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(C) Let the thermal resistance of each rod be $R$. The circuit can be simplified by considering the thermal resistances in series and parallel.The cen

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$140$

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(C) Let the thermal resistance of each rod be $R$. The circuit can be simplified by considering the thermal resistances in series and parallel.The central diamond part consists of two parallel branches,each containing two rods in series. The resistance of each branch is $R + R = 2R$. Since these two branches are in parallel,their equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{2R} = \frac{1}{R}$,so $R_{eq} = R$.Now,the total circuit consists of three resistances in series: the rod $AB$ (resistance $R$),the equivalent resistance of the diamond part $(R_{eq} = R)$,and the rod $CD$ (resistance $R$).The total resistance is $R_{total} = R + R + R = 3R$.The total temperature drop is $\Delta T = 200^{\circ}C - 20^{\circ}C = 180^{\circ}C$.The heat current $H$ flowing through the circuit is $H = \frac{\Delta T}{R_{total}} = \frac{180}{3R} = \frac{60}{R}$.For junction $B$,the temperature $	heta_B$ is given by $200 - 	heta_B = H 	imes R = \frac{60}{R} 	imes R = 60$.Therefore,$	heta_B = 200 - 60 = 140^{\circ}C$.

Six identical conducting rods are joined as shown in the figure. Points $A$ and $D$ are maintained at temperatures $200\,^{\circ}C$ and $20\,^{\circ}C$,respectively. The temperature of junction $B$ will be ......... $^{\circ}C$.

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