A uniform vertical rod has density 6000 kg/m³, cross-sectional area 10⁻⁴ m², length 10 m and Young's modulus 2 × 10¹¹ Pa. It hangs from its upper end under its own weight. Taking g = 10 m/s², the total strain energy stored is N × 10⁻⁴ J. Find N.
Physics · Properties of solid and fluid · Elastic Properties of Solids: Stress, Strain and Elastic Moduli
JEE Mains NEET JEE Advanced

Question

A uniform vertical rod has density 6000 kg/m³, cross-sectional area 10⁻⁴ m², length 10 m and Young's modulus 2 × 10¹¹ Pa. It hangs from its upper end under its own weight. Taking g = 10 m/s², the total strain energy stored is N × 10⁻⁴ J. Find N.

Physics · Properties of solid and fluid · Elastic Properties of Solids: Stress, Strain and Elastic Moduli

G1 Physics · Accepted Answer

At a section x from the top, T(x) = ρAg(L−x). Elemental strain energy is dU = [T(x)² dx] / [2AY]. Therefore, U = [ρ²Ag²/(2Y)] ∫₀ᴸ(L−x)²dx. Now, ∫₀ᴸ(L−x)²dx = L³/3. Hence, U = [ρ²Ag²L³] / [6Y]. Substitute: ρ = 6000, A = 10⁻⁴, g = 10, L = 10, Y = 2×10¹¹. Therefore, U = 3×10⁻⁴ J. Hence, N = 3.