$n$ small water drops of same size (radius $r$) fall through air with constant terminal velocity $V$. They coalesce to form a big drop of radius $R$. The terminal velocity of the big drop is

$A$ spherical ball of radius $1 \times 10^{-4} \,m$ and density $10^4 \,kg \,m^{-3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If the velocity of the ball does not change after entering the water, then the value of $h$ is:

An air bubble of radius $1 \ cm$ rises from the bottom portion through a liquid of density $1.5 \ g/cc$ at a constant speed of $0.25 \ cm \ s^{-1}$. If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (in $Pa \ s$):

Find the viscosity of glycerine $($having density $1.3 \ g \ cm^{-3})$ if a steel ball of $2 \ mm$ radius $($density $8 \ g \ cm^{-3})$ acquires a terminal velocity of $4 \ cm \ s^{-1}$ in falling freely in the tank of glycerine. $(g = 980 \ cm \ s^{-2})$ (in $\text{poise}$)

In an experiment to calculate the coefficient of viscosity of a liquid by measuring terminal velocity, it was observed that a ball of mass $(\frac{\pi}{2}) \ gm$ and radius $0.5 \ cm$ takes $5 \ s$ to fall steadily through a height of $50 \ cm$ inside a long column of liquid of density $1.2 \ gm/cc$. The coefficient of viscosity will be given as: (in $\text{poise}$)

What is the rate at which a trapped bubble of $2 \,mm$ diameter rises through a solution of density $13.6 \times 10^3 \,kg/m^3$ and coefficient of viscosity $1.5 \,cP$ (in $\,m/s$)? Assume the density of air is negligible and $g = 10 \,m/s^2$.