Category: CBSE Class 12

NCERT Solutions for Class 12 Biology Chapter 9 Strategies for Enhancement in Food Production

These Solutions are part of NCERT Solutions for Class 12 Biology. Here we have given NCERT Solutions for Class 12 Biology Chapter 9 Strategies for Enhancement in Food Production

Question 1.
Explain in brief the role of animal husbandry in human welfare.
Solution:

• Animal husbandry evolves new techniques and technologies for the management of livestock like buffaloes, cows, pigs, horses, cattle, sheep, camels, goats, etc., that are useful to humans.
• These methods can also be applied to rearing animals like bees, silkworms, prawns, crabs, fishes birds, pigs, cattle, sheep, and camels for their products like milk, eggs, meat, wool, silk, honey, etc.

Role of animal husbandry in human welfare is discussed as follows:

1. Milk is an important product of farm animals that are consumed as such, in the form of curd, cheese, butter, ice cream, etc. Milk is the only source of animal protein for vegetarians and is a complete food. Most of the milk is obtained from cows and buffalo. Other milk-yielding animals are goat, sheep, camel, and yak.
2. Egg, like milk, is also a complete food. Chicken and duck are the two major sources of the egg.
3. Meat is a protein-rich diet that is obtained from all types of livestock, e.g., goat, sheep, pig, cattle, chicken, fish, etc.
4. Honey is a sweet syrup obtained from the hives of the honey bee. Honey is used in sweetening various preparations.
5. Fibers like wool and silk are two high-quality fibres which we get from animals. Wool is the hair of sheep, some goats, and rabbits. Silk is a product of silkworms.
6. The skins of many animals are converted into hides and leather.
7. Drought animals are trained to carry men and materials besides other functions, e.g., buffalo, bullock, horse, camel, ass, elephant, reindeer, yak.
8. The rearing of animals provides employment to many persons.
9. Animal byproducts like horns, feathers, bone, dung, and droppings are all used in developing useful products.

Question 2.
If your family owned a dairy farm, what measures would you undertake to improve the quality and quantity of milk production?
Solution:
Some of the measures to be followed for proper management of dairy farm are :

1. Selection of good breeds having high milk yielding potential according to the climatic conditions of the area.
2. The shed under which the cattle are kept should be well ventilated with an adequate water supply for drinking as well as for washing. Shed should have pucca floor and proper drainage channel.
3. The feed of the animals should be a balanced diet with right proportions of carbohydrates, fats, proteins, and roughage and it should be given timely in good quantity.
4. Cleanliness and hygiene comes first for maintaining the livestock’s health and productivity. So, washing cattle and taking precautionary measures while milking are a must.
5. Inspection, keeping records of the activities and consulting a veterinary doctor for regular checkups of the livestock should be undertaken.

Question 3.
What is meant by the term ‘breed’? What are the objectives of animal breeding?
Solution:
A group of animals which are related by descent to each other and possess similar characteristics like appearance, size, features etc. are said to belong to a breed. The purpose of animal breeding is to produce animals with increased yield, faster growth, improved reproductive rate.

Question 4.
Name the methods employed in animal breeding. According to you which of the methods is best? Why?
Solution:
Animal breeding is producing improved breeds of domesticated animals by improving their genotypes through selective mating. There are two methods of animal breeding, natural breeding which includes inbreeding, out-breeding, cross-breeding, out-crossing, etc., and artificial breeding which involves artificial insemination and multiple ovulation embryo transfer technology (MOET). It involves inseminating the native cows with the semen of superior bulls of exotic or indigenous breeds. Artificial breeding is the best method of breeding because of the following reasons:

• Semen collected from males may be used immediately or can be frozen and used later.
• The semen of desired bulls is collected under hygienic conditions, preserved, and sent to all insemination centres throughout the country.
• Semen collected from one bull can be used to inseminate many cows as fewer sperms are required to achieve conception when semen is deposited artificially. Hence, artificial insemination is very economical.
• It is healthier as the spread of sexually transmitted diseases can be controlled by this technique.

Question 5.
What is apiculture? How is it important in our lives?
Solution:
Apiculture is the practice of bee-keeping for the production of various products such as honey-bee’s wax, etc. Honey is a highly nutritious food source and is used as an indigenous system of medicines. Other commercial products obtained from honeybees include bee’s wax and bee pollen. Bee’s wax is used for making cosmetics, polishes and is even used in several medicinal preparations. Therefore, to meet the increasing demand of honey, people have started practicing bee-keeping on a large scale. It has become an income-generating activity for farmers since it requires low investment and is labour intensive.

Question 6.
Discuss the role of fishery in the enhancement of food production.
Solution:
Fishery is the rearing, breeding, catching & marketing of fishes and other aquatic animals. Fishes are important food for a large portion of human population. Meat of fishes is a rich source of proteins and other useful substances like polyunsaturated fatty acids (PUFA). The meat of other aquatic animals like prawn, crab is also consumed as food by human beings.

Question 7.
Briefly describe various steps involved in plant breeding.
Solution:
The major steps in breeding a new genetic variety of a crop are as follows:

1. Collection of variability.
2. Evaluation and selection of parents.
3. Cross-hybridization among the selected parents.
4. Selection and testing of superior recombinants.
5. Testing, release, and commercialization of new cultivars.

Question 8.
Explain what is meant by biofortification.
Solution:
Biofortification is method of breeding crops with higher levels of vitamins and minerals, or higher proteins and healthier fats in view to improve public health. E.g., iron-fortified rice containing five times more iron than other varieties, wheat variety, Atlas 66 having high protein content, maize varieties having high lysine and tryptophan are produced.

Question 9.
Which part of the plant is best suited for making virus-free plants and why?
Solution:
The terminal bud having apical meristem are the best-suited parts of the plant for making a virus-free plant because they are not infected by a virus.

Question 10.
What is the major advantage of producing plants by micropropagation?
Solution:
Micropropagation is the tissue culture technique used for rapid vegetative multiplication of ornamental plants and fruit trees by using small-sized explants. Because of the minute size of the propagules in the culture, the propagation technique is named micropropagation. This method of tissue culture produces several plants. Each of these plants will be genetically identical to the original plant from which explants were taken. Plants obtained by vegetative propagation of a single plant constitute a somaclonal. The members of a single somaclonal have the same genotype. It is the only process adopted by Indian plant biotechnologists in different industries mainly for the commercial production of ornamental plants like lily, orchids, Eucalyptus, Cinchona, blueberry, etc., and fruit trees like tomato, apple, banana, grapes, potato, Citrus, palm, etc.

Question 11.
Find out what the various components of the medium used for propagation of an explant in vitro are.
Solution:
The major components of the medium for in-vitro propagation are:

• Water
• Agar-agar
• Sucrose
• Inorganic salts
• Vitamins
• Amino acids
• Growth hormones like Auxin, Cytokinins.

Question 12.
Name any five hybrid varieties of crop plants which have been developed in India.
Solution:
Some of the hybrid varieties of plants in India are:

• Pusa Gaurav
• Pusa Sem 2
• Pusa Sem 3
• Pusa Sawani
• Pusa A-4

We hope the NCERT Solutions for Class 12 Biology Chapter 9 Strategies for Enhancement in Food Production help you. If you have any query regarding NCERT Solutions for Class 12 Biology Chapter 9 Strategies for Enhancement in Food Production, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction

NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Physics
Chapter	Chapter 6
Chapter Name	Electromagnetic Induction
Number of Questions Solved	16
Category	NCERT Solutions

Question 1.
Predict the direction of induced current in the situations described by the following Fig. 2(a) to (f).


Answer:
(a) By Lenz’s law, the face of the coil towards the south pole of the magnet opposes the south pole. So this face should behave as the south pole. Hence the current flows along qrpq.
(b) Similar to the above reason the current flows along yzxy and along prqp.
(c) When the coil is energised, with a cell the increasing current produces an inverse current in the nearby coil along yzxy.
(d) Similar to the above reason the current flows along zyvz.
(e) By Lenz’s law current is along xryx.
(f) Field lines being along the plane of the loop, there is no induced current.

Question 2.
Use Lenz’s law to determine the direction of induced current in the situations described by Figure:
(a) A wire of irregular shape turning into a circular shape;
(b) A circular loop being deformed into a narrow straight wire.


Answer:
When a wire of irregular shape turns into a circular (loop the magnetic flux linked with the loop increases due to an increase in area) The irregular shape. The induced e.m.f. will cause current to flow in such a direction, so that the wire forming the loop is pulled inward from all sides. It requires that the current should flow.
By applying Lenz’s Law, it follows that the current will flow in the direction a, d, c, b, a.

Question 3.
A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1s, what is the induced e.m.f. in the loop while the current is changing?
Answer:


Question 4.
A rectangular wire loop of sides 8 cm and 2 cm with a small cut ¡s moving out of a region of the uniform magnetic field of magnitude O.3 T directed normal to the mop. What ¡s the e.m.f. developed across the cut if the velocity of the loop is 1 cm s^-2 in a direction normal to the
(a) longer side,
(b) the shorter side of the loop? For how long does the induced voltage last in each case?
Answer:
e = Blυ

Question 5.
A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s^_1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the e.m.f. developed between the center and the ring.
Answer:

Question 6.
A circular coil of radius 8.0 cm and 20 turns rotates about its vertical diameter with an angular speed of 500 rad s^_1 a uniform horizontal magnetic field of magnitude 3.0 x 10^-2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?
Answer:

(v) Source of power loss is the external rotor which provides the necessary torque to rotate the coil.

Question 7.
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 ms^-1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 x 10^-4 Wb m^-2.
(a) What is the instantaneous value of the e.m.f. induced in the wire?
(b) What is the direction of the e.m.f.?
(c) Which end of the wire is at the higher electrical potential?
Answer:

(b) Using Fleming’s Right-hand rule, the direction of induced e.m.f. is from West to East.
(c) Since the rod will act as a source, the Western end will be at a higher electrical potential.

