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Mechanical Properties of Fluids – Class 11 Physics notes, important questions and MCQs
NCERT Class 11 Physics, chapter 9. Notes, key terms, important questions with answers and MCQs.
Chapter overview
Fluids exert pressure, buoyancy and viscous forces, and show surface tension. Pascal's law, Bernoulli's principle and Stokes' law explain many everyday effects.
Pressure and flow
- Pressure at depth h: P = P₀ + ρgh; Pascal's law: pressure applied to an enclosed fluid is transmitted equally (hydraulic lift).
- Equation of continuity: A₁v₁ = A₂v₂.
- Bernoulli: P + ½ρv² + ρgh = constant; Torricelli: speed of efflux v = √(2gh).
Viscosity and surface tension
- Viscous force F = ηA dv/dx; Stokes' law F = 6πηrv; terminal velocity vₜ = 2r²(ρ − σ)g/9η.
- Reynolds number decides streamline vs turbulent flow.
- Surface tension S = F/L; excess pressure in a drop 2S/r, in a soap bubble 4S/r; capillary rise h = 2S cos θ/ρrg.
Key terms
- Viscosity
- internal friction in a fluid opposing relative motion of layers
- Terminal velocity
- constant maximum speed of a body falling through a fluid
- Buoyant force
- upward force equal to the weight of fluid displaced (Archimedes' principle)
Formulas
- P + ½ρv² + ρgh = constant
- vₜ = 2r²(ρ − σ)g/9η
- h = 2S cos θ/ρrg
Important questions with answers
Why does a liquid drop take a spherical shape?2 marks
Surface tension tends to minimise surface area, and for a given volume a sphere has the least surface area.
Derive Torricelli's law of efflux using Bernoulli's principle.3 marks
Apply Bernoulli between the open top (pressure P₀, speed ≈ 0, height h) and the hole (pressure P₀, speed v, height 0): P₀ + ρgh = P₀ + ½ρv². So v = √(2gh), the same as a body falling freely through h.
Derive the expression for terminal velocity of a sphere falling through a viscous liquid. Find it for a 2 mm radius steel ball (ρ = 8000 kg/m³) in glycerine (σ = 1300 kg/m³, η = 0.83 Pa s), g = 10 m/s².5 marks
Forces on the sphere: weight (4/3)πr³ρg downward; buoyancy (4/3)πr³σg upward; viscous drag 6πηrv upward. At terminal velocity the net force is zero: 6πηrvₜ = (4/3)πr³(ρ − σ)g, so vₜ = 2r²(ρ − σ)g/9η. Here vₜ = 2 × (2 × 10⁻³)² × 6700 × 10/(9 × 0.83) = 2 × 4 × 10⁻⁶ × 67,000/7.47 ≈ 0.072 m/s (about 7 cm/s).
Quick revision
- Terminal velocity ∝ r².
- Soap bubble has two surfaces (4S/r).
- Mercury shows capillary fall.
Exam tips
- Narrower tube → faster flow (continuity) → lower pressure (Bernoulli): explains lift on aeroplane wings.
- Surface tension decreases with temperature and with impurities like soap (detergents help water spread and clean).
Practice MCQs
Hydraulic lift works on
- Bernoulli's principle
- Pascal's law
- Archimedes' principle
- Stokes' law
Show answer
Answer: (b) Pascal's law. Pressure transmission.
SI unit of surface tension is
- N/m²
- N/m
- J
- Pa s
Show answer
Answer: (b) N/m. Force per length.
Water flows through a pipe with area 4 cm² at 2 m/s into a part with area 2 cm². New speed is
- 4 m/s
- 1 m/s
- 8 m/s
- 2 m/s
Show answer
Answer: (a) 4 m/s. A₁v₁ = A₂v₂.
Excess pressure inside a soap bubble of radius r is
- 4S/r
- 2S/r
- S/r
- 8S/r
Show answer
Answer: (a) 4S/r. Two surfaces.
A hydraulic lift has pistons of area 10 cm² and 1000 cm². A force of 50 N on the small piston can lift
- 500 N
- 5000 N
- 50 N
- 5 N
Show answer
Answer: (b) 5000 N. F₂ = F₁ × A₂/A₁.
Terminal velocity of a sphere is proportional to
- r
- r²
- 1/r
- r³
Show answer
Answer: (b) r². vₜ ∝ r².
Speed of efflux from a hole 5 m below the water surface (g = 10) is
- 5 m/s
- 10 m/s
- 50 m/s
- 100 m/s
Show answer
Answer: (b) 10 m/s. √(2 × 10 × 5).
Two drops of radius r each coalesce into one drop. The surface energy
- increases
- decreases (energy is released)
- stays the same
- becomes zero
Show answer
Answer: (b) decreases (energy is released). Total surface area falls.
All chapters of Class 11 Physics
Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.