SATHEE: Thermal Properties of Matter - Result Question 36

Thermal Properties of Matter - Result Question 36

37. 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/2002]

(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) $A_{1} S_{1} = A_{2} S_{2}$

Show Answer

Answer:

Correct Answer: 37. (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, $\frac{K_{1} A_{1} (T_{1} - T_{2})}{d}$

$= \frac{K_{2} A_{2} (T_{1} - T_{2})}{d} \Rightarrow K_{1} A_{1} = K_{2} A_{2}$

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