ICSE 2014 Physics Question Paper Solved for Class 10

Solving ICSE Class 10 Physics Previous Year Question Papers ICSE Class 10 Physics Question Paper 2014 is the best way to boost your preparation for the board exams.

ICSE Class 10 Physics Question Paper 2014 Solved

Time: 2 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) A force is applied on (i) a non-rigid body and (ii) a rigid body. How does the effect of the force differ in the above two cases? [2]
Answer:
A force when applied on a non-rigid body can change the dimensions of the body as well as can bring motion in the body while force applied on rigid body cannot change the dimensions as it cannot change the interspacing between the constituent particles but can cause motion in the body.

(b) A metallic ball is hanging by a string from a fixed support. Draw a neat labelled diagram showing the forces acting on the ball and the string. [2]
Answer:

(c) (i) What is the weight of a body placed at the centre of the earth?
(ii) What is the principle of an ideal machine? [2]
Answer:
(i) Weight of a body placed at the centre of earth will be zero.
(ii) Work done on machine = Work done by machine
Or Work input = Work output is the principle of an ideal machine where there is no dissipation of energy and efficiency of machine is 100%.

(d) Is it possible to have an accelerated motion with a constant speed? Explain. [2]
Answer:
Yes, when a body is moving in a circular motion, its direction keeps on changing but speed remains same. The continuous change in direction of motion means that the velocity is not uniform or body moving with variable velocity, so motion is called accelerated.

(e) (i) When does a force do work?
(ii) What is the work done by the moon when it revolves around the earth? [2]
Answer:
(i) When a force is able to cause displacement in the body work is done. But when force and displacement make an angle of 90° with each other, then no work is done.

(ii) Work done is zero when moon revolves around the earth, the force is the centripetal force and displacement at all instants is along the tangent to the circular path i. e., normal to the direction of force.

Question 2.
(a) Calculate the change in the Kinetic energy of a moving body if its velocity is reduced to 1/3rd of the initial velocity. [2]
Answer:
Let a body of mass ‘m’ is moving with ‘u’ m/s. So, initial K.E. = \(\frac{1}{2}\)mu². If velocity is reduced to \(\frac{1}{3}\) rd. Final K.E. = \(\frac{1}{2}\)m \(\left(\frac{u}{3}\right)^2=\frac{1}{2} \frac{m u^2}{9}\)

So K.E. becomes \(\frac{1}{9}\) th of its initial K.E.

(b) State the energy changes in the following devices while in use :
(i) A loud speaker.
(ii) A glowing electric bulb. [2]
Answer:
(i) A loud speaker converts electrical energy into sound energy.
(ii) A glowing electric bulb converts electrical energy into light energy and heat energy.

(c) (i) What is nuclear energy?
(ii) Name the process used for producing electricity using nuclear energy. [2]
Answer:
(i) Nuclear energy is the energy released by nuclear fission and nuclear fusion processes in which mass of products is less than mass of reactants, so loss of mass gets converted into energy by Einstein formula E = Δmc², where E is energy released, Δm is loss in mass and c is speed of light which is 3 ×10⁸ ms^-1.

(ii) Nuclear fission in a nuclear power plant.

(d) State one important advantage and disadvantage each of using nuclear energy for producing electricity. [2]
Answer:
A very small amount of nuclear fuel can produce a tremendous amount of energy. But its disadvantage is that along with energy it produces highly energetic and penetrating harmful nuclear radiations which are very harmful to human body and also causes environmental pollution.

(e) (i) The conversion of part of the energy into an undesirable form is called
(ii) For a given height h, the length 1 of the inclined plane, lesser will be the effort required. [2]
Answer:
(i) Dissipation of energy.
(ii) More.

Question 3.
(a) Draw the diagram given below and clearly show the path taken by the emergent ray. [2]

Answer:

(b) (i) What is consumed using different electrical appliances, for which electricity bills are paid?
(ii) Name a common device that uses electromagnets.
Answer:
(i) Energy.
(ii) Electric bell, motor of fan, grinder etc.

(c) (i) A ray of light passes from water to air. How does the speed of light change?
(ii) Which colour of light travels fastest in any medium except air?
Answer:
(i) Speed of light increases when it goes from water to air.
(ii) Red colour.

