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The elastic energy stored in a write of young's modulus Y is

$(a) y \times \frac{stain^2}{volume} \\ (b) stress \times strain \times \;volume \\(c) y\large\frac{ stain^2 \times volume}{2} \\(d) \frac{1}{2}y stress \times strain \times \;volume $

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Elastic potential energy V is stored in the wire when a force F is applied and the extension $\Delta l$
$V= \large\frac{1}{2}$$ F \Delta l$
$\quad= \large\frac{1}{2} \frac{F}{A}$$ \times A \times \large\frac{\Delta l}{l}$$ \times l$
$\quad=\large\frac {1}{2}$ stress $\times$ strain $\times$ volume $(A \times l=volume)$
$\quad=\large\frac{1}{2}$$ y \;strain ^2 \times volume$ $\qquad[Y=\large\frac{stress}{strain}]$
Hence d is the corrwect answer. 


answered Oct 2, 2013 by meena.p
edited Feb 18, 2014 by meena.p

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