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 } $
$\left\{\text { Force }=\frac{\Delta L}{L} \cdot A Y\right.$
$W=\frac{1}{2} \times \frac{\Delta L}{L} \times A \times Y \times \Delta L$
$W=\frac{1}{2} \frac{A Y}{L} \times \Delta L^{2}$
$W \propto \Delta L^{2}$