(d) Name the factors affecting the critical angle for the pair of media.
Answer:
Critical angle for a given pair of media depends on the refractive index of that pair and it is affected by

colour or wavelength of light – critical angle is least for violet light and most for red light
temperature – critical angle increases with increase in temperature.

(e) (i) Name a prism required for obtaining a spectrum of Ultraviolet light.
(ii) Name the radiations which can be detected by a thermopile.
Answer:
(i) Quartz prism.
(ii) Infrared radiation.

Question 4.
(a) Why is the colour red used as a sign of danger? [2]
Answer:
Red colour has longest wavelength so scatters least and can penetrate to a longer distance. Thus, can be seen from large distance.

(b) (i) What are mechanical waves?
(ii) Name one property of waves that do no change when the wave passes from one medium to another. [2]
Answer:
(i) Mechanical waves require a material medium to travel. These waves travel in the medium through the vibrations of the medium particles about their mean positions.

(ii) Frequency of wave remains same.

(c) Find the equivalent resistance between points A and B. [2]

Answer:
Resistance 3 Ω, 3 Ω, and 3 Ω are parallel to each other (in CD).

So, \(\frac{1}{\mathrm{R}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{3}{3}\) ; R = 1 Ω
So, resistance from A to E = 1 + 5 = 6
Resistance across EB
\(\frac{1}{\mathrm{R}}=\frac{1}{4}+\frac{1}{6}=\frac{3+2}{12}=\frac{5}{12}\) ;
R = \(\frac{12}{5}\) = 2.4 Ω
∴ Total resistance across AB = 6 + 2.4 = 8.4 Ω

(d) 50 g of metal piece at 27 °C requires 2400 J of heat energy so as to attain a temperature of 327 °C. Calculate the specific heat capacity of the metal. [2]
Answer:
Mass of metal = 50 g
t₁ = 27°C
t₂ = 327°C
∴ Δt = 327 – 27 = 300°C
Q = 2400 J (Heat energy)
C = ?
Q = mCΔt
C = \(\frac{\mathrm{Q}}{m \Delta t}=\frac{2400}{50 \times 300}\)
= 0.16 J/g/°C

(e) An electron emitter must have work function and melting point. [2]
Answer:
Low work function and high melting point.

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

Question 5.
(a) (i) A man having a box on his head, climbs up a slope and another man having an identical box walks the same distance on a levelled road. Who does more work against the force of gravity and why?
(ii) Two forces each of 5N act vertically upwards and downwards respectively on the two ends of a uniform metre rule which is placed at its mid-point as shown in the diagram. Determine the magnitude of the resultant moment of these forces about the midpoint. [4]

Answer:
(i) Work done by a man having a box on his head and climbing a rope is more as moving on the level road does no work (f × d cos 90° = 0).

(ii) Magnitude of resultant moment of forces about the mid-point
= 5 N × \(\frac{50}{100}\) m + 5 N × \(\frac{50}{100}\) m
= 2.5 Nm + 2.5 Nm
= 5 Nm (anticlockwise)

(b) (i) A body is thrown vertically upwards. Its velocity keeps on decreasing. What happens to its kinetic energy as its velocity becomes zero?
(ii) Draw a diagram to show how a single pulley can be used so as to have its ideal M.A. = 2. [3]
Answer:
(i) Its kinetic energy keeps on getting converted into potential energy when the body is thrown vertically upwards as it gains height.

(ii) L = T + T = 2T
E = T
∴ M.A. = \(\frac{2 \mathrm{~T}}{\mathrm{~T}}\) = 2
So it acts as a force multiplier.

(c) Derive a relationship between mechanical advantages, velocity ratio and efficiency of a machine. [3]
Answer:
M.A., V.R. and η
Efficiency η = \(\frac{\text { Work output }}{\text { Work input }}\)
Work output = Load × Load Arm
Work input = Effort × Effort Arm
η = \(\frac{L \times \text { L.A. }}{E \times \text { E.A }}\), But \(\frac{\text { E.A. }}{\text { L.A. }}\) = V.R. and \(\frac{L}{E}\) = M.A.
η = M.A. × \(\frac{1}{\text { V.R. }}\) or M.A. = V.R. × η

Question 6.
(a) (i) Light passes through a rectangular glass slab and through a triangular glass prism. In what way does the direction of the two emergent beams differ and why?
(ii) Ranbir claims to have obtained an image twice the size of the object with a concave lens. Is he correct? Give a reason for your Answer. [4]
Answer:
(i)

Light when passes through a rectangular glass slab suffers refraction at two surfaces Q and R as it goes from rarer to denser and from denser to rarer. The same bending takes place at Q and R of glass prism also. But ray of light while passing through a glass prism always deviates towards its base i.e., towards surface BC only which is not in case of glass slab. In each refraction through glass prism the ray always bends towards the bases of prism.

