Ncert Exercises & Solutions

Question 6.13:

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 $10 m s^{- 1}$ ?

Given,

r = 2 mm = 2 x 10^-3m.
Distance moved in each half of the journey, s=500/2= 250 m.
Density of water, p = 10³ kg/ m³
Mass of rain drop = volume of drop x density
$m = \frac{4}{3} π r^{3} \times ρ$
$= \frac{4}{3} \times \frac{22}{7} \times {(2 \times 10^{- 3})}^{3} \times 10^{3} = 3.35 \times 10^{- 5} k g$
$\Rightarrow W = m g s = 3.35 \times 10^{- 5} \times 9.8 \times 250 = 0.082 J$
(Whether the drop moves with decreasing acceleration or with uniform speed, work done by the gravitational force on the drop remains the same.)

If there were no resistive forces, energy of drop on reaching the ground.
E₁= mgh = 3.35 x 10^-5 x 9.8 x 500 = 0.164 J
Actual energy, E₂ = 1/2mv² = 1/2 x 3.35 x 10^-5 (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.

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