An air bubble is lying just below the surface of water. The surface tension of water is `70xx10^(-3)Nm^(-1)` and atmospheric pressure is `1.013xx10^(5)Nm^(-2)`. If the radius of bubble is 1mm, then the pressure inside the bubble will be-

Options

A. `1.0270xx10^(5)Pa`
B. `1.0160xx10^(5)Pa`
C. `1.0144xx10^(5)Pa`
D. `1.0131xx10^(5)Pa`

Answer

To find the pressure inside the air bubble lying just below the surface of water, we can use the formula for the pressure difference due to surface tension. The formula is given by:

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

where:
- \(\Delta P\) is the change in pressure,
- \(T\) is the surface tension of the liquid,
- \(r\) is the radius of the bubble.

### Step-by-Step Solution:

1. **Identify the given values**:
- Surface tension of water, \(T = 70 \times 10^{-3} \, \text{N/m}\)
- Atmospheric pressure, \(P_{\text{atm}} = 1.013 \times 10^{5} \, \text{N/m}^2\)
- Radius of the bubble, \(r = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m}\)

2. **Calculate the change in pressure (\(\Delta P\))**:
Using the formula for pressure difference:
\[ \Delta P = \frac{2T}{r} \]
Substitute the values:
\[ \Delta P = \frac{2 \times (70 \times 10^{-3})}{1 \times 10^{-3}} = \frac{140 \times 10^{-3}}{1 \times 10^{-3}} = 140 \, \text{N/m}^2 \]

3. **Determine the pressure inside the bubble (\(P_{\text{inside}}\))**:
The pressure inside the bubble is the sum of the atmospheric pressure and the change in pressure:
\[ P_{\text{inside}} = P_{\text{atm}} + \Delta P \]
Substitute the values:
\[ P_{\text{inside}} = 1.013 \times 10^{5} + 140 = 1.0144 \times 10^{5} \, \text{N/m}^2 \]

4. **Final Result**:
The pressure inside the bubble is:
\[ P_{\text{inside}} = 1.0144 \times 10^{5} \, \text{N/m}^2 \]

### Conclusion:
The pressure inside the bubble is \(1.0144 \times 10^{5} \, \text{N/m}^2\), which corresponds to option C.