Question

A metallic rod of length $l$ and cross-sectional area $A$ is made of a material of Young's modulus $Y$. If the rod. is elongated by an amount $y$, then the work done is proportional to

• A. $y$
• B. $\frac{1}{y}$
• C. $y^2$
• D. $\frac{1}{y^2}$

Answer 1

Answer: (C)

Work done $=$ energy stored
$W=\frac{1}{2}\times \text{ force }x\text{ elongation }$
$\{\text{ Force }=\frac{\Delta L}{L}\cdot AY$
$W=\frac{1}{2}\times \frac{\Delta L}{L}\times A\times Y\times \Delta L$
$W=\frac{1}{2}\frac{AY}{L}\times \Delta L^2$
$W\propto \Delta L^2$