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Two wires A and B of the same material have their lengths in the ratio of 1 : 2 and their diameters in the ratio of 2: 1. If they are stretched with the same force, the ratio of the increase in the length of A to that of B will be

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1 : 2

4: 1

1 : 8

1 : 4

answer is C.

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Detailed Solution

Here $l_{1} = \frac{F L_{1}}{{πr}_{1}^{2} y}$ and $l_{2} = \frac{F L_{2}}{{πr}_{2}^{2} y}$ . Therefore,

$\frac{l_{1}}{l_{2}} = \frac{L_{1}}{L_{2}} \times {(\frac{r_{2}}{r_{1}})}^{2}$

Given L₁= 2L₁ and r₂ = r₁/2. Thus $\frac{l_{1}}{l_{2}} = \frac{1}{2} \times \frac{1}{{(2)}^{2}} = \frac{1}{8}$ .

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