A \( 20-\mathrm{m} \) long steel wire (cross-section \( 1 \mathrm{~cm}^{2} \), Young's modulus \( 2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2} \) ), is subjected to a load of \( 25,000 \mathrm{~N} \). How much will the wire stretch under the load? \( \begin{array}{lllll}\text { A) } 0.25 & \text { B) } & 2.5 \mathrm{~cm} & \text { C) } 12.5 \mathrm{~cm} & \text { D) } 25 \mathrm{~cm}\end{array} \) \( \begin{array}{llll} & \text { B) } 1.25 \mathrm{~cm}\end{array} \)

To calculate the stretch of the steel wire, we can use the formula for elongation due to tensile stress: \[ \Delta L = \frac{F L_0}{A E} \] where: - \( F \) is the force applied (25,000 N), - \( L_0 \) is the original length of the wire (20 m), - \( A \) is the cross-sectional area (1 cm² = \( 1 \times 10^{-4} \) m²), - \( E \) is Young's modulus ( \( 2 \times 10^{11} \, \text{N/m}^2 \)). Plugging these values into the formula gives: \[ \Delta L = \frac{25000 \, \text{N} \times 20 \, \text{m}}{1 \times 10^{-4} \, \text{m}^2 \times 2 \times 10^{11} \, \text{N/m}^2} \] This simplifies to: \[ \Delta L = \frac{500000}{2 \times 10^{7}} = 0.025 \, \text{m} = 2.5 \, \text{cm} \] So, the wire will stretch **2.5 cm** under the load. **Answer: B) 2.5 cm** **Fun Fact about Steel:** Did you know that steel is one of the most recycled materials on the planet? In fact, it can be recycled infinitely without losing its strength, making it a fantastic choice for sustainability in construction and manufacturing! **Real-World Tips:** When dealing with materials like steel, it's essential to consider factors such as temperature, as thermal expansion can also affect elasticity. Ensuring your structures account for these changes can prevent unwanted stress and potential failure.