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

` alpha = 6`

Explanation

` alpha = 6 `
From mass conservation,
` rho .(4)/(3) pi R^3 = rho . K . (4)/(3) pi r ^3 rArr R= K ^(1//3 ) r therefore Delta U = T Delta A = T( K. 4pi r ^2 - 4 pi R^2 ) `
` = T (K . 4pi R^2 K ^ (-2//3) - 4pi R^2 ) , Delta U = 4pi R^2 T [ K^(1//3) - 1 ] `
Putting the values, we get
` 10 ^(-3) = (10^(-1))/(4 pi ) xx 4pi xx 10 ^(-4) [ K ^(1//3) - 1 ], 100 = K ^(1//3) -1 rArr K ^(1//3) cong 100 = 1 `
Given that ` K = 10 ^(alpha ) therefore 10 ^(alpha //3) = 10 ^(2) rArr (alpha ) /(3) = 2 rArr alpha = 6 `