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).

$A$ spherical ball of density $\rho$ and radius $0.003 \ m$ is dropped into a tube containing a viscous fluid filled up to the $0 \ cm$ mark as shown in the figure. The viscosity of the fluid $\eta = 1.260 \ N \cdot s \cdot m^{-2}$ and its density $\rho_L = \rho/2 = 1260 \ kg \cdot m^{-3}$. Assume the ball reaches a terminal speed by the $10 \ cm$ mark. The time taken by the ball to traverse the distance between the $10 \ cm$ and $20 \ cm$ mark is $(g = 10 \ m \cdot s^{-2})$

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).

What is the terminal velocity of an air bubble of $1.0 \ mm$ in diameter rising in a liquid of viscosity $0.85 \ N \ s \ m^{-2}$ and density $900 \ kg \ m^{-3}$? (Air density $= 1.293 \ kg \ m^{-3}$, $g = 9 \ m \ s^{-2}$)

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:

An object falling through a fluid is observed to have an acceleration given by $a = g - bv$,where $g$ is the gravitational acceleration and $b$ is a constant. After a long time of release,it is observed to fall with a constant speed. The value of this constant speed is: