A capillary tube of inner diameter 0.5 mm is dipped in a liquid of specific gravity 13.6 and having surface tension 545 dyn/cm(angle of contact `130^(@)`) . Find the depression or elevation in the tube .

Options

A. depressed 2.11 cm
B. elevated 2.11 cm
C. depressed , 3.71 cm
D. elevated , 3.71 cm

Answer

To solve the problem of finding the depression or elevation in a capillary tube dipped in a liquid, we will use the formula for capillary rise (or depression) given by:

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

where:
- \( h \) = height of capillary rise (or depression)
- \( S \) = 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: Convert the given values to appropriate units
- Inner diameter of the capillary tube = 0.5 mm = 0.05 cm
- Therefore, radius \( r = \frac{0.5 \text{ mm}}{2} = 0.025 \text{ cm} \)
- Specific gravity of the liquid = 13.6, so density \( \rho = 13.6 \text{ g/cm}^3 \)
- Surface tension \( S = 545 \text{ dyn/cm} \)
- Angle of contact \( \theta = 130^\circ \)
- Acceleration due to gravity \( g = 980 \text{ cm/s}^2 \)

### Step 2: Calculate the height of depression
Using the formula for capillary rise (or depression):

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

First, we need to calculate \( \cos(130^\circ) \):
- \( \cos(130^\circ) = -\cos(50^\circ) \) (since \( 130^\circ \) is in the second quadrant)
- \( \cos(50^\circ) \approx 0.6428 \)
- Thus, \( \cos(130^\circ) \approx -0.6428 \)

Now substituting the values into the formula:

\[ h = \frac{2 \times 545 \times (-0.6428)}{13.6 \times 980 \times 0.025} \]

Calculating the numerator:

\[ 2 \times 545 \times (-0.6428) \approx -700.56 \]

Calculating the denominator:

\[ 13.6 \times 980 \times 0.025 \approx 332.0 \]

Now substituting these values into the equation for \( h \):

\[ h = \frac{-700.56}{332.0} \approx -2.11 \text{ cm} \]

### Conclusion
The negative sign indicates that there is a depression in the liquid level in the capillary tube. Therefore, the depression in the tube is approximately:

\[ h \approx 2.11 \text{ cm} \]