Answer:
The ratio reisistivity of the wires = 15.72 : 1 : 31.44
Explanation:
ρ = RA/L................. Equation 1
Where ρ = resistivity of the wire, R = resistance, A = cross sectional area, L = length of the wire.
Note: All the resistance are the same.
For the wire 1,
L = 0.5 m, A =πd²/4, d = 0.5 cm = 0.005 m, π = 3.143
A = 3.143(0.005)²/4 = 0.0000196 m²
ρ₁ = R(0.0000196)/0.5
ρ₁ = 0.0000393R
For Wire 2,
L = 2 m, d = 0.25 cm = 0.0025 m
A =πd²/4 = 3.143(0.0025)²/4
A = 0.0000049 m²
ρ₂ = R(0.0000049)/2
ρ₂ = 0.0000025R
For Wire 3
L = 1 m, d = 1 cm = 0.01 m
A = πd²/4 = 3.143(0.01)²/4
A = 0.0000786 m²
ρ₃ = R(0.0000786)/1
ρ₃ = 0.0000786R
Therefore,
ρ₁ : ρ₂ : ρ₃ = 0.0000393R : 0.0000025R : 0.0000786R
ρ₁ : ρ₂ : ρ₃ = 3.93 : 0.25 : 7.86
ρ₁ : ρ₂ : ρ₃ = 15.72 : 1 : 31.44
Thus the ratio reisistivity of the wires = 15.72 : 1 : 31.44
