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A metal wire of length $'L'$ is suspended vertically from a rigid support. When a body of mass $M$ is attached to the lower end of wire, the elongation in wire is $'l'$, consider the following statements
$(I)$ the loss of gravitational potential energy of mass $M$ is $Mgl$
$(II)$ the elastic potential energy stored in the wire is $Mgl$
$(III)$ the elastic potential energy stored in wire is $\frac{1}{2}\, Mg l$
$(IV)$ heat produced is $\frac{1}{2}\, Mg l$
Correct statement are :-
Only $I$
$I$ and $II$
Only $III$
$I, III$ and $IV$
Solution
Energy stored per unit volume $=\frac{1}{2} \times$ stress $\times$ strain
$\mathrm{u}=\frac{1}{2} \times \frac{\mathrm{Mg}}{\mathrm{A}} \times \frac{\ell}{\mathrm{L}}$
Energy stored in total value
$\mathrm{U}=\mathrm{ALu}=\frac{1}{2} \mathrm{mg} \ell$ and work done by weight
$W=\mathbf{M g \ell}$
So heat produced $=$ loss in energy
$=\mathrm{Mg} \ell-\frac{1}{2} \mathrm{Mg} \ell$
$=\frac{1}{2} \mathrm{Mg} \ell$
