1. 0≤θ₁ θ₁>θ₂>θ₃>π/2 4. π/2>θ₁>θ₂>θ₃≥0

Answer: 1. 0≤θ₁ 0 , so each θ ρ₂>ρ₃ gives cosθ₁>cosθ₂>cosθ₃ . Cosine decreases as the angle grows, so θ₁<θ₂<θ₃ . Therefore 0≤θ₁<θ₂<θ₃<π/2 . Option (1). Options (2) and (3) have obtuse angles (liquid would fall).

Topic: Surface energy, surface tension, angle of contact, excess pressure, drops, bubbles, capillary rise· Properties of Solids and Liquids

Three liquids of densities ρ_1,ρ_2 and ρ_3 (with ρ_1>ρ_2>ρ_3), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact θ_1,θ_2 and θ_3 obey

1. \(0\le\theta_1<\theta_2<\theta_3<\frac{\pi}{2}\)

2. \(\frac{\pi}{2}<\theta_1<\theta_2<\theta_3<\pi\)

3. \(\pi>\theta_1>\theta_2>\theta_3>\frac{\pi}{2}\)

4. \(\frac{\pi}{2}>\theta_1>\theta_2>\theta_3\ge0\)

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 2 Practice test neet Save WhatsApp Report

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Answer: 1. \(0\le\theta_1<\theta_2<\theta_3<\frac{\pi}{2}\)

Capillary rise is \(h=\dfrac{2T\cos\theta}{r\rho g}\).

Here \(h\), \(T\) and \(r\) are the same for all three liquids, so \(\cos\theta\propto\rho\).

Since the liquids rise, \(\cos\theta>0\), so each \(\theta<\pi/2\).
\(\rho_1>\rho_2>\rho_3\) gives \(\cos\theta_1>\cos\theta_2>\cos\theta_3\). Cosine decreases as the angle grows, so \(\theta_1<\theta_2<\theta_3\).

Therefore \(0\le\theta_1<\theta_2<\theta_3<\pi/2\). Option (1). Options (2) and (3) have obtuse angles (liquid would fall).

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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Three liquids of densities ρ_1,ρ_2 and ρ_3 (with ρ_1>ρ_2>ρ_3), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact θ_1,θ_2 and θ_3 obey

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