Question

A spherical liquid drop of radius \(R\) acquires the terminal velocity \(v_1\) when falls through a gas of viscosity \(\eta\). Now the drop is broken into 64 identical droplets and each droplet acquires terminal velocity \(v_2\) falling through the same gas. The ratio of terminal velocities \(v_1/v_2\) is ________.

• A. 4
• B. 0.25
• C. 32
• D. 16

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When a large drop breaks into smaller identical droplets, the total volume of the liquid remains conserved. This volume conservation allows us to find the radius of the smaller droplets. Terminal velocity of a falling sphere in a viscous fluid is directly proportional to the square of its radius.
Step 2: Key Formula or Approach:
1. Conservation of volume: $\frac{4}{3}\pi R^3 = N \times \frac{4}{3}\pi r^3$, where $N=64$.
2. Terminal velocity formula: $v = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta} \implies v \propto r^2$.
Step 3: Detailed Explanation:
Equating the volume of the original large drop to the total volume of the 64 smaller droplets:
$V_{initial} = V_{final}$
$\frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3$.
Cancel out $\frac{4}{3}\pi$ from both sides:
$R^3 = 64 r^3$.
Take the cube root of both sides:
$R = 4r \implies r = \frac{R}{4}$.
The terminal velocity $v$ of a drop is proportional to the square of its radius ($v \propto r^2$).
For the large drop: $v_1 \propto R^2$.
For the small droplet: $v_2 \propto r^2$.
Taking the ratio of the two velocities:
$\frac{v_1}{v_2} = \frac{R^2}{r^2} = \left(\frac{R}{r}\right)^2$.
Substitute $r = R/4$:
$\frac{v_1}{v_2} = \left(\frac{R}{R/4}\right)^2 = (4)^2 = 16$.
Step 4: Final Answer:
The ratio of terminal velocities $v_1/v_2$ is 16.