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ICSE Class 10 Physics Solved Question Paper 2005

Short Answer Type

31.
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.

622 Views

Long Answer Type

32.

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.

i)

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

ii)

The load will be raised by a distance x/2

iii)

Distance moved by load d_L= x/2

Distance moved by the effort, d_E= x

Load (L) = 20 kgf

Effort = ?

Load x Load arm = effort x effort arm

$20 space cross times space straight x over 2 space equals space straight x space cross times space straight E rightwards double arrow space italic space italic space italic space italic space italic space straight E space equals space 10 space kgf$

586 Views

Short Answer Type

33.

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

ii) State any one use of radioisotopes.

i)

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.

361 Views

Long Answer Type

34.

If there is no heat loss to the surroundings, the heat released by the condensation of m₂g of ice at 0^oC into water at 0^o C.

i) Find:

1. The heat lost by the steam in terms of m_1.

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

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

i)

1. Heat lost by the steam of m₁ = m₁L_v + m₁C_w x (100 – 0)

= m₁ x 2268 J + m₁ x 4.2 (100 – 0)

= 2688 m₁

2. Heat gained by ice in terms of m₂L_f = m₂ x 336 J g^-1

= 336 m₂ii) Now heat lost by the steam is equal to the heat gained by ice.

Therefore, we have

M₁ x 2688 = m₂ x 336

Therefore,

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

m₂: m₁= 2688:336

That is, m₂: m₁ = 8: 1

318 Views

35.

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 = $fraction numerator weight straight space of straight space body straight space in straight space air over denominator upthrust end fraction$

Therefore,

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

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

ii)

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

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

780 Views

36.

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:

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

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37.

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.

Calculate:

i) The readings in the ammeter A and C.

ii) The total resistance of the circuit.

i) Total resistance in B and C,

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

Therefore,

R_p=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

782 Views

38.

i) What is meant by refraction?

ii) Express the refractive index n of a medium,

In terms of the velocity of light.
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.