$A$ spherical metal ball of radius $r$ falls through a viscous liquid with terminal velocity $V$. Another metal ball of the same material but of radius $\frac{r}{3}$ falls through the same liquid. What will be its terminal velocity?

The terminal velocity of a liquid drop of radius $r$ falling through air is $v$. If two such drops are combined to form a bigger drop,the terminal velocity with which the bigger drop falls through air is (ignore any buoyant force due to air).

Two solid spheres $S_{1}$ and $S_{2}$ of same uniform density fall from rest under gravity in a viscous medium and after some time, reach terminal velocities $v_{1}$ and $v_{2}$ respectively. If the ratio of masses $\frac{m_{1}}{m_{2}}=8$, then $\frac{v_{1}}{v_{2}}$ will be equal to

Two solid spheres of radii $2 \ mm$ and $4 \ mm$ are tied to the two ends of a light string and released in a liquid of specific gravity $1.3$ and coefficient of viscosity $1 \ Pa \cdot s$. The string is just taut when the two spheres are completely submerged in the liquid. If the density of the materials of the two spheres is $2800 \ kg \cdot m^{-3}$,then the terminal velocity of the system of the spheres is (take $g = 10 \ m \cdot s^{-2}$):

Suppose the average mass of raindrops is $3.0 \times 10^{-5} \ kg$ and their average terminal velocity is $9 \ m/s$. Calculate the energy transferred by rain to each square metre of the surface at a place which receives $100 \ cm$ of rain in a year.

Two spheres of the same material, but of radii $R$ and $3R$ are allowed to fall vertically downwards through a liquid of density $\rho$. The ratio of their terminal velocities is