A large number of liquid drops each of radius 'a' coalesce to form a single spherical drop of radish b. The energy released in the process is converted into kinetic energy of the big drops formed. The speed of big drop will be

Options

A. `sqrt(T/p (1/r-1/R)`
B. `sqrt((2T)/(p) (1/r-1/R))`
C. `sqrt((4T)/(p) (1/r-1/R))`
D. `sqrt((6T)/(p) (1/r-1/R))`

Answer

D

Explanation

When n drops of radius r combine to form a single drop of radius R.
`n=R^(3)/r^(3) ........(i)`
Then energy released, `E=T[n4pir^(2)-4piR^(2)]=T(4piR^(2)) [R/r-1]` from equation (I)
If this energy is converted into kinetic energy then `1/2mv^(2)=T(4piR^(2)) [R/r-1]`
`1/2 xx [4/3 pi R^(3) p]v^(2)=T(4pi R^(2)) [R/r-1]`
`v^(2)=(6T)/(p) [1/r-1/R]`
`v=sqrt((6T)/(p) [1/r-1/R])`