d) [CoF6]3-
Answer:
(a) Hexacyanoferrate(II) ion [Fe(CN)6]4-: Iron in this complex is in +2 oxidation state. Iron achieves + 2 oxidation state by the loss of two 4s electrons. The resulting Fe2+ ion has outer electronic configuration of 3d6.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 24

(b) Hexafluoriodoferrate(II) ion [FeF6]3-: Iron in this complex is in +2 oxidation state. Iron achieves + 2 oxidation state by the loss of two 4s electrons. The resulting Fe2+ ion has outer electronic configuration of 3d6.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 25
(c)

NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 26

(d) Hexafluorocobaltate(III) [CoF6]3-: Cobalt ion in the complex is in + 3 oxidation state. Cobalt achieves + 3 oxidation state by the loss of two 4s electrons and one 3d electron. The resulting Co3 + ion has outer electronic configuration of 3d6.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 27

Question 16.
Draw figure to show the splitting of d-orbitals in octahedral crystal field.
Answer:
Let us assume that the six ligands are positioned symmetrically along the Cartesian axes, with a metal atom at the origin.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 28

As the ligands approach, first there is an increase in energy of d-orbitals relative to that of the free ion just as would be the case in a spherical field.

The orbitals lying along the axes (dz2 and dx2 – y2) get repelled more strongly than dxy, dyz and dzx orbitals which have lobes directed between the axes.

The dz2 and dx2-y2 orbitals get raised in energy and dxy dyz, dxz orbitals are lowered in energy relative to the average energy in the spherical crystal field. Thus, the degenerate set of d-orbitals get split into two sets: the lower energy orbitals set, t2g, and the higher energy orbitals set, eg. The energy is separated by ∆0

Question 17.
What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Answer:
Spectrochemical series: The arrangement of ligands in order of their increasing field strengths i.e. increasing crystal field splitting energy (CFSE) values is called spectrochemical series, which is as follows:
I < Br < SCN < Cl < S2- < F < OH < C2O2-4 < H2O < NCS < edta-4 < NH3 < en < CN < CO.
Difference between weak field ligand and a strong field ligand: The ligand with a small value of CFSE (∆0) are called weak field ligands whereas those with a large value of CFSE are called strong field ligands.

Question 18.
What is crystal field splitting energy ? How does the magnitude of ∆0 decide the actual configuration of d- orbitals in a coordination entity ?
Answer:
The degenerate d-orbitals (in a spherical field environment) split two-level i.e. eg and t2g in the presence of ligands. The splitting of the degenerate orbitals in the presence of ligands is called crystal field splitting and the energy difference between the two levels (e and t2g) is called the crystal field splitting energy. It is denoted by ∆o. After the orbitals have split, the filling of the electrons takes place. After 1 electron (each) has filled in the three t2g orbitals, the filling of the electrons takes place in 2 ways.

It can enter the orbital (giving) rise to t3g eg, like electronic configuration on the pairing of the electrons can take place in the t2g orbitals (giving rise to t42g eg0 like electronic configuration). If the ∆o value of a ligand is less than the pairing energy, then the electrons enter the eg orbital. On the other hand, if the ∆o value of a ligand is more than the pairing energy, then the electrons enter the t2g orbitals.

Question 19.
[Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2- is diamagnetic. Explain why?
Answer:
(i) Hexaamminechromium(III) ion [Cr(NH3)6]3+: Outer electronic configuration of chromium (Z=24) in ground state is 3d24s1 and in this complex, it is in the +3 oxidation state. Chromium achieves +3 oxidation state by the loss of one 4s electron and two 3d-electrons. The resulting Cr3+ ion has outer electronic configuration of 3d3.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 29
(ii) Tetracyanonickelate (II) ion [Ni(CN)4]2-: Outer electronic configuration of nickel (Z = 28) in ground state is 3d84s2. Nickel in this complex is in + 2 oxidation state. It achieves + 2 oxidation state by the loss of the two 4s-electrons. The resulting Ni2+ ion has outer electronic configuration of 3d8.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 30

Question 20.
A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2- is colourless. Explain. (C.B.S.E. Delhi 2017)
Answer:
Formation of [Ni(H2O)6]2+
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 31
Since H2P molecules represent weak field ligands, they do not cause any electron pairing. As a result the complex has two unpaired electrons and is coloured. The d-d transitions absorb radiations corresponding to a red light and the complementary colour emitted is green.
Formation of [Ni(CN)4]2-. For the details of the structure of the complex,
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 32
Since the complex has no unpaired electrons there is no scope for any d-d transition. The complex is therefore, colourless.

Question 21.
[Fe(CN)6]4- and [Fe(H2O)6]2+ are of different cdlours in dilute solutions. Why?
Answer:
In both the complexes, Fe is in +2 state with the configuration 3d6 i.e., it has four unpaired electrons. As the ligand H2O and CN possess different crystal field splitting energy (∆0), they absorb different components of the visible light (VIBGYOR) for the transition. Hence, the transmitted colours are different.

