A rain drop of radius $2\; mm$ falls from a helght of $500 \;m$ above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original hetght, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey ? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is $10\; m s ^{-1} ?$

Radius of the rain drop, $r=2 mm =2 \times 10^{-3} m$

Volume of the rain drop, $V=\frac{4}{3} \pi r^{3}$ $=\frac{4}{3} \times 3.14 \times\left(2 \times 10^{-3}\right)^{3} m ^{-3}$

Density of water, $\rho=10^{3} kg m ^{-3}$

Mass of the rain drop, $m=\rho V$

$=\frac{4}{3} \times 3.14 \times\left(2 \times 10^{-3}\right)^{3} \times 10^{3} kg$

Gravitational force, $F=m g$

$=\frac{4}{3} \times 3.14 \times\left(2 \times 10^{-3}\right)^{3} \times 10^{3} \times 9.8 N$

The work done by the gravitational force on the drop in the first half of its journey

$W_{ I }=F_{S}$

$=\frac{4}{3} \times 3.14 \times\left(2 \times 10^{-3}\right)^{3} \times 10^{3} \times 9.8$

$=0.082 J$

This amount of work is equal to the work done by the gravitational force on the drop in the second half of its journey, i.e., $W_{ II },=0.082 J$

As per the law of conservation of energy, if no resistive force is present, then the total energy of the rain drop will remain the same.

.Total energy at the top:

$E_{ T }=m g h+0$

$=\frac{4}{3} \times 3.14 \times\left(2 \times 10^{-3}\right)^{3} \times 10^{3} \times 9.8{ } \times 500 \times 10^{-5}$

$=0.164 J$

Due to the presence of a resistive force, the drop hits the ground with a velocity of 10 $m / s$

$\therefore$ Total energy at the ground:

$E_{ G }=\frac{1}{2} m v^{2}+0$

$=\frac{1}{2} \times \frac{4}{3} \times 3.14 \times\left(2 \times 10^{-3}\right)^{3} \times 10^{3} \times 9.8 \times(10)^{2}$

$=1.675 \times 10^{-3} J$

$\therefore$ Resistive force $=E_{ G }-E_{ T }=-0.162 J$