A long steel rod of uniform cross sectional area is suspended from one end. If $x$ be the elongation of lower half of the rod, the elongation of upper half is 

 

 $$\begin{gathered} \mathrm{Let} \mathrm{m} \mathrm{be} \mathrm{the} \mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{rod}, l \mathrm{be} \mathrm{its} \mathrm{length} \mathrm{and} \mathrm{S} \mathrm{be} \mathrm{its} \mathrm{cross} \mathrm{sectional} \mathrm{area} \mathrm{and} \mathrm{Y} \mathrm{be} \mathrm{the} \mathrm{Young}\text{'}\mathrm{s} \mathrm{modulus} \mathrm{of} \mathrm{steel}, \\ \mathrm{then} \mathrm{elongation} \mathrm{of} \mathrm{lower} \mathrm{half} \mathrm{BC}, \\ x=\frac{\left(\raisebox{1ex}{$mg$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\raisebox{1ex}{$l$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{SY}=\frac{mgl}{8SY} \end{gathered}$$

Elongation of upper half AC, $y=\frac{\left(\raisebox{1ex}{$mg$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\raisebox{1ex}{$l$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{S.Y}+\frac{\left(\raisebox{1ex}{$mg$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\raisebox{1ex}{$l$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{SY}$

$\Rightarrow y=\frac{3}{8}\frac{mgl}{SY}\Rightarrow y=3x$