Question

When a big drop of water is formed from n small drops of water, the energy loss is 4 E , where E is the energy of the bigger drop. If \(R\) and \(r\) is the radius of the bigger and the smaller drop respectively then number of smaller drops \(n\) is

• A. \(\frac{5R^{2}}{r^{2}}\)
• B. \(\frac{2R^{2}}{r}\)
• C. \(\frac{5R}{r}\)
• D. \(\frac{5R^{2}}{r}\)

Hint:

Surface energy of a drop is surface tension times area. Equate the energy released to four times the surface energy of the big drop.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Surface energy of a drop of radius \(r\) is \(T\times 4\pi r^2\). When n small drops merge, surface area decreases and the difference is released as energy.

Step 2: Energy Loss
Energy loss \(= T\,4\pi\left(n r^2 - R^2\right)\)
Energy of the bigger drop \(E = T\,4\pi R^2\).

Step 3: Use the Given Condition
Loss \(= 4E\), so \(n r^2 - R^2 = 4R^2\)
\(n r^2 = 5R^2\)

Step 4: Solve for n
\(n = \dfrac{5R^2}{r^2}\), which is option A.

Final Answer:
The number of small drops is 5R squared divided by r squared. \[ \boxed{\text{(A) }\dfrac{5R^{2}}{r^{2}}} \]