Question

When a big drop of water is formed from n small drops of water, the energy loss is 3E, where E is the energy of the big drop. If R is the radius of big drop and r is the radius of the smaller drop then number of smaller drops n is

• A. \(\frac{4R}{r^{2}}\)
• B. \(\frac{4R}{r}\)
• C. \(\frac{2R^{2}}{r}\)
• D. \(\frac{4R^{2}}{r^{2}}\)

Hint:

The energy released equals surface tension times the decrease in surface area, and this is set equal to 3E.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
When small drops merge, surface area decreases and energy is released: \(\text{energy loss}=T\,(\text{initial area}-\text{final area})\).

Step 2: Energy of the Big Drop
\(E=T\cdot 4\pi R^{2}\).

Step 3: Write the Energy Loss
\(T\left(n\cdot 4\pi r^{2}-4\pi R^{2}\right)=3E=3T\cdot 4\pi R^{2}\)

Step 4: Solve for n
\(nr^{2}-R^{2}=3R^{2}\), so \(nr^{2}=4R^{2}\) and \(n=\frac{4R^{2}}{r^{2}}\), option (D).

Final Answer:
The number of small drops is 4R squared over r squared. \[ \boxed{\text{(D) }\dfrac{4R^{2}}{r^{2}}} \]