Question 22.
Discuss the nature of bonding in metal carbonyls.
Answer:
The metal-carbon bond in metal carbonyls possess both s and p character. The M-C σ bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. The M-C π bond is formed by the donation of a pair of electrons from a filled d-orbital of metal into the vacant antibonding π* orbital of carbon monoxide. The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal.
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 33

Question 23.
Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes :
(a) K3[CO(C2O4)3]
(b) (NH4)2[CoF4]
(C) Cis – [CrCl22(en)2]Cl
(d) [Mn(H2O)6]SO4
Answer:
(a) OS = + 3, CN = 6, d-orbital occupation is 3d6 \({ t }_{ 2g }^{ 6 }{ e }_{ g }^{ 0 }\),
(b) OS = + 2, CN = 4, 3d7 (\({ t }_{ 2g }^{ 5 }{ e }_{ g }^{ 2 }\)),
(c) OS = + 3, CN = 6, 3d3 (\({ t }_{ 2g }^{ 3 }\)),
(d) OS = + 2, CN = 6, 3d6 (\({ t }_{ 2g }^{ 3 }{ e }_{ g }^{ 2 }\)).

Question 24.
Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex :
(a) K[Cr(H2O)2(C2O4)2]3H2O
(b) [CrCl3(py)3]
(c) K4[Mn(CN)6]
(d) [Co(NH3)5Cl]Cl2
(e) Cs[FeCl4]
Answer:
(a) IUPAC name : potassium diaquadioxalatochromate (III) hydrate.
O.S. of Cr = + 3 ; 3d3 (\({ t }_{ 2g }^{ 3 }{ e }_{ g }^{ 0 }\)) CN = 6 ; shape = octahedral, three unpaired electrons.
Magnetic moment (μ) = \(\sqrt { n\left( n+2 \right) } =\sqrt { 3\times 5 } =\sqrt { 15 } =3\cdot 87BM\)
(b) IUPAC name: trichloridotripyridinechromium (III) O.S. of Cr = + 3; 3 d3 (\({ t }_{ 2g }^{ 3 }{ e }_{ g }^{ 0 }\)) CN = 6
shape = octahedral ; three unpaired electrons.
Magnetic moment (μ) = \(\sqrt { n\left( n+2 \right) } =\sqrt { 3\times 5 } =3\cdot 87BM\)
(c) IUPAC name : potassiumhexacyanomanganate (II) O.S. of Mn = + 2 ; 3d5 (\({ t }_{ 2g }^{ 5 }{ e }_{ g }^{ 0 }\)), CN = 6, shape = octahedral; one unpaired electron.
Magnetic moment (μ) = \(\sqrt { n\left( n+2 \right) } =\sqrt { 1\times 3 } =\sqrt { 3 } =1\cdot 73BM\)
(d) IUPAC name : pentaamminechloridocobalt (III) chloride
O.S. of Co = + 3 ; 3d6 (\({ t }_{ 2g }^{ 6 }{ e }_{ g }^{ 0 }\)), CN = 6
shape = octahedral; zero unpaired electron. Magnetic moment (μ) = 0
(e) IUPAC name : cesium tetrachloridoferrate (III)
O.S. of Fe = + 3 ; 3d5 (\({ e }^{ 2 }{ t }_{ 2 }^{ 3 }\)), CN = 4.
shape = tetrahedral ; five unpaired electrons.
Magnetic moment (μ) = \(\sqrt { n\left( n+2 \right) } =\sqrt { 5\times 7 } =\sqrt { 35 } =5\cdot 92BM\)

Question 25.
What is meant by stability of a coordination compound in solution ? State the factors which govern stability of complexes.
Answer:
The stability of a complex in solution refers to the degree of association between the two species involved in the state of equilibrium. The magnitude of the (stability or formation) equilibrium constant for the association, quantitatively expresses the stability. Thus, if we have a reaction of the type :
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 34
then, the larger the stability constant, the higher is the proportion of ML4 that exists in the solution. Free metal ions rarely exist in the solution so that M will usually be surrounded by solvent molecules which will compete with the ligand molecules, L, and be successively replaced by them. For simplicity, we generally ignore these solvent molecules and write four stability constants as follows :
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 35
Factors affecting stability of complexes :

  1. The smaller the size of the cation, the greater will be the stability of the complex e.g., Fe3+ forms a more stable complex than Fe2+.
  2. The greater the charge on the central metal ion, the more stable will be the complex e.g., Pt4+ forms a more stable complex than Pt2+.
  3. Stronger the ligand, more stable will be the complex formed e.g., CN forms more stable complex then NH3.

Question 26.
What is meant by chelate effect ? Give an example.
Answer:
When a ligand attaches to the metal ion in a manner that form’s a ring, then the metal-ligand association is found to be more stable. In other words, we can say that complexes containing chelate ring more stable than complexes without rings. This is known as the chelate effect.
Examples: EDTA, DMG, etc.

Question 27.
Discuss briefly giving an example in each case the role of coordination compounds in:
(i) biological systems
(ii) medicinal chemistry
(iii) analytical chemistry and
(iv) extraction/ metallurgy of metals.
Answer:
(i) Role of coordination compounds in biological systems:
We know that photosynthesis is possible by the presence of chlorophyll pigment. This pigment is a coordination compound of magnesium. In the human biological system, several coordination compounds play important roles. For example, the oxygen – carrier of blood, i.e hemoglobin is a coordination compound of iron.

(ii) Role of coordination compounds in Medicinal chemistry: Certain coordination compounds of platinum (for example cis-platin) are used for inhibiting the growth of tumors.

(iii) Role of coordination compounds in analytical chemistry: During salt analysis, a number of basic radicals are detected with the help of the colour changes they exhibit with different reagents. These colour changes are a result of the coordination compounds or complexes that the basic radicals form with different ligands.

(iv) Role of coordination compounds in interaction or metallurgy of metals: The process of extraction of some of the metals from their ores involves the formation of complexes. For example in an aqueous solution, gold combines with cyanide ions to form [Au (CN)2]. From this solution, gold is later extracted by the addition of Zn metal.

Question 28.
How many ions are produced from the complex Co(NH3)6Cl2 in solution?
(a) 6
(b) 4
(c) 3
(d) 2.
Answer:
The complex will dissociate in aqueous solution to give three ions
NCERT Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds 36
Therefore, (c) is the correct answer.

Question 29.
Amongst the following ions which one has the highest magnetic movement value?
(a) [Cr(H2O)6]3+
(b) [Fe(H2O)6]2+
(c) [Zn(H2O)6]2+
Answer:
The oxidation states of the metals in the complexes along with the electronic configuration are given:
(a) Cr3+ : 3d3 configuration ; unpaired electrons = 3
(b) Fe2+ : 3d6 configuration ; unpaired electrons = 4
(c) Zn2+ : 3d10 configuration ; unpaired electrons = 0
The complex (b) with maximum number of unpaired electrons has the highest magnetic moment. Therefore, (b) is the correct answer.

Question 30.
The oxidation number of cobalt in K[Co(CO)4] is
(a) + 1
(b) + 3
(c) – 1
(d) – 3.
Answer:
O.N. of Co : x + 4(0) = -1 or x = -1. Therefore, (c) is the correct answer.

Question 31.
Amongst the following, the most stable complex is :
(a) [Fe(H2O)6]3+
(b) [Fe(NH3)6]3+
(c) [Fe(C2O4)3]3-
(d) [FeCl6]3-.
Answer:
In all the complexes, Fe is in + 3 oxidation state. However, the complex (c) is a chelate because three \({ C }_{ 2 }{ O }_{ 4 }^{ 2- }\) ions act as the chelating ligands. Thus, the most stable complex is (c).

Question 32.
What will be the correct order for the wavelengths of absorption in the visible region for the following : [Ni(NO2)6]4-,[Ni(NH3)6]2+,[Ni(H2O)6]2+.
Answer:
In all the complexes, the metal ion is the same (Ni2+). The increasing field strengths of the ligands present as per electrochemical series are in the order :
H2O < NH3 < \({ NO }_{ 2 }^{ – }\)
The energies absorbed for excitation will be in the order :
[Ni(H2O)6]2+ < [Ni(NH3)6]2+ < [Ni(NO2)6]4-
As E = hc/λ i.e., E ∝ 1/λ; the wavelengths absorbed will be in the opposite order.

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NCERT Solutions for Class 12 History Chapter 9 Kings and Chronicles The Mughal Courts

NCERT Solutions for Class 12 History Chapter 9 Kings and Chronicles The Mughal Courts are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 9 Kings and Chronicles The Mughal Courts.

Board CBSE
Textbook NCERT
Class Class 12
Subject History
Chapter Chapter 9
Chapter Name Kings and Chronicles The Mughal Courts
Number of Questions Solved 9
Category NCERT Solutions

NCERT Solutions for Class 12 History Chapter 9 Kings and Chronicles The Mughal Courts

Question l.
Describe the process of manuscript production in the Mughal court.
Solution :
The process of manuscript production in the Mughal court involved a variety of tasks as mentioned below :

  • Paper makers prepared the folios of the manuscript.
  • Scribes or calligraphers used to copy the text.
  • Gliders illuminated the pages of the manuscript.
  • Painters illustrated the scenes from the text to describe it in a visual form.
  • Bookbinders gathered the individual folios and set them within ornamental covers.

At the end of above tasks, the finished manuscript was seen as a precious object, a work of intellectual wealth and beauty. It was an example of power of its patron, the Mughal emperor, to bring such beauty into being.

Question 2.
In what ways would the daily routine and special festivities associated with the Mughal court have conveyed a sense of the power of the emperor?
Solution :
The daily routine and special festivities associated with the Mughal court observed the following:

  1. By representing the status of the court.
  2. In form of salutation of emperor.
  3. Jharokha Darshan.
  4. Meeting held by sultan in Diwan-i-Am and Diwan-i-Khas.
  5. By honouring the mansabdar on special occasion with special gifts and jagir.

Question 3.
Assess the role played by women of the imperial household in the Mughal Empire.
Solution :
The role played by women of the imperial household was significant:

  1. Nur Jahan took an important part in the administration during the period of Jahangir.
  2. After Nur Jahan, Mughal queens and princesses began to control significant financial resources. Shah Jahan’s daughters Jahanara and Roshanara enjoyed an annual income often equal to that of high imperial mansabdars. Jahanara, in addition, received revenues from the port city of Surat, which was a lucrative centre of overseas trade.
  3. The princess Jahanara played an important role in the arrangement of marriage of Dara Shukoh and Nadira.
  4. Control over resources enabled women to commission buildings and gardens. Jahanara participated in many architectural projects of Shah Jahan’s new capital, Shahjahanabad (Delhi) which included an imposing double-storeyed caravanserai with a courtyard and garden. The bazaar of Chandni Chowk was too designed by Jahanara.
  5. Besides taking part in different activities, some women such as Gulbadan Begum, daughter of Babur, wrote the Humayun Nama. She was Akbar’s aunt and when Akbar commissioned Abu’l Fazl to write history of his reign, he requested her to record her memoirs of earlier times under Babur and Humayun. She described the conflicts and tensions among the princes and kings and the important mediating role elderly women of the family played in resolving some of these conflicts.

