An inverted bell lying at the bottom of a lake `47.6 m` deep has `50 cm^3` of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be (atmospheric pressure `= 70 cm` of `Hg` and density of `Hg = 13.6 g//cm^3`).

Options

A. `350 cm^3`
B. `300 cm^3`
C. `250 cm^3`
D. `22 cm^3`

Explanation

To solve the problem of finding the new volume of the trapped air in the inverted bell when it is brought to the surface of the lake, we can use the ideal gas law, which states that \( P_1 V_1 = P_2 V_2 \). Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the Given Data
- Depth of the lake, \( h = 47.6 \, \text{m} \)
- Initial volume of trapped air, \( V_1 = 50 \, \text{cm}^3 \)
- Atmospheric pressure, \( P_0 = 70 \, \text{cm} \, \text{Hg} \)
- Density of mercury, \( \rho_{Hg} = 13.6 \, \text{g/cm}^3 \)

### Step 2: Calculate the Pressure at the Bottom of the Lake (Position 1)
The pressure at the bottom of the lake (Position 1) can be calculated using the formula:
\[ P_1 = P_0 + \rho_{water} \cdot g \cdot h \]
Where:
- \( \rho_{water} = 1 \, \text{g/cm}^3 \) (density of water)
- \( g \) is the acceleration due to gravity, which will cancel out later.

Convert \( h \) into centimeters:
\[ h = 47.6 \, \text{m} = 4760 \, \text{cm} \]

Now substituting the values:
\[ P_1 = P_0 + \rho_{water} \cdot g \cdot h = 70 \, \text{cm} \, \text{Hg} + (1 \, \text{g/cm}^3) \cdot g \cdot 4760 \, \text{cm} \]

### Step 3: Convert Atmospheric Pressure to the Same Units
To convert atmospheric pressure from cm of Hg to the equivalent in g/cm²:
\[ P_0 = 70 \, \text{cm} \, \text{Hg} = 70 \cdot \rho_{Hg} \cdot g = 70 \cdot 13.6 \, \text{g/cm}^3 \cdot g \]

### Step 4: Calculate the Total Pressure at Position 1
Now substituting the values into the pressure equation:
\[ P_1 = 70 \cdot 13.6 + 1 \cdot 4760 \]
Calculating:
\[ P_1 = 952 + 4760 = 5712 \, \text{g/cm}^2 \]

### Step 5: Calculate the Pressure at the Surface of the Lake (Position 2)
At the surface of the lake (Position 2), the pressure is simply the atmospheric pressure:
\[ P_2 = P_0 = 70 \cdot 13.6 = 952 \, \text{g/cm}^2 \]

### Step 6: Apply the Ideal Gas Law
Using the ideal gas law:
\[ P_1 V_1 = P_2 V_2 \]
Substituting the known values:
\[ 5712 \cdot 50 = 952 \cdot V_2 \]

### Step 7: Solve for \( V_2 \)
Rearranging the equation to find \( V_2 \):
\[ V_2 = \frac{5712 \cdot 50}{952} \]
Calculating:
\[ V_2 = \frac{285600}{952} \approx 300 \, \text{cm}^3 \]

### Final Answer
The volume of the trapped air when the bell is brought to the surface of the lake is approximately \( 300 \, \text{cm}^3 \).

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