Q 11-10-024JEE MainAIEEE 2008Medium
Two walls of thicknesses $l_1$ and $l_2$ and thermal conductivities $K_1$ and $K_2$ are in contact. In the steady state, if the temperatures at the outer faces are $\theta_1$ and $\theta_2$, the temperature at the common wall is
Answer: (A) $\dfrac{K_1\theta_1l_2 + K_2\theta_2l_1}{K_1l_2 + K_2l_1}$
Same heat current through both walls:
$$\frac{K_1A(\theta_1 - \theta)}{l_1} = \frac{K_2A(\theta - \theta_2)}{l_2}$$
$$\theta = \frac{K_1\theta_1l_2 + K_2\theta_2l_1}{K_1l_2 + K_2l_1}$$
Solution by Sreeraj P, M.Sc Physics