Q.
Temperature difference of $120^{\circ}C$ is maintained between two ends of a uniform rod $AB$ of length $2L$. Another bent rod $PQ$, of same cross section as $AB$ and length $\frac{3L}{2}$ , is connected across $AB$ (See figure). In steady state, temperature difference between $P$ and $Q$ will be close to :
Solution:
$\frac{\Delta T}{R_{eq}} = I = \frac{\left(120\right)5}{8R} = \frac{120\times5}{8R} $
$ \Delta T_{PQ} = \frac{120 \times5}{8R} \times\frac{3}{5}R = \frac{360}{8} = 45^{\circ}C$