JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Current Electricity

A conducting wire of length '

ℓ

', area of cross-section A and electric resistivity

ρ

is connected between the terminals of a battery. A potential difference V is developed between its ends, causing an electric current.
If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be :

(A) $4 \frac{VA}{ρ ℓ}$
(B) $\frac{1}{4} \frac{ρ ℓ}{VA}$
(C) $\frac{3}{4} \frac{VA}{ρ ℓ}$
(D) $\frac{1}{4} \frac{VA}{ρ ℓ}$ ✓ Correct

Answer: (D)

Solution

The resistance of the wire is

R = \frac{ρ ℓ}{A}

, so at a potential difference

V

it carries

I = \frac{V A}{ρ ℓ}

.
Doubling the length and halving the cross-section gives

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

The same potential difference across four times the resistance drives a quarter of the current, so the resulting current is

\frac{1}{4} \frac{V A}{ρ ℓ}

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Current Electricity →

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Current Electricity

Solution

More JEE Main questions on Current Electricity