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"The wire is stretched to twice its length. Find the new resistance." Most students answer 2R, and lose the mark. The wire also gets thinner, and that matters just as much. This post explains the volume argument, gives the rules for length and radius, and works four examples, including the small-stretch shortcut.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists "electrical resistance" and "electrical resistivity and conductivity" in Current Electricity, and the NEET (UG) 2026 syllabus lists electrical resistivity and conductivity. The relation R = ρl/A is equation 3.9 in the NCERT Class 12 Physics chapter, Current Electricity (2026-27 reprint). Check the documents for your year.
Why does stretching change resistance twice over?
NCERT gives R = ρl/A and says the resistivity ρ depends on the material of the conductor but not on its dimensions. So stretching leaves ρ alone and changes l and A.
When a wire is stretched or drawn thinner, the amount of metal does not change, so its volume V = Al stays the same. That links the two:
Stretch to n times the length: l' = nl, so A' = A/n.
New resistance: R' = ρ(nl)/(A/n) = n²R.
Writing A = V/l gives R = ρl²/V, so at fixed volume R ∝ l². Writing l = V/A gives R = ρV/A², and since A = πr², R ∝ 1/r⁴.
Worked example 1: a real copper wire
A copper wire is 1.0 m long with diameter 0.50 mm. Take ρ = 1.72 × 10⁻⁸ Ω m, the copper value in OpenStax College Physics 2e, Table 20.1.
Original resistance. r = 0.25 mm, A = π(0.25 × 10⁻³)² = 1.963 × 10⁻⁷ m². R = 1.72 × 10⁻⁸ × 1.0 / 1.963 × 10⁻⁷ ≈ 0.0876 Ω.
Stretched to 1.2 m. n = 1.2, so R' = 1.44 × 0.0876 ≈ 0.126 Ω.
Check directly. A' = A/1.2 = 1.636 × 10⁻⁷ m². R' = 1.72 × 10⁻⁸ × 1.2 / 1.636 × 10⁻⁷ ≈ 0.126 Ω. Same answer.
Worked example 2: drawn to half the diameter
A 3.0 Ω wire is drawn through a die until its diameter is half the original. Find the new resistance and length.
Radius halves, so R ∝ 1/r⁴ gives R' = 2⁴ × 3.0 = 16 × 3.0 = 48 Ω. Area falls to a quarter, so at constant volume the length becomes four times as long. Check with the length rule: n = 4, n² = 16. Both routes agree.
Worked example 3: a small stretch
A wire is stretched by 1 percent. By what percentage does its resistance change?
Exactly: R' = (1.01)²R = 1.0201R, an increase of 2.01 percent. For small stretches, use ΔR/R ≈ 2Δl/l, giving 2 percent. For a 2 percent stretch the exact value is 4.04 percent, and the shortcut gives 4 percent.
Worked example 4: same mass, different lengths
Two wires of the same material and the same mass have lengths l and 2l. Same mass and material means same volume, so R ∝ l². The ratio of resistances is 1² : 2² = 1 : 4.
Which mistakes cost marks?
Answering nR. You changed length but forgot area.
Changing ρ. Resistivity belongs to the material; stretching does not change it in these problems.
Confusing stretching with cutting. Cutting a wire into n equal pieces gives each piece R/n, with area unchanged.
Using diameter as radius. The 1/r⁴ rule works with either, as long as you take the ratio, but mixing them inside A = πr² gives a factor of 4 error.
Squaring the wrong way. Halving the radius increases R; it does not reduce it to R/16.
These problems also assume temperature is constant and the wire stays uniform.
A practice routine
Derive R' = n²R and R ∝ 1/r⁴ from V = Al, without looking.
Work out the factor for stretching to 3 times the length (9) and for reducing the radius by a third (about 5.06).
Do one same-mass question and one cutting question back to back, and say aloud what is fixed in each.
Read our posts on drift velocity and resistivity and the resistivity experiment, which use the same formula from other angles.
If you keep picking the wrong quantity to hold fixed, the Fermi Problem Library has a step-by-step solver that explains the reasoning, so you can see where each step comes from.
Tonight, solve Example 2 again for a diameter reduced to one-third.
