$A$ 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:

$A$ 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:

$A$ copper ball of radius $3.0 \,mm$ falls in an oil tank of viscosity $1 \,kg / ms$. Then, the terminal velocity of the copper ball will be (Density of oil $= 1.5 \times 10^3 \,kg / m^3$, Density of copper $= 9 \times 10^3 \,kg / m^3$ and $g = 10 \,m / s^2$.)

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
$(A)$ $\frac{|\overrightarrow{V}_{P}|}{|\overrightarrow{V}_{Q}|}=\frac{\eta_1}{\eta_2}$
$(B)$ $\frac{|\overrightarrow{V}_{P}|}{|\overrightarrow{V}_{Q}|}=\frac{\eta_2}{\eta_1}$
$(C)$ $\overrightarrow{V}_{P} \cdot \overrightarrow{V}_{Q} > 0$
$(D)$ $\overrightarrow{V}_{P} \cdot \overrightarrow{V}_{Q} < 0$

$A$ 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 $\,m$)

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?