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 will analyze the situation using the principles of fluid mechanics and surface tension. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Problem
We have a vessel with a small hole at the bottom and we need to find the maximum height of water (h) such that water does not flow out of the hole. The surface tension of water is given as \( T = 70 \, \text{dyne/cm} \).

### Step 2: Convert Units
The diameter of the hole is given as \( d = 0.1 \, \text{mm} \). We need to convert this to centimeters:
\[ d = 0.1 \, \text{mm} = 0.1 \times 0.1 \, \text{cm} = 0.01 \, \text{cm} \]
The radius \( r \) of the hole is:
\[ r = \frac{d}{2} = \frac{0.01 \, \text{cm}}{2} = 0.005 \, \text{cm} \]

### Step 3: Apply the Pressure Balance Equation
For the water to not flow out of the hole, the pressure inside the bubble (due to surface tension) must balance the pressure due to the height of the water column. The pressure at the depth of the water is given by:
\[ P_{\text{water}} = P_0 + \rho g h \]
Where:
- \( P_0 \) is the atmospheric pressure,
- \( \rho \) is the density of water (approximately \( 1 \, \text{g/cm}^3 \)),
- \( g \) is the acceleration due to gravity (approximately \( 980 \, \text{cm/s}^2 \)),
- \( h \) is the height of the water column.

The pressure inside the bubble due to surface tension is given by:
\[ P_{\text{bubble}} = P_0 + \frac{2T}{r} \]

### Step 4: Set the Pressures Equal
For the water to not flow out, we set the pressures equal:
\[ P_0 + \rho g h = P_0 + \frac{2T}{r} \]
Cancelling \( P_0 \) from both sides, we have:
\[ \rho g h = \frac{2T}{r} \]

### Step 5: Solve for Height \( h \)
Rearranging the equation to solve for \( h \):
\[ h = \frac{2T}{\rho g r} \]

### Step 6: Substitute Values
Substituting the known values:
- \( T = 70 \, \text{dyne/cm} = 70 \, \text{g/(cm \cdot s^2)} \) (since \( 1 \, \text{dyne} = 1 \, \text{g} \cdot \text{cm/s}^2 \)),
- \( \rho = 1 \, \text{g/cm}^3 \),
- \( g \approx 980 \, \text{cm/s}^2 \),
- \( r = 0.005 \, \text{cm} \).

Substituting these values into the equation for \( h \):
\[ h = \frac{2 \times 70}{1 \times 980 \times 0.005} \]
Calculating the denominator:
\[ h = \frac{140}{980 \times 0.005} = \frac{140}{4.9} \approx 28.57 \, \text{cm} \]

### Step 7: Final Answer
Thus, the maximum height of the water level \( h \) at which the water does not flow out is approximately:
\[ h \approx 28 \, \text{cm} \]