A vessel whose bottom has round holes with a diameter of `d=0.1mm` is filled with water. The maximum height of the water level h at which the water does not flow out, will be (The water does not wet the bottom of the vessel). `[ST of water=70 "dyne"//cm]`

Options

A. `h=24.0 cm`
B. `h=25.0 cm`
C. `h=26.0 cm`
D. `h=28.0 cm`

Explanation

To solve the problem, we need to determine the maximum height of water (h) in a vessel with a small hole at the bottom, where water does not flow out. We will use the principles of hydrostatics and the concept of excess pressure due to surface tension.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Diameter of the hole, \( d = 0.1 \, \text{mm} = 0.01 \, \text{cm} \)
- Radius of the hole, \( r = \frac{d}{2} = \frac{0.1 \, \text{mm}}{2} = 0.005 \, \text{cm} \)
- Surface tension of water, \( T = 70 \, \text{dyne/cm} \)
- Density of water, \( \rho \approx 1 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 980 \, \text{cm/s}^2 \)

2. **Understand the Condition for No Flow:**
- The pressure at the depth of the water (at height \( h \)) must equal the excess pressure due to surface tension at the hole.
- The pressure at the bottom of the water column is given by:
\[ P = \rho g h + P_0 \]
- The excess pressure due to surface tension at the hole is given by:
\[ P = P_0 + \frac{2T}{r} \]

3. **Set Up the Equation:**
- For no flow condition, set the two pressures equal:
\[ \rho g h + P_0 = P_0 + \frac{2T}{r} \]
- Cancel \( P_0 \) from both sides:
\[ \rho g h = \frac{2T}{r} \]

4. **Rearrange to Solve for \( h \):**
\[ h = \frac{2T}{\rho g r} \]

5. **Substitute Known Values:**
- Substitute \( T = 70 \, \text{dyne/cm} \), \( \rho = 1 \, \text{g/cm}^3 \), \( g = 980 \, \text{cm/s}^2 \), and \( r = 0.005 \, \text{cm} \):
\[ h = \frac{2 \times 70 \, \text{dyne/cm}}{1 \, \text{g/cm}^3 \times 980 \, \text{cm/s}^2 \times 0.005 \, \text{cm}} \]

6. **Calculate \( h \):**
- Calculate the denominator:
\[ \rho g r = 1 \times 980 \times 0.005 = 4.9 \, \text{g cm/s}^2 \]
- Now substitute back into the equation:
\[ h = \frac{140}{4.9} \approx 28.57 \, \text{cm} \]

7. **Final Result:**
- The maximum height of the water level \( h \) at which the water does not flow out is approximately:
\[ h \approx 28 \, \text{cm} \]