A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1

A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1

These Solutions are part of A New Approach to ICSE Physics Part 2 Class 10 Solutions. Here we have given A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1.

SECTION I (40 MARKS)

Attempt all questions from this section
Question 1.
(a) It is possible to have an accelerated motion with a constant speed? Explain. [2]
(b)

  1. When does a force do work ?
  2. What is the work done by the moon when it revolves around the earth ? [2]

(c) A boy of mass 30 kg is sitting at a distance of 2 m from the middle of a see-saw. Where should a boy of mass 40 kg sit so as to balance the see-saw? [2]
(d)

  1. What is meant by the term ‘moment of force’?
  2. If the moment of force is assigned a negative sign then will the turning tendency of the force be clockwise or anticlockwise? [2]

(e) A ball is placed on a compressed spring. When the spring is released, the ball is observed to fly away.

  1. What form of energy does the compressed spring posses?
  2. Why does the ball fly away? [2]
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .1

Answer:
(a) Yes. Motion of a stone tied to a strong thread in a circular path at constant speed in the horizontal plane is an accelerated motion with constant speed.
Explanation : Consider a stone of mass (m) tied to a strong thread of radius (r), whirled around the point ‘O’ at a constant speed in the horizontal plane, in the clockwise direction.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .2
When the stone is at point A, then its motion is directed towards the North. When the stone reaches the point B, then its motion is directed towards East along the tangent drawn to the circumference of the circular path at point B.
From the above two positions, it is clear that when the stone is moving along a circular path, the direction of motion with respect to time (speed) is not the same. In fact the direction changes from point to point. Thus, we can say that the stone may be moving with a uniform speed, but is not moving with a uniform velocity, i.e., it is moving with a variable velocity along the circular path. As the stone is moving with a variable velocity, therefore,
it must have an accelerated motion.
(b)

  1. Work is said to be done only when a force or its component causes a displacement in its own direction.
  2. Work done by the moon when it revolves around earth is Zero. (In one complete revolution net displacement is zero)

(c)

  1.  Let the distance of 40 kg from the mean position be ‘x’m. Using principle of moments,
    30 x 2 = 40 x X
    x = 1.5m

(d)

  1. Turning effect of force acting on a body about an axis is called moment of force.
  2. Clockwise.

(e)

  1. Potential Energy
  2.  Since P.E. is converted into K.E. and is also transferred to the ball. Hence, the ball flies away.

Question 2.
(a) Samir exerts a force of 150 N in pulling a cart at a constant speed of 10 m/s. Calculate the power exerted. [2]
(b) A body of mass 0.2 kg falls from a height of 10 m to a height of 6 m above the ground. Find the loss in potential energy taking place in the body. [g = 10 m s-2] [2]
(c)

  1. Define the term refractive index of a medium in terms of velocity of light.
  2. A ray of light moves from a rare medium to a dense medium as shown in the diagram below. Write down the number of the ray which represents the partially
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .3

(d) You are provided with a printed piece of paper. Using this paper how will you differentiate between a convex lens and a concave lens? [2]
(e) A ray of light incident at an angle of incidence T passes through an equilateral glass prism such that the refracted ray inside the prism is parallel to its base and emerges from the prism at an angle of emergence V.

  1. How is the angle of emergence ‘e’ related to the angle of incidence ‘i’?
  2. What can you say about the value of the angle of deviation in such a situation ? [2]

Answer:
(a) Let Time (t) = 1s,
∴ Speed (s) = 10 m
Force (F) = 150 N
Work Done = F x s = 150 x 10 = 1500 Nm = 1500 J
Power exerted = 1500J/1s= 1500 W
(b) Mass, m = 0.2 kg, h = 10 m, g = 10 m s-2 Loss in P.E. = mgh
= 0.2 x 10 x (10 – 6)
= 0.2 x 10 x 4 = 8 J.
(c)

  1. The absolute refractive index of a medium is defined as the ratio of the speed of light in vacuum
    (or air) to the speed of light in that medium i.e., μ = c/v
  2. Ray 2

(d) On moving the printed piece of paper before the lens, if the image is always diminished then it is a concave lens otherwise it is a convex lens.
(e)

  1. Angle of incident ∠i = Angle of emergence∠e
  2. The angle of deviation is the angle of minimum deviation
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .4

Question 3.
(a)

  1.  What is meant by Dispersion of light’ ?
  2.  In the atmosphere which colour of light gets scattered the least ? [2]

(b) Which characteristic of sound will change if there is a change in

  1. its amplitude
  2. its waveform [2]

(c)

  1. Name one factor which affects the frequency of sound emitted due to vibrations in an air column.
  2. Name the unit used for measuring the sound level. [2]

(d) An electrical appliance is rated at 1000 kVA, 220V. If the appliance is operated for 2 hours, calculate the energy consumed by the appliance in :
(1)  kWh (2) joule [2]
(e) Calculate the equivalent resistance between P and Q from the following diagram : [2]

A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .5
Answer:
(a)

  1. The phenomenon of splitting of white light of a prism into its constituent colours is called dispersion of light.
  2. Red colour.

(b)

  1. Loudness
  2. Quality or Timbre.

(c)

  1. Length of air column.
  2.  Decibel (dB)

(d) Energy consumed in KWh = Power x Time
= 1000 x 2 = 2000 kWh
(e) Energy consumed in J = 2000 x 3.6 x 105:J.
= 7.2 x 109 J.
The two 10 Ω resistors are in series and they are parallel to the 5Ω resistor.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .5
Net resistance between P and Q = 3 + 4 + 2 = 9Ω.

Question 4.
(a)
(1) What is an a.c. generator or Dynamo used for?
(2) Name the principle on which it works. [2]
(b) Differentiate between heat capacity and specific heat capacity. [2]
(c) A hot solid of mass 60 g at 100°C is placed in 150 g of water at 20°C. The final steady temperature recorded is 25°C.
Calculate the specific heat capacity of the solid. [Specific heat capacity of water = 4200 J kg-1 °C-1]. [2]
(d) (1) What is the value of the speed of gamma radiations in air or vacuum?
(2) Name a material which exhibits fluorescence when cathode rays fall on it. [2]
(e) Which of the radioactive radiations :
(1) can cause severe genetic disorders.
(2) are defected by an electric field. [2]
Answer:
(a)

  1. An a.c. generator or dynamo is used to convert mechanical energy to electrical energy.
  2. Faraday’s law of Electromagnetic induction.

(b)

  1. Specific heat capacity is defined on unit mass but heat capacity is defined on entire mass.
  2. Specific heat capacity depends on the mass of the body but heat capacity does not depend on the mass of the body.

(c)  Heat gained Heat lost (Principle of calorimetry)
150 x 4.2 x (25 – 20) = 60 x C x (100 – 25)
150 x 4.2 x 5 = 60 x C x 75
C = 0.7 Jg-1C-1
(d)

  1. Speed of gamma radiations in air = 3 x 108 m s-1.
  2.  Zinc suiphide.

(e)

  1. Radioactive radiations which can cause severe genetic disorders : Gamma (γ )-radiations
  2. Radioactive radiation which are deflected on electric field : Alpha (α) and beta (β)- radiations.

SECTION II (40 MARKS)

Attempt any four questions from this section
Question 5.
(a)
(1) Which of the following remains constant in uniform circular motion : Speed or Velocity or both?
(2) Name the force required for uniform circular motion. State its direction. [3]
(b)
(1) State the class of levers and the relative positions of
load (L), effort (E) and fulcrum (F) in each of the following cases. 1. A bottle opener 2. Sugar tongs [3]
(2) Why is less effort needed to lift a load by single fixed pulley as compared to
lifting the load directly?[3]
(c)
(1) A moving body weighing 400 N possesses 500 J of kinetic energy. Calculate the velocity with which the body is moving, (g = 10 m s-2).
(2) In what way does an ideal machine differ from a practical machine? [4]
Answer:
(a)

  1.  Speed remains constant,
  2. Centripetal force directed towards the centre.

(b)

  1. A bottle opener — Class II lever Load between Fulcrum and Effort
  2. Sugar Tongs — Class II lever
    Effort between Fulcrum and Load.Less effort is required because mechanical advantage increases.

(c)

  1.  Let W = mg ; m=400/10 =40kg
    We know that K.E.
    500 =1/2 mv2
    500 =1/2 ×40×v2
    v2= 25
    v=5m s-2
  2.  For an ideal machine efficiency is 1 but for a practical machine efficiency is always less than 1. This is due to the following two reasons :
    • A part of the input is wasted in moving the parts of the machine.
    • A part of the input is wasted in overcoming friction between the various parts of the machine.

Question 6.
(a).

  1. What is meant by the term ‘critical angle’?
  2. How is it related to the refractive index of the medium?
  3. Does the depth of a tank of water appear to change or remain the same when viewed normally from above? [3]

(b) A ray of light PQ is incident normally on the hypotenuse of a right angled prism ABC as shown in the given diagram
A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .7

  1. Copy the diagram and complete the path of the ray PQ till it emerges from the prism.
  2. What is the value of the angle of deviation of the ray?
  3. Name an instrument where this action of the prism is used.

(c) A converging lens is used to obtain an image of an object placed in front of it.The inverted image is formed between F2 and 2F2 of the lens.
(1) Where is the object placed ?
(2) Draw a ray diagram to illustrate the formation of the image obtained.
Answer:
(a)

  1. Critical angle is the angle of incidence in the denser medium corresponding to which the angle of refraction in the rarer medium is 900.
  2. aμg = cosec ic where ic is the critical angle.
  3. Depth remain the same. Apparent depth = Actual depth.

(b) 


  1. A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .8
  2. Angle of deviation (δ) = 180°
  3. Prism binocular.

(c)

  1. Object is placed Beyond 2Ft
  2. A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .9

Question 7.
(a)

  1. What is meant by Resonance?
  2.  State two ways in which Resonance differs from Forced vibrations. [3]

(b)

  1.  A man standing between two cliffs produces a sound and hears two successive echoes at intervals of 3 s and 4s respectively. Calculate the distance between the two cliffs.
    The speed of sound in the air is 330 m s-2.
  2. Why will an echo not be heard when the distance between the source of sound and the reflecting surface is 10 m?

(c) The diagram given below shows the displacement-time graph for a vibrating body.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .10

  1. Name the type of vibrations produced by the vibrating body.
  2. Give one example of a body producing such vibrations.
  3. Why is the amplitude of the wave gradually decreasing?
  4. What will happen to the vibrations of the body after some time?    [4]

Answer:
(a)
(1) When the frequency of an externally applied periodic force on a body is equal to the natural frequency of the body, the body readily begins to vibrate or free to vibrate with an increased amplitude. This phenomenon is known as resonance.
Resonance

  1. Vibration are of large amplitude.
  2. Frequency of an external applied periodic force is equal or an integral multiple to the natural frequency of the body.

Forced Vibrations

  1. Vibrations are usually small amplitude
  2. Frequency need not be equal or integral multiple to the natural frequency of the body.

(b)

  1. d1 = 495m and d2 = 660m
    Dist. between the two cliffs = 495 + 660 = 1155 m.
  2. No echo will be heard because least distance of distinct hearing in 17 m.

(c)

  1. Damped vibrations.
  2. Tuning fork vibrating in air.
  3. Due to friction, energy is continuously lost.
  4. After some time, the amplitude gradually decreases finally stops.

