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 n drops = Volume of one drop
` :. (4/3 pia^(3))=4/3 pib^(3)" " n = (b^(3))/(a^(3))`
Decrease in area `= n (4pia^(2)) - 4pib^(2)`
` = 44pi (na^(2)-b^(2))`
`=4pi ((b^(3))/(a^(3))xxa^(2)-b^(2))`
` = 4 pib^(3) (1/a-1/b)`
`:. ` Energy released = (S.T) `xx` (decrease in area )
` = T. 4pib^(3) (1/a -1/b) ` ....(1)
This energy released gives K.E to the the big drop .
` :. K.E =1/2 mv^(2) = 1/2 (4/3 pib^(3)) rho xx v^(2)" "` ...(2)
From (1) and (2 ) , we get
` 1/2 (4/3 pib^(3))rhov^(2)=T.4pib^(3)(1/a-1/b)`
` v^(2) =(6T)/rho (1/a-1/b)`
` :. v = sqrt((6T)/rho (1/a-1/b))`