When a load is attached to the wire, the ratio of increase in elastic energy of the wire to the decrease in gravitational potential energy when load moves downwards by 1 mm, will be

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$\frac{1}{4}$

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$\frac{1}{3}$

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$\frac{1}{2}$

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Solution

The correct option is D
$\frac{1}{2}$

Ratio of work done = $\frac{\frac{1}{2}}{F l} = \frac{1}{2}$

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When a load is attached to the wire, the ratio of increase in elastic energy of the wire to the decrease in gravitational potential energy when load moves downwards by 1 mm, will be

The correct option is D 12 Ratio of work done = 12Fl=12

The correct option is D
$\frac{1}{2}$

Ratio of work done = $\frac{\frac{1}{2}}{F l} = \frac{1}{2}$