Question

You need to produce a set of cylindrical copper wires 3.10 m long that will have a resistance of 0.110 Ω each.

What will be the mass of each of these wires? m = ___ g

Answer 1

first, this formula : R=ρL/S
where R is resistance in ohms Ω, S is the cross sectional surface in m², L is the length in m, ρ is the resistivity of the material (copper in this case) in Ω⋅m.
for copper : ρ=1.68×10⁻⁸ Ω⋅m (googled it)
and we have : V=SL , and m=dV ; whre V is the volume in m³, m is the mass in kg, d the density in kg/m³, and S & L are above.
you get : m=dSL
then : S=m/Ld (we need S because S is the unknown in the question)
in the original formula then : R=ρL/(m/Ld)
R=ρL²d/m
then : m=ρL²d/R
=1.68×10⁻⁸ Ω⋅m * (3.1m)²*(8960 kg/m³)/0.11Ω ( you can google the copper density as well)
≈ 0.0132 kg = 13.2 g