Work done in forming a liquid drop of radius R is `W_(1)` and that of radius 3R is `W_(2)`. The ratio of work done is

Answer

10

Explanation

As we know, work done ` W = T xx 4 pi R^2 `
` rArr (W_1)/(W_2) = (T xx 4 pi R ^ 2 )/( T xx 4pi (3R ) ^ 2 ) = (T xx 4pi R ^ 2 )/( T xx 36 pi R ^ 2 ) = (1)/(9) therefore W_1 : W _ 2 = 1 : 9 `