Two identical drops of water are falling through air with steady velocity $V$. If the two drops coalesce to form a single drop,what is the new terminal velocity of the single drop?

The velocity of a small ball of mass $m$ and density $\rho$ when dropped in a container filled with glycerine of density $\sigma$ becomes constant after some time. The viscous force acting on the ball in the final stage 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

An iron sphere of mass $20 \times 10^{-3} \ kg$ falls through a viscous liquid with terminal velocity $0.5 \ ms^{-1}$. The terminal velocity (in $ms^{-1}$) of another iron sphere of mass $54 \times 10^{-2} \ kg$ is (in $.5$)

$A$ spherical body of mass $m$ and radius $r$ is allowed to fall in a medium of viscosity $\eta$. The time in which the velocity of the body increases from zero to $0.63$ times the terminal velocity $(v)$ is called the time constant $(\tau)$. Dimensionally,$\tau$ can be represented by:

Two uniform solid balls of same density and of radii $r$ and $2r$ are dropped in air and fall vertically downwards. The terminal velocity of the ball with radius $r$ is $1 \, cm \, s^{-1}$. Find the terminal velocity of the ball of radius $2r$ (neglect buoyant force on the balls). ........... $cm \, s^{-1}$