Question 8.
(a)

  1. A cell is sending current in an external circuit How the terminal voltage compare with the e.m.f. of the cell?
  2. What is the purpose of using a fuse in an electrical circuit?
  3. What are the characteristic properties of fuse wire? [3]

(b)

  1.  Write an expression for the electrical energy spent in the flow of current through an electrical appliance in terms of I, R and t.
  2. At what voltage is the alternating current supplied to our houses?
  3. How should the electric lamps in a building be connected?     [3]
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .11

(c) Three resistors are connected to a 6V battery as shown in the figure given alongside.Calculate :

  1. the equivalent resistance of the circuit.
  2. total current in the circuit.
  3. potential difference across the 7.2 Q resistor. [4]

Answer:
(a)

  1. Terminal voltage < e.m.f.
  2. An electric fuse is a safety device which limits the current flowing in an electric circuit.
    High resistance and Low melting point.

(b)

  1. We know that E = Vlt = I2R/
  2. 220 V – 240 V
  3. In a building, electrical lamps are connected in parallel.

(c)
A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .12

Question 9.
(a)

  1. Write an expression for the heat energy liberated by a hot body.
  2. Some heat is provided to a body to raise its temperature by 25°C. What Will be the corresponding rise in temperature of the body as shown on the kelvin scale ?
  3. What happens to the average kinetic energy of the molecules as    ice melts at 0°C? [3]

(b) A piece of ice at 0°C is heated at a constant rate and its temperature recorded at regular intervals till steam is formed at 100°C. Draw a temperature – time graph to represent the change in phase. Label the different parts of your graph.    [3]
(c)
40 g of ice at 0°C is used to bring down the temperature of a certain mass of water at 60°C to 10°C. Find the mass of water used. [Specific heat capacity of water = 4200 J kg-1 C-1 [Specific latent heat of fusion of ice = 336 x 103 J kg-1] [4]
Ans.
(a)

  1. ΔQ = me Δt
    where m → mass, c → specific heat capacity; Δt → change in temperature.
  2. Since ΔQ α Δt
    Hence the corresponding rise in temperature of the body in kelvin 25 K.
  3. Average kinetic energy of the molecules increases.

(b)
A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .13
(c)
A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .14

Question 10.
(a) The diagram given below shows a current carrying loop or a circular coil passing through a sheet of cardboard at a points M and N.
The sheet of cardboard is sprinkled uniformly with iron filings.

  1. Copy the diagram and draw an arrow on the circular coil to show the direction of current flowing through it.
  2.  Draw the pattern of arrangement of the iron filings when current is passed through the loop. [3]

    A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .15

(b)

  1. Define nuclear fusion.
  2. Which of the two, fusion or fission is a nuclear chain reaction? Explain.  [3]

(c) A certain nucleus X has a mass number 14 and atomic . number 6. The nucleus X change to after the loss of a particle.

  1. Name the particle emitted.
  2.  Represent this change in the form of an equation.
  3. A radioactive substance is oxidized. What change would you expect to take place in the nature of its radioactivity? Give a reason for your answer. [4]

Answer:
(a)

  1.  M to N
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .16
  2. Anticlockwise around M
    Clockwise around N

(b)

  1. Nuclear fusion : The process of combining lighter nuclei (atomic mass less than 20) into heavier nuclei is called nuclear fusion.
  2. Fission is a nuclear chain reaction. As one atom of U235 can liberate three neutrons when the atom, of
  3. U235 is bombarded with a slow moving neutron. If all of these neutrons (or more than one) are utilised to break further U235 atoms, a kind of chain reaction starts with spontaneous release of a large amount of heat energy. Thus nuclear fission is chain reaction.

(c)

  1. 3-particle
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Model Test Paper -1 .17
  2. No change because radioactivity is a nuclear phenomenon and oxidation is concerned with outermost orbit.

 

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A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics

A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics

These Solutions are part of A New Approach to ICSE Physics Part 2 Class 10 Solutions. Here we have given A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics.

Exercise-1

Question 1.
Give at least two differences between a chemical change and nuclear change.
Answer:
Chemical change

  1. Change in number of orbital electrons takes place.
  2. Requires energy of few eV for a chemical reaction to take place.
  3. Number of atoms of each kind is conserved in reactants and products

Nuclear change

  1. Change in number of nucleons takes place.
  2. Nuclear change require much higher energy of the order of 106 times as compared to chemical change.
  3. Atomic number and mass number is conserved.

Question 2.
State Rutherford and Soddy’s Laws of natural radioactive decay for
(1) alpha emission
(2) beta emission.
Answer:
(1) Rutherford and Soddy’s Law of Alpha Emission : “When a radio-active nuclide ejects on alpha particle (α) i.e. 42H, its mass number decreases by 4 and atomic number decreases by 2 such that the position of daughter nuclide is two places behind in the predictable as compared to the parent nuclide ?

A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 1
(2) Rutherford and Soddy’s Law of Beta Emission : “When a radio-active nuclide ejects a beta particle, its mass number remain Unaffected but Atomic Number Increases by one such that the position of the daughter nuclide is one place Ahead in the periodic table as compared to the parent nuclei.” The daughter product is Isobar.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 2
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 3

Question 3.
Thorium isotope 22390Th undergoes two successive β-decays. Find the mass number and atom C number after the decay. Also represent the above decays in the form of a nuclear equation.
Answer:
After first p decay mass number remains same i.e. 223 but Atomic number increases by 1 and becomes 91
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 4
Two successive P radiations cannot take place. He a radiation takes place which does not change mass number or atomic number.
Now second β decay

A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 5
mass number remains 223 atomic number becomes 92.

Question 4.
How is the ionising and penetrating powers of α, β and γ radiations compared with each other ?
Answer:
Comparison of Ionising power of α, β,γ Heavier particle has high ionising power.
a → 10,000 times of α and 100 times of β
P → 100 times of α
∴ Maximum ionising power a, minimum ionising power γ.
Comparison of Penetrating power of α, β and γ
Light particle has maximum speed hence maximum penetrating power.
α has very large penetrating power up to a few hundred meter in air.
α has small penetrating power (being very heavy) 3 to 8 cm in air
β- has large penerating power up to few meter in air α < β < γ.

Question 5.
When does the nucleus of an atom tend to be radioactive?
Answer:
Nucleus of an atom become radio-active when increase in nuclear force.
PACE WITH INCREASE IN THE REPULSIVE FORCE :
i.e.when there are too many neutrons compared to protons or too many protons compared to neutrons in the nucleus of atom. Or The nucleus of the atom becomes radioactive if it is a radio-isotope, i.e. number of neutrons in the nucleus exceeds the number of protons inside it.
For example :
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 6
Question 6.
Radioactive sodium 2411Na changes to stable 2412Th Which particle does it eject ?
Answer:
P-particle  (0-1e )
∴ There is no change in mass number but atomic number of daughter nucleus magnesium increases by one.

Question 7.
A radioactive element AzX loses two successive β-particles and then an alpha particle, such that the resulting nuclide PQY  is Calculate the values of P and Q.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 7

When AzX loses first β-particle atomic number of daughter nucleus Y increases by one and mass number remains the same
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 8
When Az+1X  loses another β-particle changes to daughter nucleus Z and atomic number again increases by one and mass number still remains the same A.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 9
Question 8.
(a) An imaginary radioactive particle 23592X decays to form elements X1, X2, X3, X4, X5 and X6 by ejecting 2 beta particles, followed by an alpha particle and again 2 beta particles followed by an alpha particle. Represent the above in the form of nuclear equations. What is the mass number of X6 ?
(b) List the isotopes and isobars formed in the above nuclear reactions.
Answer:
(a) 23592X when α β particle is emitter mass number does not change but atomic number increases by 1 (isobar)
∴ When 2 successive β particles decay mass number does not change (isobar) but atomic number increases by 2.
Then α-particle decays, mass number decreases by 4 and atomic number decreases by 2.
Ultimately when 2 β particles decays and la-particle decays
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 10
This means to change 23592X to 23592X1 here, atomic number remains same but mass number
decreases by 4
∴ In 6 steps [X → X1 → X2 → X3 →X4 →X5 →X5]
Atomic number remains same but mass number decreases by
6 x 4 = 24
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 11
(b) After every β decay isobar is formed and after α-decay isotope is formed
∴ Total 12 isobars and 6 isotopes are formed.

Question 9.
Give one example of nuclear fission.
Answer:
Nuclear power station is example of nuclear fission. Controlled chain reaction takes place and electricity is produced.

Question 10.
Indicate the missing particle in the following reaction :
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 12
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 13
Question 11.
Complete the statement given below :
Splitting of nucleus into two nearly lighter nuclei is called
Answer:
Splitting of nucleus into two nearly lighter nuclei is called nuclear fission.

Question 12.
Name the fuel generally used in nuclear reactors.
Answer:
Enriched Uranium 23592U is used as fuel in nuclear reactor.

Question 13.
Give one example of a controlled and uncontrolled nuclear fission reaction.
Answer:
Example of:     

  1. Controlled nuclear fission reaction is to generate electricity in nuclear thermal plant.
  2. Uncontrolled nuclear fission reaction is production of ATOM BOMB.

Question 14.
Name the isotopes of an element which are used in fusion
Answer:
ISOTOPES of ELEMENT HYDROGEN are :
11H protium
21H deuterium
31H Tritium

Question 15.
What is meant by nuclear chain reaction ? What happens, if this reaction goes out of control ?
Answer:
NUCLEAR CHAIN REACTION : A piece of URANIUM consists of millions of uranium atoms. When a slow moving neutron is bombarded, it produces three neutrons, if these neutrons are utilised to break further uranium atoms, 9 neutrons are produced and in this way a kind of chain reaction takes place with release of large amount of energy and reaction becomes uncontrolled. This results in atom bomb.

A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 14

Question 16.
The mass numbers of three elements A, B and C are 3, 180 and 235 respectively. Which one is suitable for making atomic bomb?
Answer:
Element C with atomic number 235 is suitable for making atom bomb.

Question 17(a).
Define nuclear fusion.
Answer:
NUCLEAR FUSION : The process of combining LIGHTER NUCLEI (Atomic weight less than 20) into heavier nuclei is called nuclear fusion
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 15
Question 17(b).
Which of the two, fission or fusion is a nuclear chain reaction?
Answer:
Fission is nuclear chain reaction.

Question 18.
Why can nuclear fusion and possible to generate electricity?
Answer:
It is not easy to start a fusion reaction as to fuse HYDROGEN atom minimum temperature of 1000000 °C is required which is not possible to create in laboratory. Controlled fusion reaction is not possible so far.

Question 19.
Give any two differences between nuclear fusion and nuclear fission.
Answer:
Nuclear Fusion

  1. Two light nuclei fuse to form a heavy nucleus.
  2. Cannot be controlled to generate electricity position.

Nuclear Fission

  1. A heavy nucleus splits to form two smaller nuclei.
  2. Can be controlled and is used to generate electricity.

Question 20.
Write nuclear equations for the fusion of
(a)   2 deuterium atoms
(b)   one hydrogen and one tritium atom.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 16

Multiple Choice Questions

Tick ( ✓) the most appropriate option.

