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?

The correct answer is B, P = P₁/2. Wire B has half the resistance of wire A, and at the same voltage power is V²/R, so B dissipates twice the power: P₁ = 2P, which means P = P₁/2. Resistance of a wire is R = ρl/A, where ρ is the resistivity of the material and A = πr² is the cross-section. For wire A, R = ρl/πr². For wire B the length doubles to 2l but the area becomes π(2r)² = 4πr², so its resistance is ρ(2l)/(4πr²) = R/2. Both wires are copper, so ρ is the same. With the same voltage V, halving the resistance doubles the current and doubles the power. Option A is wrong because P = 2P₁ would need wire B to have twice the resistance, not half. Option C is wrong because P = 4P₁ would need B's resistance to be four times A's. Option D is wrong because equal power would need equal resistance, but the thicker wire B has less resistance. Exam tip: R ∝ l/r²; at fixed voltage, P = V²/R, so lower resistance means more power.