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

Radius of the rain drp , r = 2 mm = ` 2xx10^(-3) m`
Distance moved in hafl kjourney , `h = 500/2 = 250 m`
Density of water = `1000 kg//m^(3)`
Mass of the rain drop = volume `xx `density
` 4/3 pir^(3)rho`
`= 4/3 xx pi (2xx10^(-3))^(3) xx 1000`
` = (32pi)/3 xx10^(-6) kg `
Work done by gravitational force in each half journey ` = W = F xx h = mgh`
` (32pi)/3 xx10^(-6) xx 9.8 xx250 `
` = 0.082 J `
If there are no resistive forces then energy of the rain drop on reaching the ground should be
`E_(1) = mgh `
` = (32pi)/3 xx10^(-6) xx 9.8 xx500`
` = 0.164 J `
`E_(2) = ` Actual energy on raching ground = `1/2 mv^(2)`
` = 1/2 xx (32pi)/3 xx10^(-6) xx (10)^(2)`
`= 1.675 xx10^(-3) J`
Therefore work done by resistive force ` = E = E_(1) - E_(2)`
` = 0.164 - 1.675 xx10^(3)`
` = 0.1623 J `