Previous Year Paper year:2019 : ICSE-X-Physics

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ICSE-X-Physics

Previous Year Paper year:2019

with Solutions - page 7

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#8-c-i
the resistance of the circuit when the key k is crpen.
Ans : R_S = 5 + 0.5 = 5.5 Ω

#8-c-ii
the current drawn from the cell when the key k is open,
Ans : I=``\frac { V }{ R }``=``\frac { 3.3 }{ 5.5 }``= 0.6A

#8-c-iii
the resistance of the circuit when the key k is closed.
Ans : R_P = ``\frac{5 \times 5}{5+5}`` == 2.5 Ω, thus total resistance is R_T = 2.5 +0.5 = 3.0

#8-c-iv
the current drawn from the cell when the key k is closed.
Ans : I= ``\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{T}}}`` = ``\frac { 3.3 }{ 3.0 }``

= 1.1 A

#9

#9-a

#9-a-i
Define Calorimetry.
Ans : Calorimetry is the measurement of heat,

#9-a-ii
Name the material used for making a Calorimeter.
Ans : Copper.

#9-a-iii
Why is a Calorimeter made-up of thin sheets of the above material answered in (ii) ?

[3]

Ans : Copper has small specific heat capacity. The thin sheets ensure that the box has small heat capacity.

#9-b
The melting point of naphthalene is 80°C and the room temperature is 30°C. A sample of

liquid naphthalene at 100°C is cooled down to the room temperature. Draw a temperature¬time graph to represent this cooling. In the graph, mark the region which corresponds to the freezing process.

[3]

Ans : The graph is as shown :

#9-c
104 g of water at 30°C is taken in a calorimeter made of copper of mass 42 g. When a certain mass of ice at 0°C is added to it, the final steady temperature of the mixture after the ice has melted, was found to be 10°C. Find the mass of ice added.

[Specific heat capacity of water = 4.2 J g^-1 °C^-1 : Specific latent heat of fusion of ice - 336 J g^-1 ; Specific heat capacity of copper = 0.4 J g^-1 °C^-1 ]

[4]

Ans : Mass of water m₁ = 104 g

Temperature of water T₁ = 30°C

Final temperature T₂ = 10°C

Mass of calorimeter m₂ = 42 g

Temperature of ice T₃= 0°C

Let the mass of ice added be = x g

Total heat energy lost = ``(104 \times 4.2 \times 20)+(42 \times 0.4 \times 20) = 8736 + 336 = 9072 J``

Total heat energy gained = ``(336 \times x)+(x \times 4.2 \times 10) = 336x + 42x = 378x``

By principle of calorimetry,

Heat lost = Heat gained

378x =9072

x = ``\frac { 9072 }{ 378 }`` = 24g

#10

#10-a
Draw a neat labelled diagram of an AC generator.
Ans : The diagram is as shown:

#10-b

#10-b-i
Define nuclear fission.
Ans : It is the splitting of a heavy nudeus into two or more smaller nudei with the

release of tremendous energy.