The length of a wire is 1.0 , m and the area | vedclass

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(C) The work done in stretching a wire is given by the formula: $W = \frac{1}{2} 	imes 	ext{Stress} 	imes 	ext{Strain} 	imes 	ext{Volume} = \fra

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$2.0 	imes 10^{11} \, N/m^2$

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(C) The work done in stretching a wire is given by the formula: $W = \frac{1}{2} 	imes 	ext{Stress} 	imes 	ext{Strain} 	imes 	ext{Volume} = \frac{1}{2} \frac{YA \Delta l^2}{L}$.Given:Length $L = 1.0 \, m$Area $A = 1.0 	imes 10^{-2} \, cm^2 = 1.0 	imes 10^{-2} 	imes 10^{-4} \, m^2 = 1.0 	imes 10^{-6} \, m^2$Extension $\Delta l = 0.2 \, cm = 0.2 	imes 10^{-2} \, m = 2 	imes 10^{-3} \, m$Work done $W = 0.4 \, J$Substituting the values into the formula:$0.4 = \frac{1}{2} 	imes \frac{Y 	imes (1.0 	imes 10^{-6}) 	imes (2 	imes 10^{-3})^2}{1.0}$$0.4 = 0.5 	imes Y 	imes 10^{-6} 	imes 4 	imes 10^{-6}$$0.4 = 2 	imes 10^{-12} 	imes Y$$Y = \frac{0.4}{2 	imes 10^{-12}} = 0.2 	imes 10^{12} = 2.0 	imes 10^{11} \, N/m^2$.

The length of a wire is $1.0 \, m$ and the area of cross-section is $1.0 \times 10^{-2} \, cm^2$. If the work done for an increase in length by $0.2 \, cm$ is $0.4 \, J$,then the Young's modulus of the material of the wire is:

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