First slide

Conduction

Question

Heat is flowing through two cylindrical rods of the same material. The diameters of the rods are in the ratio 1 : 2 and the lengths in the ratio 2 : 1. If the temperature difference between the ends is same, then ratio of the rate of flow of heat through them will be

Moderate

A

1 : 1
B

1 : 8
C

2 : 1
D

8 : 1

Solution

$\frac{dQ}{dt} = \frac{KAΔT}{L} = \frac{Kπ (D / 2)^{2} ΔT}{L} \begin{aligned} ∴ \frac{(dQ / dt)_{1}}{(dQ / dt)_{2}} = \frac{Kπ {(D_{1} / 2)}^{2} ΔT \times L_{2}}{L_{1} \times Kπ {(D_{2} / 2)}^{2} ΔT} \\ = \frac{D_{1}^{2} \times L_{2}}{D_{2}^{2} \times L_{1}} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \end{aligned} or (dQ / dt)_{1} : (dQ / dt)_{2} = 1 : 8$

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