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A wall has two layers A and B, each made of a different material. Both the layers have the same thickness. The thermal conductivity of the material of A is twice that of B ; if under steady state the temperature difference across the wall is $36^{°} C$ , the temperature difference across the layer A is

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$12^{°} C$

$6^{°} C$

$18^{°} C$

$24^{°} C$

answer is B.

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Detailed Solution

$\frac{Q}{t} = \frac{K A Δ θ}{l} = const.$

$Δ θ \propto \frac{1}{K} \Rightarrow \frac{Δ θ_{A}}{Δ θ_{B}} = \frac{K_{B}}{K_{A}} = \frac{K}{2 K}$

$\frac{Δ θ_{A}}{Δ θ_{B}} = \frac{1}{2}$

$Δ θ_{A} = \frac{1}{3} Δ θ = \frac{1}{3} \times 36$

$= 12^{\circ} C$

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