Question

You want to produce three 1.00-mmmm-diameter cylindrical wires, each with a resistance of 2.00 ΩΩ at room temperature. One wire is gold, one is copper, and one is aluminum. Refer to Table 25.1 in the textbook for the resistivity values.

C.) What will be the length of the aluminum wire? Include units

D.) Gold has a density of 1.93 ×× 1044 kg/m3kg/m3. What will be the mass of the gold wire? Include Units

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

Answer 1

(c)

R = pL / A

where p is resistivity

so,

L = RA / p

for Aluminium

L = 2 * pi * 0.5e-3² / 2.77e-8

L = 56.7 m ( length of aluminium wire)

___________________________

(d)

for gold

L = 2 * pi * 0.5e-3² / 2.44e-8

L = 64.37 m

so,

m = PAL

where P is density of gold

m = 1.93e4 * pi * 0.5e-3² * 64.37

m = 0.975 kg

______________________

(d)

m = 0.975 kg = 975 grams

1 grams = $ 40

so,

975 grams = $ 39033 ( very expensive)