A spherical drop of water has a diameter of 2 mm . If the surface tension of water is `70 xx 10^(3)` N/m , then the difference of pressures between the inside and outside of the spherical drop is

Options

A. `70 N//m^(2)`
B. `90 N//m^(2)`
C. `110 N//m^(2)`
D. `140 N//m^(2)`

Explanation

To solve the problem of finding the difference of pressures between the inside and outside of a spherical drop of water, we will use the formula for excess pressure due to surface tension.

### Step-by-Step Solution:

1. **Identify the given values:**
- Diameter of the spherical drop, \( D = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Surface tension of water, \( T = 70 \times 10^{-3} \, \text{N/m} \)

2. **Calculate the radius of the spherical drop:**
- The radius \( R \) is half of the diameter:
\[ R = \frac{D}{2} = \frac{2 \times 10^{-3}}{2} = 1 \times 10^{-3} \, \text{m} \]

3. **Use the formula for excess pressure:**
- The formula for the excess pressure \( P \) inside a spherical drop due to surface tension is given by:
\[ P = \frac{2T}{R} \]

4. **Substitute the known values into the formula:**
- Substitute \( T \) and \( R \) into the formula:
\[ P = \frac{2 \times (70 \times 10^{-3})}{1 \times 10^{-3}} \]

5. **Calculate the excess pressure:**
- Perform the calculations:
\[ P = \frac{140 \times 10^{-3}}{1 \times 10^{-3}} = 140 \, \text{N/m}^2 \]

6. **Final result:**
- The difference of pressures between the inside and outside of the spherical drop is:
\[ P = 140 \, \text{N/m}^2 \]

### Summary:
The difference of pressures between the inside and outside of the spherical drop is \( 140 \, \text{N/m}^2 \).