Question 8.
Current in a circuit falls from 5.0 A to 0.0 As. If an average emf of 200 V is induced, give an estimate of the self-inductance of the circuit.
Answer:

Question 9.
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Answer:
dΦ_B = M.dl = 1.5 x (20 – 0) = 30 Wb

Question 10.
A jet plane is travelling towards the west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of 5 x 10^-4 T and the dip angle is 30°?
Answer:

Question 11.
Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s^-1. If the cut is joined and the loop has a resistance of 1.6 Ω, how much power is dissipated by the loop as heat? What is the source of this power?
Answer:

The Source of this power is the external agency which brings a change in a magnetic field.

Question 12.
A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s^_1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10^-3 T cm^-1 along the negative x-direction (that is it increases by 10^-3 T cm^-1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10^-3 T s^_1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.
Answer:


The direction of the induced current is such that it increases the magnetic flux linking with the loop in positive 2-direction.

Question 13.
It is desired to measure the magnitude of the field between the poles of a powerful loudspeaker magnet. A small flat search coil of area 2 cm² with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction. The total charge flew in the coil (measured by a ballistic galvanometer connected to the coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of the magnet.
Answer:

Question 14.
The figure shows a metal rod PQ resting on the rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutually perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed-loop containing the rod = 9.0 mΩ. Assume the field to be uniform.

(a) Suppose K is open and the rod is moved with a speed of 12 cm s^-1 in the direction shown. Give the polarity and magnitude of the induced emf.
(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?
(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience a magnetic force due to the motion of the rod. Explain.
(d) What is the retarding force on the rod when K is closed?
(e) How much power is required (by an external agent) to keep the rod moving at the same speed (= 12 cm s^_1) when K is closed? How much power is required when K is open?
(f) How much power is dissipated as heat in the closed-circuit? What is the source of this power?
(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
Answer:
Here, B = 0.50 T; 1 = 15 cm = 15 x 10^-2 m;
R = 9.0m Q = 9.0 x 10³ fl
(a) Now, e = Bvl
Here, V = 12 cm s^{-1 =} 12 x 10^-2 ms^-1
∴ e = 0.50 x 12 x 10-²x 15 x 10 ² = 9 x 10^-3V
If q is a charge on as electron, then the electrons in the rod will experience magnetic Lorentz force \(-q[\vec { v } +\vec { B } ]\) P. Q. Hence, the end P of the rod will become positive and the end Q will become negative.

(b) When the switch K is open, the electron collects at the end Q. Therefore, excess change is built up at the end Q. However, when the switch K is closed, the accumulated charge at the end Q flows through the circuit,

(c) The magnetic Lorentz force on the electron is cancelled by the electronic force acting on it due to the electronic field set up across the two ends due to the accumulation of positive and negative charges at the ends P and Q respectively.

(d) Retarding force, F = BIl =B \(\frac { e }{ R } \)

Question 15.
An air-cored solenoid with length 30 cm, area of cross-section 25 cm² and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10^-3 s. How much is the average back emf induced across the ends of the open switch in the circuit ? Ignore the variation in magnetic field near the ends of the solenoid.
Answer:

Question 16.
(a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Figure

(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, υ = 10 m/s. Calculate the induced e.m.f. in the loop at the instant when x = 0.2 m. Take a = 0.1 m and assume that the loop has a large resistance.
Answer:
(a) Consider a small portion of the coil of thickness dt at a distance t from the current-carrying wire. Then the magnetic field strength experienced by this portion

Question 17.
A line charge X per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Figure.) A uniform magnetic field extends over a circular region within the rim. It is given by

What is the angular velocity of the wheel after the field is suddenly switched off?

Answer:
Change in magnetic field is given by,

We hope the NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction, help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 6 Electromagnetic Induction, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Physics
Chapter	Chapter 8
Chapter Name	Electromagnetic Waves
Number of Questions Solved	15
Category	NCERT Solutions

Question 1.
Figure shows a capacitor made of two circular plates each of radius 12 cm and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in figure). The charging current is constant and equal to 0.15 A.
(a) Calculate the capacitance and the rate of change of potential difference between the plates.
(b) Obtain the displacement current across the plates.
(c) Is Kirchhoffs first rule valid at each plate of the capacitor ? Explain.

Answer:

Yes. Because conduction current entering one plate is equal to the displacement current leaving that plate.

Question 2.
A parallel plate capacitor (shown in the figure) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V a.c. supply with an (angular) frequency of 300 rad s^_1.
(a) What is the r .m.s. value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of \( \overrightarrow { B } \) at a point 3.0 cm from the axis between the plates.

Answer:

Question 3.
What physical quantity is the same for X-rays of g g wavelength 10^-10 m, the red light of wavelength 6800 A, and radiowaves of wavelength 500 m?
Answer:
The speed in a vacuum is the same for all are c = 3 x 10⁸ ms^-1. (Electromagnetic waves)

Question 4.
A plane electromagnetic wave travels in a vacuum along Z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
Answer:
The electric field vector \( \overrightarrow { E } \) and magnetic field vector\( \overrightarrow { B } \) are in xy plane. They are normal to each other.

Question 5.
A radio can tune into any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band ?
Answer:

=> The corresponding wavelength band is 40 m to 25 m.

Question 6.
A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
Answer:
The frequency of the electromagnetic waves produced is the same as that of the oscillating charged particle. Hence the frequency of the electro­magnetic waves produced is, υ= 10⁹ Hz.

Question 7.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in a vacuum is B₀ =510 nT. What is the electromagnetic waves produced by the oscillator?
Answer:

Question 8.
Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is v = 50.0 MHz.
(a) Determine, B₀, ω, k and λ.
(b) Find expressions for E and B.
Answer:

Question 9.
The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hv (for the energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?
Answer:
Using the relation for photon energy,


Conclusion. The above result indicates that the different wavelengths in the electromagnetic spectrum can be obtained by multiplying roughly the powers often.
The visible wavelengths are spaced by a few eV.
The nuclear energy levels (from y rays) are spaced about 1 MeV.

Question 10.
In-plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 x 10¹⁰ Hz and amplitude 48 Vm^_1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field. [c= 3 x 10⁸ ms^-1].
Answer:

Question 11.
Suppose that the electric field part of an electromagnetic wave in a vacuum is
E = {(3.1 N/C) cos[(1.8 rad/m)y + (5.4 x 10⁶ rad/s)f]}\(\hat { i } \).
(a) What is the direction of propagation?
(b) What is the wavelength X?
(c) What is the frequency v?
(d) What is the amplitude of the magnetic field part of the wave?
(e) Write an expression for the magnetic field part of the wave.
Answer:

Question 12.
About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation?
(a) at a distance of 1 m from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect reflection.
Answer:

Question 13.
Use the formula λ_mT = 029 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?
Answer:

These numbers tell us the range of temperature required to obtain the different parts of the spectrum. For example, to obtain a wavelength of 1 μm, a temperature of 2900 K is required.

Question 14.
Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).
(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift.)
(c) 2.7 K (temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big-bang’ origin of the universe.)
(d) 5890 Å-5896 Å (double lines of sodium)
(e) 14.4 keV [energy of a particular transition in ⁵⁷Fe nucleus associated with a famous high-resolution spectroscopic method (Mossbauer spectroscopy).]
Answer:
(a) Radio waves (short-wavelength end)
(b) Radio waves (short-wavelength end)
(c)

(d) Given wavelength is of the order of 10^-7 m i.e visible radiations(yellow light)

Question 15.
Answer the following questions:

1. Long-distance radio broadcasts use shortwave bands. Why?  (C.B.S.E. 2005)
   
2. It is necessary to use satellites for long-distance TV transmission. Why? (C.B.S.E. 2005)
   
3. Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why? (C.B.S.E. 2009)
   
4. The small ozone layer on top of the stratosphere is crucial for human survival. Why ?(C.B.S.E. 2005, 2009)
   
5. If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?
6. Some scientists have predicted that a global nuclear war on the earth would be followed by a severe nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction? (C.B.S.E. 1995)
   

Answer:

1. The ionosphere reflects waves in these bands.
2. Television signals are of >30 MHz penetrate the ionosphere. Therefore, reflection is effected by satellites.
3. The atmosphere absorbs X-rays. while visible and radio waves can penetrate it.
4. It absorbs ultraviolet radiation from the sun and prevents it from reaching the earth’s surface and causing damage to life.
5. The temperature of the earth would be lower because the Greenhouse effect of the atmosphere would be absent.
6. The clouds produced by global nuclear war would perhaps cover substantial parts of the sky preventing solar light from reaching many parts of the globe. This would cause a ‘winter’. against which life on earth cannot withstand.

We hope the NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves, help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves, drop a comment below and we will get back to you at the earliest.

Here we are providing NCERT Solutions for Class 12 English Flamingo Poem 5 A Roadside Stand. Students can get Class 12 English A Roadside Stand NCERT Solutions, Questions and Answers designed by subject expert teachers.

A Roadside Stand NCERT Solutions for Class 12 English Flamingo Poem 5

A Roadside Stand NCERT Text Book Questions and Answers

A Roadside Stand Think it out 

Question 1.
The city folk who drove through the countryside hardly paid any heed to the roadside stand or to the people who ran it. If at all they did, it was to complain. Which lines bring this out? What was their complaint about?
Answer:
The lines that bring out the irritation of the passers-by are:
Or if ever aside a moment, the out of sorts
At having the landscape marred….
They complained that the disfigured paint of the stall spoilt the beauty of the landscape, the signposts pointed the wrong way and the stalls were not maintained.

Question 2.
What was the plea of the folk who had put up the roadside stand?
Answer:
The people of the roadside stand sat in prayer that some city traffic should stop by and buy their wares so that they could make some money to improve their life beyond mere survival.

