$A$ steel ball of radius $6 \ mm$ has a terminal speed of $12 \ cm s^{-1}$ in a viscous liquid. What will be the terminal speed of a steel ball of radius $3 \ mm$ in the same liquid (in $cm s^{-1}$)?

An air bubble of $1\, cm$ radius is rising at a steady rate of $2.00\, mm/sec$ through a liquid of density $1.5\, g/cm^3$. Neglect the density of air. If $g = 1000\, cm/sec^2$,then the coefficient of viscosity of the liquid is:

If the terminal velocity of a metal sphere of mass $8 \ g$ falling through a liquid is $3 \ cm s^{-1}$, then the terminal velocity of another sphere of mass $64 \ g$ made of the same metal falling through the same liquid is (in $cm s^{-1}$)

$A$ water drop whose radius is $0.0015 \, mm$ is falling through the air. If the coefficient of viscosity of air is $1.8 \times 10^{-5} \, kg/(m \cdot s)$,then assuming the buoyancy force is negligible,the terminal velocity of the drop will be:

Two solid spheres of the same metal but of mass $M$ and $8M$ fall simultaneously in a viscous liquid. If their terminal velocities are $v$ and $nv$, then the value of $n$ is:

Two spheres of equal masses but radii $r_1$ and $r_2$ are allowed to fall in a liquid of infinite column. The ratio of their terminal velocities is