The ratio of the areas of cross-section of two rods of different materials is $1 : 2$ , and the ratio of the thermal conductivities of their materials is $4 : 3$ . On keeping equal temperature difference between the ends of these rods, the rate of conduction of heat are equal. The ratio of the lengths of the rods is

2 : 3

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2 : 4

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4 : 2

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4 : 5

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Solution

The correct option is A $2 : 3$
Given,
$\frac{k_{1}}{k_{2}} = 1 : 2$

$\frac{A_{1}}{A_{2}} = 4 : 3$

$\frac{d Q_{1}}{d t} = \frac{d Q_{2}}{d t}$

$Δ T_{2} = Δ T_{2}$

Now,

$\frac{k_{1} A_{1} Δ T_{1}}{l_{1}} = \frac{k_{2} A_{2} Δ T_{2}}{l_{2}}$
$\frac{l_{1}}{l_{2}} = \frac{k_{1} A_{1}}{k_{2} A_{2}} = \frac{k_{1}}{k_{2}} \times \frac{A_{1}}{A_{2}}$

$\frac{l_{1}}{l_{2}} = \frac{2}{3}$

Hence option $A$ is the correct answer.

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The correct option is A 2:3Given,k1k2=1:2A1A2=4:3dQ1dt=dQ2dtΔT2=ΔT2Now,k1A1ΔT1l1=k2A2ΔT2l2l1l2=k1A1k2A2=k1k2×A1A2l1l2=23Hence option A is the correct answer.