A capillary tube whose inside radius is `0.5mm` is dipped in water having surface tension `7.0xx10^(-2) N//m`. To what height is the water raised above the normal water level? Angle of contact of water with glass is `0^(@)`. Density of water is `10^(3)kg//m^(3) and g=9.8 m//s^(2)`.

Explanation

To solve the problem of how high water is raised in a capillary tube, we can use the formula for capillary rise:

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

Where:
- \( h \) = height of the liquid column
- \( T \) = surface tension of the liquid
- \( \theta \) = angle of contact
- \( R \) = radius of the capillary tube
- \( \rho \) = density of the liquid
- \( g \) = acceleration due to gravity

### Step-by-Step Solution:

1. **Identify the given values:**
- Surface tension, \( T = 7.0 \times 10^{-2} \, \text{N/m} \)
- Radius of the capillary tube, \( R = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} \)
- Angle of contact, \( \theta = 0^\circ \)
- Density of water, \( \rho = 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

2. **Convert the angle of contact to cosine:**
- Since \( \theta = 0^\circ \), we have:
\[ \cos \theta = \cos(0^\circ) = 1 \]

3. **Substitute the values into the formula:**
\[ h = \frac{2 \times (7.0 \times 10^{-2}) \times 1}{(0.5 \times 10^{-3}) \times (10^3) \times (9.8)} \]

4. **Calculate the denominator:**
- First, calculate \( R \rho g \):
\[ R \rho g = (0.5 \times 10^{-3}) \times (10^3) \times (9.8) = 0.5 \times 9.8 = 4.9 \]

5. **Calculate the numerator:**
\[ 2T \cos \theta = 2 \times (7.0 \times 10^{-2}) \times 1 = 0.14 \]

6. **Now, substitute back into the equation for \( h \):**
\[ h = \frac{0.14}{4.9} \]

7. **Calculate \( h \):**
\[ h \approx 0.02857 \, \text{m} \]

8. **Convert \( h \) to centimeters:**
\[ h \approx 0.02857 \, \text{m} \times 100 = 2.857 \, \text{cm} \approx 2.86 \, \text{cm} \]

### Final Answer:
The height to which water is raised in the capillary tube is approximately \( 2.86 \, \text{cm} \).