A large number of droplets, each of radius a, coalesce to form a bigger drop of radius `b`. Assume that the energy released in the process is converted into the kinetic energy of the drop. The velocity of the drop is `sigma =` surface tension, `rho =` density)

Options

A. `[(sigma)/(rho)((1)/(a)-(1)/(b))]^(1)/(2)`
B. `[(2sigma)/(rho)((1)/(a)-(1)/(b))]^((1)/(2))`
C. `[(3sigma)/(rho)((1)/(a)-(1)/(b))]^((1)/(2)`
D. `[(6sigma)/(rho)((1)/(a)-(1)/(b))]^((1)/(2))`

Answer

D

Explanation

Energy released `[nxx4pia^(2)-4pib^(2)]sigma`
Now, `4xx(4)/(3)pir^(3)=(4)/(3)pib^(3)` or `n=(b^(3))/(a^(3))`
Therefore, energy released is
`[(b^(3))/(a^(3))xx4pia^(2)-4pib^(2)]sigma=4pib^(2)((b)/(a)-1)sigma`
Now, `(1)/(2)((4)/(3)pib^(2))rhov^(2)=4pib^(2)[(b)/(a)-1]sigma`
or `v=[(6sigma)/(rho)((1)/(a)-(1)/(b))]^((1)/(2))`