Question

When a mercury drop of radius $R$ breaks into $n$ droplets of equal size, the radius $r$ of each droplet is

• A. $r = \dfrac{R}{\sqrt{n}}$
• B. $r = \dfrac{R}{n}$
• C. $r = \dfrac{R}{n^{1/3}}$
• D. $r = R n^{1/3}$

Hint:

In breakup problems of liquid drops, always use conservation of volume.

Answer 1

The Correct Option is C

Step 1: Apply conservation of volume.
When a mercury drop breaks into smaller droplets, total volume remains conserved.

Step 2: Write volume expressions.
Volume of original drop:
\[ V = \frac{4}{3}\pi R^3 \] Volume of one small droplet:
\[ v = \frac{4}{3}\pi r^3 \]

Step 3: Equate total volumes.
\[ \frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3 \] \[ R^3 = n r^3 \]

Step 4: Solve for $r$.
\[ r = \frac{R}{n^{1/3}} \]

Step 5: Conclusion.
The radius of each droplet is $\dfrac{R}{n^{1/3}}$.