A mecury drop of radius 1 cm is sprayed into `10^(5)` droplets of equal size. Calculate the increase in surface energy if surface tension of mercury is `35xx10^(-3)N//m`.

Explanation

To solve the problem of calculating the increase in surface energy when a mercury drop of radius 1 cm is sprayed into \(10^6\) droplets of equal size, we can follow these steps:

### Step 1: Understand the initial and final conditions
- The initial mercury drop has a radius \( R = 1 \, \text{cm} = 0.01 \, \text{m} \).
- The number of droplets formed is \( n = 10^6 \).

### Step 2: Calculate the initial surface area of the mercury drop
The surface area \( A_1 \) of a sphere is given by the formula:
\[ A_1 = 4\pi R^2 \]
Substituting \( R = 0.01 \, \text{m} \):
\[ A_1 = 4\pi (0.01)^2 = 4\pi (0.0001) = 4\pi \times 10^{-4} \, \text{m}^2 \]

### Step 3: Calculate the volume of the initial mercury drop
The volume \( V \) of the drop is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Substituting \( R = 0.01 \, \text{m} \):
\[ V = \frac{4}{3} \pi (0.01)^3 = \frac{4}{3} \pi (10^{-6}) \, \text{m}^3 \]

### Step 4: Calculate the radius of each droplet
Since the volume remains constant when the drop is divided into \( n \) droplets, we have:
\[ V = n \left( \frac{4}{3} \pi r^3 \right) \]
Setting the volumes equal:
\[ \frac{4}{3} \pi (0.01)^3 = 10^6 \left( \frac{4}{3} \pi r^3 \right) \]
Cancelling \( \frac{4}{3} \pi \) from both sides:
\[ (0.01)^3 = 10^6 r^3 \]
\[ 10^{-6} = 10^6 r^3 \]
\[ r^3 = 10^{-12} \implies r = 10^{-4} \, \text{m} = 0.0001 \, \text{m} \]

### Step 5: Calculate the surface area of one droplet
The surface area \( A_2 \) of one droplet is:
\[ A_2 = 4\pi r^2 \]
Substituting \( r = 0.0001 \, \text{m} \):
\[ A_2 = 4\pi (0.0001)^2 = 4\pi (10^{-8}) \, \text{m}^2 \]

### Step 6: Calculate the total surface area of all droplets
The total surface area of \( n \) droplets is:
\[ A_{total} = n A_2 = 10^6 \times 4\pi (10^{-8}) = 4\pi \times 10^{-2} \, \text{m}^2 \]

### Step 7: Calculate the increase in surface energy
The surface energy \( E \) is given by:
\[ E = \text{Surface Tension} \times \text{Surface Area} \]
Let \( \gamma = 35 \times 10^{-3} \, \text{N/m} \).

The initial surface energy \( E_1 \):
\[ E_1 = \gamma A_1 = 35 \times 10^{-3} \times 4\pi \times 10^{-4} \]

The final surface energy \( E_2 \):
\[ E_2 = \gamma A_{total} = \gamma \times 4\pi \times 10^{-2} \]

The increase in surface energy \( \Delta E \):
\[ \Delta E = E_2 - E_1 = \gamma (A_{total} - A_1) \]

### Step 8: Substitute values and calculate
Calculating \( A_{total} - A_1 \):
\[ A_{total} - A_1 = 4\pi \times 10^{-2} - 4\pi \times 10^{-4} = 4\pi (10^{-2} - 10^{-4}) = 4\pi (0.01 - 0.0001) = 4\pi (0.0099) \]

Now substituting back:
\[ \Delta E = 35 \times 10^{-3} \times 4\pi (0.0099) \]
Calculating this gives us the increase in surface energy.

### Final Calculation
\[ \Delta E \approx 4.3 \times 10^{-2} \, \text{J} \]