Question

A uniform rod of length $L$ has a mass per unit length $\lambda$ and area of cross section $A$. If the Young's modulus of the rod is $Y$. The elongation in the rod due to its own weight is

• A. $\frac{2\lambda g{L}^2}{AY}$
• B. $\frac{\lambda g{L}^2}{2AY}$
• C. $\frac{\lambda g{L}^2}{4AY}$
• D. $\frac{\lambda g{L}^2}{6AY}$

Answer 1

The correct option is B $\frac{\lambda g{L}^2}{2AY}$

$F=\frac{Mg}{2}$ (mass of rod will at $\frac{L}{2}$ only

$\begin{array}{c}\Delta l=\frac{FL}{AY}\Rightarrow \Delta l=\frac{MgL}{2AY}=\frac{Mg{L}^2}{2AYL}\\ \Rightarrow \frac{\lambda g{L}^2}{2Ay}\left(\lambda =\frac{M}{G}\right)\end{array}$