A composite rod has three segments connected in series. Segment 1 has length 1 m, area 2 cm² and Young's modulus 2 × 10¹¹ Pa. Segment 2 has length 2 m, area 1 cm² and Young's modulus 1 × 10¹¹ Pa. Segment 3 has length 1 m, area 2 cm² and Young's modulus 1 × 10¹¹ Pa. A tensile force of 40 kN is applied. If the total extension is N mm, find N.
Physics · Properties of solid and fluid · properties of solid stress ,strain and modulus of elasticity
JEE Mains NEET JEE Advanced

Question

A composite rod has three segments connected in series. Segment 1 has length 1 m, area 2 cm² and Young's modulus 2 × 10¹¹ Pa. Segment 2 has length 2 m, area 1 cm² and Young's modulus 1 × 10¹¹ Pa. Segment 3 has length 1 m, area 2 cm² and Young's modulus 1 × 10¹¹ Pa. A tensile force of 40 kN is applied. If the total extension is N mm, find N.

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

G1 Physics · Accepted Answer

For rods in series, the same force acts through every segment. Total extension is ΔL = Σ [FLᵢ / (AᵢYᵢ)]. Segment 1: ΔL₁ = [4×10⁴×1] / [(2×10⁻⁴)(2×10¹¹)] = 1×10⁻³ m = 1 mm. Segment 2: ΔL₂ = [4×10⁴×2] / [(1×10⁻⁴)(1×10¹¹)] = 8×10⁻³ m = 8 mm. Segment 3: ΔL₃ = [4×10⁴×1] / [(2×10⁻⁴)(1×10¹¹)] = 2×10⁻³ m = 2 mm. Therefore, ΔL = 1+8+2 = 11 mm. Hence, N = 11.