Question 3.
The government and other social service agencies appear to help the poor rural people, but actually do them no good. Pick out the words and phrases the poet uses to show their double standards.
Answer:
The poet uses the word ‘greedy good-doers, beneficent beasts of prey’ and ‘enforcing benefits that are calculated’.

Question 4.
What is the ‘childish longing’ that the poet refers to? Why is it in vain?
Answer:
The poet refers to the tireless longing of the stall owners for some car to stop by and give them an opportunity to make some money. But they wait in vain because the cars just pass by without thinking of the hope and longing of the sad faces peeping from the windows. If at all they stop, it is to ask the way or to take turn.

Question 5.
Which lines tell us about the insufferable pain that the poet feels at the thought of the plight of the rural people?
Answer:
The lines that express the poet’s insufferable pain are:
I wonder how I should like you to come to me
And offer to put me gently out of my pain.

A Roadside Stand Extra Questions and Answers

A Roadside Stand Short Answer Questions

Question 1.
What is the untold sorrow of the owners of the roadside stand?
Answer:
The untold sorrow of the roadside stand owners is that nobody pays attention to the efforts of the country folk to make some money. The city folk just pass by their stalls without helping them to maike some money. Their lives have not progressed at all as they merely earn to survive.

Question 2.
What is the poet’s complaint in the poem?
Answer:
The poet does not complain like passers-by that the landscape has been marred. He is complaining about the lack of opportunity and encouragement to these people in the countryside. He is upset about the sorrow of those who had set up the roadside stall in the hope that people would stop by and some money would tickle into their palms.

Question 3.
Why do country people ask for money?
Answer:
The country people ask for money to improve their lives. They set up stalls on (he roadside in the hope that they would make some money by selling goods of daily use and make their life better, as they had seen in movies and as had been promised by the party in power.

Question 4.
What was the news that was doing the rounds?
Answer:
There was news that the people in power were planning to move all these rural people to the city next to the theatre and the big stores. Their lives would be secured and they would not have to worry about themselves any longer. They were promised that they would soon be pulled out of their poverty.

Question 5.
How would the innocent be soothed out of their wits?
Answer:
The selfish good-doers would outwit the simple innocent people into believing that their intentions and efforts were for their improvement, while they would be seeking their own profits from the labour of these folks.

Question 6.
Why are the cars called ‘selfish’?
Answer:
The poet has used a transferred epithet here. He actually means to call the car owners selfish as they just pass by without a thought for the plight of the owners of the roadside stands and if at all they do stop, it is either to complain or to turn their car round.

Question 7.
What is the sadness that lurks near the open window there?
Answer:
The poet is referring to the disappointed faces that wait in vain at their stall windows for someone to ask for their wares and drop some money in their palm. But their hopes for a better living are belied.

Question 8.
What is the open prayer made by the country folk?
Answer:
The country folk make an open appeal to the city dwellers that they should not be selfish. They expectantly pray for the city cars to stop at their roadside stand and help them lead a better life.

Question 9.
What is the trusting sorrow? What remains unsaid?
Answer:
The country folk trust their rich brethren in the city to come to their help but they feel sad when their trust is breached by the city people through their indifference. Although the city people have said nothing but their silence speaks volumes about their cold and indifferent attitude to the rural poor, who feel hurt by it.

Question 10.
Which things irritated those passers-by who stopped at the roadside stand?
Answer:
The passers-by got irritated by the tastelessly painted roadside stand. The thought that the artless decor of the stand was in disharmony with their surroundings and it had destroyed the scenic beauty of the landscape. Even their ‘N’ and ‘S’ on the signboards was wrongly presented. They did not approve of the things offered for sale.

Question 11.
Why did the people driving along the highway think that the landscape was marred?
Answer:
The people driving along the highway objected to the tastelessly painted roadside stand. They thought that the artless decor of the stand was in disharmony with the surroundings and had destroyed the scenic beauty of the landscape. Although the shed had been recently renovated but it could never impress the city dwellers. They were always critical and felt that these unhygienically maintained roadside stands marred the beautiful mountain scene.

Question 12.
Who actually stopped near the sheds put up by the farmers at the edges of the road?
Answer:
The poet states clearly that three cars stopped but none inquired about the prices of the farmer’s produce. One car stopped to reverse and another asked the way to where it was bound. The third foolishly asked if they could sell it a gallon of gas.

Question 13.
What would be the great relief for the poet in reference to these village folks?
Answer:
The poet says loudly that he would be happy to own the great relief if the pains of these people were removed at one stroke. Obviously, he is much moved by their pathetic plight of life. He wants something to be done to improve their lives economically.

Question 14.
What hope does the poet nurture about himself when he asks that these people should be put at one stroke out of their pain?
Answer:
The poet hopes that these people are put at one stroke out of their pain. The poet wants that the authorities should come to him and offer to put him ‘gently out of my pain’. The poet identifies himself with the village folks as far as their economic conditions are concerned.

Question 15.
What is the poet’s attitude to the good-doers and why is it so?
Answer:
The poet condemns the good-doers for they actually take away the villagers’ freedoms to think for themselves. They force benefits on them which lull them into doing nothing and destroy their peace of mind and their lives. He criticizes them for exploiting the villagers for their own gains.

Question 16.
What different attitudes do the city dwellers display to the country people?
Answer:
The city dwellers are indifferent to the plight of the country people and ignore the stands selling their goods. They get irritated with them for spoiling the landscape with their wrong signboards. They also exploit them for their selfish gains by offering them hollow charity which spoils their lives.

Question 17.
On what occasions do the country people express their anger at the city elite?
Answer:
The country people get angry with the city elite when, despite having money, they do not buy any of their goods. Again when a car stops and asks for gas which they obviously do not have, but does not ask the price of what they are actually selling.

Question 18.
What do the country people want?
Answer:
The country people want a share in the wealth enjoyed by the city people which they also have a right to, so that they can improve their conditions and lead better lives just as those promised by the movies and which the government has denied them.

Question 19.
Why are the country folks disappointed?
Answer:
The country folks have put up a roadside stand to sell their wares to the city dwellers. They desperately hope to earn some city money so that they could support their lives with it. They are disappointed because the city dwellers rush away in their polished cars with their minds focused only on their destination. If ever they pause, they are rather critical in their comments. They complain that the roadside stand had marred the scenic beauty of the landscape.

Question 20.
Bring out the contrast between the urban rich and the rural poor.
Answer:
The urban rich are on the move, they are in a hurry, they are speeding looking ahead. They have no time to inquire about the goods put up by rural poor for sale. On the other hand, the rural poor are standing and pleading for help.

Question 21.
How does the poet describe the double standards of the government and other social service agencies towards the poor rural people?
Answer:
The poet is sad that the government which came into power had many promises for the wellbeing of rural poor folks. But it and other social agencies did nothing for that. These poor rural people put up their roadside stands to sell what they produce. But no passer-by buys them. The poet feels much pain at their poor plight.

NCERT Solutions for Class 12 Physics Chapter 15 Communication Systems

NCERT Solutions for Class 12 Physics Chapter 15 Communication Systems are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 15 Communication Systems.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Physics
Chapter	Chapter 15
Chapter Name	Communication Systems
Number of Questions Solved	8
Category	NCERT Solutions

Question 1.
At which of the following frequency/frequencies the communication will not be reliable for a receiver situated beyond the horizon:
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz
Answer:
(b) is correct. Here (c) and (d) frequencies have high penetration power so the earth will absorb them. Radiation (a) of 10 kHz will suffer from the problem of the size of the antenna.

Question 2.
Frequencies in the UHF range normally propagate by means of
(a) ground waves
(b) sky waves
(c) surface waves
(d) space waves.
Answer:
(d) space waves.

Question 3.
Digital signals (i) do not provide a continuous set of values, (ii) represent values as discrete steps, (Hi) can utilize the only binary system, and (iv) can utilize decimal as well as a binary system. Which of the following options is true :
(a) Only (i) and (ii).
(b) Only (ii) and (iii).
(c) Only (i), (ii) and (iii), but not (iv).
(d) AH the above (i) to (iv).
Answer:
(c) is correct because the decimal system is concerned with continuous values (i) to (iii).

Question 4.
Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication? A TV transmitting antenna is 81 m tall. How much service area can it cover if the receiving antenna is at the ground level?
Answer:
For line-of-sight communication, it is necessary that the transmitting antenna and receiving antenna should be eye to eye but it is not necessary that they should be at the same height.

Question 5.
A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%?
Answer:

Question 6.
A modulation signal is a square wave as shown in the figure. The carrier wave is given by
C(t) = 2 sin(8πt) V

(a) Sketch the amplitude modulated waveform.
(b) What is the modulation index?
Answer:


Accordingly, the amplitude modulated waveform is shown ahead:

Question 7.
For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is found to be 2 V. Determine the modulation index μ. What would be the value of μ if the minimum amplitude is zero V?
Answer:

Question 8.
Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.
Answer:
Let there be two signals represented by
A_c cos ω_ct and A₀ cos(ω_c + ω_m)t where A_c is the
amplitude, ω_c is the angular frequency of a carrier wave at the receiving end and A₀ is the amplitude, (ω_c+ ω_m) is the angular velocity of the modulated wave.
Multiplying these signals, we get

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

NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Physics
Chapter	Chapter 2
Chapter Name	Electrostatic Potential and Capacitance
Number of Questions Solved	37
Category	NCERT Solutions

Question 1.
Two charges 5 x 10^-8 C and -3 x 10^-8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. Ans. Let the potential be zero at 0, then
Answer:
Let the potential be zero at 0, then v_A + V_B= 0, where V_A is electric potential due to charge q_A and V_B is the electric potential due to charge q_B.

