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(D) From the graph,the slope of the extension-load curve is $	an(45^{\circ}) = 1$.Thus,$\frac{\Delta L}{F} = 1\,m/N$.Young's modulus $Y$ is given by

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$5$

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(D) From the graph,the slope of the extension-load curve is $	an(45^{\circ}) = 1$.Thus,$\frac{\Delta L}{F} = 1\,m/N$.Young's modulus $Y$ is given by the formula $Y = \frac{F L}{A \Delta L}$,where $L$ is the length,$A$ is the cross-sectional area,and $\Delta L$ is the extension.Rearranging,we get $Y = \frac{L}{A} 	imes \frac{F}{\Delta L} = \frac{L}{A} 	imes \frac{1}{1} = \frac{L}{A}$.Given $L = 62.8\,cm = 0.628\,m$ and diameter $d = 4\,mm = 4 	imes 10^{-3}\,m$.The radius $r = 2 	imes 10^{-3}\,m$.The area $A = \pi r^2 = 3.14 	imes (2 	imes 10^{-3})^2 = 3.14 	imes 4 	imes 10^{-6} = 12.56 	imes 10^{-6}\,m^2$.Substituting the values: $Y = \frac{0.628}{12.56 	imes 10^{-6}} = \frac{628000}{12.56} = 50000 = 5 	imes 10^4\,N/m^2$.Comparing this with $x 	imes 10^4\,N/m^2$,we get $x = 5$.

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