$F=\mathrm{Ay} \cdot \frac{\Delta \ell}{\ell}$
$\Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{Ay}}=\frac{4 \mathrm{F} \ell}{\pi \mathrm{d} 2 \mathrm{y}}$
$\Delta \ell \propto \frac{\ell}{\mathrm{d}^{2}}$

$(I)$ 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 :-