JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Current Electricity

An aluminium wire is stretched to make its length, 0.4% larger. Then percentage change in resistance is:

Answer: (C)

Solution

For a wire, resistance

R = \frac{ρ ℓ}{A}

. When stretched, volume remains constant:

ℓ A = k

(constant). Taking differentials:

\frac{Δ ℓ}{ℓ} + \frac{Δ A}{A} = 0

, so

\frac{Δ A}{A} = - \frac{Δ ℓ}{ℓ}

. The fractional change in resistance is

\frac{Δ R}{R} = \frac{Δ ℓ}{ℓ} - \frac{Δ A}{A} = \frac{Δ ℓ}{ℓ} + \frac{Δ ℓ}{ℓ} = 2 \frac{Δ ℓ}{ℓ}

. Given

\frac{Δ ℓ}{ℓ} = 0.4%

, we get

\frac{Δ R}{R} = 2 \times 0.4% = 0.8%

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Single correct option