Step 1: Understanding the Concept:
The total surface energy of a liquid drop is the product of its surface area and the surface tension \(T\).
When smaller drops coalesce into a larger one, the total surface area decreases, leading to a release (loss) of surface energy.
Step 2: Key Formula or Approach:
Energy of a drop: \(U = T \cdot (4\pi R^2)\).
Given: Energy of bigger drop \(E = T(4\pi R^2)\).
Total initial energy (from \(n\) drops): \(U_{\text{total, initial}} = n \cdot T(4\pi r^2)\).
Step 3: Detailed Explanation:
Energy loss = Initial energy \(-\) Final energy.
Given energy loss is \(3E\):
\[ U_{\text{total, initial}} - E = 3E \]
\[ U_{\text{total, initial}} = 4E \]
Substitute the expressions for energy:
\[ n \cdot T(4\pi r^2) = 4 \cdot T(4\pi R^2) \]
Cancel common terms \(T\) and \(4\pi\):
\[ n r^2 = 4 R^2 \]
Solve for \(n\):
\[ n = \frac{4R^2}{r^2} \]
(Note: From volume conservation, we also know \(n = (R/r)^3\). This allows us to find numerical values if needed, e.g., \(R/r = 4 \implies n = 64\)).
Step 4: Final Answer:
The value of \(n\) is \(\frac{4R^2}{r^2}\).