A copper wire has diameter 0.5mm and resistivity of 1.6×10⁻⁸ Ωm. What will be the length of this wire to make its resistance 10Ω?

Given, diameter, d = 0.5 mm
resistivity, ρ = 1.6×10⁻⁸ Ωm
Resistance, R = 10 Ω
Let the length of wire be l.
A = πd²/4
R = ρl/A = ρl/(πd²/4)
=> l = Rπd²/4ρ
l = 10 × 3.14 × (0.5 × 10⁻³)² / (4 × 1.6 × 10⁻⁸)
l = 122.7 m
R ∝ 1/d²
If the diameter is doubled, resistance will be one-fourth.
Hence, new resistance = 2.5Ω