Why do rain drops not possess a velocity greater than a certain limit? Explain.

$A$ ball of mass $m$ and radius $r$ is gently released in a viscous liquid. The mass of the liquid displaced by it is $m'$ such that $m > m'$. The terminal velocity is proportional to

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 $.... \ cm/s$.

$A$ metal sphere of mass $m$ and density $\sigma_{1}$ falls with terminal velocity through a container containing liquid. The density of the liquid is $\sigma_{2}$. The viscous force acting on the sphere is

What is the terminal velocity of a rain drop of radius $0.02 \ mm$ (in $cm \ s^{-1}$)? [Note that the coefficient of viscosity of air is $1.8 \times 10^{-5} \ N \ s \ m^{-2}$, density of water is $1000 \ kg \ m^{-3}$. Use $g = 10 \ m \ s^{-2}$ and density of air can be neglected in comparison with the density of water.]

$A$ 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 $= 10\,ms^{-2}$)