Question

A metallic rod of length $l$ and cross-sectional area $A$ is made of a material of Young modules $Y$. 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)

$y^2$

$Volume=A\times L$ or $V=Al$
$strain=\frac{Elongation}{\text{Original length}}=\frac{Y}{l}$
Young's modulus $=\frac{stress}{strain}$
Work done, $W=\frac{1}{2}\times stress\times strain\times volume$
$W=\frac{1}{2}\times Y\times (strain{)}^2\times Al$
$=\frac{1}{2}\times Y{\left[\frac{y}{l}\right]}^2\times Al=\frac{1}{2}\left[\frac{YA}{l}\right]y^2\Rightarrow W\propto y^2$