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(B) Let $T$ be the temperature at the junction $X$.Let $L$ and $A$ be the length and area of cross-section of each rod respectively,and $K$ be the the

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$60$

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(B) Let $T$ be the temperature at the junction $X$.Let $L$ and $A$ be the length and area of cross-section of each rod respectively,and $K$ be the thermal conductivity.Heat current from $Y$ to $X$ is $H_1 = \frac{KA(90^{\circ}C - T)}{L}$.Heat current from $Z$ to $X$ is $H_2 = \frac{KA(90^{\circ}C - T)}{L}$.Heat current from $X$ to $W$ is $H_3 = \frac{KA(T - 0^{\circ}C)}{L}$.At the junction $X$,by the principle of conservation of energy (steady state),the sum of incoming heat currents equals the outgoing heat current:$H_1 + H_2 = H_3$$\frac{KA(90 - T)}{L} + \frac{KA(90 - T)}{L} = \frac{KA(T - 0)}{L}$$(90 - T) + (90 - T) = T$$180 - 2T = T$$3T = 180^{\circ}C$$T = 60^{\circ}C$

Three metal rods of the same material and identical in all respects are joined as shown in the figure. The temperatures at the ends are maintained as indicated. Assuming no loss of heat from the curved surfaces of the rods,the temperature at the junction $X$ would be ....... $^oC$

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