A structural steel rod has a radius of 10 mm and length of 1.0 m. A 100 kN force stretches it along its length. Young's modulus of structural steel is $$2 \times 10^{11}\ \text{Nm}^{-2}$$. The percentage strain is about

The percentage strain is obtained from the definition

$$\text{strain} \; \varepsilon = \frac{\text{stress}}{\text{Young’s modulus}}$$

Hence we first find the longitudinal stress produced in the rod.

Step 1: Cross-sectional area
Radius $$r = 10\ \text{mm} = 0.01\ \text{m}$$
$$A = \pi r^{2} = \pi (0.01)^{2} = \pi \times 10^{-4}\ \text{m}^{2}$$

Step 2: Stress
Applied force $$F = 100\ \text{kN} = 100 \times 10^{3}\ \text{N}$$
$$\sigma = \frac{F}{A} = \frac{100 \times 10^{3}}{\pi \times 10^{-4}} \approx 3.18 \times 10^{8}\ \text{N\,m}^{-2}$$

Step 3: Strain
Young’s modulus $$Y = 2 \times 10^{11}\ \text{N\,m}^{-2}$$
$$\varepsilon = \frac{\sigma}{Y} = \frac{3.18 \times 10^{8}}{2 \times 10^{11}} = 1.59 \times 10^{-3}$$

Step 4: Percentage strain
$$\varepsilon(\%) = \varepsilon \times 100 = 1.59 \times 10^{-3} \times 100 = 0.159\% \approx 0.16\%$$

Therefore, the percentage strain is about 0.16 %.
Option A which is: 0.16 %.