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:

  1. (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 $l=[L]$

$\therefore H \propto L \Rightarrow \frac{H_2}{H_1}=\frac{L_2}{L_1}=2 \Rightarrow H_2=2 H_1$