Question

You want to produce three 2.00-mm-diametercylindrical wires, each with a resistance of 1.00 Ω at room temperature. One wire is gold (rhog=2.44×10^−8Ω⋅m), one is copper (rhoc=1.72×10^−8Ω⋅m), and one is aluminum (rhoa=2.75×10−8Ω⋅m).
Required:

a. What will be the length of the gold wire?
b. What will be the length of the copper wire?
c. What will be the length of the aluminum wire?
d. Gold has a density of 1.93 × 10^4 kg/m^3. What will be the mass of the gold wire?
e. If gold is currently worth $40 per gram, what is the cost of the gold wire?

Answer 1

Answer:

(a) L =  128.75 m

(b) L =  182.56 m

(c) L =  114.28 m

(d) Mass of Gold = 7.68 kg = 7680 gram

(e) Cost of Gold Wire = $ 307040

Explanation:

The resistance of the wire is given as:

R = ρL/A

where,

R = Resistance

ρ = resistivity

L = Length

A = cross-sectional area

(a)

For Gold Wire:

ρ = 2.44 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)

L =  128.75 m

(b)

For Copper Wire:

ρ = 1.72 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(1.72 x 10⁻⁸ Ω.m)

L =  182.56 m

(c)

For Aluminum Wire:

ρ = 2.75 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.75 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)

L =  114.28 m

(d)

Density = Mass/Volume

Mass = (Density)(Volume)

Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³

Therefore,

Mass of Gold = (1.9 x 10⁴ kg/m³)(4.04 x 10⁻⁴ m³)

Mass of Gold = 7.68 kg = 7680 gram

(e)

Cost of Gold Wire = (Unit Price of Gold)(Mass of Gold)

Cost of Gold Wire = ($ 40/gram)(7680 grams)

Cost of Gold Wire = $ 307040