Calculate the capillary rise in a glass tube of 3.0 mm diameter when immersed vertically in (a) water, and (b) mercury. Take surface tensions for water as 0.0725 N/m and for mercury as 0.52 N/m in contact with air. Specific gravity for mercury is given as 13.6.

Capillary Rise and Depression Calculation

Problem Statement

Calculate the capillary rise in a glass tube of 3.0 mm diameter when immersed vertically in (a) water, and (b) mercury. Take surface tensions for water as 0.0725 N/m and for mercury as 0.52 N/m in contact with air. Specific gravity for mercury is given as 13.6.

Given Data

  • Tube Diameter, \(d = 3.0 \, \text{mm} = 3.0 \times 10^{-3} \, \text{m}\)
  • Surface Tension of Water, \(\sigma_w = 0.0725 \, \text{N/m}\)
  • Surface Tension of Mercury, \(\sigma_m = 0.52 \, \text{N/m}\)
  • Specific Gravity of Mercury, \(S.G._m = 13.6\)
  • Contact Angle for Water, \(\theta_w = 0^\circ\) (Assumed for clean glass)
  • Contact Angle for Mercury, \(\theta_m = 130^\circ\) (Standard value)

Solution

The general formula for capillary rise or depression (\(h\)) is:

$$ h = \frac{4\sigma \cos\theta}{\rho g d} $$

(a) Capillary Rise for Water

For water, the density \(\rho_w = 1000 \, \text{kg/m}^3\) and \(\theta_w = 0^\circ\).

$$ h_w = \frac{4 \sigma_w \cos\theta_w}{\rho_w g d} $$ $$ h_w = \frac{4 \times 0.0725 \, \text{N/m} \times \cos(0^\circ)}{1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 3.0 \times 10^{-3} \, \text{m}} $$ $$ h_w = \frac{0.29}{29430} $$ $$ h_w \approx 0.00985 \, \text{m} $$ $$ h_w \approx 9.85 \, \text{mm} $$

(b) Capillary Depression for Mercury

First, find the density of mercury: \(\rho_m = 13.6 \times 1000 = 13600 \, \text{kg/m}^3\). The contact angle \(\theta_m = 130^\circ\).

$$ h_m = \frac{4 \sigma_m \cos\theta_m}{\rho_m g d} $$ $$ h_m = \frac{4 \times 0.52 \, \text{N/m} \times \cos(130^\circ)}{13600 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 3.0 \times 10^{-3} \, \text{m}} $$ $$ h_m = \frac{2.08 \times (-0.6428)}{400248} $$ $$ h_m \approx \frac{-1.337}{400248} $$ $$ h_m \approx -0.00334 \, \text{m} $$ $$ h_m \approx -3.34 \, \text{mm} $$

The negative sign correctly indicates a capillary depression (the level falls).

Final Results:

Capillary Rise in Water: \( h_w \approx 9.85 \, \text{mm} \)

Capillary Depression in Mercury: \( h_m \approx -3.34 \, \text{mm} \)

Explanation of Capillarity

Capillary action is the movement of a liquid along the surface of a solid caused by the attraction of molecules of the liquid to molecules of the solid. This effect is a result of the balance between two forces:

  • Adhesive Forces: The forces of attraction between unlike molecules (e.g., between liquid and glass).
  • Cohesive Forces: The forces of attraction between like molecules (e.g., within the liquid itself).

The contact angle (\(\theta\)) is the angle formed by the liquid at the three-phase (liquid-gas-solid) boundary. It indicates which force is dominant.

Physical Meaning of Results

(a) Water (Wetting Fluid): For water in a glass tube, the adhesive forces between water and glass are stronger than the cohesive forces within the water. This causes the water to “wet” the glass and climb its walls, resulting in a concave meniscus and a capillary rise. The contact angle is acute (\(\theta < 90^\circ\)), making \(\cos\theta\) positive.

(b) Mercury (Non-Wetting Fluid): For mercury, the strong cohesive forces between mercury atoms are much greater than the weak adhesive forces with glass. The mercury pulls itself inward, avoiding contact with the glass. This results in a convex meniscus and a capillary depression, where the liquid level inside the tube is lower than outside. The obtuse contact angle (\(\theta > 90^\circ\)) makes \(\cos\theta\) negative, leading to a negative height.

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