A rain drop of radius 2mm, falls from a height of 500 m above the ground. It falls with decreasing acceleration due to viscous resistance of air until half its original height. It attains its maximum (terminal ) speed, and moves with uniform speed there after. What is the work done by the gravitational force on the drop in the first half and second half of its journey ? Take density of water `=10^(3)kg//m^(3)`. What is the work done by the resistive force in the entire journey if its speed on reaching the ground is `10ms^(-1)` ?

Explanation

Here, r=2 mm `=2 xx 10^(-3)m`
Distance moved in each half ot the journey, S=`500//2` = 250 m
Density of water , p=`10^(3)kg//m^(3)`
Mass of rain drop = volume of drop x density
`m=4//3pir^(2)xx p= 4//3 xx 22//7(2 xx 10^(-3))^(3) xx 10^(3)= 3.35 xx 10^(-5)kg`
`therefore W= mg xx s = 3.35 xx 10^(-5) xx 9.8 xx 250= 0.082` J
Note, whether the drop moves with decreasing acceleration or with uniform speed, work done by the gravitational force on the drop on reaching the ground.
`E_(1) = mgh= 3.35 xx 10^(-5) xx 9.8 xx 500 = 0.164`J
Actual energy, `E_(2) = 1//2mv^(2)= 1//2 xx 3.35 xx 10^(-5)(10^(2)= 1.675 xx 10^(-3)J`
Work done by the resistive forces, `W=E_(1)-E_(2) = 0.164 - 1.675 xx 10^(-3)` W
= 0.1623 Joule.