Calculate the work done when a spherical drop of mercury of radius 1 mm falls from some height and breaks into 27droplets.each of the the same size. The surface tension of mermcury is T= 0.47 N/m.

Explanation

Data : R = 1mm = ` 1 xx 10^(-3) m. n = 27` , T = 0.47 N/m
Volume of the drop = volume of n droplets
` 4/3 pi R^(3) = n xx 4/3 pir^(3)`
` r= R/root(3)(N)= R/(root(3) (27)) = R/3`
Surface area of the drop = ` 4 pi R^(2)`
Surface area of n droplets = `n xx 4 pir^(2)`
Increase in surface area = ` 4pi (n r^(2) - R^(2)) = 4 pi (27 xx R^(2)/9 - R^(2))`
` 2 xx 4 pi R^(2)`
The work done = surface tension` xx ` increase in surface area
` T (8 pi R^(2)) = 0.47 xx 8 xx 4 xx 3.142(1xx 10^(-3))^(2)`
` = 1.181 xx 10^(-5)` J