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

1:2

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4:1

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1:8

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1:4

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Solution

The correct option is C
1:8

Here, $l_{1} = \frac{F L_{1}}{π r_{1}^{2} Y}$ and $l_{2} = \frac{F L_{2}}{π r_{2}^{2} Y}$ .
Therefore, $\frac{l_{1}}{l_{2}} = \frac{L_{1}}{L_{2}} \times {(\frac{r_{2}}{r_{1}})}^{2}$
Given that, $L_{2} = 2 L_{1}$ and $r_{2} = \frac{r_{1}}{2}$
Thus, $\frac{l_{1}}{l_{2}} = \frac{1}{2} \times \frac{1}{(2)^{2}} = \frac{1}{8}$ .
Hence, the correct choice is (c).

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The correct option is C 1:8 Here, l1=FL1πr21Y and l2=FL2πr22Y. Therefore, l1l2=L1L2×(r2r1)2 Given that, L2=2L1 and r2=r12 Thus, l1l2=12×1(2)2=18. Hence, the correct choice is (c).