Question 4.
What were the concerns that shaped Mughal policies and attitudes towards regions outside the subcontinent ?
Solution :
The following concerns shaped Mughal policies and attitudes towards regions outside the subcontinent:

  1. The Safavids and Qandahar : Qandahar was a bone of contention between the Safavids and the Mughals due to its strategic importance because all conquerors who sought to make their way into the Indian subcontinent had to cross the Hindukush to have access to north India. It was, therefore, a constant aim of Mughal policy to ward off this potential danger by controlling strategic outposts — notably Qandahar and Kabul. It was under these circumstances that the Mughals tried to have control over Qandahar. It was under the possession of Humayun but later the Mughals lost control of it. Akbar reconquered it in 1595. The Safavids, however, maintained diplomatic relations with the Mughals, but they always continued to stake claims to Qandahar. In 1622, the Persian army besieged Qandahar and defeated the Mughals. The city once again came under the control of the Safavids. Thus, due to its strategic importance Qandahar remained a bone of contention between the Mughals and the Safavids.
  2. The Ottomans : Pilgrimage and trade :
    • The relationship between the Mughals and the Ottomans was to ensure free movement for merchants and pilgrims in the territories under Ottoman control, particularly in Hijaz (Ottoman Arabia) where the important centres of Mecca and Madina were located.
    • The Mughal emperors combined religion and commerce by exporting valuable merchandise to Aden and Mokha, both Red Sea ports. They distributed the proceeds of the sales in charity to the keepers of shrines and religious men there.
  3. Mughals and the Portuguese : After the discovery of a direct sea route to India, the Portuguese king was interested in the propagation of Christianity with the help of missionaries. Akbar too was curious about Christianity. So, Jesuits mission came to India in 1580, 1591 and 1595. At public assemblies, Jesuits were assigned places in close proximity to Akbar’s throne.

Question 5.
Discuss the major features of Mughal provincial administration. How did the centre control the provinces?
Solution :

  1. The head of the provincial administration was the governor (subadar). He reported directly to the emperor.
  2. Each suba was divided into sarkar,
  3. The local administration was looked after at the level of the pargana (sub-district) by three semi-hereditary officers, the qanungo (keeper of revenue records), the chaudhur (in charge of revenue collection) and the qazi.
  4. Each department of administration maintained a large support staff of clerks, accountants, auditors, messengers, and other functionaries who were technically qualified officials, functioning in accordance with standardised rules and procedures, and generating copious written orders and records.

Question 6.
Discuss, with examples, the distinctive features of Mughal chronicles.
Solution :
The distinctive features of Mughal chronicles were as given below :

  1. They projected a vision of enlightened kingdom to all those who were under it and conveyed a message to the resistors that they could not be successful in their objects.
  2. The chronicles were commissioned by the Mughal rulers to ensure that there was an account of their rule for posterity.
  3. The chronicles were written by the courtiers who focused on events centred on the ruler, his family, the court and nobles, wars and administrative arrangements because for them the history of the empire and the court was synonymous with that of the emperor. Their titles were therefore, after the name of emperors such as Akbar Naina, Shahjahan Nama and Alamgir Nama.
  4. Mughal chronicles were written in Persian e.g., Akbar Nama. Babur’s memoirs was translated from the Turkish into Persian Babur Nama.
  5. The chronicles were manuscripts and included paintings to enhance their beauty.
  6. Chronicles showed the power of the Mughal kings came directly from God. The emperors were portrayed wearing the halo to symbolise the light of God.
  7. Mughal chronicles presented the empire as compromising many different ethnic and religious communities. Abu’l Fazl described the ideal of sulh-ikul as the comer stone of enlightened rule.

Question 7.
To what extent do you think the visual material presented in this chapter corresponds with Abu’l Fazl’s description of the taswir (Source 1)?
Ans:

  1. Drawing the likeness of anything is called taswir. His Majesty from his earliest youth, has shown a great predilection for this art, and gives it every encouragement, as he looks upon it as a means both of study and amusement.
  2. A very large number of painters set to work.
  3. Each week, several supervisors and clerks of the imperial workshop submit before the emperor the work done by each artist, and his Majesty gives a reward.
  4. Paintings served not only to enhance the beauty of a book, but were believed to possess special powers of communicating ideas about the kingdom and the power of kings in ways that the written medium could not.
  5. The historian Abu’l Fazl described painting as a ‘magical art’ in his view it had the power to make inanimate objects look as if they possessed life.

Question 8.
What were the distinctive features of the Mughal nobility ? How was their relationship with the emperor shaped ?
Solution :

  1. The distinctive features of the Mughal nobility were as given below :
    • The nobility was recruited from diverse ethnic and religious groups. No faction was large enough to challenge the authority of the state.
    • There were Turani, Iranian, Rajputs, and Indian Muslims (Shaikhzadas).
    • Nur Jahan was an Iranian and so Iranians gained high offices under Jahangir.
    • Aurangzeb appointed Rajputs to high offices.
    • All holders of government offices held mansabs comprising two numerical designations zat and sawar.
    • The nobles participated in military campaigns with their armies. They also served as officers of the empire in the provinces.
    • Akbar established spiritual relationships with a select band of his nobility by treating them as his disciples (murid).
    • Nobles stationed at the court were a reserve force to be deputed to a province or military campaign.
  2. Their relationships with the emperor were shaped in different ways :
    • As the nobility was recruited from diverse ethnic and religious groups, no faction was strong enough to challenge the authority of the state.
    • The emperor personally reviewed their rank, titles and postings.
    • Akbar maintained spiritual relationship with some nobles and treated them as the disciples.
    • The influence of different groups of nobility changed from time to time. For example, Iranians were influential in the time of Jahangir but Rajputs and Marathas became influential dining the time of Aurangzeb.
    • The emperor used to award them with office or titles to maintain their loyality towards
      him.
    • According to the Jesuit priest Father Antonio Monserrate, resident at the court of Akbar, in order to prevent the great nobles becoming insolent through the unchallenged enjoyment of power, the emperor used to summon them to court and give them imperious commands, as though they were his slaves.

Question 9.
Identify the elements that went into the making of the Mughal ideal of kingship.
Solution :
The following elements went into the making of the Mughal ideal of kingship :

  1. Divine light – Court chronicles considered that the power of the Mughal kings came directly from God. Abu’l Fazl placed Mughal kingship as the highest station in the hierarchy of objects receiving light emanating from God. King was the source of spiritual guidance for his subjects. Thus, Mughal artists portrayed emperors wearing the halo to symbolise the light of God.
  2. A unifying force – The emperor was source of peace and stability. He stood above all religious and ethnic groups, mediated among them, and ensured that justice and peace prevailed. Abu’l Fazl describes the ideals of sulh-i kul as the comer stone of enlightened rule. Under the sulh-i kul, the nobility of the Mughals consisted of Iranis, Turanis, Afghans, Rajputs and Deccanis — all of whom were given positions on the basis of their merits and loyalty to the king. Akbar abolished pilgrimage tax in 1563 and Jizya in 1564.
  3. Just sovereignty as social contract : Under Mughal ideal of kingship, Abu’l Fazl defined sovereignty as a social contract. The emperor protected the life (jan), property (mal), honour (namus) and faith (din) of his subjects and in return demanded their obedience and a share of resources.

We hope the NCERT Solutions for Class 12 History Chapter 9 Kings and Chronicles The Mughal Courts help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 9 Kings and Chronicles The Mughal Courts, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current

NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current

NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current

Board CBSE
Textbook NCERT
Class Class 12
Subject Physics
Chapter Chapter 7
Chapter Name Alternating Current
Number of Questions Solved 26
Category NCERT Solutions

Question 1.
A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle?
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 1
Question 2.
(a) The peak voltage of an a.c. supply is 300 V. What is the rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current?
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 2
Question 3.
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit.
Answer:
Here, reactance XL = 2 Πv L = 2Π X 50 x 44 x 10-3
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 3

Question 4.
A 60 μF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the RMS value of the current in the circuit.
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 4

Question 5.
In Exercise 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle? Explain your answer.
Answer:
In the case of an ideal inductor or capacitor, there is no power loss.

Question 6.
Obtain the resonant frequency ωr of a series LCR circuit with L =2.0 H, C = 32 μF and R = 10 Ω. What is the Q-value of this circuit?
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 5

Question 7.
A charged 30 μF capacitor is connected to a 27 mil inductor. What is the angular frequency of free oscillations of the circuit ?
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 6

Question 8.
Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 7
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 8
At a later time, energy is shared between capacitor and inductor, However, the total energy remains the same, provided there is no loss of energy.

Question 9.
A series LCR circuit with R = 20 Ω, L = 1.5 H and C = 35 μF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?
Answer:
At natural frequency
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 9

Question 10.
A radio can tune over the frequency range of a portion of the MW broadcast band : (800 kHz to 1200 kHz). If its LC circuit has an effective inductance of 200 pH, what must be the range of its variable capacitor?
[Hint. For tuning, the natural frequency i.e., the frequency of free oscillations for the LC circuit should be equal to the frequency of the radio wave.]
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 10
Thus, the range of the variable capacitor must be 88 pF to 198 pF.

Question 11.
Figure shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80 μF, R = 40 Cl.
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 11
(a) Determine the source frequency which drives the circuit in resonance.
(b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
(c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency. (C.B.S.E. 1994, 1998, 2006)
Answer:
(a) Resonant angular frequency,
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 12

Question 12.
An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0.
(a) What is the total energy stored initially? Is it conserved during LC oscillations?
(b) What is the natural frequency of the circuit?
(c) At what time is the energy stored:
(i) completely electrical (i.e., stored in the capacitor)?
(ii) completely magnetic (i.e., stored in the inductor)?
(d) At what times is the total energy shared equally between the inductor and the capacitor?
(e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat? (C.B.S.E. Sample Paper 1998)
Answer:
(a) Total initial energy
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 13
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 14
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 15
(e) Total initial energy of 1 J will be lost as heat due to Joule’s heating effect in the resistor.

Question 13.

A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply.
(a) What is the maximum current in the coil?
(b) What is the time lag between the voltage maximum and the current maximum?
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 16
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 17

Question 14.

Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high-frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a dc circuit after the steady-state?
Answer:
For the given high frequency, ω = 2 Π v = 2 Π x 104 rad s-1
I0, in this case, is too small, so it can be concluded that at high frequencies an inductor behaves as on an open circuit.
In a steady d.c. circuit v = 0, so the inductor acts as a simple conductor.

Question 15.
A 100 μF capacitor in series with a 40 Ω resistance is connected to a 110 V, 60 Hz supply.
(а)    What is the maximum current in the circuit?
(b)  What is the time lag between the current maximum and the voltage maximum?
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 18
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 20
Question 16.
Obtain the answers to (a) and (b) in Exercise 7.15 if the circuit is connected to a 110 V, 12 kHz supply. Hence, explain the statement that a capacitor is a conductor at very high frequencies. Compare this behavior with that of a capacitor in a dc circuit after the steady-state.
Answer
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 21

embed=”true”>or n is nearly zero at high frequency. In part (a) C term is negligible at a high frequency so it acts like a resistor. For a steady d. c. we have v like

Question 17.
Keeping the source frequency equal to the resonating frequency of the series LCR circuit, if the three elements. L, C, and R are arranged in parallel, show that the total current in the parallel LCR circuit is minimum at this frequency. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7.11 for this frequency.
Answer:
In the case of parallel LCR circuit, impedance is given by,
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 22
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 23

Question 18.
A circuit containing an 80 mF inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.
(a) Obtain the current amplitude and rms values.
(b) Obtain the rms values of potential drops across each element.
(c) What is the average power transferred to the inductor?
(d) What is the average power transferred to the capacitor?
(e) What is the total average power absorbed by the circuit?
[‘Average’ implies ‘averaged over one cycle’.]
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 24
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 25

Question 19.
Suppose the circuit in Exercise 7.18 has a resistance of 15 Ω. Obtain the average power transferred to each element of the circuit, and the total power absorbed.
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 26

Question 20.
A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 V variable frequency supply.
(a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.
(b) What is the source frequency for which the average power absorbed by the circuit is maximum. Obtain the value of this maximum power.
(c) For which frequencies of the source are the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?
(d) What is the Q-factor of the given circuit? (C.B.S.E. 1992 )
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 27
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 28
Question 21.
Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 μF, and R = 7.4 Ω. It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. Suggest a suitable way.
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 29
Question 22.
Answer the following questions :
(a) In any ac circuit, is then applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?
(b) A capacitor is used in the primary circuit of an induction coil.
(c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L.
(d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.
(e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil?
Answer:
(a) Yes, the applied instantaneous voltage is equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit. It is because voltages across different elements are not in phase.
It is not true for rms voltages. It is because rms voltages across different elements are not in phase with each other.

(b) At the break, a large induced emf is produced. In case the capacitor is not connected, sparking will take place. But when the capacitor is used, the large induced emf produced at break is used up in charging the capacitor and no sparking takes place.

(c)
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 30
For d.c., XL = 0, therefore, XL = 0 and Xc = oo. Hence, d.c. signal appears across capacitor.For high frequency a.c., XL → High and Xc → 0. Hence, a.c. signal appears across inductor.

(d) When a choke coil in series with a lamp is connected to a d.c. line, L has no effect on the steady value of the current. Therefore, the brightness of the lamp is not affected by the insertion of the iron core in the choke. On a.c. line, the lamp will shine dimly due to the impedance of the choke coil. The brightness of the lamp will further go dim on the insertion of an iron core, which increases the impedance of the choke coil.

(e) The choke coil is used to reduce the current. As its power factor is zero, it reduces the current without wasting the power. If an ordinary resistor is used instead of a choke coil, ft will waste power in the form of heat.

Question 23.
A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary in order to get output power at 230 V? (C.B.S.E. 1997)
Answer:

NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 31

Question 24.
At a hydroelectric power plant, the water pressure head is at a height of 300 m and the water flow available is 100 m3s-1. If the turbine generator efficiency is 60%, estimate the electric power available from the plant (g = 9.8 ms-2).
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 32
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 33

Question 25.
A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wirelines carrying power is 0.5 Q per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
(c) Characterise the step-up transformer at the plant.(C.B.S.E. Sample Paper 2003)
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 34

Question 26.
Do the same exercise as above with the replacement of the earlier transformer by a 40,000 – 220 V step-down transformer (Neglect, as before, leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Hence, explain why high voltage transmission is preferred.
Answer:
NCERT Solutions for Class 12 Physics Chapter 7 Alternating Current 35

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NCERT Solutions for Class 12 History Chapter 3 Kinship, Caste and Class Early Societies

NCERT Solutions for Class 12 History Chapter 3 Kinship, Caste and Class Early Societies are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 3 Kinship, Caste and Class Early Societies.

Board CBSE
Textbook NCERT
Class Class 12
Subject History
Chapter Chapter 3
Chapter Name Kinship, Caste and Class Early Societies
Number of Questions Solved 9
Category NCERT Solutions

NCERT Solutions for Class 12 History Chapter 3 Kinship, Caste and Class Early Societies

Question 1
Explain why patriliny may have been particularly important among elite families.
Solution :
Patriliny means tracing descent from father to son, grandson and so on. While ; patriliny had existed prior to the composition of the Mahabharata, its central story reinforced the idea that it was valuable. It was important among elite and ruling families to avoid conflicts over land, power and resources including the throne in the case of kings of their fathers when the latter died as had happened between the Kauravas and the Pandavas, who belonged to a single ruling family.

There were, however, variations in practice. Sometimes if there were no sons, brothers succeeded one another. Sometimes other kinsmen claimed the throne, and in very exceptional ; circumstances, women such as Prabhavati Gupta exercised power.

