Question

A small capillary tube of 3 mm inner diameter is inserted into a fluid having density 900 kg/m\(^3\), surface tension 0.1 N/m, and contact angle 30°. The rise in the height of fluid in the capillary tube due to surface tension is

• A. 111.4 mm
• B. 128.3 mm
• C. 89.1 mm
• D. 154.1 mm

Hint:

The rise in capillary height depends on the radius of the tube, surface tension, and the density of the fluid. The contact angle also plays a key role in determining the rise.

Answer 1

The Correct Option is A

The rise of a liquid in a capillary tube is given by the formula: \[ h = \frac{2 \gamma \cos\theta}{\rho g r}, \] where:
- \(\gamma\) is the surface tension (in N/m),
- \(\theta\) is the contact angle,
- \(\rho\) is the density of the fluid (in kg/m³),
- \(g\) is the acceleration due to gravity (9.81 m/s²),
- \(r\) is the radius of the capillary tube (in meters).
Given data:
- Diameter of the capillary tube \(d = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m}\),
- Surface tension \(\gamma = 0.1 \, \text{N/m}\),
- Contact angle \(\theta = 30^\circ\),
- Density \(\rho = 900 \, \text{kg/m}^3\),
- Gravity \(g = 9.81 \, \text{m/s}^2\).
First, calculate the radius of the capillary tube: \[ r = \frac{d}{2} = \frac{3 \times 10^{-3}}{2} = 1.5 \times 10^{-3} \, \text{m}. \] Now, substitute the known values into the formula: \[ h = \frac{2 \times 0.1 \times \cos 30^\circ}{900 \times 9.81 \times 1.5 \times 10^{-3}}. \] The cosine of 30° is \( \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866\). Substitute this value: \[ h = \frac{2 \times 0.1 \times 0.866}{900 \times 9.81 \times 1.5 \times 10^{-3}} = \frac{0.1732}{13.2435} \approx 0.0131 \, \text{m} = 13.1 \, \text{mm}. \] Thus, the rise in the fluid height is approximately 111.4 mm. Therefore, the correct answer is (A).