A thin - walled cylinder of mass (m), height (h) and cross- sectional area (A) is filled with a gas and floats on the surface of water. As a result of leakage from the lower part of the cylinder, the depth of its submergence has increased by`Delta h`. Find the initial pressure `p_(1)` of the gas in the cylinder if the atmospheric pressure is `p_(0)` and the temperature remains constant.

Explanation

To solve the problem step by step, we will analyze the situation of the floating cylinder and apply the principles of fluid mechanics and gas laws.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a thin-walled cylinder of mass \( m \), height \( h \), and cross-sectional area \( A \) filled with gas. It floats on the surface of water.
- Due to leakage, the depth of submergence increases by \( \Delta h \).

2. **Initial and Final Conditions**:
- Initially, the height of the gas column in the cylinder is \( h \).
- After leakage, the height of the gas column becomes \( h - \Delta h \).

3. **Volume of Gas**:
- The initial volume of gas in the cylinder is given by:
\[ V_1 = A \cdot h \]
- The final volume of gas after leakage is:
\[ V_2 = A \cdot (h - \Delta h) \]

4. **Applying Boyle's Law**:
- Since the temperature remains constant, we can apply Boyle's Law, which states that \( P_1 V_1 = P_2 V_2 \).
- Let \( P_1 \) be the initial pressure of the gas and \( P_2 \) be the final pressure of the gas.
- Thus, we have:
\[ P_1 \cdot (A \cdot h) = P_2 \cdot (A \cdot (h - \Delta h)) \]
- Simplifying this, we get:
\[ P_1 \cdot h = P_2 \cdot (h - \Delta h) \]

5. **Finding Final Pressure \( P_2 \)**:
- The pressure \( P_2 \) at the bottom of the cylinder (where the water is) can be expressed as:
\[ P_2 = P_0 + \frac{mg}{A} \]
- Here, \( P_0 \) is the atmospheric pressure, and \( \frac{mg}{A} \) is the pressure due to the weight of the cylinder.

6. **Substituting \( P_2 \) into the Equation**:
- Substitute \( P_2 \) into the equation derived from Boyle's Law:
\[ P_1 \cdot h = \left(P_0 + \frac{mg}{A}\right) \cdot (h - \Delta h) \]

7. **Solving for \( P_1 \)**:
- Rearranging the equation to solve for \( P_1 \):
\[ P_1 = \frac{\left(P_0 + \frac{mg}{A}\right) \cdot (h - \Delta h)}{h} \]

8. **Final Expression**:
- Thus, the initial pressure \( P_1 \) of the gas in the cylinder is given by:
\[ P_1 = P_0 + \frac{mg}{A} \cdot \left(1 - \frac{\Delta h}{h}\right) \]