Answer:
2×10⁻⁷ Ωm
Explanation:
From the question,
R = 4ρL/πd².................... Equation 1
Where R = Resistance of the wire, ρ = resistivity of the wire, L = length of the wire, d = diameter of the wire.
Therefore,
ρ = Rπd²/4L............. Equation 2
Given: R = 1 ohm, ρ = 1×10⁻⁷Ωm
1×10⁻⁷ = πd²/4L.................... Equation 3
If the length and the diameter are doubled, and the resistance remaining constant
ρ' = π(2d)²/4(2L)
Where ρ' = new resistivity
ρ' = 4πd²/8L
ρ' = 2πd²/4L = 2ρ
ρ' = 2(1×10⁻⁷)
ρ' = 2×10⁻⁷ Ωm
