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MHT CET 2021 (23 Sep, Shift 2) · Heat Conduction

Heat Conduction question from MHT CET 2021 (23 Sep, Shift 2)

Q34MHT CET 2021 · 23 SEP · SHIFT 2Medium

Heat Conduction

A cylindrical rod has temperatures ' $T_1$ ' and ' $T_2$ ' at its ends. The rate of flow of heat is ' $Q_1$ ' cal $\text{s}^{-1}$ . If length and radius of the rod are doubled keeping temperature constant, then the rate of flow of heat ' $Q_2$ ' will be

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Answer: (d)

$Q = \frac{K\pi r^2\Delta T}{L} \Rightarrow Q \propto \frac{r^2}{L}$
$\frac{Q_2}{Q_1} = \frac{(2)^2}{2} = 2$
So the answer is (d) $Q_2 = 2Q_1$ .

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