Two equal drops of water falling through air with a steady velocity `5 cm//s`. If the drops combire to from a single drop, what will be new terminal velocity?

Explanation

Here, `v=5 cms^(-1)` .Let r be the radius of each drop. Terminal velocity of each drop is
`v=(2pi r^(2)(rho-sigma)g)/(9 eta)=5` ……(i)
Let R be the radius of by drop formed when two small drop coalese together. Then
`(4)/(3) piR^(3) =2xx(4)/(3) pi r^(3)` or `R=2^(1//3) r`
Terminal velocity of big drop is
`v=(2R^(2)(rho- sigma)g)/(9 eta)` .....(ii)
Dividing (ii) by (i), we get
`(V)/(5) =(R^(2))/(r^(2))` or `V=5xx(R^(2))/(r^(2))`
`V=5xx((2^(1//3)r)^(2))/(r^(2))=5xx2^(2//3)=5xx1.5874`
`=7.937 cms^(-1)`