A long cylinderical glass vessel has a small hole of radius r at its bottom. The depth to which the vessel can be lowered vertically in a deep water (surface tension S) without any water entering inside is

Options

A. `(4S)/(r rho g)`
B. `(3S)/(r rho g)`
C. `(2 S)/(r rho g)`
D. `(S)/(r rho g)`

Answer

To solve the problem of determining the depth to which a cylindrical glass vessel with a small hole at its bottom can be submerged in water without allowing water to enter, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Pressure**:
- When the vessel is submerged in water, the hydrostatic pressure at the depth \( h \) is given by the formula:
\[ P_{\text{hydrostatic}} = \rho g h \]
where \( \rho \) is the density of the water, \( g \) is the acceleration due to gravity, and \( h \) is the depth of the water above the hole.

2. **Excess Pressure Due to Surface Tension**:
- The excess pressure \( P_{\text{excess}} \) created by the surface tension \( S \) at the hole is given by:
\[ P_{\text{excess}} = \frac{2S}{r} \]
where \( r \) is the radius of the hole.

3. **Condition for Water Not Entering**:
- For water to not enter the vessel, the hydrostatic pressure must be less than or equal to the excess pressure. Thus, we set up the equation:
\[ P_{\text{hydrostatic}} = P_{\text{excess}} \]
This gives us:
\[ \rho g h = \frac{2S}{r} \]

4. **Solve for Depth \( h \)**:
- Rearranging the equation to solve for \( h \):
\[ h = \frac{2S}{\rho g r} \]

5. **Final Expression**:
- Therefore, the depth \( h \) to which the vessel can be submerged without water entering is:
\[ h = \frac{2S}{\rho g r} \]