A raindrop 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 ms-1?

Asked by Abhisek | 1 year ago |  65

##### Solution :-

Radius of the drop  = 2 mm = 2 x 10-3 m.
Height from which the raindrops fall, S=500 m.
The density of water, ρ= 103 kg/ m3
Mass of rain drop = volume of drop x density
m =($$\dfrac{4}{3}$$)π r3x ρ

=($$\dfrac{4}{3}$$) x ($$\dfrac{22}{7}$$)x (2 x 10-3)3 x 103

= 3.35 x 10-5 kg

The gravitational force experienced by the rain drop

F = mg

= ($$\dfrac{4}{3}$$) x ($$\dfrac{22}{7}$$)x (2 x 10-3)3 x 103 x 9.8 N

The work done by gravity on the drop is

W = mg x s = 3.35 x 10-5 x 9.8 x 250 = 0.082 J

The work done on the drop in the second half of the journey will remain the same.

The total energy of the raindrop will be conserved during the motion

Total energy at the top
E1= mgh = 3.35 x 10-5  x 9.8 x 500 = 0.164 J

Due to resistive forces, energy of drop on reaching the ground.
E2 = $$\dfrac{1}{2}$$mv2 = $$\dfrac{1}{2}$$x (10)2

= 1.675 x 10-3J

Work done by the resistive forces, W =E1 – E2

= 0.164 – 1.675 x 10-3 W
= 0.1623 joule.

Answered by Pragya Singh | 1 year ago

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