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(C) In a steady state,the rate of heat flow through both sections of the rod must be equal.Let $T$ be the temperature at the interface.The rate of hea

Vedclass · Accepted Answer

$(K_1 l_2 T_1 + K_2 l_1 T_2) / (K_1 l_2 + K_2 l_1)$

Vedclass · Answer

(C) In a steady state,the rate of heat flow through both sections of the rod must be equal.Let $T$ be the temperature at the interface.The rate of heat flow $H$ is given by $H = \frac{KA(T_{high} - T_{low})}{l}$.Since the cross-sectional area $A$ is the same for both sections,we have:$\frac{K_1 A (T_1 - T)}{l_1} = \frac{K_2 A (T - T_2)}{l_2}$$\frac{K_1 (T_1 - T)}{l_1} = \frac{K_2 (T - T_2)}{l_2}$$K_1 l_2 (T_1 - T) = K_2 l_1 (T - T_2)$$K_1 l_2 T_1 - K_1 l_2 T = K_2 l_1 T - K_2 l_1 T_2$$K_1 l_2 T_1 + K_2 l_1 T_2 = T (K_1 l_2 + K_2 l_1)$$T = \frac{K_1 l_2 T_1 + K_2 l_1 T_2}{K_1 l_2 + K_2 l_1}$Thus,the correct option is $C$.

One end of a thermally insulated rod is kept at a temperature $T_1$ and the other at $T_2$. The rod is composed of two sections of lengths $l_1$ and $l_2$ and thermal conductivities $K_1$ and $K_2$ respectively. The temperature at the interface of the two sections is

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