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

Volume of one big drop
= Volume of n identical droplets
` (4pi)/3 b^(3) = n (4pi)/3 a^(3)`
`b = n^(1//3)a`
By the law conversation of energy ,
KE= Energy released
`1/2 mv^(2) = 4pib^(2) T [ n^(1//3)-1]`
` :. v^(2) = (8pib^(2)T)/m [ n^(1//3) - 1]`
` v = sqrt((8pib^(2)T)/(rho (4pi)/3b^(3))[n^(1//3)-1])`
` sqrt((6T)/rho xx1/b [ b/a-1])`
` v = sqrt((6T)/rho [1/a-1/b])`