A mercury drop of radius `1.0 cm` is sprayed into `10^(8)` droplets of equal size. Calculate the energy expanded. (Surface tension of mercury `= 32 xx 10^(-2) N//m`).

Answer

To solve the problem of calculating the energy expanded when a mercury drop of radius 1.0 cm is sprayed into \(10^8\) droplets of equal size, we can follow these steps:

### Step 1: Calculate the volume of the original mercury drop
The volume \(V\) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
where \(R\) is the radius of the drop.

Given that the radius \(R = 1.0 \text{ cm} = 0.01 \text{ m}\):
\[ V = \frac{4}{3} \pi (0.01)^3 = \frac{4}{3} \pi (1 \times 10^{-6}) \approx 4.19 \times 10^{-6} \text{ m}^3 \]

### Step 2: Calculate the volume of one droplet
If the original drop is divided into \(n = 10^8\) droplets of equal size, the volume of one droplet \(v\) is:
\[ v = \frac{V}{n} = \frac{4.19 \times 10^{-6}}{10^8} = 4.19 \times 10^{-14} \text{ m}^3 \]

### Step 3: Calculate the radius of one droplet
Using the volume formula for a sphere, we can find the radius \(r\) of one droplet:
\[ v = \frac{4}{3} \pi r^3 \]
Rearranging gives:
\[ r^3 = \frac{3v}{4\pi} \]
Substituting \(v\):
\[ r^3 = \frac{3 \times 4.19 \times 10^{-14}}{4\pi} \approx 1.00 \times 10^{-14} \]
Taking the cube root:
\[ r \approx (1.00 \times 10^{-14})^{1/3} \approx 2.15 \times 10^{-5} \text{ m} \approx 0.0215 \text{ mm} \]

### Step 4: Calculate the change in surface energy
The change in surface energy \(\Delta E\) when the drop is divided into smaller droplets can be calculated using:
\[ \Delta E = \gamma \Delta A \]
where \(\gamma\) is the surface tension and \(\Delta A\) is the change in surface area.

The initial surface area \(A_i\) of the original drop is:
\[ A_i = 4 \pi R^2 = 4 \pi (0.01)^2 \approx 1.26 \times 10^{-4} \text{ m}^2 \]

The final surface area \(A_f\) of \(n\) droplets is:
\[ A_f = n \cdot 4 \pi r^2 = 10^8 \cdot 4 \pi (2.15 \times 10^{-5})^2 \approx 10^8 \cdot 4 \pi (4.62 \times 10^{-10}) \approx 5.79 \times 10^{-2} \text{ m}^2 \]

The change in surface area \(\Delta A\) is:
\[ \Delta A = A_f - A_i \approx 5.79 \times 10^{-2} - 1.26 \times 10^{-4} \approx 5.79 \times 10^{-2} \text{ m}^2 \]

### Step 5: Calculate the energy expanded
Now substituting the values into the energy formula:
\[ \Delta E = \gamma \Delta A = (32 \times 10^{-2}) \cdot (5.79 \times 10^{-2}) \approx 1.85 \times 10^{-3} \text{ J} \]

### Final Answer
The energy expanded is approximately:
\[ \Delta E \approx 1.85 \times 10^{-3} \text{ J} \]

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