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MHT CET 2024 (2 May, Shift 2) · Heat Conduction

Heat Conduction question from MHT CET 2024 (2 May, Shift 2)

Q20MHT CET 2024 · 2 MAY · SHIFT 2Medium

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, the rate of flow of heat becomes ' $H_2$ '. Then

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Answer: (b)

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

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