ICSE 2012 Physics Question Paper Solved for Class 10

Solving ICSE Class 10 Physics Previous Year Question Papers ICSE Class 10 Physics Question Paper 2012 is the best way to boost your preparation for the board exams.

ICSE Class 10 Physics Question Paper 2012 Solved

Time: 1½ hours
Maximum Marks : 80

General Instructions:

• Answers to this paper must be written on the paper provided separately.
• You will not be allowed to write during the first 15 minutes.
• This time is to be spent in reading the Question Paper.
• The time given at the head of this paper is the time allowed for writing the answers.
• Section I is compulsory. Attempt any four questions from Section II.
• The intended marks for questions or parts of questions are given in brackets [].

SECTION – I (40 Marks)
(Attempt all questions from this section)

Question 1.
(a) (i) Define 1 kgf.
(ii) How is it related to the S.I. unit of force ? [2]
Answer.

1. One kilogram force is the force due to gravity on a mass of 1 kilogram.
2. 1 kgf = 9.8 N.

(b) (i) What are non-contact forces ?
(ii) How does the distance of separation between two bodies affect the magnitude of the non-contact force between them ? [2]
Answer.

1. The forces which act on bodies without being physically touched are called the non-contact forces.
2. It decreases with the increase in separation and increases as the separation decreases. These forces vary inversely as the square of distance of separation.

(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 sea-saw? [2]
Answer.
According to principle of moments of force

W₁l₁ = W₂l₂
⇒ m₁l₁ = m₂l₂
⇒ 30 × 2 = 40 × l₂
⇒ l₂ = \(\frac{30 \times 2}{40}\)
= 1.5 m
(from fulcrum on the other side of see saw)

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

1. It is the turning effect of a force. The moment of a force is equal to the product of the magnitude of the force and the perpendicular distance of the line of action of force from the axis of rotation.
2. Clockwise.

(e) A ball is placed on a compressed spring. When the spring is released, the ball is observed to fly away.
(i) What form of energy does the compressed spring possess ?
(ii) Why does the ball fly away ? [2]

Answer.

1. Potential energy.
2. Because of change of potential energy of spring into the kinetic energy of the ball.

Question 2.
(a) (i) State the energy conversion taking place in a solar cell.
(ii) Give one disadvantage of using a solar cell. [2]
Answer:

1. Light energy converted into electrical energy.
2. The efficiency of conversion of solar energy to electrical energy is low.

(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 ms^-2] [2]
Answer:
Loss of potential energy = mgh
= 0.2 × 10 × (10-6)
= 0.2 × 10 × 4
= 8 Joule

(c) (i) Define the term refractive index of a medium in terms of velocity of light.
(ii) 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 reflected ray.

Answer:

1. Refractive index is defined the ratio of the speed of light in vacuum (or air) to the speed of light in the medium.
2. Ray 2.

(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]
Answer:
Look at the printed piece of paper through both the lenses one after the other, holding the lens close to the paper. The lens through which letters appear to be magnified is convex lens.

(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.
(i) How is the angle of emergence V related to the angle of incidence ?
(ii) What can you say about the value of the angle of deviation in such a situation ? [2]
Answer:

1. ∠e = ∠i
2. The deviation suffered by the incident ray is minimum.

Question 3.
(a) (i) What is meant by ‘Dispersion of light’ ?
(ii) In the atmosphere which colour of light gets scattered the least ? [2]
Answer:

1. The phenomenon of splitting of white light into its constituent colours is known as dispersion of light.
2. Red, because of its large wavelength.