1. The atoms of same element having same atomic number, but different atomic masses are called :
(a) isotopes
(b) isobars
(c) isotones
(d) both (a) and (b)

2. When an element gives out high energy radiations on its own, the change which takes place is :
(a) physical change
(b) chemical change
(c) nuclear change
(d) none of these

3. The atoms of different elements having same mass number, but different atomic numbers are called :
(a) isotopes
(b) isotones
(c) isobars
(d)  none of these

4. The radiations given out by radioactive elements :
(a) affect photographic plates
(b) ionise the gases
(c) are affected by electrostatic and magnetic fields
(d) all of these

5. A radioactive substances emits :
(a) simultaneously α, β and γ radiations
(b) α-radiations or β-radiations
(c) in the order of α, β and γ particles
(d) X-rays and γ -rays

6. During α-emission :
(a) the mass number and atomic number of an atom decrease by 2 a mu.
(b) the mass number decreases by 4 amu and atomic number decreases by 2 amu.
(c) the mass number remains unchanged, but atomic number decreases by 2 amu.
(d) the mass number decreases by 4 amu and atomic number remains unchanged.

7. During β-emmision :
(a) the mass number remains unchanged, but atomic number increases by 1 amu.

(b) the mass number remains unchanged, but atomic number decreases by 1 amu.
(c) the mass number increases by 1 amu, but atomic number remains same.
(d) the mass number and atomic number decrease by 1 amu.

8. During β-emission an electron is ejected from the atom of radioactive substance. The electron is ejected from the:
(a) the outermost orbit of atom
(b) the innermost orbit of atom
(c) the nucleus of the atom
(d) none of these

9. Which of the following radiation is most ionising ?
(a) α-particles
(b) β-particles
(c) -radiation
(d) X-rays

10. Which of the following radiation is most penetrating ?
(a) αparticles
(b) β-particles
(c) ϒ -radiation
(d) X-rays

11. Which of the following radiation gets deflected most in electric or magnetic field ?
(a) αparticles
(b) β-particles
(c) ϒ -radiation
(d) X-rays

Questions From ICSE Examination Papers

2003
Question 1.
A small cube of lead is embedded in a big cube of aluminium metal It is placed in the path of a powerful radioactive emmision, such that on the opposite side of the cube is placed a fluorescent screen. It is observed that the shadow formed by aluminium metal is lighter than the lead metal. State one reason for the above phenomenon.
Answer:
The shadow formed by aluminium is lighter as α, β particles are stopped by aluminium and ϒ -radiations are not stopped by it.
The shadow formed by lead is dark because all the three α, β, ϒ are stopped by lead.

Question 2.
(a) Explain why a paint, containing of zinc sulphide and a trace of radium salt, glows in the dark.
(b) An isotope of 92U238 decays into thorium (Th) by the emission to an alpha-particle. The nucleus of thorium then decay into Protactinium (Pa) by the emission of a beta-paticle. Write two nuclear equations to illustrate the above changes.
Answer:
(a)
It makes a flourescent material and contains radioactive material like radium.
(b)
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 17

2004

Question 3.
(a) The diagram below shows a thick lead cube having a cavity in the middle. In the cavity is placed some radioactive substance. Copy the diagram and trace the paths of α particles,β particles and ϒ radiations as they pass through powerful electric field.

A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 18
(b) Name the radiations which have the least penetrating power.
Answer:
(a) 
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 19
(b) The alpha radiations being heaviest have the least penetrating power.

Question 4.
Copy and complete the following nuclear equations by filling in the correct values in the blanks
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 20

2005

Question 5(a).
How many alpha and beta particles are emitted when 23592U Uranium nucleus  decays to Lead 20682 Pb ?
Answer:
When 23592U decays to 23592U, the mass number decreases from 238 to 206 i.e. it decreases by 32. Since emission of beta particle (β) does not change the mass number and with the emission of one a (alpha) particle, mass number decreases 4, so total number
of alpha particles emitted will be 32/4 = 8.
In the decay of 23592U to lead 20682 Pb,the atomic number has decreased by 10. But due to emission of 8 α-particles, the atomic mass would have decreased by 2 x 8 = 16. Thus there is an increase in atomic number by 6, hence 6 Beta (β) particles will be emitted. (Because in emission of one beta particle, atomic number increases by 1). Thus 23592U decays to 23592U with emission of 8- α particles and 6 β-particles.

Question 5(b).
Mention two important precautions that should be taken while handling radioactive materials.
Answer:

  1. Storage of radio active source should be done with great care; they should be enclosed by lead blocks and further surrounded by concrete walls.
  2. Artificial, long mechanical ‘arms’ should be used to handle the more dangerous sources.

Question 5(c).
State one use of radioisotopes.
Answer:
They are used to study the function of fertilizers for different plants. They have also been used for developing new species of a plant by causing genetic mutations. They are also used in treatment of cancer

2006

Question 6.
A certain radioactive nucleus emits a particle that leaves its mass unchanged but increases its atomic number by 1. Identify the the particle and write its symbols.
Answer:
Particle is β(beta) i.e. electron  0-1e

Question 7.
State three properties that are common to and shown by beta rays and cathode rays.
Answer:

  1. Particle of both are electrons and are NEGATIVELY CHARGED.
  2. Both are deflected by electric and magnetic field.
  3. Both cause FLUORESCENCE.
  4. Both produce X-RAYS when stopped by metal of high ATOMIC NUMBER and high melting point (m.p.) such as, TUNGSTEN.

2007

Question 8.
What will an alpha particle change  into when it absorbs:
(a) One electron ?
(b) Two electrons?
Answer:
(a) On absorbing one electron, an a — particle (2He4 nucleus) will change into an helium ¡on which is singly ionised.
(b) On absorbing two electrons, it will change into an helium atom.

Question 9.
(a) What happens to the atomic number of an element when it emits:
1. an alpha particle
2. a beta particle.
(b) Explain why alpha and beta particles are deflected in an electric or a magnetic field but gamma rays are not deflected in such a field.
Answer:
(a)
1. When an a-particle is emitted, its atomic number decreases by 2.
2. When a beat particle is emitted, the atomic number increases by 1.
(b)
α-particles and beta particles when subjected to an electric field are deflected. This is so because both the α -particles and beta particles are charged. Electric field exerts a force on them which deflects them.
However, γ rays are not deflected because they are not charged and do not experience any force. Hence they are not deflected.

2008

Question 10.
(a) What is radioactivity ?
(b) Mention any two differences between nuclear energy and chemcial energy.
Answer:
(a) Radioactivity is a self spontaneous disintegration of a heavy nucleus into α, β, and γ radiations.
(b) Two differences between nuclear energy and chemical energy are as below:

Nuclear energy

  1. 1. Nuclear energy is liberated from the nucleus of the atom in which protons and neutrons take part.
  2. 2. Nuclear energy produced is tremendous as 1 kg mass gets converted into 9 x 1016 J of energy.

Chemical energy

  1. Chemical energy is liberated because of electrons which combine or release.
  2. A very small amount of energy is released in chemical reactions. It can be absorbed or released.

Question 11(a).
(1) When does the nucleus of an atom become radio active
(2) How is the radioactivity of an element affected when it undergoes a chemical change to form a chemical compound ?
(3) Name the product of nuclear fission which is utilized to bring about further fission of 23592U
Answer:

  1. Nucleus of an atom become radioactive if
    • n/p ratio is more than 1.5 i.e., number of neutrons are much more than number of protons in the nucleus
    • Electrostatic force is “more than nuclear force.
  2. No change (:. Any physical or chemical change do not alter the rate of disintegration of radioactive substance).
  3. Slow moving neutron.

Question 11(b).
(1) Mention one use and one harmful effect of radioactivity.
(2) Give one source of background radiation.
Answer:

  1. Use : Radiations from cobalt – 60 are used to treat cancer.
    Harmful Effect : The persons who are exposed to radiations these harmful radiations can kill the living tissues and cause damage.
  2. Cosmic rays from high altitudes and internal radiations given out by radioactive element present in the earth’s crust.

2009
Question 12(a).
Give two important precautions that should be taken while handling radioactive materials.
Answer:
Two precautions to be taken while handling radioactive substance :

  1. Nuclear must be kept in thick lead container with narrow mouth and handled with mechanical tongs, wearing lead lined glasses.
  2. The workers must wear special film badges which can absorb nuclear radiation and should under go compulsory medical check up from time to time.

Question 12(b).
(1) What is the name given to atoms of a substance which have the same atomic number but different mass numbers ?
(2) What is the difference in the atomic structures of such atoms ?
Answer:
(1) Isotopes.
(2) Difference in the number of neutrons only.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 21

Question 13.
A nucleus AZX emits an alpha particle followed.by ϒemission; thereafter it emits two β particles to form X3.
(a) Copy and complete the values of A and Z for X3 :
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 22
(b)

  • Out of alpha (α), beta (β ) and gamma (ϒ) radiations –
  • Which radiation is the most penetrating ?
  • Which radiations are negatively charged ?

Answer:
(a)  A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 23
∴ X3 has mass no A-4
Atomic no. Z i.e, it is an isotope of X
(b)

  1. Most penetrating radiation isϒ.
  2. β radiations are negatively charged.

2010

Question 14.
(1) Name the radioactive radiations which have the least penetrating power.
(2) Give one use of radioisotopes.
(3) What is meant by background radiation ?
Answer:

  1. α rays have the least penetrating power.
  2. Radioisotopes are used to study the function of fertilizer for different plants. They have also been used for developing new species of a plant by causing genetic changes.
  3. Background radiations are present at all places even in the absence of any radioactive source. The radiations are present in the atmosphere even when there is no source nearby.
    The source of background radiations are :
    (a) cosmic radiation
    (b) rocks in the earth which contain traces of radioactive substances
    (c) naturally occurring radioisotopes.

Question 15(a).
Complete the following nuclear changes :
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 24
Answer:
1. The completed nuclear changes are as below (use the fact that total mass number and charge numbers are always conserved in a nuclear change.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 25
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 26

2
.A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 27

Question 15(b).
(1) Which radiation produce maximum biological damage?
(2)What happens to the atomic number of an element when the radiation named by you in part
(i) above, are emitted.
Answer:

  1. α-radiation being heaviest and slowest causes most biological damage.
  2. When a-radiation is emitted, the atomic number decreases by 2.

2011

Question 16(a).
Fill in the blank with appropriate words
During the emission of a beta particle, the Mass number remains the same.

Question 16(b).
A mixture of radioactive substances gives off three types of radiations.

  1. Name the radiation which travels with the speed of light.
  2. Name the radiation which has the highest ionizing power.
  3. When an alpha particle gains two electrons it becomes neutral and becomes an atom of an element which is a rare gas. Name of this rare gas ?

Answer:

  1. γ-rays travels with the speed of light (3 Χ 108 m/s)
  2. Alpha particle
  3. Rare gas is Helium (He)

Question 16(c).
1. Define radioactivity.
2.What happens inside a nucleus that causes emission of beta particle ?
3. Express the above change in the form of an equation.
Answer:

  1. Radioactivity is a phenomenon self spontaneous disintegration of a heavy nucleus in α, β, and γ radiation.
  2. During Beta emission, one of the neutrons breaks into an electron and a proton, with the release of energy. The proton remain in the nucleus (atomic number increases by 1), but electron is ejected out.
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 28

Question 16(d).
(1) The nucleus 202 84X emits an alpha particle and forms.The nucleus Y. Represent this change in the form of an equation.
(2) What changes will take place in the mass number and atomic number of nucleus Y, if it emits gamma radiations ?
Answer:
(1)A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 29
(2) There will be no change

2012

Question 17(a).
(1) What is the value of the speed of gamma radiations in air or vacuum ?
(2) Name any two important sources of background radiation.
Answer:

  1. The speed of gamma rays (ϒ) in air and vaccum is 3 x 108 ms-1.
  2. (a) The radioactive emissions given out by the earth.
    (b) Sources are K-40, C-14 and Radium contained inside our body.

