$A$ 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.

$A$ spherical body of radius $r$ and density $\sigma$ falls freely through a viscous liquid having density $\rho$ and viscosity $\eta$ and attains a terminal velocity $v_0$. The estimated maximum error in the quantity $\eta$ is: (Ignore errors associated with $\sigma, \rho$ and $g$, gravitational acceleration)

An air bubble of radius $1 \ cm$ rises from the bottom portion through a liquid of density $1.5 \ g/cc$ at a constant speed of $0.25 \ cm \ s^{-1}$. If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (in $Pa \ s$):

Two solid spheres of radii $2 \ mm$ and $4 \ mm$ are tied to the two ends of a light string and released in a liquid of specific gravity $1.3$ and coefficient of viscosity $1 \ Pa \cdot s$. The string is just taut when the two spheres are completely submerged in the liquid. If the density of the materials of the two spheres is $2800 \ kg \cdot m^{-3}$,then the terminal velocity of the system of the spheres is (take $g = 10 \ m \cdot s^{-2}$):

What is the terminal velocity of a rain drop of radius $0.02 \ mm$ (in $cm \ s^{-1}$)? [Note that the coefficient of viscosity of air is $1.8 \times 10^{-5} \ N \ s \ m^{-2}$, density of water is $1000 \ kg \ m^{-3}$. Use $g = 10 \ m \ s^{-2}$ and density of air can be neglected in comparison with the density of water.]

$A$ solid sphere falls with a terminal velocity of $10 \, cm/s$ in the Earth's gravitational field. If it is allowed to fall freely in a region outside the gravitational field of the Earth,the terminal velocity will be: