A thick rope of density $\mathrm{\rho }$ and length L is hung from a rigid support. The increase in length of the rope due to its own weight is (Y is the Young's modulus)

As the weight of wire acts at centre of gravity.

Only half the length of wire gets extended.

Now Y = $\frac{\mathrm{F}}{\mathrm{A}}.\frac{\left({\displaystyle \frac{\mathrm{L}}{2}}\right)}{∆\mathrm{l}} = \frac{\mathrm{Mg}\left({\displaystyle \frac{\mathrm{L}}{2}}\right)}{\mathrm{A}∆\mathrm{l}}$

$\Rightarrow ∆\mathrm{l} = \frac{\mathrm{MgL}}{2\mathrm{AY}} \Rightarrow ∆\mathrm{l} = \frac{\mathrm{AL\rho gL}}{2\mathrm{AY}}$

$∆\mathrm{l} = \frac{{\mathrm{\rho L}}^2\mathrm{g}}{2\mathrm{Y}}$

So the correct choice is (2)