NEET Mechanical Properties of Fluids PYQs & Practice Questions

10 free NEET Physics questions on Mechanical Properties of Fluids, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Surface Tension · Pressure and Pascal's Law · Bernoulli's Theorem · Viscosity and Stokes' Law · Surface Tension and Capillarity · Archimedes' Principle and Buoyancy · Equation of Continuity and Streamline Flow · Fluid Flow · Equation of Continuity · Surface Tension and Angle of Contact · Surface Tension - Force to Remove Disc from Water Surface · Venturimeter Principle

Q1NEET 2026Terminal Velocity and Viscosity
In the measurement of viscosity of liquids using the terminal velocity experiment, spherical balls of the same radius but having different densities are used. The variation of the terminal velocity (v) with the ratio of density of the spherical ball (σ) to density of the liquid (ρ), is best represented by a graph that is:
- A. a straight line through the origin with positive slope
- B. a straight line with a positive y-intercept at σ/ρ=0
- C. zero for σ/ρ < 1 with a sudden jump and a flat plateau beyond σ/ρ = 1
- D. zero (flat along the axis) for σ/ρ ≤ 1 and then rising linearly for σ/ρ > 1
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Answer: C — zero for σ/ρ < 1 with a sudden jump and a flat plateau beyond σ/ρ = 1
Terminal velocity is given by v = (2r²g/9η)(σ-ρ) = (2r²gρ/9η)(σ/ρ - 1). For σ/ρ ≤ 1, the ball does not sink (it floats or is neutrally buoyant), so there is no downward terminal (sinking) velocity to measure, giving v = 0 in that region. For σ/ρ > 1, v increases linearly with σ/ρ, with the line originating from the point (1, 0). This flat-then-linear-rising-from-(1,0) shape matches option (4).
Q2NEET 2026Bernoulli's Theorem and Continuity Equation
Water flows in a streamline motion through a horizontal pipe of circular cross-section. The pressure difference of water between P and Q is 15 Nm⁻². The area of cross-section at P and Q are 40 cm² and 20 cm², respectively. The rate of flow of water through the pipe, in cm³s⁻¹, is: [Take density of water = 1000 kg m⁻³]
- A. 200
- B. 300
- C. 400
- D. 100
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Answer: B — 300
By continuity, v_P A_P = v_Q A_Q, so v_Q = 2v_P (since A_P=2A_Q). By Bernoulli's theorem: P_P - P_Q = (1/2)ρ(v_Q² - v_P²) = (1/2)ρ(4v_P² - v_P²) = (3/2)ρv_P². So 15 = (3/2)(1000)v_P², giving v_P² = 0.01, v_P = 0.1 m/s = 10 cm/s. Flow rate = A_P × v_P = 40 cm² × 10 cm/s = 400 cm³/s.
Q3NEET 2026Pressure and depth
A submarine is designed to withstand an absolute pressure of 100 atm. How deep can it go below the water surface? (density of water=1000 kg/m³, 1 atm=1×10⁵ Pa, g=10 m/s²)
- A. 9900 m
- B. 990 m
- C. 9000 m
- D. 99 m
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Answer: A — 9900 m
P=P0+ρgh → (100-1)×10⁵=1000×10×h → h=990 m.
Q4NEET 2025Surface Tension
Consider a water tank shown in the figure. It has one wall at x = L and can be taken to be very wide in the z direction. When filled with a liquid of surface tension S and density ρ, the liquid surface makes angle θ0 (θ0 << 1) with the x-axis at x = L. If y(x) is the height of the surface then the equation for y(x) is - (take θ(x) = sin θ(x) = tan θ(x) = dy/dx, g is the acceleration due to gravity)
- A. d²y/dx² = (ρg/S)x
- B. d²y/dx² = (ρg/S)y
- C. d²y/dx² = √(ρg/S)
- D. dy/dx = √(ρg/S)x
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Answer: A — d²y/dx² = (ρg/S)x
Radius of curvature R = [1+(dy/dx)²]^(3/2) / (d²y/dx²) ≈ 1/(d²y/dx²) for small θ. Excess pressure ΔP = S/R = S·(d²y/dx²). Hydrostatic pressure at height y is ρgy = ΔP = S·(d²y/dx²). Therefore d²y/dx² = ρgy/S.

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Q5NEET 2024Surface Tension - Force to Remove Disc from Water Surface
A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is 0.07 N m⁻¹, then the excess force required to take it away from the surface is:
- A. 1.98 mN
- B. 99 N
- C. 19.8 mN
- D. 198 N
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Answer: B — 99 N
F = T × 2πr = 0.07 × 2 × 3.14 × 4.5×10⁻² = 19.8 mN.
Q6NEET 2023Venturimeter Principle
The venturi-meter works on:
- A. Bernoulli's principle
- B. The principle of parallel axes
- C. The principle of perpendicular axes
- D. Huygen's principle
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Answer: —
A venturimeter is used to measure the flow speed of a fluid through a pipe, and its working is based on Bernoulli's principle, which relates pressure and velocity changes in a flowing fluid.
Q7NEET 2023Energy to Form Soap Bubble
The amount of energy required to form a soap bubble of radius 2 cm from a soap solution is nearly: (surface tension of soap solution = 0.03 N/m)
- A. 5.06 × 10^-4 J
- B. 3.01 × 10^-4 J
- C. 50.1 × 10^-4 J
- D. 30.16 × 10^-4 J
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Answer: A — 5.06 × 10^-4 J
Work W = T(ΔA), and since a soap bubble has two surfaces, ΔA = 2×4πr². W = 0.03 × 4π×(0.02)² × 2 = 96π×10^-6 ≈ 300×10^-6 = 3.01×10^-4 J.
Q8NEET 2022Terminal Velocity in Viscous Medium
A spherical ball is dropped in a long column of a highly viscous liquid. Among curves A, B, C, D shown in a v-t graph (where A rises slowly and stays low, B rises and levels off to a constant terminal value, C and D show other increasing/peaking patterns), the curve which represents the speed of the ball (v) as a function of time (t) is:
- A. A
- B. B
- C. C
- D. D
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Answer: A — A
When a ball falls through a highly viscous liquid, it accelerates initially due to gravity, but as viscous drag force increases with speed, the net force decreases until it reaches zero at terminal velocity, after which speed remains constant. This matches curve B, which rises and then levels off to a constant (terminal) value.

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Q9NEET 2022Pressure Inside a Soap Bubble
If a soap bubble expands, the pressure inside the bubble:
- A. Decreases
- B. Increases
- C. Remains the same
- D. Is equal to the atmospheric pressure
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Answer: —
Excess pressure inside a soap bubble is given by P = 4T/r, where T is surface tension and r is the radius. As the bubble expands, r increases, so the excess pressure (and hence the pressure inside) decreases.
Q10NEET 2022Viscosity, Terminal Velocity
A spherical ball is dropped in a long column of a highly viscous liquid. The curve in the graph shown (v vs t, with curves A, B, C, D), which represents the speed of the ball (v) as a function of time (t) is
- A. A
- B. B
- C. C
- D. D
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Answer: A — A
The ball accelerates initially and then reaches terminal velocity, asymptotically approaching a constant value — represented by curve B.

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