Question 17(b).
A certain nucleus X has a mass number 14 and atomic number 6. The nucleus X changes to 7ϒ14 after the loss of a particle.

  1. Name the particle emitted.
  2. Represent this change in the form of an equation.
  3. A radioactive substance is oxidized. What change would you expect to take place in the nature of its radioactivity? Give a reason for you answer.

Answer:
(1) The particle emitted is beta particle.
(2)
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 30
(3) No change will take place in its rate of activity. It is because oxidation is a chemical change which takes at electron level. It has nothing to do with nucleus of the atom.

2013
Question 18.
(a) Which of the radioactive radiations –

  1. can cause severe genetical disorders.
  2. are deflected by an electric field ?

(b)  A radioactive nucleus undergoes a series of decays according to the sequence
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 31

If the mass number and atomic number of X are 172 and 69 respectively, what is the mass number and atomic number of X ?
(C)
(1) What is meant by Radioactivity ?
(2) What is meant by nuclear waste ?
(3) Suggest one effective way for the safe disposal of nuclear waste.
Answer:
(a)
(1) Gamma radiations
(2) Alpha and beta radiations

(b)
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 32

Thus, mass number of X is 180 and atomic number 72 ;

(c)

  1. The phenomenon due to the nucleus of certain elements decays on its own, giving out harmful radiations, such as a alpha particles, beta particles and gamma radiations is called radioactivity.
  2. The residual material left in the nuclear reactors after generating heat energy is called nuclear waste. The nuclear waste is radioactive and very harmful to the environment.
  3. The nuclear waste should be stored in stainless steel containers, lined from within with thick sheets of lead, so that no radioactive rays come out of it. The containers should be stored in safe and well guarded place, so that they do not fall in the hands of criminal elements.

2014
Question 19.
A nucleus 11Na24 emits a beta particle to change into Magnesium (Mg)

  1. Write the symbolic equation of the process.
  2. What are numbers 24 and 11 called?
  3. What is the general name of 24 11Mg with respect to n Na ?

Answer:
(1)A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 33
(2) The number 24 is mass number and 11 is atomic number.
(3) They are isobars.

2015
Question 20(a).
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 34

  1. Complete the diagram as given below by drawing the deflection of radioactive radiations in an electric filed.
  2. State any two precautions to be taken while handling radio – active substances.

Answer:
(1) Deflection of radioactive radiations α, β and Y in an elec­tric field is as shown below:

A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 35

(2) The two safety precautions to be taken while handling radio ac­tive substances are (any two):

  1. Radioactive substances should be kept in thick lead containers with a very narrow opening so as to restrict the radiations com­ing out from other directions.
  2. Radioactive materials should be handled with long lead tongs.
  3. People working with radioactive substances should put on special lead lined aprons and lead gloves.

Question 20(b).
An atomic nucleus A is composed of 84 protons and 128 neutrons.

  1. The nucleus A emits an alpha particle and is transformed into nucleus B. What is the composition of nucleus B?
  2. The nucleus B emits a beta particle and is transformed into nucleus C. What is the composition of nucleus C?
  3. Does the composition of nucleus C change if it emits gamma radiations?

Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 36

2016

Question 21.
(a) An element zSA decays to SSR222 after emitting 2 alpha particles and 1 beta particle. Find the atomic number and atomic mass of the element S.
(b) A radioactive substance is oxidized. Will there be any change in the nature its radioactivity? Give a reason for your answer.
Answer:
(a) The decay will follow the following sequence
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 37
Therefore, we have
Z – 3 = 85 or Z = 85+ 3 = 88
And A- 8 = 222 or A = 222 + 8 = 230 Z = 88, A = 230
(b) Radioactivity is the nuclear phenomenon, i.e., radioactive radiation are emitted from the nucleus of the radioactive substance, so any physical change like oxidation or any chemical change does not affect its nature of radioactivity.

Question 22.
(a) Arrange α,β, and γ rays in ascending order with respect to their

  1. Penetrating power.
  2. Ionising power.
  3. Biological effect.

(b) (1) Represent the change in the nucleus of a radioactive element when α β particle is emitted.
(2) What is the name given to elements with same mass number and different atomic number.
(3) Under which conditions does the nucleus of an atom tend to radioactive ?
Answer:
(a)

  1. Penetrating power : α,β,γ
  2. Ionising power : γ, β, α
  3. Biological effect = α,β,γ

(b) (1) In an unstable nucleus, the neutron is changed into a proton by emitting a beta particle. This is represented as
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 38
(2) Elements with the same mass number but different atomic numbers are called isobars.
(3) The nucleus of an atom tends to be radioactive

  • When its atomic number exceeds 82.
  • There is an imbalance of protons and neutrons as compared to normal stable atom.

Additional Questions

Question 1.
Define nucleons, nuclide, neutrino, antineutrino.
Answer:
NUCLEONS : “The protons and neutrons inside the nucleus are called NUCLEONS.”
NUCLIDE : “Representation of an atom along with its ATOMIC NUMBER and MASS NUMBER is called NUCLIDE” e.g. ZXA is the nuclide of atom X.
ANTINEUTRINO : “An uncharged particle having negligible small mass emitted along with P-particle is called ANTI NEUTRINO.”

  1.  ATOMIC NUMBER : “The number of protons in the nucleus is called ATOMIC NUMBER -Z”.
  2. MASS NUMBER : “The number of protons and neutrons present in nucleus is called MASS NUMBER-A.”

Question 2.
What are isotopes ? Give one example.
Answer:
ISOTOPES : “Atoms of the same element having same ATOMIC NUMBER and different mass number are called ISOTOPES.”
Example:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 39

Question 3.
What are isobars ? Give one example.
Answer:
ISOBARS : “Atoms of different elements having same MASS NUMBER and different ATOMIC NUMBER are called ISOBARS.”
Example :
Sodium and Magnesium both have atomic Mass 23 but different Atomic numbers of 11 and 12.
11Na23 12Mg23

Question 4.
What is radio-activity ? Name two radioactive substances.
Answer:
“The phenomenon of some substances to give out spontaneous emission of invisible radiation which can penetrate through some thickness is called RADIOACTIVITY” and these substances are called RADIOACTIVE SUBSTANCES.
Name of RADIOACTIVE SUBSTANCES : URANIUM, RADIUM, THORIUM, POLONIUM and ACTINIUM.

Question 5.
Fig. shows a radioactive source S in a thick lead container. The radiations pass through an electric field between the plates A and B. Complete the diagram to show the paths of α, β and γ radiations
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 40
Answer:

A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 41

Question 6.
State the penetrating range of α, β and γ radiations.
Answer:
The penetrating range of a-particles is 2.7 cm to 8.62 cm of air, of [β-particles is 5 mm of aluminium or 1 mm of lead and of γ -radiations is about 30 cm of iron.

Question 7.
A radioactive source emits three types of radiations. Name them.

  1. Name the radiations which are charged.
  2. Name the radiations which are most penetrating.
  3. Name the radiations which travel with the speed of light.

Answer:
α, β and γ radiations.

  1. α and β radiations are charged.
  2. γ (Gamma Radiations) are the most penetrating.
  3. γ radiations travel with the velocity of light.

Question 8.
How do infrared and y-rays differ in their :
(1) wavelength
(2) penetrating power ?
Answer:

  1. Although both the infrared and y -rays are electromagnetic radiations, y-rays have much shorter wavelength (10-13 m) as compared to infrared whose wavelength is nearly 10-6 m or more.
  2. γ -rays are much more penetrating as compared to the infrared radiations.

Question 9.
State two similarities and two dissimilarities between the y-rays and X-rays.
Answer:
Similarities :

  1.  Both y -rays and X-rays are the electromagnetic waves.
  2. Both travel with speed 3 ×108 m s-1 in air (or vacuum).

Dissimilarities :

  1.  The wavelength of y -rays is shorter than that of X-rays.
  2.  y -rays are more penetrating than X-rays.

Question 10.
Is it possible to detect γ radiation in the way that a and β-particles can be deflected using the electric or magnetic field? Give reasons.
Answer:
γ-radiations are electromagnetic waves like light and are not charged particles. Hence can not be deflected by ELECTRIC or MAGNETIC field and cannot be detected.

Question 11.
State the penetrating range of α, β and γ radiations.
Answer:
The penetrating range of a-particles is 2.7 cm to 8.62 cm of air, of P-particles is 5 mm of aluminium or 1 mm of lead and . of Y -radiations is about 30 cm of iron.

Question 12.
A mixture of radioactive substance gives off three types of radiations.

  1. Name the three types of radiations.
  2. Name the type consisting of the same kind of particles as the beam of electrons.
  3. One of the radiations is similar to light. Name the radiation.
  4. Name the radiations which have the lowest ionising power.
  5. Name the radiations which have the lowest penetrating power.
  6. Give the charge and mass of particles composing the radiations in (V).
  7. Explain why radiations in (V) have the lowest penetrating power.
  8. When the particle referred to in (V) becomes neutral, they are found to be the atoms of a rare gas. Name this rare gas and draw a model of its neutral atom.
  9. From which part of the atom do these radiations come ?

Answer:

  • α, β and γ radiations.
  • Beta (β) are fast moving electrons.
  • Gamma (γ) radiations travel with the velocity of light.
  • Gamma (γ) radiations.
  • Alpha (α) radiations.
    Charge on α-particle is POSITIVE and mass 4 times that of proton,
    .’. α = 2He4
  • α-particles have lowest penetrating power because they are massive i.e. mass in more and velocity is least.
  • Rare gas is Helium.
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 42
  • Model of neutral atom is
  • From the nucleus.

Question 13.
A radioactive substance emits radiations :
(a) α, β and γ simultaneously .
(b) in the order α, β and γ one by one
(c) at one time α and β and then γ
(d) α and γ or β and γ

Question 14.
A radioactive sample is kept at the center of a large evacuated sphere. How safe will it be ?
Answer:
For safety, the radiations {α, β and γ emitted by the radioactive sample should not come out of the sphere, α-particles have a less penetrating power and therefore, the walls of the sphere easily stop them. β-particles will not he stopped by the walls and there will be no absorption of β particles inside the sphere as the air has been withdrawn from it. The number of β particles reaching per unit area at the surface of sphere will depend on the radius of sphere. This number will be reduced to 1/4, if the radius is doubled. So some safety is obtained if the sphere is large.
γ-radiations are also not absorbed by the walls.Thus for safety, the container should have lead walls and it should not be evacuated. The air will help in absorbing the radiations.Thus for safety, the container should have lead walls and it should not be evacuated. The air will help in absorbing the radiations.

Question 15.
Which of the radiations α, β and γ is similar to a beam of electrons?
Answer:
β-RADIATION is similar to a beam of electrons.

Question 16.
Give the relative ionising power of α, β and γ radiations.
Answer:
α-particles being the heaviest have maximum ionisation power. It is about 102 times that of β-particles and 10000 times that of gamma ( γ ) radiation.

Question 17.
A mass of lead is embedded in a block of wood. Radiations from a radioactive source incident on the side of block produce a shadow on a fluorescent screen placed beyond the block. The shadow of wood is faint but the shadow of lead is dark. Give reason for this difference.
Answer:
The shadow of wood is faint because only the α-radiations are stopped by wood (since α-radiations are least penetrating). The shadow of lead is dark because β and γ -radiations are also stopped by lead. (If wood is replaced by aluminium or any other light metal, the appearance on screen and reason will be the same), i.e. shadow will be dark.

