# ICSE Class 10 For Physics Question Paper Solved Set G

Students should refer to ICSE Class 10 Physics Question Paper solved Set G given below which will help them to prepare for the upcoming ICSE Physics exams. Students should read ICSE Physics Class 10 Books to make sure they are completely prepared and should also refer to ICSE Class 10 Physics Solutions to understand all questions and their answers.

ICSE Class 10 Physics Question Paper solved Set G

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 [ ].

ICSE Class 10 Physics Question Paper solved Set G
SECTION – I (40 Marks)
(Attempt all Questions)

Question 1
(a) Give any two effects of a force on a non-rigid body.
(b) One end of a spring is kept fixed while the other end is stretched by a force as shown in the diagram.

(i) Copy the diagram and mark on it the direction of the restoring force.
(ii) Name one instrument which works on the above principle. (2)
(c) (i) Where is the centre of gravity of a uniform ring situated ?
(ii) ‘The position of the centre of gravity of a body remains unchanged even when the body is deformed.’ State whether the statement is true or false. (2)
(d) A force is applied on a body of mass 20 kg moving with a velocity of 40 ms–1. The body attains a velocity of 50 ms–1 in 2 seconds. Calculate the work done by the body. (2)
(e) A type of single pulley is very often used as a machine even though it does not give any gain in mechanical advantage.
(i) Name the type of pulley used.
(ii) For what purpose is such a pulley used ? (2)

(a) (i) A force can bring about change in the state of rest or uniform motion of a body.
(ii) A force can change the shape of dimensions of a body.

(c) (i) The centre of gravity of a uniform ring is its centre. (ii) False
(d) Acceleration of body (a) = v-u / t = (50-40)ms-1 / 2s = 5 ms-2
Distance covered by body while accelerating,

Force possessed by body, F = m.a = 20 (kg) × 5 ms–2 = 100 N
∴ Work done by the body, W = F × S = 100 N × 90 m = 9000 J
(e) (i) Single fixed pulley. (ii) It helps in changing the direction of applying efforts.

Question 2.
(a) (i) In what way does an ‘Ideal machine’ differ from a ‘Practical machine’ ?
(ii) Can a simple machine act as a force multiplier and a speed multiplier at the same time?(2)
(b) A girl of mass 35 kg climbs up from the first floor of a building at a height 4 m above the ground to the third floor at a height 12 m above the ground. What will be the increase in her gravitational potential energy ? (g = 10 ms–2).
(c) Which class of lever found in the human body is being used by a boy –
(i) when he holds a load on the palm of his hand.
(ii) when he raises the weight of his body on his toes ?
(d) A ray of light is moving from a rarer medium to a denser medium and strikes a plane mirror placed at 90o to the direction of the ray as shown in the diagram.

(i) Copy the diagram and mark arrows to show the path of the ray of light after it is reflected from the mirror.
(ii) Name the principle you have used to mark the arrows to show the direction of the ray.(2)
(e) (i) The refractive index of glass with respect to air is 1.5. What is the value of the refractive index of air with respect to glass ?
(ii) A ray of light is incident as a normal ray on the surface of separation of two different mediums. What is the value of the angle of incidence in this case ? (2)

(a) (i) An ideal machine is the one whose parts are frictionless as well as weightless, such that its mechanical advantage is equal to its velocity ratio.
A practical machine is the one whose parts are neither frictionless nor weightless. Furthermore, its mechanical advantage is always less than its velocity ratio.
(ii) No, it can be either speed multiplier or force multiplier.
(b) Mass of girl (m) = 35 kg
Height gained by girl (h) = (12 – 4) = 8 m
∴ Increase in gravitational potential
energy = mgh = 35 (kg) × 10 ms–2 × 8 m
= 2800 J
(c) (i) Lever of third order.
(ii) Lever of second order
(d) (i)

(ii) The principle is the law of reversibility of light

(ii) Angle of incidence is zero.

