The adjacent graph shows the extension (Delta | vedclass

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(A) From the graph,we can select a point where the load $W = 20 \ N$ and the corresponding extension $\Delta l = 1 	imes 10^{-4} \ m$.Given:Length of

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

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(A) From the graph,we can select a point where the load $W = 20 \ N$ and the corresponding extension $\Delta l = 1 	imes 10^{-4} \ m$.Given:Length of the wire,$L = 1 \ m$Cross-sectional area,$A = 10^{-6} \ m^2$Young's modulus $Y$ is given by the formula:$Y = \frac{F \cdot L}{A \cdot \Delta l}$Substituting the values:$Y = \frac{20 	imes 1}{10^{-6} 	imes 1 	imes 10^{-4}}$$Y = \frac{20}{10^{-10}}$$Y = 20 	imes 10^{10} = 2 	imes 10^{11} \ N/m^2$Therefore,the correct option is $A$.

The adjacent graph shows the extension $(\Delta l)$ of a wire of length $1 \ m$ suspended from the top of a roof at one end with a load $W$ connected to the other end. If the cross-sectional area of the wire is $10^{-6} \ m^2$,calculate the Young's modulus of the material of the wire.

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