A capillary tube of diameter 4 mm is dipped into a liquid of specific gravity 0.6. The liquid rises in the tube by 20 mm, making an angle of contact of 0° with the tube. Determine the surface tension of the liquid in contact with air and the glass tube.
Try it first: pick an option, then read the explanation.
Asked in SSC JE 09/10/2023 (10am-12pm)
Detailed Explanation
Capillary rise occurs when adhesive forces between the liquid and tube wall exceed cohesive forces within the liquid. The vertical component of surface tension around the meniscus circumference supports the weight of the elevated liquid column.
Given: \(d = 4\text{ mm} = 0.004\text{ m}\), \(h = 20\text{ mm} = 0.02\text{ m}\), \(\theta = 0^\circ\), \(S = 0.6 \implies \rho = 0.6 \times 1000 = 600\text{ kg/m}^3\), and taking \(g = 10\text{ m/s}^2\):
\[ h = \frac{4T\cos\theta}{\rho g d} \;\implies\; T = \frac{h \cdot \rho g d}{4\cos\theta} \]
\[ T = \frac{0.02 \times 600 \times 10 \times 0.004}{4 \times \cos 0^\circ} = \frac{0.48}{4} = 0.12\text{ N/m} \]
Why “0.12 N/m” Is Correct
Rearranging the capillary rise equation gives T = 0.12 N/m, which matches option (b).
Why the Other Options Are Wrong
- A. 1 N/m: 1 N/m is much too large for typical liquid surface tension values.
- C. 0.08 N/m: 0.08 N/m results from omitting the factor of 1.5 in density or using d = 6 mm.
- D. 0.5 N/m: 0.5 N/m is an incorrect distractor.
Key Point
Capillary rise h = 4T cosθ / (ρgd); T = hρgd / (4 cosθ).
Key Concepts for Students
- Surface tension T has units of N/m or J/m².
- For complete wetting (pure water on clean glass), contact angle θ ≈ 0° (cos θ = 1).
- Capillary rise is inversely proportional to tube diameter d.
Practice more: this is question 14 in Properties of Fluid MCQs – Part 1. Try the whole set and check every answer instantly.