(b) Which characteristic of sound will change if there is a change in
(i) its amplitude
(ii) waveform. [2]
Answer:

1. Loudness
2. Pitch of a note or quality of a musical sound.

(c) (i) Name one factor which affects the frequency of sound emitted due to vibrations in an air column.
(ii) Name the unit used for measuring the sound level. [2]
Answer:

1. Frequency is inversely proportional to the length of the air column.
2. Decibel (db).

(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 :
(i) kWh
(ii) joule
Answer:
Power = 1000 kVA
= 1000 kW
(as VA = W) (Power = VI)
V = 220 V
t = 2 hours
(i) Electric energy consumed
= 1000 kW x 2 h
= 2000 kWh

(ii) 2000 kWh = 3.6 × 10⁶ × 2000
= 7.2 × 10⁹ Joule

(e) Calculate the equivalent resistance between P and Q from the following diagram : [2]

Answer.
The equivalent circuit is

In series :
R_S = R₁ + R₂
= 10+ 10 = 20 Ω
In parallel:
\(\frac{1}{\mathrm{R}_{\mathrm{P}}}=\frac{1}{\mathrm{R}_{\mathrm{S}}}+\frac{1}{\mathrm{R}_2}\)
⇒ \(\frac{1}{R_P}=\frac{1}{20}+\frac{1}{5}\)
⇒ \(\frac{1}{\mathrm{R}_{\mathrm{P}}}=\frac{1+4}{20}\)
⇒ R_P = 4 Ω
Now, equivalent resistance = 3 + 4 + 2 = 9Ω

Question 4.
(a) (i) What is an a.c. generator or Dynamo used for ?
(ii) Name the principle on which it works.
Answer:

1. a.c. generator produces alternating voltage. It converts mechanical energy into electrical energy.
2. A generator works on principle of electromagnetic induction.

(b) Differentiate between heat capacity and specific heat capacity. [2]
Answer:

Heat capacity	Specific heat capacity
(i) It depends on the mass of the body.	(i) It does not depend on the mass of the body.
(ii) Its unit is JK-1. Heat capacity C’ = \(\frac{\mathrm{Q}}{\theta}\)	(ii) Its unit is Jkg-1 K1. S.H.C. or C = \(\frac{\mathrm{Q}}{m \theta}\)

(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]
Answer:
Heat energy given by hot solid of specific heat capacity (c)
= 60 × c × (100° – 25°)
= 60 × c × 75
= 4500 c
Heat energy taken by water
= 150 × 4.2 × (25 – 20)
= 150 × 4.2 × 5
(c of water = 4200 Jkg^-1 k^-1. So given c = 4.2 Jg^-1K^-1)
If there is no loss of heat energy, then heat energy given = Heat energy taken
⇒4500 c = 150 × 4.2 × 5
⇒ c = \(\frac{150 \times 4.2 \times 5}{4500}\)
= 0.7 Jg^-1 k^-1

(d) (i) What is the value of the speed of gamma radiations in air or vacuum ?
(ii) Name a material which exhibits fluorescence when cathode rays fall on it.
Answer:

1. 3 × 10⁸ ms^-1.
2. Screen is coated with fluorescent material such as zinc sulphide containing a trace of manganese.

(e) Give any two important sources of background radiation.
Answer:
Sources of background radiation are :

1. Naturally occurring radio isotopes.
2. Radiation produced in the emission from nuclear power plants.

SECTION – II (40 Marks)
(Attempt any four questions from this Section)

Question 5.
(a) (i) Which of the following remains constant in uniform circular motion : Speed or Velocity or both ?
(ii) Name the force required for uniform circular motion. State its direction ? [3]
Answer:

1. Constant speed in a circular path.
2. Centripetal force. The direction of the force is along the radius towards the centre.

(b) (i) 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.
(ii) Why is less effort needed to lift a load over an inclined plane as compared to lifting the load directly ? [3]
Answer:

1. 1. Class II lever. Here, load is between the effort and the fulcrum.
   2. Class III lever. Here, effort is between load and fulcrum.
2. Because inclined plane is a simple machine and acts as a force multiplier. Hence, less effort is needed.

