A steel wire of length 3.2 m(Yₛ=2.0 × 10¹¹ Nm⁻²) and a copper wire of length 4.4 m(Y_c=1.1 × 10¹¹ Nm⁻²), both of radius 1.4 mm are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm. The load applied, in Newton, will be: ( Given π=22/7)

Solution

Physics sir · 4 steps

The wires are joined end to end, so the same load pulls both; add their stretches and solve for F.

1. Same force, add stretches. In series, each wire carries the full load F, and both have the same area A. Each wire stretches by Δ L = FL/AY, and the two stretches add up to the total.

   Δ Lₙₑₜ = F/A(Lₛ/Yₛ + L_c/Y_c)

2. Area and bracket. A = π r² = 22/7 × (1.4 × 10⁻³)² = 6.16 × 10⁻⁶ m². The bracket is 3.2/(2.0 × 10¹¹) + 4.4/(1.1 × 10¹¹) = 1.6 × 10⁻¹¹ + 4.0 × 10⁻¹¹ = 5.6 × 10⁻¹¹ m³ N⁻¹.
3. Solve for F. F = (Δ L × A)/(5.6 × 10⁻¹¹) = (1.4 × 10⁻³ × 6.16 × 10⁻⁶)/(5.6 × 10⁻¹¹) = (8.624 × 10⁻⁹)/(5.6 × 10⁻¹¹) = 154 N.
4. So. The load applied is 154 N, option (d).

The idea to keep. Wires joined end to end behave like springs in series: the same force acts on each, and the elongations add. Each stretch is FL/AY.

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