A wire of cross sectional area A, modulus of elasticity 2 × 10¹¹ Nm⁻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 is ________ × 10⁻⁴ m² (consider x<<L ).
(given : g=10 m / s²)

Solution

Physics sir · 4 steps

Get the tension from the force balance at the mass, then get the strain from the geometry of the sag.

1. Tension from balance. The total length is 2L = 2 m, so each half has L = 1 m. At the midpoint, the two halves pull up with 2T sin θ = mg. For small θ, sin θ ≈ θ, so T = mg/2θ = 20/0.02 = 1000 N.
2. Strain from geometry. Each half stretches from L to √(L² + x²), where x = Lθ is the sag. Because x ≪ L, the extension is √(L² + x²) - L ≈ x²/2L.

   strain = (x²/2L)/L = θ²/2 = (0.01)²/2 = 5 × 10⁻⁵

3. Use Young's modulus. Y = (T/A)/strain gives A = T/(Y × strain) = 1000/(2 × 10¹¹ × 5 × 10⁻⁵) = 1000/10⁷ = 10⁻⁴ m².
4. So. A = 1 × 10⁻⁴ m², and the blank is 1.

The common slip: 2. Balancing the weight against only one half of the wire, T sin θ = mg, gives T = 2000 N and A = 2. Both halves pull up, so the balance is 2T sin θ = mg.

The idea to keep. In a sagging wire, a tiny angle needs a large tension, T = mg/2θ, while the strain is only second order, θ²/2. Get the force from the balance and the strain from the geometry.

Answer from the NTA final answer key. Solution by PrepWiser.