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Class 11 Physics · NCERT Chapter 9

Mechanical Properties of Fluids Class 11 Physics Notes

Complete, exam-ready notes on the mechanical properties of fluids: pressure and its variation with depth, Pascal's law and hydraulic machines, Archimedes' principle, streamline and turbulent flow, the equation of continuity, Bernoulli's principle and its applications, viscosity and Stokes' law, and surface tension and capillary rise — every NCERT topic with MCQs, mark-wise questions and solved numericals for CBSE, JEE and NEET.

ChapterMechanical Properties of FluidsClassClass 11SubjectPhysicsBoardCBSEExamsCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

State Bernoulli's principle and write its equation.

For an incompressible, non-viscous fluid in steady flow, the total mechanical energy per unit volume is constant along a streamline: P + ½ρv² + ρgh = constant. Where the speed of the fluid is higher, the pressure is lower.

Pressure and Its Variation with Depth

Pressure

Pressure is the normal force acting per unit area: P = F/a. Its SI unit is the pascal (1 Pa = 1 N m⁻²). In a fluid at rest, pressure is the same in all directions about any point and acts normally to any surface.

P=P0+ρghP = P_0 + \rho g h
Pressure at a depth h below the free surface

The pressure at any point in a fluid at rest equals the atmospheric pressure P₀ plus ρgh, where ρ is the fluid density, g the gravitational acceleration and h the depth below the free surface. Hence pressure increases uniformly with depth and is the same in all directions at a given depth.

  • The pressure does not depend on the shape of the container — only on depth.
  • Two vessels of different shapes connected at the bottom reach the same liquid level.
  • Gauge pressure = P − P_atm = ρgh; the manometer and barometer work on this principle.

Why dams are thicker at the bottom

Because pressure increases with depth, the walls of a dam experience the greatest force at the bottom and are therefore built thicker there.

MCQ Questions

1The SI unit of pressure is —

2Pressure at a depth h in a liquid of density ρ is —

3The pressure at a point inside a liquid at rest —

4Gauge pressure is —

1 Mark Questions

  1. Define pressure. Write its SI unit.
  2. Write the expression for pressure at a depth h in a liquid.
  3. Define gauge pressure.

2 Mark Questions

  1. Find the pressure at a depth of 10 m below the surface of water. (ρ = 1000 kg m⁻³, g = 10 m/s², atmospheric pressure = 10⁵ Pa.)
  2. Why is a dam wall thicker at the bottom than at the top?
  3. Distinguish between absolute pressure and gauge pressure.

3 Mark Questions

  1. Derive an expression for the pressure at a depth h inside a liquid at rest, taking atmospheric pressure into account.
  2. A rectangular tank is filled with water up to a height of 2 m. Find the total force on a wall of dimensions 2 m × 3 m due to the water. (ρ = 1000 kg m⁻³, g = 10 m/s².)

Pascal's Law and Hydraulic Machines

Pascal's law

For a fluid at rest, the pressure at any point within the fluid and at the walls of its container is transmitted undiminished in all directions. A change in pressure applied to an enclosed fluid is transmitted equally to every part of the fluid and the walls of its container.

F2=F1 A2A1,F1A1=F2A2F_2 = F_1\,\frac{A_2}{A_1}, \qquad \frac{F_1}{A_1} = \frac{F_2}{A_2}
Hydraulic amplification by Pascal's law

The hydraulic lift, hydraulic press and hydraulic brakes all exploit Pascal's law. A small force F₁ applied on a small piston of area A₁ produces the same pressure throughout, giving a much larger force F₂ on a large piston of area A₂, since F₂ = F₁(A₂/A₁). The mechanical advantage is the ratio of the areas.

  • Hydraulic brakes use Pascal's law to transmit the driver's pedal force to all four wheels.
  • A small input force can lift a heavy car on a hydraulic lift.
  • The principle works because liquids are nearly incompressible.

Big area, big force

The larger the ratio of the piston areas A₂/A₁, the greater the force amplification. Work is conserved — the small piston moves a larger distance while the large piston moves a smaller one.

MCQ Questions

1Pascal's law is valid for —

2A hydraulic lift multiplies force in proportion to the —

3Hydraulic machines work because liquids are —

4If the area of the output piston is five times that of the input piston, a force of 50 N produces an output force of —

1 Mark Questions

  1. State Pascal's law.
  2. Name two machines that work on Pascal's law.
  3. Write the formula for the force amplification in a hydraulic press.

