A raindrop of radius ‘r’ falls in air with terminal speed vt. What is the terminal speed of a raindrop of radius 2r

The terminal speed of a falling object is determined by the balance between the force of gravity pulling it down and the drag force acting upward. For a raindrop, these forces are given by:

1. Gravity force (Fg) = mass * gravity = volume * density * gravity = 4/3 * π * r³ * ρ * g
2. Drag force (Fd) = 1/2 * drag coefficient * air density * cross-sectional area * speed² = 1/2 * Cd * ρ_air * π * r² * v²

At terminal speed, these two forces balance each other, so we have:

4/3 * π * r³ * ρ * g = 1/2 * Cd * ρ_air * π * r² * v²

Solving for v (terminal speed), we get:

v = sqrt((8/3) * (r * ρ * g) / (Cd * ρ_air))

Now, if we double the radius (r -> 2r), the terminal speed becomes:

v' = sqrt((8/3) * (2r * ρ * g) / (Cd * ρ_air))

Since the radius is under the square root, the terminal speed increases by a factor of sqrt(2), not 2. So, the terminal speed of a raindrop of radius 2r is sqrt(2) times the terminal speed of a raindrop of radius r.