Step 1: Understanding the Concept:
When a large liquid drop breaks into multiple smaller droplets, the total volume is conserved, but the total exposed surface area increases significantly.
Because surface energy is directly proportional to surface area, the total surface energy of the system increases after fragmentation.
Step 2: Key Formula or Approach:
1. Volume Conservation: $V_{\text{large\_drop}} = n \times V_{\text{small\_droplet}}$
\[ \frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3 \implies R = n^{1/3} r \]
2. Surface Energy: $E = T \times A$, where $T$ is surface tension and $A = 4\pi R^2$ is the spherical surface area.
Step 3: Detailed Explanation:
Let the initial radius of the large drop be $R$ and its surface tension be $T$.
Initial surface energy, $E = T \times 4\pi R^2$.
It is divided into $n = 729$ equal droplets of radius $r$.
Using volume conservation to relate $R$ and $r$:
\[ R^3 = 729 r^3 \]
Taking the cube root of both sides (since $9^3 = 729$):
\[ R = 9 r \implies r = \frac{R}{9} \]
Now, calculate the final total surface energy $E_{\text{final}}$ of the 729 droplets:
\[ E_{\text{final}} = 729 \times (\text{Surface energy of one small droplet}) \]
\[ E_{\text{final}} = 729 \times \left( T \times 4\pi r^2 \right) \]
Substitute $r = R/9$ into the equation:
\[ E_{\text{final}} = 729 \times T \times 4\pi \left(\frac{R}{9}\right)^2 \]
\[ E_{\text{final}} = 729 \times T \times 4\pi \frac{R^2}{81} \]
\[ E_{\text{final}} = \left(\frac{729}{81}\right) \times \left( T \times 4\pi R^2 \right) \]
Since $729 / 81 = 9$:
\[ E_{\text{final}} = 9 \times E \]
Step 4: Final Answer:
The final surface energy of the droplets is $9\text{ E}$.