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 :

Solution

Physics sir · 3 steps

Doubling length and halving area each double R, so R becomes four times and the current drops to a quarter.

1. Original wire. Resistance is R = (ρ l)/A, so with voltage V across it the current is I = V/R = VA/(ρ l).
2. New wire. Same material, so ρ stays the same. The length becomes 2l and the area becomes A/2.

   R' = (ρ (2l))/(A/2) = 4 (ρ l)/A = 4R

3. So. With the same V, the current is I' = V/4R = 1/4VA/(ρ l), which is option (c).

The idea to keep. For a wire, R ∝ l/A. Track how each change scales R and multiply the factors. At fixed V, the current scales the opposite way.

Answer from the NTA final answer key. Solution by PrepWiser.