The surface tension of water is `7xx10^(-2)` N/m . The work required to break a drop of water of radius 0.5 cm into identical drops , each of radius 1 mm is

Options

A. `8.8xx10^(-4)`J
B. `8.8xx10^(-5)` J
C. `4.4xx10^(-4)` J
D. `4.4xx10^(-5)` J

Answer

To solve the problem, we need to calculate the work required to break a drop of water of radius 0.5 cm into identical drops, each of radius 1 mm. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the volume of the original drop
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
For the original drop with radius \( R = 0.5 \) cm:
\[ V_{\text{original}} = \frac{4}{3} \pi (0.5 \, \text{cm})^3 \]
Converting \( 0.5 \) cm to meters:
\[ 0.5 \, \text{cm} = 0.005 \, \text{m} \]
Now calculating the volume:
\[ V_{\text{original}} = \frac{4}{3} \pi (0.005)^3 = \frac{4}{3} \pi (1.25 \times 10^{-7}) \approx 5.24 \times 10^{-7} \, \text{m}^3 \]

### Step 2: Calculate the volume of one smaller drop
For the smaller drops with radius \( r = 1 \) mm:
\[ r = 1 \, \text{mm} = 0.001 \, \text{m} \]
Calculating the volume of one smaller drop:
\[ V_{\text{small}} = \frac{4}{3} \pi (0.001)^3 = \frac{4}{3} \pi (1 \times 10^{-9}) \approx 4.19 \times 10^{-9} \, \text{m}^3 \]

### Step 3: Calculate the number of smaller drops formed
Since the total volume remains the same, we can find the number of smaller drops \( n \):
\[ n = \frac{V_{\text{original}}}{V_{\text{small}}} = \frac{5.24 \times 10^{-7}}{4.19 \times 10^{-9}} \approx 125 \]

### Step 4: Calculate the change in surface area
The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi r^2 \]
Calculating the surface area of the original drop:
\[ A_{\text{original}} = 4 \pi (0.005)^2 = 4 \pi (2.5 \times 10^{-5}) \approx 3.14 \times 10^{-4} \, \text{m}^2 \]
Calculating the surface area of one smaller drop:
\[ A_{\text{small}} = 4 \pi (0.001)^2 = 4 \pi (1 \times 10^{-6}) \approx 1.26 \times 10^{-5} \, \text{m}^2 \]
Calculating the total surface area of all smaller drops:
\[ A_{\text{total small}} = n \cdot A_{\text{small}} = 125 \cdot 1.26 \times 10^{-5} \approx 1.575 \times 10^{-3} \, \text{m}^2 \]

### Step 5: Calculate the change in surface area
\[ \Delta A = A_{\text{total small}} - A_{\text{original}} = 1.575 \times 10^{-3} - 3.14 \times 10^{-4} \approx 1.26 \times 10^{-3} \, \text{m}^2 \]

### Step 6: Calculate the work done
The work done \( W \) is given by:
\[ W = \text{Surface Tension} \times \Delta A \]
Given that the surface tension \( \gamma = 7 \times 10^{-2} \, \text{N/m} \):
\[ W = 7 \times 10^{-2} \times 1.26 \times 10^{-3} \approx 8.82 \times 10^{-5} \, \text{J} \]

### Final Answer
The work required to break the drop is approximately:
\[ W \approx 8.82 \times 10^{-5} \, \text{J} \]

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