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

Answer 1

Answer: (D)

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