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(B) Let the temperature of the junction be $	heta$. Since rods $B$ and $C$ are connected in parallel to rod $A$,we can calculate the equivalent therm

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$60$

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(B) Let the temperature of the junction be $	heta$. Since rods $B$ and $C$ are connected in parallel to rod $A$,we can calculate the equivalent thermal resistance of the parallel combination of $B$ and $C$. Let the thermal resistance of each rod be $R$. Since rods $B$ and $C$ are in parallel,their equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}$,which implies $R_p = \frac{R}{2}$. Now,the system acts as two resistances $R$ and $\frac{R}{2}$ in series. The rate of heat flow $\frac{dQ}{dt}$ through the junction must be conserved. Thus,the heat flowing from the $90^{\circ}C$ ends through the parallel combination must equal the heat flowing through rod $A$ to the $0^{\circ}C$ end. Using the formula $\frac{dQ}{dt} = \frac{\Delta T}{R_{eq}}$: $\frac{90 - 	heta}{R/2} = \frac{	heta - 0}{R}$ $2(90 - 	heta) = 	heta$ $180 - 2	heta = 	heta$ $3	heta = 180$ $	heta = 60^{\circ}C$.

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