To what height does a liquid of density `0.4 xx 10^(3)` kg `m^(-3)` and surface tension 0.05 `Nm^(-1)` rise in a capillary tube of radius 0.2 mm when dipped in ? [Given `cos theta = 0.4 g = 10 m s^(-2)`]

Options

A. 25 cm
B. 15 cm
C. 5 cm
D. 20 cm

Answer

To find the height to which a liquid rises 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 1: Write down the given values
- Density of the liquid, \( \rho = 0.4 \times 10^3 \, \text{kg/m}^3 \)
- Surface tension, \( T = 0.05 \, \text{N/m} \)
- Angle of contact, \( \cos \theta = 0.4 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Radius of the capillary tube, \( R = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m} \)

### Step 2: Substitute the values into the formula
Now, we substitute the values into the formula for \( h \):

\[ h = \frac{2 \times 0.05 \times 0.4}{0.4 \times 10^3 \times 10 \times 0.2 \times 10^{-3}} \]

### Step 3: Simplify the equation
Calculating the numerator:

\[ 2 \times 0.05 \times 0.4 = 0.04 \]

Calculating the denominator:

\[ 0.4 \times 10^3 \times 10 \times 0.2 \times 10^{-3} = 0.4 \times 10^3 \times 10 \times 0.2 \times 10^{-3} = 0.4 \times 10 \times 0.2 = 0.008 \]

### Step 4: Calculate \( h \)
Now, substituting back into the equation for \( h \):

\[ h = \frac{0.04}{0.008} = 5 \]

### Step 5: Convert to centimeters
Since the height is in meters, we convert it to centimeters:

\[ h = 5 \, \text{m} = 5 \, \text{cm} \]

### Final Answer
The height to which the liquid rises in the capillary tube is \( 5 \, \text{cm} \).

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