Question 2.
A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the center of the hexagon.
Answer:
Total potential at O is given by,

Question 3.
Two charges 2μC and -2μC are placed at points A and B, 6 cm apart.
(a) Identify an equipotential surface of the system. % What is the direction of the electric field at every through the mid-point. On this plane, the potential is zero everywhere.
(b) The direction of the electric field is from positive to negative charge i.e. A to B, which is in fact perpendicular to the equipotential plane.
Answer:
(a) A surface containing an equatorial line and a perpendicular line.
(b) Towards the side of – ve charge, parallel to the axis.

Question 4.
A spherical conductor of a radius of 12 cm has a charge of 1.6 x 10^-7 C distributed uniformly on its surface. What is the electric field
(a) inside the sphere
(b) just outside the sphere
(c) at a point 18 cm from the center of the sphere?
Answer:
(a) inside a conductor, the electric field is zero because the charge resides on the surface of a conductor.
(b) Electric field just outside the sphere is given by

Question 5.
A parallel plate capacitor with air between the plates has a capacitance of 8 pF
(1 pF = 10^-12 F.) What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Answer:

Question 6.
Three capacitors each of capacitance 9 pF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply ?
Answer:

Question 7.
Three capacitors of capacitances 2 pF, 3 pF and 4pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Answer:
(a) Total capacitance
= c₁ + c₂ + c₃
= 2 + 3+ 4 = 9pF.
(b) Using C = \(\frac { q }{ v } \) we get q = CV
∴ q_x = C₁V = 2 x 10-¹² x 100
= 2 x 10-¹⁰ C = 200 pC
q₂ = c₂V
= 3 x 10^-12 x 100
= 3 x 10-¹⁰ C = 300 pC
q₃ = c₃v
= 4 x 10-¹² x 100
= 4 x 10-¹⁰ C = 400 pC

Question 8.
In a parallel plate capacitor with air between the plates, each plate has an area of
6 x 10^-3 in² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor ?
Answer:

Question 9.
Explain what would happen if in the capacitor given in Q. 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
(a) while the voltage supply remained connected.
(b) after the supply was disconnected.
Answer:

Question 10.
A 12 pF capacitor is connected to a 50 V battery.How much electrostatic energy is stored in the capacitor ?
Answer:
E = \(\frac { 1 }{ 2 } \) CV² = \(\frac { 1 }{ 2 } \) x 12 x 10-¹² x 50 x 50 = 1.5 2 2 x 10-⁸J.

Question 11.
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from die supply and is connected  to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process ?
Ans.
Here, C₁ = 600pF = 6 x 10^-10 F, C₂ =
6 x 10-¹⁰ F, V₁ = 200V, V₂ = 0

Question 12.
A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of -2 x 10^-9 C from a point P(0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point
R(0, 6 cm, 9 cm.)
Answer:
The work done by electrostatic force on a charge is independent of the path followed by the charge. It depends only on the initial and final positions of the charge.

Question 13.
A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the center of the cube.
Answer:
(1) Distance of the center of the cube from vertex is half of the diagonal of the cube

(2) From symmetry, it is clear that electric field at center of the cube is zero.

Question 14.
Two tiny spheres carrying charges 1.5 μc and 2.5 μc are located 30 cm apart. Find the potential and electric field :
(a) at the mid point of the line joining the two charges,
(b) at a point 10 cm from this mid-point in a plane normal to the line and passing through the mid­point.
Answer:

Question 15.
A spherical conducting shell of inner radius r_l and outer radius r₂ has a charge Q.
(a) A charge q is placed at the center of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
(b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.
Answer:
(a) Charge Q appears on the outer surface.

When charge q is placed at the center, it induces – q charge on the inner surface and +q on the outer surface.
.’. charge density of the inner surface,

and charge density of the outer surface,

Consider a cavity of an irregular shape with the net charge to be zero inside it. Let a closed-loop be partially inside and the rest outside the cavity. The field inside the conductor is zero, so some work is done by the field to carry a test charge in the closed-loop, but this is against the provisions of an electrostatic field because as per Gauss’s law, the net charge inside a Gaussian surface must be zero. Thus, there cannot be field lines inside the cavity irrespective of its shape.

Question 16.
(a) Show that the normal component of the electrostatic field has a discontinuity from one side of a charged surface to another given by

where \(\hat { n } \) is a unit vector normal to the surface at a point and a is the surface charge density at that point. (The direction of h is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σ\(\hat { n } \) /ε₀
(b) Show that the tangential component of the electrostatic field is continuous from one side of a charged surface to another.[Hint. For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]
Answer:
Consider a sheet of charge having charge density a. E on either side of the sheet, perpendicular to the plane of sheet, has same magnitude at all points equidistant from the sheet.
Electric field intensity on the left side of the sheet,

The electric field tangential to the plate is continuous throughout.

Question 17.
a long charged cylinder of linear charged density A. is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?
Answer:
A cylinder P has linear charge density, λ, length l, and radius r₁
The charge on cylinder P, q = XL A hollow co-axial conducting cylinder of length / and radius r₂ surrounds the cylinder P. Charge on cylinder Q = – q.

Consider a Gaussian surface in the form of a cylinder of radius r and length l. The electric flux through the curved surface of the Gaussian surface,

Question 18.
In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:
(a) Estimate the potential Energy of the system in eV, taking the zero of the potential energy at an infinite separation of the electron from the proton.
(b) What is the minimum work required to free the electron, given that it’s kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 A separation?
Answer:

Question 19.
If one of the two electrons of an H₂ molecule is removed, we get a hydrogen molecular ion H⁺₂. In the ground state of an H⁺₂, the two protons are separated by roughly 1.5 Å, and the electron is roughly Å from each proton. Determine the potential energy of the system. Specify your choice of zero potential energy.
Answer:

Question 20.
Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of the electric field at the surfaces of the two spheres? Use the result obtained to explain why the charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
Answer:
Two charged conducting spheres of radii a and b connected by a wire will reach to the same potential.

Clearly, electric charge density for the pointed surface will be more because a flat surface can be equated to a spherical surface of large radius and a pointed portion to a spherical surface of small radius.

Question 21.
Two charges -q and + q are located at points (0, 0, -a) and (0, 0, a), respectively.
(a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0)?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a > > 1.
(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (-7, 0, 0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?
Answer:

The point (x, y, 0) is perpendicular to Z-axis, there­fore, the potential at (x, y, 0) is zero.
(b) Consider P to be the point of observation at a distance r from the center (O) of the electric dipole.
Let OP make an angle 0 with the dipole moment \( \overrightarrow { p } \)
and r_1, r₂ be the distances of point P from – q charge and + q charge respectively. Potential at P due to – q charge,



The answer does not change because, in electrostatics, the work done does not depend upon the actual path, it simply depends upon the initial and final positions.

Question 22.
The figure shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/a > > l, and contrast your results with that due to an electric dipole, and an electric monopole
(i.e., a single charge.)

Answer:

(2) Due to electric dipole, the potential is of 1/r² type.
(3) Due to an electric monopole, the potential is of 1/r type.

Question 23.
An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1 kV. A large number of 1 μF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
Answer:
Let N capacitors be used in m rows when each row has n capacitor i.e. N = mn
In series

Question 24.
What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm ? (You will realise from your answer why ordinary capacitors are in the range of μF or less. However, electrolytic capacitors do have a much larger capacitance ((0.1 F) because of very minute separation between the conductors.)
Answer:

Question 25.
Obtain the equivalent capacitance of the network in figure. For a 300 V supply, determine the charge and voltage across each capacitor. (C.B.S.E. 2008)

Answer:
The equivalent circuit is as shown below :

Potential difference across C₄ is in the ratio 2:1
i.e., 200 V
.’. Charge on C₄ = C₄V₄
= 100 x 200 x 10^-12 = 2 x 10^-8 C
Potential difference across C₁= 100 V
Charge on C₁ = C₁x V₁
=100 x 100 x 10-¹² = 1 x 10^-8 C
Potential difference across C₂ and C₃ is 50 v each
∴ Charge on C₂ or C₃ = C₂V₂
= 200 x 50 x 10^-12 = 10-⁸ C.

Question 26.
The plates of a parallel plate capacitor have an area of 90 cm² each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume Hence arrive at a relation between u and the magnitude of electric field E between the plates.
Answer:

Question 27.
A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 2 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
Answer:

Question 28.
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2) QE, where Q is the charge on the capacitor and E is the magnitude of the electric field between the plates. Explain the origin of the factor (1/2).
Answer:
Let F be the force on each plate of the capacitor. If the distance between the plates of the capacitor is increased by dx, then work done = F dx. This work done is stored as the potential energy of the capacitor. The increase in the volume of capacitor = A dx

Question 29.
A spherical capacitor consists of two concentric spherical conductors held in position by suitable insulating supports (Figure.) Show that the capacitance of a spherical capacitor is given by

where r_x and r₂ are the radii of outer and inner spheres, respectively.
Answer:
It consists of two concentric spherical shells A and B of radii a and b with charge +q and charge -q respectively. (Outer sphere is grounded)

Question 30.
A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 pC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.
(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere?
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
Answer:

Question 31.
Answer carefully:
(a) Two large conducting spheres carrying charges Q₁ and Q₂ are brought close to each other. Is the magnitude of the electrostatic force between them exactly given by Q₁Q₂/4Πε₀r², where r is the distance between their centers?
(b) If Coulomb’s law involved 1/r³ dependence (instead of 1/r²), would Gauss’s law be still true?
(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?
(e) We know that the electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?
(f) What meaning would you give to the capacitance of a single conductor?
(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (=6).
Answer:
(a) No. The given relation is Coulomb’s law which is true for point charges. In the present case, as the spheres are brought closer, the distribution of charge on them becomes nonuniform.

(b) No. The surface area in space varies as r₂ so that field varies as \(\frac { 1 }{ { r }^{ 2 } }\). Hence \(\frac { 1 }{ { r }^{ 2 } }\) dependence is essential.

