help me with this physics problem it's urgent?

Question

the mercury drop of 1 radius is sprayed into 10^6 droplets of equal size. calculate work done

tanx
tom

Answer 1

If the area of the liquid surface has to be increased work has to be done against the force of surface tension. The work done to form a film is stored as potential energy in the surface and the amount of this energy per unit area of this surface under isothermal condition is the "intrinsic surface energy" or free surface energy density.
work done in small displacement dX is
dW = f x dX
therefore, W=2TLX=TA
therefore T = W/A

the problem
R=1cm = 10^-2
M = 4/3 (pi) R^3 p where p is the surface density
This is broken into 10^6 small drops, each of radius r and mass m
M = 10^6 m
or 4/3 (pi) R^3 p = 10^6 . 4/3 (pi) r^3 p
R = 100r
r = R/100=10^-2/100 = 1 x 10^-4
energy spent in spraying droplets = TA
= T(10^6 x 4(pi) r^2 - 4 (pi) R^2)
= 4(pi) T (10^6 r^2 - R ^-3)
=4.356 x 10^-3.
surface tension of mearcury = T = 35 x 10^-3

Answer 2

the formula for work is force times distance, i just dont know how to use the information you gave to solve for work