Q.
One end of a thermally insulated rod is kept at a temperature $T_{1}$ and other at $T_{2}$. The rod is composed of two sections of lengths $l_{1}$ and $l_{2}$ and thermal conductivities $K_{1}$ and $K_{2}$ respectively. The temperature at the interface of the two sections is
A) ${(K}_{2}l_{2}T_{1} + K_{1}l_{1}T_{2}) \slash (K_{1}l_{1} + K_{2}l_{2})$
B) ${(K}_{2}l_{1}T_{1} + K_{1}l_{2}T_{2}) \slash (K_{2}l_{1} + K_{1}l_{2})$
C) ${(K}_{1}l_{2}T_{1} + K_{2}l_{1}T_{2}) \slash (K_{1}l_{2} + K_{2}l_{1})$
D) ${(K}_{1}l_{1}T_{1} + K_{2}l_{2}T_{2}) \slash (K_{1}l_{1} + K_{2}l_{2})$
Explanation
(c) Let temperature at the interface is T.
For part AB,
[image:11_image69.emf]
$\frac{Q_{1}}{t} \propto \frac{\left( T_{1} - T \right)K_{1}}{l_{1}}$
For part $BC,$
$\frac{Q_{2}}{t} \propto \frac{\left( T - T_{2} \right)K_{2}}{l_{2}}$
At equilibrium, $\frac{Q_{1}}{t} = \frac{Q_{2}}{t}$
$\therefore$ $\frac{\left( T_{1} - T \right)K_{1}}{l_{1}} = \frac{\left( T - T_{2} \right)K_{2}}{l_{2}}$
$\Rightarrow$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T = \frac{T_{1}K_{1}l_{2} + T_{2}K_{2}l_{1}}{K_{1}l_{2} + K_{2}l_{1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.