(c) Not necessarily. The motion of charged particles need not be along the line of the field. It does so in the uniform field. The field gives the direction of acceleration and not that of velocity in general.

(d) Zero. For any complete path in the electrostatic field (the shape does not matter), it is zero.

(e) No. Potential is continuous there.

(f) The single conductor can form a condenser with the other conductor at infinity. Hence the meaning of storage of charge retains.

(g) Water molecules are polar molecules.

Question 32.
A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 pC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects, (i.e., bending of field lines at the end)
Answer:
The capacitance of a cylindrical capacitor is given by


Question 33.
A parallel plate capacitor is to be designed with a voltage rating of 1 kV, using a material of dielectric constant 3 and dielectric strength of about 10⁷ Vm^-1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionization.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
Answer:
10% of the given field i.e. 10⁷ V m¹ gives E = 0.1 X 10⁷ Vnr¹

Question 34.
Describe schematically the equipotential surfaces corresponding to
(a) a constant electric field in the z-direction,
(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction,
(c) a single positive charge at the origin, and
(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
Answer:
(a) A plane parallel to XY plane.
(b) Plane parallel to XY plane but the planes having different fixed potential will become closer with the increase in field intensity.
(c) Concentric spheres with origin as the center.
(d) A time-dependent changing shape nearer to the grid which slowly becomes planar and parallel to the grid at far off distances from the grid.

Question 35.
In a van de Graaff type generator a spherical metal shell is to be a 15 X 10⁶ V electrode. The dielectric strength of the gas surrounding the electrode is 5 x 10⁷ vm^-1. What is the minimum radius of the spherical shell required? (You will learn from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.) (C.B.S.E. 2008)
Answer:
The minimum radius of the shell of the van de Graaff generator is given by the relation

Question 36.
A small sphere of radius r₁ and charge q₁ is enclosed by a spherical shell of radius r₂ and charge q₂ . Show that if q₁ is positive, the charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge
q₂ on the shell is.
Answer:
Charge resides on the outer surface of a conductor. So the charge on the inner sphere will flow towards the shell through the conducting wire. Moreover, from Gauss’s law, no electric field exists inside a Gaussian surface, and also the charges enclosed by a closed surface only contribute towards the field. So q₂ does not matter in this case. It is positive, a potential difference is also positive.

Question 37.
Answer the following:
(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decrases with altitude. Near the surface of the earth, the field is about 100 Vm^_1. Why then do we not get an electric shock as we step out of our house into the open ? (Assume the house to be a stell cage so there is no field inside !)
(b) A man fixes outside his house one evening a two meter high insulating slab carrying on its top a large aluminium sheet of area 1 m². Will he get an electric shock if he touches the metal sheet next morning ?
(c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral ? In other words, what keeps the atmosphere charged ?
(d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning ?
[Hint. The earth has an electric field of about 100 Vm^-1 at its surface in the downward direction, corresponding to a surface charge density = -10^-9 C m^-2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about +1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.]
Answer:
(a) Our body and the ground form an equipotential surface. As we step out into the open, the original equipotential surfaces of open-air change, keeping our head and the ground at the same potential.

(b) Yes. The steady discharging current in the atmosphere charges up the aluminum sheet gradually and raises its voltage to an extent depending on the capacitance of the capacitor (formed by the sheet, slab, and the ground).

(c) The atmosphere is continually being charged by thunderstorms and lightning all over the globe and discharged through regions of ordinary weather. The two opposing currents are, on average, in equilibrium.

(d) Light energy involved in lightning; heat and sound energy in the accompanying thunder.

We hope the NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance, help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 History Chapter 13 Mahatma Gandhi and the Nationalist Movement Civil Disobedience and Beyond are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 13 Mahatma Gandhi and the Nationalist Movement Civil Disobedience and Beyond.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	History
Chapter	Chapter 13
Chapter Name	Mahatma Gandhi and the Nationalist Movement Civil Disobedience and Beyond
Number of Questions Solved	9
Category	NCERT Solutions

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

Question 1.
How did Mahatma Gandhi seek to identify with the common people ?
Solution :
Mahatma Gandhi was appreciated by the people because he dressed like them, lived like them, and spoke their languages. He tried to identify with them in the following ways :

• Gandhiji dressed himself in simple dhoti or loin cloth.
• He spent part of each day working on the charkha (spinning wheel) and encouraged other nationalists to do likewise.
• His new appearance i.e., shaved head and wearing a loincloth, came to symbolise asceticism and abstinence – the qualities he had in opposition to the consumerist culture of the modem world.

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

Question 3.
Why did the salt laws become an important issue of struggle ?
Solution :

1. The salt laws became an important issue of stmggle because these laws gave the state a monopoly in the manufacture and sale of salt.
2. For in every Indian household, salt was indispensable, yet people were forbidden from making salt even for domestic use, compelling them to buy from shops at a high price.
3. The salt tax at times was even fourteen times its value.
4. According to Gandhiji, the government not only prevented the public from manufacturing but also destroyed what nature manufactured without effort. The salt officers were posted for carrying on destruction of natural salt.
5. Gandhiji stated that the salt monopoly was a fourfold curse :
   • It deprived the people of a valuable easy village industry,
   • It involved wanton destruction of property that nature produced in abundance,
   • Its destruction meant more national expenditure, and
   • Tax of more than 1000 per cent was exacted from a starving people.

Question 4.
Why are newspapers an important source for the study of the national i movement ?
Solution :
Newspapers, published in English as well as in the different Indian languages are important contemporary sources because they tell us about daily movements of leaders. They report on their activities and their views. Newspapers publish the views of the ordinary people in different parts of the country and the ways in which they participate in the various movements. The newspapers tell us about the objectives of the people and their expectations from their leaders. The newspapers publish all types of views which are generally different from each other. For example, a newspaper published in London gave the view point and reaction of the British public while an Indian newspaper gave the reaction of the people of India. Thus, newspapers are an important source for the study of the national movement.

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

Question 6.
How was non-cooperation a form of protest?
Solution :
Non-cooperation was a form of protest in the following ways :

1. Indians were asked to adhere to a “renunciation of all voluntary association with the government to end colonialism”.
2. Gandhiji had joined hands with the Khilafat Movement to restore the Caliphate, a symbol of Pan-Islamism which had been abolished. It was also hoped that Britain would impose a harsh treaty on Turkey after its defeat in World War I.

Thus, it was a protest against the British policies in India and towards Turkey. Gandhiji hoped that if non-cooperation was effectively carried out, India would win Swaraj within a year.

Question 7.
Why were the dialogues at the Round Table Conference inconclusive?
Solution :
The dialogues at the Round Table Conference were inconclusive due to the following factors :

1. The First Conference was held in November 1930 when the Civil Disobedience Movement was being organised by the Congress. So, none of its leaders was present in the conference and without Congress participation it could not succeed.
2. Second Round Table Conference was held in the latter part of 1931. Gandhiji represented as the sole representative of

Congress. But his representation was challenged by the Muslim League, the Princes, and B.R. Ambedkar.
Under these circumstances, the Conference was inconclusive and could not take any decision.

Question 8.
In what way did Mahatma Gandhi transform the nature of the national movement?
Solution :
Before Gandhiji came to India, the national movement was limited to few sections of society. During the Swadeshi Movement of 1905-07, Bal Gangadhar Tilak, Bipin Chandra Pal and Lala Lajpat Rai tried to make it an all-India phenomenon but even then the participation of ordinary people remained limited. In his speech at Banaras in February, 1916, even Gandhi said that Indian nationalism was an elite phenomenon – a creation of lawyers, doctors and landlords. He reminded that the peasants and workers were not represented there. He desired to make Indian nationalism more properly representative of the Indian people as a whole. And thereafter whenever he got a chance he tried to implement his desire into action. Thus, in 1917 and 1918, he took initiatives at Champaran, Ahmedabad and Kheda which marked Gandhiji as a nationalist with a deep sympathy for poor.

In 1919, he called for a countrywide campaign against the “Rowlatt Act”. The campaign in the Punjab led to Jallianwala Bagh massacre. This satyagraha made Gandhiji a truly nationalist leader. Gandhiji was encouraged by the success of these satyagrahas and decided to start “non¬cooperation” movement which totally changed the nature of the national movement because it was the first mass movement in which all classes of people — students, lawyers, peasants, etc. took part. Thus, by 1922 he had transformed Indian nationalism, thereby redeeming the promise he made in his BHU speech of February 1916. It was no longer a movement of professionals and intellectuals but of hundreds of thousands of peasants, workers and artisans.

Question 9.
What do private letters and autobiographies tell us about an individual? How are these sources different from official accounts ?
Solution :
(a)

• Private letters give us a glimpse of individual’s thoughts.
• In letters, a person expresses his anger and pain, his dismay and anxiety, his hopes and frustrations in ways in which he may not express himself in public statements.
• Sometimes an individual cannot express his opinion in letters due to fear that a letter may be printed in future.

(b)

• Autobiographies give us an account of the past of the individual. It is often rich in human detail.
• Autobiographies tell us what an individual recollect from his memory. It could be important from the point of view of an individual.

(c) These sources – private letters and autobiographies – are different from official accounts because private letters and autobiographies are written according to the wishes of an individual. On the other hand, official accounts are written while performing official duty. These represents the view point of the government. For example, the fortnightly reports that were prepared by the Home Department were based on police information and expressed what the higher officials saw or wanted to believe.

We hope the NCERT Solutions for Class 12 History Chapter 13 Mahatma Gandhi and the Nationalist Movement Civil Disobedience and Beyond help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 13 Mahatma Gandhi and the Nationalist Movement Civil Disobedience and Beyond, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism

NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 4 Moving Charges and Magnetism

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Physics
Chapter	Chapter 4
Chapter Name	Moving Charges and Magnetism
Number of Questions Solved	28
Category	NCERT Solutions

Question 1.
A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the center of the coil?
Answer:

Question 2.
A long straight wire carries a current of 35 A. What is the magnitude of the Held B at a point 20 cm from the wire ?
Answer:

Question 3.
A long straight wire in the horizontal plane carries a current of 50 A in the north to south direction. Give the magnitude and direction of R at a point 2.5 m east at the wire.
Answer:

Question 4.
A horizontal overhead power line carries a current of 90A in the east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?
Answer:

Question 5.
What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?
Answer:

Question 6.
A 3.0 cm wire carrying a current of 10A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?
Answer:
I = 10 A, B = 0.27 T
1 = 3cm = 0.3m
and θ = 90°
F = BI 1 sin θ
=0.27×10 × 0.03 × sin90°
= 0.27×10 x 0.03 x 1
= 8.1 × 10^-2N.

Question 7.
Two long and parallel straight wires A and B carrying currents of 8*0 A and 5*0 A in the same direction are separated by a distance of 4-0 cm. Estimate the force on a 10 cm section of wire A.
Answer:

Question 8.
A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its center.
Answer:

Question 9.
A square coil of side 10 cm consists for 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Answer:
N = 20, A = 100 x 10^-4 m²
I = 12 A
B = 0.8 T
θ = 30°
Torque, τ = NI(\(\bar { A }\) x \(\bar { B }\))
= 20 x 12 x 100 x 10^-4 x 0.8 x sin 30°1
= 20 x 12 x 100 x 10^-4 x 0.8 x \(\frac { 1 }{ 2 }\)
= 0.96 Nm

Question 10.
Two moving .coil meters, M₁and M₂ has the following particulars:
R₁ = 10 Ω, N₁ = 30, A₁= 3.6 x 10^-3m², B₁= 0.25 T
R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 x 10^-3 m², B₂ = 0.50 T (The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M₁.
Answer:
Using Current sensitivity = NBA/k
For M₁ Current Sensitivity


Question 11.
In a chamber, a uniform magnetic field of 6.5G (1G = RHT) is maintained. An electron is shot into the field with a speed of 4.8 x 10⁶ ms^-1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit,
(e =1.6 x 10^-19 C, m = 9.1 x 10^-31 kg).
Answer:

Question 12.
In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
Answer:

Question 13.
(a) A circular coil of 30 turns and radius 8-0 cm carrying a current of 6-0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter-torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar toil of some irregular shape that encloses the same area? (All other particulars are also unaltered)  (C.B.S.E. 1998 C )
Answer:
(a) Using τ = NBIA sin 9, we get
τ= 30 X 1 x 6 x n (8 X 10^-2)² sin 60
= 180 x it (8 x 10-²)² 0.866
= 3.13 N m
The magnitude of the counter-torque is 3 .13 N m
(b) Answer will not change because torque does not depend upon the shape of the coil provided it encloses the same area.

Question 14.
Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A: coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their center.
Answer:
For Coil X

Question 15.
A magnetic field of 100 G (1 G = 10^-4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10^-3 m². The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m^-1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.
Answer:

We may take I = 10 A and n = 800. The given solenoid may have a length of 50 cm having 400 turns and area of cross-section = 5 x 10^-3m² (five times the given value.)

Question 16.
For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its center is given by,

(a) Show that this reduces to the familiar result for field at the center of the coil.
(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,

(Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.)
Answer:

(b) Let there be two coils as mentioned in the statement. The magnetic field in a small region of length 2d about the mid-point of the space between the two coils is given by,


Question 17.
A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field
(i) outside the toroid
(ii) inside the core of the toroid, and
(iii) in the empty space surrounded by the toroid?
Answer:
(i) Zero

Question 18.
Answer the following questions:
(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?

(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?

(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in the north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight-line path.
Answer:
(a) The force on a charged particle moving inside the magnetic field is given by
\({ \vec { F } }_{ m }=q(\vec { v } \times \vec { B } )\)
The force on the charged particle coil bd zero (will remain undeflected), if v x B is zero. Therefore, either the initial velocity v is parallel or anti-parallel to the magnetic field B.

(b) The magnetic field exerts force on the charged particle, which is always perpendicular to its motion and hence does no work. Therefore, charged particle will have its final speed equal to its initial speed, provided it suffered no collision with the environment.

(c) Under the action of the electrostatic field, the electron will be deflected towards north (towards the positive plate). It will remain undeflected if the force due to the magnetic field is towards south. As the velocity: y of the electron is from west to east, the expression for the magnetic Lorentz force i.e. \({ \vec { F } }_{ m }=-e(\vec { v } \times \vec { B } )\) it tells that the magnetic field \(\vec { B } \) should be applied along the vertical and in downward direction. The direction of the magnetic field may be found by applying Fleming’s left-hand rule.

Question 19.
An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30° with the initial velocity.
Answer:
K.E, acquired by electron while passing through V

(b) When electron moves with velocity r making an angle of 30° with the direction of magnetic field, then r cos θ is


Question 20.
A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 x 10⁵ V m^-1, make a simple guess as to what the beam contains. Why is the answer not unique.
Answer:

The given particle may be deutron.
The result is not unique because this e/m ratio can be true for He^{+ +} , Li ^{+ + +} etc.

Question 21.
A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.”
(a) What magnetic field should be set up normally to the conductor in order that the tension in the wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field the same as before? (Ignore the mass of the wires.) g = 9.8 ms^-2.
Answer:
(a) The tension in the wire is zero if the force on the current-carrying wire due to current is equal and opposite to the weight of the wire. This is, BIl= mg

(b) In case the current is reversed, the tension is equal to the force acting on the wire due to the magnetic field plus the weight of the wire. This is,
T = BlL + mg
= 0.26 x 5 x 0.45 + 60 x 10^-3 x 9.8
= 1.18 N.

Question 22.
The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive (H.S.E.B.2001)
Answer:

Since the currents in two wires are in opposite directions so the force is repulsive.

Question 23.
A uniform magnetic field of 1.5 T exists in a cylindrical region of a radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying a current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,
(a) the wire intersects the axis,
(b) the wire is turned from N-S to the northeast-northwest direction?
(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?
Answer:

Question 24.
A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Figure? What is the force on each case? Which case corresponds to stable equilibrium?

Answer:

Question 25.
A circular coil of 20 turns and a radius of 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the
(a) total torque on the coil,
(b) the total force on the coil,
(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area 10^-5 m², and the free electron density in copper is given to be about 10²⁹ m³)
Answer:

Question 26.
A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its center) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with an appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? g = 9.8 ms^-2.
Answer:

Question 27.
A galvanometer coil has a resistance of 12 Q and the meter shows full-scale deflection for a current of 3 mA. How will you convert the meter into a voltmeter of range 0 to 18 V?
Answer:
Here 1=3 mA = 3 x 10^-3 A
Galvanometer resistance, G = 12 Ω The galvanometer can be converted into the voltmeter of range 0 to V (here V = 18 V) by connecting a high series resistance R given

Question 28.
A galvanometer coil has a resistance of 15Ω and the meter shows full-scale deflection for a current of 4 mA. How will you convert the meter into an ammeter of range 0 to 6?
Answer:

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

NCERT Solutions for Class 12 History Chapter 5 Through the Eyes of Travelers Perceptions of Society are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 5 Through the Eyes of Travellers Perceptions of Society.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	History
Chapter	Chapter 5
Chapter Name	Through the Eyes of Travelers Perceptions of Society
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for Class 12 History Chapter 5 Through the Eyes of Travellers Perceptions of Society

Question 1.
Write a note on the Kitab-ul-Hind.
Solution :

1. The Kitab-ul-Hind is written in Arabic and is simple and lucid.
2. It has 80 chapters on subjects such as religion, and philosophy, festivals, astronomy, alchemy, manners and customs, social life, weights and measures, iconography, laws and metrology.
3. Al-Biruni adopted a distinctive structure in each chapter, beginning with a question, following this up with a description based on Sanskritic traditions, and concluding with a comparison with other cultures. This almost geometric structure is remarkable for its precision and predictability.
4. He probably intended his work for peoples living along the frontiers of the subcontinent.

Question 2.
Compare and contrast the perspectives from which Ibn Battuta and Bernier wrote their accounts of their travels in India.
Solution :
Both have written them accounts in their different prospectives. While Ibn Battuta describe everything that impressed and excited him because of his novelty, Bernier had followed a different intellectual tradition. He wrote whatever he saw in India.
Bernier wanted to pin point the weakness of the Indian society and considered the Mughal India Inferior to European society. In his description Ibn Battuta recorded his observation about new culture, people, believes and values.

Question 3.
Discuss the picture of urban centres that emerges from Bernier’s account.
Solution :
During the seventeenth century, about 15 per cent of the population in India lived in cities. This was higher than the urban population in Western Europe in the same period. In spite of this Bernier described Mughal cities as “Camp towns”. These towns owed their existence, and depended for their survival on the imperial camp. He believed that these camp towns came into existence when the imperial court moved in and rapidly declined when it moved out. He stated that these “camp towns” did not have viable social and economic foundations. They were dependent on imperial patronage.

The above picture of urban centres that emerges from Bernier’s account does not seem to be correct because it is an oversimplified picture. There were all kinds of urban centres or towns i.e., manufacturing towns, trading towns, port towns, sacred centres and pilgrimage towns.

