Eight equal drops of water each of radius r = 2mm are falling through air with a terminal velocity of 16 cm/s. The eight drops combine to form a big drop. Calculate the square root of magnitude of terminal velocity (in cm/s) of big drop.

Sure, let's solve this step by step.

1. First, we need to understand that the terminal velocity of a droplet is proportional to the square of its radius. This is derived from Stokes' Law, which states that the force experienced by small particles suspended in a fluid medium is proportional to the radius squared.

2. Given that the terminal velocity of the small droplets is 16 cm/s and their radius is 2 mm, we can write the proportionality as: V1 ∝ r1^2, where V1 is the terminal velocity of the small droplets and r1 is their radius.

3. Now, when the eight droplets combine to form a larger droplet, the volume of the larger droplet is eight times the volume of a small droplet. Since the volume of a sphere (which we can consider a droplet to be) is proportional to the cube of its radius, we can write: r2^3 = 8*r1^3, where r2 is the radius of the larger droplet.

4. Solving this equation for r2, we get: r2 = 2*r1.

5. Now, we can find the terminal velocity of the larger droplet using the proportionality: V2 ∝ r2^2, where V2 is the terminal velocity of the larger droplet.

6. Since r2 = 2r1, we can write: V2 ∝ (2r1)^2 = 4*r1^2.

7. Therefore, the terminal velocity of the larger droplet is four times the terminal velocity of the small droplets, or V2 = 4*V1.

8. Substituting the given value of V1 = 16 cm/s, we get: V2 = 4*16 cm/s = 64 cm/s.

9. Finally, the square root of the terminal velocity of the larger droplet is √V2 = √64 cm/s = 8 cm/s.