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

Two small spherical metal balls,having equal masses,are made from materials of densities $\rho_{1}$ and $\rho_{2}$ (where $\rho_{1} = 8 \rho_{2}$) and have radii of $1 \; mm$ and $2 \; mm$,respectively. They are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals $\eta$ and whose density is $0.1 \rho_{2}$. The ratio of their terminal velocities would be:

If the terminal speed of a sphere of gold (density $= 19.5 \times 10^3 \ kg/m^3$) is $0.2 \ m/s$ in a viscous liquid (density $= 1.5 \times 10^3 \ kg/m^3$), find the terminal speed (in $m/s$) of a sphere of silver (density $= 10.5 \times 10^3 \ kg/m^3$) of the same size in the same liquid.

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.]

If the average terminal velocity of a rain drop is $2 \,m/s$, then the energy transferred by rain to each square metre of the surface at a place which receives $100 \,cm$ of rain in a year is

Sixty-four rain drops of radius $1 \ mm$ each, falling down with a terminal velocity of $10 \ cm/s$, coalesce to form a bigger drop. The terminal velocity of the bigger drop is . . . . . . $cm/s$.