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Q.

Two wires made of the same material have lengths 3 m & 4 m and masses 18 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

1 : 2

b

4 : 3

c

16 : 18

d

9 : 4

answer is A.

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Detailed Solution

$y = \frac{F ℓ}{A e} = \frac{F ℓ^{2}}{V e} = \frac{F ℓ^{2} d}{m e}$

$e \propto \frac{ℓ^{_{2}}}{m}$

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

$\frac{e_{1}}{e_{2}} = {(\frac{3}{4})}^{2} (\frac{16}{18}) = \frac{1}{2}$

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Two wires made of the same material have lengths 3 m & 4 m and masses 18 gm and 16 gm respectively. When they are subjected to the same force, then the ratio of their elongations is