Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6 \ cm \ s^{-1}$. If they coalesce to form one big drop,what will be the terminal speed of the bigger drop (in $cm \ s^{-1}$)? (Neglect the buoyancy of the air)

$A$ spherical solid ball of volume $V$ is made up of a material of density $\rho$. It is falling through a liquid of density $\sigma$ $(\sigma < \rho)$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of the terminal speed $v_{T}$,$F = -K v_{T}^2$ $(K > 0)$. Then,the terminal speed of the ball is ($g =$ acceleration due to gravity).

$A$ steel ball of radius $0.05 \,cm$ and density $7.8 \,g \,cm^{-3}$ is dropped into a tank of water. The terminal velocity of the steel ball is (Density of water $= 1 \,g \,cm^{-3}$ and viscosity of water $= 0.001 \,Pa \,s$) (in $\,m/s$)

If the terminal speed of a sphere of gold (density $= 19.5 \times 10^3 \ kg/m^3$) is $0.2 \ m/s$ in a viscous liquid (density $= 1.5 \times 10^3 \ kg/m^3$), find the terminal speed (in $m/s$) of a sphere of silver (density $= 10.5 \times 10^3 \ kg/m^3$) of the same size in the same liquid.

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 an air bubble of diameter $2 \text{ mm}$ rises steadily through a liquid of density $2000 \text{ kg/m}^3$ at a rate of $0.5 \text{ cm/s}$, then the coefficient of viscosity of the liquid is . . . . . . $\text{Poise}$. (Take $g = 10 \text{ m/s}^2$)