TS File:JEE Main:Elastic Moduli: Set-10

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🔬 Part 1: Problem Solving

1. A steel wire of length 2.0 m and cross-sectional area 1.5 × 10⁻⁶ m² is stretched by a force of 300 N. If the extension is 0.8 mm, calculate Young's modulus for steel.

2. A copper cube with each side 10 cm is subjected to a uniform pressure of 5 × 10⁶ Pa. If the bulk modulus of copper is 1.4 × 10¹¹ Pa, calculate the change in volume of the cube.

3. A cylindrical steel shaft of diameter 50 mm and length 1.2 m is twisted by a torque of 800 N⋅m. If the angle of twist is 2.5°, calculate the modulus of rigidity for steel. (Use the formula: θ = TL/GJ, where J = πd⁴/32)

📊 Part 2: Analysis Questions

4. A rectangular aluminium beam (20 cm × 15 cm × 3 m) supports a load that causes a stress of 80 MPa. If Young's modulus for aluminium is 70 GPa, determine:
a) The strain in the beam
b) The change in length of the beam
c) The elastic energy stored per unit volume

5. A solid sphere of radius 8 cm made of brass (Bulk modulus = 6.1 × 10¹⁰ Pa) is submerged to a depth where the pressure increases by 2 × 10⁶ Pa. Calculate:
a) The volumetric strain
b) The change in radius
c) The percentage change in density

🔑 Answer Key

1. E = FL₀/(A × ΔL) = (300 × 2.0)/(1.5 × 10⁻⁶ × 0.8 × 10⁻³) = 5.0 × 10¹¹ Pa

2. ΔV/V₀ = -P/K = -(5 × 10⁶)/(1.4 × 10¹¹) = -3.57 × 10⁻⁵
ΔV = -3.57 × 10⁻⁵ × (0.1)³ = -3.57 × 10⁻⁸ m³ = -0.0357 cm³

3. J = π(0.05)⁴/32 = 6.14 × 10⁻⁷ m⁴; θ = 2.5° = 0.0436 rad
G = TL/(θJ) = (800 × 1.2)/(0.0436 × 6.14 × 10⁻⁷) = 3.58 × 10¹⁰ Pa

4. a) ε = σ/E = 80 × 10⁶/(70 × 10⁹) = 1.14 × 10⁻³
b) ΔL = ε × L = 1.14 × 10⁻³ × 3 = 3.43 mm
c) U = σ²/(2E) = (80 × 10⁶)²/(2 × 70 × 10⁹) = 45.7 kJ/m³

5. a) ΔV/V₀ = -P/K = -(2 × 10⁶)/(6.1 × 10¹⁰) = -3.28 × 10⁻⁵
b) Δr/r = (1/3)(ΔV/V₀) = -1.09 × 10⁻⁵; Δr = -8.75 × 10⁻⁴ mm
c) Δρ/ρ = -ΔV/V₀ = 3.28 × 10⁻⁵ = 0.00328%