A liquid drop of diameter 4 mm breaks into 1000 droplets of equal size. Calculate the resultant change in surface energy, the surface tension of the liquid is `0.07 Nm^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the initial radius of the liquid drop
Given the diameter of the liquid drop is 4 mm, we can find the radius (R) using the formula:
\[ R = \frac{\text{Diameter}}{2} = \frac{4 \text{ mm}}{2} = 2 \text{ mm} = 2 \times 10^{-3} \text{ m} \]

### Step 2: Calculate the radius of the smaller droplets
When the original drop breaks into 1000 smaller droplets of equal size, the volume of the original drop must equal the total volume of the smaller droplets.

The volume of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]

Let \( r \) be the radius of the smaller droplets. The volume of the original drop is:
\[ V_{\text{original}} = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (2 \times 10^{-3})^3 \]

The volume of the 1000 smaller droplets is:
\[ V_{\text{small}} = 1000 \times \frac{4}{3} \pi r^3 \]

Setting the two volumes equal gives:
\[ \frac{4}{3} \pi (2 \times 10^{-3})^3 = 1000 \times \frac{4}{3} \pi r^3 \]

Cancelling \( \frac{4}{3} \pi \) from both sides:
\[ (2 \times 10^{-3})^3 = 1000 r^3 \]

Now, solving for \( r^3 \):
\[ r^3 = \frac{(2 \times 10^{-3})^3}{1000} \]
\[ r^3 = \frac{8 \times 10^{-9}}{1000} = 8 \times 10^{-12} \]

Taking the cube root:
\[ r = (8 \times 10^{-12})^{1/3} = 2 \times 10^{-4} \text{ m} \]

### Step 3: Calculate the change in surface area
The change in surface area (\( \Delta A \)) when the drop breaks into smaller droplets can be calculated as:
\[ \Delta A = 1000 \times 4 \pi r^2 - 4 \pi R^2 \]
\[ \Delta A = 4 \pi (1000 r^2 - R^2) \]

Substituting the values of \( r \) and \( R \):
\[ \Delta A = 4 \pi \left(1000 \times (2 \times 10^{-4})^2 - (2 \times 10^{-3})^2\right) \]
\[ = 4 \pi \left(1000 \times 4 \times 10^{-8} - 4 \times 10^{-6}\right) \]
\[ = 4 \pi \left(4 \times 10^{-5} - 4 \times 10^{-6}\right) \]
\[ = 4 \pi \left(3.6 \times 10^{-5}\right) \]

Calculating:
\[ \Delta A = 4 \times \frac{22}{7} \times 3.6 \times 10^{-5} \approx 4.57 \times 10^{-4} \text{ m}^2 \]

### Step 4: Calculate the change in surface energy
The change in surface energy (\( \Delta E \)) is given by:
\[ \Delta E = \text{Surface Tension} \times \Delta A \]
\[ \Delta E = 0.07 \, \text{N/m} \times 4.57 \times 10^{-4} \text{ m}^2 \]
\[ \Delta E \approx 3.199 \times 10^{-5} \text{ J} \]

### Final Answer
The resultant change in surface energy is approximately \( 3.199 \times 10^{-5} \text{ J} \).

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