Q. If two metallic plates of equal thicknesses and thermal conductivities $K_{1}$ and $K_{2}$ are put together face to face and a common plate is constructed, then the equivalent thermal conductivity of this plate will be
A) $\frac{K_{1}K_{2}}{K_{1} + K_{2}}$
B) $\frac{2K_{1}K_{2}}{K_{1} + K_{2}}$
C) $\frac{\left( K_{1}^{2} + K_{2}^{2} \right)^{3/2}}{K_{1}K_{2}}$
D) $\frac{\left( K_{1}^{2} + K_{2}^{2} \right)^{3/2}}{2K_{1}K_{2}}$
Explanation
(b) In series, $R_{eq} = R_{1} + R_{2} \Rightarrow \frac{2l}{K_{eq}A} = \frac{l}{K_{1}A} + \frac{l}{K_{2}A}$
$\Rightarrow \frac{2}{K_{eq}} = \frac{1}{K_{1}} + \frac{1}{K_{2}} \Rightarrow K_{eq} = \frac{2K_{1}K_{2}}{K_{1} + K_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.