2 Mark Questions

  1. A hydraulic lift has pistons of areas 10 cm² and 200 cm². What force is needed on the small piston to lift a car of weight 2000 N?
  2. Explain the working of hydraulic brakes using Pascal's law.
  3. State Pascal's law and explain why it is the basis of the hydraulic press.

3 Mark Questions

  1. State and explain Pascal's law, and derive the force amplification in a hydraulic lift.
  2. The small piston of a hydraulic press has a diameter of 2 cm and the large piston a diameter of 20 cm. If a force of 10 N is applied to the small piston, find the force on the large piston.

Archimedes' Principle and Floatation

Archimedes' principle

A body wholly or partially immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by it. The buoyant force acts through the centre of buoyancy and depends only on the volume of fluid displaced.

FB=ρfluidVsubg=(weight of displaced fluid)F_B = \rho_{\text{fluid}} V_{\text{sub}} g = \text{(weight of displaced fluid)}
Buoyant force

For floatation, the weight of the floating body equals the weight of the fluid displaced. A body sinks if its density exceeds that of the fluid and floats if its density is lower. This is why a steel ship floats — its average density (including the air inside) is less than water.

  • The fraction of a floating body submerged equals the ratio of its density to the fluid's density.
  • A hydrometer measures the density of a liquid by the depth it sinks.
  • An iceberg floats with about 90% of its volume submerged because ice is slightly less dense than water.

The ship that floats

A solid block of steel sinks because its density exceeds water's. But shaped as a hollow hull, its average density is less than water, so the buoyant force matches its weight and it floats.

MCQ Questions

1The buoyant force on a body immersed in a fluid equals —

2A body floats if its average density is —

3The fraction of a floating body submerged is —

4An iceberg floats because —

1 Mark Questions

  1. State Archimedes' principle.
  2. What is the condition for a body to float?
  3. State the condition for a body to sink in a liquid.

2 Mark Questions

  1. A body of volume 2 × 10⁻³ m³ is fully immersed in water. Find the buoyant force on it. (ρ = 1000 kg m⁻³, g = 10 m/s².)
  2. Explain why a solid steel ball sinks in water but a steel ship floats.
  3. A body floats with one quarter of its volume above water. Find the ratio of its density to that of water.

3 Mark Questions

  1. State Archimedes' principle and use it to show the condition for floatation.
  2. A cube of side 5 cm floats in water with 60% of its volume submerged. Find the density of the cube. (ρ_water = 1000 kg m⁻³.)

Equation of Continuity and Bernoulli's Principle

Equation of continuity

For an incompressible fluid in steady flow, the mass flowing past any cross-section per unit time is constant. Since density is constant, A₁v₁ = A₂v₂ — the product of area and speed is constant. Where the tube narrows, the fluid speeds up.

A1v1=A2v2,Q=Av=constantA_1 v_1 = A_2 v_2, \qquad Q = Av = \text{constant}
Equation of continuity (volume flow rate)

Bernoulli's principle

For an incompressible, non-viscous fluid flowing steadily, the sum of the pressure energy, kinetic energy per unit volume and potential energy per unit volume is constant along a streamline: P + ½ρv² + ρgh = constant. It is a statement of energy conservation for fluid flow.

P+12ρv2+ρgh=constantP + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}
Bernoulli's equation

The most important consequence is that where the fluid speed is higher, the pressure is lower. This explains the lift on an aeroplane wing (faster air over the top, lower pressure), the working of a venturi meter, the atomiser and blow, and the curved flight of a spinning ball (Magnus effect).

Assumptions of Bernoulli's equation

Bernoulli's equation holds only for steady, incompressible, non-viscous (frictionless) flow. Real viscous fluids dissipate energy, so the equation is an idealisation.

MCQ Questions

1For an incompressible fluid in steady flow, A₁v₁ = A₂v₂ is the —

2Where the cross-section of a pipe narrows, the speed of an incompressible fluid —

3Bernoulli's principle states that where fluid speed is higher, the pressure is —

4The lift on an aeroplane wing is explained by —

1 Mark Questions

  1. State the equation of continuity for an incompressible fluid.
  2. Write Bernoulli's equation for a horizontal pipe.
  3. State Bernoulli's principle.

2 Mark Questions

  1. Water flows in a pipe of area 40 cm² at 2 m/s. Find the speed in a section of the pipe of area 10 cm².
  2. State and explain Bernoulli's principle, and give one application.
  3. Derive the equation of continuity for an incompressible fluid.

