1. Young's modulus relation

For a wire,

$Y=\frac{\text{stress}}{\text{strain}}=\frac{F/A}{\Delta L/L}=\frac{FL}{A\,\Delta L}$

So,

$\Delta L=\frac{FL}{AY}$

This means the slope of the extension-load graph is

$\frac{\Delta L}{F}=\frac{L}{AY}$

Hence,

$Y=\frac{L}{A\left(\frac{\Delta L}{F}\right)}$

2. Interpret the graph

The graph is of extension vs load and is a straight line through origin making an angle $45^\circ$ with the load axis.

Therefore, its slope is

$\tan 45^\circ=1$

So,

$\frac{\Delta L}{F}=1$

(in SI units, i.e. m/N)

3. Substitute wire dimensions

Given:

• Length: $L=62.8\text{ cm}=0.628\text{ m}$
• Diameter: $d=4\text{ mm}=4\times10^{-3}\text{ m}$
• Radius: $r=2\times10^{-3}\text{ m}$

Cross-sectional area:

$A=\pi r^2=\pi(2\times10^{-3})^2=4\pi\times10^{-6}\text{ m}^2$

4. Calculate Young's modulus

Since $\dfrac{\Delta L}{F}=1$,

$Y=\frac{L}{A}$

Thus,

$Y=\frac{0.628}{4\pi\times10^{-6}}$

Using $\pi\approx 3.14$,

$4\pi\times10^{-6}=12.56\times10^{-6}$

So,

$Y=\frac{0.628}{12.56\times10^{-6}}=5\times10^4\ \text{N m}^{-2}$

Thus,

$Y=x\times10^4\ \text{N m}^{-2}$

so,

$x=5$

5. Comparison with stored answer

Stored correct answer = $5$

Our derived answer = $5$

So they agree.