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Resistance of wire after drawing

Short answer

Volume constant: l2/l1 = (d1/d2)². R ∝ l/A ∝ (d1/d2)⁴. R2 = 0.031 × (11.7/5)⁴ ≈ 0.93 Ω.

A 1km length of wire having a diameter of 11.7mm and a resistance 0.031Ω is drawn so that its diameter becomes 5mm. What does its resistance become?

When a wire is drawn out, its volume stays constant, so length rises as area falls. Resistance R = ρl/A.

Step 1: Area and length change

Volume constant: A₁l₁ = A₂l₂

A ∝ d², so A₂/A₁ = (5/11.7)² = 0.1826

l₂/l₁ = A₁/A₂ = (11.7/5)² = 5.476

l₂ = 1000 × 5.476 = 5476 m (new length)

Step 2: New resistance

ρ is unchanged, so

R₂/R₁ = (l₂/l₁) × (A₁/A₂) = (11.7/5)² × (11.7/5)² = (11.7/5)⁴ = 29.98

R₂ = 0.031 × 29.98 = 0.929 Ω

Answer: New resistance ≈ 0.93 Ω (about 30 times larger)

✓ Answer checked by Claude Opus 5.5 (high effort)

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