JEE Main 2021 · 26 February · evening · Single correct · +4 / −1
A wire of 1Ω has a length of 1 m. It is stretched till its length increases by 25%. The percentage change in resistance to the nearest integer is :
- (a) 76%
- (b) 12.5%
- (c) 25%
- (d) 56%
Answer and solution
Answer (d) 56%
Solution
Physics sir · 3 steps
Stretching keeps the volume fixed, so the wire gets thinner too, and that makes R grow as L².
- Volume stays fixed. Stretching does not add or remove metal, so V = AL is constant. If the length goes up, the cross-section area must go down: A = V/L.
R = (ρ L)/A = (ρ L²)/V ⇒ R ∝ L²
- Put in numbers. A 25% increase gives L' = 1.25L. So R'/R = (1.25)² = 1.5625, and R' = 1.5625 Ω.
- So. The resistance rises by 56.25%, which is 56% to the nearest integer. That is option (d).
The idea to keep. When a wire is stretched, its volume does not change, so R ∝ L². Square the length ratio. For small stretches, the percentage change in R is about twice the percentage change in L.
Answer from the NTA final answer key. Solution by PrepWiser.