(c) (i) A moving body weighing 400 N possesses 500 J of kinetic energy.
Calculate the velocity with which the body is moving, (g = 10 ms^-2)
(ii) Under what condition will a set of gears produce
1. a gain in speed
2. a gain in torque. [4]
Answer:
(i) We know
W = mg
∴ m = \(\frac{\text { Weight }}{g}=\frac{400}{10}\) = 40 kg
K.E. = \(\frac{1}{2}\) mv²
⇒ v² = \(\frac{2 \times \mathrm{K} . \mathrm{E}}{m}\)
⇒ v² = \(\frac{2 \times 500}{40}\)
∴ v = \(\sqrt{\frac{100}{2}}=\frac{10}{2}\) = 5 ms^-1

(ii)

1. The driving gear has more number of teeth than the driven gear.
2. For a gain in torque, the number of teeth in the driven gear should be more than the number of teeth in the driver gear.

Question 6.
(a) (i) What is meant by the term ‘critical angle’ ?
(ii) How is it related to the refractive index of the medium ?
(iii) Does the depth of a tank of water appear to change or remain the same when viewed normally from above ? [3]
Answer:

1. Critical angle is the angle of incidence in the denser medium corresponding to which the angle of refraction in the rarer medium is 90°.
2. The refractive index p and critical angle C are related as μ = \(\frac{1}{\sin C}\)
3. Remains the same.

(b) A ray of light PQ is incident normally on the hypotenuse of a right-angled prism ABC as shown in the diagram given alongside :

(i) Copy the diagram and complete the path of the ray PQ till it emerges from the prism.
(ii) What is the value of the angle of deviation of the ray ?
(iii) Name an instrument where this action of the prism is used. [3]
Answer:

1. The path of the ray is as shown in the fig.
   
2. 180°
3. Periscope

(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.
(i) Where is the object placed ?
(ii) Draw a ray diagram to illustrate the formation of the image obtained. [4]
Answer.

1. Beyond 2F₁
2. 

Question 7.
(a) (i) What is meant by Resonance?
(ii) State two ways in which Resonance differs from Forced vibrations. [3]
Answer:

1. Resonance is a special case of the forced vibrations. When the frequency of an externally applied periodic force on a body is equal to its natural frequency, the body readily begins to vibrate with an increased amplitude. This phenomenon is resonance.

2. Forced vibration	Resonance
   (i) The amplitude of vibration is usually small.	(i) The amplitude of vibration is very large.
   (ii) The vibrations of the body are not in phase with the external periodic force.	(ii) The vibrations of the body are in phase with the external periodic force.

(b) (i) A man standing between two cliffs produces a sound and hears two successive .echoes at intervals of 3 s and 4 s respectively. Calculate the distance between the two cliffs. The speed of sound in the air is 330 ms^-1.
(ii) Why will an echo not be hears when the distance between the source of sound and the reflecting surface is 10 m ? [3]
Answer:
(i) Let d₁ and d₂ is the distance of the two cliffs from the observer, then
3 = \(\frac{2 d_1}{v}=\frac{2 d_1}{330}\)
∴ d₁ = \(\frac{3 \times 330}{2}\)
= 495 m
Similarly, for the second cliff
4 = \(\frac{2 d_2}{330}\)
d₁ = \(\frac{4 \times 330}{2}\)
= 660 m
∴ Distance between the two cliffs
= 495 + 660 = 1155 m

(ii) No echo will be heard when the distance between source and reflector is 10 m because it is less than the minimum required distance of 17 m.

(c) The diagram alongside shows the displacement-time graph for a vibrating body.

(i) Name the type of vibrations produced by the vibrating body.
(ii) Give one example of a body producing such vibrations.
(iii) Why is the amplitude of the wave gradually decreasing ?
(iv) What will happen to the vibrations of the body after some time ?
Answer:

1. Damped vibrations.
2. When a slim branch of a tree is pulled and then released.
3. The amplitude of motion decreases due to the frictional force which the surrounding medium exerts on the body vibrating in it.
4. Body will stop vibrating.

Question 8.
(a) (i) A cell is sending current in an external circuit. How does the terminal voltage compare with the e.m.f. of the cell ?
(ii) What is the purpose of using a fuse in an electrical circuit ?
(iii) What are the characteristic properties of fuse wire ? [3]
Answer:

1. Terminal voltage < emf i.e., E > V in closed circuit, emf is equal to terminal voltage when no current is drawn.
2. Fuse is used to limit the current in an electric circuit. Its purpose is to provide safeguards to the circuit and the appliances connected in that circuit from being damaged.
3. Fuse wire is made up of a material of high resistivity and of low melting point so that it may easily melt due to overheating when current in excess of the prescribed limit passes through it.

