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(A) Young's modulus $Y$ is defined as $Y = \frac{F/A}{l/L}$,where $F$ is the force,$A$ is the cross-sectional area,$l$ is the extension,and $L$ is the

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(A) Young's modulus $Y$ is defined as $Y = \frac{F/A}{l/L}$,where $F$ is the force,$A$ is the cross-sectional area,$l$ is the extension,and $L$ is the original length.From this,the force required to stretch the wire is $F = \frac{Y A l}{L}$.The work done $W$ in stretching the wire by a small distance $dl$ is $dW = F \, dl$.Integrating from $0$ to $l$,we get $W = \int_{0}^{l} \frac{Y A}{L} l \, dl = \frac{1}{2} \frac{Y A l^2}{L}$.Given values are: $L = 1 \, m$,$A = 1 \, mm^2 = 10^{-6} \, m^2$,$l = 1 \, mm = 10^{-3} \, m$,and $Y = 2 	imes 10^{11} \, N/m^2$.Substituting these values into the formula:$W = \frac{1}{2} 	imes \frac{(2 	imes 10^{11} \, N/m^2) 	imes (10^{-6} \, m^2) 	imes (10^{-3} \, m)^2}{1 \, m}$$W = \frac{1}{2} 	imes 2 	imes 10^{11} 	imes 10^{-6} 	imes 10^{-6} \, J$$W = 10^{11} 	imes 10^{-12} \, J = 0.1 \, J$.

The work done in increasing the length of a $1 \, m$ long wire of cross-sectional area $1 \, mm^2$ by $1 \, mm$ is (Given $Y = 2 \times 10^{11} \, N/m^2$): (in $, J$)

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