Question

The energy needed for breaking a liquid drop of radius 'R' into 216 droplets, each of radius 'r' is 'x' times $TR^2$. The value of 'x' is [ T = surface tension of the liquid].

• A. $4\pi$
• B. $12\pi$
• C. $180\pi$
• D. $20\pi$

Hint:

If $n$ droplets are formed, $r = R / n^{1/3}$ and the new area is $n^{1/3}$ times the old area.

Answer 1

The Correct Option is D

Step 1: Volume Conservation
Volume of big drop = $216 \times$ Volume of small droplet.
$\frac{4}{3}\pi R^3 = 216 \times \frac{4}{3}\pi r^3 \Rightarrow R = \sqrt[3]{216} \cdot r \Rightarrow R = 6r \Rightarrow r = \frac{R}{6}$.
Step 2: Surface Energy Formula
Energy $= T \cdot \Delta A = T(A_{final} - A_{initial})$.
Step 3: Area Calculation
$A_{initial} = 4\pi R^2$.
$A_{final} = 216 \times 4\pi r^2 = 216 \times 4\pi (\frac{R}{6})^2 = 216 \times 4\pi \frac{R^2}{36} = 6 \times 4\pi R^2 = 24\pi R^2$.
Step 4: Finding x
Energy $= T(24\pi R^2 - 4\pi R^2) = 20\pi TR^2$.
Comparing with $x TR^2$, we get $x = 20\pi$.
Final Answer:(D)