Question

A cylindrical rod having temperature $T_1$ and $T_2$ at its end. The rate of flow of heat $Q_1$ cal/sec. If all the linear dimension are doubled keeping temperature constant, then rate of flow of heat $Q_2$ will be

• A. $4Q_1$
• B. $2Q_1$
• C. $\frac{Q_1}{4}$
• D. $\frac{Q_1}{2}$

Answer 1

Answer: (B)

$2Q_1$

Answer 2

Heat flow rate $\frac{dQ}{dt} = \frac{KA(T_1\, - \, T_2)}{L} = Q$
When linear dimensions are double.
$A_1 \propto r^2_1, \, L_1 = L$
$A_2 \propto 4r^2_1, \, L_1 = 2L$ so $Q_2 = 2Q_1$.