Using the same copper, what diameter must the wire have if its resistance is 9.0 Ω and its length is determined by the mass? (density 8900 kg/m^3, copper resistivity 1.68×10^-8 Ω·m)

• A. 1.35×10^-3 m
• B. 1.35×10^-5 m
• C. 1.35×10^-6 m
• D. 1.35×10^-4 m

Answer: (D)

Resistance in a uniform conductor scales with resistivity, length, and cross-sectional area as R = ρ L / A. When the copper amount is fixed by mass, the length and area are linked through m = ρ_density × A × L, so L = m/(ρ_density A). Substituting into the resistance expression gives R = ρ m /(ρ_density A^2). Solving for the cross-sectional area yields A = sqrt(ρ m /(ρ_density R)). For a circular wire, A = π d^2/4, so d = sqrt(4A/π).

Plugging in ρ = 1.68×10^-8 Ω·m, ρ_density = 8900 kg/m^3, R = 9.0 Ω, and m = 0.001 kg (1 g) gives A ≈ 1.45×10^-8 m^2, and d ≈ sqrt(4×1.45×10^-8 / π) ≈ 1.36×10^-4 m. So the diameter is about 1.35×10^-4 m, which matches the given result.