Two wires of same material and same diameter have lengths in the ratio 2 : 5. They are stretched by the same force. The ratio of work done in stretching them is :

5 : 2

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2 : 5

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1 : 3

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3 : 1

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Solution

The correct option is A 2 : 5
$W o r k D o n e = \frac{1}{2} \times F o r c e \times E l o n g a t i o n Y = \frac{F l}{A △ l} △ l = \frac{F l}{A Y}$
$W o r k D o n e = \frac{1}{2} \times F \times \frac{F l}{A Y} = \frac{F^{2} L}{2 A Y}$
Since, the Force, Area and young's modulus is same,
$∴ W o r k D o n e \propto l e n g t h$
$\frac{W_{1}}{W_{2}} = \frac{l_{1}}{l_{2}} = \frac{2}{5}$
Hence, the answer is OPTION B.

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Two wires of same material and same diameter have lengths in the ratio 2 : 5. They are stretched by the same force. The ratio of work done in stretching them is :

The correct option is A 2 : 5WorkDone=12×Force×ElongationY=FlA△l△l=FlAYWorkDone=12×F×FlAY=F2L2AYSince, the Force, Area and young's modulus is same,∴WorkDone∝length W1W2=l1l2=25 Hence, the answer is OPTION B.