The change is energy when a big drop is split in small n droplets is

Options

A. `4R^(2)(n^(2//3)-1)T`
B. `4piR^(2)(n^(1//3)-1)T`
C. `4piR^(2)(n^(-1//3)-1)T`
D. `4piR^(2)[n^(-2//3)-1]T`

Answer

B

Explanation

When a drop of radius R split into n smaller drops ( each of radius r ) , then surface area of liquid increases. Hence, the work is to be done against surface tension.
Since , the volume of liquid remains constant
`:. (4)/(3) pi R^(3)=n""(4)/(3)pir^(3)" ":. R^(3)=nr^(3)`
Work done ` =T xx DeltaA`
`=T[n4 pir^(2)-4pi R^(2)]`
`=4 pi R^(2)T(n^(1//3)-1)`