Q. Two rods of same length and cross section are joined along the length. Thermal conductivities of first and second rod are $K_{1}$ and $K_{2}$. The temperature of the free ends of the first and second rods are maintained at $\theta_{1}$ and $\theta_{2}$ respectively. The temperature of the common junction is
A) $\frac{\theta_{1} + \theta_{2}}{2}$
B) $\frac{K_{2}K_{1}}{K_{1} + K_{2}}(\theta_{1} + \theta_{2})$
C) $\frac{{K_{1}\theta}_{1} + K_{2}\theta_{2}}{K_{1} + K_{2}}$
D) $\frac{{K_{2}\theta}_{1} + K_{1}\theta_{2}}{K_{1} + K_{2}}$
Explanation
(c) At steady state, rate of heat flow for both blocks will be same,
$i.e.,\ \frac{K_{1}A(\theta_{1} - \theta)}{l_{1}} = \frac{K_{2}A(\theta - \theta_{2})}{l_{2}}\lbrack Given\ l_{1} - l_{2}\rbrack$
$\Rightarrow K_{1}A\left( \theta_{1} - \theta \right) = K_{2}A\left( \theta - \theta_{2} \right) \Rightarrow \theta = \frac{K_{1}\theta_{1} + K_{2}\theta_{2}}{K_{1} + K_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.