The length of a given cylindrical wire is increased by 100% . Due to the consequent decrease in diameter the change in the resistance of the wire will be

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L 1 ​ = 2 l or ( π r 2 l ) = ( π r 2 2 ​ ) ( 2 l ) ⇒ r 2 ​ = 2 ​ r ​ ; R = ρ π 2 1 ​ R new ​ = ( ρ ) ( π ) ( r / 2 ​ ) 2 21 ​ = ( π ) r 2 ( ρ ) 4 l ​ = 4 × R ∴ Δ R = 4 R − R = 3 R R Δ R ​ % = R 3 R ​ × 100 = 300%

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