A cylindrical rod is having temperatures θ₁ and θ₂ at its ends. The rate of heat flow is ' Q ' J s⁻¹. All the linear dimensions of the rod are doubled by keeping the temperatures constant. What is the new rate of flow of heat? (a) Q/2 (b) Q/4 (c) 2Q (d) 3Q/2

Answer: (c). Q = (Kπ r²Δθ)/(L) ⇒ Q ∝ r²/L Doubling r and L: factor 4/2 = 2 New rate = 2Q, so the answer is (c).

A cylindrical rod is having temperatures θ₁ and θ₂ at its ends. The… | MHT CET 2021 (22 Sep, Shift 1)

Q35MHT CET 2021 · 22 SEP · SHIFT 1Medium

Heat Conduction

A cylindrical rod is having temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is ' $Q$ ' $\text{J s}^{-1}$ . All the linear dimensions of the rod are doubled by keeping the temperatures constant. What is the new rate of flow of heat?

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Answer: (c)

$Q = \frac{K\pi r^2\Delta\theta}{L} \Rightarrow Q \propto \frac{r^2}{L}$
Doubling $r$ and $L$ : factor $\frac{4}{2} = 2$
New rate $= 2Q$ , so the answer is (c).

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Heat Conduction question from MHT CET 2021 (22 Sep, Shift 1)

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