Question

You want to produce three 1.00-mm-diameter cylindrical wires, each with a resistance of 5.00 Ω at room temperature. One wire is gold, one is copper, and one is aluminum.

Part A: What will be the length of the gold wire? Express your answer with appropriate units.

Part B: What will be the length of the copper wire? Express your answer with appropriate units.

Part C: What will be the length of the aluminum wire? Express your answer with appropriate units.

Part D: Gold has a density of 1.93x10^4 kg/m^3. What will be the mass of the gold wire? Express your answer with the appropriate units.

Part E: If gold is currently worth $40 per gram, what is the cost of the gold wire?

Answer 1

here,

diameter of wire , d = 1 mm = 0.001 m

radius , r = d/2 = 5 * 10^-4 m

a)

resistivity of gold , p1 = 2.44 * 10"^-8 ohm.m

let the length of wire be l1

resistance , R = p1 *l1 /(area)

5 = 2.44 * 10^-8 * l1 /(pi * (5 * 10^-4)^2)

solving for l1

l1 = 160.9 m

the length of gold wire is 160.9 m

b)

resistivity of copper , p2 = 21.72 * 10"^-8 ohm.m

let the length of wire be l2

resistance , R = p2 *l2 /(area)

5 = 1.72 * 10^-8 * l2 /(pi * (5 * 10^-4)^2)

solving for l2

l2 = 228.2 m

the length of copper wire is 228.2 m

c)

resistivity of alumunium , p3 = 2.65 * 10"^-8 ohm.m

let the length of wire be l3

resistance , R = p3 *l3 /(area)

5 = 2.65 * 10^-8 * l3 /(pi * (5 * 10^-4)^2)

solving for l3

l3 = 148.1 m

the length of alumunium wire is 148.1 m

d)

the mass of gold wire , m = density of gold * volume

m = 1.93 * 10^4 * ( pi * (5 * 10^-4)^2 * 160.9) kg

m = 2.43 kg

e)

m = 2.43 kg = 2430 g

the cost of gold wire , C = 2430 * 40 dollars

C = 9.75 * 10^4 dollars