A cylindrical rod is having temperatures θ₁ and θ₂ at its ends. The rate of heat flow is Q J/s. All the linear dimensions of the rod are doubled by keeping the temperature constant. The new rate of flow of heat is (a) 4Q (b) 2Q (c) Q/2 (d) Q/4

Answer: (b). Q = (KAΔθ)/(L), with A = π r². Doubling all linear dimensions: A → 4A, L → 2L. New rate = 4/2Q = 2Q, so the answer is (b).

A cylindrical rod is having temperatures θ₁ and θ₂ at its ends. The… | MHT CET 2024 (15 May, Shift 2)

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Heat Conduction

A cylindrical rod is having temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is $Q\text{ J/s}$ . All the linear dimensions of the rod are doubled by keeping the temperature constant. The new rate of flow of heat is

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Answer: (b)

$Q = \frac{KA\Delta\theta}{L}$ , with $A = \pi r^2$ .
Doubling all linear dimensions: $A \to 4A$ , $L \to 2L$ .
New rate $= \frac{4}{2}Q = 2Q$ , so the answer is (b).

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Heat Conduction question from MHT CET 2024 (15 May, Shift 2)

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