Question 2.
Discuss whether kings in early states invariably were Kshatriyas.
Solution :
As per the Dharmashastra, only Kshatriyas were supposed to be the kings. But it was also to be noted that many important ruling lineages perhaps had different origins. Mauryas were considered Kshatriyas by many people. Some Brahmanical texts described Mauryas as of low origin. The Shungas and Kanvas who were immediate successors of the Mauryas were Brahmanas. In fact those sections of the society controlled the political power which enjoyed support and resources. It did not depend on the question of being bom as Kshatriya.

There were other rulers like Shakas who came from Central Asia. But the Brahmanas considered them as mlechchhas, barbarians and outsiders. Similarly, Gotami-putra Satkami, the best known ruler of the Satavahana dynasty, became a destroyer of the pride of kshatriyas. This we see that the Satavahanas claimed to be Brahmanas whereas the Brahmanas were of the opinion that the king should be Kshatriyas.

Question 3.
Compare and contrast the dharma or norms mentioned in the stories of Drona, Hidimba and Matanga.
Solution :
(a) The following principles of dharma or norms have been mentioned in the stories of Drona, Hidimba and Matanga:

  • Story of Drona : To teach Kshatriyas only, to take fee or guru dakshina from the pupils, and to keep his words.
  • Story of Hidimba : Marriage of Bhima with Hidimba against the principles of marriage.
  • Story of Matanga : To treat chandalas at the very bottom of the hierarchy of vamas due to handling of corpses and dead animals and treating them as “polluting” by those who claimed to be at the top of the social order.

(b) In all the three stories the dharma or norms as mentioned in the Dharmasutras and Dharmashastras have been violated in one way or the other. In case of Drona, he refused to have Ekalavya as his pupil because he was a forest dwelling nishada who did not fit into the fourfold varna system. It was considered a jati. Ekalavya acquired great skill in archery that was perhaps better than Arjuna before the image of Drona. He acknowledged Drona as his teacher. So, when Drona approached him and asked for his right thumb as his fee, he did not hesitate and offered it to his teacher. This shows that Drona followed a double standard towards Ekalavya. He refused to treat him as his pupil but in order to keep his words that no one would be better than Aijuna, he demanded the right thumb from Ekalavya as his fee. It was against the principles of morality.

In case of Hidimba, Yudhisthira agreed to the marriage conditionally. After giving birth to a rakshasa son, the mother and son left the Pandavas. This was in violation of the norms for marriages because rakshasa were the people whose practices differed from those laid down in Brahmanical texts.

In case of Matanga, on his first encounter with Dittha Mangalika, he was beaten as she had seen something inauspicious. Later on Matanga attained spiritual powers and married her. They had a son named Mandavya Kumara who treated Matanga unworthy of alms and treated him badly. When Mangalika learnt about the incident, she sought his forgiveness. This story shows that chandalas were regarded as “polluting”. At the same time this proves that they did not accept the life of degradation prescribed in the Shastra. Matanga attained spiritual powers. He gave a bit of the leftover from his bowl to Dittha Mangalika and asked her to give it to Mandavya and Brahmans which hint that occasionally the social realities were different from the Brahmanical texts. The above stories reflect the realities of social condition of that period.

Question 4.
In what ways was the Buddhist theory of a social contract different from the Brahmanical view of a society derived from the Purusha Sukta?(VBQ)
Solution :
The Purusha Sukta of the Rig Veda says that the four Vamas emerged because of the sacrifice of Purusha, the primeval man. The four vamas were Brahmanas, Kshatriyas, Vaishyas and Shudras. These Vamas had different jobs. The Brahmanas had supreme position in the society. They were also considered as teachers. Kshatriyas were considered warriors. They also ran the administration. The Vaishyas were the masters of trade. The Shudras were at the lowest strata. Their duty was to serve the above three vamas. Under this Brahmanical system, birth was the only criteria to judge the status and prestige in the society.

But the Buddhisftheory of a social contract was different. As per the Buddhist concept, there was inequality in society. But they also opined that this inequality was neither natural nor permanent. They did not favour the idea of birth being the criteria of social status.

Question 5.
The following is an excerpt from the Mahabharata, in which Yudhisthira, the eldest Pandava, speaks to Sanjaya, a messenger :
Sanjaya, convey my respectful greetings to all the Brahmanas and the chief priest of the house of Dhritarashtra. I bow respectfully to teacher Drona… I hold the feet of our preceptor Kripa … (and) the chief of the Kurus, the great Bhishma. I bow respectfully to the old king (Dhritarashtra). I greet and ask after the health of his son Duryodhana and his younger brother .. Also greet all the young Kuru warriors who are our brothers, sons and grandsons … Greet above all him, who is to us like father and mother, the wise Vidura (bom of a slave woman)… bow to the elderly ladies who are known as our mothers. To those who are our wives you say this, “I hope they are well-protected” … Our daughters-in-law bom of good families and mothers of children greet on my behalf. Embrace for me those who are our daughters … The beautiful, fragrant, well-dressed courtesans of ours you should also greet. Greet the slave women and their children, greet the aged, the maimed (and) the helpless …

Try and identify the criteria used to make this list – in terms of age, gender, kinship ties. Are there any other criteria ? For each category, explain why they are placed in a particular position in the list.
Solution :
(a) The list has been prepared on the following basis and order. The criteria used for each category has been mentioned against them :
(b) From the above list it is clear that each category has been placed keeping in view the age, gender and kinship ties as well as their varnas. Slave women and aged, the maimed and the helpless have been placed in the last two categories.

