When a glass capilary tube of the radius 0.015 cm is dipped in water, the water rises to a heigth of 15 cm within it. Assuming the contact angle between water and glass to be `0^@`, the surface tension of water is
`[rho_("water") = 1000 kg m^(-3), g = 9.81 m s^(-2)`]

Options

A. `0.11 N m^(-1)`
B. `0.7 N m^(-1)`
C. `0.072 N m^(-1)`
D. None of these

Answer

To solve the problem of finding the surface tension of water when it rises in a glass capillary tube, we can use the formula for capillary rise:

\[ h = \frac{2S \cos \theta}{\rho g r} \]

Where:
- \( h \) = height of the liquid column (in meters)
- \( S \) = surface tension of the liquid (in N/m)
- \( \theta \) = contact angle (in degrees)
- \( \rho \) = density of the liquid (in kg/m³)
- \( g \) = acceleration due to gravity (in m/s²)
- \( r \) = radius of the capillary tube (in meters)

### Step 1: Convert given values to appropriate units
- Radius \( r = 0.015 \, \text{cm} = 0.015 \times 10^{-2} \, \text{m} = 0.00015 \, \text{m} \)
- Height \( h = 15 \, \text{cm} = 15 \times 10^{-2} \, \text{m} = 0.15 \, \text{m} \)
- Density \( \rho = 1000 \, \text{kg/m}^3 \)
- Acceleration due to gravity \( g = 9.81 \, \text{m/s}^2 \)
- Contact angle \( \theta = 0^\circ \) (which means \( \cos \theta = 1 \))

### Step 2: Rearrange the formula to solve for surface tension \( S \)
From the formula, we can rearrange it to find \( S \):

\[ S = \frac{h \cdot \rho \cdot g \cdot r}{2 \cos \theta} \]

### Step 3: Substitute the known values into the equation
Substituting the known values into the rearranged formula:

\[ S = \frac{0.15 \, \text{m} \cdot 1000 \, \text{kg/m}^3 \cdot 9.81 \, \text{m/s}^2 \cdot 0.00015 \, \text{m}}{2 \cdot 1} \]

### Step 4: Calculate the surface tension
Now, calculate the value:

\[ S = \frac{0.15 \cdot 1000 \cdot 9.81 \cdot 0.00015}{2} \]

Calculating the numerator:

\[ 0.15 \cdot 1000 = 150 \]
\[ 150 \cdot 9.81 = 1471.5 \]
\[ 1471.5 \cdot 0.00015 = 0.220725 \]

Now divide by 2:

\[ S = \frac{0.220725}{2} = 0.1103625 \, \text{N/m} \]

### Step 5: Round the answer
Rounding to two decimal places, we get:

\[ S \approx 0.11 \, \text{N/m} \]

Thus, the surface tension of water is approximately **0.11 N/m**.

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