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. \( 4\pi T R^2 [\frac{R}{r} + 1] \)
• B. \( 4\pi T R^2 [\frac{R}{r} - 1] \)
• C. \( 4\pi T R^2 [\frac{r}{R} + 1] \)
• D. \( 4\pi T r^2 [\frac{R}{r} - 1] \)

Hint:

Work done in splitting a drop: $W = 4\pi R^2 T (n^{1/3} - 1)$.

Answer 1

The Correct Option is B

Step 1: Surface Area Change
Initial Area $A_1 = 4\pi R^2$.
Final Area $A_2 = n(4\pi r^2)$.
From volume conservation: $\frac{4}{3}\pi R^3 = n(\frac{4}{3}\pi r^3) \implies n = \frac{R^3}{r^3}$.
Step 2: Substitute n
$A_2 = \frac{R^3}{r^3} (4\pi r^2) = 4\pi \frac{R^3}{r}$.
Step 3: Work Done
$W = T \Delta A = T(A_2 - A_1) = T (4\pi \frac{R^3}{r} - 4\pi R^2)$.
$W = 4\pi T R^2 \left( \frac{R}{r} - 1 \right)$.
Final Answer: (B)