The terminal velocity of a small-sized spherical body of radius $r$ falling vertically in a viscous liquid is given by the proportionality:

Two spheres of different material are falling through a liquid of density $\rho_L = 2 \times 10^3 \ kg/m^3$ with the same uniform terminal speed. The density of sphere $1$ is $\rho_1 = 8 \times 10^3 \ kg/m^3$ and the density of sphere $2$ is $\rho_2 = 11 \times 10^3 \ kg/m^3$. The ratio of their radii $(r_1/r_2)$ is:

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6 \,cm \,s^{-1}$. If they coalesce to form one big drop, what will be the terminal speed of the bigger drop? (Neglect the buoyancy of the air)

$A$ spherical solid ball of volume $V$ is made up of a material of density $\rho$. It is falling through a liquid of density $\sigma$ $(\sigma < \rho)$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of the terminal speed $v_{T}$,$F = -K v_{T}^2$ $(K > 0)$. Then,the terminal speed of the ball is ($g =$ acceleration due to gravity).

The diameter of an air bubble,which was initially $2\,mm$,rises steadily through a solution of density $1750\,kg\,m^{-3}$ at the rate of $0.35\,cm\,s^{-1}$. The coefficient of viscosity of the solution is (in poise,nearest integer). (The density of air is negligible).

$A$ sphere of radius $r$ and density $\rho$ is dropped from a height $h$. When it falls into water,it attains a terminal velocity. If the coefficient of viscosity of water is $\eta$,then $h =$