Question 4.
Analyse the evidence for slavery provided by Ibn Battuta.
Solution :
Travellers who left written accounts, sometimes took special inequalities for granted as a “natural” state of affairs. For example, is the evidence for slavery provided by Ibn Battuta that is as given below :

1. Slaves were openly sold in markets, like any other commodity and were regularly exchanged as gifts.
2. When Ibn Battuta reached Sind, he purchased “horses, camels and slaves” as gifts for Sultan Muhammad bin Tughlaq.
3. When he reached Multan, he presented the governor with, “a slave and horse together with raisins and almonds”.
4. Muhammad bin Tughlaq, informs Ibn Battuta, was so happy with the sermon of a preacher named Nasiruddin that he gave him “a hundred thousand tankas (coins) and two hundred slaves”.
5. According to Ibn Battuta’s accounts there were female slaves in the service of the Sultan. They were experts in music and dance. Female slaves were also employed by the Sultan to keep a watch on his nobles.
6. Slaves were generally used for domestic labour, particularly for carrying women and men on palanquins or dola.

Question 5.
What were the elements of the practice of sati that drew the attention of Bernier ?
Solution :
Bernier chose the practice of sati for detailed description. The elements of the practice of sati that drew the attention of Bernier were that while some women seemed to embrace death cheerfully, others were, forced to die. For example at Lahore, he saw a most beautiful young widow sacrificed. She was hardly twelve years of age and was forced by the Brahmanas and others towards the pyre and was burnt alive.

Question 6.
Discuss Al-Biruni’s understanding of the caste system.
Solution :
Al-Biruni’s description of the caste system in India was as given below :

1. He tried to explain the caste system by looking for parallels in other societies. For example, he noted that in Ancient Persia, there were four categories i.e., knights and princes; monks, fire-priests and lawyers; physicians, astronomers and other scientists; and finally, peasants and artisans. Thus, he stated that social divisions were not unique to India. He, however, pointed out that within Islam, Ml men were considered equal, differing only in their observance of piety.
2. He accepted the Brahmanical description of the caste system but disapproved of the notion of pollution. The conception of social pollution was, according to him, contrary to the laws of nature.

Thus, Al-Biruni’s understanding of the caste system was deeply influenced by his study of normative Sanskrit texts which laid down the rules governing the system from the point of view of the Brahmanas.

Question 7.
Do you think Ibn Battuta’s account is useful in arriving at an understanding of life in contemporary urban centres? Give reasons for your answer.
Solution :
Battuta’s observation about the cities of India.
(i)  According to him, Indian cities had many exciting opportunities and are useful for those who had the necessary drive, skill and resources.
(ii) The Indian cities were prosperous and densely populated.
(iii) These cities had colourful market trading in different kinds of goods.
(iv) Delhi was the largest city of India and had a lot of population. Daultabad was an another important city of India which challenged Delhi in size.
(v) The cities were not only the centre of economic transactions but also the centres of! social and cultural activities.
(vii) Most of the bazars in the cities had temple and mosques.
(viii) Cities also had fixed places for public performances by dancer, musicians and singer. He found that many towns derived their wealth and prosperity through the appropriation of surplus from villages.
(ix) Indian goods were in great demand in west Asia and South-east Asia. So the artisans and merchants earned huge profit.

Question 8.
Discuss the extent to which Bernier’s account enables historians to reconstruct contemporary rural society.
Solution :
Bernier’s account does not enable historians much to reconstruct contemporary rural society. His accounts contain discussions trying to place the history of the Mughals within some sort of a universal framework. He constantly compared Mughal India with contemporary Europe, generally emphasising the superiority of the latter.
His description of rural society was far from truth. For example, he thought that in the Mughal Empire, the Empire owned all the land and distributed it among his nobles. This had disastrous consequences for the economy and society. Owning to crown ownership of land, argued Bernier, landholders could not pass on their land to their children. So, they were averse to any long-term investment in the sustenance and expansion of production. This had resulted in uniform ruination of agriculture, excessive oppression of peasantry and a continuous decline in the living standards of all sections of society, except the ruling aristocracy. He also stated that there was no middle state in India.

The above description does not give us a true picture of rural society. None of the Mughal official document suggest that the state was the sole owner of land. Abul Fazl describes the land revenue as “remunerations of sovereignty”. In fact, during the sixteenth and seventeenth centimes, rural society was characterised by considerable social and economic differentiation. At one end were the big zamindars and on the other were the “untouchable” landless labourers. In between was the big peasant who used hired labour and engaged in commodity production, and the smaller peasant who could barely produce for his subsistence.

Question 9.
Read this excerpt from Bernier :
Numerous are the instances of handsome pieces of workmanship made by persons destitute of tools, and who can scarcely be said to have received instruction from a master. Sometimes they imitate so perfectly articles of European manufacture that the difference between the original and copy can hardly be discerned. Among other things, the Indians make excellent muskets, and fowling-pieces, and such beautiful gold ornaments that it may be doubted if the exquisite workmanship of those articles can be exceeded by any European goldsmith. I have often admired the beauty, softness, and delicacy of their paintings.

List the crafts mentioned in the passage. Compare these with the descriptions of artisanal activity in the chapter.
Solution :
(a) The following crafts have been mentioned in the passage :

• muskets;
• fowling-pieces;
• gold ornaments;
• paintings.

(b) There were imperial Karkhanas or workshops for the artisans where embroiderers, goldsmiths, painters, varnishers, joiners, turners, tailors and shoe-makers, manufacturer of silk, brocade and fine muslins were employed. They worked the whole day and in the evening they returned to their homes. The artisans were employed in manufacturing carpets, gold and silver cloths and various sorts of silk and cotton goods. Bernier also stated that the Indian artisans were expert in copying goods that it was difficult to differentiate between the original and the duplicate.

We hope the NCERT Solutions for Class 12 History Chapter 5 Through the Eyes of Travellers Perceptions of Society help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 5 Through the Eyes of Travellers Perceptions of Society, drop a comment below and we will get back to you at the earliest.

Here we are providing NCERT Solutions for Class 12 English Flamingo Poem 3 Keeping Quiet. Students can get Class 12 English Keeping Quiet NCERT Solutions, Questions and Answers designed by subject expert teachers.

Keeping Quiet NCERT Solutions for Class 12 English Flamingo Poem 3

Keeping Quiet NCERT Text Book Questions and Answers

Keeping Quiet Think it out 

Question 1.
What will counting up to twelve and keeping still help us achieve?
Answer:
The poet advises his readers to count up to twelve and keep still to achieve a sense of togetherness. The poet advises on introspection by taking a break from the mundane activities of life. He compares this with the earth that seems quiet and yet nurtures so much life on it.

Question 2.
Do you think the poet advocates total inactivity and death?
Answer:
No, the poet does not advocate total inactivity and death. He feels that peace and tranquillity should*not be confused with lethargy. He compares lethargy to death itself. He wants a perceptive silence in which people are not obsessed with apparent progress. The self- introspection might also lead to less violence and destruction.

Question 3.
What is the “sadness” that the poet refers to in the poem?
Answer:
The “sadness” that the poet talks about refers to the ceaseless activity that leaves little time for self-analysis and finally leads to doom. This mindless activity leads men to destruction. They kill whales and injure themselves by chemicals or fire. He also pleads with mankind to end environmental degradation. Such activities lead to a situation that is a mere illusion of victory, but in actuality, is fatal.

Question 4.
What symbol from nature does the poet invoke to say that there can be life under apparent stillness?
Answer:
The poet invokes a powerful symbol of the Earth to educate that there can be life under apparent stillness. He says that Earth is a great teacher. In its quietness, its strong message resounds. It seems silent and yet nurtures so much life on it. The poet wants to begin his mission of giving our lives a thought. He wants us to be productive in our silence.

Keeping Quiet Extra Questions and Answers

Keeping Quiet Short Answer Questions

Question 1.
How are the first two lines different from the last two lines of the poem?
Answer:
The first two lines appeal to the reader to count till twelve and keep silent. The speaker expresses his desire for silence and peace as an answer to end mindless activity. Whereas, in the last two lines, he wants the people to begin the exercise at their own level. He moves on to initiate this activity elsewhere.

Question 2.
What is the kind of a world that Neruda dreams of?
Answer:
Neruda dreams of a calm, quiet and a peaceful world where the hallmark would be togetherness. He dreams of a world which is free of destructive activity.

Question 3.
According to the poet, silence is profound. Justify.
Answer:
According to the poet, silence is the hope for a peaceful world. Silence is opposed to lethargy as it is the time for introspection. In the present world, the poet feels that people are obsessed with seeming progress that . leads to their destruction.

Question 4.
Earth is the best teacher. Why does the poet feel so?
Answer:
The poet feels that Earth is the best teacher because it is quiet and calm, yet nurtures so much life on it. He implies that such silence is productive as serenity and tranquillity leads to progress.

Keeping Quiet Value Based Question

Question 1.
The poem “Keeping Quiet” is based on the theme of universal peace and tranquillity. Justify.
Answer:
Neruda begins the poem by urging his readers to observe a moment of stillness and silence. He longs for a moment sans communication and activity where man is at complete peace with himself and his surroundings. _ This “exotic moment” would be a moment of universal peace. He feels, humans are preoccupied with their own progress which deprives them of the true pleasure of living. They merely exist and indulge in meaningless activities like that of the fishermen harming the whales. Man through his meaningless trysts endangers the environment and his own life.

The poet thus urges people to cleanse their souls, wear clean clothes and walk peacefully in universal brotherhood. The poet however, does not want the readers to confuse tranquillity with total inactivity, which meant sluggishness and death. He urges people to be unselfish and wants them to allow a moment of silence to interject their fast-moving, useless lives. He wants people to look upon Earth as a teacher, to fulfil a greater purpose in life.

Give examples from the poem of the following poetic devices.

