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Asked in GSEB Board July 2015 · Resistance of a wire and stretching
Given: the length is increased by 100%, so ℓ' = 2ℓ; the volume of the wire does not change.
Idea: R = (ρℓ)/A, and constant volume means Aℓ = A'ℓ', so A' = A/2.
R' = (ρ(2ℓ))/(A/2) = 4 (ρℓ)/A = 4R.
Change = (R' - R)/R×100% = (4 - 1)×100% = 300%.
So the resistance increases by 300%.
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