Question

A glass capillary of radius 0.15 mm is dipped into a liquid of density and surface tension 1600 kg/m³ and 0.12 Nm^-1,respectively. The liquid in the capillary rises by a height of 5.0 cm. The contact angle between liquid and glass will be:(Take g=10 ms^-2)

• A. 30°
• B. 0°
• C. 45°
• D. 75°
• E. 60°

Answer 1

Answer: (E)

Given:

• Capillary radius, \( r = 0.15 \, \text{mm} = 0.15 \times 10^{-3} \, \text{m} \)
• Liquid density, \( \rho = 1600 \, \text{kg/m}^3 \)
• Surface tension, \( \gamma = 0.12 \, \text{Nm}^{-1} \)
• Capillary rise, \( h = 5.0 \, \text{cm} = 0.05 \, \text{m} \)
• Gravitational acceleration, \( g = 10 \, \text{ms}^{-2} \)

Step 1: Capillary Rise Formula

The height of liquid rise in a capillary tube is given by:

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

where \( \theta \) is the contact angle.

Step 2: Solve for \( \cos \theta \)

Rearrange the formula:

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

Substitute the given values:

\[ \cos \theta = \frac{0.05 \times 1600 \times 10 \times 0.15 \times 10^{-3}}{2 \times 0.12} \]

\[ \cos \theta = \frac{0.12}{0.24} = 0.5 \]

Step 3: Determine \( \theta \)

\[ \theta = \cos^{-1}(0.5) = 60^\circ \]

Conclusion:

The contact angle between the liquid and glass is 60°.

Answer: \(\boxed{E}\)

Answer 2

1. Define variables and given information:

• r = 0.15 mm = 1.5 × 10⁻⁴ m (radius of the capillary)
• ρ = 1600 kg/m³ (density of the liquid)
• S = 0.12 N/m (surface tension)
• h = 5.0 cm = 0.05 m (height of the liquid rise)
• g = 10 m/s² (acceleration due to gravity)
• θ = ? (contact angle)

2. Recall the formula for capillary rise:

The height (h) to which a liquid rises in a capillary tube is given by:

\[h = \frac{2S \cos θ}{rρg}\]

where:

• S is the surface tension
• θ is the contact angle
• r is the radius of the capillary
• ρ is the density of the liquid
• g is the acceleration due to gravity

3. Rearrange the formula to solve for cos θ:

\[\cos θ = \frac{hrρg}{2S}\]

4. Substitute the given values and calculate cos θ:

\[\cos θ = \frac{(0.05 \, m)(1.5 \times 10^{-4} \, m)(1600 \, kg/m^3)(10 \, m/s^2)}{2(0.12 \, N/m)}\]

\[\cos θ = \frac{0.12}{0.24} = \frac{1}{2}\]

5. Calculate θ:

Since \(\cos θ = \frac{1}{2}\), we have:

\[θ = \arccos(\frac{1}{2}) = 60^\circ\]