Question 3
(a) A bucket kept under a running tap is getting filled with water. A person sitting at a distance is able to get an idea when the bucket is about to be filled.
(i) What change takes place in the sound to give this idea ?
(ii) What causes the change in the sound ? (2)
(b) A sound made on the surface of a lake takes 3 s to reach a boatman.
How much time will it take to reach a diver inside the water at the same depth ?
[Velocity of sound in air = 330 ms–1 ; Velocity of sound in water = 1450 ms–1] (2)
(c) Calculate the equivalent resistance between the points A and B for the following combination of resistors : (2)

(d) You have been provided with a solenoid AB.

(i) What is the polarity at end A ?
(ii) Give one advantage of an electromagnet over a permanent magnet. (2)
(e) (i) Name the device used to protect the electric circuits from overloading and short circuits.
(ii) On what effect of electricity does the above device work ? (2)

(a) (i) The sharp pitched sound slowly changes to low pitched sound as the bucket gets filled. The sound almost dies when the bucket is completely filled.
(ii) As the length of vibrating air column decreases due to the water, the frequency of the sound changes.
(b) Distance covered by the sound to reach boatman = 330 ms–1 × 3 s = 990 m
∴ Distance of diver from the source of sound = 990 m
∴ Time taken by the sound to reach diver = 990m/1450ms-1 = 990/145 s = 0.68 s (Appox.).
(c) Resistance of three 4Ω resistors in series = 4 × 3 = 12Ω
Resistance of three 2Ω resistors in series = 2 × 3 = 6Ω
∴ Equivalent resistance of 12Ω , 6Ω and 4Ω in parallel.

∴ Equivalent resistance of 5Ω , 2Ω and 6Ω in series.
Rs = (5 + 2 + 6)Ω = 13Ω
(d) (i) Polarity at the end A is NORTH.
(ii) 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.
(e) (i) Electric fuse.
(ii) It works on the heating effect of electric current.

Question 4
(a) Define the term ‘Heat capacity’ and state its S.I. unit. (2)
(b) What is meant by Global warming ? (2)
(c) How much heat energy is released when 5 g of water at 20oC changes to ice at 0oC?
[Specific heat capacity of water = 4.2 Jg–1 oC–1 ; Specific latent heat of fusion of ice = 336 g–1] (2)
(i) can cause severe genetical disorders. (ii) are deflected by an electric field ?
(e) A radioactive nucleus undergoes a series of decays according to the sequence

If the mass number and atomic number of X3 are 172 and 69 respectively, what is the mass number and atomic number of X ?

(a) Heat capacity : The amount of heat energy required to raise the temperature of a given mass of a substance through 1 K (or 1oC) is called its heat capacity.
S.I. unit of heat capacity is JK–1.
(b) The rise in average temperature of the atmosphere around the earth, due to the trapping of radiant solar heat due to carbon dioxide and chloroflurocarbons is called global warming.
(c) Heat energy released in cooling water to 0oC = mcθf = 5 × 4.2 × 20 = 420J
Heat energy released in freezing water = mL = 5 × 336 = 1680 J
∴ Total heat energy released = (1680 + 420) J = 2100 J

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

ICSE Class 10 Physics Question Paper solved Set G
SECTION II (40 Marks)
Attempt any four questions from this Section.

Question 5.
(a) (i) With reference to their direction of action, how does a centripetal force differ from a centrifugal force ?
(ii) State the Principle of conservation of energy.
(iii) Name the form of energy which a body may possess even when it is not in motion. (3)
(b) A coolie is pushing a box weighing 1500 N up an inclined plane 7.5 m long on to a platform, 2.5 m above the ground.
(i) Calculate the mechanical advantage of the inclined plane.
(ii) Calculate the effort applied by the coolie.
(iii) In actual practice, the coolie needs to apply more effort than what is calculated. Give one reason why you think the coolie needs to apply more effort. (3)
(c) A block and tackle system of pulley’s a velocity ratio 4.
(i) Draw a labelled diagram of the system indicating clearly the points of application and directions of a load and effort.
(ii) What is the value of the mechanical advantage of the given pulley system if it is an ideal pulley system ? (4)

