A mercury drop of radius 1 cm is sprayed into `10^6` drops of equal size. The energy expended in joule is (Surface tension of mercury is `460 xx10 N m^-1` )

Options

A. 0.057
B. 5.7
C. `5.7xx10^-4`
D. `5.7xx10^-3`

Answer

A

Explanation

Volume of big drop = Sum of volume small drops
`4/3piR^(3) = 10^(6) 4/3 pi r^(3)`
`therefore r=R/10^(2)`
Energy consumed, E= work done,
`W= Delta A xxT=(n4pi^(2) - 4piR^(2)]T`
`= 4pi[n(R/10^(2))^(2) - R^(2)] T = 4piR^(2) [ n/10^(n) -1]T`
0.057J