In: Physics
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. Refer to Table 25.1 in the textbook for the resistivity values.
Part A
What will be the length of the gold wire?
Part B
What will be the length of the copper wire?
Part C
What will be the length of the aluminum wire?
Part D
Gold has a density of 1.93 × 104 kg/m3. What will be the mass of the gold wire?
Part E
If gold is currently worth $40 per gram, what is the cost of the gold wire?
We know that resistance of a cylindrical wire is given by:
R = rho*L/A
rho = resistivity of wire
L = length of wire & A = Cross-sectional area of wire
Part A
If value of resistivity are different than used, then let me know.
Resistivty of gold = 2.44*10^-8 ohm.m
A = Cross-sectional area = pi*d^2/4
R = resistance of wire = 5.00 ohm
d = diameter = 1.00 mm = 1.00*10^-3 m
So,
L = R*A/rho
L = 5.00*pi*(1.00*10^-3)^2/(4*2.44*10^-8)
L = 160.94 m = 161 m = length of gold wire
Part B.
Resistivty of copper = 1.72*10^-8 ohm.m
A = Cross-sectional area = pi*d^2/4
R = resistance of wire = 5.00 ohm
d = diameter = 1.00 mm = 1.00*10^-3 m
So,
L = R*A/rho
L = 5.00*pi*(1.00*10^-3)^2/(4*1.72*10^-8)
L = 228.31 m = 228 m = length of copper wire
Part C.
Resistivty of aluminum = 2.75*10^-8 ohm.m
A = Cross-sectional area = pi*d^2/4
R = resistance of wire = 5.00 ohm
d = diameter = 1.00 mm = 1.00*10^-3 m
So,
L = R*A/rho
L = 5.00*pi*(1.00*10^-3)^2/(4*2.75*10^-8)
L = 142.80 m = 143 m = length of aluminum wire
Part D.
Mass is given by:
Mass = Volume*density
Volume of wire = length*cross-sectional area = L*A = pi*d^2*L/4
density of gold = 1.93*10^4 kg/m^3
Mass = density*pi*d^2*L/4
Using given values:
Mass = 1.93*10^4*pi*(1.00*10^-3)^2*160.94/4
Mass = 2.43956 kg
Mass of gold wire = 2.44 kg
Part E.
Mass of 1 gram gold = $40
Mass of 1 kg gold = $40000
So mass of 2.44 kg gold = $40000*2.44 = $97600
Let me know if you've any query.