Q. A cylinder of radius $r$ and thermal conductivity $K_{1}$ is surrounded by a cylindrical shell of linear radius $r$ and outer radius$2r$, whose thermal conductivity is $K_{2}.$ There is no loss of heat across cylindrical surfaces, when the ends of the combined system are maintained at temperatures $T_{1}\ and\ T_{2}$. The effective thermal conductivity of the system, in the steady state is
A) $\frac{K_{1}K_{2}}{K_{1} + K_{2}}$
B) $K_{1} + K_{2}$
C) $\frac{K_{1} + 3K_{2}}{4}$
D) $\frac{3K_{1} + K_{2}}{4}$
Explanation
(c) $\left( \frac{\mathrm{\Delta}Q}{\mathrm{\Delta}t} \right)_{inner} + \left( \frac{\mathrm{\Delta}Q}{\mathrm{\Delta}t} \right)_{outer} = \left( \frac{\mathrm{\Delta}Q}{\mathrm{\Delta}t} \right)_{total}$
$\frac{K_{1}\pi r^{2}(T_{2} - T_{1})}{l} + \ \frac{K_{2}\pi\left\lbrack (2r)^{2} - r^{2} \right\rbrack(T_{2} - T_{1})}{l} = \frac{K\pi(2r)^{2}\left( T_{2} - T_{1} \right)}{l}$
or $\left( K_{1} + 3K_{2} \right)\frac{\pi r^{2}\left( T_{2} - T_{1} \right)}{l}\ = \ \frac{K\pi\ 4\ r^{2}(T_{2} - T_{1})}{l}$
$\therefore\ K = \frac{K_{1} + 3K_{2}}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.