The work done in splitting a drop of water of 1 mm radius into `10^(6)` drops is (S.T. of water `=72xx10^(-3)J//m^(2)`)

Options

A. `8.95xx10^(-5)J`
B. `10.5xx10^(-5)J`
C. `6.5xx10^(-5)J`
D. `8xx10^(-4)J`

Answer

A

Explanation

In splitting, the volume remains constant.
i.e., `(4)/(3)piR^(3)=(4)/(3)pir^(3)xxn`
or `R=(n^(1//3))rimpliesr=(R)/((n^(1//3)))`
Work done=S.T.`xx` change in area
change in area`=-4piR^(2)+n4pir^(2)`
`implies=4pi(-R^(2)+(nR^(2))/(n^(2//3)))`
`implies=4piR^(2)(n^(1//3)-1)implies` As `n=10^(6)`
so, change in area
`=4piR^(2)[(10^(6))^(1//3)-1]=4piR^(2)(10^(6)-1)`
`=396piR^(2)`
so,work done `=72xx10^(-3)xx396piR^(2)`
`=72xx10^(-3)xx396xx3.14xx(1xx10^(-4))^(2)`
`=8.95xx10^(-5)J`