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

We are given that radius of the drop, `r = 2 mm = 2 xx 10^(-3) m`,
Total length of the journey = 500 m.
The force of gravity against which the work is to be done remains constant throughout the journey. Therefore, the work done (W) in each half of the journey (= 250 m) is the same, i.e.,
`W = mgh = (4pi)/3 r^3 rho gh`
or `W = 4.19 xx (2 xx 10^(-3))3 xx 103 xx 9.8 xx 250 = 0.082 J` (as density of water, `rho = 1000 kg//m^3)`
PE of the drop before reaching the ground `= 2W = 2 xx 0.082 J = 0.164 J`
KE of the drop on reaching the ground =
`1/2 mv^2 = 1/2[((4pi)/(3)) r^3 rho]v^2`
`=1/2 (4.19)( 2xx10^(-3))^(3) xx 10^(3) xx (10)^(2) = 1.676 × 10^(–3) J`
Work done by the resistive force `= 1.676 xx 10^(-3) J – 0.164 J = – 0.1623 J`.