(a) (i) With reference to action, the centripetal force is directed towards the centre of circular path and the centrifugal force is directed away from the centre of circular path, such that they are acting opposite to one another.
(ii) Law of conservation of energy : Energy in a system cannot be created, nor it can be destroyed and the sum total of energy remains constant, no matter it can change its form.
(iii) Potential energy.
(b) (i) For a perfect machnine ; Mechanical advantage = Velocity ratio

(iii) Inclined plane is not a perfect machine. It offers certain amount of resistance to the load. Thus, in actual practice coolie has to apply more effort.
E = T
∴ V.R. = 4
Block and tackle system of pulley having V.R. = 4
(ii) Mechanical advantage will be equal to the number of pulleys engaged in a given pulley system.

Question 6
(i) that are used for photography at night. (ii) used for detection of fracture in bones.
(ii) whose wavelength range is from 100 Å to 4000 Å (or 10 nm to 400 nm). (3)
(b) (i) Can the absolute refractive index of a medium be less than one ?
(ii) A coin placed at the bottom of a beaker appears to be raised by 4.0 cm.
If the refractive index of water is 4/3, find the depth of the water in the beaker.
(3) (c) An object AB is placed between 2F1 and F1 on the principal axis of a convex lens as shown in the diagram.

Copy the diagram and using three rays starting from point A, obtain the image of the object formed by the lens. (4)

(b) (i) No, absolute refractive index of a medium is always greater than 1, as speed of light in any medium is always less than that in vacuum.
(ii) Let the real depth of water = x
Apparent depth of water = (x – 4) cm.
Now, refractive index of water = Real depth / Apparent depth ⇒ 4/3 = x/x-4 or (4x – 16) = 3x
∴ Real depth = x = 16 cm

Question 7.
(a) (i) What is the principle on which SONAR is based.
(ii) An observer stands at a certain distance away from a cliff and produces a loud sound. He hears the echo of the sound after 1.8 s. Calculate the distance between the cliff and the observer if the velocity of sound in air is 340 ms–1.
(b) A vibrating tuning fork is placed over the mouth of a burette filled with water. The tap of the burette is opened and the water level gradually starts falling. It is found that the sound from the tuning fork becomes very loud for a particular length of the water column.
(i) Name the phenomenon taking place when this happens.
(ii) Why does the sound become very loud for this length of the water column ? (3)
(c) (i) What is meant by the terms (1) amplitude (2) frequency of a wave ?
(ii) Explain why stringed musical instruments, like the guitar, are provided with a hollow box. (4)

(a) (i) Echo depth sounding
Ultrasonic waves have the same speed as of audible sound but are not absorbed in the medium. So transmitter sends these waves receiver receives the waves back after striking the rigid obstacle so time taken is recorded. And we can calculate distance d = vt/2
(ii) Distance between cliff and source of sound = Velocity of sound x time / 2 = 340ms–1 x 1.8s / 2 =306m
(b) (i) The phenomenon is called ‘resonance of sound’.
(ii) A some particular length of air column the natural frequency of air column corresponds the frequency of tuning fork. At this moment the sound waves reinforce to produce loud sound.
(c) (i) The maximum displacement of vibrating particle about its mean position is called its amplitude.The number of waves which pass through a point in a medium in one second is called frequency.
(ii) The air trapped in the hollow box starts vibrating with forced vibrations, thereby producing lounder sound.

