$A$ solid steel ball of diameter $3.6 \ mm$ acquires a terminal velocity of $2.45 \times 10^{-2} \ m/s$ while falling under gravity through an oil of density $925 \ kg/m^3$. Take the density of steel as $7825 \ kg/m^3$ and $g$ as $9.8 \ m/s^2$. The viscosity of the oil in $SI$ units is:

The terminal velocity of a liquid drop of radius $r$ falling through air is $v$. If two such drops are combined to form a bigger drop,the terminal velocity with which the bigger drop falls through air is (ignore any buoyant force due to air).

Which falls faster,big rain drops or small rain drops?

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.]

Sixty-four rain drops of radius $1 \ mm$ each, falling down with a terminal velocity of $10 \ cm/s$, coalesce to form a bigger drop. The terminal velocity of the bigger drop is . . . . . . $cm/s$.

Which of the following diagrams correctly shows the relation between the terminal velocity $v_{T}$ of a spherical body falling in a liquid and viscosity $\eta$ of the liquid?