A long cylindrical vessel has a small hole of diameter D at its bottom. This vessel can be lowered vertically in water to a depth h without any water entering the vessel. Given : A=surface tension, B=density of liquid, C=acceleration due to gravity. The value of h is

Options

A. `(A)/(DBC)`
B. `(2A)/(DBC)`
C. `(3A)/(DBC)`
D. `(4A)/(DBC)`

Answer

To solve the problem, we need to analyze the forces acting on the water at the hole of the cylindrical vessel when it is submerged in water to a depth \( h \).

### Step-by-step Solution:

1. **Understand the Scenario**:
- A long cylindrical vessel has a small hole at its bottom.
- The vessel can be submerged to a depth \( h \) without water entering it.
- We need to find the value of \( h \) in terms of surface tension \( A \), density of the liquid \( B \), acceleration due to gravity \( C \), and the diameter of the hole \( D \).

2. **Identify the Forces**:
- When the vessel is submerged, the hydrostatic pressure \( P_H \) at the hole due to the water column above it is given by:
\[ P_H = \rho g h \]
where \( \rho \) is the density of the liquid (given as \( B \)), \( g \) is the acceleration due to gravity (given as \( C \)), and \( h \) is the depth of the water above the hole.

3. **Surface Tension Effect**:
- The surface tension \( T \) acts at the hole and creates a force that opposes the hydrostatic pressure. The force due to surface tension can be expressed as:
\[ F_T = T \cdot \text{circumference of the hole} \]
- The circumference of the hole (with diameter \( D \)) is:
\[ \text{Circumference} = \pi D \]
- Therefore, the force due to surface tension is:
\[ F_T = T \cdot \pi D \]

4. **Setting Up the Equation**:
- For the water to not enter the vessel, the hydrostatic pressure must equal the force due to surface tension:
\[ P_H \cdot A = T \cdot \text{Circumference} \]
- Here, \( A \) is the area of the hole, which can be expressed as:
\[ A = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4} \]
- Substituting this into the equation gives:
\[ \rho g h \cdot \frac{\pi D^2}{4} = T \cdot \pi D \]

5. **Simplifying the Equation**:
- Cancel \( \pi \) from both sides:
\[ \rho g h \cdot \frac{D^2}{4} = T \cdot D \]
- Rearranging gives:
\[ h = \frac{4T}{\rho g D} \]

6. **Substituting Given Variables**:
- Substitute \( T \) with \( A \), \( \rho \) with \( B \), and \( g \) with \( C \):
\[ h = \frac{4A}{BC}D \]

### Final Result:
The value of \( h \) is:
\[ h = \frac{4A}{BC}D \]