Thermal Properties Of Matter Ques 31
31. Two rods of thermal conductivities $K_1$ and $K_2$, crosssections $A_1$ and $A_2$ and specific heats $S_1$ and $S_2$ are of equal lengths. The temperatures of two ends of each rod are $T_1$ and $T_2$. The rate of flow of heat at the steady state will be equal if
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(a) $\frac{K_1}{A_1 S_1}=\frac{K_2}{A_2 S_2}$
(b) $K_1 A_1=K_2 A_2$
(c) $K_1 S_1=K_2 S_2$
(d) $A_1 S_1=A_2 S_2$
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Answer:
Correct Answer: 31.(b)
Solution:
- (b) Rate of heat flow for one rod
$=\frac{K_1 A_1(T_1-T_2)}{d}(d \to$ Length $)$
Rate of heat flow for other rod
$=\frac{K_2 A_2(T_1-T_2)}{d}$
In steady state, $\frac{K_1 A_1(T_1-T_2)}{d}$
$=\frac{K_2 A_2(T_1-T_2)}{d} \Rightarrow K_1 A_1=K_2 A_2$
BETA