Question 18.
What is meant by nuclear energy ?
Answer:
NUCLEAR ENERGY : “Energy obtained from the nucleus of atom i.e. from defect mass is called NUCLEAR ENERGY.”

Question 19.
What is Einstein’s mass-energy relation ?
Answer:
E = mc2 is Einstein’s mass-energy relation.

Question 20.
Calculate the energy released when a mass of 1 kg is completely converted into energy.
Answer:
According to Einstein’s mass-energy equivalence
E = mc2
E = 1 kg x [3 x 108 ms-1]2
Energy released E = 1 x 3 x 108 x 3 x 108 = 9 x 1016 J

Question 21.
State the unit in which the mass of nuclear particles is expressed. How is it related to kg ?[V.Imp.]
Answer:
Mass of nuclear particles is expressed in
u → Unified atomic mass unit
1 u = 1.6603 x 10-27 kg = 931 MeV in terms of energy

Question 22.
Calculate the energy released in the following fission reaction:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 43
Answer:

A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 44

Question 23.
Uranium nucleus 23592U decays to lead nucleus 20682 Pb .How many alpha and beta particles are emitted ?
Answer:
When 23592U decays to 20682 Pb, the mass number decreases from 238 to 206 i.e., it decreases by 32. Since with the emission of one alpha particle, mass number decreases by 4, so total number of alpha particles emitted will be 32/4 = 8
In decay of 23592U to 20682 Pb, the atomic number has decreased by 10. But due to emission of 8 alpha particles, the atomic number would have decreased by 2 x 8 = 16. Thus there is an increase in atomic number by 6, hence 6 beta particles will be emitted (because in emission of one beta particle, atomic number increases by 1). Thus 23592U decays to 20682 Pb with the emission of 8 a-particles and 6 P-particles.

Question 24.
If the loss in mass in fission of a uranium nucleus is 205 u, find the energy released. Take lu = 931 MeV.
Answer:
Energy released = 0.205 x 931 MeV
= 190.86 MeV

Question 25.
What is nuclear fission ?
Answer:
NUCLEAR FISSION : “The process in which an unstable nucleus of a heavy atom (like uranium -235) by bombardment with slow neutrons, splits up into two daughter medium weight nuclei with liberation of an enormous amount of energy is called NUCLEAR FISSION.”
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 45

Question 26.
What is nuclear fusion ?
Answer:
NUCLEAR FUSION : “When two light nuclei combine to form a heavier nucleus, the process is called nuclear fusion.”
A New Approach to ICSE Physics Part 2 Class 10 Solutions Nuclear Physics 46

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A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism

A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism.

These Solutions are part of A New Approach to ICSE Physics Part 2 Class 10 Solutions. Here we have given A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism.

Exercise – 1

Question 1.
When can an electric charge give rise to a magnetic field?
Answer:
When electric charge is in motion e. charge flows, it gives rise to magnetic field.

Question 2.
Describe Oersted’s experiment to show that a conductor carrying current produces a magnetic field around it
Answer:
Oersted’s Experiment :
Set up the apparatus as shown in fig. when switch is in open circuit
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 1
magnetic needle points in north direction. Showing there is no magnetic field around it. Now when switch is closed and current flows through the wire, the magnetic needle gets deflected from north. This shows that conductor carrying current produces magnetic field around it and deflects the magnetic needle.

Question 3.
(a) How will you plot the magnetic field lines around a straight conductor carrying current ?
(b) State two rules by which you can determine the direction of the magnetic field around a straight conductor.
Answer:
(a) To plot the magnetic field lines around a straight conductor carrying current: Pass straight conductor through a cardboard or glass-plate. And pass current.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 2
From A to B direction in the upward direction. Sprinkle iron filings on glass plate. Tap the glass plate, the iron filing will arrange themselves in circles around the conductor along the magnetic lines of force. The magnetic fields are circular in nature. The direction of magnetic field can be detected with the help of magnetic compass and direction is found anti clock wise.

(b) Two rules are :

Right hand thumb rule: “Imagine you are holding the current carrying wire in your right-hand so that your thumb points in the direction of current, then the direction in which your fingers encircle the wire will give the direction of magnetic field lines around the wire.”

A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 3
Max Well’s Screw Rule : “Imagine driving a cork screw in the direction of current, then the direction in which we turn its handle is the direction of magnetic field (or magnetic field lines).

A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 4

Question 4.
(a) Draw a set up for plotting magnetic field around a circular coil carrying current.
(b) State the properties of the magnetic field in 4(a).
Answer:
(a)
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 5
(b) Properties of the magnetic field :

  1. The magnetic field lines are circular near current carrying loop.
  2. At the center of the circular loop the magnetic field lines are in the same direction and strength of magnetic field increases.

Question 5.
What is a solenoid ? Draw a magnetic field around a solenoid, when direct current flows through it. How will you find the magnetic polarity of a solenoid without using a magnetic needle ?
Answer:
Solenoid : “Is a long coil containing a large number of close turns of insulated copper wire.”
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 6
Magnetic polarity of solenoid can be determined like a bar magnet i.e. when it is suspended freely, it will come to rest pointing in North and South direction or polarity can be checked by bringing north pole of a bar magnet. The pole repelled by north pole must be North pole of solenoid.

Question 6.
How does the magnetic field set up in a solenoid changes when:
(a) number of turns are increased ?
(b) diameter of the solenoid is increased ?
(c) strength of the current is increased ?
(d) a soft iron core is placed in it ?

Answer:

(a) When number of turns are increased magnetic field will be stronger.
(b) When diameter of the solenoid is increased but diameter should be less than the length of solenoid, so that parallel lines should add up to give a stronger field.
(c) When strength of current is increased stronger will be the magnetic field produced.
(d) When soft iron core is placed in the solenoid very strong magnetic field is produces.

Question 7.
Give four differences between an electromagnet and a permanent magnet.
Answer:
Differences between electromagnet and permanent magnet:
Electromagnet:

  1. It is temporary magnet
  2. It strength can be changed, by changing the current.
  3. Polarity can be changed by changing the direction of current.
  4. Produces very strong magnetic force.

Permanent Magnet

  1. It is permanent magnet.
  2. It strength cannot be changed.
  3. Polarity cannot be changed.
  4. Produces weak magnetic force.

Question 8.
State four practical applications of electromagnets.
Answer:
Four applications of electromagnet :

  1. In electric bell.
  2. In magnetising steel bars.
  3. For scanning machines (MRI)
  4. In electric motor, generator.

Question 9.
(a) On what factors does the force experienced by a straight conductor placed in a magnetic field depend
(b) State the law which determines the force experienced by a conductor.
Answer:
(a) Factors on which force experienced by a straight conductor depend :

  1. Current passing : direct proportional to current passing.
  2. Inversely proportional to the distance of that point from the wire.

(b) When a conductor carrying current is placed in a magnetic field in a direction other them the direction of magnetic field, experiences a force called Lorentz force. This force is perpendicular to both, the direction of current I and the direction of magnetic fields B.
This law is called :
Fleming’s left hand rule : “Stretch the fore finger, control finger and the thumb of your left hand mutually perpendicular to each other. If the fore-finger points in the direction of magnetic field, central finger indicates the direction of current, then the thumb will indicate the direction of motion of conductor (i.e., force on conductor)

Question 10.
(a) Draw a neat and labelled diagram of a d.c. motor and explain its construction and working.
(b) How can you make a d.c. motor more powerful ?
(c) How can you convert its jerky motion into uniform circular motion ?
Answer:
(a) D.C. Motor
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 7

Construction : A d.c. motor consists of :

  1. Rectangular coil ABCD of insulated copper wire moved between two Horse shoe
  2. Permanent magnet M such that AB and CD are perpendicular to the magnetic field.
  3. Two half rings (commutators) of copper X, Y are soldered to ends A and D of coil to change the direction of current flowing after every half rotation of the coil.
  4. Two carbon brushes P and Q fixed to the base of motor keep pressing highly against commutators. Battery to supply the current to coil is connected to the coil. The function of brushes is to make contact with the rotating rings of  the commutator and through them to supply current to the coil.

Multiple Choose Questions

Tick (✓) the most appropriate option.

1. A wire carrying a current is held over a freely suspended magnetic needle, such that the current in the wire flows from south to north. The direction in which the north end of freely suspended magnetic needle will point towards.
(a) West
(b) East
(c)South
(d) North

2. In an electric motor :
(a) mechanical energy changes to heat energy
(b) mechanical energy changes to electric energy
(c) electric energy changes to mechanical energy.
(d) electric energy changes to magnetic energy

3. By reversing the direction of current in an electromagnet, the magnetic field produced by it
(a) increases in strength
(b) remains unchanged in strength and direction
(c) gets reversed in direction
(d) decreases in strength

4. The power of a d.c. motor can be increased :
(a) by increasing number of turns in its coil
(b) by laminating its soft iron core
(c) by increasing the strength of current flowing through it
(d) all of these

5. Commutator is a device in a d.c. motor which :
(a) increases the power
(b) reverses direction of current coil after full rotation of coil
(c) reverses direction of current after half rotation of coil
(d) increases the strength of electromagnet

6. Which is not the use of an electromagnet ?
(a) Used in electric appliances such as electric bell and electric fans.
(b) Used for magnetising steel bars.
(c) Used for making sensitive magnetic compass.
(d) Used in separating iron particles from a scrap of iron and other metals.

7. Which is not the property of a solenoid ? The magnetic field of solenoid can be increased
(a) by increasing the number of turns in the solenoid.
(b) by increasing the strength of current flowing through the solenoid.
(c) by placing a stainless steel core within the solenoid.
(d) by placing a laminated soft iron core within the solenoid.

Exercise – 2

Question 1.
State Faraday’s laws of electromagnetic induction.
Answer:
Faraday’s Laws of Electromagnetic Induction :

  1.  When ever there is a change in the magnetic flux linked with a coil an e.m.f. is induced. The induced e.m.f. lasts as long as the change lasts (i-e-, there is a change in the magnetic flux linked with the coil.
  2. The magnitude of the e.m.f. induced is directly proportional to the rate of change of the magnetic flux linked with the coil. If the magnetic flux changed at a fixed rate, a stready e.m.f. is produced.

Question 2.
State
(a) Fleming’s right hand rule,
(b) Lenz’s law for finding the direction of induced current. Which of the above laws is most suitable for finding the direction of current in

  1. straight conductor
  2.  coiled conductor ?

Answer:
(a) Fleming’s right hand rule:

Or

(1) Generator rule : “Stretch the thumb, fore finger and central finger of right hand mutually perpendicular to each other. If the fore finger indicates the direction of magnetic field and thumb indicates the direction of motion of the conductor then central finger will indicate the direction of induced current.”
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 8
(2) Lenz’s Law : “The direction of induced e.m.f. (or induced current) always tends to oppose the cause which produces it.”

(a) In straight conductor — Fleming’s Right Hand Rule
(b) in coiled conductor — Lenz’s Law.

