Question

Two identical rain drops are falling through air each with a terminal velocity \(V\). If the two drops coalesce to form one single big drop, then the terminal velocity of the big drop is

• A. \(\dfrac{V}{2^{2/3}}\)
• B. \(2^{2/3}V\)
• C. \(2^{1/3}V\)
• D. \(\dfrac{V}{2^{1/3}}\)

Hint:

Under Stokes' law, \[ \boxed{ V_t\propto r^2 } \] and when \(n\) identical drops coalesce, \[ \boxed{ R=n^{1/3}r. } \]

Answer 1

The Correct Option is B

Step 1: Relate the radii of the drops. When two identical drops coalesce, \[ \frac{4}{3}\pi R^3 = 2\left(\frac{4}{3}\pi r^3\right), \] which gives \[ R=2^{1/3}r. \]

Step 2: Use the expression for terminal velocity. According to Stokes' law, \[ V_t\propto r^2. \] Hence, \[ \frac{V_{\text{big}}}{V} = \left(\frac{R}{r}\right)^2 = \left(2^{1/3}\right)^2 = 2^{2/3}. \] Therefore, \[ V_{\text{big}} = 2^{2/3}V. \] Thus, \[ \boxed{2^{2/3}V} \] Hence, \[ \boxed{(B)} \] is the correct answer.