A spherical drop of water has `1mm` radius. If the surface tension of water is `70xx10^(-3)`N//m, then the difference of pressure between inside and outside of the spherical drop is:

Options

A. `140 N//m^(2)`
B. `14 N//m^(2)`
C. `35 N//M^(2)`
D. None of the above

Explanation

To find the difference of pressure between the inside and outside of a spherical drop of water, we can use the formula:

\[ \Delta P = \frac{2T}{R} \]

where:
- \(\Delta P\) is the difference of pressure,
- \(T\) is the surface tension of the liquid,
- \(R\) is the radius of the drop.

### Step-by-Step Solution:

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

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

3. **Calculate the numerator**:
\[ 2T = 2 \times 70 \times 10^{-3} = 140 \times 10^{-3} \text{ N/m} \]

4. **Calculate the pressure difference**:
\[ \Delta P = \frac{140 \times 10^{-3}}{1 \times 10^{-3}} = 140 \text{ N/m}^2 \]

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