Two materials having coefficients of thermal conductivity $3K$ and $K$ and thickness $d$ and $3d$ respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are $\theta_2$ and $\theta_1$ respectively, ($\theta_2 > \theta_1$). The temperature at the interface is
Answer: (D) $\dfrac{\theta_1}{10} + \dfrac{9\theta_2}{10}$
Thermal resistances (per unit area): $R_1 = \dfrac{d}{3K}$ and $R_2 = \dfrac{3d}{K} = 9R_1$.
The same heat current flows through both, so the temperature drop divides in the ratio $1:9$. Across the first layer the drop is $\frac{1}{10}(\theta_2 - \theta_1)$:
$$\theta = \theta_2 - \frac{\theta_2 - \theta_1}{10} = \frac{\theta_1}{10} + \frac{9\theta_2}{10}$$
Solution by Sreeraj P, M.Sc Physics