What mass of copper must be taken to make a wire with a resistance of 0.25 Ohm and a cross-sectional area of 90 mm?

Given:
R = 0.25 Ohm – the required resistance of the copper wire;
s = 90 mm2 (square millimeters) is the cross-sectional area of ​​the copper wire;
r = 0.017 Ohm * mm2 / m – copper resistivity;
ro = 8900 kg / m3 (kilogram per cubic meter) – copper density.
It is required to determine m (kilogram) – what mass of copper must be taken to make a wire with resistance R and section s.
Find the length of the copper wire:
l = R * s / r = 0.25 * 90 / 0.017 = 1324 meters (accurate to whole).
Then the volume of copper will be equal to:
V = l * s = 1324 * 90 * 10-6 = 0.12 m3 (where 10-6 is the conversion factor from mm2 to m2).
Then the mass of copper will be equal to:
m = V * ro = 0.12 * 8900 = 1068 kilograms.
Answer: to make a wire, you need 1068 kilograms of copper.