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

Answer

A

Explanation

Suppose n number of liquid are merged. Equating initial final volume we have,
`V_(i) = V_(f) rArr n xx (4)/(3) pi a^(3) = (4)/(3)pi b^(3)`
`rArr b = (n)^(1//3)a`
`rArr n = ((b)/(a))^(3)`
Initial surface energy, `U_(i) = n xx 4pi a^(2) xx S`
Final surface energym `U_(f) = 4pi b^(2) xx S`
Change in surface energy,
`Delta U = U_(i) - U_(f) = 4pi s[n a^(2) - b^(2)]`
`= 4piT[a^(2)((b)/(a))^(3) - b^(2)] = 4pi T[(b^(3))/(a)-b^(2)]`
Kinetic energy of the bigger drop,
`K = (1)/(2) mv^(2) = (1)/(2) xx rho xx (4)/(3)pi b^(3) xx v^(2)`
According to the question, `K = Delta U`
`rArr rho xx (4)/(3)pi b^(3) v^(2) = 4pi T[(b^(3))/(a) - b^(2)]`
`rArr rho b^(3) v^(2) = 2 xx 3T[(b^(3))/(a)-b^(2)]`
`rArr v^(2) = (6T)/(rho)[(1)/(a)-(1)/(b)]^(1//2)`
Speed of the bigger drop, `v = [(6T)/(rho)((1)/(a)-(a)/(b))]^(1//2)`