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,$g = 10\,m/s^2$).

The terminal velocity of a small sphere of radius $a$ in a viscous liquid is proportional to

$A$ water drop of radius $1\,\mu m$ falls in a situation where the effect of buoyant force is negligible. The coefficient of viscosity of air is $1.8 \times 10^{-5}\,Nsm^{-2}$ and its density is negligible as compared to that of water $10^{3}\,kgm^{-3}$. The terminal velocity of the water drop is . . . . . . $\times 10^{-6}\,ms^{-1}$. (Take acceleration due to gravity $= 10\,ms^{-2}$)

$A$ 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.

On which factors does terminal velocity depend? Explain.

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