An air bubble of diameter 3.0 mm is rising steadily through a long column of a viscous liquid. The density of the liquid is 2.0 g/cm ^3 and its coefficient of viscosity is 10 Poise. Neglecting the density of air, the terminal velocity of the air bubble is (Take g = 10 m/s ^2 )

Step-by-step solution

For a rising air bubble, the driving force is the buoyant force minus its weight. Since the density of air is negligible, the net upward force is approximately the buoyant force: F_B = 4 3 r^3 _l g . At terminal velocity, this is balanced by the downward viscous drag force (Stokes' law): F_v = 6 r v_t . Equating the two forces: 4 3 r^3 _l g = 6 r v_t v_t = 2r^2g _l 9 Given: Diameter d = 3.0 mm radius r = 1.5 mm = 0.15 cm Density of liquid _l = 2.0 g/cm ^3 Viscosity = 10 Poise = 10 g/(cm s) Acceleration due to gravi