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).

Calculate the amount of work done in spliting a water droplet of radiu

1.	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).
Answer» Solution :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`