What will be the approximate terminal velocity of a rain drop of diameter $1.8 \times 10^{-3} \ m$, when density of rain water $\approx 10^{3} \ kg \ m^{-3}$ and the coefficient of viscosity of air $\approx 1.8 \times 10^{-5} \ N \ s \ m^{-2}$ (in $m \ s^{-1}$)? (Neglect buoyancy of air)

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:

$Assertion :$ Falling raindrops acquire a terminal velocity.
$Reason :$ $A$ constant force in the direction of motion and a velocity dependent force opposite to the direction of motion,always result in the acquisition of terminal velocity.

$A$ ball is thrown downward with some initial velocity into a viscous liquid. Which of the following curves represents the correct variation of velocity versus time?

An iron sphere of mass $20 \times 10^{-3} \ kg$ falls through a viscous liquid with terminal velocity $0.5 \ ms^{-1}$. The terminal velocity (in $ms^{-1}$) of another iron sphere of mass $54 \times 10^{-2} \ kg$ is (in $.5$)

$A$ small drop of water falls from rest through a large height $h$ in air; the final velocity is ................