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)

$A$ small ball of mass $m$ and density $\rho$ is dropped in a viscous liquid of density $\rho_0$. After some time,the ball falls with constant velocity. The viscous force on the ball is:

$A$ small spherical ball of radius $0.1 \,mm$ and density $10^{4} \,kg \,m^{-3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering the water the velocity of the ball does not change and it continues to fall with the same constant velocity inside the water,then the value of $h$ will be $m$. (Given $g = 10 \,m \,s^{-2}$,viscosity of water $\eta = 1.0 \times 10^{-5} \,N \,s \,m^{-2}$,density of water $\rho_w = 10^3 \,kg \,m^{-3}$)

$n$ 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

The graph between terminal velocity (along $y-$ axis) and square of radius (along $x-$ axis) of a spherical body of density $\rho$ allowed to fall through a fluid of density $\sigma$ is a

Two rain drops falling through air have radii in the ratio $1: 2$. They will have terminal velocity in the ratio