Question

Energy needed in breaking a liquid drop of radius \( R \), into \( n \) smaller drops each of radius \( r \), is
[ \( T \) - Surface tension of the liquid ]

• A. \( (4\pi r^2 n - 4\pi R^2)T \)
• B. \( \left(\frac{4}{3}\pi r^3 n - \frac{4}{3}\pi R^3\right)T \)
• C. \( (4\pi R^2 - 4\pi r^2)nT \)
• D. \( \frac{(4\pi R^2 - n4\pi r^2)}{T} \)

Hint:

Always equate initial and final surface energies when dealing with droplet division. Surface energy = surface area × surface tension.

Answer 1

The Correct Option is A

The energy required to break a large drop into smaller drops is equal to the increase in surface energy.
Surface energy \( E = \text{Surface area} \times T \)
- Surface area of the original drop = \( 4\pi R^2 \)
- Total surface area of \( n \) small drops = \( n \cdot 4\pi r^2 \)
So the increase in surface area: \[ \Delta A = 4\pi r^2 n - 4\pi R^2 \] Hence, the required energy: \[ E = T \cdot \Delta A = (4\pi r^2 n - 4\pi R^2)T \]