A drop of radius r is broken into n equal drips. Calulate the work done if surface tension of water is T.

Options

A. `4rpiR^(2)nT`
B. `4rpiR^(2)T(n^(2//3-1))`
C. `4rpiR^(2)T(n^(1//3-1))`
D. None of the above

Explanation

The volume of n smaller drop =volume of bigger drop
`:. " " n(4)/(3)pir^(3)=(4)/(3)piR^(2)`
`:. " " R=n^(1//3)r`
`:. " " r=(R)/(n^(1//3))`
Here, R=radius of bigger drop
r=radius of smallar drop
`:. W=4pi(nr^(2)-R^(2))T`
`4pi[n((R)/(n^(1//2)))^(2)-R^(2)]T=4pi{(nR^(2))/(n^(2//3))-R^(2)}T`
`=4piR^(2)(n^(1-2//3)-1)T=4piR^(2)T(n^(1//3)-L)`