Question 5 A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, 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 = 2 mm = 2 * 10^-3m

Volume of the drop = 4/3 π r³ = 4/3 * 22/7 * (2 * 10^-3)³

Density of the water = ρ = 10³ kg / m³

Mass of water = ρV = 10³ * 4/3 * 22/7 * (2 * 10^-3)³ Kg

Gravitational force = mg = * 4/3 * 22/7 * (2 * 10^-3)³ * 9.8 N

The work done by the gravitational force on the first half of the journey W = Fs

As we take half the journey of drop.
Distance s=500/2=250m

so we get 
W in first half =.0.082 J

The work done by the gravitational force on the first half of the journey  = The work in the second half of the journey

W in second half =.0.082 J

As per the law of conservation, if there are no resistive forces, the total energy of the rain drop will remain the same

Total E=mgh + 0
here h is completely distance 
i.e
h=2s
Energy of drop when it is top 
E=2w
E=2*.082= 0.164J

Due to the presence of resistive force, when drop falls from the top with velocity of 10m/s

Here is potential energy is zero only kinetic energy is exist 
Ground level E=1/2mv²  =.0016J

Hence
Resistive force = Energy at ground – Total  energy at top
=.0016-.164
=  -0.162J