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Work done in increasing the length of $1\;m$ long wire of cross-sectional area $1\;mm^2$ through $1\;mm$ is $y=2 \times 10^{11}N/m^2$

\[\begin {array} {1 1} (a)\;250\;J \\ (b)\;10\;J \\ (c)\;5\;J \\ (d)\;0.1\;J \end {array}\]
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1 Answer

Work done $= \large\frac{1}{2} $$ \times$ Young's module $ \times$ strain $^2$ $ \times$Volume
$W=\large\frac{1}{2}$$ \times Y \times \large \frac{\Delta l^2}{L^2} \times$$ AL$
$\qquad= \large\frac{Y A \Delta L^2}{2L}$
$\qquad= \large\frac{2 \times 10^{11} \times 10^{-6} \times 10^{-6}}{2 \times 1}$
$\qquad=0.1\;J$
Hence d is the correct answer.

 

answered Oct 4, 2013 by meena.p
edited Feb 18, 2014 by meena.p
 

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