Mechanical Properties of Fluids – Surface Tension and Capillarity: NEET MCQs with Answers

15 free questions on Mechanical Properties of Fluids – Surface Tension and Capillarity for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Surface Tension and Capillarity: Surface energy and surface tension, angle of contact, excess of pressure across a curved surface, application of surface tension - drops, bubbles, and capillary rise.

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Question 1

Previous-year question
A wire of length L and radius r (r ≪ L) is kept floating on the surface of a liquid of density ρ. The maximum radius of the wire for which it may not sink is: (surface tension of liquid = T) [NEET 2024]
  1. √(2T/ρg)
  2. √(2Tρ/πg)
  3. √(2T/πρg)
  4. √(T/ρg)
Show answer
Answer: (C) √(2T/πρg)
Explanation: Surface tension supports the wire along two lines of length L: 2TL = weight = πr²Lρg. r = √(2T/(πρg)).

Question 2

Previous-year question
A soap bubble in vacuum has radius r₁ and another has radius r₂. If the bubbles coalesce under isothermal conditions, the radius of the new bubble is: [NEET 2021]
  1. (r₁ + r₂)/2
  2. (r₁² + r₂²)^(1/2)
  3. r₁r₂/(r₁ − r₂)
  4. r₁r₂/(r₂ − r₁)
Show answer
Answer: (B) (r₁² + r₂²)^(1/2)
Explanation: Isothermal: P₁V₁ + P₂V₂ = PV with P = 4T/r (vacuum) gives r₁² + r₂² = r². r = (r₁² + r₂²)^(1/2).

Question 3

Previous-year question
A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is [NEET 2020]
  1. 5.0 g
  2. 10.0 g
  3. 20.0 g
  4. 2.5 g
Show answer
Answer: (B) 10.0 g
Explanation: Since m ∝ r, doubling the radius doubles the mass: 10 g.

Question 4

Previous-year question
A soap bubble, having a radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5 × 10⁻² N/m. The pressure inside the bubble equals that at a point Z₀ below the free surface of water in a container. Taking g = 10 m/s² and the density of water = 10³ kg/m³, the value of Z₀ is [NEET 2019]
  1. 100 cm
  2. 10 cm
  3. 1 cm
  4. 0.5 cm
Show answer
Answer: (C) 1 cm
Explanation: The excess pressure is 4T/r = 4 × 2.5 × 10⁻²/10⁻³ = 100 Pa, and h = P/ρg = 100/10⁴ = 0.01 m = 1 cm.

Question 5

Previous-year question
The cylindrical tube of a spray pump has radius R, one end of which has n fine holes, each of radius r. If the speed of the liquid in the tube is V, the speed of the ejection of the liquid through the holes is [AIPMT 2015]
  1. VR²/nr²
  2. VR²/n³r²
  3. V²R/nr
  4. VR²/n²r²
Show answer
Answer: (A) VR²/nr²
Explanation: Continuity gives πR²V = nπr²v, so v = VR²/nr².

Question 6

Previous-year question
Water rises to a height h in a capillary tube. If the length of the capillary tube above the surface of water is made less than h, then [AIPMT 2015]
  1. water rises up to the top of the capillary tube and stays there without overflowing
  2. water rises up to a point a little below the top and stays there
  3. water does not rise at all
  4. water rises up to the tip of the capillary tube and then starts overflowing like a fountain
Show answer
Answer: (A) water rises up to the top of the capillary tube and stays there without overflowing
Explanation: Water cannot overflow: the meniscus flattens (its radius of curvature increases) just enough to support the shorter column, so the water stops at the top of the tube.

Question 7

Previous-year question
A certain number of spherical drops of a liquid of radius r coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then [AIPMT 2014]
  1. energy = 4VT(1/r – 1/R) is released
  2. energy = 3VT(1/r + 1/R) is absorbed
  3. energy = 3VT(1/r – 1/R) is released
  4. energy is neither released nor absorbed
Show answer
Answer: (C) energy = 3VT(1/r – 1/R) is released
Explanation: The decrease in surface area gives ΔE = 3VT(1/r – 1/R), and since the area decreases this energy is released.
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Question 8

Previous-year question
A certain number of spherical drops of a liquid, each of radius r, coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then [AIPMT 2014]
  1. energy = 4VT(1/r − 1/R) is released
  2. energy = 3VT(1/r + 1/R) is absorbed
  3. energy = 3VT(1/r − 1/R) is released
  4. energy is neither released nor absorbed
Show answer
Answer: (C) energy = 3VT(1/r − 1/R) is released
Explanation: Volume is conserved, so n(4/3)πr³ = (4/3)πR³ = V, giving n = R³/r³. Initial total area, A_i = n × 4πr² = (R³/r³)(4πr²) = 4πR³/r = 3V/r (using 4πR³ = 3V). Final area, A_f = 4πR² = 3V/R. The area decreases, so ΔA = A_i − A_f = 3V(1/r − 1/R). Energy released = T ΔA = 3VT(1/r − 1/R). Since R > r, this is positive — coalescing reduces surface area, so energy is given out and the drop warms slightly.

