শক্তির নিত্যতা

When a small mass mm is suspended at the lower end of the elastic wire, with the upper end fixed to the ceiling. There is a loss in gravitational potential energy. let it be x, due to extension of wire. Mark correct option, 

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Work energy theorem

wspring +wgravity +wloss =0−12kx2+mgx+wloss =0−12mgx+mgx+wloss =0wloss=−mgx2 \begin{array}{l} \mathrm{w}_{\text {spring }}+\mathrm{w}_{\text {gravity }}+\mathrm{w}_{\text {loss }}=0 \\ -\frac{1}{2} \mathrm{kx}^{2}+\mathrm{mgx}+\mathrm{w}_{\text {loss }}=0 \\ -\frac{1}{2 \mathrm{mgx}}+\mathrm{mgx}+\mathrm{w}_{\text {loss }}=0 \\ \mathrm{w}_{\mathrm{loss}}=-\frac{\mathrm{mgx}}{2} \end{array}

Since there is a loss so not recoverable half of it is lost rest is recoverable

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