Calculate the amount of work done in spliting a water droplet of radius 0.5 mm into 8 million identical droplets. Given surface tension of water `=0.072 Nm^(-2)`.

Explanation

Given, `R=0.5 xx 10^(-3)m" "n=8 xx 10^(6)`
From the law of conservation of mass,
mass of single big drop=mass of 8 million small droplets.
i.e, `(4/3piR^(3))rho=8 xx 10^(6) (4/3pir^3rho)`
i.e. `R^(3)=8 xx 10^(6) r^(3)`
i.e, `r=(R)/(2 xx 10^(2))m`
Increase in S.A. `=8 xx 10^(6) (4pir^(2))=4pi (8 xx 10^(6) (R^(2))/(4 xx 10^(4)) -R^(2))`
`=4pi R^(2)(200-1)=4 xx 3.142 xx 199 xx (5 xx 10^(-4))^(2)`
`DeltaA=6.252 xx 10^(-4)m^(2)`
Hence work done=(surface tension) `(DeltaA)`
`=0.072 xx 6.252 xx 10^(-4) =4.50 xx 10^(-5)J`