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 $Y$. Then elongation in the rod due to its own weight is

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

Answer 1

Answer: (D)

Total mass can be assumed to be concentrated at center of mass at distance $\frac{L}{2}$ from top
$\frac{M}{L}=\lambda $
$M=\lambda L$
$\Delta x=\frac{F L / 2}{A Y}=\frac{1}{2} \times \frac{\lambda L^{2} g}{A Y}$