Answer: 2.6*10^-8 Ωm
Step-by-step explanation:
given
Length of the wire, l = 12.2 m
width of the wire, w = 1.8 mm
thickness of the wire, T = 0.11 mm
Potential difference of the battery, v = 12 V
Current in the battery, I = 7.5 A
Remember, Ohms law says, V = IR.
So that, R = V/I
R = 12/7.5 = 1.6 Ω
Resistivity if a material is
ρ = RA/l
ρ = [1.6 * 1.8*10^-3 * 0.11*10^-3] / 12.2
ρ = (3.168*10^-7) / 12.2
ρ = 2.596*10^-8
Therefore, the resistivity of the wire is 2.60*10^-8 Ωm