If a copper wire is stretched to increase its length by 20%. The percentage increase in resistance of the wire is __________%.

Solution

Physics sir · 4 steps

Stretching keeps the volume fixed, so the wire thins as it lengthens and R grows as length squared.

1. Volume stays fixed. Stretching does not add or remove copper, so V = Aℓ stays constant. If the length becomes 1.2ℓ, the area must drop to A/1.2.
2. Write R with volume. Put A = V/ℓ into R = ρℓ/A. With ρ and V both fixed, the resistance depends only on the length squared.

   R = (ρ ℓ²)/V ⇒ R₂/R₁ = (ℓ₂/ℓ₁)²

3. Put in numbers. R₂/R₁ = (1.2)² = 1.44, so the new resistance is 1.44R₁.
4. So. The resistance goes up by 0.44R₁, which is an increase of 44%.

The common slip: 40. Using the small-change rule Δ R/R = 2 Δℓ/ℓ gives 40%. That shortcut drops the (0.2)² term, and you need it here because 20% is not small.

The idea to keep. When a wire is stretched, its volume is fixed, so R ∝ ℓ². For small changes this gives Δ R/R ≈ 2 Δℓ/ℓ, but use the exact square when the change is 20% or more.

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