Question

\({}'n'\) small drops of water having same size fall through air with constant velocity \(4\) cm/s. They coalesce to form a big drop. The terminal velocity of the big drop is

• A. \(2n^{2/3}cm/s\)
• B. \(12n^{2/3}cm/s\)
• C. \(4n^{2/3}cm/s\)
• D. \(8n^{2/3}cm/s\)

Hint:

Terminal velocity is proportional to the square of the radius, and the coalesced drop radius is n to the power one third times the small radius.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Terminal velocity of a spherical drop is \(v\propto r^{2}\).

Step 2: Radius of the big drop
Volume is conserved: \(\frac{4}{3}\pi R^{3}=n\cdot\frac{4}{3}\pi r^{3}\)
So \(R=n^{1/3}r\).

Step 3: Calculation
\(\frac{V}{v}=\frac{R^{2}}{r^{2}}=n^{2/3}\)
\(V=4\,n^{2/3}\) cm/s.

Step 4: Compare the options
This is option (C).

Final Answer:
The big drop has terminal velocity 4 n to the power two thirds cm/s. \[ \boxed{\text{(C) }4n^{2/3}\,\text{cm/s}} \]