Two drops of equal radius are falling with a terminal velocity of $5 \, cm/s$. If they coalesce to form a single larger drop,what will be the terminal velocity of the new drop?

$A$ small ball of mass $m$ and density $\rho$ is dropped in a viscous liquid of density $\rho_0$. After some time,the ball falls with constant velocity. The viscous force on the ball is:

If the average terminal velocity of a rain drop is $2 \,m/s$, then the energy transferred by rain to each square metre of the surface at a place which receives $100 \,cm$ of rain in a year 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
$(A)$ $\frac{|\overrightarrow{V}_{P}|}{|\overrightarrow{V}_{Q}|}=\frac{\eta_1}{\eta_2}$
$(B)$ $\frac{|\overrightarrow{V}_{P}|}{|\overrightarrow{V}_{Q}|}=\frac{\eta_2}{\eta_1}$
$(C)$ $\overrightarrow{V}_{P} \cdot \overrightarrow{V}_{Q} > 0$
$(D)$ $\overrightarrow{V}_{P} \cdot \overrightarrow{V}_{Q} < 0$

$A$ sphere is dropped under gravity through a fluid of viscosity $\eta$. If the average acceleration is half of the initial acceleration,the time to attain the terminal velocity is ($\rho$ = density of sphere; $r$ = radius).

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 $= 1.293 \ kg \ m^{-3}$, $g = 9 \ m \ s^{-2}$)