The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be _____%.

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answer is 300.

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Detailed Solution

$\begin{aligned} V^{'} = V \\ ℓ^{'} A = ℓA \\ 2 {ℓA}^{'} = ℓA \\ A^{'} = \frac{A}{2} \\ R = ρ \frac{ℓ}{A} \dots (i) \\ ℓ^{'} = 2 ℓ \\ A^{'} = \frac{A}{2} \\ R^{'} = \frac{{ρℓ}^{'}}{A} = \frac{ρ 2 ℓ}{A} \\ R^{'} = \frac{4 ρℓ}{A} \end{aligned}$

R' = 4R from equation (i)
% increase in resistance

$\begin{aligned} = \frac{R^{'} - R}{R} \times 100 = \frac{4 R - R}{R} \times 100 \\ = 300 % \end{aligned}$

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