Radius of the drop is given as $2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$.

Height from which the raindrops fall is given as $\mathrm{S}=500 \mathrm{~m}$.

The density of water is given as $\rho=10^{3} \mathrm{~kg} / \mathrm{m}^{3}$

Mass of rain drop $=$ volume of drop $\times$ density

$m=(4 / 3) \pi r^{3} \times \rho=(4 / 3) \times(22 / 7) \times\left(2 \times 10^{-3}\right)^{3} \times 10^{3}=3.35 \times 10^{-5} \mathrm{~kg}$

The gravitational force experienced by the rain drop can be calculated as,

$F=m g$

$=(4 / 3) \times(22 / 7) \times\left(2 \times 10^{-3}\right)^{3} \times 10^{3} \times 9.8 \mathrm{~N}$

The work done by gravity on the drop is calculated as,

$\left.W=m g \times S=3.35 \times 10^{-5} \times 9.8 \times 250=0.082\right\rfloor$

In the second half of the journey, the work on the drop will remain the same.

The total energy of the raindrop will be conserved during the motion according to conservation of energy

Total energy at the top will be,

$E_{1}=m g h=3.35 \times 10^{-5} \times 9.8 \times 500=0.164 \mathrm{~J}$

Due to resistive forces, energy of drop on reaching the ground can be caculated as,

$E_{2}=1 / 2 m v^{2}=1 / 2 \times(10)^{2}=1.675 \times 10^{-3} \mathrm{~J}$

Work done by the resistive forces will be,

$W=E_{1}-E_{2}=0.164-1.675 \times 10^{-3} \mathrm{~W}$

$=0.1623$ joule.