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 :

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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$

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