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# A wire of length L and cross-sectional area A is made of material of Young's modulus Y. The work done in stretching the wire by an amount x is given by

$\begin {array} {1 1} (a)\;\frac{YAx^2}{L} \\ (b)\;\frac{YAx^2}{2L} \\ (c)\;\frac{YAL^2}{x} \\ (d)\;\frac{YAL^2}{2x} \end {array}$
Can you answer this question?

## 1 Answer

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Work done = $\large\frac{1}{2} \times$ stress $\times$ strain $\times$ volume
$\qquad= \large\frac{1}{2}$$\times Y \times (strain )^2 \times Volume \qquad =\large\frac{1}{2}$$ \times Y \times \bigg( \large\frac{x}{L}\bigg)^2$$\times AL$
$\qquad= \large\frac{Yx^2A}{2L}$
Hence b is the correct answer.

answered Oct 3, 2013 by
edited Feb 18, 2014 by meena.p

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