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 realeased `=[nxx4pia^(2)=-4pib^(2)]sigma`
Now `nxx4/3pir^(3)=4/3pib^(3)` or `n=(b^(3))/(a^(3))`
Therefore energy released is
`=[(b^(3))/(a^(3))=4pia^(2)-4pib^(2)]sigma=4pib^(2)[b/a-1]sigma`
Now `1/2(4/3pib^(3))rhov^(2)=4pib^(2)[b/a-1]sigma`
or `v=[(6sigma)/(rho)(1/a-1/b)]^(1/2)`