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The length of a wire is $1.0 \mathrm{~m}$ and the area of cross-section is $1.0 \times 10^{-2} \mathrm{~cm}^2$. If the work done for increase in length by 0.2 $\mathrm{cm}$ is 0.4 joule, then Young's modulus of the material of the wire is
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The correct answer is:
$2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
$W=\frac{1}{2} \frac{Y A l^2}{L}=0.4=\frac{1}{2} \times \frac{Y \times 10^{-6} \times\left(0.2 \times 10^{-2}\right)^2}{1}$
$\therefore \gamma=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
$\therefore \gamma=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$
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