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)`.

Options

A. `2.55 cm`
B. `2.86 cm`
C. `3.76 cm`
D. `3.42 cm`

Explanation

To solve the problem of determining the height to which water is raised in a capillary tube, we can use the formula for capillary rise:

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

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

### Step-by-Step Solution:

1. **Identify the given values:**
- Inside radius of the capillary tube, \( R = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} \)
- Surface tension of water, \( T = 7.0 \times 10^{-2} \, \text{N/m} \)
- 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:**
\[ H = \frac{2 \times (7.0 \times 10^{-2}) \times 1}{(10^{3}) \times (9.8) \times (0.5 \times 10^{-3})} \]

4. **Calculate the denominator:**
\[ \text{Denominator} = (10^{3}) \times (9.8) \times (0.5 \times 10^{-3}) = 10^{3} \times 9.8 \times 0.5 \times 10^{-3} = 4.9 \]

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

6. **Calculate the height \( H \):**
\[ H = \frac{0.14}{4.9} \approx 0.02857 \, \text{m} \]

7. **Convert the height from meters to centimeters:**
\[ H \approx 0.02857 \, \text{m} \times 100 \approx 2.857 \, \text{cm} \approx 2.86 \, \text{cm} \]

### Final Answer:
The height to which the water is raised in the capillary tube is approximately **2.86 cm**.