The average mass of rain drops is $3.0 \times 10^{-5} \, kg$ and their average terminal velocity is $9 \, m/s$. Calculate the energy transferred by rain to each square metre of the surface at a place which receives $100 \, cm$ of rain in a year.

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

$A$ metal sphere of radius $R$ and density $\varrho_{1}$ moves with terminal velocity $v_{1}$ through a liquid of density $\sigma$. Another sphere of the same radius but of density $\varrho_{2}$ moves through the same liquid. Its terminal velocity will be:

$A$ spherical ball of radius $1 \ mm$ and density $10.5 \ g/cc$ is dropped in glycerine of coefficient of viscosity $9.8 \ \text{poise}$ and density $1.5 \ g/cc$. The viscous force on the ball when it attains constant velocity is $3696 \times 10^{-x} \ N$. The value of $x$ is $\text{(Given, } g = 9.8 \ m/s^2 \text{ and } \pi = \frac{22}{7}\text{)}$.

$A$ tiny spherical oil drop carrying a net charge $q$ is balanced in still air, with a vertical uniform electric field of strength $\frac{81}{7} \pi \times 10^5 \,V / m$. When the field is switched $OFF$, the drop is observed to fall with terminal velocity $2 \times 10^{-3} \,m / s$. Here $g=9.8 \,m / s^2$, viscosity of air is $\eta = 1.8 \times 10^{-5} \,N s / m^2$ and density of oil is $\rho = 900 \,kg / m^3$. The magnitude of $q$ 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