If an air bubble of diameter $2 \text{ mm}$ rises steadily through a liquid of density $2000 \text{ kg/m}^3$ at a rate of $0.5 \text{ cm/s}$, then the coefficient of viscosity of the liquid is . . . . . . $\text{Poise}$. (Take $g = 10 \text{ m/s}^2$)

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6 \,cm \,s^{-1}$. If they coalesce to form one big drop, what will be the terminal speed of the bigger drop? (Neglect the buoyancy of the air)

Two drops of same radius are falling through air with steady velocity of $v \ cm/s$. If the two drops coalesce,what would be the terminal velocity?

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).

$A$ ball of radius $r$ and density $\rho$ falls freely under gravity through a distance $h$ before entering water. The velocity of the ball does not change even on entering the water. If the viscosity of water is $\eta$ and the density of water is $\sigma$ (assumed to be $1$ unit for calculation),the value of $h$ is given by:

In an experiment to verify Stokes' law,a small spherical ball of radius $r$ and density $\rho$ falls under gravity through a distance $h$ in air before entering a tank of water. If the terminal velocity of the ball inside water is the same as its velocity just before entering the water surface,then the value of $h$ is proportional to: (ignore viscosity of air)