Question

Class 10PhysicsElectricity

Resistance of second copper wire (same mass, double length)

Verified Answer

Same mass + same material → same volume: A₁L = A₂(2L) → A₂ = A₁/2

R₂/R₁ = (ρ · 2L / A₂) / (ρ · L / A₁) = (2L · A₁) / (L · A₁/2) = 4

R₂ = 4 × 0.5 = 2 Ω

(Length doubles AND area halves → resistance quadruples)