**Worked Example – Determining Young's Modulus of a Steel Rod** 1. **Given**: - Material: Steel rod - Original length, L = 1.00 m - Cross‑sectional area, A = 1.00 cm² = 1.00 × 10⁻⁴ m² - Applied load, F = 500 N - Measured extension, ΔL = 0.10 mm = 1.0 × 10⁻⁴ m 2. **Formula (Searle’s method)**: E = (F × L) / (A × ΔL) 3. **Plug in the numbers**: E = (500 N × 1.00 m) / (1.00 × 10⁻⁴ m² × 1.0 × 10⁻⁴ m) = 500 / (1.0 × 10⁻⁸) = 5.0 × 10¹⁰ Pa 4. **Result**: The Young’s modulus of the steel rod is 5.0 × 10¹⁰ Pascals (Pa). This is close to the typical value for steel (~2 × 10¹¹ Pa), indicating the rod behaves elastically within the measured range.