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Q.

Two wires made of the same material have lengths 2m & 4m and masses 8 gm and 16 gm respectively. When they are subjected to the same force, then the ratio of their elongations is

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a

4 : 1

b

2 : 1

c

1 : 2

d

1 : 4

answer is A.

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Detailed Solution

$e = \frac{F L}{A Y}$

Here $A = \frac{V}{L} = \frac{m}{ρ L}$

$e = \frac{F L^{2} ρ}{m Y}$

$∴ e \propto \frac{L^{2}}{m}$

$\frac{e_{1}}{e_{2}} = {(\frac{L_{1}}{L_{2}})}^{2} (\frac{m_{2}}{m_{1}})$

$= {(\frac{2}{4})}^{2} (\frac{16}{8}) = \frac{1}{4} \times 2 = \frac{1}{2}$

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