R=5.2±0.1 Ω, d=0.36±0.01 mm and L=0.800±0.002 m. Calculate ρ=LRA and its uncertainty, and suggest an improvement.
A=π(0.18×10−3)2=1.018×10−7 m2, so ρ=0.8005.2×1.018×10−7=6.6×10−7 Ω m.
Percentage uncertainties: R 1.9%; d 2.8%, which doubles to 5.6% for A; L 0.25%. The total is 7.7%.
Answer: ρ=(6.6±0.5)×10−7 Ω m.
Improvement: the diameter dominates. Measure d at several points along the wire, in two perpendicular directions at each point, and average. Check the micrometer’s zero error. A wire of larger diameter would also lower %d.