Question 6.
This is what a famous historian of Indian literature, Maurice Wintemitz, wrote about the Mahabharata : “just because the Mahabharata represents more of an entire literature … and contains so much and so many kinds of things. … (it) gives(s) us an insight into the most profound depths of the soul of the Indian folk.” Discuss.
Solution :
The above statement of Maurice Wintemitz about the Mahabharata seems to be correct because it is one of the richest texts of the subcontinent. It is a colossal epic running in

its present form into over 100,000 verses with depictions of a wide range of social categories and situations. It was composed over a period of about 1000 years and some of the stories it contains may have been in circulation even earlier. The text also contains sections laying down norms of behaviour for various social groups. Occasionally, the principal characters seem to follow these norms. Over the centuries, versions of the epic were written in a variety of languages. Several stories that originated in specific regions or circulated among certain people found their way into the epic. The central story of the epic was often retold in different ways. Its episodes have been depicted in sculpture and painting. They also provided themes for a wide range of performing arts – plays, dance, and other kinds of narrations. Its central story describes a feud over land and power between two groups of cousins. It reinforces the principle of patriliny. It mentions rules of marriage too. Thus, it is correct that it contains so many things and gives us an insight into the most profound depths of the soul of the Indian folk.

Question 7.
Discuss whether the Mahabharata could have been the work of a single author.
Solution :

  1. The Mahabharata could not have been the work of a single author because a text which initially perhaps had less than 10,000 verses grew to comprise about 100,000 verses. This enormous composition is traditionally attributed to a sage named Vyasa. There is also a tradition that Vyasa dictated the text to the deity Ganesha.
  2. The original story was probably composed by charioteer-bards known as sutas who generally accompanied Kshatriya warriors to the battlefields. They used to compose poems celebrating their victories and other achievements. These compositions circulated orally.
  3. Then from the fifth century BCE, Brahmanas began to commit it to writing.
  4. This was the time when the Kurus and Panchalas became kingdoms and perhaps the new kings wanted their itihas to be recorded and preserved systematically. Social values were often replaced by new norms and these have been mentioned in the Mahabharata.
  5. The next stage was between 200 BCE and 200 CE when worship of Vishnu grew in importance and Krishna was identified with Vishnu.
  6. Between 200 and 400 CE, large didactic sections resembling Manusmriti were added. Thus, the epic was neither written by one author nor was it written during one period.

Question 8.
How important were gender differences in early societies? Give reasons for your answer.
Solution :
It is seen that in early societies families were generally patriliny. Patriliny means tracing descent from father to son and to grandson and so on. Matriliny family was k generally not in use. But exception was also available. As exception, Satavahanas of Andhra can be mentioned. Historical sources mention the name of some rulers from inscriptions associated with the names of the mothers of the king. As Gotami-putra means’ son of Gotami’. Gotami and Vasistha are the feminines of Gotama and Vasistha. Sons were considered important for the continuity of the family. Attitudes towards daughter were different. They had no claims towards the resources of the household. But marrying them into the families outside the kin was considered desirable. This system of marriage was called exogamy. According to this system, the lives of the young ‘girls and women belonged to those families which claimed that high status were often carefully regulated to ensure that they were married at the right time and to the right person. This gave rise to the tradition that in marriage Kanyadana was an important religious duty of the father.

After marriage women were supposed to give up their father’s gotra and adopt their husband’s.As per Manusmriti, the paternal state was to be divided equally amongst sons after the death of parents, with a special share for the eldest. Women were not given any share
in this state.

But women were allowed to keep the gifts with themselves which they received at the [ time of their marriage. This was called stridhana. This could be inherited by. their children and the husband had no claim over it. But at the same time Manusmriti also told women not to hoard family property or even their own valuables without the permission of their husband.
In fact, social differences were sharpened because ofthe differences in access of resources.Many texts suggest that while upper class women may have access to resources but l and, cattle, money were generally controlled by the men. Vakataka queen Prabhavati Gupta was a rich woman.

Question 9.
Discuss the evidence that suggests that Brahmanical prescriptions about kinship and marriage were not universally followed.
Solution :
The following evidence suggests that Brahmanical prescriptions about kinship and marriage were not universally followed :

  1. Change in kinship relations : There was change in kinship relations. For example, the Mahabharata is a story of a feud over land and power between two groups of cousins, the Kauravas and the Pandavas. Ultimately, the conflict ended in a battle, in which the Pandavas emerged victorious.
  2. Patriliny : Patriliny means tracing descent from father to son, grandson and so on. But there were variations in practice. Sometimes, if there were no sons, brothers succeeded one another. Sometimes other kinsmen claimed the throne. In very exceptional circumstances, women such as Prabhavati Gupta exercised power.
  3. Marriages : The Dharmasutras and Dharmashastras recognised as many as eight forms of marriage. Of these, the first four were considered as “good” while the remaining were condemned. These were perhaps practised by those who did not accept Brahmanical norms.
  4. Gotra of women : According to Brahmanical practice, women were expected to give up their father’s gotra and adopt that of their husband on marriage and members of the same gotra could not marry. This was not followed universally. For example, names of many women who married Satavahana rulers had been derived from gotras such as Gotama and Vasistha, their father’s gotras. They did not adopt their husband’s gotra. Not only this, some of them belonged to the same gotra that was against exogamy. The system of endogamy or marriage within the kin group was too prevalent among several communities in south India.

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