ICSE 2017 Physics Question Paper Solved for Class 10

Solving ICSE Class 10 Physics Previous Year Question Papers ICSE Class 10 Physics Question Paper 2017 is the best way to boost your preparation for the board exams.

ICSE Class 10 Physics Question Paper 2017 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) A brass ball is hanging from a stiff cotton thread. Draw a neat labelled diagram showing the forces acting on the brass ball and the cotton thread. (2)
Answer:
The diagram is as shown.

(b) The distance between two bodies is doubled. How is the magnitude of gravitational force between them affected? (2)
Answer:
It becomes one-fourth.
This is because F = \(\frac{G m_1 m_2}{r^2}\)

(c) Why is a jack screw provided with a long arm? (2)
Answer:
As a large mechanical advantage is required by the jack screw used for cutting metal, hence its handle should be long and the blades short. (MA = effort arm / load arm)

(d) If the power of a motor be 100 kW, at what speed can it raise a load of 50,000 N? (2)
Answer:
P = F ν, therefore, ν = P/F = 100000 / 50000 = 2 m s^-1.

(e) Which class of level will always have MA > I and why? (2)
Answer:
Class II lever, because the effort arm is always greater than the load arm.

Question 2.
(a) Define heat capacity and state its SI unit. (2)
Answer:
It is defined as the amount of heat required in raising the temperature of a unit mass of a substance through 1°C. Its SI unit is J kg^-1 K^-1.

(b) Why is the base of a cooking pan generally made thick? (2)
Answer:
It is because thicker pans can have better heat distribution and heat retention due to thermal conductivity and thermal mass.

(c) A solid of mass 50 g at 150°C is placed in 100 g of water at 11°C, when the final temperature recorded is 20°C. Find the specific heat capacity of the solid.
(Specific heat capacity of water = 4.2 J/g°C)
Answer:
Given mass of solid m_s = 50 × 10^-3 kg
Mass of water m_w = 100 × 10^-3 kg
Temperature of solid = 1500°C
Temperature of water = 110°C
Final temperature = 200°C
Specific heat of water = 4.2 J kg^-1 °C^-1
Now,heat lost by the solid = heat gained by water
50 × 10^-3 × c × (1500 – 200)
= 100 x 10^-3 x 4200 x (200 – 110)
Or c = 581 Jkg^-1 °C^-1

(d) How is the refractive index of a material related to :
(i) real and apparent depth?
(ii) velocity of light in vacuum or air and the velocity of light in a given medium? (2)
Answer:
(i) μ = \(\frac{\text { Real depth }}{\text { Apparent depth }}\) and
(ii) μ = \(\frac{c}{v}\)

(e) State the conditions required for total internal reflection of light to take place. (2)
Answer:
(i) The incident ray should travel from the denser to the rarer medium.
(ii) The angle of incidence in the denser medium should be greater than the critical angle for the given pair of media.

Question 3.
(a) Draw a ray diagram to show the refraction of a monochromatic ray through a prism when it suffers minimum deviation. (2)
Answer:
The ray diagram is as shown :

(b) The human ear can detect continuous sounds in the frequency range from 20 Hz to 20,000 Hz.
Assuming that the speed of sound in air is 330 ms^-1 for all frequencies, calculate the wavelengths corresponding to the given extreme frequencies of the audible range. (2)
Answer:
The corresponding wavelengths are
λ_lowest = \(\frac{\mathrm{V}}{v}=\frac{330}{20000}\) = 0.0165 m and
λ_highest = \(\frac{\mathrm{V}}{v}=\frac{330}{20}\) = 16.5 m.

(c) An enemy plane is at a distance of 300 km from a radar. In how much time will the radar be able to detect the plane? Take velocity of radiowaves as 3 × 10⁸ ms^-1. (2)
Answer:
Given S = 300 km = 300000 m,
c = 3 × 10⁸ m s^-1, t = ?
Using \(\frac{2 \mathrm{~S}}{v}=\frac{2 \times 300000}{3 \times 10^8}\) = 2 × 10^-3 s

(d) How is the frequency of a stretched string related to :
(i) its length?
(ii) its tension? (2)
Answer:
(i) Inversely proportional to its length and
(ii) Directly proportional to the square root of tension.

(e) Define specific resistance and state its SI unit. (2)
Answer:
It is the resistance of unit cube of the material. Its SI unit is ohm-m.

Question 4.
(a) An electric bulb of resistance 500 Q draws a current of 0.4 A. Calculate the power of the bulb and the potential difference at its end. (2)
Answer:
Given R = 500 W, I = 0.4 A, V = ?, Using V = IR,
we have V = 0.4 × 500 = 200 V

(b) State two causes of energy loss in a transformer. (2)
Answer:
Copper loss and iron loss.

Should have high melting point and
Should have low value of work function.

