A wire of length L and radius r is clamped rigidly at one end. When the other end of the wire is pulled by a force F, its length increases by 5 cm. Another wire of the same material of length 4 L and radius 4 r is pulled by a force 4 F under same conditions. The increase in length of this wire is __________________ cm.

Solution

Physics sir · 3 steps

Same material means same Y, so write Δ L = FL/(π r² Y) and see how each factor scales.

1. Extension formula. From Y = (F/A)/(Δ L/L), the extension of a wire is Δ L = FL/AY with A = π r². Both wires are of the same material, so Y is the same for both.

   (Δ L₂)/(Δ L₁) = F₂/F₁ × L₂/L₁ × (r₁/r₂)² = 4 × 4 × 1/16

2. Put in numbers. Force and length each grow 4 times, which multiplies Δ L by 16. The radius grows 4 times, so the area grows 4² = 16 times, which divides Δ L by 16. The ratio is 1.
3. So. Δ L₂ = 1 × 5 cm, so the increase in length is 5 cm.

The common slip: 20. Using radius instead of area, as if Δ L ∝ FL/r, gives 5 × 4 × 4 ÷ 4 = 20 cm, not 5.

The idea to keep. In elasticity ratio questions, the radius enters through the area, so it counts squared. Write Δ L ∝ FL/r² and multiply the factors.

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