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)

Two drops of same radius are falling through air with steady velocity of $v \ cm/s$. If the two drops coalesce,what would be the terminal velocity?

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

What is the terminal velocity of a rain drop of radius $0.02 \ mm$ (in $cm \ s^{-1}$)? [Note that the coefficient of viscosity of air is $1.8 \times 10^{-5} \ N \ s \ m^{-2}$, density of water is $1000 \ kg \ m^{-3}$. Use $g = 10 \ m \ s^{-2}$ and density of air can be neglected in comparison with the density of water.]

$A$ solid metal sphere released in a vertical liquid column has attained terminal velocity in the downward direction. The magnitudes of viscous force, buoyant force, and gravitational force acting on it are $F_{v}$, $F_{B}$, and $F_{W}$ respectively. Then, the correct relation between them is:

Consider two solid spheres $P$ and $Q$ each of density $8 \ g \ cm^{-3}$ and diameters $1 \ cm$ and $0.5 \ cm$,respectively. Sphere $P$ is dropped into a liquid of density $0.8 \ g \ cm^{-3}$ and viscosity $\eta = 3 \ \text{poiseuille}$. Sphere $Q$ is dropped into a liquid of density $1.6 \ g \ cm^{-3}$ and viscosity $\eta = 2 \ \text{poiseuille}$. The ratio of the terminal velocities of $P$ and $Q$ is: