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MHT CET 2026 (16 Apr, Shift 1) · Heat Conduction

Heat Conduction question from MHT CET 2026 (16 Apr, Shift 1)

Q5MHT CET 2026 · 16 APR · SHIFT 1Medium

Heat Conduction

Rate of flow of heat through a cylindrical rod is ' $H_1$ '. The temperature at the ends of the rod are $T_1$ and $T_2$ . If all the dimensions of the rod become double and the temperature difference remains the same and if the rate of flow of heat becomes ' $H_2$ ' then $H_2 =$

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Answer: (c)

$H = \frac{KA\Delta T}{L}$ , with $A = \pi r^2$ .
Doubling all dimensions: $A \to 4A$ , $L \to 2L$ .
$H_2 = \frac{4}{2}H_1 = 2H_1$ , so the answer is (c).

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