A metallic rod of length $I$ 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 ......
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(c)
Work done = energy stored
$W=\frac{1}{2} \times \text { force } \times \text { elongation }$ $\left\{\right.$ Force $=\frac{\Delta L}{L} \cdot A Y$
$W=\frac{1}{2} \times \frac{\Delta L}{L} \times A \times Y \times \Delta L$
$W=\frac{1}{2} \frac{A Y}{L} \times \Delta L^2$
$W \propto \Delta L^2$
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