(d) State two factors upon which the rate of emission of thermions depends. (2)
Answer:

The nature of the metal and
Temperature of the metal surface.

(e) When does the nucleus of an atom tend to be radioactive?
Answer:
The nucleus tends to be radioactive when it becomes unstable.

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

Question 5.
(a) A uniform half-metre rule balances horizontally on a knife edge at the 29 cm mark when a weight of 20 gf is suspended from one end.
(i) Draw a diagram of the arrangement.
(ii) What is the weight of the half-metre rule? (3)
Answer:
Let the weight of the half-meter scale be W.

To balance the half-metre scale at 29 cm marie, the 20 gf weight should be suspended from the 50 cm end as shown.
Moments due to W in anticlockwise direction
= W gf × 4 cm
Moments due to 20 gf in clockwise direction
= 20 gf × 21 cm
By the principle of moments we have
W gf × 4 = 20 gf × 21
Or W = \(\frac{20 \times 21}{4}\) = 105 gf

(b) (i) A boy uses a single fixed pulley to lift a load of 50 kgf to some height. Another boy uses a single movable pulley to lift the same load to the same height. Compare the efforts applied by them. Give a reason to support your answer.
(ii) How does uniform circular motion differ from uniform linear motion?
(iii)Name the process used for producing electricity using nuclear energy. (3)
Answer:
(i) A single fixed pulley simply changes the direction of the action of the force. Therefore, for this pulley E = L = T = 50 kgf. Thus, it has a mechanical advantage 1.

For a single movable pulley, we have L = T + T = 2T and E = T = L/2 = 50/2 = 25 kgf, therefore, its mechanical advantage is 2.

(ii) In uniform circular motion, velocity is not constant, whereas in uniform linear motion velocity is constant. (iii) Nuclear fission

(c) A pulley system with VR = 4 is used to lift a load of 175 kgf through a vertical height of 15 m. The effort required is 50 kgf in the downward direction (g = 10 N kg^-1). (4)
(i) Distance moved by the effort.
(ii) Work done by the effort.
(iii) MA of the pulley system.
(iv) Efficiency of the pulley system.
Answer:
Given VR = 4, L = 175 kgf, distance moved by load = d_L= 15 m, E = 50 kgf, d_E = ?, W_E = ?, MA = ?, h = ?
Using the formula VR
= \(\frac{\text { distance moved by effort }}{\text { distance moved by load }}=\frac{d_{\mathrm{E}}}{d_{\mathrm{L}}}\) ,
therefore, we have
d_E= VR x d_L= 4 × 15 = 60 m
W_E = Effort × d_E= 50 × 10 × 60 = 30000 J
Now, MA = \(\frac{\text { load }}{\text { effort }}=\frac{175}{50}\) = 3.5
Now, η = \(\frac{\mathrm{MA}}{\mathrm{VR}}=\frac{3.5}{4}\) = 0.875 = 87.5%

Question 6.
(a) (i) How is the transference of heat energy by radiation prevented in a calorimeter?
(ii) You have a choice of three metals, A, B and C, of specific heat capacities 900 Jkg^-1C^-1, 380 Jkg^-1C^-1 and 460 Jkg^-1C^-1 respectively, to make a calorimeter. Which material will you select? Justify your answer? (3)
Answer:
(i) This is done by polishing the inner and outer surface of the copper calorimeter and the space between the copper vessel and the insulating container is filled with some poor conductor like wood wool or glass wool.

(ii) A, as it will gain less energy in a given time.

(b) Calculate the mass of ice needed to cool 150 g of water contained in a calorimeter of mass 50 g at 30°C such that the final temperature is 5°C. (3)
Specific heat capacity of calorimeter 0.4 J/g°C
Specific heat capacity of water 4.2 J/g°C
Latent heat capacity of ice 330 J/g
Answer:
Given m_w = 150 g, m_i = ?, m_C = 50 g, T_i = 0 °C, T_w = 32 °C, T_f= 5 °C
Specific heat capacity of calorimeter = 0.4 J g^-1 °C^-1
Specific heat capacity of water = 4.2 J g^-1 °C^-1
Latent heat capacity of ice 330 J g^-1
Heat lost by 150 g water + 50 g of calorimeter in falling form 32 °C to 5 °C
= 150 × 4.2× 27 + 50 × 0.4 × 27 = 17550 J
Heat needed to melt ice = m_iL = m_i × 330
= 330 m_i
So 330m_i = 17550 or m_i= 17550/330 = 53.2 g

(c) (i) Name the radiations which are absorbed by greenhouse gases in the earth’s atmosphere.
(ii) A radiation X is focused by a particular device on the bulb of a thermometer and mercury in the thermometer shows a rapid increase. Name the radiation X.
(iii)Name two factors on which the heat energy liberated by a body depends.
Answer:
(i) Infrared
(ii) Infrared
(iii) Temperature and surface area.

