1. **Problem statement:** We need to find the minimum diameter of a steel rod that can hold a tensile force of 30000 N (30 kN) without elongating more than 0.005 m (5 mm). The rod is 10 m long and has a Young's modulus $E = 200 \times 10^9$ Pa. 2. **Relevant formula:** The elongation $\Delta L$ of a rod under tensile force $F$ is given by $$\Delta L = \frac{F L}{A E}$$ where $L$ is the original length, $A$ is the cross-sectional area, and $E$ is Young's modulus. 3. **Cross-sectional area:** For a circular cross-section, $$A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}$$ where $d$ is the diameter. 4. **Rearranging the elongation formula to solve for $A$:** $$A = \frac{F L}{E \Delta L}$$ 5. **Substitute $A$ in terms of $d$ and solve for $d$:** $$\frac{\pi d^2}{4} = \frac{F L}{E \Delta L} \implies d^2 = \frac{4 F L}{\pi E \Delta L} \implies d = \sqrt{\frac{4 F L}{\pi E \Delta L}}$$ 6. **Plug in the values:** $$F = 30000\,\text{N},\quad L = 10\,\text{m},\quad E = 200 \times 10^9\,\text{Pa},\quad \Delta L = 0.005\,\text{m}$$ $$d = \sqrt{\frac{4 \times 30000 \times 10}{\pi \times 200 \times 10^9 \times 0.005}}$$ 7. **Calculate the value inside the square root:** $$= \sqrt{\frac{1,200,000}{\pi \times 1,000,000,000}} = \sqrt{\frac{1,200,000}{3,141,592,653.59}} \approx \sqrt{0.00038197}$$ 8. **Final diameter:** $$d \approx 0.01955\,\text{m} = 19.55\,\text{mm}$$ **Answer:** The minimum diameter of the steel rod is approximately **19.55 mm**.