Last Activity: 10 Years ago
Sol. ℓ ‘ = 2 ℓ volume of the wire remains constant. A ℓ = A’ ℓ ‘ ⇒ A ℓ = A’ * 2 ℓ ⇒ A’ = A/2 f = Specific resistance R = fℓ/A : R’ = fℓ’/A’ 100 Ω = f2ℓ/A / 2 = 4fℓ/A = 4R ⇒ 4 * 100 Ω = 400 Ω
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Last Activity: 3 Years ago