The terminal velocity $(v_{t})$ of a spherical raindrop depends on the radius $(r)$ of the raindrop as:

In an experiment to calculate the coefficient of viscosity of a liquid by measuring terminal velocity, it was observed that a ball of mass $(\frac{\pi}{2}) \ gm$ and radius $0.5 \ cm$ takes $5 \ s$ to fall steadily through a height of $50 \ cm$ inside a long column of liquid of density $1.2 \ gm/cc$. The coefficient of viscosity will be given as: (in $\text{poise}$)

$A$ metal sphere of radius $R$,density $\rho_1$ moves with terminal velocity $V_1$ through a liquid of density $\sigma$. Another sphere of same radius but density $\rho_2$ moves through the same liquid. Its terminal velocity is $V_2$. The ratio $V_1: V_2$ is

$A$ liquid disturbed by stirring comes to rest after some time due to its property of

Two spherical rain drops reach the surface of the earth with terminal velocities having ratio $16: 9$. The ratio of their surface area is

Why do rain drops not possess a velocity greater than a certain limit? Explain.