The diameter of one drop of water is 0.2 cm. The work done in breaking one drop into 1000 droplets will be (Given, `S_("water") = 7 xx 10^(-2) Nm^(-1)`)

Options

A. `7.9 xx 10^(-6) J`
B. `5.92 xx 10^(-6) J`
C. `2.92 xx 10^(-6) J`
D. `1.92 xx 10^(-6)J`

Explanation

To solve the problem of calculating the work done in breaking one drop of water into 1000 smaller droplets, we will follow these steps:

### Step 1: Calculate the initial radius of the drop
Given the diameter of the drop is 0.2 cm, we can find the radius (R) as follows:
\[ R = \frac{\text{Diameter}}{2} = \frac{0.2 \, \text{cm}}{2} = 0.1 \, \text{cm} = 0.1 \times 10^{-2} \, \text{m} = 1 \times 10^{-3} \, \text{m} \]

### Step 2: Calculate the radius of the smaller droplets
Since the original drop is broken into 1000 smaller droplets, we can find the radius (r) of each droplet using the volume conservation principle. The volume of the original drop is equal to the total volume of the 1000 smaller droplets:
\[ V_{\text{original}} = V_{\text{small}} \times 1000 \]
The volume of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Thus, we have:
\[ \frac{4}{3} \pi R^3 = 1000 \times \frac{4}{3} \pi r^3 \]
Cancelling \(\frac{4}{3} \pi\) from both sides, we get:
\[ R^3 = 1000 \times r^3 \]
From this, we can express \(r\):
\[ r = \frac{R}{10} \]
Substituting \(R = 1 \times 10^{-3} \, \text{m}\):
\[ r = \frac{1 \times 10^{-3}}{10} = 1 \times 10^{-4} \, \text{m} \]

### Step 3: Calculate the change in surface area
The change in surface area (\(\Delta A\)) when breaking the drop can be calculated as:
\[ \Delta A = A_{\text{final}} - A_{\text{initial}} \]
Where:
- \(A_{\text{initial}} = 4 \pi R^2\)
- \(A_{\text{final}} = 1000 \times 4 \pi r^2\)

Substituting the values:
\[ A_{\text{initial}} = 4 \pi (1 \times 10^{-3})^2 = 4 \pi \times 10^{-6} \, \text{m}^2 \]
\[ A_{\text{final}} = 1000 \times 4 \pi (1 \times 10^{-4})^2 = 1000 \times 4 \pi \times 10^{-8} \, \text{m}^2 = 4 \pi \times 10^{-5} \, \text{m}^2 \]

Now, calculating \(\Delta A\):
\[ \Delta A = 4 \pi (4 \times 10^{-5} - 10^{-6}) = 4 \pi (3.9 \times 10^{-5}) = 15.6 \pi \times 10^{-5} \, \text{m}^2 \]

### Step 4: Calculate the work done
The work done (W) in breaking the drop is given by:
\[ W = S \times \Delta A \]
Where \(S\) is the surface tension of water, given as \(7 \times 10^{-2} \, \text{N/m}\):
\[ W = 7 \times 10^{-2} \times 15.6 \pi \times 10^{-5} \]
Calculating this:
\[ W \approx 7 \times 10^{-2} \times 15.6 \times 3.14 \times 10^{-5} \approx 7 \times 10^{-2} \times 49.02 \times 10^{-5} \]
\[ W \approx 3.43 \times 10^{-6} \, \text{J} \]

### Final Answer
Thus, the work done in breaking one drop into 1000 droplets is approximately:
\[ W \approx 7.92 \times 10^{-6} \, \text{J} \]