Question

A uniform heavy rod of length $L$ and area of cross-section area $A$ is hanging from a fixed support. If Young's modulus of the material of the rod is $Y$, then the increase in the length of the rod is ($\rho$ is a density of the material of the rod) :

• A. $\frac{L^{2}Y}{2\rho g}$
• B. $\frac{L^2\rho g}{2Y}$
• C. $\frac{L^{2}g}{2Y\rho }$
• D. $\frac{L^{2}g}{3Y\rho }$

Answer 1

Answer: (B)

$\frac{L^2\rho g}{2Y}$

Consider an element dx at a distance x from the top.

Let the wt. of rod is W.
The force acting on this because of a part which is below dx is given as $\frac{(L-x)W}{L}$

The elongation in element $dx$ is $=\left(\frac{L-x}{L}\right)\frac{Wdx}{AY}$$=\frac{(L-x)Wdx}{LAY}$

Total elongation, $\Delta L={\int }_O^L\frac{(L-x)Wdx}{LAY}$$=\frac{WL}{2AY}$

$W=AL\rho g$

$\Delta L=\frac{AL\rho gL}{2AY}$$=\frac{L^2\rho g}{2Y}$