Two drops of same radius are falling through air with steady velocity of $v \ cm/s$. If the two drops coalesce,what would be the 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).

$125$ small water drops of same size fall through air with constant terminal velocity $4 \,cm/s$. They coalesce to form a big drop. The terminal velocity of the big drop is: (in $\,m/s$)

Consider two solid spheres $P$ and $Q$ each of density $8 \ g \ cm^{-3}$ and diameters $1 \ cm$ and $0.5 \ cm$,respectively. Sphere $P$ is dropped into a liquid of density $0.8 \ g \ cm^{-3}$ and viscosity $\eta = 3 \ \text{poiseuille}$. Sphere $Q$ is dropped into a liquid of density $1.6 \ g \ cm^{-3}$ and viscosity $\eta = 2 \ \text{poiseuille}$. The ratio of the terminal velocities of $P$ and $Q$ is:

$A$ small drop of water falls from rest through a large height $h$ in air; the final velocity is ................

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: