Question

Energy needed in breaking a drop of radius $R$ into $n$ drops of radii $r$ is given by

• A. $4\pi T(nr^2 - R^2)$
• B. $\frac{4}{3}\pi (r^3 n - R^2)$
• C. $4\pi T(R^2 - nr^2)$
• D. $4\pi T(nr^2 + R^2)$

Hint:

Energy required to break a drop = increase in surface energy.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Energy needed = increase in surface energy.
Step 2: Detailed Explanation:
Initial surface energy = $4\pi R^2 T$. Final surface energy = $n \times 4\pi r^2 T$. Energy needed = final - initial = $4\pi T(nr^2 - R^2)$.
Step 3: Final Answer:
The energy needed is $4\pi T(nr^2 - R^2)$.