(ii) No, a concave lens can always form a virtual, erect, diminished image as it has a virtual focus.

(b) A lens forms an erect, magnified and virtual image of an object.
(i) Name the lens.
(ii) Draw a labelled ray diagram to show the image formation. [3]
Answer:
(i) Convex lens.
(ii) When object is between O and/of a convex lens. A’

(c) (i) Define the power of a lens.
(ii) The lens mentioned in 6(b) above is of focal length 25 cm. Calculate the power of the lens. [3]
Answer:
(i) Power of a lens is a measure of deviation produced by it in the path of rays refracted through it and is measured in Dioptres.
P (in D) = \(\frac{1}{f(m)}\) or \(\frac{100}{f(cm)}\)

(ii) Focal length of convex lens is taken positive.
f = +25 cm
P (in D) = 100/25 = +4 D

Question 7.
(a) The adjacent diagram shows three different modes of vibrations P, Q and R of the same string.
(i) Which vibrations will produce a louder sound and why?
(ii) The sound of which string will have maximum shrillness?
(iii) State the ratio of wavelengths of P and R. [4]

Answer:
(i) R, because its amplitude is more.
(ii) P, as frequency is more.
(iii) Ratio of wavelengths of P and R is 1 : 3.

(b) A type of electromagnetic wave has wavelength 50 A.
(i) Name the wave.
(ii) What is the speed of the wave in vacuum?
(iii) State one use of this type of wave. [3]
Answer:
(i) X – rays.
(ii) 3 × 10⁸ ms-¹.
(iii) They can penetrate through human flesh but are stopped by bones, hence used for detection of fracture and for diagnostic purpose.

(c) (i) State one important property of waves used for echo depth sounding.
(ii) A radar sends a signal to an aircraft at a distance of 30 km away and receives it back after 2 × 10^-4 second. What is the speed of the signal? [4]
Answer:
(i) They (ultrasonic waves) can travel undeviated through long distances and can be confined to a narrow beam. They are not absorbed easily.

(ii) Distance = 30 km = 30000 m
Time = 2 × 10^-4 s (for echo to be heard)
Speed = 2 d/t
= \(\) = 30000 × 10⁴
= 3 × 10⁸ m/s

Question 8.
(a) Two resistors of 4 Ω and 6 Ω are connected in parallel to a cell to draw 0.5 A current from the cell.
(i) Draw a labelled circuit diagram showing the above arrangement.
(ii) Calculate the current in each resistor. [3]
Answer:
(i)

(ii) Resistance in parallel
\(\frac{1}{4}+\frac{1}{6}=\frac{1}{R}\) ; \(\frac{3+2}{12}=\frac{5}{12}=\frac{1}{R}\) ; R = 2.4 Ω
So, PD in parallel = Total I × R (only in parallel)
= 0.5 × 2.4 = 1.2 V
I₁ = \(\frac{3+2}{12}=\frac{5}{12}=\frac{1}{R}\) = 0.3 A ;
I₂ = \(\frac{3+2}{12}=\frac{5}{12}=\frac{1}{R}\) = 0.2 A

(b) (i) What is an Ohmic resistor?
(ii) Two copper wires are of the same length, but one is thicker than the other.
(1) Which wire will have more resistance?
(2) Which wire will have more specific resistance? [3]
Answer:
(i) Ohmic resistor obeys Ohm’s law i.e., PD is directly proportional to current keeping resistance constant in that conductor i.e., at constant temperature, e.g., all metallic conductors are Ohmic.

(ii)

Resistance is inversely proportional to thickness, so thick wire will have less resistance and thin wire more resistance when lengths are same.
Specific resistance does not depend on length and area of wire so it will be same for both as only depends on material and temperature of wire.

