The terminal velocity of a small-sized spherical body of radius $r$ falling vertically in a viscous liquid is given by the proportionality:

What is the velocity $v$ of a metallic ball of radius $r$ falling in a tank of liquid at the instant when its acceleration is one-half that of a freely falling body? (The densities of metal and of liquid are $\rho$ and $\sigma$ respectively,and the viscosity of the liquid is $\eta$).

$A$ small metal sphere of mass $M$ and density $d_{1}$,when dropped in a jar filled with liquid,moves with terminal velocity after some time. The viscous force acting on the sphere is ($d_{2} =$ density of liquid,$g =$ gravitational acceleration).

$A$ metal sphere of mass $m$ and density $\sigma_1$ falls with terminal velocity through a container containing liquid. The density of the liquid is $\sigma_2$. The viscous force acting on the sphere is:

$A$ small metallic sphere of diameter $2 \ mm$ and density $10.5 \ g/cm^3$ is dropped in glycerine having viscosity $10 \ \text{Poise}$ and density $1.5 \ g/cm^3$. The terminal velocity attained by the sphere is . . . . . . $cm/s$. $(\pi = \frac{22}{7}$ and $g = 10 \ m/s^2)$

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 (in $cm \ s^{-1}$)? (Neglect the buoyancy of the air)