A wire of resistance R and radius r is stretched till its radius became r / 2. If new resistance of the stretched wire is x R, then value of x is ________.

Solution

Physics sir · 3 steps

Stretching keeps the volume fixed, so halving the radius quarters the area and makes the wire 4 times longer.

1. Volume stays fixed. Stretching only reshapes the wire. No metal is added or removed, so V = Aℓ stays the same. If the area drops, the length must grow by the same factor.
2. New area and length. Area goes as r², so radius r/2 gives area A/4. To keep Aℓ fixed, the length becomes 4ℓ.

   R' = ρ4ℓ/(A/4) = 16 ρℓ/A = 16R

3. So. The new resistance is 16R, so x = 16.

The common slip: 4. Using only the smaller area and forgetting that the length also grows gives R' = 4R, so x = 4 instead of 16.

The idea to keep. When a wire is stretched at constant volume, R ∝ ℓ² ∝ 1/r⁴. Shrinking the radius by a factor n raises the resistance by n⁴.

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