Question

A water droplet falls in air and attains terminal velocity \(v_1\). If it splits into \(64\) identical droplets each having terminal velocity \(v_2\). Find \( \dfrac{v_2}{v_1} \).

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{16} \)
• D. \( \frac{1}{32} \)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Terminal velocity of a spherical body falling in a viscous fluid depends on its radius.
When a droplet splits into multiple identical droplets, the total volume (and thus mass) is conserved, which alters the radius of the new droplets.
Step 2: Key Formula or Approach:
Terminal velocity $v_T = \frac{2 r^2 g}{9 \eta} (\rho - \sigma)$, implying $v_T \propto r^2$.
Volume conservation: $V_{initial} = n \times V_{final}$.
Step 3: Detailed Explanation:
Let the radius of the initial large droplet be $R$.
Let the radius of each of the 64 small droplets be $r$.
Using volume conservation:
\[ \frac{4}{3}\pi R^3 = 64 \times \left(\frac{4}{3}\pi r^3\right) \]
\[ R^3 = 64r^3 \implies R = 4r \implies r = \frac{R}{4} \]
Since terminal velocity is directly proportional to the square of the radius ($v \propto r^2$), we have:
\[ \frac{v_2}{v_1} = \left(\frac{r}{R}\right)^2 \]
Substituting $r/R = 1/4$:
\[ \frac{v_2}{v_1} = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \]
Step 4: Final Answer:
The ratio $v_2/v_1$ is $\frac{1}{16}$.