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Thermal Properties of Matter MCQs for NEET — Physics Questions with Answers

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To keep constant time, watches are fitted with balance wheel made of -

Invar

Stainless steel

Tungsten

Platinum

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Explanation

(a)Because dimension of invar does not varies with temperature.

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A metal bar of length L and area of cross-section A is clamped between two rigid supports. For the material of the rod, its Young’s modulus is Y and coefficient of linear expansion is $α$ . If the temperature of the rod is increased by $∆ t ° C$ , the force exerted by the rod on the supports is

$Y A L ∆ t$

$Y A α ∆ t$

$\frac{Y L α ∆ t}{A}$

$Y α A L ∆ t$

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Explanation

(b)

$∆ L = L α ∆ T Y = \frac{σ}{ε} = \frac{\frac{F}{A}}{\frac{∆ L}{L}} F = Y A \frac{∆ L}{L} = Y A \frac{L α ∆ T}{L} = Y A α ∆ T$

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The coefficient of linear expansion of brass and steel are $α_{1}$ and $α_{2}$ . If we take a brass rod of length $l_{1}$ and steel rod of length $l_{2}$ at 0°C, their difference in length $(l_{2} - l_{1})$ will remain the same at a temperature if

$α_{1} l_{2} = α_{2} l_{1}$

$α_{1} l_{2}^{2} = α_{2} l_{1}^{2}$

$α_{1}^{2} l_{1} = α_{2}^{2} l_{2}$

$α_{1} l_{1} = α_{2} l_{2}$

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Explanation

(d)

$L_{2} = l_{2} (1 + α_{2} ∆ θ) a n d L_{1} = l_{1} (1 + α_{1} ∆ θ) \Rightarrow (L_{2} - L_{1}) = (l_{2} - l_{1}) + ∆ θ (l_{2} α_{2} - l_{1} α_{1}) N o w (L_{2} - L_{1}) = (l_{2} - l_{1}) s o, l_{2} α_{2} - l_{1} α_{1} = 0$

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Under steady state, the temperature of a body

Increases with time

Decreases with time

Does not change with time and is same at all the points of the body

Does not change with time but is different at different points of the body

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Explanation

(d) In steady state there is no absorption of heat in any position. Therefore, in steady state the temperature of the body does not change with time but can be different at different points of the body.

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The coefficient of thermal conductivity depends upon

Temperature difference of two surfaces

Area of the plate

Thickness of the plate

Material of the plate

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Explanation

(d) It is the property of material.

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The ratio of thermal conductivity of two rods of different material is 5 : 4. The two rods of same area of cross-section and same thermal resistance will have the lengths in the ratio

4 : 5

9 : 1

1 : 9

5 : 4

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Explanation

(d) Given $A_{1} = A_{2}$ and $\frac{K_{1}}{K_{2}} = \frac{5}{4}$

$∴ R_{1} = R_{2} \Rightarrow \frac{l_{1}}{K_{1} A} = \frac{l_{2}}{K_{2} A} \Rightarrow \frac{l_{1}}{l_{2}} = \frac{K_{1}}{K_{2}} = \frac{5}{4}$

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In variable state, the rate of flow of heat is controlled by

Density of material

Specific heat

Thermal conductivity

All the above factors

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Explanation

(d) In variable state $\frac{Q}{t} \propto K$ and $\frac{Q}{t} \propto \frac{1}{ρc} \Rightarrow \frac{Q}{t} \propto \frac{K}{ρc}$

(K = thermal conductivity, $ρ$ = density, c = specific heat)

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The dimensions of thermal resistance are

$M^{- 1} L^{- 2} T^{3} K$

${ML}^{2} T^{- 2} K^{- 1}$

${ML}^{2} T^{- 3} K$

${ML}^{2} T^{- 2} K^{- 2}$

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Explanation

(a) Thermal resistance

$= \frac{l}{KA} = [\frac{L}{{MLT}^{- 3} K^{- 1} \times L^{2}}] = [M^{- 1} L^{- 2} T^{3} K]$

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Two walls of thicknesses $d_{1}$ and $d_{2}$ and thermal conductivities $k_{1}$ and $k_{2}$ are in contact. In the steady state, if the temperatures at the outer surfaces are $T_{1}$ and $T_{2}$ , the temperature at the common wall is -

$\frac{k_{1} T_{1} d_{2} + k_{2} T_{2} d_{1}}{k_{1} d_{2} + k_{2} d_{1}}$

$\frac{k_{1} T_{1} + k_{2} d_{2}}{d_{1} + d_{2}}$

$(\frac{k_{1} d_{1} + k_{2} d_{2}}{T_{1} + T_{2}}) T_{1} T_{2}$

$\frac{k_{1} d_{1} T_{1} + k_{2} d_{2} T_{2}}{k_{1} d_{1} + k_{2} d_{2}}$

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A slab consists of two parallel layers of copper and brass of the same thickness and having thermal conductivities in the ratio 1 : 4. If the free face of brass is at 100°C and that of copper at 0°C, the temperature of interface is

80°C

20°C

60°C

40°C

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Explanation

(a) Temperature of interface

$θ = \frac{k_{1} θ_{1} + k_{2} θ_{2}}{k_{1} + k_{2}} (∵ \frac{k_{1}}{k_{2}} = \frac{1}{4} \Rightarrow If k_{1} = k then k_{2} = 4 k) \Rightarrow θ = \frac{k \times 0 + 4 k \times 100}{5 k} = 80 ° C$

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