Two walls of thickness $d_{1}$ and $d_{2}$ , thermal conductivities $K_{1}$ and $K_{2}$ are in contact. In the steady state, if the temperature at the outer surfaces are $T_{1}$ and $T_{2}$ , the temperature at the contact of the walls will be

\frac{K_{1} T_{1} + K_{2} T_{2}}{d_{1} + d_{2}}

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\frac{K_{1} T_{1} d_{2} + K_{2} T_{2} d_{1}}{K_{1} d_{2} + K_{2} d_{1}}

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\frac{(K_{1} d_{1} + K_{2} d_{2}) T_{1} T_{2}}{T_{1} + T_{2}}

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\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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Solution

The correct option is B $\frac{K_{1} T_{1} d_{2} + K_{2} T_{2} d_{1}}{K_{1} d_{2} + K_{2} d_{1}}$

Heat flow across both the walls will be same.
$- \frac{K_{1} A}{d_{1}} (T - T_{1}) = - \frac{K_{2} A}{d_{2}} (T_{2} - T)$

$\Rightarrow K_{1} T d_{2} - K_{1} T_{1} d_{2} = K_{2} T_{2} d_{1} - K_{2} T d_{1}$

$\Rightarrow T = \frac{K_{1} T_{1} d_{2} + K_{2} T_{2} d_{1}}{K_{1} d_{2} + K_{2} d_{1}}$

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