A drop of water of radius 1 mm is to be divided into `10^6` point drops of equal size. How much mechanical work should be done ? The surface tension of water = 72 dyn. `cm^-1`.

Explanation

Let the radius of each point drop be r
`therefore (4)/(3)pir^3xx10^6=(4)/(3)pi((1)/(10)) or, r=0.001 cm`
The surface area of the original drop `=4pi((1)/(10))^2 cm^2` and the total surface area of `10^6` point drops
`=10^6xx4pi(0.001)^2=4picm^2`
`therefore ` Increase in surface area
`=4pi-4pi((1)/(10))^2=3pixx0.99 cm^2`
`therefore` Mechanical work done = increases in surface area `xx` surface tension
`=72xx4pixx0.99=895.73 erg.`