Question 3.
What do you understand by the term mutual induction ? Describe an experiment in support of your answer.
Answer:
Mutual induction: “The phenomenon of production of induced e.m.f. in a closed coil, by varying the magnetic flux in another coil is called mutual induction.”
Experiment :
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 9
P(primary coil) behaves as electro magnet when current passed by opening and closing the switch and pointer of the secondary coil (s) shows deflection proving that induced e.m.f. is produce in secondary coil. As soon as the switch is off (open circuit) deflection stops in secondary coil.
Or
The induced e.m.f. can be generated in the secondary coil, by placing the primary coil permanently in the secondary coil, and rapidly closing and opening the switch closing switch amounts to increases in magnetic flux in the primary coil, and hence in the secondary coil. Opening the switch amounts to decrease in magnetic flux in the primary and the secondary coil. Thus induced e.m.f. is generated in secondary coil.

Question 4.
What do you understand by the terms (a) self induction (b) eddy current ?
Answer:
(a) Self induction: “The phenomenon due to which a current flowing through a part of coil, induces an e.m.f. in the rest of coil due to change in magnetic flux is called self induction.”

(b) Eddy currents : “The current produced in any metallic conductor when a magnetic flux is changed around it is called Eddy Current.

Question 5.
(a) Describe with the help of a clear diagram the structure of a.c. transformer, suitable for lighting 12 V lamp from 240 V mains.
(b) Explain how a transformer reduces emf ?
(c) Why are transformers so important for the transmission of energy ?
Answer:
(a) We are to transformer for lighting 12V lamp from 240 V mains.
Therefore step down transformer is needed.
The magnitude of induced e.m.f. is produced by the formula.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 10
Primary coil:

  • It should have 20 times more number of turns as compared to secondary coil.
  • It is made of thinner copper wire.
  • It should be more heavily insulated secondary coil.

Secondary coil :

  • It should be made of thicker copper wire.
  • It should be less heavily in solated.

(b) The number of turns in secondary 1/20 of the number of turns in primary coil reduces the e.m.f. to 1/20  th e.m.f. of primary coil.
(c) Transformers are important for transmission of energy. The power from the generating station is transmitted over long distances at a voltage higher than 11 kv to minimise the loss of energy in form of heat in the line wires used for transmission for this step up transformer is used which step up 11 kv to 132 kv at the generating station. To transmit if to industries it is stepped down to 33 kv using step down transformer. Further stepped down to 11 kv to transmit to light industry. It is further stepped down to 220 V by step down transformer to supply it to domestic consumers.

Question 6.
(a) Draw a labelled diagram of an a.c. generator.
(b) Why is the emf produced by an a.c. generator zero at a certain instant and reaches the maximum value at the other instant?
(c) The efficiency of a generator is never 100%. Explain.
Answer:
(a)
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 11

(b) The e.m.f. produced by a.c. generator is zero when magnetic flux linked with coil is maximum. This happens when the plane of coil is normal to the magnetic field.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 12
(c) The efficiency of a generator is never 100% because

  1. a part of energy is wasted due to magnetic hysteresis.
  2. A part of energy is wasted an account of the resistance of primary and secondary coils wound around the core
  3. A part of current is wasted on account of the Eddy currents formed with in the soft iron core.

A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 13

Question 7(a) .
The adjacent diagram 10.44 shows a movable permanent magnet and fixed copper coil of many turns, connected to a center zero galvanometer.
Describe your observations when:
Answer:

  1. The magnet is rapidly moved in the direction of the arrow.
  2. The magnet is stopped within the coil
  3. The magnet is then pulled out rapidly.
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 14

Question 7(b).
What will be your observations, if a more powerful magnet is used ?
Answer:
(1)

(1) A momentary deflection of pointer of galvanometer towards right side is seen.
(2) More the speed, more is the deflection.
(3) When magnet is stopped with in the coil, the pointer of galvanometer comes to zero position.
(4) Deflection of pointer of galvanometer is in opposite direction to towards the left.

(2)  Then more powerful magnet is used deflection of pointer will be more.

Multiple Choose Questions

Tick (✓) the most appropriate option.

1. The direction of current in a conductor can be obtained by ?
(a) Fleming’s right hand rule
(b) Fleming’s left hand rule
(c) Right hand thumb rule
(d) Maxwell’s cork-screw rule

2. In a step down transformer :
(a) number of turns in primary coil are less than the secondary coil.
(b) number of turns in primary coil are more than the secondary coil.
(c) number of turns in primary coil are equal to the secondary coil
(d) the primary and secondary coils are wound on separate steel cores.

3. Step up transformers are used:
(a) for long distance transmission of power
(b) for distribution power in localities.
(c) for saving sensitive appliances, such as T.V andA.C.
(d) any of these

4. Which is the incorrect statement ? In a step down transformer
(a) number of turns in primary are more than the secondary coil
(b) the primary coil is thinner as compared to the secondary coil.
(c) the primary coil is thicker as compared to secondary coil.
(d) the primary coil is more heavily insulated as compared to secondary coil

5. In an cue. generator the magnitude of induced current can be increased by :
(a) increasing number of turns in the coil
(b) increasing the area of cross-section of the coil
(c) increasing the strength of field magnets
(d) all of these

Questions from ICSE Examination Papers 

2001

Question 1.
Draw a diagram of a D.C. motor, labelling clearly the following parts:
(a) The field magnet
(b) The armature
(c) Commutator
(d) Wire brushes
What energy changes are involved in the D.C. motor ?
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 15

Energy changes involved in D.C. Motor is Mechanical energy is converted into electrical energy.

2002

Question 2(a).
Draw a sketch of an electric bell with the electrical connections and label its main parts. Why is the armature made of soft iron and not steel ?
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 16

Question 2(b).
State two dissimilarities between a d c motor and an a.c. generator.
Answer:
Different between D.C. motor and A.C. generator
D.C. Motor

  1. It converts electrical energy to mechanical
  2. It is based on Fleming’s left hand rule energy.
  3. It has split rings.
  4. DC motor works on the principle of force acting on current carrying conductor placed in a magnetic field.

A.C. Generator

  1. It converts mechanical energy to electrical energy.
  2. It is based on Fleming’s right hand rule.
  3. It has slip rings.
  4. AC generation works on the principle of electromagnetic induction

2003
Question 3.
Fig. shows a coil connected to a center zero galvanometer G The galvanometer shows a deflection to the right when the North — pole of a powerful magnet is moved to the right as shown:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 17
(a) Explain why the deflection occurs in the galvanometer.
(b) Does the direction of the current (clockwise or anticlockwise) when viewed from the end A ?
(c) State observation in G when the coil AB is moved away from N.
(d) State the observation in G when, both the coil and the magnet, are moved towards the right at the same speed.
Answer:

(a) Due to motion of magnet, magnet flux linked with coil changes. As a result, e.m.f. is induced across its ends, due to which induced current fluxes in coil so galvanometer shows deflection.
(b) clockwise (c) It deflects to left (d) No deflection is observed as there is no change in magnetic flux linked with coil.
(c) Deflection in G is towards left of zero.
(d) Relative motion is zero and there is no deflection e. pointer is at 0 (rest position).

2004

Question 4(a).
Draw a neat and labelled diagram of a step-down transformer.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 18

Question 4(b).
What is the main difference between a step-up and step- down transformer.
Answer:
Difference between step Up and Step down Transformer :
Step up Transformer

  1. It incrases the a.c. voltage and decreases the current i e. Es >Ep and Is <Ip
  2. The wire of primary coil is thicker than that of the secondary coil.
  3. NS/NP> I
  4. Used at power generating station, in X-ray, T.V. etc.

Step Down Transformer

  1. It decreases the a.c. voltage and increases the current. i.e. Es >Ep and Is <Ip
  2. The wire of the secondary coil is thicker than that the primary coil.
  3. Ns /Np <1
  4. Used at power stations electric bell, night electric bulb.

2004
Question 5(a).
How can you demagnetise a bar magnet by employing alternating current ?
Answer:
When current starts flowing in the coil and magnet is with drawn from the coil in East-West direction and is kept far away, the magnetic field produced changes its direction continously, magnetisation becomes weak and weak .untill demagnetised.

Question 5(b).
State two ways by which the emf in an A.C. generator can be increased.
Answer:
The e.m.f. in an a.c. generator can be increased by

  1. increasing the number of turns in the armature (or coil).
  2. increasing the speed of rotation of the armature about its axis (or coil).

2005

Question 6.
State the energy change which takes place when a magnet is moved inside a coil having a galvanometer at its ends. Name this phenomenon.
Answer:
When a magnet is moved in side a coil MECHANICAL ENERGY changes into ELECTRICAL ENERGY i.e. magnetic flux (Magnetic lines of force) is converted into e.m.f. or electric current.
The phenomenon is called INDUCED CURRENT.

Question 7.
Draw a labelled diagram of an A.C. generator.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 19

Question 8.
(1) State the function of a split ring in a D.C. motor.
(2) Mention two reasons why a soft iron core is used within the coil of a moving coil galvanometer.
Answer:

  1. Function of split ring in a D.C. Motor to change the direction of current flowing after eveiy half rotation the coil.
  2. Soft iron core provides STRONG MAGNETIC FIELD .when a current flows through the coil wound around it.

2006

Question 9.
State two advantages of an electromagnet over a permanent magnet.
Answer:
Two ADVANTAGES OF ELECTROMAGNETIC OVER PERMANENT MAGNET are :

  1. Its strength can be increased or decreased by increasing or decreasing current.
  2. It loses magnetism as soon as current is stopped and acquires magnetism as soon as current is passed.
  3. The polarity can be changed by changing the direction of current.
  4. It produces stronger magnetic field.

Question 10.
(a) What will happen to a compass needle when the compass is placed below a wire and a current is made to flow through the wire ? Give a reason to justify your answer.
(b) What energy conversion takes place during the working of a d.c. motor ?
Answer:
(a) The needle of compass will deflect as the wire carrying current is associated with magnetic field.
(b) ELECTRICAL energy is transformed to MECHANICAL energy.

2007

Question 11.
(1) State two factors on which the strength of an induced current depends.
(2) When a solenoid that is carrying current is freely suspended, it comes to rest along a particular direction. Why does this happen ?
Answer:
(1) Two factors on which strength of induced current depends are:

(a) The magnitude of current is directly proportional to “The rate of change of magnetic flux with in the closed coil.
(b) It depends upon the number of turns and the area of cross-section of the coil.

(2) This means the solenoid carrying current behaves like a bar magnet and it has come to rest along a particular direction N­S N-end of this solenoid points towards north pole of earth and S-end points towards south-pole of earth.

2008

Question 12.
State one point of similarity and one point of difference between an a.c. generator and a d.c. motor.
Answer:
Difference between a.c. generator and d.c. motor
In a.c. generator mechanical energy gets converted into electrical energy and induced e.m.f. is alternating in nature. In d.c. motor electrical energy gets converted into mechanical energy and electricity used in direct current.
Similarity between a.c. generator and d.c. motor is that both use the principle of electromagnetic induction and use a solenoid on which copper wire is would.

Question 13.
(a) What is the name given to a cylindrical coil whose diameter is less in comparison to its length ?
(b) If a piece of soft iron is placed inside a current carrying coil, what is the name given to the device ?
(c) Give one use of the device named by you in 13 (b) above.
Answer:

(a) It is called Solenoid.
(b) Aarmature
(c) Aarmature using soft iron core becomes Magnetised and Increases the strength of magnetic field.Armature makes the motor more powerful.

