An inverted vessel (ball) lying at the bottom of a lake, 47.6 m deep, has 50 c.c. of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will now be (Atmospheric pressure is 70 cm of Hg, density of HG - 13.6 g `cm^(-3)`)

Options

A. 350 c.c.
B. 300 c.c.
C. 250 c.c.
D. 200 c.c.

Explanation

To solve the problem step by step, we will use the principles of fluid mechanics and the ideal gas law.

### Step 1: Calculate the Pressure at the Bottom of the Lake (P1)

The pressure at the bottom of the lake can be calculated using the formula:
\[ P_1 = P_0 + h \cdot \rho \cdot g \]

Where:
- \( P_0 \) = Atmospheric pressure
- \( h \) = Depth of the lake = 47.6 m
- \( \rho \) = Density of water = \( 10^3 \, \text{kg/m}^3 \)
- \( g \) = Acceleration due to gravity = \( 9.8 \, \text{m/s}^2 \)

Given that the atmospheric pressure is 70 cm of Hg, we first convert it to Pascals:
\[ P_0 = 70 \, \text{cm} \, \text{Hg} = 0.7 \, \text{m} \cdot 13.6 \times 10^3 \, \text{Pa} \]

Calculating \( P_0 \):
\[ P_0 = 0.7 \times 13.6 \times 10^3 = 9512 \, \text{Pa} \]

Now, calculate the pressure due to the water column:
\[ P_{\text{water}} = h \cdot \rho \cdot g = 47.6 \cdot 10^3 \cdot 9.8 \]

Calculating \( P_{\text{water}} \):
\[ P_{\text{water}} = 47.6 \cdot 10^3 \cdot 9.8 = 466,088 \, \text{Pa} \]

Now, add both pressures to find \( P_1 \):
\[ P_1 = 9512 + 466088 = 475600 \, \text{Pa} \]

### Step 2: Use the Ideal Gas Law to Find the Volume at the Surface (V2)

According to Boyle's Law, for a given mass of gas at constant temperature:
\[ P_1 V_1 = P_2 V_2 \]

Where:
- \( V_1 = 50 \, \text{cm}^3 = 50 \times 10^{-6} \, \text{m}^3 \)
- \( P_2 \) is the atmospheric pressure at the surface, which is \( P_0 = 9512 \, \text{Pa} \)

Rearranging the equation to find \( V_2 \):
\[ V_2 = \frac{P_1 V_1}{P_2} \]

Substituting the values:
\[ V_2 = \frac{475600 \cdot 50 \times 10^{-6}}{9512} \]

Calculating \( V_2 \):
\[ V_2 = \frac{23780 \times 10^{-6}}{9512} \approx 2.49 \times 10^{-3} \, \text{m}^3 \]

Converting \( V_2 \) to cm³:
\[ V_2 = 2.49 \times 10^{-3} \, \text{m}^3 = 2490 \, \text{cm}^3 \]

### Final Answer
The volume of the trapped air when the vessel is brought to the surface of the lake is approximately **2490 cm³**.