Step 1: Understanding the Concept:
Surface energy is the energy required to create a certain area of a liquid surface. It is directly proportional to the surface area.
When a large drop is sprayed into many smaller droplets, the total volume remains constant, but the total surface area increases. Because the surface area increases, the total surface energy also increases.
Step 2: Key Formula or Approach:
1. Conservation of Volume: $V_{\text{initial}} = n \times V_{\text{final\_droplet}}$, where $n$ is the number of droplets.
$\frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3 \implies R = n^{1/3} r \implies r = \frac{R}{n^{1/3}}$.
2. Surface Energy Formula: $E = T \times A$, where $T$ is surface tension and $A = 4\pi R^2$ is the surface area.
Step 3: Detailed Explanation:
Let the initial radius of the large drop be $R$.
Its initial surface energy is given as $V$:
\[ V = T \times 4\pi R^2 \]
The drop is broken into $n = 1000$ equal droplets of radius $r$.
Using volume conservation to find the relationship between $R$ and $r$:
\[ R^3 = 1000 r^3 \]
Taking the cube root of both sides:
\[ R = 10 r \implies r = \frac{R}{10} \]
Now calculate the total surface energy of the new 1000 droplets, let's call it $V_{\text{new}}$:
\[ V_{\text{new}} = 1000 \times (\text{Surface energy of one small droplet}) \]
\[ V_{\text{new}} = 1000 \times \left( T \times 4\pi r^2 \right) \]
Substitute $r = R/10$:
\[ V_{\text{new}} = 1000 \times T \times 4\pi \left(\frac{R}{10}\right)^2 \]
\[ V_{\text{new}} = 1000 \times T \times 4\pi \frac{R^2}{100} \]
\[ V_{\text{new}} = \left(\frac{1000}{100}\right) \times \left( T \times 4\pi R^2 \right) \]
\[ V_{\text{new}} = 10 \times (V) \]
Step 4: Final Answer:
The total surface energy of all the droplets is $10\text{ V}$.