Two metallic wires A and B are made using copper. The radius of wire A

Two metallic wires A and B are made using copper. The radius of wire A is r while its length is l. A dc voltage V is applied across the wire A, causing power dissipation, P. The radius of wire B is 2r and its length is 2l and the same dc voltage V is applied across it causing power dissipation, P₁. Which one of the following is the correct relationship between P and P₁?

P = 2P₁
P = P₁/2
P = 4P₁
P = P₁
This question was previously asked in
UPSC NDA-1 – 2019
The correct relationship is P = P₁/2.
The resistance of a wire is given by R = ρ * (l / A), where ρ is resistivity, l is length, and A is the cross-sectional area (A = πr²). Power dissipated is P = V²/R.
For wire A: R_A = ρ * (l / (πr²)). Power P = V² / R_A = V² / [ρ * (l / (πr²))] = (V²πr²) / (ρl).
For wire B: radius is 2r, length is 2l. R_B = ρ * (2l / (π(2r)²)) = ρ * (2l / (4πr²)) = ρ * (l / (2πr²)).
Power P₁ = V² / R_B = V² / [ρ * (l / (2πr²))] = (2V²πr²) / (ρl).
Comparing P and P₁: P₁ = 2 * [(V²πr²) / (ρl)].
Since P = (V²πr²) / (ρl), we have P₁ = 2P. This relationship can be rewritten as P = P₁/2.
The resistance of wire B is half the resistance of wire A (R_B = (ρl / (2πr²)) compared to R_A = (ρl / (πr²))). Since power is inversely proportional to resistance for a constant voltage (P = V²/R), lower resistance leads to higher power dissipation. Therefore, wire B dissipates more power than wire A when the same voltage is applied.