Fog consists of a huge amount of tiny water droplets hanging motionless in the air. A drop of water

Fog consists of a huge amount of tiny water droplets hanging motionless in the air. A drop of water with a radius of R0 = 1 mm begins to fall to the ground from a height of 5 m, “absorbing” the encountered droplets. Assuming that the drop retains the shape of a ball, find its smallest possible diameter before falling to the ground. Consider n = 7 × 106 fog droplets per unit volume on average. The mass of one droplet is 1 mg. the volume of the ball is calculated by the formula V = (4/3) πR3, the density of water is 10 ^ 3 kg / m3.

Since the concentration of fog droplets is n = 7 * 10 ^ 6, the number of droplets per 1 m will be: n ^ (1/3) = (7 * 10 ^ 6) ^ 1/3 = 191, then the total number of fog droplets is: N = 5 * 191 = 955.
The mass of the initial drop is m0 = ρ * V = ρ * 3 / 4πRo ^ 3 = 1000 * 3 / 4π * 10 ^ -9 = 2.36 * 10-6kg = 2.36mg.
Drop weight when falling: mo + N * mt = 2.36 + 955 * 1 = 957.4 mg,
and its radius:
R = m * 4 / 3πρ = (957.4 * 10 ^ -6) * 4/3000 * π = 1.27 * 10 ^ -6m = 1.27mm
D = 2R = 2 * 1.27 = 2.6mm