Step 1: A steel ball (radius r = 1.00 mm, density ρ_s = 7800 kg/m³) falls through glycerine (ρ_l = 1260 kg/m³) and reaches a terminal velocity v₀ = 2.20 cm/s between two marks 40 cm apart. Find η. Step 2: Use η = 2r²g(ρ_s − ρ_l)/(9v₀). Step 3: r² = (1.00×10⁻³)² = 1.00×10⁻⁶ m²; ρ_s − ρ_l = 6540 kg/m³; v₀ = 0.0220 m/s. Step 4: Numerator: 2 × 1.00×10⁻⁶ × 9.81 × 6540 = 0.1283. Step 5: Denominator: 9 × 0.0220 = 0.198. Step 6: η = 0.1283/0.198 ≈ 0.648 Pa·s. Result: η ≈ 0.65 Pa·s, close to the accepted glycerine value (~1.4 Pa·s at 20°C for pure glycerine; lab samples with absorbed water read lower — a good discussion point).