Problem Statement
Distilled water stands in a glass tube of 10mm diameter at a height of 25mm. What is the true static height? Take surface tension of water \( \sigma = 0.074 \, \text{N/m} \) and angle of contact \( \theta = 0^\circ \).
- Diameter of the tube (\( d \)) = 10 mm = 0.01 m
- Surface tension of water (\( \sigma \)) = 0.074 N/m
- Angle of contact (\( \theta \)) = \( 0^\circ \)
- Capillary height change (\( h \)) and true static height = ?
Solution
Step-by-Step Calculation:
Capillary height change is given by:
\( h = \frac{4 \sigma \cos\theta}{\rho g d} \)
Substitute the known values:
\( h = \frac{4 \cdot 0.074 \cdot \cos 0^\circ}{1000 \cdot 9.81 \cdot 0.01} \)
Simplify:
\( h = \frac{0.296}{98.1} = 0.003 \, \text{m} = 3 \, \text{mm} \)
The true static height is the observed height minus the capillary rise:
\( \text{True static height} = 25 \, \text{mm} – 3 \, \text{mm} = 22 \, \text{mm} \)
Final Result:
The true static height of water is \( 22 \, \text{mm} \).
Explanation
This calculation involves the following concepts:
- Surface tension and capillary rise: The surface tension of water causes it to climb up the sides of the glass tube, resulting in a rise above the true static level. The capillary height is inversely proportional to the diameter of the tube.
- Capillary rise formula: The formula accounts for the surface tension (\( \sigma \)), contact angle (\( \theta \)), density of water (\( \rho = 1000 \, \text{kg/m}^3 \)), gravitational acceleration (\( g = 9.81 \, \text{m/s}^2 \)), and tube diameter (\( d \)).
- True static height: By subtracting the capillary rise from the observed height, the true static height is determined.
This problem demonstrates how surface tension affects the measured height of liquids in narrow tubes, with practical applications in capillarity and fluid mechanics.