Question 8.
(a) (i) It is observed that the temperature of the surrounding starts falling when the ice in a frozen lake starts melting. Give a reason for the observation.
(ii) How is the heat capacity of the body related to its specific heat capacity ? (3)
(b) (i) Why does a bottle of soft drink cool faster when surrounded by ice cubes than by ice cold water, both at 0o C ?
(ii) A certain amount of heat Q will warm 1 g of material X by 3oC and 1 g of material Y by 4oC. Which material has a higher specific heat capacity.
(c) A calorimeter of mass 50 g and specific heat capacity 0.42 J g–1 oC–1 contains some mass of water at 20oC. A metal piece of mass 20 g at 100 oC is dropped into the calorimeter. After stirring, the final temperature of the mixture is found to be 22oC. Find the mass of water used in the calorimeter.
[specific heat capacity of the metal piece = 0.3 Jg–1 oC–1 ; specific heat capacity of water = 4.2 Jg–1 oC–1] (4)

(a) (i) Every kilogram of ice at 0oC on melting to form water at 0oC needs 336 × 103J of heat energy as its specific latent heat is 336 × 103J. This heat energy is supplied by the surrounding of the lake, which in turn results in the fall in temperature.
(ii) Specific heat capacity of a body =  Heat capacity of the body / Mass of the body
(b) (i) Every gram of ice surrounding the soft drink extracts out 336 J of heat energy from it and the temperature of surrounding the soft drink remains at 0oC. However, in case of cold water, it will extract out only 4.2 J of heat energy per gram. Furthermore, the temperature of surrounding water starts rising. Thus, soft-drink bottle cools better in case of ice.
(ii) Material X has higher specific heat capacity compared to material Y.
It is because for the mass amount of heat its temperature rises less than Y.
(c) Data

Heat absorbed by cold water = mc θR = x × 4.2 × 2 = 8.4x
Heat absorbed by calorimeter = mc θR = 50 × 0.42 × 2 = 42
Total heat absorbed = 8.4x + 42.
Heat given out by metal mc θF = 20 × 0.3 × 78 = 468.
By the principle of calorimetry, Total heat absorbed = total heat lost
8.4x + 42 = 468 ⇒ 8.4x = 468 – 42 = 426 ⇒ x = 426/8.4 = 50.7 g

Question 9.
(a) (i) State Ohm’s law.
(ii) A metal wire of resistance 6 Ω is stretched so that its length is increased to twice its original length. Calculate its new resistance. (3)
(b) (i) An electrical gadget can give an electric shock to its user under certain circumstances.
Mention any two of these circumstances.
(ii) What preventive measure provided in a gadget can protect a person from an electric shock ?
(c) The figure shows a circuit

When the circuit is switched on, the ammeter reads 0.5 A.
(i) Calculate the value of the unknown resistor R.
(ii) Calculate the charge passing through the 3 Ω resistor in 120s.
(iii) Calculate the power dissipated in the 3 Ω resistor. (4)

(a) (i) Ohm’s Law : It states, all physical conditions of a conductor remaining same, the current flowing through it is directly proportional to the potential difference at its ends.
(ii) Let the original length be (l) and area of cross-section (a), such that its resistance is 6 Ω .
Applying, R = K 1/α ⇒ 6 = Kl/a …(i)
When the length 2l, its area of cross-section becomes a/2. If R is the new resistance of conductor then :
R = K 2l / a/2 =  Kl/a ….(ii)
Dividing (ii) by (i) R/6 = 4    ∴ R = 24Ω
(b) (i) (1) The electrical gadget may be short circuited i.e., its live or neutral wire is touching its metallic body directly.
(2) The hands of the user may be wet, such that water dripping from his hands makes contact with the live wire.
(ii) The body of the electric gadget is connected to the earth terminal by means of earth wire.
In case of short circuit a huge surge of current flows through the earth terminal. This in turn melts fuse in the live wire and hence the flow or current stops in the gadget.

(ii) Charge Q = I × t = 0.5 × 120 = 60 Coulombs
(iii) Power dissipated in 3Ω resistor = I2R = 0.5 × 0.5 × 3 = 0.75 W

Question 10.
(a) Name the three main parts of a Cathode Ray Tube.
(b) (i) What is meant by Radioactivity ? (3)
(ii) What is meant by nuclear waste ? (3)
(iii) Suggest one effective way for the safe disposal of nuclear waste.
(c) (i) Draw a simple labelled diagram of d.c. electric motor.
(ii) What is the function of the split rings in a d.c. motor ?
(iii) State one advantage of a.c. over d.c. (4)