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 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)
30°
0°
45°
75°
60°
The Correct Option is
Approach Solution - 1
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: \(\boxed{E}\)
Approach Solution -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:
\[θ = \arccos(\frac{1}{2}) = 60^\circ\]
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Concepts Used:
Surface Tension
The amount of energy required to increase the liquid's surface area by one unit area is known as surface tension. In other words, it is a property of the liquid surface to resist force.
Surface tension is defined as,
The ratio of the surface force F to the length L along which the force acts.
Mathematically, the surface tension formula can be expressed as follows:
T=F/L
Where,
- F is the force per unit length
- L is the length in which force act
- T is the surface tension of the liquid
Read More: Detergents and Surface Tension
Factors affecting surface tension:
- Impurities: The surface tension decreases with the addition of impurities.
- Surfactants: Adding surfactants in liquids lowers the tension of water making it interrupt aside or get susceptible.
- Temperature: The surface tension of a liquid reduces as the temperature rises.
The Unit of Surface Tension:
The SI unit of Surface Tension is Newton per Meter or N/m.