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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

\[\begin {array} {1 1} (a)\;\frac{YAx^2}{L} \\ (b)\;\frac{YAx^2}{2L} \\ (c)\;\frac{YAL^2}{x} \\ (d)\;\frac{YAL^2}{2x} \end {array}\]
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1 Answer

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 meena.p
edited Feb 18, 2014 by meena.p
 

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