Question

Eight equal drops of water are falling through air with a steady speed of \( 10 \, \text{cm/s} \). If the drops coalesce, the new velocity is:

• A. \( 10 \, \text{cm/s} \)
• B. \( 40 \, \text{cm/s} \)
• C. \( 16 \, \text{cm/s} \)
• D. \( 5 \, \text{cm/s} \)

Hint:

When drops merge, the radius of the combined drop increases as the cube root of the total volume. Since terminal velocity depends on the square of the radius, the new velocity scales quadratically with the radius increase.

Answer 1

The Correct Option is B

When small drops merge, their combined volume is conserved. Let \( r \) represent the radius of an individual small drop, and \( R \) denote the radius of the resulting larger drop.

Step 1: Volume Conservation The total volume of eight small drops equals the volume of the combined drop: \[ 8 \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3. \] Simplifying this equation yields: \[ R^3 = 8r^3 \quad \implies \quad R = 2r. \] 

Step 2: Terminal Velocity Relationship The terminal velocity \( v_T \) of a drop is directly proportional to the square of its radius: \[ v_T \propto r^2. \] Given that \( v_1 = 10 \, \text{cm/s} \) is the terminal velocity of the small drops, and \( v_2 \) is the terminal velocity of the combined drop, the following ratio holds: \[ \frac{v_1}{v_2} = \left( \frac{r}{R} \right)^2. \] Substituting \( R = 2r \) into the ratio: \[ \frac{v_1}{v_2} = \left( \frac{r}{2r} \right)^2 = \frac{1}{4}. \] 

Step 3: Calculate \( v_2 \) Solving for \( v_2 \): \[ v_2 = 4v_1 = 4 \cdot 10 = 40 \, \text{cm/s}. \] 

Final Answer: \[ \boxed{40 \, \text{cm/s}} \]