Question 14(a).
(1) Why does a magnetic needle show a deflection when brought close to a current carrying conductor ?
(2) A wire bent into a circle carries current in an anticlockwise direction. What polarity does this face of the coil exhibit ?
Answer:

  1. When an electric current is passed through a conducting wire, a magnetic field is produced around it. A compass needle, if freely suspended, rests in North South direction but due to magnetic field produced around wire, the compass needle gets attracted or repelled and changes its direction.
  2. North polarity is formed.

Question 14(b).
Draw a simple sketch of a step-down transformer. Label the different parts in the diagram.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 20

2009

Question 15.
The figure below shows an electromagnet.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 21
(a) What will be the polarity at the end X ?
(b) Suggest a way by which the strength of the elctromagnet referred to in the question, may be increased.
Answer:
(a) At end X – North polarity will be formed.
(b) The strength of the electromagnet formed can be increased by

  1. Increasing the number of turns of the winding
  2.  Increasing the current through it and decreasing the gap between them.

Question 16.
(1) Draw a neat and labelled diagram to show the structure of an a.c. generator.
(2) State the energy conversion taking place in the generator when it is working.
Answer:
(1)
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 22
(2) Mechanical energy gets connected into electrical energy when generator is working.

2010
Question 17.
A device is used to transform 12V a.c. to 200V a.c.
(1)  What is the name of this device ?
(2) Name the principle on which it works.
Answer:

 

  1.  Since the device is being used to step up 12 V a.c. to 200 V a.c., it is called a step up transformer.
  2. A transformer works on the principle of mutual induction.
  3. “When an alternating current is passed in the primary coil wound on the soft iron core, an induced emf- is produced in the secondary coil, wound on a soft iron

Question 18(a).

  1. A straight wire conductor passes vertically through a piece of cardboard sprinkled with iron filings. Copy the diagram and show the setting of iron filings when a current is passed through the wire in the upward direction and the cardboard is tapped gently. Draw arrows to represent the direction of the magnetic field lines.
  2. Name the law which helped you to find the direction of the magnetic field lines
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 23

Answer:

  1. The direction of the magnetic lines of force will be anticlockwise. The iron filings will align themselves in circular lines of force (By Right hand thumb rule.)
  2. Right hand thumb rule or Maxwell’s cork screw rule.

Question 18(b).

  1. State two ways by which the magnetic field of a solenoid can be made stronger.
  2. What material is used for making the armature of a electric bell ? Give a reason for using this material.

Answer:

  1. The magnetic field along the axis of a solenoid is given by B = μ0ni
    Clearly, the magnetic field can be increased by .

    • Increasing the current.
      A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 24
    • Increasing the number of tums/length.
  2. The armature of the electric bell is made of soft iron. This is because soft iron when magnetised is quickly demagnetised when the circuit is broken. Steel armature might lock up the circuit and the bell may not work.

2012

Question 19(a).
(1) What is an a.c. generator or dynamo used for?
(2) Name the principle on which it works.
Answer:
(a)

  1. An a.c. generator or dynamo is used to produce electric city by mutual induction i.e. rotating a coil in magnetic field.
  2. It works on the principle of electromagnetic Induction that is a coil is rotated in a magnetic field and magnetic flux linked with the coil changes therefore an e.m.f. is induced between the ends of the coil.

Question 19(b).
The diagram alongside below a current carrying loop or a circular coil passing through a sheet of cardboard at the points M and N. The sheet of cardboard is sprinkled uniformly with iron filings.

  1. Copy the diagram and draw an arrow on the circular coil to show the direction of current flowing through it.
  2. Draw the pattern of arrangement of the iron filings when current is passed through the loop.
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 25

Answer:
(1) and (2)
Direction of current flowing is shown by arrows
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 26

2013

Question 20(a).
You have been provided with a solenoid AB.

  1. What is the polarity at end A ?
  2. Give one advantage of an electromagnet over a permanent magnet.
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 27

Answer:
(a) 

  1. Polarity at the end A is NORTH pole .
  2. An electromagnet’s strength can be increased by increasing the flow of current in the coil, which is not possible in case of a permanent magnet.

Question 20(b).

  1. Draw a simple labelled diagram of d.c. electric motor,
  2. What is the function of the split rings in a d.c. motor ?
  3. State one advantage of a.c. over d.c.

Answer:
(1)
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 28

(2) The split rings alter the direction of current in the coil after every half rotation. This in turn helps the coil to move in the same direction, i.e., clockwise or anticlockwise direction.
(3) The alternating current can be easily stepped up or down and can be transmitted over long distance cable wires. This is not possible in case of direct current.

2014

Question 21.
(a) Name two factors on which the magnitude of an induced e.m.f. in the secondary coil depends.
(b)
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 29
In the following diagram an arrow shows the motion of the bar magnet.
(1) State in which direction the current flows A to B or B to A?
(2) Name the law used  to come to the conclusion
Ans:
(a) The magnittude of induction e.m.f in the secondary coil depends on the following two factors

  1. Strengh of current in primary
  2. Speed of motion of coil
  3.  Area of coil and number of turns

(b)

  1. Current flows from A and B as N-pole is formed at end B and S-pole is is formed at end A
  2. Name of the law is Lenz’s law  i.e such polarity is formed at coil which opposes the cause which produces

2015
Question 22 .
(a)Why does a current carrying freely suspended solenoid rest along a particular direction
(b)State the dircection in which it rest
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 30
(c) Give two similartyies between an Ac generation and DC motor
(d) fill in the blanks
on reversing the direction of the current in a wire the magnetic field produced by its gets……..
Answer:
(a) A current-carrying freely suspended solenoid acts as a bar mag­net, and thus, due to the Earth’s magnetic field, it rests along a particular direction.
(b) In rests in the North -South Direction
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 31

(c) Two similarities between an AC generator and a DC motor are

  • A coil rotates in a magnetic field between the pole pieces of a powerful electromagnet.
  • The external circuit is connected to two carbon brushes B and B.,.

(d) On reversing the direction of the current in a wire, the magnetic field produced by it gets reversed.

2016

Question 23(a).
Which coil of a step up transformer is made thicker and why?
Answer
(1) The primary coil of step up transformer is made thicker. it is because a current of higher magnitude flows in this coil, which in turn made melt it on account of resistance. Thus, in order to lower resistance it is made thicker.

Question 23(b).
(1) Name the transformer used in the power transmitting station of a power plant.
(2) What type of current is transmitted from the power station?
(3) At what voltage is this current available to our household?
Answer

  1. A step-up transformer is used in the power transmitting station of a power plant.
  2. An alternating current is transmitted from the power station.
  3. The current is available to our household at a voltage of 220 V.

 Additional Questions.

Question 1.
Draw a diagram showing the magnetic field lines due to a straight wire carrying current.
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 32

Question 2.
A straight wire lying in a horizontal plane carries a current from North to South. What will be the direction of magnetic field at a point just beneath it ? Name the rule used to arrive at answer and state it.
Answer:s.
Towards EAST.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 33

OERSTED RULE. “If a wire is held ABOVE magnetic needle, such that current is flowing NORTH to SOUTH then north pole of magnetic needle will deflect towards. EAST.”

Question 3.
State a law, which determines the direction of magnetic field around a current carrying wire.
Answer:
It is RIGHT HAND THUMB RULE which states “If we hold a current carrying conductor in the right hand such that thumb points in the direction of the current, the tips of fingers pointing the direction of magnetic field lines.”

Question 4.
What is Solenoid ?
Answer:
SOLENOID : “An insulated copper wire wound on some cylindrical cardboard or plastic tube, such that its length is greater than its diameter and it behaves like a magnet when current flows through it is called SOLENOID.”

Question 5.
Draw a labelled diagram to make an electromagnet from a soft iron bar. Mark the polarity at its ends. What precautions would you observe ?
Answer:
The labelled diagram is shown in Fig.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 34
Precautions :

  1. The source of current must be the d.c. source.
  2. The bar must be placed in north-south direction.

Question 6.
The adjacent diagram shows a small magnet placed near a solenoid. When current is switched on in the solenoid, will the magnet be attracted or repelled ? Give a reason for your answer.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 35
Answer:
The magnet will be REPELLED as current flowing is ANTI CLOCK-WISE and N-pole is formed near magnet.

Question 7.
A straight conductor passes vertically through a cardboard sprinkled with iron filings. Show the setting of iron filings when a weak current is passed in the downward direction and then the card board is gently tapped.What changes occur if,

  1. The strength of current is increased ?
  2. The single conductor is replaced by several parallel conductor with current flowing in the same direction ?

Answer:
Fig. shows the setting of iron filings :

  1. The arrangement of iron filings remains unchanged but they get arranged upto a larger distance.
  2. The magnetic field strength increases.
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 36

Question 8.
The following diagram shows two straight wires A and B carrying equal currents. Draw the pattern of magnetic field lines around them and mark their directions. What will be the resultant magnetic field at a point K equidistant from the wires A and B ? (V.V.Imp)
Answer:
Pattern of magnetic field lines is shown.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 37
Resultant magnetic field at K is zero
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 38

Question 9.
Draw a diagram representing the magnetic field inside and outside a solenoid through which a current is flowing and mark with arrows, the direction of current in the solenoid and the direction of magnetic field lines. Also mark the polarity at the faces of solenoid. A bar of soft-iron is then placed parallel to its length inside the solenoid. Describe what happens. (V.Imp)
Answer:
Fig. shows the magnetic field lines due to a current carrying solenoid.
The direction in the solenoid at the face A is clockwise, so it will have the south (S) polarity and the face B of solenoid will have the north (N) polarity.

A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 39
If a bar of soft iron is placed parallel to its length inside the solenoid, the magnetic field inside the solenoid becomes closer i.e. the magnetic field inside the solenoid is increased.

Question 10.
Name and state the rule by which tlv direction of magnetic field in a current carrying solenoid is determined
Answer:
Name of the rule is CLOCK RULE.
Statement : “Looking at the face of loop, if the current around that face is ANTI-CLOCK wise, the face has the NORTH POLARITY and if the current at that face is in CLOCK wise direction, the face has the SOUTH polarity.

A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 40

Question 11.
Complete the following sentences :

  • When current flows in a wire, it creates magnetic field around it
  • A current carrying solenoid behaves like a bar magnet
  • When a current is passed through a conductor,  is set up around the conductor a magnetic field..

Question 12.
A soft iron bar is introduced inside a current carrying solenoid. The magnetic field inside the solenoid :
(a) will become zero
(b) will decrease
(c) will increase    
(d) will remain unaffected.

Question 13.
The diagram below show a current carrying loop passing through a sheet of stiff cardboard. Draw three magnetic field lines on the card board. State two factors on which the magnitude of magnetic field at the center of the loop, depends.

A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 41
Answer:
Fig. represents the magnetic field lines due to the current carrying loop using right hand thumb rule.

A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 43

The magnitude of magnetic field at the center depends on
(1) the strength of current in loop and
(2) the CURRENT radius of loop.

Question 14.
The magnetic field lines inside a current carrying solenoid, are
(a) along the axis and are parallel to each other

(b) perpendicular to the axis and equidistant from each other
(c) circular and do not intersect each other
(d) circular at the ends but they are parallel to the axis inside the solenoid.

