ICSE Class 10

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type


State the functions of the following in a nuclear reactor:-

i) Moderator ii) Control rods iii) Coolant 

i) Moderator – It shows down the speed of moving neutrons to allow the nuclear fission of uranium.

ii) Control rods – They absorb the extra neurons which are released during the nuclear fission reaction. They absorb that much neurons so that only one neuron is left to carry out controlled nuclear fission. They are made of boron and cadmium.

iii) Coolant – It absorbs the heat released during nuclear fission reaction. It is used to carry the energy of the fission reaction from the core of the reactor and take it to a place of utilization in steam generator. 


 Multiple Choice QuestionsLong Answer Type


The combination of a movable pulley P1 with a fixed pulley P2 is used for lifting up a load W.

i) State the function of the fixed pulley P2.

ii) If the free end of the string moves through a distance C, find the distance by which the load W is raised.

iii) Calculate the force to be applied at C to just raise the load W = 20 kgf, neglecting the weight of the pulley P1and friction. 


The fixed pulley P2 only changes the direction of the force. i.e., force can be applied downwards which is more convenient.


The load will be raised by a distance x/2


Distance moved by load dL= x/2

Distance moved by the effort, dE= x

Load (L) = 20 kgf

Effort = ?

Load x Load arm  = effort x  effort arm

20 space cross times space straight x over 2 space equals space straight x space cross times space straight E

rightwards double arrow space italic space italic space italic space italic space italic space straight E space equals space 10 space kgf


 Multiple Choice QuestionsShort Answer Type


i) Mention two important precautions that should be taken while handling radioactive materials.

ii) State any one use of radioisotopes. 


1. Storage of rad  ioactive substance 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.

ii) 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. 


 Multiple Choice QuestionsLong Answer Type


If there is no heat loss to the surroundings, the heat released by the condensation of m2 g of ice at 0oC into water at 0o C.

i) Find:

1. The heat lost by the steam in terms of m1.

2. the heat gained by ice in terms of m2.

ii) Form a heat equation and find the ratio of m2: m1 from it.


1. Heat lost by the steam of m1 = m1Lv + m1Cw x (100 – 0)

                                              = m1 x 2268 J + m1 x 4.2 (100 – 0)

                                              = 2688 m1

2. Heat gained by ice in terms of m2Lf = m2 x 336 J g-1

                                                  = 336 m2

ii) Now heat lost by the steam is equal to the heat gained by ice.

Therefore, we have

M1 x 2688 = m2 x 336


m subscript 2 over m subscript 1 space equals space 2688 over 336 

m2: m1 = 2688:336

That is, m2: m1 = 8: 1



A solid body weighs 2.10 N in air. Its relative density is 8.4. How much will the body weigh if placed:-

i) in water,

ii) in a liquid of relative density 1.2?

Weight in air = 2.10 N

R.D = 8.4 

i) RD = fraction numerator weight straight space of straight space body straight space in straight space air over denominator upthrust end fraction 


Upthrust = fraction numerator weight straight space of straight space body straight space in straight space air over denominator RD end fraction = Error converting from MathML to accessible text.

Weight of the body in water = 2.10 – 0.25 = 1.85 N


R.Dequals fraction numerator l o s s space o f space w e i g h t space i n space l i q u i d over denominator l o s s space o f space w e i g h t space i n space w space a t e r end fraction

Loss of weight in liquid = RD x loss of weight in water

                                 = 1.2 x 0.25

                                 = 1.85 N

Therefore, weight of the body in liquid = 2.1 – 0.3

                                                        = 1.8 N



i) Draw a labeled diagram of a hot cathode ray tube.

ii) Why are materials of low work function preferred as thermionic cathode materials?

iii) Write an equation to show the fission of a nucleus of U235 with the production of three neutrons. 

i) The labeled diagram of Cathode Ray Tube is as sown below: 


ii) It is because they require less energy for the electrons to be emitted from them.

iii) The fission of U235 is as represented below:

U presubscript 92 presuperscript 235 space plus space straight n presubscript 0 presuperscript 1 space rightwards arrow space Ba presubscript 56 presuperscript 141 space plus Kr presubscript 36 presuperscript 92 space plus space 3 straight n presubscript 0 presuperscript 1 space plus space Energy



In the figure given alongside, A, B and C are three ammeters. The ammeter B reads 0.5 A. All the ammeters have negligible resistance.



i) The readings in the ammeter A and C.

ii) The total resistance of the circuit.

i) Total resistance in B and C,

1 over r space equals space 1 over 6 plus space 1 third equals space fraction numerator 1 plus 2 over denominator 6 end fraction equals space 3 over 6


Rp =2 ohm

Resistance of B = 6 ohm

Current in B = 0.5 A

Volatge = ?

V = IR = 0.5 x 6 = 3 V

Current in C = ?

I = V/R = 3/3 = 1 A

Currnt in A = 1 + 0.5 = 1.5 A  

ii) Total resistance = 2 + 2 = 4 ohm



i) What is meant by refraction?

ii) Express the refractive index n of a medium,

  1. In terms of the velocity of light.
  2. In terms of the angle of incidence 'i' in air and the angle of refraction 'r' in a denser medium.
iii) If  a ray of light passes from medium 1 to medium II without any change of direction, what can be said about the refractive indices of these media (angle i is not 0)? 

i) Refraction is the phenomenon in which a ray of light bends from its path as it attempts to move from one medium into another. 

ii) 1. n = fraction numerator S p e e d space o f space l i g h t space i n space v a c u u m over denominator S p e e d space o f space l i g h t space i n space m e d i u m end fraction  

2. n = fraction numerator sin space function application i over denominator sin space function application r end fraction 
iii) It means that the refractive index of both the medium is the same.