JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Current Electricity

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

____

Answer: 300

Solution

When the wire is stretched, its volume remains constant:

V^{'} = V \Rightarrow ℓ^{'} A^{'} = ℓ A

With

ℓ^{'} = 2 ℓ

A^{'} = \frac{A}{2}

Resistance is given by

R = ρ \frac{ℓ}{A}

. The new resistance:

R^{'} = ρ \frac{ℓ ^{'}}{A ^{'}} = ρ \frac{2 ℓ}{A /2} = 4 ρ \frac{ℓ}{A} = 4 R

Percentage increase:

\frac{R ^{'} - R}{R} \times 100 = \frac{4 R - R}{R} \times 100 = 300%

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Integer answer