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(C) Let the elongation in the wire be $\Delta l$. The loss in gravitational potential energy of the mass $m$ is given by $\Delta U = mg \Delta l$. Giv

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Only $\frac{x}{2}$ amount of energy is recoverable.

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(C) Let the elongation in the wire be $\Delta l$. The loss in gravitational potential energy of the mass $m$ is given by $\Delta U = mg \Delta l$. Given that this loss is $x$,we have $x = mg \Delta l$.The elastic potential energy stored in the stretched wire is given by $U_e = \frac{1}{2} 	imes 	ext{Force} 	imes 	ext{Elongation} = \frac{1}{2} (mg) \Delta l$.Substituting $mg \Delta l = x$ into the expression for elastic potential energy,we get $U_e = \frac{1}{2} x$.Since the elastic potential energy is stored in the wire and can be recovered,only $\frac{x}{2}$ amount of the lost gravitational potential energy is recoverable.

When a small mass $m$ is suspended at the lower end of an elastic wire having its upper end fixed to a ceiling,there is a loss in gravitational potential energy,let it be $x$,due to the extension of the wire. Mark the correct option.

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