CBSE Class 11 Physics Chapter 10 Notes: Mechanical Properties of Fluids

Chapter Introduction

Unlike solids, fluids (liquids and gases) do not have a fixed shape — they flow and adapt to the container. Despite this, fluids have well-defined mechanical properties: they exert pressure, exhibit viscosity, and possess surface tension. Chapter 10 of NCERT Physics for Class 11 covers these properties with powerful results like Pascal's law, Bernoulli's theorem, Stokes' law, and the phenomenon of capillarity. These concepts explain everything from hydraulic brakes to aircraft lift, from the flow of blood to the rise of sap in trees.

Key Definitions

Detailed Concepts

Pressure and Pascal's Law

Pressure at a depth h in a fluid of density ρ: P = ρgh + P₀ (where P₀ is atmospheric pressure).

Pascal's Law: A change in pressure applied to an enclosed incompressible fluid is transmitted equally to every element of the fluid and to the walls of the container.

Applications of Pascal's Law: 1. Hydraulic lift: A small force applied on a small piston creates a much larger force on a large piston: F₂/F₁ = A₂/A₁ 2. Hydraulic brakes: Small force on the brake pedal is amplified to apply large force on brake pads 3. Hydraulic press: Used in industry for pressing, forging, and lifting heavy loads

Streamline Flow and Equation of Continuity

Streamline flow (laminar flow): Each particle follows the same path as the previous particle. Streamlines never cross each other.

Equation of continuity (conservation of mass for incompressible fluids):

A₁v₁ = A₂v₂

where A is the cross-sectional area and v is the flow velocity. This means: where the pipe narrows, velocity increases, and vice versa.

Bernoulli's Theorem and Applications

For an ideal fluid (incompressible, non-viscous, irrotational) in steady flow:

P + ½ρv² + ρgh = constant

This is the conservation of energy per unit volume: pressure energy + kinetic energy + potential energy remains constant along a streamline.

Viscosity

Viscosity is the internal friction in a fluid that opposes relative motion between its layers.

Newton's law of viscosity: The viscous force F between two parallel layers of area A separated by a velocity gradient dv/dx is:

F = –η A (dv/dx)

where η is the coefficient of viscosity. The negative sign indicates that viscous force opposes relative motion.

Poiseuille's formula for the volume flow rate of a liquid through a capillary tube of radius r and length l under pressure difference ΔP:

V/t = πr⁴ΔP / 8ηl

Note the r⁴ dependence — a small change in radius greatly affects flow rate (relevant in blood vessels).

Stokes' Law: When a small sphere of radius r moves with velocity v through a viscous medium of viscosity η, the viscous drag force is:

F = 6πηrv

Terminal velocity: When a sphere falls through a viscous medium, it reaches a constant velocity where weight = buoyant force + viscous force:

v_t = 2r²(ρ – σ)g / 9η

where ρ is the density of the sphere and σ is the density of the medium.

Reynolds Number

Re = ρvD / η

where ρ = fluid density, v = flow velocity, D = characteristic length (pipe diameter), η = coefficient of viscosity.

Surface Tension

Surface tension arises because molecules at the surface experience a net inward pull (unlike bulk molecules which are pulled equally in all directions).

Surface tension T = Force/Length = Work/Area (surface energy)

Key phenomena explained by surface tension: 1. Spherical shape of drops and bubbles: Surface tension minimises surface area for a given volume; a sphere has the smallest surface-to-volume ratio. 2. Excess pressure inside a drop: ΔP = 2T/r (for a liquid drop); ΔP = 4T/r (for a soap bubble, which has two surfaces) 3. Angle of contact: The angle between the tangent to the liquid surface and the solid surface inside the liquid. Concave meniscus (θ < 90°) when adhesive forces > cohesive forces; convex meniscus (θ > 90°) when cohesive forces > adhesive forces.

Capillarity and Jurin's Law

The rise or fall of a liquid in a capillary tube of radius r:

h = 2T cos θ / rρg (Jurin's law)

• θ < 90° (water in glass): cos θ is positive → liquid rises (concave meniscus)
• θ > 90° (mercury in glass): cos θ is negative → liquid falls (convex meniscus)
• θ = 90°: no capillary rise or fall

Practical examples of capillarity:

Important Derivations

1. Bernoulli's Theorem

Consider a fluid flowing through a pipe of varying cross-section and height. Applying the work-energy theorem to a small element of fluid:

Work done by pressure forces = Change in KE + Change in PE

P₁A₁Δx₁ – P₂A₂Δx₂ = ½m(v₂² – v₁²) + mg(h₂ – h₁)

Since m = ρA₁Δx₁ = ρA₂Δx₂ (by continuity), dividing throughout by volume V:

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

2. Terminal Velocity (Stokes' Law)

For a sphere of radius r and density ρ falling in a medium of density σ:

At terminal velocity: Weight = Buoyant force + Viscous drag

(4/3)πr³ρg = (4/3)πr³σg + 6πηrv_t

v_t = [(4/3)πr³(ρ – σ)g] / 6πηr = 2r²(ρ – σ)g / 9η

3. Excess Pressure Inside a Liquid Drop

Consider half of a spherical drop of radius r. Force due to excess pressure ΔP acts on the cross-section πr². Surface tension T acts along the circumference 2πr.

