The velocity of a small spherical ball of mass $m$ and density $d_1$,when dropped in a container filled with glycerine,becomes constant after some time. The viscous force acting on the ball if the density of glycerine is $d_2$ is

$n$ 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

In Millikan's oil drop experiment, a charged drop falls with terminal velocity $V$. If an electric field $E$ is applied in a vertically upward direction, then it starts moving in an upward direction with terminal velocity $2V$. If the magnitude of the electric field is decreased to $E/2$, then the terminal velocity will become

Two raindrops reach the earth with different terminal velocities having the ratio $9: 4$. The ratio of their volumes is

$A$ spherical ball of radius $1 \ mm$ and density $10.5 \ g/cc$ is dropped in glycerine of coefficient of viscosity $9.8 \ \text{poise}$ and density $1.5 \ g/cc$. The viscous force on the ball when it attains constant velocity is $3696 \times 10^{-x} \ N$. The value of $x$ is $\text{(Given, } g = 9.8 \ m/s^2 \text{ and } \pi = \frac{22}{7}\text{)}$.

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.]