1. Problem statement. Problem: A brass rod of length 1.5 m and diameter 1.2 m is stretched by 1.2 m under a load of 25 N; find the Young's modulus of the brass. 2. Formula and important rules. Young's modulus is the ratio of stress to strain. Display formula: $$E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L / L} = \frac{F L}{A \Delta L}$$ Rule: Stress is force per unit area and strain is relative extension (dimensionless). 3. Cross-sectional area calculation. Radius: $r = d/2 = 1.2/2 = 0.6\ \text{m}$. Area formula: $$A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2$$ Substitute numbers: $$A = \pi \cdot 0.6^2 = 0.36\pi\ \text{m}^2$$ Numeric area: $$A \approx 0.36\cdot 3.141592653589793 = 1.130973355\ \text{m}^2$$ 4. Strain calculation. $$\text{strain} = \frac{\Delta L}{L} = \frac{1.2}{1.5} = 0.8$$ 5. Stress calculation. $$\text{stress} = \frac{F}{A} = \frac{25}{0.36\pi}\ \text{N m}^{-2}$$ Numeric stress: $$\text{stress} \approx \frac{25}{1.130973355} = 22.11804573\ \text{N m}^{-2}$$ 6. Young's modulus calculation with intermediate work. Using the formula: $$E = \frac{F L}{A \Delta L}$$ Substitute values: $$E = \frac{25\times 1.5}{0.36\pi \times 1.2}$$ Simplify denominator: $0.36\pi\times 1.2 = 0.432\pi \approx 1.357168026$. Compute numerator: $25\times 1.5 = 37.5$. Division: $$E \approx \frac{37.5}{1.357168026} = 27.629336\ \text{N m}^{-2}$$ 7. Final answer. The Young's modulus of the brass (from the given data) is approximately $E \approx 27.63\ \text{N m}^{-2}$.