What will be the difference of pressure inside and outside a drop of water of radius 1.0 mm ? (The surface tension of water is 73 dyne/cm-)

Options

A. `146N//m^(2)`
B. `73N//m^(2)`
C. `7.3N//m^(2)`
D. `200N//m^(2)`

Answer

To solve the problem of finding the difference of pressure inside and outside a drop of water with a radius of 1.0 mm and a surface tension of 73 dyne/cm, we can follow these steps:

### Step-by-Step Solution

1. **Identify the formula for excess pressure**:
The difference in pressure (ΔP) between the inside and outside of a droplet due to surface tension is given by the formula:
\[ \Delta P = \frac{2S}{R} \]
where \( S \) is the surface tension and \( R \) is the radius of the droplet.

2. **Convert the surface tension from dyne/cm to SI units (N/m)**:
Given that the surface tension \( S = 73 \) dyne/cm, we convert this to SI units:
\[ 1 \text{ dyne/cm} = 10^{-3} \text{ N/m} \]
Therefore,
\[ S = 73 \text{ dyne/cm} = 73 \times 10^{-3} \text{ N/m} = 0.073 \text{ N/m} \]

3. **Convert the radius from mm to meters**:
The radius \( R = 1.0 \text{ mm} \) can be converted to meters:
\[ R = 1.0 \text{ mm} = 1.0 \times 10^{-3} \text{ m} \]

4. **Substitute the values into the formula**:
Now, we can substitute the values of \( S \) and \( R \) into the formula for ΔP:
\[ \Delta P = \frac{2 \times 0.073 \text{ N/m}}{1.0 \times 10^{-3} \text{ m}} \]

5. **Calculate the difference in pressure**:
Performing the calculation:
\[ \Delta P = \frac{0.146 \text{ N/m}}{1.0 \times 10^{-3} \text{ m}} = 146 \text{ N/m}^2 \]

6. **Conclusion**:
The difference in pressure inside and outside the drop of water is:
\[ \Delta P = 146 \text{ N/m}^2 \]

### Final Answer
The difference of pressure inside and outside a drop of water of radius 1.0 mm is **146 N/m²**.