Which of the following laws is used in Millikan's oil drop experiment for the determination of the charge of an electron?

$A$ small spherical ball of radius $0.1 \,mm$ and density $10^{4} \,kg \,m^{-3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering the water the velocity of the ball does not change and it continues to fall with the same constant velocity inside the water,then the value of $h$ will be $m$. (Given $g = 10 \,m \,s^{-2}$,viscosity of water $\eta = 1.0 \times 10^{-5} \,N \,s \,m^{-2}$,density of water $\rho_w = 10^3 \,kg \,m^{-3}$)

$A$ spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown, which represents the speed of the ball $(v)$ as a function of time $(t)$ is

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).

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.]

$A$ solid sphere falls with a terminal velocity of $10 \, cm/s$ in the Earth's gravitational field. If it is allowed to fall freely in a region outside the gravitational field of the Earth,the terminal velocity will be: