A wall is made up of two layers A and B. The thickness of the two layers is the same, but materials are different. The thermal conductivity of A is double than that of B. In thermal equilibrium the temperature difference between the two ends is 36º C . Then the difference of temperature at the two surfaces of A will be

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6º C

12º C

18º C

24º C

answer is B.

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Detailed Solution

Suppose thickness of each wall is x then ${\left( {\frac{Q}{t}} \right)_{combination}} = {\left( {\frac{Q}{t}} \right)_A} \Rightarrow \frac{{{K_S}A({\theta _1} - {\theta _2})}}{{2x}} = \frac{{2KA({\theta _1} - \theta )}}{x}$
$\because {K_S} = \frac{{2 \times 2K \times K}}{{(2K + K)}} = \frac{4}{3}K\,and\,({\theta _1} - {\theta _2}) = 36^\circ$
$\Rightarrow \frac{{\frac{4}{3}KA \times 36}}{{2x}} = \frac{{2KA({\theta _1} - \theta )}}{x}$
Hence temperature difference across wall A is $({\theta _1} - \theta ) = {12^o}C$
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