A steel rod of length 10 m, density 8000 kg/m³ and Young's modulus 2 × 10¹¹ Pa hangs vertically from its upper end. Take g = 10 m/s². If its elongation due to its own weight is N × 10⁻² mm, find N.

Question

G1 Physics · Accepted Answer

For a uniform rod hanging under its own weight, ΔL = [ρgL²] / [2Y]. Substitute: ρ = 8000 kg/m³, g = 10 m/s², L = 10 m, Y = 2×10¹¹ Pa. Therefore, ΔL = [8000×10×10²] / [2×2×10¹¹] = 2×10⁻⁵ m. Convert to millimetres: 2×10⁻⁵ m = 2×10⁻² mm. Hence, N = 2.