alleryTwo wires of same material, same length, but cross-sections 1 mm² and 2 mm². Same load: ratio of extensions:
A1:1
B2:1
C1:2
DΔL ∝ 1/A → ratio 2:1 (thinner stretches more)
Answer & Solution
Correct answer: D. ΔL ∝ 1/A → ratio 2:1 (thinner stretches more)
ΔL = FL/(AY).
At same F, L, Y: ΔL ∝ 1/A. So thinner wire (A = 1 mm²) extends 2× more than thicker (A = 2 mm²). Ratio 2:1.
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