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(C) Let the temperature of the junction be $	heta$. Since the rods are of the same material,cross-section,and length,their thermal resistance $R = \f

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$45$

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(C) Let the temperature of the junction be $	heta$. Since the rods are of the same material,cross-section,and length,their thermal resistance $R = \frac{l}{KA}$ is the same for all.According to the principle of conservation of energy (steady state),the sum of heat currents flowing into the junction must equal the sum of heat currents flowing out of the junction.Assuming heat flows from the higher temperature ends towards the junction $	heta$:$\frac{KA(90 - 	heta)}{l} + \frac{KA(60 - 	heta)}{l} = \frac{KA(	heta - 0)}{l} + \frac{KA(	heta - 30)}{l}$Since $\frac{KA}{l}$ is common,we can cancel it from both sides:$(90 - 	heta) + (60 - 	heta) = 	heta + (	heta - 30)$$150 - 2	heta = 2	heta - 30$$180 = 4	heta$$	heta = 45^{\circ}C$

Four rods of the same material and having the same cross-section and length have been joined as shown in the figure. The temperature of the junction of the four rods will be . . . . . . $^oC$.

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