For a steel wire of length 1.80 m and diameter 0.75 mm, two points on the best straight line of…

Step 1. The slope. $\dfrac{1.14\ \text{mm} - 0.38\ \text{mm}}{6.0\ \text{kg} - 2.0\ \text{kg}}$ $= \dfrac{0.76\ \text{mm}}{4.0\ \text{kg}}$ $= 0.19\ \text{mm/kg}$.

Step 2. The modulus. $Y = \dfrac{4gL}{\pi d^2 \times \text{slope}}$. The numerator is $4gL = 4 \times 9.8\ \text{m/s}^2$ $\times 1.80\ \text{m}$ $= 70.6\ \text{m}^2/\text{s}^2$. The denominator is $\pi \times (0.75 \times 10^{-3}\ \text{m})^2$ $\times 0.19 \times 10^{-3}\ \text{m/kg}$ $= 3.36 \times 10^{-10}\ \text{m}^3/\text{kg}$. So $Y = \dfrac{70.6\ \text{m}^2/\text{s}^2}{3.36 \times 10^{-10}\ \text{m}^3/\text{kg}}$ $= 2.10 \times 10^{11}\ \text{N/m}^2$.

Step 3. The prediction. The line passes through the origin, so $\Delta L = 0.19\ \text{mm/kg} \times 5.0\ \text{kg}$ $= 0.95$ mm.