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:

Thermal resistance formula: $$R_{\text{th}} = \frac{l}{kA} \propto l$$

Let resistance per unit length be $$r$$.

Segment $$AP$$ length $$= \frac{L}{2} \implies R_{AP} = \frac{1}{2}rL = 0.5R$$

Segment $$PQ$$ length $$= L \implies R_{PQ} = rL = R$$

Segment $$QB$$ length $$= 2L - \left(\frac{L}{2} + L\right) = \frac{L}{2} \implies R_{QB} = \frac{1}{2}rL = 0.5R$$

Bent rod $$PQ$$ length $$= \frac{3L}{2} \implies R_{\text{bent}} = \frac{3}{2}rL = 1.5R$$

Equivalent resistance between $$P$$ and $$Q$$:

$$R_{PQ, \text{eq}} = \frac{R_{PQ} \times R_{\text{bent}}}{R_{PQ} + R_{\text{bent}}} = \frac{R \times 1.5R}{R + 1.5R} = \frac{1.5R^2}{2.5R} = 0.6R$$

Total thermal resistance of the network:

$$R_{\text{total}} = R_{AP} + R_{PQ, \text{eq}} + R_{QB} = 0.5R + 0.6R + 0.5R = 1.6R$$

Temperature difference across $$P$$ and $$Q$$:

$$\frac{\Delta T_{PQ}}{\Delta T_{\text{total}}} = \frac{R_{PQ, \text{eq}}}{R_{\text{total}}}$$

$$\Delta T_{PQ} = 120 \times \frac{0.6R}{1.6R} = 120 \times \frac{3}{8} = 45^\circ\text{C}$$