where $A$ is the area of $cross-section$ of the wire.
$Young's\,modulus,\,Y = \frac{{\left( {F/A} \right)}}{{\left( {\Delta l/l} \right)}} = \frac{{Fl}}{{A\Delta l}}$
$\Delta l = \frac{{Fl}}{{YA}} = \frac{{F{l^2}}}{{YV}}$ $(Using(i))$
$\Delta l \propto {l^2}$
Hence, the graph between $\Delta l$ and ${l^2}$ is a straight line.
