Question

Eight spherical raindrops of same mass and radius are falling down with a terminal speed of \(6\,cm\,s^{-1}\). If they coalesce to form one big drop, what will be the terminal speed of bigger drop? (Neglect buoyancy of air)

• A. \(1.5\,cm\,s^{-1}\)
• B. \(6\,cm\,s^{-1}\)
• C. \(24\,cm\,s^{-1}\)
• D. \(32\,cm\,s^{-1}\)

Hint:

When drops merge, radius increases as cube root of volume. Terminal velocity varies as \(r^2\).

Answer 1

The Correct Option is C

Step 1: Relation of terminal velocity with radius.
For a spherical drop (Stokes’ law region):
\[ v_t \propto r^2 \]
Step 2: Coalescence of 8 drops.
If 8 identical drops merge, volume becomes 8 times.
\[ \frac{4}{3}\pi R^3 = 8 \cdot \frac{4}{3}\pi r^3 \Rightarrow R^3 = 8r^3 \Rightarrow R = 2r \]
Step 3: Compare terminal velocities.
\[ \frac{v_2}{v_1} = \left(\frac{R}{r}\right)^2 = (2)^2 = 4 \]
Given \(v_1 = 6\,cm\,s^{-1}\):
\[ v_2 = 4 \times 6 = 24\,cm\,s^{-1} \]
Final Answer: \[ \boxed{24\,cm\,s^{-1}} \]