An air bubble of radius 1 mm is located at a depth of 20 cm below water level. The excess pressure inside the bubble above the atmospheric pressure is [Given, the surface tension of water is `"0.075 Nm"^(-1)` and density is `"1000 kg m"^(-3)`]

Options

A. 2110 Pa
B. 2210 Pa
C. 2260 Pa
D. 2310 Pa

Answer

To find the excess pressure inside the air bubble located at a depth of 20 cm below the water level, we can use the formula for excess pressure in a bubble, which is given by:

\[ P_e - P_1 = \frac{2S}{R} \]

where:
- \( P_e \) is the pressure inside the bubble,
- \( P_1 \) is the pressure at the depth of the bubble,
- \( S \) is the surface tension of the liquid,
- \( R \) is the radius of the bubble.

### Step 1: Convert the radius of the bubble to meters
The radius of the bubble is given as 1 mm. We need to convert this to meters:

\[ R = 1 \text{ mm} = 1 \times 10^{-3} \text{ m} \]

### Step 2: Calculate the pressure at the depth of the bubble (\( P_1 \))
The pressure at a depth \( h \) in a fluid is given by:

\[ P_1 = P_0 + \rho g h \]

where:
- \( P_0 \) is the atmospheric pressure (approximately \( 101325 \text{ Pa} \)),
- \( \rho \) is the density of water (\( 1000 \text{ kg/m}^3 \)),
- \( g \) is the acceleration due to gravity (\( 9.8 \text{ m/s}^2 \)),
- \( h \) is the depth (20 cm = 0.2 m).

Substituting the values:

\[ P_1 = 101325 + (1000)(9.8)(0.2) \]

Calculating the second term:

\[ P_1 = 101325 + (1000 \times 9.8 \times 0.2) = 101325 + 1960 = 103285 \text{ Pa} \]

### Step 3: Substitute values into the excess pressure formula
Now we can substitute \( P_1 \) into the excess pressure formula:

\[ P_e - P_1 = \frac{2S}{R} \]

Substituting the known values:

\[ P_e - 103285 = \frac{2 \times 0.075}{1 \times 10^{-3}} \]

Calculating the right side:

\[ P_e - 103285 = \frac{0.15}{1 \times 10^{-3}} = 150 \]

### Step 4: Solve for \( P_e \)
Now we can solve for \( P_e \):

\[ P_e = 103285 + 150 = 103435 \text{ Pa} \]

### Step 5: Calculate the excess pressure
The excess pressure \( \Delta P \) is given by:

\[ \Delta P = P_e - P_0 \]

Substituting the values:

\[ \Delta P = 103435 - 101325 = 2110 \text{ Pa} \]

### Final Answer
The excess pressure inside the bubble above the atmospheric pressure is:

\[ \Delta P = 2110 \text{ Pa} \]