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

$A$ small spherical ball of radius $r$,falling through a viscous medium of negligible density has terminal velocity $v$. Another ball of the same mass but of radius $2r$,falling through the same viscous medium will have terminal velocity:

In an experiment to calculate the coefficient of viscosity of a liquid by measuring terminal velocity, it was observed that a ball of mass $(\frac{\pi}{2}) \ gm$ and radius $0.5 \ cm$ takes $5 \ s$ to fall steadily through a height of $50 \ cm$ inside a long column of liquid of density $1.2 \ gm/cc$. The coefficient of viscosity will be given as: (in $\text{poise}$)

$A$ large drop of oil (density $0.8 \, g/cm^3$ and viscosity $\eta_0$) floats up through a column of another liquid (density $1.2 \, g/cm^3$ and viscosity $\eta_L$). Assuming that the two liquids do not mix,the velocity with which the oil drop rises will depend on.

$A$ solid metal sphere released in a vertical liquid column has attained terminal velocity in the downward direction. The magnitudes of viscous force, buoyant force, and gravitational force acting on it are $F_{v}$, $F_{B}$, and $F_{W}$ respectively. Then, the correct relation between them is:

Eight identical drops of same radii are falling through air with steady velocity $6 \ cm/s$. If these drops coalesce,then the terminal velocity would be .......... $cm/s$.