A temperature difference of 120,^oC is mainta | vedclass

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(A) Let $R_0$ be the thermal resistance of a rod of length $L$. The resistance of segment $AP$ (length $L/2$) is $R_0/2$,segment $PQ$ (length $L$) is

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(A) Let $R_0$ be the thermal resistance of a rod of length $L$. The resistance of segment $AP$ (length $L/2$) is $R_0/2$,segment $PQ$ (length $L$) is $R_0$,and segment $QB$ (length $L/2$) is $R_0/2$. The bent rod $PQ$ has a total length of $3L/2$ (two vertical segments of $L/4$ and one horizontal segment of $L$). Its resistance is $R_{bent} = R_0/4 + R_0 + R_0/4 = 1.5R_0$. The segment $PQ$ of the main rod has resistance $R_0$. These two are in parallel,so their equivalent resistance $R_p = \frac{R_0 	imes 1.5R_0}{R_0 + 1.5R_0} = \frac{1.5}{2.5}R_0 = 0.6R_0$. The total resistance of the circuit is $R_{eq} = R_{AP} + R_p + R_{QB} = 0.5R_0 + 0.6R_0 + 0.5R_0 = 1.6R_0$. The temperature drop across $PQ$ is $\Delta T_{PQ} = \Delta T_{total} 	imes \frac{R_p}{R_{eq}} = 120 	imes \frac{0.6R_0}{1.6R_0} = 120 	imes \frac{6}{16} = 120 	imes 0.375 = 45\,^oC$.

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