Find the pressure inside an air bubble of radius 1 mm at a depth of 20 m in water. (Assume ` P _ 0 " " 10 ^5 N//m^ 2`, surface tension of water ` 0.072 N//m` )

Options

A. 300000 Pa
B. 300144 Pa
C. 300288 Pa
D. 300072 Pa

Explanation

To find the pressure inside an air bubble of radius 1 mm at a depth of 20 m in water, we can follow these steps:

### Step 1: Understand the pressure at the depth
The pressure at a depth \( h \) in a fluid is given by the formula:
\[ P_L = P_0 + \rho g h \]
where:
- \( P_0 \) is the atmospheric pressure (given as \( 10^5 \, \text{N/m}^2 \)),
- \( \rho \) is the density of water (approximately \( 10^3 \, \text{kg/m}^3 \)),
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)),
- \( h \) is the depth (given as \( 20 \, \text{m} \)).

### Step 2: Calculate the hydrostatic pressure at 20 m depth
Substituting the values into the equation:
\[ P_L = 10^5 + (10^3)(10)(20) \]
Calculating the term \( (10^3)(10)(20) \):
\[ (10^3)(10)(20) = 200000 \, \text{N/m}^2 \]
Now, substituting this back into the equation for \( P_L \):
\[ P_L = 10^5 + 200000 = 300000 \, \text{N/m}^2 \]

### Step 3: Calculate the excess pressure due to surface tension
The excess pressure inside the bubble due to surface tension is given by:
\[ \Delta P = \frac{2T}{R} \]
where:
- \( T \) is the surface tension of water (given as \( 0.072 \, \text{N/m} \)),
- \( R \) is the radius of the bubble (given as \( 1 \, \text{mm} = 0.001 \, \text{m} \)).

Calculating the excess pressure:
\[ \Delta P = \frac{2 \times 0.072}{0.001} = \frac{0.144}{0.001} = 144 \, \text{N/m}^2 \]

### Step 4: Calculate the total pressure inside the bubble
The total pressure inside the bubble \( P \) is the sum of the hydrostatic pressure \( P_L \) and the excess pressure \( \Delta P \):
\[ P = P_L + \Delta P \]
Substituting the values:
\[ P = 300000 + 144 = 300144 \, \text{N/m}^2 \]

### Final Answer
The pressure inside the air bubble at a depth of 20 m in water is:
\[ P = 300144 \, \text{N/m}^2 \]

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