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) $\dfrac{K_1\theta_1l_2 + K_2\theta_2l_1}{K_1l_2 + K_2l_1}$
Concept. Walls in series carry the same heat current: $\dfrac{K_1A(\theta_1-\theta)}{l_1}=\dfrac{K_2A(\theta-\theta_2)}{l_2}$.
Step 1. Solve. $K_1l_2(\theta_1-\theta)=K_2l_1(\theta-\theta_2)\Rightarrow\theta=\dfrac{K_1l_2\theta_1+K_2l_1\theta_2}{K_1l_2+K_2l_1}$.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
