Surface tension of water is `0.072 N m^(-1)`. The excess pressure inside a water drop of diameter 1.2 mm is

Options

A. `240 Nm^(-2)`
B. `120 Nm^(-2)`
C. `60 Nm^(-2)`
D. `72 Nm^(-2)`

Explanation

To find the excess pressure inside a water drop, we can use the formula for excess pressure in a droplet, which is given by:

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

where:
- \(\Delta P\) is the excess pressure,
- \(T\) is the surface tension,
- \(R\) is the radius of the droplet.

### Step-by-Step Solution:

1. **Identify the given values**:
- Surface tension of water, \(T = 0.072 \, \text{N/m}\)
- Diameter of the water drop, \(d = 1.2 \, \text{mm}\)

2. **Convert the diameter to radius**:
- The radius \(R\) is half of the diameter:
\[ R = \frac{d}{2} = \frac{1.2 \, \text{mm}}{2} = 0.6 \, \text{mm} \]
- Convert millimeters to meters:
\[ R = 0.6 \, \text{mm} = 0.6 \times 10^{-3} \, \text{m} = 0.0006 \, \text{m} \]

3. **Substitute the values into the excess pressure formula**:
\[ \Delta P = \frac{2T}{R} = \frac{2 \times 0.072 \, \text{N/m}}{0.0006 \, \text{m}} \]

4. **Calculate the excess pressure**:
- First, calculate the numerator:
\[ 2T = 2 \times 0.072 = 0.144 \, \text{N/m} \]
- Now, substitute this value into the formula:
\[ \Delta P = \frac{0.144 \, \text{N/m}}{0.0006 \, \text{m}} = 240 \, \text{N/m}^2 \]

5. **Final answer**:
- The excess pressure inside the water drop is:
\[ \Delta P = 240 \, \text{N/m}^2 \]