A wire of cross sectional area A, modulus of elasticity 2 × 10¹¹ N m⁻² and length 2 m is stretched between two vertical rigid supports. When a mass of 2 kg is suspended at the middle it sags lower from its original position making angle θ = 1/100 radian on the points of support. The value of A i s _____ × 10⁻⁴ m² (consider x << L). (given : g = 10 m/s²)
Physics · Properties of solid and fluid · propertes of solid stress ,strain and modulus of elasticity
JEE Mains 2024 06 Apr Evening Shift

Question

A wire of cross sectional area A, modulus of elasticity 2 × 10¹¹ N m⁻² and length 2 m is stretched between two vertical rigid supports. When a mass of 2 kg is suspended at the middle it sags lower from its original position making angle θ = 1/100 radian on the points of support. The value of A i s _____ × 10⁻⁴ m² (consider x << L). (given : g = 10 m/s²)

Physics · Properties of solid and fluid · propertes of solid stress ,strain and modulus of elasticity

JEE Mains 2024 06 Apr Evening Shift

G1 Physics · Accepted Answer

Image For vertical equilibrium: 2T sinθ = 20 Using small-angle approximation: sinθ ≈ θ = 1/100 Therefore: 2T(1/100) = 20 T = 1000 N If each half has horizontal length L and sag x: tanθ ≈ x/L So: x = Lθ Increase in total length: ΔL(total) = 2√(L² + x²) - 2L For x << L: ΔL(total) ≈ x²/L Strain: = ΔL(total)/(2L) = x²/(2L²) = θ²/2 Young's modulus: Y = Stress/Strain = (T/A)/(θ²/2) Therefore: A = 2T/(Yθ²) = [2 × 1000]/[2 × 10¹¹ × (10⁻²)²] = 1 × 10⁻⁴ m² Hence: answer = 1