Hello there!

Greetings!

It's quite easy;

Let n be the number of little droplets which coalesce to form single drop. Then,
Volume of n little droplets = Volume of single large drop
So, n 4/3 r = 4/3 R or nr = R
Decrease in surface area = n 4r - 4R
= 4 [ nr - R] = 4 [ nr/r - R]
= 4 [ R/r - R]
= 4R [ 1/r - 1/R] As nr = R

Now; energy released,
E = Surface tension decrease in surface area
= 4SR [ 1/r - 1/R]


The mass of bigger drop,
M = 4/3 R 1 = 4/3R
E = 4/3 SR 3 [ 1/r - 1/R]
= 3SM [ 1/r - 1/R] { M = 4/3 R }

KE of bigger droplets =heat energy released

= 1/2 MV = 3SM [ 1/r - 1/R]

V = 6S(R - r/ Rr).


Thankyou