A wire 6.90 m long with diameter of 2.15 mm has a resistance of 0.0320 Ω. Find the resistivity of the material of the wire. rho = Ω⋅m

1.68 x 10⁻⁸Ωm

The resistance (R) of a wire is related to its length(L), its material resistivity(ρ) and its crossectional area(A) as follows;

R = ρL/A               (i)

Where;

A = πd² / 4              [where d = diameter of the wire]

From the question;

L = 6.90m

d = 2.15mm = 0.00215m

R = 0.0320Ω

First calculate the crossectional area (A) of the wire as follows;

A = πd² / 4      

[Take π = 3.142]

d = 0.00215m

∴ A = 3.142 x (0.00215)² / 4

∴ A = 0.000003631m²

Now, substitute the values of A, L, and R into equation (i) as follows;

R = ρL/A

0.0320 = ρ x 6.90 / 0.000003631

0.0320 = 1900302.95 x ρ

Solve for ρ;

=> ρ = 0.0320 / 1900302.95

=> ρ = 1.68 x 10⁻⁸Ωm

Therefore, the resistivity of the material of the wire is 1.68 x 10⁻⁸Ωm

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b) phosphorus pentoxide