3 Mark Questions

  1. Derive Bernoulli's equation for the steady flow of an ideal incompressible fluid and state the assumptions made.
  2. Water flows through a horizontal pipe of cross-section 30 cm² where its speed is 2 m/s and pressure 2 × 10⁵ Pa. In a narrower section of area 10 cm² the speed rises; find the pressure there. (ρ = 1000 kg m⁻³.)

Viscosity and Stokes' Law

Viscosity

Viscosity is the property of a fluid by which it resists relative motion between its layers — it is the internal friction of a fluid. The viscous force between two layers separated by distance dy and moving with velocity difference dv is F = ηA(dv/dy), where η is the coefficient of viscosity.

F=ηAdvdyF = \eta A \frac{dv}{dy}
Viscous force (Newton's law of viscosity)

The SI unit of viscosity is the poiseuille (Pa s); the cgs unit is the poise. Viscosity of liquids decreases with temperature, while that of gases increases. Honey is far more viscous than water.

Stokes' law

When a sphere of radius r falls through a viscous fluid with terminal velocity vₜ, the viscous drag opposing its motion is F = 6πηrvₜ. The terminal velocity is vₜ = (2/9)(r²/η)(ρ_sphere − ρ_fluid)g.

F=6πηrv,vt=29r2η(ρ−σ)gF = 6\pi \eta r v, \qquad v_t = \frac{2}{9}\frac{r^2}{\eta}(\rho - \sigma)g
Stokes' law and terminal velocity

Why does a parachute fall steadily?

As a body falls through a fluid, the viscous drag grows with speed until it balances the weight. From then on the body falls at a constant terminal velocity — no further acceleration.

MCQ Questions

1The property of a fluid that resists relative motion between its layers is called —

2The SI unit of coefficient of viscosity is —

3Stokes' law for a falling sphere gives the viscous force as —

4Terminal velocity is the constant speed at which —

1 Mark Questions

  1. Define viscosity.
  2. State Stokes' law.
  3. Define terminal velocity.

2 Mark Questions

  1. A sphere of radius 1 mm falls through water of viscosity 0.001 Pa s at a terminal velocity of 2 cm/s. Find the viscous force on it.
  2. Why does the viscosity of liquids decrease with temperature but that of gases increase?
  3. Explain how a parachute achieves a constant terminal velocity.

3 Mark Questions

  1. State Stokes' law and derive an expression for the terminal velocity of a sphere falling through a viscous fluid.
  2. A rain drop of radius 0.1 mm falls through air with viscosity 1.8 × 10⁻⁵ Pa s. Estimate its terminal velocity. (ρ_water = 1000 kg m⁻³, ρ_air ≈ 0, g = 10 m/s².)

Surface Tension and Capillarity

Surface tension

Surface tension is the property of a liquid surface by which it behaves like a stretched elastic membrane, tending to minimise its area. It is the force per unit length acting along the free surface: S = F/L. Its SI unit is N m⁻¹.

S=FL,h=2Scos⁡θρgrS = \frac{F}{L}, \qquad h = \frac{2S\cos\theta}{\rho g r}
Surface tension and capillary rise

Surface tension arises because molecules at the surface have fewer neighbours and hence higher potential energy — the liquid pulls them inward. It explains why a needle floats on water, why drops are spherical, and the rise (or fall) of a liquid in a narrow capillary tube: h = 2S cosθ/(ρgr). Water rises in a glass tube because it wets it (θ < 90°).

  • Small liquid drops and soap bubbles are spherical because a sphere has minimum surface area.
  • Surface tension decreases with temperature.
  • Capillary rise is greater in narrower tubes.

Narrower tube, higher rise

The height of liquid rise in a capillary tube is inversely proportional to the radius (h ∝ 1/r). A thinner tube pulls the liquid higher — think of a wick drawing up oil.

MCQ Questions

1Surface tension is the —

2The SI unit of surface tension is —

3A small liquid drop is spherical because —

4In a capillary tube, the height of liquid rise is —

1 Mark Questions

  1. Define surface tension. Write its SI unit.
  2. Write the expression for the capillary rise of a liquid.
  3. Why are small liquid drops spherical?

2 Mark Questions

  1. Water rises to 4 cm in a capillary tube of radius 0.5 mm. Find the surface tension of water. (ρ = 1000 kg m⁻³, cosθ ≈ 1, g = 10 m/s².)
  2. Explain why a needle can float on the surface of water although steel is denser than water.
  3. State the factors on which the capillary rise of a liquid depends.

3 Mark Questions

  1. Explain the phenomenon of surface tension and derive the ascent of a liquid in a capillary tube.
  2. A liquid of surface tension 0.072 N m⁻¹ rises to 2 cm in a capillary tube of radius 0.36 mm. Find the density of the liquid. (cosθ = 1, g = 10 m/s².)

