Mechanical Properties of Fluids NEET PYQReal NEET & AIPMT 2013–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Mechanical Properties of Fluids question from the NEET & AIPMT papers on this site (2013–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Mechanical Properties of Fluids previous year questions
11 questions. Tap an option to check your answer.
Q1NEET 2021
The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is d/2, then the viscous force acting on the ball will be :
Not quite — the answer is A.
At terminal velocity the net force is zero: weight = buoyant force + viscous force. Ball volume = M/d, so buoyancy = (M/d)(d/2)g = Mg/2, giving viscous force = Mg − Mg/2 = Mg/2. Taking viscous force equal to the full weight Mg ignores buoyancy.
Q2NEET 2020 · Q178
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 :
Not quite — the answer is C.
Jurin's law gives h ∝ 1/r, and the mass of the raised column m = ρπr²h ∝ r²/r = r. Doubling the radius doubles the mass, from 5 g to 10 g. Assuming the height stays the same would give 20 g.
Q3NEET 2019 · Q7
A soap bubble, having 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 at a point Z₀ below the free surface of water in a container. Taking g = 10 m/s², density of water = 10³ kg/m³, the value of Z₀ is:
Not quite — the answer is C.
Excess pressure in a soap bubble (two surfaces) = 4T/r = 4 × 2.5 × 10⁻²/10⁻³ = 100 Pa. Setting ρgZ₀ = 100 gives Z₀ = 100/(10³ × 10) = 0.01 m = 1 cm. Using 2T/r (single surface) would give 0.5 cm.
Q4NEET 2019 · Q38
A small hole of area of cross-section 2 mm² is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s², the rate of flow of water through the open hole would be nearly:
Not quite — the answer is A.
Torricelli: v = √(2gh) = √40 ≈ 6.32 m/s. Flow rate = Av = 2 × 10⁻⁶ × 6.32 ≈ 12.6 × 10⁻⁶ m³/s. 6.4 × 10⁻⁶ m³/s is roughly the speed alone without multiplying by the 2 mm² area.
Q5NEET 2018 · Q42
A small sphere of radius 'r' falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to
Not quite — the answer is A.
At terminal velocity the heating rate equals F·v = 6πηrv·v. Since v_t ∝ r², power ∝ r × r⁴ = r⁵. Taking only the force (∝ r³ at terminal speed) misses the extra factor of velocity.
Q6NEET 2016 Phase 2 · Q13
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
Not quite — the answer is B.
Area increases from 8 cm² to 20 cm², i.e. by 12 × 10⁻⁴ m², and a film has two surfaces, so the total new surface is 24 × 10⁻⁴ m². T = W/ΔA = 3 × 10⁻⁴/24 × 10⁻⁴ = 0.125 N/m. Forgetting the two surfaces gives 0.250 N/m.
Q7NEET 2016 Phase 2 · Q14
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
Not quite — the answer is B.
From h = 2T cosθ/(rρg), equal h, T and r means cosθ ∝ ρ. The densest liquid has the largest cosθ and hence the smallest angle, so θ₁ < θ₂ < θ₃; since the liquids rise, all angles are below π/2. Angles above π/2 would mean capillary depression.
Q8NEET 2016 Phase 1 · Q33
Two non-mixing liquids of densities ρ and nρ (n > 1) are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL (p < 1) in the denser liquid. The density d is equal to:
Not quite — the answer is D.
Length pL is in the denser liquid and (1 − p)L in the lighter one. Weight = buoyancy: dLAg = nρ(pL)Ag + ρ(1 − p)LAg, so d = ρ[1 + (n − 1)p]. Using (n + 1) comes from adding instead of subtracting the lighter-liquid share.
Q9AIPMT 2015
A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m². Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be: (ρ_air = 1.2 kg/m³)
Not quite — the answer is D.
By Bernoulli the pressure above the roof is lower by ½ρv² = 0.5 × 1.2 × 40² = 960 Pa. Force = 960 × 250 = 2.4 × 10⁵ N, acting upwards because the higher pressure is inside the house.
Q10AIPMT 2014
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:
Not quite — the answer is B.
With n = R³/r³ drops, the area lost is 4π(nr² − R²) = 4πR³(1/r − 1/R) = 3V(1/r − 1/R). Since surface area decreases, surface energy 3VT(1/r − 1/R) is released, not absorbed.
Q11NEET 2013
The wettability of a surface by a liquid depends primarily on -
Not quite — the answer is D.
Whether a liquid wets a surface is decided by the angle of contact: an acute angle means wetting, an obtuse angle means non-wetting. Surface tension alone does not decide this, since the contact angle depends on the balance of adhesive and cohesive forces.
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