The work done in increasing the length of a $1$ $metre$ long wire of cross-section area $1\, mm^2$ through $1\, mm$ will be ....... $J$ $(Y = 2\times10^{11}\, Nm^{-2})$
A$0.1$
B$5$
C$10$
D$250$
Medium
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A$0.1$
a $\mathrm{W}=\frac{1}{2} \frac{\mathrm{YA}}{\ell}(\Delta \ell)^{2}$
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