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Mechanical Properties of Fluids

NCERT Class 11 Physics Chapter 9

Mechanical Properties of Fluids for Class 11 Physics, JEE and NEET: pressure, Pascal’s law, Bernoulli’s principle, viscosity, Stokes’ law, surface tension.

7 topics Updated 8 Nov 2026

For: Class 11, JEE and NEET, unless a topic says otherwise

In the syllabus: JEE Class 11 syllabus, NEET Class 11 syllabus

With your textbook

NCERT Physics Part II (Class 11), Chapter 9 “Mechanical Properties of Fluids”, reprint 2026-27.

Open the official NCERT chapter (PDF, 2.2 MB, ncert.nic.in)

Checked on 19 Oct 2026.

Learn in this order

  1. Pressure in fluids

    Pascal’s law, pressure at depth P = P + ρgh, gauge and absolute pressure, barometers and the hydraulic lift.

    NCERT Section 9.2 17 min read 6 solved examples 6 practice questions

  2. Streamline and turbulent flow

    Steady and turbulent flow, the equation of continuity Av = Av, critical velocity and the Reynolds number.

    NCERT Section 9.3 12 min read 4 solved examples 6 practice questions

  3. Bernoulli’s principle

    The equation and its derivation, Torricelli’s law, dynamic lift and the venturi meter.

    NCERT Section 9.4 16 min read 5 solved examples 6 practice questions

  4. Viscosity

    The coefficient of viscosity, its unit and dimensions, Stokes’ law F = 6πηav and terminal velocity, solved.

    NCERT Section 9.5 15 min read 4 solved examples 6 practice questions

  5. Experiment: viscosity from terminal velocity

    The falling-ball method, Stokes’ law, terminal velocity and its graph against a, precautions and errors.

    JEE and NEET only 12 min read 3 solved examples 6 practice questions

  6. Surface tension

    Surface energy, angle of contact, excess pressure in drops and bubbles, capillary rise and detergents, solved.

    NCERT Section 9.6 20 min read 6 solved examples 6 practice questions

  7. Experiment: surface tension by capillary rise

    Capillary rise with a travelling microscope, S = rhρg/2, the effect of detergents, precautions and errors.

    JEE and NEET only 11 min read 4 solved examples 6 practice questions

Key formulas in this chapter

Pressure in fluids

  • P = F/A (normal force; unit Pa)
  • P − P = ρgh
  • P = P + ρgh; gauge pressure = ρgh
  • P = ρgh (barometer; about 76 cm of mercury)
  • F = FA/A; AL = AL (hydraulic lift)

Streamline and turbulent flow

  • ρAv = ρAv (conservation of mass)
  • Av = Av, Av = constant (incompressible fluid)
  • Volume flow rate = Av (m s)
  • Re = ρvd/η (laminar below about 2000 in a pipe; beyond NCERT)

Bernoulli’s principle

  • P + ½ρv + ρgh = constant (along a streamline)
  • P − P = ½ρ(v − v) + ρg(h − h)
  • v = √(2gh) (Torricelli’s law, open tank)
  • v = √(2gh + 2(P − P)/ρ) (closed tank at pressure P)
  • ΔP = ½ρ(v − v) (lift per unit wing area; v over the top)

Viscosity

  • η = (F/A)/(v/l) = Fl/(vA) (unit Pa s)
  • F = 6πηav (Stokes’ law)
  • vt = 2a2(ρ − σ)g/(9η) (terminal velocity)

Experiment: viscosity from terminal velocity

  • vt = 2a2(ρ − σ)g/(9η); vt = d/t
  • η = 2a2(ρ − σ)g/(9vt)
  • η = 2(ρ − σ)g/(9 × slope), slope of vt against a2
  • Δη/η ≈ 2Δa/a + Δt/t + Δd/d (first-order worst-case; densities and g exact)

Surface tension

  • S = F/(2l) (film with two surfaces); S = surface energy per unit area
  • Sla cos θ + Ssl = Ssa (angle of contact)
  • Pi − Po = 2S/r (drop; air bubble in a liquid)
  • Pi − Po = 4S/r (soap bubble)
  • h = 2S cos θ/(ρga) (capillary rise)

Experiment: surface tension by capillary rise

  • S = rhρg/(2 cos θ) ≈ rhρg/2 (water in clean glass)
  • S ≈ r(h + r/3)ρg/2 (hemispherical meniscus correction; θ ≈ 0)
  • hr = 2S/(ρg) (constant for a liquid)
  • ΔS/S ≈ Δr/r + Δh/h (first-order worst-case; other inputs exact)

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