(c) (i) Two sets A and B, of the three bulbs each, are glowing in two separate rooms. When
one of the bulbs in set A is fused, the other two bulbs, alos cease to glow. But in set B, when one bulb fuses, the other two bulbs continue to glow. Explain why this phenomenon occurs.
(ii) Why do we prefer arrangement of Set B for house circuiting? [3]
Answer:
(i) In Set A, the bulbs are arranged in series while in Set B, the bulbs are arranged in parallel. In parallel, even if one bulb fuses other paths are working.

(ii)

In parallel, all appliances work independently at same voltage so glow of bulb (or any other appliance) is not affected.
In parallel, resistance so current does not decrease, so current does not decrease.
All appliances work independently and do not depend on other so each has a separate switch

Question 9.
(a) Heat energy is supplied at a constant rate to lOOg of ice at 0 °C. The ice is converted into
water at 0 °C in 2 minutes. How much time will be required to raise the temperature of water from 0 °C to 20 °C? [Given : sp. heat capacity of water = 4.2 J g^-1 °C^-1, sp. latent heat of ice = 336 J g^-1]. [4]
Answer:
Mass of ice = 100 g
L of ice = 336 J/g
mL (of ice) = P × time
100 × 336 = P × 2 × 60
P = 280 W
mCQ (of water) = P × time
As same appliance with same power is used
100 × 4.2 × 20 = P × time
8400 = 280 × time
time = 30 s
So, time required to raise the temperature of water from 0°C to 20°C is 30 s.

(b) Specific heat capacity of substance A is 3.8 J g^-1 K^-1 whereas the Specific heat capacity of substance B is 0.4 J g^-1 K^-1.
(i) Which of the two is a good conductor of heat?
(ii) How is one led to the above conclusion?
(iii) If substances A and B are liquids then which one would be more useful in car radiators? [3]
Answer:
(i) B is a good conductor of heat.

(ii) Specific heat capacity is the amount of heat energy required to raise the temperature of unit mass of a substance by 1 K. So the one which uses less amount of heat energy is a good conductor.

(iii) A would be useful in car radiators as it can extract large amount of heat energy without raising its own temperature much.

(c) (i) State any two measures to minimize the impact of global warming.
(ii) What is the Greenhouse effect? [3]
Answer:
(i)

Cutting of trees must be banned and plantation should be increased.
Burning of fossil fuels should be for limited purposes only. We should find and use some alternative sources of energy.

(ii) Trapping of infrared radiations which causes global heating is known as greenhouse effect. It can be done by mainly carbon dioxide and methane gases.

Question 10.
(a) (i) Name two factors on which the magnitude of an induced e.m.f. in the secondary coil depends.
(ii) In the following diagram an arrow shows the motion of the coil towards 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.

Answer:
(i) Change in magnetic flux : More the change in magnetic flux in a given time, more is the e.m.f. induced.
Time in which the magnetic flux changes: More rapid the flux changes, more is the e.m.f. induced.

(ii)

Induced current is always opposite to the cause which produces it, so current will flow from B to A.
Lenz’s law and Faraday’s laws.

(b) A nucleus ₁₁Na²⁴ emits a beta particle to change into Magnesium (Mg)
(i) Write the symbolic equation for the process.
(ii) What are numbers 24 and 11 called?
(iii) What is the general name of ₁₂²⁴ Mg with respect to ₁₁²⁴Na? [3]
Answer:
(i) \({ }_{11}^{24} \mathrm{Na}\) → \({ }_{11}^{24} \mathrm{Mg}\) + \({ }_{-1}^0 e\) + (where, \({ }_{-1}^0 e\) is β)

(ii) 24 is mass number, where 11 is atomic number.

(iii) Isobars.

(c) In a cathode ray tube state :
(i) The purpose of covering cathode by thorium and carbon.
(ii) The purpose of the fluorescent screen.
(iii) How is it possible to increase the rate of emission of electrons. [3]
Answer:
(i) As a good electron emitter because thorium and carbon lower work function.

(ii) Electrical signal applied on the deflecting plates change into visual patterns on the screen.

(iii) By increasing temperature and surface area of metal and by using metal with a low work function – rate of emission of electrons can be increased.