(b) (i) Write an expression for the electrical energy spent in the flow of current through an electrical appliance in terms of I, R and t.
(ii) At what voltage is the alternating current supplied to our houses ?
(iii) How should the electric lamps in a building be connected ? [3]
Answer:

1. W = I² Rt.
2. 220 V.
3. In parallel.

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

Calculate:
(i) the equivalent resistance of the circuit.
(ii) total current in the circuit.
(iii) potential difference across the 7.2 Ω resistor.
Answer:
(i) In parallel.
\(\frac{1}{R_P}=\frac{1}{R_1}+\frac{1}{R_2}\)
⇒ \(\frac{1}{R_P}=\frac{1}{8}+\frac{1}{12}\)
⇒ \(\frac{1}{R_P}=\frac{3+2}{24}\)
⇒ R_P = \(\frac{24}{5}\) = 4.8 Ω
Equivalent resistance (R)
= 7.2+ 4.8
= 12 Ω
(ii) Current = \(\frac{\mathrm{V}}{\mathrm{R}}\)
= \(\frac{6}{12}\) = 0.5 A
(iii) Potential difference across the 7.2 Ω
= 7.2 × 0.5
= 3.6 V

Question 9.
(a) (i) Write an expression for the heat energy liberated by a hot body.
(ii) 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 ?
(iii) What happens to the average kinetic energy of the molecules as ice melts at 0°C ?
Answer:

1. Q = mass (m) × specific heat × fall of temperature.
2. 25 K
3. Average kinetic energy remains the same.

(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]
Answer:
Since temperature remains the same.

1. AB represents the change of phase when ice at 0°C changes to water at 0°C (although time increases from 0 to t₁)
2. BC represents the change in temperature from 0°C to 100°C when water increases its temperature and time reaches from t₁ to t₂.
3. CD represents the change of phase when water at 100°C changes to steam at 100°C (no change in temp.) but time changes from t₂ to t₄ as L of vaporization of steam is 2268 J/g.

(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 × 10³ J kg^-1] [4]
Answer:
According to principle of calorimetry
Heat lost by water = Heat gained by ice to melt + Heat gained by water to reach to 10°C.
Heat Lost by water = m × 4.2 × (60 – 10)
= 210 m
(where mass of water = m)
Heat gained by ice to melt,
mL= 40 × 336= 13440 J
Heat gained by water to raise its temperature to 10°C
= 40 × 4.2 × 10 = 1680 J
So 210 m = 13440+ 1680
(Given c of water = 4200 J kg/k = 4.2 J/g/k,
= 15120
L of ice = 336 × 10³ J/kg = 336 J/g)
m = \(\frac{15120}{210}\) = 72 g

Question 10.
(a) The diagram alongside shows 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.

(i) Copy the diagram and draw an arrow on the circular coil to show the direction of current flowing through it.
(ii) Draw the pattern of arrangement of the iron filings when current is passed through the loop. [3]
Answer:

(b) (i) Draw a simplified labelled diagram of a hot cathode ray tube.
(ii) Name a common device where a hot cathode ray tube is used. [3]
Answer:

1. 
2. A hot cathode ray is used as a picture tube in television and in a cathode ray oscilloscope.

(c) A certain nucleus X has a mass number 14 and atomic number 6. The nucleus X changes to ₇Y¹⁴ after the loss of a particle.
(i) Name the particle emitted.
(ii) Represent this change in the form of an equation.
(iii) A radioactive substance is oxidized. What change would you expect to take place in the nature of its radioactivity 1 Give a reason for you answer. [4]
Answer.

1. The emitted particle is an electron.
2. The required equation is
   6 X¹⁴ → ₇Y¹⁴ + \({ }_{-1}^0 e\)
3. Oxidation does not alter the radioactivity since the nucleus remains the same.