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Exam level3 past questions

Capillarity and contact angle

Focuses on capillary action, meniscus formation, and the role of contact angle in wetting phenomena.

Why this shows up in the exam

NEET tests your ability to predict and calculate capillary rise and understand surface interactions.

How NEET tests this

Statement analysis · 3 QsSequence / ordering

Learn the idea

Capillary rise is governed by h = 2T cosθ / (ρ g r); the unlocking insight is that for the same rise, cosθ varies directly with liquid density, so a denser liquid must have a smaller contact angle (all acute).

🧠 Memory hook: Higher‑density liquids need a tighter ‘hug’ on the wall → larger cosθ → smaller angle, just like a firm handshake pulls the wall closer.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Capillary rise h = (2 T cosθ)/(ρ g r)
  • T is surface tension (force per unit length)
  • θ is the angle between the liquid surface and the tube wall measured inside the liquid
  • For a liquid to rise, θ < 90° so that cosθ is positive
  • r is the radius of the capillary tube
  • g ≈ 9.8 m s⁻² is the acceleration due to gravity

How to approach it

  1. 1Write the rise formula h = 2T cosθ/(ρ g r)
  2. 2Since h, T, r and g are identical for the three liquids, isolate cosθ = (h ρ g r)/(2T)
  3. 3Recognise that cosθ ∝ ρ; larger ρ gives larger cosθ and therefore a smaller θ (because cos decreases from 0° to 90°)
  4. 4Order the contact angles using the given density order, remembering all θ must be acute for rise

Worked example — watch it click

Three liquids of densities ρ₁, ρ₂ and ρ₃ (with ρ₁>ρ₂>ρ₃), having the same value of surface tension T, rise to the same height in three identical capillaries. The angles of contact θ₁, θ₂ and θ₃ obey:

  • A)π/2>θ₁>θ₂>θ₃≥0
  • ✅0≤θ₁<θ₂<θ₃<π/2
  • C)π/2<θ₁<θ₂<θ₃<π/2
  • D)π>θ₁<θ₂>θ₃>π/2

The concept behind this problem

The question links the same capillary height for three liquids to their contact angles; using h = 2T cosθ/(ρ g r) forces you to relate density ordering to angle ordering.

Step by step

  1. 1Capillary rise h = (2T cosθ)/(ρgr).
  2. 2For same h and T, cosθ ∝ ρ.
  3. 3Since ρ₁>ρ₂>ρ₃, we have cosθ₁>cosθ₂>cosθ₃, which means θ₁<θ₂<θ₃ (all acute for rise).

Watch out

Treating ρ cosθ as a constant instead of recognizing that cosθ is directly proportional to ρ for equal rise.

Common slip-ups that cost marks

  • •Assuming the contact angle can be obtuse when the liquid actually rises
  • •Reversing the proportionality and concluding larger density gives larger θ
  • •Neglecting that T and r are the same for all three capillaries

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 3NEET 2015

Water rises to height 'h' in a capillary tube. If the length of the capillary tube above the surface of water is made less than 'h', then

Push further

More challenging

3 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 3

A liquid rises to a height of 10 cm in a capillary tube. If the capillary tube is replaced by another tube of the same radius but only 8 cm long, what will happen?

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