The work done to get 'n' smaller equal size spherical drops from a bigger size spherical drop of water is proportional to :

Options

A. `1//n^(2//3)`-1
B. `1//n^(1//3)-1`
C. `n^(1//3)-1`
D. `n^(4//3)-1`

Answer

C

Explanation

Here volume is conserved.
`(4piR^(3))/3=n(4pir^(3))/3rArrR^(3)=nr^(3)rArrr=n^(-1//3)R`
Now increase in area=`n4pir^(2)-4piR^(2)=4pi[nr^(2)-R^(2)]`
`=4pi[n.n^(-2//3)R^(2)-R^(2)]`
`=4piR^(2)[n^(1//3)-1]`
`therefore` Work done `= T xx` increase in area
`W=T.4piR^(2)[n^(1//3)-1]`
`rArrWprop(n^(1//3)-1)`
Hence the correct choice is (c).