A capillary of radius 0.15mm is dipped in liquid of density 'rho' = 667 kg/m^3. If surface tension of liquid is `1/20 N/m` then find height upto which liquid rises in capillary. Angle of contact between liquid and capillary tube is `60^@`. `(g=10m/s^2)`

Options

A. `0.01m`
B. `0.02m`
C. `0.04m`
D. `0.05m`

Answer

To find the height to which the liquid rises in the capillary tube, we can use the formula derived from the balance of forces acting on the liquid column. The height \( h \) of the liquid column can be calculated using the formula:

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

Where:
- \( T \) is the surface tension of the liquid,
- \( \theta \) is the angle of contact,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( r \) is the radius of the capillary tube.

### Step-by-step Solution:

1. **Identify the given values:**
- Radius of the capillary tube, \( r = 0.15 \, \text{mm} = 0.15 \times 10^{-3} \, \text{m} \)
- Density of the liquid, \( \rho = 667 \, \text{kg/m}^3 \)
- Surface tension of the liquid, \( T = \frac{1}{20} \, \text{N/m} \)
- Angle of contact, \( \theta = 60^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Convert the radius to meters:**
\[ r = 0.15 \, \text{mm} = 0.15 \times 10^{-3} \, \text{m} = 0.00015 \, \text{m} \]

3. **Calculate \( \cos \theta \):**
\[ \cos 60^\circ = \frac{1}{2} \]

4. **Substitute the values into the formula:**
\[ h = \frac{2 \times \left(\frac{1}{20}\right) \times \left(\frac{1}{2}\right)}{667 \times 10 \times 0.00015} \]

5. **Simplify the numerator:**
\[ 2 \times \left(\frac{1}{20}\right) \times \left(\frac{1}{2}\right) = \frac{1}{20} \]

6. **Calculate the denominator:**
\[ 667 \times 10 \times 0.00015 = 1.0005 \]

7. **Now substitute back into the height formula:**
\[ h = \frac{\frac{1}{20}}{1.0005} \]

8. **Calculate the height \( h \):**
\[ h \approx \frac{0.05}{1.0005} \approx 0.0499 \, \text{m} \]

9. **Convert to centimeters for clarity:**
\[ h \approx 0.0499 \, \text{m} \approx 4.99 \, \text{cm} \approx 5 \, \text{cm} \]

### Final Answer:
The height up to which the liquid rises in the capillary is approximately \( 0.05 \, \text{m} \) or \( 5 \, \text{cm} \).