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

$A$ spherical solid ball of volume $V$ is made of a material of density $\rho_1$. It is falling through a liquid of density $\rho_2$ $(\rho_2 < \rho_1)$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$,i.e.,$F_{viscous} = -kv^2$ $(k > 0)$. The terminal speed of the ball is:

$A$ steel ball of radius $0.05 \,cm$ and density $7.8 \,g \,cm^{-3}$ is dropped into a tank of water. The terminal velocity of the steel ball is (Density of water $= 1 \,g \,cm^{-3}$ and viscosity of water $= 0.001 \,Pa \,s$) (in $\,m/s$)

$n$ small drops of the same size fall through air with a constant terminal velocity of $5 \ cm/s$. If they coalesce to form a single big drop,what is the terminal velocity of the big drop?

Two drops of the same radius are falling through air with a steady velocity of $5 \text{ cm/s}$. If the two drops coalesce,the terminal velocity would be

What will be the approximate terminal velocity of a rain drop of diameter $1.8 \times 10^{-3} \ m$,when density of rain water $\approx 10^{3} \ kg \ m^{-3}$ and the coefficient of viscosity of air $\approx 1.8 \times 10^{-5} \ N \ s \ m^{-2}$ (in $m \ s^{-1}$)? (Neglect buoyancy of air)