The diameter of an air bubble,which was initially $2\,mm$,rises steadily through a solution of density $1750\,kg\,m^{-3}$ at the rate of $0.35\,cm\,s^{-1}$. The coefficient of viscosity of the solution is (in poise,nearest integer). (The density of air is negligible).

The terminal velocity of a copper ball of radius $5\,mm$ falling through a tank of oil at room temperature is $10\,cm\,s^{-1}$. If the viscosity of oil at room temperature is $0.9\,kg\,m^{-1}s^{-1}$,the viscous drag force is:

If a ball of steel (density $\rho = 7.8 \; g \cdot cm^{-3}$) attains a terminal velocity of $10 \; cm \cdot s^{-1}$ when falling in a tank of water (coefficient of viscosity $\eta_{\text{water}} = 8.5 \times 10^{-4} \; Pa \cdot s$),then its terminal velocity in glycerine (density $\rho_{\text{gly}} = 1.2 \; g \cdot cm^{-3}$,coefficient of viscosity $\eta_{\text{gly}} = 13.2 \; Pa \cdot s$) would be nearly:

What will be the approximate terminal velocity of a rain drop of diameter $1.8 \times 10^{-3} \ m$, when density of rain water $\approx 10^{3} \ kg \ m^{-3}$ and the coefficient of viscosity of air $\approx 1.8 \times 10^{-5} \ N \ s \ m^{-2}$ (in $m \ s^{-1}$)? (Neglect buoyancy of air)

$A$ large drop of oil (density $0.8 \, g/cm^3$ and viscosity $\eta_0$) floats up through a column of another liquid (density $1.2 \, g/cm^3$ and viscosity $\eta_L$). Assuming that the two liquids do not mix,the velocity with which the oil drop rises will depend on.

Write the equation of terminal velocity.