NCERT Ques6.13: A rain drop of radius 2mm falls from a height of 500m above the ground. It falls with decreasing acceleration until 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 in the first and second half of its journey? My query is that in the second half of its journey, the raindrop falls with a uniform speed, so there must be zero work done on it. But according to the solution, 0.082J of work is done on it. How is this possible?

Asked by Avijit Gupta | 8th Feb, 2012, 07:37: PM

Expert Answer:

Radius of the rain drop, r = 2 mm = 2 × 10–3 m

Volume of the rain drop, 

Density of water, ? = 103 kg m–3

Mass of the rain drop, m = ?V

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 = E– ET = –0.162 J

Answered by  | 8th Feb, 2012, 11:01: PM

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