Question 7.
(a) A lens forms an upright and diminished image of an object when the object is placed at the focal point of the given lens.
(i) Name the lens.
(ii) Draw a ray diagram to show the image formation. (3)
Answer:
(i) Concave lens.
(ii) The diagram is shown.

(b) A ray of light travels from water to air as shown in the diagram given below:

(i) Copy the diagram and complete the path of the ray. Given that the critical angle for water is 48°.
(ii) State the condition so that total internal reflection occurs in the above diagram. (3)
Answer:
(i) The redrawn diagram is as shown :

(ii) The angle of incidence should be greater than the critical angle.

(c) The diagram below shows a point source P inside a water container. Four rays A, B, C, D starting from the source P are shown up to the water surface.

(i) Show in the diagram the path of these rays after striking the water surface. The critical angle for water-air surface is 48°.
(ii) Name the phenomenon which the rays B and D exhibit. (4)
Answer:
(i) The diagram is as shown.

(ii) Ray B— Refraction, Ray D —Total internal reflection.

Question 8.
(a) Name the factor that determines : (3)
(i) Loudness of the sound heard.
(ii) Quality of the note.
(iii) Pitch of the note.
Answer:
(i) Intensity/Amplitude.
(ii) Frequency of source and relative motion between the source of sound and the listener.
(iii) The number and the frequency of the overtones present in a given sound.

(b) (i) What are damped vibrations? (3)
(ii) Give one example of damped vibrations.
(iii) Name the phenomenon that causes a loud sound when the stem of a vibrating tuning fork is kept pressed on the surface of a table.
Answer:
(i) Those vibrations which die out with time.
(ii) Vibrations of a simple pendulum.
(iii) Resonance

(c) (i) A wire of length 80 cm has a frequency of 256 Hz. Calculate that length of a similar wire ‘ under similar tension, which will have a frequency of 1024 Hz. (4)
(ii) A certain sound has a frequency of 256 hertz and a wavelength of 1.3 m.
(1) Calculate the speed with which this sound travels.
(2) What difference would be felt by a listener between the above sound and another sound travelling at the same speed, but of wavelength 2.6 m?
Answer:
(i) Given L₁ = 80 cm, v₁ = 256 Hz, L₂ = ?, v₂ = 1024 Hz
We know that for a vibrating string
Lv = constant. Therefore, we have
L₂ = \(\frac{\mathrm{L}_1 v_1}{v_2}=\frac{80 \times 256}{1024}\) = 20 m

(ii)

v = 256 Hz, λ = 1.3 m, V = ?
Using V = vλ, we have
V = 256 × 1.3 = 332.8 ms^-1
He will hear a dull sound as the frequency of the sound wave of wavelength 2.6 m will be half the frequency (256/2 = 128 Hz) of the sound of wavelength 1.3 m.

Question 9.
(a) (i) Name the colour code of the wire which has connected to the metallic body of an appliance,
(ii) Draw the diagram of a dual control switch when the appliance is switched ‘ON’. (3)
Answer:
(i) Green or yellow.

(ii)

(b) (i) Which particles are responsible for current in conductors?
(ii) To which wire of a cable in a power circuit should the metal case of a geyser be connected?
(iii) To which wire should the fuse be connected? (3)
Answer:
(i) Electrons
(ii) Earth wire
(iii) Live wire

(c) (i) Explain the meaning of the statement ‘current rating of a fuse is 5A’.
(ii) In the transmission of power the voltage of power generated at the generating stations is stepped up from 11 kV to 132 kV before it is transmitted. Why? (4)
Answer:
(i) It means that it can withstand a maximum current of 5 A before it “blows”.
(ii) Due to this the current flowing in the circuit decreases. As a result heat losses I²R decreases.

Question 10.
(a) Answer the. following questions based on a hot cathode ray tube. (3)
(i) Name the charged particles.
(ii) State the approximate voltage used to heat the filament.
(iii) What will happen to the beam when it passes through the electric field?
Answer:
(i) Electrons.
(ii) 5 to 6 V.
(iii) It is deflected by the electric field.

(b) State three factors on which the rate of emission of electrons from a metal surface depends. (3)
Answer:

Nature of the metallic surface
Temperature of the metal surface and
Surface area of the metal surface.

(c) (i) What are the free electrons?
(ii) Why do they not leave the metal surface on their own?
(iii) How can they be made to leave the metal surface? (State any two ways) (4)
Answer:
(i) These are the electrons which are not bound to the nucleus and are free to move in the metal.
(ii) Because it finds itself being pulled by the positive charge left behind as it becomes.
(iii) They can be made to leave the metal surface by

Thermionic emission (heating the metal surface)
Photoelectric emission (By shining a radiation of sufficient frequency on the metal surface).