A steel ball (ρ = 7800 kg/m³) of radius 1 mm falls through glycerine (η = 1.49 Pa·s, ρ = 1260 kg/m³). Find its terminal velocity. (g = 9.8 m/s²) Step 1: v_t = (2/9)·r²(ρ_ball − ρ_fluid)g/η. Step 2: r² = (10⁻³)² = 10⁻⁶ m²; density difference = 7800 − 1260 = 6540 kg/m³. Step 3: v_t = (2/9) × 10⁻⁶ × 6540 × 9.8 / 1.49 = 0.2222 × 10⁻⁶ × 6540 × 6.577 ≈ 9.56 × 10⁻³ m/s. Answer: ≈ 9.6 mm/s — barely a centimetre per second. In water (η 1500× smaller) it would plummet far faster.