$A$ metal sphere of mass $m$ and density $\sigma_{1}$ falls with terminal velocity through a container containing liquid. The density of the liquid is $\sigma_{2}$. The viscous force acting on the sphere is

$A$ 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:

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$: $A$ spherical body of radius $(5 \pm 0.1) \ mm$ having a particular density is falling through a liquid of constant density. The percentage error in the calculation of its terminal velocity is $4\,\%$.
Reason $R$: The terminal velocity of the spherical body falling through the liquid is inversely proportional to its radius.
In the light of the above statements,choose the correct answer from the options given below:

Eight equal drops of water,each of radius $r = 2 \ mm$,are falling through air with a terminal velocity of $16 \ cm/s$. The eight drops combine to form a single big drop. The terminal velocity of the bigger drop will be $.... \ cm/s$.

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.]

$A$ small steel ball is dropped from a height of $1.5 \,m$ into a glycerine jar. The ball reaches the bottom of the jar $1.5 \,s$ after it was dropped. If the retardation in the glycerine is $2.66 \,m/s^2$, the height of the glycerine in the jar is about (acceleration due to gravity $g = 9.8 \,m/s^2$) (in $\,m$)