Two drops of same radius are falling through air with steady velocity of $v \ cm/s$. If the two drops coalesce,what would be the terminal velocity?

If a ball of steel (density $\rho = 7.8 \; g \cdot cm^{-3}$) attains a terminal velocity of $10 \; cm \cdot s^{-1}$ when falling in a tank of water (coefficient of viscosity $\eta_{\text{water}} = 8.5 \times 10^{-4} \; Pa \cdot s$),then its terminal velocity in glycerine (density $\rho_{\text{gly}} = 1.2 \; g \cdot cm^{-3}$,coefficient of viscosity $\eta_{\text{gly}} = 13.2 \; Pa \cdot s$) would be nearly:

$A$ 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)

$A$ sphere of mass $M$ and radius $R$ is falling in a viscous fluid. The terminal velocity attained by the falling object will be proportional to

In Millikan's oil drop experiment,what is the viscous force acting on an uncharged drop of radius $2.0 \times 10^{-5} \, m$ and density $1.2 \times 10^{3} \, kg/m^3$? Take the viscosity of air $= 1.8 \times 10^{-5} \, Nsm^{-2}$. (Neglect buoyancy due to air).

An air bubble of diameter $6\,mm$ rises steadily through a solution of density $1750\,kg/m^3$ at the rate of $0.35\,cm/s$. The coefficient of viscosity of the solution (neglect density of air) is $..........\,Pa\cdot s$ (given,$g = 10\,m/s^2$).