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Two wires of the same material and length but diameters in the ratio 1:2 are stretched by the same force. The elastic potential energy per unit volume for the wires when stretched by the same force will be in the ratio

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16 : 1

1 : 1

2 : 1

4:1

answer is A.

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Detailed Solution

We know that the elastic energy per unit volume is,

$E = \frac{1}{2} \times S t r e s s \times s t r a i n$

using Hooke's law

$s t r a i n = \frac{s t r e s s}{Y} E = \frac{{(S t r e s s)}^{2}}{2 Y} = \frac{F^{2}}{2 Y A^{2}} E = \frac{8 F^{2}}{(π^{2} D^{4} Y)} \frac{E_{1}}{E_{2}} = {(\frac{D_{2}}{D_{1}})}^{4} \frac{E_{1}}{E_{2}} = {(\frac{2}{1})}^{4} = \frac{16}{1}$

Hence the correct answer is 16:1.

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