What is the terminal velocity of an air bubble of $1.0 \ mm$ in diameter rising in a liquid of viscosity $0.85 \ N \ s \ m^{-2}$ and density $900 \ kg \ m^{-3}$? (Air density $= 1.293 \ kg \ m^{-3}$, $g = 9 \ m \ s^{-2}$)

$A$ 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 $\,m/s$)

$A$ copper ball of radius $r$ travels with a uniform speed $v$ in a viscous fluid. If the ball is changed with another ball of radius $2r$,then the new uniform speed will be

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

We have three beakers $A, B$,and $C$ containing glycerine,water,and kerosene respectively. They are stirred vigorously and placed on a table. The liquid which comes to rest at the earliest is

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.