$A$ 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

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

$A$ large drop of oil (density $0.8 \, g/cm^3$ and viscosity $\eta_0$) floats up through a column of another liquid (density $1.2 \, g/cm^3$ and viscosity $\eta_L$). Assuming that the two liquids do not mix,the velocity with which the oil drop rises will depend on.

$A$ small sphere of mass $m$ is dropped from a great height. After it has fallen $100 \; m$,it has attained its terminal velocity and continues to fall at that speed. The work done by air friction against the sphere during the first $100 \; m$ of fall is

An air bubble of $1\, cm$ radius is rising at a steady rate of $2.00\, mm/sec$ through a liquid of density $1.5\, g/cm^3$. Neglect the density of air. If $g = 1000\, cm/sec^2$,then the coefficient of viscosity of the liquid is:

$A$ 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: