Q. A partition wall has two layers $A$ and $B$ in contanct, each made of a different material. They have the same thickness but the thermal conductivity of layer $A$ is twice that of layer $B$. If the steady state temperature difference across the wall is $60K$, then the corresponding difference across the layer $A$ is
A) $10\ K$
B) $20\ K$
C) $30\ K$
D) $40\ K$
Explanation
(b) Suppose conductivity of layer $B$ is $K$, then it is $2K$ for layer $A$. Also conductivity of combination layers $A$ and $B$ is $K_{S} = \frac{2 \times 2K \times K}{(2K + K)} = \frac{4}{3}K$
Hence $\left( \frac{Q}{t} \right)_{Combination} = \left( \frac{Q}{t} \right)_{A}$
$\Rightarrow \frac{4}{3}\frac{KA \times 60}{2x} = \frac{2K.A \times (\mathrm{\Delta}\theta)_{A}}{x} \Rightarrow (\mathrm{\Delta}\theta)_{A} = 20K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.