From amongst the following curves, which one shows the variation of the velocity v with time t for a small sized spherical body falling vertically in a long column of a viscous liquid

Write $\mathrm{SI}$ and $\mathrm{CGS}$ unit of coefficient of viscosity.

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

$(A)$ $\frac{\left|\overrightarrow{ V }_{ P }\right|}{\left|\overrightarrow{ V }_{ Q }\right|}=\frac{\eta_1}{\eta_2}$ $(B)$ $\frac{\left|\overrightarrow{ V }_{ P }\right|}{\left|\overrightarrow{ V }_{ Q }\right|}=\frac{\eta_2}{\eta_1}$

$(C)$ $\overrightarrow{ V }_{ P } \cdot \overrightarrow{ V }_{ Q } > 0$ $(D)$ $\overrightarrow{ V }_{ P } \cdot \overrightarrow{ V }_{ Q } < 0$

The diameter of an air bubble which was initially $2\,mm$, rises steadily through a solution of density $1750\,kg\,m\,m ^{-3}$ at the rate of $0.35\,cms ^{-1}$. The coefficient of viscosity of the solution is poise (in nearest integer). (the density of air is negligible).

If a ball of steel (density $\rho=7.8 \;gcm ^{-3}$) attains a terminal velocity of $10 \;cms ^{-1}$ when falling in a tank of water (coefficient of viscosity $\eta_{\text {water }}=8.5 \times 10^{-4} \;Pa – s$ ) then its terminal velocity in glycerine $\left(\rho=12 gcm ^{-3}, \eta=13.2\right)$ would be nearly