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Question

(C) Let the thermal resistance of each identical rod be $R$. The circuit consists of a rod $AB$ in series with a parallel combination of two branches,

Vedclass · Accepted Answer

$140$

Vedclass · Answer

(C) Let the thermal resistance of each identical rod be $R$. The circuit consists of a rod $AB$ in series with a parallel combination of two branches,each containing two rods in series,followed by a rod $CD$ in series. The equivalent resistance of the diamond-shaped part between $B$ and $C$ is $R_{eq} = \frac{(R+R)(R+R)}{(R+R)+(R+R)} = \frac{2R 	imes 2R}{2R+2R} = R$. Thus,the total thermal resistance between $A$ and $D$ is $R_{total} = R_{AB} + R_{BC} + R_{CD} = R + R + R = 3R$. The total heat current is $H = \frac{T_A - T_D}{R_{total}} = \frac{200 - 20}{3R} = \frac{180}{3R} = \frac{60}{R}$. The temperature at junction $B$ $(	heta_1)$ can be found using the heat current through rod $AB$: $H = \frac{T_A - 	heta_1}{R} \implies \frac{60}{R} = \frac{200 - 	heta_1}{R}$. Solving for $	heta_1$: $60 = 200 - 	heta_1 \implies 	heta_1 = 140^{\circ}C$.

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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