The rise in the water level in a capillary tube of radius 0.07 cm when dipped veryically in a beaker containing water of surface tension `0.07 N m^(-1)` is (g = `10 m s^(-2)`)

Options

A. 2 cm
B. 4 cm
C. 1.5 cm
D. 3 cm

Explanation

To solve the problem of finding the rise in water level in a capillary tube, we can use the formula for capillary rise:

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

Where:
- \( h \) = height of the liquid column (rise in water level)
- \( S \) = surface tension of the liquid
- \( \theta \) = angle of contact (for water in a glass tube, this is typically 0 degrees)
- \( r \) = radius of the capillary tube
- \( \rho \) = density of the liquid (for water, approximately \( 1000 \, \text{kg/m}^3 \))
- \( g \) = acceleration due to gravity

### Step-by-step Solution:

1. **Identify the given values:**
- Surface tension, \( S = 0.07 \, \text{N/m} \)
- Radius of the capillary tube, \( r = 0.07 \, \text{cm} = 0.07 \times 10^{-2} \, \text{m} = 7 \times 10^{-4} \, \text{m} \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Angle of contact, \( \theta = 0^\circ \) (thus, \( \cos 0^\circ = 1 \))

2. **Substitute the values into the formula:**
\[ h = \frac{2 \times 0.07 \, \text{N/m} \times \cos(0^\circ)}{(7 \times 10^{-4} \, \text{m}) \times (1000 \, \text{kg/m}^3) \times (10 \, \text{m/s}^2)} \]

3. **Calculate the numerator:**
\[ \text{Numerator} = 2 \times 0.07 = 0.14 \, \text{N/m} \]

4. **Calculate the denominator:**
\[ \text{Denominator} = (7 \times 10^{-4}) \times (1000) \times (10) = 7 \times 10^{-4} \times 10000 = 7 \, \text{N/m} \]

5. **Now, substitute the numerator and denominator into the equation:**
\[ h = \frac{0.14}{7} = 0.02 \, \text{m} \]

6. **Convert meters to centimeters:**
\[ h = 0.02 \, \text{m} = 2 \, \text{cm} \]

### Final Answer:
The rise in the water level in the capillary tube is **2 cm**.