A drop of liquid of radius `R=10^(2)`m having surface tension `S=(0.1)/(4pi)N//m^(-1)` divides itself into K identical drop. In this process the total change in the surface energy `Delta U =10^(-3)` J. If `K=10^(alpha)` , then the value of `alpha` is :

Answer

6

Explanation

Let r be the radius of each small drop. Then volume of big drop = K x volume of small drop.
or `(4)/(3)pi R^(3) =Kxx(4)/(3)pi r^(3)` or `r=R K^(-1//3)`
Change in surface energy
`Delta U =S.T.` x change in surface area
`=Sxx[K 4pi r^(2)-4 pi R^(2)]`
`=Sxx[K 4 pi R^(2) K^(-2//3) - 4pi R^(2)]`
`=4 pi R^(2) S[K^(1//3) - 1]`
`:. 10^(-3) =4 pi (10^(-2))^(2)xx(0.1)/(4pi)[(10^(alpha))^(1//3) - 1]`
or `100=10^(alpha//3) - 1` or `10^(alpha//3) =100+1 ~~10^(2)`
or `(alpha)/(3)= 2` or `alpha =3xx2=6`