A capillary tube of 2mm diameter. What is the height to which water rises, if surface tension of water is `7.4 xx 10^(-9) Nm^(-1)` and angle of contact is `0^(@)` ?

Options

A. 0.25mm
B. 1.51mm
C. 0.4mm
D. 0.5mm

Explanation

To solve the problem of how high water rises in a capillary tube of 2 mm diameter, given the surface tension of water and the angle of contact, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Diameter of the capillary tube, \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)
- Radius of the capillary tube, \( r = \frac{d}{2} = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Surface tension of water, \( \sigma = 7.4 \times 10^{-9} \, \text{N/m} \)
- 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. **Use the Capillary Rise Formula:**
The height \( h \) to which the liquid rises in a capillary tube is given by the formula:
\[ h = \frac{2\sigma \cos \theta}{\rho g r} \]

3. **Substitute the Values:**
Since \( \cos 0^\circ = 1 \), we can substitute the values into the formula:
\[ h = \frac{2 \times (7.4 \times 10^{-9}) \times 1}{(10^3) \times (9.8) \times (1 \times 10^{-3})} \]

4. **Calculate the Numerator:**
\[ \text{Numerator} = 2 \times 7.4 \times 10^{-9} = 14.8 \times 10^{-9} \, \text{N/m} \]

5. **Calculate the Denominator:**
\[ \text{Denominator} = (10^3) \times (9.8) \times (1 \times 10^{-3}) = 9.8 \]

6. **Calculate the Height:**
\[ h = \frac{14.8 \times 10^{-9}}{9.8} \approx 1.51 \times 10^{-3} \, \text{m} \]

7. **Convert to Millimeters:**
\[ h \approx 1.51 \, \text{mm} \]

### Final Answer:
The height to which water rises in the capillary tube is approximately **1.51 mm**.