A glass capillary sealed at the upper end is of length `0.11 m` and internal diameter `2xx10^(5)`m. The tube is immersed vertically into a liquid of surface tension `5.06xx10^(-2)N//m`. To what length the capillary has to be immersed so that the liquid level inside and outside the capillary becomes the same. What will happen to water level inside the capillary if the seal is now broken?

Explanation

To solve the problem step by step, we will break it down into two parts as stated in the question.

### Part 1: Finding the length to which the capillary must be immersed

1. **Understanding the Setup**:
- We have a sealed glass capillary of length \( L = 0.11 \, m \) and internal diameter \( d = 2 \times 10^{-5} \, m \).
- The capillary is immersed in a liquid with surface tension \( T = 5.06 \times 10^{-2} \, N/m \).
- We need to find the length \( x \) to which the capillary must be immersed so that the liquid level inside and outside the capillary becomes the same.

2. **Using Pressure Balance**:
- The pressure at the bottom of the capillary (inside) must equal the atmospheric pressure (outside).
- The pressure inside the capillary can be expressed as:
\[ P_{\text{inside}} = P_0 + \frac{2T}{R} \]
- Where \( R \) is the radius of the capillary, which is half of the diameter:
\[ R = \frac{d}{2} = \frac{2 \times 10^{-5}}{2} = 1 \times 10^{-5} \, m \]

3. **Applying Boyle's Law**:
- The volume of liquid inside the capillary can be expressed as \( V_{\text{inside}} = A \cdot (L - x) \), where \( A \) is the cross-sectional area of the capillary.
- The area \( A \) can be calculated as:
\[ A = \pi R^2 = \pi (1 \times 10^{-5})^2 \]
- The volume outside the capillary is \( V_{\text{outside}} = A \cdot L \).

4. **Setting Up the Equation**:
- According to Boyle's Law:
\[ P_0 \cdot V_{\text{outside}} = P_{\text{inside}} \cdot V_{\text{inside}} \]
- Substituting the expressions for pressure and volume:
\[ P_0 \cdot A \cdot L = \left(P_0 + \frac{2T}{R}\right) \cdot A \cdot (L - x) \]
- Canceling \( A \) from both sides:
\[ P_0 \cdot L = \left(P_0 + \frac{2T}{R}\right) \cdot (L - x) \]

5. **Solving for \( x \)**:
- Rearranging gives us:
\[ P_0 \cdot L = P_0 \cdot L - P_0 \cdot x + \frac{2T}{R} \cdot L - \frac{2T}{R} \cdot x \]
- Simplifying leads to:
\[ P_0 \cdot x + \frac{2T}{R} \cdot x = \frac{2T}{R} \cdot L \]
- Factoring out \( x \):
\[ x \left(P_0 + \frac{2T}{R}\right) = \frac{2T}{R} \cdot L \]
- Finally, solving for \( x \):
\[ x = \frac{\frac{2T}{R} \cdot L}{P_0 + \frac{2T}{R}} \]

6. **Substituting Values**:
- Given \( P_0 = 1.01 \times 10^5 \, Pa \), \( T = 5.06 \times 10^{-2} \, N/m \), \( R = 1 \times 10^{-5} \, m \), and \( L = 0.11 \, m \):
\[ x = \frac{\frac{2 \times 5.06 \times 10^{-2} \times 0.11}{1 \times 10^{-5}}}{1.01 \times 10^5 + \frac{2 \times 5.06 \times 10^{-2}}{1 \times 10^{-5}}} \]

7. **Calculating the Final Value**:
- After performing the calculations, we find:
\[ x \approx 0.01 \, m \text{ or } 1 \, cm \]

### Part 2: Effect of Breaking the Seal

1. **Understanding the Effect of Breaking the Seal**:
- When the seal is broken, the pressure inside the capillary becomes equal to the atmospheric pressure.
- This will allow the liquid to rise in the capillary.

2. **Calculating the New Height**:
- The height \( h \) to which the liquid will rise can be calculated using:
\[ h = \frac{2T}{\rho g R} \]
- Where \( \rho \) is the density of water (\( \approx 1000 \, kg/m^3 \)) and \( g \) is the acceleration due to gravity (\( \approx 9.8 \, m/s^2 \)).

3. **Substituting Values**:
- Substituting the known values:
\[ h = \frac{2 \times 5.06 \times 10^{-2}}{1000 \times 9.8 \times 1 \times 10^{-5}} \]

4. **Calculating the Height**:
- Performing the calculations gives:
\[ h \approx 1.03 \, m \]

5. **Conclusion**:
- Since the original length of the capillary is \( 0.11 \, m \) and the liquid can rise \( 1.03 \, m \), the liquid will rise to the top of the capillary when the seal is broken.

### Summary
- The capillary must be immersed to a length of **1 cm** for the liquid levels to equalize.
- If the seal is broken, the liquid will rise to the top of the capillary, reaching a height of approximately **1.03 m**.