Young's modulus of a rod is $\frac{{\mathrm{AgL}}^2}{2\mathrm{l}}$ for which elongation is $\mathrm{\lambda }$ due to its own weight when suspended from the ceiling. L is the length of the rod and A is a constant, which is:

Elongation in (dx) at distance x from lower end

= $\left(\frac{\mathrm{W}}{{\mathrm{A}}_1\mathrm{Y}}\right)\mathrm{dx} = \left(\frac{\mathrm{mgx}}{\mathrm{L}}\right)\left(\frac{\mathrm{dx}}{{\mathrm{A}}_1\mathrm{Y}}\right) (\mathrm{where} {\mathrm{A}}_1 \mathrm{is} \mathrm{the} \mathrm{area})$

Increase in length due to own weight

$\mathrm{I} = {\int }_0^{\mathrm{L}}\frac{\mathrm{mgxdx}}{{\mathrm{LA}}_1\mathrm{Y}} = \frac{\mathrm{mgL}}{2{\mathrm{A}}_1\mathrm{Y}}$

$\mathrm{Y} = \frac{\mathrm{mgL}}{2{\mathrm{A}}_1\mathrm{l}} = \frac{{\mathrm{\lambda gL}}^2}{2{\mathrm{A}}_1\mathrm{l}}$         (Since, m = $\mathrm{\lambda L}$)

 = $\left(\frac{\mathrm{\lambda }}{{\mathrm{A}}_1}\right)\frac{{\mathrm{gL}}^2}{2\mathrm{l}} = \frac{{\mathrm{AgL}}^2}{2\mathrm{l}}$

where $\mathrm{A} = \frac{\mathrm{\lambda }}{{\mathrm{A}}_1} = \mathrm{mass} \mathrm{per} \mathrm{unit} \mathrm{length} \mathrm{per} \mathrm{unit} \mathrm{area}.$