Question

If \( n \) small identical liquid drops, each having terminal velocity \( v \), merge together, then the terminal velocity of the bigger drop is:

• A. \( n^2 v \)
• B. \( n^{1/3} v \)
• C. \( \frac{v}{n} \)
• D. \( nv \)
• E. \( n^{2/3} v \)

Hint:

The terminal velocity of merging drops increases with the size of the drop raised to the \( 2/3 \) power.

Answer 1

Answer: (E)

Step 1: Understanding Terminal Velocity and Volume Conservation The terminal velocity \( v \) of a small drop is given by Stokes' law: \[ v \propto R^2 \] where \( R \) is the radius of the drop. When \( n \) small identical drops merge, the volume remains conserved: \[ n \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \] where \( r \) is the radius of each small drop and \( R \) is the radius of the bigger drop. 
Step 2: Finding the Radius of the Bigger Drop From volume conservation, \[ n r^3 = R^3 \] \[ R = n^{1/3} r \] 
Step 3: Finding the Terminal Velocity of the Bigger Drop Since terminal velocity is proportional to the square of the radius, we can write: \[ V \propto R^2 \] \[ V \propto (n^{1/3} r)^2 \] \[ V \propto n^{2/3} r^2 \] Since the terminal velocity of the small drop is \( v \propto r^2 \), we can write: \[ V = n^{2/3} v \] 
Thus, the terminal velocity of the bigger drop is \( n^{2/3} v \).