Force balance: ΔP × πr² = T × 2πr → ΔP = 2T/r

For a soap bubble (two surfaces): ΔP = 4T/r

Common Mistakes

1. Confusing coefficient of viscosity units: The CGS unit is poise (P); the SI unit is Pa·s (poiseuille). 1 Pa·s = 10 P. Always use SI units in calculations. 2. Forgetting the two surfaces of a soap bubble: A soap bubble has two liquid-air interfaces (inner and outer), so excess pressure is 4T/r, not 2T/r. 3. Applying Bernoulli's equation to viscous flows: Bernoulli's theorem assumes an ideal (non-viscous, incompressible) fluid. Real fluids with significant viscosity require corrections. 4. Ignoring atmospheric pressure in hydrostatic problems: Total pressure at depth h is P₀ + ρgh, not just ρgh. 5. Wrong sign in Jurin's law: When θ > 90°, cos θ is negative, giving capillary depression (e.g., mercury in glass). Do not take the absolute value blindly. 6. Mixing up terminal velocity formula: The density that appears in the numerator is (ρ – σ), the difference between the body's density and the medium's density. Using ρ alone is incorrect.

Numerical Practice

Example 1: Hydraulic Lift

A hydraulic lift has pistons of area 100 cm² and 5000 cm². What force on the small piston is needed to lift a 1000 kg car?

F₁/A₁ = F₂/A₂ → F₁ = F₂ × A₁/A₂ = (1000 × 9.8) × 100/5000 = 196 N

Example 2: Terminal Velocity

A raindrop of radius 0.3 mm falls through air. If η_air = 1.8 × 10⁻⁵ Pa·s, ρ_water = 1000 kg/m³, ρ_air ≈ 1.2 kg/m³, find the terminal velocity.

v_t = 2r²(ρ – σ)g / 9η = 2 × (0.3 × 10⁻³)² × (1000 – 1.2) × 9.8 / (9 × 1.8 × 10⁻⁵)

v_t = 2 × 9 × 10⁻⁸ × 998.8 × 9.8 / (1.62 × 10⁻⁴) ≈ 10.9 m/s

Example 3: Capillary Rise

A capillary tube of radius 0.5 mm is dipped in water (T = 0.075 N/m, θ ≈ 0°). Find the rise. (ρ = 1000 kg/m³, g = 9.8 m/s²)

h = 2T cos θ / rρg = 2 × 0.075 × 1 / (0.5 × 10⁻³ × 1000 × 9.8) = 0.15 / 4.9 ≈ 0.031 m = 3.1 cm

Exam Tips

1. Pascal's law and hydraulic lift: A very common 2-mark or 3-mark question. Know the ratio F₂/F₁ = A₂/A₁. 2. Bernoulli's theorem derivation: A standard 3-mark derivation. State assumptions clearly: incompressible, non-viscous, steady, irrotational flow. 3. Bernoulli's applications: Be ready to explain any of the five applications listed above in 2–3 sentences with the underlying principle. 4. Stokes' law and terminal velocity: Derivation of v_t is a frequent 3-mark question. Show the force balance clearly. 5. Surface tension and capillarity: Jurin's law derivation is important. Draw the meniscus diagram (concave/convex) and label the angle of contact. 6. Reynolds number: Usually a 1-mark factual question. Know the threshold values (1000, 2000). 7. Always draw diagrams: Flow through a venturimeter, forces on a falling sphere, meniscus in a capillary — diagrams carry marks.

NCERT Exercise Focus Areas

The following table maps the most frequently tested NCERT exercise question types for this chapter, along with their typical mark allocation and difficulty level in CBSE board exams.

Tip: NCERT exercises in this chapter lean heavily on Bernoulli's theorem applications and terminal velocity. Practise the derivations first, then work through the numericals — most follow a standard template.

Conceptual Questions for Practice

1. Why does a spinning ball curve in flight? Explain with reference to the Magnus effect and Bernoulli's principle. 2. Viscosity of gases is lower than that of liquids at ordinary temperatures. Why? 3. Two soap bubbles of different radii are in contact with each other through a narrow tube. In which direction does air flow — from the smaller bubble to the larger one, or the other way? Justify. 4. A small hole is made at the bottom of a vessel filled with water. How does the speed of efflux depend on the height of the water column? Name the theorem you used. 5. Why does the liquid rise in a capillary tube when the angle of contact is acute? What happens if the tube is tilted?

These questions require no numerical computation — they test whether you truly understand the physical reasoning behind the formulas. Try answering them in your own words before checking a reference.

Quick Revision Notes

Bernoulli applications: Venturimeter, atomiser, aircraft lift, Bunsen burner, Magnus effect

Capillarity: Rise when θ < 90° (water-glass), fall when θ > 90° (mercury-glass)