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A cylindrical rod having temperature $T_1$ and $T_2$ at its ends. The rate of flow of heat $Q_1$ cal/sec. If all the linear dimensions are doubled keeping temperature constant then rate of flow of heat $Q_2$ will be : -
A
$4Q_1$
B
$2Q_1$
C
$\dfrac{Q_1}{4}$
D
$\dfrac{Q_1}{2}$
Detailed Solution
Heat flow rate: $Q = \dfrac{KA(T_1 - T_2)}{L}$
When all linear dimensions are doubled: radius $\rightarrow 2r$, so area $A \propto r^2$ becomes 4A; length becomes 2L.
$Q_2 = \dfrac{K(4A)(T_1 - T_2)}{2L} = 2 \times \dfrac{KA(T_1 - T_2)}{L}$
So $Q_2 = 2Q_1$
When all linear dimensions are doubled: radius $\rightarrow 2r$, so area $A \propto r^2$ becomes 4A; length becomes 2L.
$Q_2 = \dfrac{K(4A)(T_1 - T_2)}{2L} = 2 \times \dfrac{KA(T_1 - T_2)}{L}$
So $Q_2 = 2Q_1$