Question

If a number of little droplets of water, each of radius r, coalesce to form a single drop of radius R, and the energy released is converted into kinetic energy then find out the velocity acquired by the bigger drop.

Answer 1

Hi

Here is your answer,

Let n be the number of little droplets which coalesce to form single drop. Then,

Volume of n little droplets = Volume of single drop

or n × 4/3 πr³ = 4/3 πR³ or  nr³ = R³

Decrease in surface area = n × 4πr² - 4πR²

      = 4π [ nr² - R²] = 4π [ nr³/r - R²]

      = 4π [ R³/r - R²] = 4πR³ [ 1/r - 1/R]
                                                                                { ∴ nr³ = R³ }

The energy released,

           E = Surface tension × decrease in surface area

 = 4πSR³ [ 1/r - 1/R]

The mass of bigger drop,

              M = 4/3 πR³ × 1 = 4/3πR³

              E = 4/3 πSR³ × 3 [ 1/r - 1/R]

                     = 3SM [ 1/r - 1/R]                      { ∴ M = 4/3 πR³ }

 ∴ KE of bigger droplets = Energy released 

         = 1/2 MV² = 3SM [ 1/r - 1/R]

                 V = √6S(R - r/ Rr)


Hope it helps you !