1. **State the problem:** We have a steel cable with a cross-sectional area of 3 cm², mass per meter 2.4 kg/m, and length 500 m. We want to find how much the cable stretches under its own weight. 2. **Given data:** - Cross-sectional area, $A = 3 \text{ cm}^2 = 3 \times 10^{-4} \text{ m}^2$ - Mass per unit length, $\mu = 2.4 \text{ kg/m}$ - Length, $L = 500 \text{ m}$ - Young's modulus for steel, $Y = 2 \times 10^{10} \text{ N/m}^2$ - Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$ 3. **Formula for elongation due to weight of cable:** $$\Delta L = \frac{\mu g L^2}{2 A Y}$$ This formula comes from integrating the tension along the cable due to its own weight. 4. **Calculate the elongation:** Substitute the values: $$\Delta L = \frac{2.4 \times 9.8 \times 500^2}{2 \times 3 \times 10^{-4} \times 2 \times 10^{10}}$$ 5. **Simplify step-by-step:** Calculate numerator: $$2.4 \times 9.8 \times 500^2 = 2.4 \times 9.8 \times 250000 = 2.4 \times 2,450,000 = 5,880,000$$ Calculate denominator: $$2 \times 3 \times 10^{-4} \times 2 \times 10^{10} = 12 \times 10^{6} = 1.2 \times 10^{7}$$ 6. **Divide numerator by denominator:** $$\Delta L = \frac{5,880,000}{1.2 \times 10^{7}} = \frac{5.88 \times 10^{6}}{1.2 \times 10^{7}} = 0.49 \text{ m}$$ 7. **Final answer:** The cable stretches approximately $0.49$ meters under its own weight.