If l is the length and A is area of cross-section of a wire having resistivity ρ, then
Resistance of the wire, R = ρ\(\frac{l}{A}\)
when its length is doubled, its area of cross-section reduces to half of its initial area of cross-section.
∴ New Resistance,
R1 = \(\frac{\rho2l}{A/2} = 4 \frac{\rho l}{A} = 4R\)
Given R = 1Ω
∴ R1 = 4 x 1 = 4Ω
