A liquid drop of diameter 4 mm breaks into 1000 droplets of equal size. Calculate the resultant change in surface energy, the surface tension of the liquid is `0.07 Nm^(-1)`

Answer

3168

Explanation

Volume of 1000 droplets = Volume of larger drop `1000 xx 4/3 pi r^3 = 4/3 piR^3`
`r=R/10 = (2 xx10^(-3)m)/10 = 2 xx10^(-4)m`
Surface area of larger drop `= 4pi R^2 = 4pi xx (2xx10^(-3))^(2) = 16pi xx10^(-6)m^2`
Surface area of 1000 droplets `= 4pir^2 xx1000 = 4pi xx (2xx10^(-4))^(2)xx1000`
`= 16pi xx 10^(-5)m^2`
`:.` Increase in surface area `= 16pi xx10^(-6) (10-1) = 144pi xx 10^(-6)m^2`
The resultant increase in surface = energy = Surface tension x increase in surface area
`=0.07 xx144 xx22/7 xx10^(-6)= 3168 xx10^(-8)J`