1. **Problem statement:** A steel rod with diameter 10 mm and length 2 m is subjected to a tensile load of 15 N. We need to calculate the stress, strain, and elongation. 2. **Given data:** - Diameter, $d = 10$ mm = $10 \times 10^{-3}$ m = 0.01 m - Length, $L = 2$ m - Load, $F = 15$ N 3. **Formulas:** - Stress, $\sigma = \frac{F}{A}$ where $A$ is the cross-sectional area - Strain, $\varepsilon = \frac{\Delta L}{L}$ where $\Delta L$ is elongation - Elongation, $\Delta L = \frac{FL}{AE}$ where $E$ is Young's modulus of steel (approx. $2 \times 10^{11}$ Pa) 4. **Calculate cross-sectional area:** $$A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.01}{2}\right)^2 = \pi (0.005)^2 = \pi \times 25 \times 10^{-6} = 7.854 \times 10^{-5} \text{ m}^2$$ 5. **Calculate stress:** $$\sigma = \frac{F}{A} = \frac{15}{7.854 \times 10^{-5}} = 190,985.93 \text{ Pa} = 1.91 \times 10^{5} \text{ Pa}$$ 6. **Calculate elongation:** Using $E = 2 \times 10^{11}$ Pa, $$\Delta L = \frac{FL}{AE} = \frac{15 \times 2}{7.854 \times 10^{-5} \times 2 \times 10^{11}} = \frac{30}{1.5708 \times 10^{7}} = 1.91 \times 10^{-6} \text{ m}$$ 7. **Calculate strain:** $$\varepsilon = \frac{\Delta L}{L} = \frac{1.91 \times 10^{-6}}{2} = 9.55 \times 10^{-7}$$ **Final answers:** - Stress $\sigma = 1.91 \times 10^{5}$ Pa - Strain $\varepsilon = 9.55 \times 10^{-7}$ (dimensionless) - Elongation $\Delta L = 1.91 \times 10^{-6}$ m