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Asked by Baidyabati Last Modified
Answer Radius of the rain drop, r = 2 mm = 2 × 10–3 m
Volume of the rain drop, 
Density of water, ρ = 103 kg m–3
Mass of the rain drop, m = ρV
= 
Gravitational force, F = mg
= 
The work done by the gravitational force on the drop in the first half of its journey:
WI= Fs
= 
× 250
= 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., WII, = 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:
ET = mgh + 0
= × 500 × 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.
∴Total energy at the ground:
∴Resistive force = EG – ET = –0.162 J
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