A water drop of radius R is split into n smaller drops , each of radius r. If T is the surface tension of water ,then the work done in the process is

Options

A. `4/3piR^(3)T(1/r-1/R)`
B. `3/4piR^(3)T(1/R-1/r)`
C. `4piR^(3)T(1/r-1/R)`
D. `6piR^(2)(1/R-1/r)`

Answer

C

Explanation

Volume of one drop = volume of n drops
` :. 4/3 piR^(3)=(4/3 pir^(3)) xx n " " R^(3) = nr^(3)`
Change in area ` = 4pir^(2) -4piR^(2) = 4pi [nr^(2)-R^(2)]`
` :. dA = 4pi [ (nr^(3))/r - R^(2)] = 4pi [(R^(3))/r- R^(2)] `
`dA = 4pi R^(3) [1/r-1/R] `
` :. ` Work done `= T xx dA = 4piR^(3) T [ 1/r -1/R]`