Question

A wire is hung between two towers and has a length of 208 m. A current of 154 A exists in the wire, and the potential difference between the ends of the wire is 0.245 V. The density of the wire material is 3610 kg/m3 and its resistivity is 4.23 x 10-8Ω·m. Find the mass of the wire.

Answer 1

Given Information:

Length of wire = L = 208 m

Current in wire = I = 154 A

Voltage across wire = V = 0.245 V

Resistivity of wire= ρ = 4.23x10⁻⁸ Ω.m

Density of wire = d = 3610 kg/m³

Required Information:

Mass of wire = m = ?

Answer:

Mass of wire = 4152.36 kg

Step-by-step explanation:

The mass of wire can be found using

m = dAL

where d is the density of wire, A is the area of wire and L is the length of wire.

First we need to find area A, we know that resistivity of wire is given by

ρ = RA/L

A = ρL/R

First we need to find R, we know know that resistance is given by

R = V/I

R = 0.245/154

R = 0.00159 Ω

So the area becomes,

A = ρL/R

A = (4.23x10⁻⁸*208)/0.00159

A = 0.00553 m²

Finally, the mass of wire is

m = dAL

m = 3610*0.00553*208

m = 4152.36 kg

Therefore, the mass of wire is 4152.36 kg

Answer 2

Mass = 4152kg

Step-by-step explanation:

Given

L = 208m

I = 154A

V = 0.245V

Density = 3610 kg/m3

ρ = 4.23 x 10-8Ω·m = resistivity of wire

Resistance R = ρL/ A

R = voltage / current = V/I = 0.245/154 = 1.59×10-³ohms

1.59×10-³ = 4.23 x 10-⁸×208/A

Rearranging,

A = 4.23 x 10-⁸×208/1.59×10-³

A = 5.53×10-³m²

Mass = density × volume

Volume = L×A = 208×5.53×10-³m³= 1.15m³

Mass = 3610×1.15 = 4152kg