Question

The energy needed for breaking a liquid drop of radius 'R' into 'n' droplets each of radius 'r' is
[ \(T\) = surface tension of the liquid]

• A. \(4πTR^2[\frac{R}{r}-1]\)
• B. \(4πTR[\frac{R}{r}-1]\)
• C. \(4πT[\frac{R^2}{r^2}-1]\)
• D. \(4πT[1+\frac{R^3}{r}]\)

Hint:

The energy needed equals surface tension times the increase in surface area.

Answer 1

The Correct Option is A

Step 1: Understand the concept
Breaking a drop into smaller droplets increases the total surface area. The work done equals the surface tension times the increase in area: \(W = T\,\Delta A\).

Step 2: Find the number of droplets
The volume is conserved: \(\frac{4}{3}\pi R^3 = n\cdot\frac{4}{3}\pi r^3\), so \(n = \dfrac{R^3}{r^3}\).

Step 3: Change in area
\[ \Delta A = n\cdot4\pi r^2 - 4\pi R^2 = 4\pi\left(\frac{R^3}{r} - R^2\right) = 4\pi R^2\left(\frac{R}{r} - 1\right) \]

Step 4: Result
\[ W = 4\pi T R^2\left[\frac{R}{r} - 1\right] \]
Option (A). The other options have wrong powers of \(R\) or \(r\) and do not reduce to zero when \(r = R\) (where no energy should be needed).

Final Answer:
The energy is 4 pi T R^2 (R/r - 1). This is option (A). \[ \boxed{\text{(A) }4\pi TR^2\left[\frac{R}{r}-1\right]} \]