Class XI · Chapter 09
Mechanical Properties of Fluids — Exam Practice
Practice questions in CBSE, NEET, and JEE Main style covering Pascal’s Law, Bernoulli’s principle, viscosity, and surface tension.
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.
1CBSE Board Exam Style
Q1. Why do small droplets of a liquid tend to take a spherical shape?
Answer
Surface tension acts to minimise a liquid surface’s area for a given volume. Among all shapes, a sphere has the least surface area for its volume, so surface tension naturally pulls small droplets into spherical form.
Surface tension acts to minimise a liquid surface’s area for a given volume. Among all shapes, a sphere has the least surface area for its volume, so surface tension naturally pulls small droplets into spherical form.
Using Stokes’ Law, derive the expression for the terminal velocity of a small sphere falling through a viscous fluid.
Answer
At terminal velocity, net force is zero: weight = buoyant force + viscous drag.
At terminal velocity, net force is zero: weight = buoyant force + viscous drag.
(4/3)πr³ρg = (4/3)πr³σg + 6πηrvt.
Solving for vt: vt = 2r²(ρ−σ)g / (9η), where ρ = sphere’s density, σ = fluid’s density, η = viscosity.
Q3. (a) State and prove Bernoulli’s theorem for horizontal fluid flow. (b) Water flows through a pipe of radius 2 cm with speed 3 m/s. Find the speed when the pipe narrows to a radius of 1 cm.
Answer
(a) Applying the work-energy theorem to a fluid element moving through a pipe of varying cross-section, and accounting for pressure-difference work and changes in kinetic and potential energy, gives: P + ½ρv² + ρgh = constant along any streamline, for an ideal (incompressible, non-viscous) fluid.
(a) Applying the work-energy theorem to a fluid element moving through a pipe of varying cross-section, and accounting for pressure-difference work and changes in kinetic and potential energy, gives: P + ½ρv² + ρgh = constant along any streamline, for an ideal (incompressible, non-viscous) fluid.
(b) By the equation of continuity: A₁v₁ = A₂v₂ → v₂ = v₁(r₁/r₂)² = 3 × (2/1)² = 3 × 4 = 12 m/s.
2NEET Style (Single-Correct MCQ)
Q4. The viscosity of a liquid, as temperature increases:
Solution
Higher temperature weakens intermolecular attraction in a liquid, allowing layers to slide past each other more easily — reducing viscosity (the opposite trend to gases, where viscosity increases with temperature).
Higher temperature weakens intermolecular attraction in a liquid, allowing layers to slide past each other more easily — reducing viscosity (the opposite trend to gases, where viscosity increases with temperature).
Q5. Reynolds number is used to predict:
Solution
Re = ρvd/η — low values indicate smooth laminar flow; high values indicate chaotic turbulent flow.
Re = ρvd/η — low values indicate smooth laminar flow; high values indicate chaotic turbulent flow.
Q6. Capillary rise of a liquid in a narrow tube is caused by:
Solution
h = 2Tcosθ/(ρgr) — surface tension pulls the liquid up along the tube walls, with narrower tubes producing greater rise.
h = 2Tcosθ/(ρgr) — surface tension pulls the liquid up along the tube walls, with narrower tubes producing greater rise.
Q7. Archimedes’ Principle relates the buoyant force on a submerged object to:
Solution
Buoyant force exactly equals the weight of the fluid the object displaces — regardless of the object’s own weight or material.
Buoyant force exactly equals the weight of the fluid the object displaces — regardless of the object’s own weight or material.
3JEE Main Style
Q8. Find the excess pressure inside a soap bubble of radius 1 mm. (Surface tension of soap solution T = 0.03 N/m)
Solution
A soap bubble has two surfaces, so Δp = 4T/r = (4 × 0.03) / 0.001 = 120 Pa.
A soap bubble has two surfaces, so Δp = 4T/r = (4 × 0.03) / 0.001 = 120 Pa.
Q9. A hydraulic press amplifies a small applied force into a much larger output force, based on:
Solution
Pressure applied to an enclosed fluid transmits undiminished to every part of it — a small force on a narrow piston produces a proportionally larger force on a wider piston.
Pressure applied to an enclosed fluid transmits undiminished to every part of it — a small force on a narrow piston produces a proportionally larger force on a wider piston.
Q10. Find the terminal velocity of a small ball of radius 1 mm and density 8000 kg/m³ falling through a fluid of density 1000 kg/m³ and viscosity 1 Pa·s. (g = 10 m/s²)
Solution
vt = 2r²(ρ−σ)g/(9η) = [2 × (0.001)² × (8000−1000) × 10] / (9×1) = 0.14/9 ≈ 0.0156 m/s.
vt = 2r²(ρ−σ)g/(9η) = [2 × (0.001)² × (8000−1000) × 10] / (9×1) = 0.14/9 ≈ 0.0156 m/s.
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