Question 9

Previous-year question
The angle of contact between pure water and pure glass is [AIPMT 2015]
  1. 0°
  2. 45°
  3. 90°
  4. 135°
Show answer
Answer: (A) 0°
Explanation: For pure water on clean glass, the adhesive forces between water and glass are strong enough that the water spreads completely along the surface. The tangent to the liquid surface then coincides with the solid surface, giving an angle of contact of 0°. This is why water climbs highest in a clean glass capillary.

Question 10

Previous-year question
Water rises to a height h in a capillary tube. If the length of the capillary tube above the surface of the water is made less than h, then [AIPMT 2015]
  1. water rises up to the top of the capillary tube and stays there without overflowing
  2. water rises up to a point a little below the top and stays there
  3. water does not rise at all
  4. water rises up to the tip of the capillary tube and then starts overflowing like a fountain
Show answer
Answer: (A) water rises up to the top of the capillary tube and stays there without overflowing
Explanation: The ascent formula h = 2σ/(ρgR) shows that the height and the radius of curvature of the meniscus are inversely related: hR = constant. If the available length L is less than h, the water simply rises to the top and the meniscus flattens out, its radius of curvature increasing to R′ such that L R′ = h R. The liquid neither overflows nor trickles back — it stops at the top with a less curved surface.

Question 11

Previous-year question
Three liquids of densities ρ₁, ρ₂ and ρ₃ (with ρ₁ > ρ₂ > ρ₃), all 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 [NEET 2016]
  1. π/2 > θ₁ > θ₂ > θ₃ ≥ 0
  2. 0 ≤ θ₁ < θ₂ < θ₃ < π/2
  3. π/2 < θ₁ < θ₂ < θ₃ < π
  4. θ₁ > θ₂ > θ₃ > π/2
Show answer
Answer: (B) 0 ≤ θ₁ < θ₂ < θ₃ < π/2
Explanation: By the ascent formula, h = 2T cos θ/(ρgr). Since h, T and r are the same for all three, cos θ must be proportional to ρ : cos θ₁/ρ₁ = cos θ₂/ρ₂ = cos θ₃/ρ₃. Given ρ₁ > ρ₂ > ρ₃, it follows that cos θ₁ > cos θ₂ > cos θ₃, and since cosine decreases over this range, θ₁ < θ₂ < θ₃. The liquids rise rather than fall, so every angle must be acute: 0 ≤ θ₁ < θ₂ < θ₃ < π/2.

Question 12

Previous-year question
A rectangular film of liquid is extended from (4 cm × 2 cm) to (5 cm × 4 cm). If the work done is 3 × 10⁻⁴ J, the value of the surface tension of the liquid is [NEET 2016]
  1. 0.250 N m⁻¹
  2. 0.125 N m⁻¹
  3. 0.2 N m⁻¹
  4. 8.0 N m⁻¹
Show answer
Answer: (B) 0.125 N m⁻¹
Explanation: A liquid film has two surfaces, so the change in total area is ΔA = 2 × (5 × 4 − 4 × 2) = 2 × 12 = 24 cm² = 24 × 10⁻⁴ m². W = T ΔA 3 × 10⁻⁴ = T × 24 × 10⁻⁴ T = 3/24 = 0.125 N m⁻¹.

Question 13

Previous-year question
A soap bubble of radius 1 mm is blown from a detergent solution having a surface tension of 2.5 × 10⁻² N/m. The pressure inside the bubble equals the pressure at a point Z₀ below the free surface of water in a container. Taking g = 10 m/s² and the density of water as 10³ kg/m³, the value of Z₀ is [NEET (National) 2019]
  1. 10 cm
  2. 1 cm
  3. 0.5 cm
  4. 100 cm
Show answer
Answer: (B) 1 cm
Explanation: Pressure inside the bubble: p₁ = p₀ + 4T/r. Pressure at depth Z₀ in the water: p₂ = p₀ + Z₀ρg. Equating the two and cancelling p₀ : 4T/r = Z₀ρg Z₀ = 4T/(rρg) = (4 × 2.5 × 10⁻²)/(10⁻³ × 10³ × 10) = 0.1/10 = 10⁻² m = 1 cm.

Question 14

Previous-year question
A capillary tube of radius r is immersed in water and the water rises in it to a height h. The mass of the water in the capillary tube is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of the water that will rise in this tube is [NEET (Sep.) 2020]
  1. 5.0 g
  2. 10.0 g
  3. 20.0 g
  4. 2.5 g
Show answer
Answer: (B) 10.0 g
Explanation: The height of rise is h = 2T cos θ/(ρgr), so h ∝ 1/r. The mass of the raised liquid is m = πr²hρ, so m ∝ r² × (1/r) = r — the mass is directly proportional to the radius. Doubling the radius therefore doubles the mass: m₂ = 2 × 5 = 10 g.

Question 15

Previous-year question
A liquid does not wet a solid surface if the angle of contact is [NEET (Oct.) 2020]
  1. equal to 45°
  2. equal to 60°
  3. greater than 90°
  4. zero
Show answer
Answer: (C) greater than 90°
Explanation: When the angle of contact is obtuse, i.e. greater than 90°, the cohesive forces within the liquid dominate over the adhesive forces between liquid and solid. The liquid then beads up rather than spreading, so it does not wet the surface — mercury on glass is the classic example. An acute angle (and especially zero, as for pure water on clean glass) means the liquid wets the surface.

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