Two water drops of equal size are falling through air with constant velocity of `2(m)/(s)`. If the two drops are allowed to coalesce to form a single drop, what would be its terminal velocity?

Explanation

Ratio of volumes of the smaller and bigger water drop `=1:2`
So, ratio of their radii `=1:2^(1//3)`
Terminal velocity,
`v=2/9(r^2(rho-sigma)g)/(eta), i.e. v prop r^2`
`therefore (v_1)/(v_2)=(r_1)/(r_2)^2`
or, `v_2=v_1(r_2/r_1)^2=2xx(2^(1//3))/(1)^2=2^(5//3)`
`root(3)(32)=3.175 m . s^-1`