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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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Two thin blankets keep more hotness than one blanket of thickness equal to these two. The reason is

Their surface area increases

A layer of air is formed between these two blankets, which is bad conductor

These have more wool

They absorb more heat from outside

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Explanation

(b) A layer of air, which is a bad conductor of heat, is formed between these two blankets.

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Ice formed over lakes has

Very high thermal conductivity and helps in further ice formation

Very low conductivity and retards further formation of ice

It permits quick convection and retards further formation of ice

It is very good radiator

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Explanation

(b) Ice has law thermal conductivity, it prevents further loss of heat from water to surroundings.

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