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 bigger drop. The radius of the bigger drop is ' R ' and that of smaller drop is ' \(r\) ' then the value of ' \(n\) ' is

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

Hint:

Energy $\propto$ surface area for liquid drops.

Answer 1

The Correct Option is B

Concept:
Surface energy: \[ E = T \times \text{surface area} \] Step 1: Initial energy. \[ E_i = n \times 4\pi r^2 T \]
Step 2: Final energy. \[ E_f = 4\pi R^2 T \]
Step 3: Energy loss. \[ E_i - E_f = 3E_f \] \[ n(4\pi r^2 T) - 4\pi R^2 T = 3(4\pi R^2 T) \]
Step 4: Solve. \[ n r^2 = 4R^2 \Rightarrow n = \frac{4R^2}{r^2} \]
Step 5: Conclusion. \[ n = \frac{4R^2}{r^2} \]