Question

A mercury drop of radius `R` is sprayed into `n` droplets of equal size. Calculate the energy expanded if surface tension of mercury is `T`.

Answer 1

If drop of radiur `R` is sprayed into `n` droplets of equla radius `r` then as a drop has only onen surface, the initial surface area will be `4piR^(2)` while final area is `n(4pir^(2))`. So the increase in area
`/_S=n(pir^(2))=4piT[nr^(2)-R^(2)]`
Now since the total volume of `n` droplets is same as that of initial drop i.e.
`4/3piR^(3)=n[(4/3)pir^(3)]`
or `r=R/n^(1//3)`.........ii
Putting the value of `r` from eqnn ii in eqn i
`W=4piR^(2)T[(n)^(1//3)1]`