Thermal Properties of Matter - Result Question 38
39. A cylindrical rod having temperature $T_1$ and $T_2$ at its end. The rate of flow of heat is $Q_1 cal /$ sec. If all the linear dimensions are doubled keeping temperature constant, then the rate of flow of heat $Q_2$ will be
[2001]
(a) $4 Q_1$
(b) $2 Q_1$
(c) $Q_1 / 4$
(d) $Q_1 / 2$
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Answer:
Correct Answer: 39. (b)
Solution:
- (b) $Q=\frac{K A(\theta_1-\theta_2) t}{l}$
Rate of heat flow,
$H=\frac{Q}{t}=\frac{K A(\theta_1-\theta_2)}{l}$ i.e., $H \propto \frac{A}{l}$
Dimensions of area $A=[L^{2}]$, dimensions of distance $l=[L]$
$\therefore H \propto L \Rightarrow \frac{H_2}{H_1}=\frac{L_2}{L_1}=2 \Rightarrow H_2=2 H_1$
