JEE Main 2022 · 26 June · morning · Single correct · +4 / −1
An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is :
- (a) 0.4%
- (b) 0.2%
- (c) 0.8%
- (d) 0.6%
Answer and solution
Answer (c) 0.8%
Solution
Physics sir · 3 steps
Stretching keeps the volume fixed, so the area shrinks as length grows: R goes as length squared.
- Volume stays fixed. Stretching does not add or remove metal, so V = lA stays constant. That means A = V/l, and the area falls as the length rises.
R = (ρ l)/A = (ρ l²)/V ⇒ (Δ R)/R = 2 (Δ l)/l
- Put in numbers. With ρ and V constant, R ∝ l². For a small change, (Δ R)/R = 2 × 0.4% = 0.8%. The exact value is (1.004)² - 1 = 0.008016, which is still about 0.8%.
- So. The resistance increases by 0.8%, which is option (c).
The idea to keep. When a wire is stretched, use constant volume. Then R ∝ l², and for small changes the percentage change in R is twice the percentage change in length.
Answer from the NTA final answer key. Solution by PrepWiser.