Step 1: Understanding the Question:
Breaking a large drop into small droplets increases the total surface area, which requires work (energy). This work is provided by multiplying surface tension with the change in surface area.
Step 2: Key Formula or Approach:
1. Volume conservation: $\frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3 \implies R^3 = n r^3$
2. Energy $W = T \cdot \Delta A = T(n \cdot 4\pi r^2 - 4\pi R^2)$
Step 3: Detailed Explanation:
Given $n = 216$:
$R = \sqrt[3]{216} \cdot r = 6r \implies r = \frac{R}{6}$.
Now, calculate the increase in surface area:
\[ \Delta A = 216 \times 4\pi r^2 - 4\pi R^2 \]
\[ \Delta A = 216 \times 4\pi \left(\frac{R}{6}\right)^2 - 4\pi R^2 \]
\[ \Delta A = 216 \times 4\pi \left(\frac{R^2}{36}\right) - 4\pi R^2 \]
\[ \Delta A = 6 \times 4\pi R^2 - 4\pi R^2 = 24\pi R^2 - 4\pi R^2 = 20\pi R^2 \]
Work done:
\[ W = T \cdot \Delta A = T(20\pi R^2) = 20\pi TR^2 \]
Comparing with the given form $x TR^2$, we get $x = 20\pi$.
Step 4: Final Answer:
The value of x is $20\pi$.