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

Two spheres $P$ and $Q$ of equal radii have densities $\rho_1$ and $\rho_2$,respectively. The spheres are connected by a massless string and placed in liquids $L_1$ and $L_2$ of densities $\sigma_1$ and $\sigma_2$ and viscosities $\eta_1$ and $\eta_2$,respectively. They float in equilibrium with the sphere $P$ in $L_1$ and sphere $Q$ in $L_2$ and the string being taut (see figure). If sphere $P$ alone in $L_2$ has terminal velocity $\overrightarrow{V}_{P}$ and $Q$ alone in $L_1$ has terminal velocity $\overrightarrow{V}_{Q}$,then
$(A)$ $\frac{|\overrightarrow{V}_{P}|}{|\overrightarrow{V}_{Q}|}=\frac{\eta_1}{\eta_2}$
$(B)$ $\frac{|\overrightarrow{V}_{P}|}{|\overrightarrow{V}_{Q}|}=\frac{\eta_2}{\eta_1}$
$(C)$ $\overrightarrow{V}_{P} \cdot \overrightarrow{V}_{Q} > 0$
$(D)$ $\overrightarrow{V}_{P} \cdot \overrightarrow{V}_{Q} < 0$

If the terminal speed of a sphere $A$ [density $\rho_A = 7.5 \ kg \ m^{-3}$] is $0.4 \ ms^{-1}$ in a viscous liquid [density $\rho_L = 1.5 \ kg \ m^{-3}$],the terminal speed of sphere $B$ [density $\rho_B = 3 \ kg \ m^{-3}$] of the same size in the same liquid is: (in $ms^{-1}$)

The velocity of a small ball of mass $m$ and density $\rho$ when dropped in a container filled with glycerine of density $\sigma$ becomes constant after some time. The viscous force acting on the ball in the final stage is

$A$ tiny spherical oil drop carrying a net charge $q$ is balanced in still air, with a vertical uniform electric field of strength $\frac{81}{7} \pi \times 10^5 \,V / m$. When the field is switched $OFF$, the drop is observed to fall with terminal velocity $2 \times 10^{-3} \,m / s$. Here $g=9.8 \,m / s^2$, viscosity of air is $\eta = 1.8 \times 10^{-5} \,N s / m^2$ and density of oil is $\rho = 900 \,kg / m^3$. The magnitude of $q$ 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: