A water drop falls in air with a uniform velocity. Find the difference between the curvature radii of the drop's surface at the upper and lower points of the drop separated by the distance `h = 2.3 mm`.

Explanation

The radius of curvature of the drop is `R_1` at the upper end of the drop and `R_2` at the lawer end. Then the pressure inside the drop is `p_0 + (2 alpha)/(R_1)` at the top end and `p_0 + (2 alpha)/(R_2)` at the bottom end Hence.
`p_0 + (2 alpha)/(R_1) = p_0 + (2 alpha) /(R_(2)) + rho gh`or `(2 alpha(R_2 - R_1))/(R_1 R_2)= rho gh`
To a first approximation `R_1 ~~ T_2 ~~ (h)/(2)` so `R_2 - R_1 ~~ (1)/(8) rho gh^3//alpha ~~ 0.20 mm`
if `h = 2.3 mm, alpha = 73 mN//m`.