Two wires of the same material and length, having diameters in the ratio 2 : 1, are stretched by the same force. The potential energy per unit volume stored in the two wires will be in the ratio:

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1 : 4

4 : 1

16 : 1

1 : 16

answer is D.

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Detailed Solution

$U = \frac{1}{2} \frac{A Y}{I_{i}} (Δ L)^{2}$

$\frac{U}{2 L} = u = \frac{1}{2} Y {(\frac{Δ L}{L})}^{2}$

$\frac{u_{1}}{{\ddot{u}}_{2}} {[\frac{(Δ L)_{1}}{(Δ L)_{2}}]}^{2} = {(\frac{A_{2}}{A_{1}})}^{2} = {[\frac{\frac{π}{4} (d)^{2}}{\frac{π}{4} (2 d)^{2}}]}^{2}$

$= \frac{1}{16}$

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