$=2 \sqrt{l^2+x^2}-2 l$.

$=2\left(\sqrt{\ell^2+x^2}-l\right)$.

$=2 l\left(\sqrt{1+x^2 /l^2}-1\right)$

$\therefore$ strain $=\frac{2 l\left[\sqrt{1+x^2/l^2}-1\right]}{2 l}$

$=\left(1+x^2 / l^2\right)^{1 / 2}-1$

$=1+\frac{1}{2} x^2 / L^2-1=\frac{x^2}{2l^2}$

Now, tension, $T=\frac{m g}{2 \cos \theta}$

$\cos \theta=\frac{x}{\left(1+\frac{x^2}{l^2}\right)^{1 / 2} l}$

$=\frac{x}{\left(1+\frac{1}{2} \frac{x^2}{l^2}\right) l}$

$=x / l(\because l>>x)$

$T=\frac{m g}{2 \cos \theta}$

$=\frac{m g l}{2 x}$

Stress $=T/ A=\frac{2 m g l}{2 A x}$

$y=\frac{m g l / 2 A x}{x^2 / 2 l^2}=\frac{m g l^3}{A x^3}$.

$\therefore x=l\left(\frac{m g}{\text { YA }}\right)^{1 / 3}$