Work done in splitting a drop of water of 1 mm radius into 10^(6) droplets is (Surface tension of water=`72xx10^(-3)J//m^(2)`)

Options

A. `8.95xx10^(-5)` erg
B. `8.95xx10^(-5)` joule
C. `17.90xx10^(-5)` joule
D. `17.90xx10^(-5)` erg

Answer

To solve the problem of calculating the work done in splitting a drop of water of radius 1 mm into \(10^6\) droplets, we will follow these steps:

### Step 1: Understand the initial and final conditions
- We have a single large drop with radius \(r = 1 \text{ mm} = 1 \times 10^{-3} \text{ m}\).
- We want to split this drop into \(n = 10^6\) smaller droplets.

### Step 2: Calculate the initial surface area of the large drop
The surface area \(A\) of a sphere is given by the formula:
\[ A = 4\pi r^2 \]
For the large drop:
\[ A_{\text{initial}} = 4\pi (1 \times 10^{-3})^2 = 4\pi \times 10^{-6} \text{ m}^2 \]

### Step 3: Determine the radius of the smaller droplets
Using the principle of conservation of volume, the volume of the large drop must equal the total volume of the \(10^6\) smaller droplets. The volume \(V\) of a sphere is given by:
\[ V = \frac{4}{3}\pi r^3 \]
For the large drop:
\[ V_{\text{initial}} = \frac{4}{3}\pi (1 \times 10^{-3})^3 = \frac{4}{3}\pi \times 10^{-9} \text{ m}^3 \]
For \(n\) smaller droplets, the total volume is:
\[ V_{\text{final}} = n \times \frac{4}{3}\pi R^3 \]
Setting these equal:
\[ \frac{4}{3}\pi (1 \times 10^{-3})^3 = 10^6 \times \frac{4}{3}\pi R^3 \]
Cancelling \(\frac{4}{3}\pi\) from both sides:
\[ (1 \times 10^{-3})^3 = 10^6 R^3 \]
Solving for \(R\):
\[ R^3 = \frac{(1 \times 10^{-3})^3}{10^6} = \frac{10^{-9}}{10^6} = 10^{-15} \]
\[ R = (10^{-15})^{1/3} = 10^{-5} \text{ m} = 0.01 \text{ mm} \]

### Step 4: Calculate the final surface area of the \(10^6\) droplets
The surface area of one smaller droplet is:
\[ A_{\text{final}} = 4\pi R^2 = 4\pi (10^{-5})^2 = 4\pi \times 10^{-10} \text{ m}^2 \]
Thus, the total surface area for \(10^6\) droplets is:
\[ A_{\text{total}} = n \times A_{\text{final}} = 10^6 \times 4\pi \times 10^{-10} = 4\pi \times 10^{-4} \text{ m}^2 \]

### Step 5: Calculate the change in surface area
The change in surface area \(\Delta A\) is given by:
\[ \Delta A = A_{\text{total}} - A_{\text{initial}} = 4\pi \times 10^{-4} - 4\pi \times 10^{-6} \]
Factoring out \(4\pi\):
\[ \Delta A = 4\pi (10^{-4} - 10^{-6}) = 4\pi (0.0001 - 0.000001) = 4\pi (0.000099) = 4\pi \times 9.9 \times 10^{-5} \]

### Step 6: Calculate the work done using surface tension
The work done \(W\) is given by:
\[ W = \text{Surface Tension} \times \Delta A \]
Given that the surface tension of water is \(72 \times 10^{-3} \text{ J/m}^2\):
\[ W = 72 \times 10^{-3} \times 4\pi \times 9.9 \times 10^{-5} \]
Calculating this gives:
\[ W \approx 72 \times 10^{-3} \times 4 \times 3.14 \times 9.9 \times 10^{-5} \approx 8.95 \times 10^{-5} \text{ J} \]

### Final Answer
The work done in splitting the drop is approximately:
\[ \boxed{8.95 \times 10^{-5} \text{ J}} \]