What should be the pressure inside a small air bubble of 0.2mm diameter situated just below the surface of water. `("Surface tension of water" = 0.072N//m)`

Options

A. `1.44xx10^(2)Pa`
B. `1.44xx10^(3)Pa`
C. `1.44xx10^(4)Pa`
D. `1.44xx10^(5)Pa`

Explanation

To find the pressure inside a small air bubble of diameter 0.2 mm situated just below the surface of water, we can follow these steps:

### Step 1: Identify the given values
- Diameter of the air bubble, \(d = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m}\)
- Radius of the air bubble, \(r = \frac{d}{2} = \frac{0.2 \times 10^{-3}}{2} = 0.1 \times 10^{-3} \, \text{m}\)
- Surface tension of water, \(T = 0.072 \, \text{N/m}\)
- Atmospheric pressure, \(P_{\text{atm}} = 1.013 \times 10^{5} \, \text{N/m}^2\)

### Step 2: Calculate the excess pressure due to surface tension
The excess pressure inside the bubble due to surface tension is given by the formula:
\[ \Delta P = \frac{2T}{r} \]
Substituting the values:
\[ \Delta P = \frac{2 \times 0.072}{0.1 \times 10^{-3}} = \frac{0.144}{0.1 \times 10^{-3}} = 0.144 \times 10^{4} \, \text{N/m}^2 \]
\[ \Delta P = 1.44 \times 10^{3} \, \text{N/m}^2 \]

### Step 3: Calculate the total pressure inside the bubble
The total pressure inside the bubble is the sum of the atmospheric pressure and the excess pressure:
\[ P = P_{\text{atm}} + \Delta P \]
Substituting the values:
\[ P = 1.013 \times 10^{5} + 1.44 \times 10^{3} \]
To add these, we can express \(1.44 \times 10^{3}\) in terms of \(10^{5}\):
\[ 1.44 \times 10^{3} = 0.0144 \times 10^{5} \]
Now,
\[ P = 1.013 \times 10^{5} + 0.0144 \times 10^{5} = (1.013 + 0.0144) \times 10^{5} \]
\[ P = 1.0274 \times 10^{5} \, \text{N/m}^2 \]

### Final Answer
The pressure inside the small air bubble is:
\[ P \approx 1.0274 \times 10^{5} \, \text{N/m}^2 \]

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