A uniform rod of mass m, length L, area of cross-section A and Young's modulus Y hangs from the ceiling. Its elongation under its own weight will be 

Let mg be the weight of the rod.

A be the area of cross section of the rod.

Thus the stress,

$\mathrm{stress}=\frac{\mathrm{mg}}{\mathrm{A}}$

Let L' be the change in length and L be the original length of the rod.

$$\begin{gathered} \mathrm{strain}=\frac{\mathrm{L}\text{'}}{\mathrm{L}} \\ \mathrm{Youngs} \mathrm{modulus}=\frac{\mathrm{stress}}{\mathrm{strain}} \\ \mathrm{Y}=\frac{{\displaystyle \frac{\mathrm{mg}}{\mathrm{A}}}}{{\displaystyle \frac{\mathrm{L}\text{'}}{\mathrm{L}}}} \\ \mathrm{Y}=\frac{\mathrm{mgL}}{\mathrm{AL}\text{'}} \\ \mathrm{L}\text{'}=\frac{\mathrm{mgL}}{\mathrm{AY}} \end{gathered}$$

Hence the correct answer is $\frac{\mathrm{mgL}}{\mathrm{AY}}.$