JEE Advanced level
A JEE Advanced-level example on Properties of Solids and Liquids
Question. A U-tube has two vertical limbs of radii r₁ = 1 mm and r₂ = 0.5 mm connected at the bottom. It contains water with surface tension σ = 0.07 N/m, contact angle θ = 0°, and density ρ = 1000 kg/m³. (a) Find the difference in water levels between the two limbs. (b) If the temperature rises by 50°C, the coefficient of volume expansion of water is β = 2 × 10⁻⁴ /°C, and surface tension decreases by 0.15% per °C, find the new difference in levels. Take g = 10 m/s².
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- Capillary rise formula: h = 2σcosθ/(ρgr). With θ = 0°, this becomes h = 2σ/(ρgr).
- For narrow limb (r₂ = 0.5 mm): h₂ = 2 × 0.07/(1000 × 10 × 0.0005) = 2.8 cm. For wider limb (r₁ = 1 mm): h₁ = 2 × 0.07/(1000 × 10 × 0.001) = 1.4 cm.
- Initial difference in levels = h₂ − h₁ = 2.8 − 1.4 = 1.4 cm.
- After heating: New surface tension σ' = 0.07 × (1 − 0.0015 × 50) = 0.06475 N/m. New density ρ' = 1000/(1 + 0.01) = 990.1 kg/m³.
- New heights: h₂' = 2 × 0.06475/(990.1 × 10 × 0.0005) = 2.62 cm and h₁' = 2 × 0.06475/(990.1 × 10 × 0.001) = 1.31 cm.
- New difference = 2.62 − 1.31 = 1.31 cm.
Answer: (a) 1.4 cm; (b) 1.31 cm
This example was solved twice independently and both solutions agree.