Personification
“face of the earth”

Synecdoche
“not move our arms so much”

Hyperbole
“victory with no survivors”

Contrasting/antithetical imagery
“wars/ walk about with their brothers in shade”

 

NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation

These Solutions are part of NCERT Solutions for Class 12 Biology. Here we have given NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation

Question 1.
Mention the advantages of selecting pea plant for experiment by Mendel.
Solution:

1. The plant shows clear-cut contrasting characters.
2. Hybrids are perfectly fertile.
3. Genes for the seven contrasting characters are located on seven separate chromosomes.
4. Easy to cultivate.
5. The floral structure is suitable for artificial pollination.
6. Short growth period and life cycle.
7. Cross-pollination is easy if self-pollination is prevented.
8. Pure breeding varieties are available

Question 2.
Differentiate between the following:

1. Dominant and Recessive
2. Homozygous and Heterozygous
3. Monohybrid and Dihybrid

Solution:

1. Differences between dominant and recessive genes are as follows :
   
2. Differences between homozygous and heterozygous are as follows :
   
3. Differences between monohybrid and dihybrid cross are as follows :
   

Question 3.
A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
Solution:
A diploid organism heterozygous for 4 loci will have the supported genetic constitution YyRr for two characters. The alleles Y-y and R-r will be present on different 4 loci. Each parent will produce four types of gametes – YR, Yr, yR, yr.

Question 4.
Explain the law of dominance using a monohybrid cross.
Solution:
The Law of dominance states that when a pair of alleles or allelomorphs are brought together in F₁ hybrid, then only one of them expresses itself, masking the expression of the other completely. Monohybrid cross was made to study the simultaneous inheritance of a single pair of Mendelian factors. The cross in which only alternate forms of a single character are taken into consideration is called a monohybrid cross. The trait which appeared in the F₁ generation was called dominant and the other which did not appear in the F₁ population was called recessive.

Thus, when a pair of alleles are brought together in an F₁ hybrid, then only one of them expresses itself masking the expression of the other completely. In the above example, in Tt – F₁ hybrid (tall) only ‘T’ expresses itself so dominant, and ‘t’ is masked so recessively. Thus, this’ proves and explains the law of dominance.

Question 5.
Define and design a test-cross.
Solution:
The crossing of F₁ individuals having dominant phenotype with its homozygous recessive parent is called test cross. The test cross is used to determine whether the individuals exhibiting dominant character are homozygous or heterozygous.
Example: When a tall plant (TT) is crossed with the dwarf plant (tt) in the F₁, generation only tall plant (Tt) appears which is then crossed with homozygous recessive (tt) in a test cross.

In the given test cross between tall heterozygous F₁ hybrid with dwarf homozygous recessive parent produces tall and dwarf progeny in equal proportion indicating that F : hybrids are heterozygous.

Question 6.
Using a Punnett square, work out the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.
Solution:
When a heterozygous male tall plant (Tt) is crossed with the homozygous dominant female tall plant (TT), we get two types of gametes in males: half with T and a half with t, and in females, we get only one type of gametes i.e., T.

From the Punnett square it is seen that all the progeny in the F generation are tall (Tt), 50% homozygous tall (TT), and 50% heterozygous tall (Tt).

Question 7.
When a cross is made between a tall plant with yellow seeds (TtYy) and a tall plant with the green seed (Ttyy), what proportions of phenotype in the offspring could be expected to be

1. tall and green
2. dwarf and green

Solution:

Phenotypes of the offsprings –
Tall Yellow : 3
Tall Green : 3
Dwarf Green: 1
Dwarf Yellow: 1
(a) Proportion of tall and green is 3/8.
(b) Proportion of dwarf and green is 1/8.

Question 8.
Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F₁ generation for a dihybrid cross?
Solution:
Two heterozygous parents (i.e. GgLl and GgLl) are crossed and the two loci are linked then the cross will be

This means, if ‘G’ represent grey body (dorhinant), ‘g’ black body (recessive), ‘L’-long (dominant) and ‘I’-dwarf (recessive) then the distribution of phenotypic features in F1 generation will be 3 : 1 i.e. 3/4 will show the dominant feature, grey and long, either in homozygous (GGLL) or in heterozygous (GgLl) condition and 1/4 will show the recessive feature, black and dwarf (ggll).

Question 9.
Briefly mention the contribution of T.H. Morgan in genetics.
Solution:
TH Morgan is a Geneticist who got Nobel Prize.

• He found fruit fly (Drosophila Melanogaster) to be an experimental material as it was easy to rear and multiply.
• The established presence of genes over the chromosomes.
• Principle of linkage and crossing over.
• Discovered sex linkage and crossing over.
• He observed mutations.
• The developed technique of chromosome mapping,
• Wrote the book “The theory of Gene”.

Question 10.
What is pedigree analysis? Suggest how such an analysis, can be useful.
Solution:
A record of inheritance of certain genetic traits for two or more generations presented in the form of a diagram of family tree is called pedigree. Pedigree analysis is study of pedigree for the transmission of particular trait and finding the possibility of absence or presence of that trait in homozygous or heterozygous state in a particular individual. Pedigree analysis is useful for the following:

• It is useful for the genetic counsellors to advice intending couples about the possibility of having children with genetic defects like haemophilia, colour blindness, alkaptonuria, phenylketonuria, thalassemia, sickle cell anaemia (recessive traits), brachydactyly and syndactyly (dominant traits).
• Pedigree analysis indicates that Mendel’s principles are also applicable to human genetics with some modifications found out later like quantitative inheritance, sex linked characters and other linkages.
• It can indicate the origin of a trait in the ancestors, e.g., haemophilia appeared in Queen Victoria and spread in royal families of Europe through marriages.
• It helps to know the possibility of a recessive allele to create a disorder in the progeny like thalassemia, muscular dystrophy, haemophilia.
• It can indicate about the harm that a marriage between close relatives, may cause.
• It helps to identify whether a particular genetic disease is due to a recessive gene or a dominant gene.
• In certain cases it may help to identify the genotypes of offspring yet to be born.

Question 11.
How is sex determined in human beings?
Solution:
In humans, there are 23 pairs of chromosomes. 22 pairs of these chromosomes do not take part in sex determination called autosomes. The 23rd pair determines the sex of an individual called allosome or sex chromosome. If it is XX then female, if XY then male. The presence of Y¹ makes a person male. Human females produce only 1 type of gamete 22 + X. In males, it could be 22 + X or 22+ Y.

Question 12.
A child has blood group O. If the father has blood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings.
Solution:
If the father has blood group A i.e., I^AI^A (homozygous) and mother has blood group B i.e., I^BI^B (homozygous) then all the offsprings will have blood group AB (I^AI^B) and not blood group O.


Thus the genotypes of the parents of child with blood group O will be I^Ai and I^Bi There is the possibility of 3 other types of blood groups of offsprings besides O blood group offspring. They are I^Ai (blood group A). I^Bi (blood group B) and I^AI^B (blood group AB).

Question 13.
Explain the following terms with an example:

1. Codominance
2. Incomplete dominance

Solution:
Codominance (1 : 2 : 1) — It is the phenomenon of two alleles (different forms of a Mendelian factor present on the same gene locus on homologous chromosomes) lacking dominant- recessive relationship and are able to express themselves independently when present together.

Example – AB blood group: Alleles for blood group A(I^A) and blood group B(I^B) are codominant so that when they come together in an individual, they produce blood group AB. It is characterized by the presence of both antigen A (from I^A) and antigen B (from I^B) over the surface of erythrocytes.

Incomplete dominance (1 : 2 : 1) – It is the phenomenon where none of the two contrasting alleles being dominant so that expression in the hybrid is intermediate between the expressions of the two alleles in the homozygous state. F₂ phenotypic ratio is 1 : 2 : 1, similar to genotypic ratio. Example-In Mirabilis jalapa (Four o’clock) and Antirrhinum majus (Snapdragon or dog flower), there are two types of flower colour generation are of three types- red, pink and white flowered in the ratio of 1 : 2 : 1. The pink colour apparently appears either due to the mixing of red and white colours (incomplete dominance).

Question 14.
What is point mutation? Give one example.
Solution:
Point mutation is a gene mutation that arises due to a change in a single base pair of DNA.
Example: Sickle-cell anaemia.
Substitution of a single nitrogen base at the sixth codon of the β- globin chain of haemoglobin molecule causes the change in the shape of the R.B.C. from biconcave disc to the elongated shaped, structure which results in sickle cell anaemia.

Question 15.
Who had proposed the chromosomal theory of inheritance?
Solution:
Sutton and Boveri proposed the chromosomal theory of inheritance. The theory believes that chromosomes are vehicles of hereditary information that possess mendelian factors or genes and it is the chromosomes which segregate and assort independently during transmission from one generation to the next.

Question 16.
Mention any two autosomal genetic disorders with their symptoms
Solution:
Cystic fibrosis is an autosomal recessive disorder of infants, children, and young adults that is due to a recessive autosomal allele present on chromosome 7. It is common in Caucasian Northern Europeans and White North Americans. The disease gets its name from the fibrous cysts that appear in the pancreas. In 70% of cases, it is due to the deletion of three bases. It produces a defective glycoprotein. The defective glycoprotein causes the formation of thick mucus in the skin, lungs, pancreas, liver, and other secretory organs. Accumulation of thick mucus in the lungs results in obstruction of airways. Because of it, the disease was also called mucoviscoides, Mucus deposition in the pancreas blocks secretion of pancreatic juice. There is a maldigestion of food with high-fat content in the stool. The liver may undergo cirrhosis and there is impaired production of bile. Vasa deferentia of males undergo atrophy.

Huntington’s disease or Huntington’s chorea is a dominantly autosomal inherited disorder in which muscle and mental deterioration occur. There is gradual loss of motor control resulting in uncontrollable shaking and dance-like movements (chorea). The brain shrinks between 20-30% in size followed by slurring of speech, loss of memory, and hallucinations. Life expectancy averages 15 years from the onset of symptoms. This disorder does not occur till the age of 25 to 55. The defective gene is dominant autosomal, located on chromosome 4. This defective gene has 42 -100 repeats of CAG instead of 10-34 repeats in the normal gene. The frequency of this disorder is 1 in 10000 to 1 in 20000.

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