Two metallic wires P and Q have same volume and are made up of same material. If their area of cross sections are in the ratio 4: 1 and force F₁ is applied to P, an extension of Δ l is produced. The force which is required to produce same extension in Q is F₂.
The value of F₁/F₂ is _________.

Solution

Physics sir · 3 steps

Same volume ties length to area: the thicker wire is also shorter, so both effects favour P.

1. Force for an extension. From Y = (F/A)/(Δ l/l), the force needed for an extension Δ l is F = (YA Δ l)/l. Here Y and Δ l are the same for both wires, so F ∝ A/l.
2. Use equal volumes. Equal volume means A₁ l₁ = A₂ l₂, so l₂/l₁ = A₁/A₂ = 4. Wire P is 4 times thicker, so it is also 4 times shorter.

   F₁/F₂ = A₁/A₂ × l₂/l₁ = (A₁/A₂)² = 4²

3. So. F₁/F₂ = 16, so the answer is 16.

The common slip: 4. Comparing only the areas and taking the lengths as equal gives 4, not 16. Equal volume forces the thicker wire to be shorter.

The idea to keep. When two wires share the same volume, l ∝ 1/A. So any quantity going as A/l, like force for a fixed extension, scales as A².

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