A long capillary tube of radius `r = 1 mm` open at both ends is filled with water and placed vertically. What will be the height (in `cm`) of the column of water left in the capillary walls is negligible. (surface tension of water is `72 "dyne"//"cm"` and `g = 1000 "cm"//"sec"^(2)`)

Explanation

To solve the problem of determining the height of the water column in a capillary tube, we will use the formula for capillary rise:

\[ h = \frac{2T}{R \rho g} \]

where:
- \( h \) is the height of the liquid column,
- \( T \) is the surface tension of the liquid,
- \( R \) is the radius of the capillary tube,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity.

### Step 1: Identify the given values
- Surface tension \( T = 72 \, \text{dyne/cm} \)
- Radius \( r = 1 \, \text{mm} = 0.1 \, \text{cm} \) (conversion from mm to cm)
- Density of water \( \rho = 1 \, \text{g/cm}^3 \)
- Acceleration due to gravity \( g = 1000 \, \text{cm/s}^2 \)

### Step 2: Substitute the values into the formula
Now we substitute the known values into the formula:

\[ h = \frac{2 \times 72 \, \text{dyne/cm}}{0.1 \, \text{cm} \times 1 \, \text{g/cm}^3 \times 1000 \, \text{cm/s}^2} \]

### Step 3: Simplify the equation
Calculating the numerator:

\[ 2 \times 72 = 144 \, \text{dyne/cm} \]

Now calculating the denominator:

\[ 0.1 \, \text{cm} \times 1 \, \text{g/cm}^3 \times 1000 \, \text{cm/s}^2 = 0.1 \times 1000 = 100 \, \text{g/cm}^2 \]

### Step 4: Calculate the height \( h \)
Now substitute the values back into the equation:

\[ h = \frac{144 \, \text{dyne/cm}}{100 \, \text{g/cm}^2} \]

Since \( 1 \, \text{dyne} = 1 \, \text{g} \cdot \text{cm/s}^2 \), we can simplify:

\[ h = \frac{144}{100} = 1.44 \, \text{cm} \]

### Final Answer
The height of the water column in the capillary tube is:

\[ h = 1.44 \, \text{cm} \]

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