A metal rod has length, cross-sectional area and Young's modulus as ' $\mathrm{L}^{\prime},{ }^{\prime}…
A metal rod has length, cross-sectional area and Young's modulus as ' $\mathrm{L}^{\prime},{ }^{\prime} \mathrm{A}^{\prime}$ and $\mathrm{Y}^{\prime}$
respectively. If the elongation in the rod produced is ' $\ell^{\prime}$ then work done is proportional to
$\ell$
$\ell^{4}$
$\ell^{2}$
$\ell^{3}$
Solution
$\mathrm{V}$ olume $=\mathrm{A} \times \mathrm{L}$ or $\mathrm{V}=\mathrm{Al}$
strain $=\frac{\text { Elongation }}{\text { Original length }}=\frac{\mathrm{Y}}{\mathrm{l}}$
Young's modulus $=\frac{\text { stress }}{\text { strain }}$
Work done, $W=\frac{1}{2} \times$ stress $\times$ strain $\times$ volume
$\begin{array}{l}
W=\frac{1}{2} \times Y \times(\operatorname{strain})^{2} \times A l \\
=\frac{1}{2} \times Y\left[\frac{y}{1}\right]^{2} \times A l=\frac{1}{2}\left[\frac{Y A}{1}\right] y^{2} \Rightarrow W \propto y^{2}
\end{array}$
.