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Pressure with depth

P=P0+ρghP = P_0 + \rho g h

Pascal's law

F1A1=F2A2\frac{F_1}{A_1} = \frac{F_2}{A_2}

Buoyant force

FB=ρfluidVsubgF_B = \rho_{\text{fluid}} V_{\text{sub}} g

Equation of continuity

A1v1=A2v2A_1 v_1 = A_2 v_2

Bernoulli's equation

P+12ρv2+ρgh=constantP + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}

Viscous force

F=ηAdvdyF = \eta A \frac{dv}{dy}

Stokes' law

F=6πηrvF = 6\pi \eta r v

Terminal velocity

vt=29r2η(ρ−σ)gv_t = \frac{2}{9}\frac{r^2}{\eta}(\rho-\sigma)g

Surface tension

S=FLS = \frac{F}{L}

Capillary rise

h=2Scos⁡θρgrh = \frac{2S\cos\theta}{\rho g r}

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • Pressure increases with depth: P = P₀ + ρgh.
  • Pascal's law: pressure transmitted undiminished in a fluid at rest.
  • Buoyant force = weight of displaced fluid; float if average density less than the fluid.
  • Continuity: A₁v₁ = A₂v₂.
  • Bernoulli: P + ½ρv² + ρgh = constant; higher speed → lower pressure.
  • Stokes' law F = 6πηrv; terminal velocity ∝ r².
  • Surface tension S = F/L; capillary rise h = 2S cosθ/(ρgr).

Solved problems

Worked examples

JEE / NEET-style numericals, solved step by step.

1

Pressure at depth

Find the pressure at a depth of 20 m in a lake. (ρ = 1000 kg m⁻³, g = 10 m/s², atmospheric pressure = 1.01 × 10⁵ Pa.)

  1. P = P₀ + ρgh = 1.01 × 10⁵ + 1000 × 10 × 20.
  2. P = 1.01 × 10⁵ + 2 × 10⁵ = 3.01 × 10⁵ Pa.

Answer

3.01 × 10⁵ Pa

2

Hydraulic lift force

A hydraulic lift has pistons of areas 25 cm² and 500 cm². What force on the small piston will lift a car of weight 10000 N?

  1. Pascal's law: F₁/A₁ = F₂/A₂.
  2. F₁ = F₂ × A₁/A₂ = 10000 × 25/500 = 10000 × 0.05 = 500 N.

Answer

500 N

3

Bernoulli pressure in a narrow pipe

Water flows in a horizontal pipe of area 20 cm² at 2 m/s with pressure 2 × 10⁵ Pa. In a section of area 10 cm², find the pressure. (ρ = 1000 kg m⁻³.)

  1. Continuity: A₁v₁ = A₂v₂ → v₂ = v₁ × A₁/A₂ = 2 × 20/10 = 4 m/s.
  2. Bernoulli (horizontal, h same): P₁ + ½ρv₁² = P₂ + ½ρv₂².
  3. P₂ = P₁ + ½ρ(v₁² − v₂²) = 2×10⁵ + ½×1000×(4 − 16) = 2×10⁵ − 6000 = 1.94×10⁵ Pa.

Answer

1.94 × 10⁵ Pa

FAQ

Common questions

Why does pressure in a liquid increase with depth?

At a greater depth the liquid above exerts a larger weight, so the pressure rises by ρgh. This is why dams are built thicker at the bottom and divers feel increased pressure as they go deeper.

How does a hydraulic lift lift a heavy car with a small force?

By Pascal's law, the pressure applied on a small piston is transmitted undiminished to a much larger piston. Since pressure = force/area, the larger area produces a larger force: F₂ = F₁(A₂/A₁).

Why can a steel ship float even though steel is denser than water?

A solid block of steel sinks, but a ship's hollow hull has a large volume and low average density (including the air inside). Its weight equals the buoyant force (weight of displaced water), so it floats.

What is the physical meaning of the equation of continuity?

It expresses conservation of mass in steady flow: the same mass must pass every cross-section per second. For an incompressible fluid this becomes A₁v₁ = A₂v₂ — narrower pipes force the fluid to flow faster.

Why does a liquid climb higher in a thinner capillary tube?

Capillary rise is h = 2S cosθ/(ρgr). Since the rise is inversely proportional to the radius r, a narrower tube holds the liquid column higher — the effect behind wicks drawing up oil and water soaking into paper.

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