Two equal drops of water are falling through air with a steady velocity v. If the drops coalesced, what will be the new velocity?

Options

A. `(2)^(1//3)v`
B. `(2)^(3//2)v`
C. `(2)^(2//3)v`
D. `(2)^(1//4)v`

Answer

C

Explanation

Let r be the radius of the each drop. The terminal velocity of drop will be given by
`v=2/9(r^2(rho-sigma)g)/(eta) " " …(i)`
where `rho` is density of drop and `sigma` is density of viscous medium of coefficient of viscosity `eta`.
When two drop each of radius r coalesce to form a new drop, then the radius of coalesced drop will be
`R=(2)^(1//3) r`
Hence, new terminal velocity of coalesced drop will be
`v'=2/9[((2^(1//3)r)^2(rho-sigma)g)/(eta)] " " ...(ii)`
From eqn , (i) and (ii) we get
`(v')/(v) = (2)^(2//3) " or " v' = (2)^(2//3) v`