The adjacent graph shows the extension (Delta | vedclass

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(A) From the graph,we observe that for a change in load $\Delta W = (40 - 20) \, N = 20 \, N$,the change in extension is $\Delta(\Delta l) = (2 - 1) \

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) From the graph,we observe that for a change in load $\Delta W = (40 - 20) \, N = 20 \, N$,the change in extension is $\Delta(\Delta l) = (2 - 1) 	imes 10^{-4} \, m = 10^{-4} \, m$.Young's modulus $Y$ is given by the formula $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F/A}{\Delta l/l} = \frac{F \cdot l}{A \cdot \Delta l}$.Using the slope of the graph,$\frac{\Delta l}{F} = \frac{10^{-4} \, m}{20 \, N} = 0.05 	imes 10^{-4} \, m/N = 5 	imes 10^{-6} \, m/N$.Given $l = 1 \, m$ and $A = 10^{-6} \, m^2$,we have:$Y = \frac{l}{A} \cdot \frac{F}{\Delta l} = \frac{1}{10^{-6}} \cdot \frac{1}{5 	imes 10^{-6}} = \frac{10^6}{5 	imes 10^{-6}} = 0.2 	imes 10^{12} \, N/m^2 = 2 	imes 10^{11} \, N/m^2$.

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 and 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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