A mercury drop of radius 1.0 cm is sprayed into `10^(6)` droplets of equal sizes. The energy expended in this process is (Given, surface tension of mercury is `32 xx 10^(-2) Nm^(-1)`)

Explanation

To solve the problem of calculating the energy expended when a mercury drop of radius 1.0 cm is sprayed into \(10^6\) droplets of equal sizes, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Radius of the large mercury drop, \( R = 1.0 \, \text{cm} = 0.01 \, \text{m} \)
- Number of droplets, \( n = 10^6 \)
- Surface tension of mercury, \( \gamma = 32 \times 10^{-2} \, \text{N/m} = 0.32 \, \text{N/m} \)

2. **Calculate the Volume of the Large Drop**:
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
Substituting \( R \):
\[ V = \frac{4}{3} \pi (0.01)^3 = \frac{4}{3} \pi (1 \times 10^{-6}) \approx 4.19 \times 10^{-6} \, \text{m}^3 \]

3. **Volume of Each Small Droplet**:
Since the large drop is divided into \( 10^6 \) smaller droplets, the volume of each droplet \( v \) is:
\[ v = \frac{V}{n} = \frac{4.19 \times 10^{-6}}{10^6} = 4.19 \times 10^{-12} \, \text{m}^3 \]

4. **Calculate the Radius of Each Small Droplet**:
Let \( r \) be the radius of each small droplet. The volume of one small droplet is:
\[ v = \frac{4}{3} \pi r^3 \]
Setting this equal to the volume calculated:
\[ \frac{4}{3} \pi r^3 = 4.19 \times 10^{-12} \]
Solving for \( r^3 \):
\[ r^3 = \frac{4.19 \times 10^{-12} \times 3}{4 \pi} \approx 1.00 \times 10^{-12} \]
Therefore,
\[ r \approx (1.00 \times 10^{-12})^{1/3} \approx 1.00 \times 10^{-4} \, \text{m} = 0.1 \, \text{cm} \]

5. **Calculate the Change in Surface Area**:
The change in surface area \( \Delta A \) when the large drop is split into smaller droplets is given by:
\[ \Delta A = n \cdot 4 \pi r^2 - 4 \pi R^2 \]
Substituting the values:
\[ \Delta A = 10^6 \cdot 4 \pi (1.00 \times 10^{-4})^2 - 4 \pi (0.01)^2 \]
Simplifying:
\[ \Delta A = 10^6 \cdot 4 \pi (1.00 \times 10^{-8}) - 4 \pi (1.00 \times 10^{-4}) = 4 \pi (10^{-2} - 10^{-4}) \approx 4 \pi (0.0099) \approx 0.124 \, \text{m}^2 \]

6. **Calculate the Work Done**:
The work done \( W \) due to surface tension is given by:
\[ W = \gamma \Delta A \]
Substituting the values:
\[ W = 0.32 \times 0.124 \approx 0.03968 \, \text{J} \]

### Final Answer:
The energy expended in the process is approximately \( 0.03968 \, \text{J} \).