Two drops of the same radius are falling through air with a steady velocity of $5 \text{ cm/s}$. If the two drops coalesce,the terminal velocity would be

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 $Pa \ s$):

$A$ small drop of water falls from rest through a large height $h$ in air; the final velocity is

$A$ water drop of radius $1\,\mu m$ falls in a situation where the effect of buoyant force is negligible. The coefficient of viscosity of air is $1.8 \times 10^{-5}\,Nsm^{-2}$ and its density is negligible as compared to that of water $10^{3}\,kgm^{-3}$. The terminal velocity of the water drop is . . . . . . $\times 10^{-6}\,ms^{-1}$. (Take acceleration due to gravity $= 10\,ms^{-2}$)

Eight equal drops of water,each of radius $r = 2 \ mm$,are falling through air with a terminal velocity of $16 \ cm/s$. The eight drops combine to form a single big drop. The terminal velocity of the bigger drop will be $.... \ cm/s$.

In Millikan's oil drop experiment,what is the terminal speed of an uncharged drop of radius $2.0 \times 10^{-5} \; m$ and density $1.2 \times 10^{3} \; kg \; m^{-3}$? Take the viscosity of air at the temperature of the experiment to be $1.8 \times 10^{-5} \; Pa \; s$. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.