Question 15.
What is LORENTZ force ?
Answer:
“A moving charge in magnetic field not parallel to the field experiences a force called LORENTZ FORCE.” and since moving charge is called current then conductor carrying a current in magnetic field experiences force F = BI/ in other words this force F α B (magnetic field)
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 44

Question 16.
State Fleming’s left hand rule. (V.Imp.)
Answer:
Fleming’s left hand Rule :
“Stretch the thumb, the force-finger and the middle finger of your left hand mutually at right-angles to each other, such that force-finger points in the direction of magnetic field and middle finger in the direction of flow of current. Then, thumb gives direction of motion of conductor.”
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 45

Question 17.
Draw a clear labelled diagram of a moving coil galvanometer. Explain why pole piece are made cylindrical?
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 46

In order to keep the plane of coil always in the direction of magnetic field lines, the poles of the horse-shoe magnet are made cylindrical.

Question 18.
State the unit of magnetic field in terms of force experienced by a current carrying conductor placed in a magnetic field.
Answer:
Unit of magnetic field is TESLA. See Page 472 for Tesla.

Question 19.
Why is a soft iron piece placed at the center of the coil of a moving coil galvanometer ?
Answer:
A soft iron core is placed at the center of the coil which intensifies the magnetic field and makes it radial by concentrating the magnetic field lines due to its high permeability as shown in next fig.

Question 20.
What is the use of radial magnetic field ?
Answer:
The coil rotates but its plane always remains parallel to the magnetic field making θ = 0°.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 47

Question 21.
How is the deflection in a moving coil galvanometer related to the current in its coil ?
Answer:
The rotation of coil depends on the magnitude of the deflecting couple which is directly proportional to the strength of current. Hence DEFLECTION OF COIL IS THE MEASURE of CURRENT PASSED in the coif

Question 22.
Give uses of moving coil galvanometer. (V.Imp.)
Answer:
USES OF GALVANOMETER :

  1. To detect the presence of current in a circuit. .
  2. To convert it into AMMETER to measure current.
  3. To convert it in VOLTMETER to measure p.d.
  4. To detect the direction of flow of current.

Question 23.
What is an electric motor ? State its principle.
Answer:
ELECTRIC MOTOR : “is a device which converts electrical energy into mechanical energy.”
PRINCIPLE : “When a current carrying conductor is placed in a magnetic field normally, it begins to move.” The direction of motion is obtained by FLEMING’S LEFT HAND RULE.

Question 24.
A rectangular coil ABCD is placed between the pole pieces of a horse-shoe magnet as shown in fig. (V.V.Imp.)

  1. What is the direction of force on each arm ?
  2. What is the effect of the forces on coil ?
  3. How is the effect of force on coil changed if the terminals of battery are interchanged ?
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 48

Answer:

  1. In Fig. the current in coil is in direction DCBA. By Fleming’s left hand rule, on the arm AB, the force is outward at right angles to the plane of coil. On the arm BC no force acts. On the arm CD, the force is inwards perpendicular to the plane of coil. On the arm DA, no force acts.
  2. The force on the arms AB and CD are equal in magnitude, but opposite in direction. They form a clockwise couple. So the coil will rotate clockwise with the arm AB coming out and the arm CD going in.
    On interchanging the terminals of battery, the direction of current in coil is reversed so the coil will rotate anticlockwise.

(V.Imp. Note)

If a conductor (coil) is moved in a magnetic field or a magnet is moved in a coil (conductor) e.m.f. is produced.

Question 25.
What is electromagnetic induction ?
Answer:
ELECTROMAGNETIC INDUCTION : “Whenever there is a change in the number of magnetic field lines associated with a conductor, an electromotive force (e.m.f.) is developed between the ends of the conductor which lasts as long as the change is taking place. This phenomenon is called electromagnetic induction.”

Question 26.
State Faraday’s laws of electromagnetic induction. (V.Imp.)
Answer:
FARADAY’S LAWS OF e.m.f:

  1. The magnitude of the e.m.f. induced is directly proportional to the rate of magnetic flux linked with the coil.
  2. Whenever there is a change in magnetic flux linked with a coil, an e.m.f. is induced. The induced e.m.f. lasts so long as there is a change in magnetic flux linked with the coil.

Question 27.
State the two factors on which the magnitude and direction of induced e.m.f depend.
Answer:
Two Factors are :

  1. The change in magnetic flux and
  2. The time in which the magnetic flux changes.
    DIRECTION of induced e.m.f depends on whether the magnetic flux increases or decreases.

Question 28.
Two coils A and B are placed as shown in Fig. The coil A is connected to a battery and a key K while the coil B is connected to a center zero galvanometer G (V.Imp.)
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 49
What will you observe in the galvanometer G when

  1. the key K is closed.
  2. the key K is opened.
  3. with the key K closed, the coil A is move rapidly towards the coil B.
  4. with the key K closed, the coil B is moved rapidly towards the coil A.
  5. with the key K closed, the coils A and B are moved away from each other ?

Answer:

  1. As the key K is closed, a deflection is observed in the galvanometer for a short while (i.e. a momentary deflection). In other words, the galvanometer needle deflects for a moment and returns to zero.
  2. As the key K is opened, again a momentary deflection (but more) is observed in the opposite direction.
  3. With the key K closed, if the coil A is moved rapidly towards the coil B, a deflection is obtained in the galvanometer in the direction as in
    • due to increase in magnetic flux through the coil B. But the deflection lasts so long as the coil A moves.
  4. With the key K closed, if the coil B is moved rapidly towards the coil A, again a deflection is observed in the galvanometer in direction as in
    • due to increase in magnetic flux through the coil B. The deflection lasts so long as the coil B moves.
  5. With the key K closed, if the coil A and B are moved away from each other, a deflection is observed in the galvanometer in direction as in
    • due to decrease in magnetic flux through the coil B. The deflection lasts so long as there is relative motion between the coils.

Question 29.
The direction of induced current is obtained by :
(a) Fleming’s left hand rule
(b) Maxwell’s cork-screw rule
(c) Right hand palm rule
(d) Fleming’s right hand rule.

Question 30.
(1) State Fleming’s right hand rule.
(2)What is Lenz’s law ?
Answer:

  1. FLEMING RIGHT HAND RULE : “Stretch the thumb, the fore finger and the middle finger of your right hand, mutually at right angles to each other, such that fore finger points in the direction of magnetic field and the thumb in the direction of motion of conductor. Then the direction in which the middle finger points, gives direction of flow of induced current.”
  2. LENZ’S LAW : “The direction of induced current is such, that it always opposes the cause (the motion of conductor) which produced it.”

Question 31.
The diagram below shows two coils X and Y. The coil X is connected to a battery S and a key K. The coil Y is connected to a galvanometer G
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 50

When the key K is closed, State the polarity.

  1. at the end B of the coil X.
  2. at the end C of the coil Y.
  3. at the end C of the coil Y if the coil Y is
    (a) moved towards the coil X,
    (b) moved away from the coil X

Answer:

  1. Current at the end B of coil X is anti-clockwise therefore at this end there is a NORTH pole.
  2. While closing the key, polarity at the end C of the coil Y will be north. There will be no polarity at the end C of the coil Y when the current becomes steady in the coil X.
  3. (a) While the coil Y is moved towards the coil X, the polarity at the end C of the coil Y is NORTH.(b) While the coil Y is moved away from the coil X, the polarity at the end C of the coil B is SOUTH.

Question 32.
What is Generator ? Draw a labelled diagram of a simple a.c. generator.(V.Imp)
Answer:
A.C. GENERATOR OR DYNAMO
“A dynamo is a device which converts mechanical energy into electrical energy using the principle of electromagnetic induction.”
Diagram :

Question 33.
(1) Name the principle on which a transformer works.
(2) What is the function of a step-up transformer ?
(3) Can a transformer work when it is connected to a d.c. source ?
(4) Draw a simple labelled diagram of a step-down transformer ?
(5) Draw a simple labelled diagram of a step-up transformer. (V.V.Imp.)
Answer:

  1. A transformer works on the principle of electromagnetic induction.
  2. The function of a step-up transformer is to convert a low a.c. voltage to a high a.c. voltage.
  3. No. A transformer cannot work when it is connected to a d.c. source.
  4. Fig. shows a simple labelled diagram of step -down transformation.
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 51
  5. Fig. shows a simple labelled diagram of step -up transformation.
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 52

Question 34.
For what purpose are transformers used ? Can they be used with a direct source ?
Answer:
Transformers are used to increase or decrease the amplitude of alternating e.m.f.
They cannot be used with direct source (D.C.) since its working is based on change of magnetic flux with the varying current.

Question 35.
The following diagram shows a coil of several turns of copper wire connected to a sensitive center-zero galvanometer G near a magnet NS. The coil is free to move.
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 53

  1. Describe the observation if the coil is rapidly moved in the direction of arrow.
  2. How would the observation be altered if
    (a) the coil has twice as many turns,
    (b) the coil is made to move three times as fast ?

Answer:

  1. Galavanometer will show deflection and N-polarity will be formed at end A of the coil repel its motion.
  2. (a) The deflection will be more but in the same direction,
    (b) The deflection will be three times in the same direction.

Question 36.
How are the e.m.f. in the primary and secondary coils of a transformer related with the number of turns in these coils ?
Answer:
A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 54
Question 37.
Complete the following diagram of a transformer and name the parts labelled A and B. Name the part you have drawn to complete the diagram. What is the material of this part ? Is this transformer step-up or step-down ? Why ?

A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 55
Answer:
Diagram is completed
Part A →  Primary Coil
Part B → Secondary Coil
Part drawn is → Input A.C. Alternating source of e.m.f. and output A.C. and core.

A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 56

Material of Core is SOFT IRON
It is STEP DOWN TRANSFORM. Since number of turns in Secondary coil is less than that of primary.

Question 38.
Draw labelled diagram to show various components of a
(1) Step up transformer
(2) Step down transformer.
Answer:

  1. STEP-UP TRANSFORMER : A transformer which increases the applied e.m.f. of an alternating current is called SET-UP TRANSFORMER.
    Number of turns in SECONDARY Coil Should be MORE.
  2. STEP-DOWN TRANSFORMER : “A transformer which DECREASES the applied e.m.f. of alternating current is called STEP-
    DOWN TRANSFORMER.”
    A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 57

Question 39.
A flat rectangular coil is rotated between the pole pieces of a horse-shoe magnet In which position of coil with respect to the magnetic field, will the e.m.f
(1) be maximum,
(2) be zero and
(3) change direction ? (V.Imp.)
Answer:

  1. The e.m.f. is maximum when the plane of coil is parallel to the magnetic field.
  2. The e.m.f. is zero when the plane of coil is normal perpendicular to the magnetic field.
  3. The e.m.f. will change direction when the plane of coil passes from the position normal to the magnetic field.

Question 40.
A dc. motor is rotating in a clockwise direction. How can the direction of rotation be reversed ?
Answer:
The direction of rotation of motor can be reversed by interchanging the terminals of the battery connected to the brushes of motor.

Question 41.
Fig. shows a rectangular coil ABCD placed in between the pole pieces of a horse-shoe magnet with its plane perpendicular to the magnetic field.

A New Approach to ICSE Physics Part 2 Class 10 Solutions Electromagnetism 58

  1. What is the direction of force on each arm of coil ?
  2. Will the coil rotate due to the forces on its arms ?

Answer:

  1. In Fig. the current in coil is in direction DCBA. On the arm AB, the force is upward in the plane of coil i.e. away from the arm CD. On the arm BC, the force is outward in the plane of coil i.e., away from the arm AD. On the arm CD,the force is downward in the plane of coil i.e., away from the arm AB. On arm DA, the force is outward in the plane of coil away from the arm BC.
  2. The coil will not rotate (because all the arms experience an outward pull.)

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