A mercury drop of radius `10^(-3)m` is broken into `125` equal size droplets.Surface tension of mercury is `0.45Nm^(-1)`.The gain in surface energy is:

Options

A. `5times10^(-5)J`
B. `17.5times10^(-5)J`
C. `2.26times10^(-5)J`
D. `28times10^(-5)J`

Answer

To solve the problem of finding the gain in surface energy when a mercury drop is broken into smaller droplets, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
We have a mercury drop of radius \( R = 10^{-3} \, \text{m} \) that is broken into \( n = 125 \) equal smaller droplets. The surface tension of mercury is given as \( \sigma = 0.45 \, \text{N/m} \).

2. **Calculate the Volume of the Initial Drop**:
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
Substituting \( R = 10^{-3} \, \text{m} \):
\[ V_{\text{initial}} = \frac{4}{3} \pi (10^{-3})^3 = \frac{4}{3} \pi \times 10^{-9} \, \text{m}^3 \]

3. **Volume of the Smaller Droplets**:
When the large drop is broken into \( n = 125 \) smaller droplets, the total volume remains the same:
\[ V_{\text{final}} = n \cdot V_{\text{small}} = 125 \cdot \frac{4}{3} \pi r^3 \]
Setting \( V_{\text{initial}} = V_{\text{final}} \):
\[ \frac{4}{3} \pi (10^{-3})^3 = 125 \cdot \frac{4}{3} \pi r^3 \]
Canceling \( \frac{4}{3} \pi \) from both sides:
\[ (10^{-3})^3 = 125 r^3 \]
\[ 10^{-9} = 125 r^3 \]
\[ r^3 = \frac{10^{-9}}{125} = 8 \times 10^{-12} \]
Taking the cube root:
\[ r = (8 \times 10^{-12})^{1/3} = 2 \times 10^{-4} \, \text{m} \]

4. **Calculate the Surface Areas**:
- Surface area of the initial drop:
\[ A_{\text{initial}} = 4 \pi R^2 = 4 \pi (10^{-3})^2 = 4 \pi \times 10^{-6} \, \text{m}^2 \]
- Surface area of the smaller droplets:
\[ A_{\text{final}} = n \cdot 4 \pi r^2 = 125 \cdot 4 \pi (2 \times 10^{-4})^2 \]
\[ A_{\text{final}} = 125 \cdot 4 \pi \cdot 4 \times 10^{-8} = 2000 \pi \times 10^{-8} \, \text{m}^2 \]

5. **Calculate the Change in Surface Area**:
\[ \Delta A = A_{\text{final}} - A_{\text{initial}} = (2000 \pi \times 10^{-8}) - (4 \pi \times 10^{-6}) \]
\[ \Delta A = (2000 \times 10^{-8} - 400 \times 10^{-8}) \pi = 1600 \times 10^{-8} \pi \, \text{m}^2 \]

6. **Calculate the Gain in Surface Energy**:
The change in surface energy \( \Delta U \) is given by:
\[ \Delta U = \sigma \cdot \Delta A \]
Substituting the values:
\[ \Delta U = 0.45 \cdot (1600 \times 10^{-8} \pi) \]
\[ \Delta U = 720 \times 10^{-8} \pi = 72 \times 10^{-7} \pi \, \text{J} \]
Approximating \( \pi \approx 3.14 \):
\[ \Delta U \approx 72 \times 10^{-7} \times 3.14 \approx 226.08 \times 10^{-7} \approx 2.26 \times 10^{-5} \, \text{J} \]

### Final Answer:
The gain in surface energy is approximately \( 2.26 \times 10^{-5} \, \text{J} \).