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(C) The work done in stretching a wire or rod is given by the formula for elastic potential energy stored in the material.Work done $(W)$ = $\frac{1}{

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$\ell^{2}$

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(C) The work done in stretching a wire or rod is given by the formula for elastic potential energy stored in the material.Work done $(W)$ = $\frac{1}{2} 	imes 	ext{stress} 	imes 	ext{strain} 	imes 	ext{volume}$.We know that Young's modulus $(Y)$ = $\frac{	ext{stress}}{	ext{strain}}$,so $	ext{stress} = Y 	imes 	ext{strain}$.Substituting this into the work formula:$W = \frac{1}{2} 	imes Y 	imes (	ext{strain})^2 	imes 	ext{Volume}$.Here,$	ext{strain} = \frac{\ell}{L}$ and $	ext{Volume} = A 	imes L$.Substituting these values:$W = \frac{1}{2} 	imes Y 	imes \left(\frac{\ell}{L}ight)^2 	imes (A 	imes L)$,$W = \frac{1}{2} 	imes Y 	imes \frac{\ell^2}{L^2} 	imes A 	imes L$,$W = \frac{1}{2} 	imes \frac{Y 	imes A}{L} 	imes \ell^2$.Since $Y$,$A$,and $L$ are constants for a given rod,we have $W \propto \ell^2$.

$A$ metal rod has length,cross-sectional area,and Young's modulus as $L$,$A$,and $Y$ respectively. If the elongation produced in the rod is $\ell$,then the work done is proportional to:

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