Question

A rain drop of radius 2mm falls from a height of 500m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) untill 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 10ms^-1?

Answer 1

r = 2 × 10^-3m,

volume = \(\frac{4}{3} \times \frac{22}{7}\) (2 × 10^-3)³ m³

p = 1000kgm^-3,

h = 250m

W= \(\frac{4}{3} \times \frac{22}{7}\) × 8 × 10^-9 × 1000 × 9.8 × 250J = 0.082J

Data remains unchanged in the next half.