Question

Determine the ratio of surface energy of \(1\) large drop to \(1\) small drop, if \(1000\) small drops combined to form \(1\) large drop.

• A. \(100:1\)
• B. \(10:1\)
• C. \(1000:1\)
• D. \(1:1\)

Hint:

Volume conservation gives \(R = 10r\); surface energy goes as radius squared.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Surface energy is surface tension times area, so it is proportional to \(r^2\). The volume stays the same when drops merge.

Step 2: Find the large drop radius:
\(1000\times\frac43\pi r^3 = \frac43\pi R^3\), so \(R^3 = 1000r^3\) and \(R = 10r\).

Step 3: Ratio:
\[ \frac{E_{\text{large}}}{E_{\text{small}}} = \frac{4\pi R^2T}{4\pi r^2T} = \frac{R^2}{r^2} = 100 \]
So the ratio is \(100:1\). Note this compares one drop with one drop. The total surface energy of the 1000 small drops is \(1000\) times that of one drop, which is \(10\) times that of the large drop, but that is a different ratio.

Final Answer:
The ratio is \(100:1\), option (A). \[ \boxed{100:1} \]