In question number 88, if an air bubble of the same dimensions were formed at a depth of 30 cm inside a container containing the soap solution of relative density 1.20, then the pressure inside the bubble is ( Take 1 atm = `1.01 xx 10^(5)` Pa)

Options

A. `1.01 xx 10^(4)` Pa
B. `1.05 xx 10^(5)` Pa
C. `2.01 xx 10^(4)` Pa
D. `3.01 xx 10^(4)` Pa

Explanation

To find the pressure inside the air bubble formed at a depth of 30 cm in a soap solution with a relative density of 1.20, we can follow these steps:

### Step 1: Convert Depth to Meters
The depth given is 30 cm. We need to convert this to meters:
\[ \text{Depth} = 30 \, \text{cm} = 30 \times 10^{-2} \, \text{m} = 0.30 \, \text{m} \]

### Step 2: Calculate the Density of the Soap Solution
The relative density (specific gravity) of the soap solution is given as 1.20. The density of water is approximately \(1000 \, \text{kg/m}^3\). Therefore, the density of the soap solution (\(\rho\)) can be calculated as:
\[ \rho = 1.20 \times 1000 \, \text{kg/m}^3 = 1200 \, \text{kg/m}^3 \]

### Step 3: Calculate the Pressure at Depth
The pressure at a certain depth in a fluid is given by the formula:
\[ P = P_0 + H \cdot \rho \cdot g \]
Where:
- \(P_0\) is the atmospheric pressure (given as \(1.01 \times 10^5 \, \text{Pa}\)),
- \(H\) is the depth (0.30 m),
- \(\rho\) is the density of the soap solution (1200 kg/m³),
- \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)).

Substituting the values:
\[ P = 1.01 \times 10^5 \, \text{Pa} + (0.30 \, \text{m}) \cdot (1200 \, \text{kg/m}^3) \cdot (9.81 \, \text{m/s}^2) \]

### Step 4: Calculate the Hydrostatic Pressure
Calculating the hydrostatic pressure:
\[ \text{Hydrostatic Pressure} = 0.30 \cdot 1200 \cdot 9.81 = 3529.2 \, \text{Pa} \]

### Step 5: Calculate Total Pressure
Now, we can calculate the total pressure inside the bubble:
\[ P = 1.01 \times 10^5 \, \text{Pa} + 3529.2 \, \text{Pa} = 1.0135292 \times 10^5 \, \text{Pa} \]
Rounding this to two decimal places:
\[ P \approx 1.05 \times 10^5 \, \text{Pa} \]

### Conclusion
Thus, the pressure inside the bubble is approximately:
\[ \boxed{1.05 \times 10^5 \, \text{Pa}} \]

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