(Young's modulus of material of track is $10^{11} \,{Nm}^{-2}$ ))
Elastic energy per unit length $=\frac{Y}{2}(\text { strain })^{2} \times$ Area
$\left(\text { strain }=\frac{\Delta \ell}{\ell}=\alpha \Delta T=10^{-5} \times 10=10^{-4}\right)$
$=\frac{10^{11}}{2} \times\left(10^{-4}\right)^{2} \times 10^{-2}=5\, {J} / {m}$
$(I)$ 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 :-
[Area of cross section of wire $=0.005 \mathrm{~cm}^2$, $\mathrm{Y}=2 \times 10^{11}\ \mathrm{Nm}^{-2}$ and $\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right]$