Question

A uniform rod of mass $m$, length $L$, area of cross-section $A$ and Young's modulus $Y$ hangs from a rigid support. Its elongation due to its own weight will be

• A. $\frac{mgL}{AY}$
• B. $\frac{mgL}{2\,AY}$
• C. $\frac{2mgL}{AY}$
• D. zero

Answer 1

Answer: (B)

$\frac{mgL}{2\,AY}$

Answer 2

When a wire of length $L$ and cross section area $A$ is stretched by a force F $\therefore$ Young?? modulus $y=\frac{FL}{A\, \Delta\,L}$ or $\Delta\,L=\frac{FL}{AY}$ In case of elongation by its own weight, $F(=mg)$ will act at centre of gravity of the wire so that length of the wire which is stretched will be $\left(\frac{L}{2}\right)$ $\therefore \Delta L=\frac{\left(mg\right)\left(L 2\right)}{AY}$ $=\frac{mgL}{2\,AY}$