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Chapter 9: Mechanical Properties of Fluids

Grade 11 Science | Chapter 9

Mechanical Properties of Fluids

Fluids press, push up and flow. This chapter develops pressure and Pascal’s law, Archimedes’ principle, the equation of continuity and Bernoulli’s principle.

Core Concepts

Key Principles

Worked Examples

Practice Sets

Contents

Introduction: What Is a Fluid

Pressure in a Fluid

Pascal’s Law

Archimedes’ Principle

Flow and the Equation of Continuity

Bernoulli’s Principle

Key Reasoning (Principles)

Worked Examples (10)

Practice Sets A to D

10.

Summary and Exam Quick-Check

1. Introduction: What Is a Fluid

A fluid is a substance that flows, that is, a liquid or a gas. Fluids cannot sustain a shearing force at rest, so they take the shape of their container. This chapter studies fluids at rest, where pressure is central, and fluids in motion, where the conservation of mass and energy lead to the equation of continuity and Bernoulli’s principle.

Core idea

Pressure in a fluid grows with depth as P = hρg, is transmitted equally in all directions (Pascal), and for moving fluids speed and pressure trade off (Bernoulli).

2. Pressure in a Fluid

Pressure is the force acting per unit area, P = F ÷ A, measured in pascals. In a fluid at rest the pressure at a depth h is P = h ρ g, where ρ is the density, so pressure increases with depth and is the same at all points on one level. This is why a dam is built thicker at its base.

Diagram 1 – Pressure Increases with Depth

A container of fluid showing pressure increasing with depth

Fig 1. In a fluid at rest the pressure grows with depth according to P equals h rho g, so it is greatest at the bottom.

3. Pascal’s Law

Pascal’s law states that a pressure applied to an enclosed fluid is transmitted equally and undiminished to every part of the fluid and the walls of its container. This is the basis of the hydraulic press and car brakes, where a small force on a small piston produces a large force on a large piston, since the pressure is the same throughout.

4. Archimedes’ Principle

Archimedes’ principle states that a body wholly or partly immersed in a fluid experiences an upward buoyant force equal to the weight of the fluid it displaces. An object floats when this buoyant force balances its weight, and sinks when its weight is greater. This explains why a heavy steel ship floats: it displaces a large weight of water.

5. Flow and the Equation of Continuity

For a fluid flowing steadily through a pipe, the mass entering equals the mass leaving. For an incompressible fluid this gives the equation of continuity: A₁ v₁ = A₂ v₂, the product of cross-sectional area and speed is constant. So a fluid speeds up where the pipe narrows, which is why water from a hose moves faster when the opening is squeezed.

Diagram 2 – Flow Through a Narrowing Pipe

Fluid flowing through a pipe that narrows, moving slowly in the wide part and fast in the narrow part

Fig 2. By continuity the fluid speeds up where the pipe narrows; by Bernoulli’s principle its pressure falls there.

6. Bernoulli’s Principle

Bernoulli’s principle follows from the conservation of energy in a flowing fluid: where the speed is high the pressure is low, and where the speed is low the pressure is high. It explains the lift on an aeroplane wing and the spin of a ball, and it pairs with continuity to describe flow through pipes of changing width.

7. Key Reasoning (Principles)

Principle 1: Pressure grows with depth

In a fluid at rest, P = hρg, so pressure depends only on depth, density and g, and is equal at all points on the same horizontal level.

Principle 2: Pressure is transmitted equally (Pascal)

An applied pressure reaches every part of an enclosed fluid undiminished, which lets a hydraulic press multiply force.

Principle 3: Speed and pressure trade off (Bernoulli)

In a moving fluid, faster flow means lower pressure. Combined with continuity, this explains lift and flow through narrowing pipes.

8. Worked Examples

Example 1

Q: Find the pressure at 5 m depth in water (ρ = 1000 kg/m³, g = 10 m/s²).

▶ Show Solution

P = h ρ g = 5 × 1000 × 10.

= 50000 Pa.

Answer: 50000 Pa.

Example 2

Q: Find the force per unit area when 100 N acts on 0.02 m².

▶ Show Solution

P = F ÷ A = 100 ÷ 0.02.

= 5000 Pa.

Answer: 5000 Pa.

Example 3

Q: Find the buoyant force on a body that displaces 0.002 m³ of water (ρ = 1000, g = 10).

▶ Show Solution

Buoyant force = weight of displaced water = 0.002 × 1000 × 10.

= 20 N.

Answer: 20 N.

Example 4

Q: Water flows at 2 m/s in a pipe of area 0.01 m². Find the speed in a 0.004 m² section.

▶ Show Solution

A1 v1 = A2 v2: 0.01 × 2 = 0.004 × v2.

v2 = 5 m/s.

Answer: 5 m/s.

Example 5

Q: In a hydraulic press, 50 N acts on 0.01 m². Find the pressure transmitted.

▶ Show Solution

P = 50 ÷ 0.01.

= 5000 Pa.

Answer: 5000 Pa.

Example 6

Q: Find the pressure at 10 m depth (ρ = 1000, g = 10).

▶ Show Solution

P = 10 × 1000 × 10.

= 100000 Pa.

Answer: 100000 Pa.

Example 7

Q: Why does a fluid speed up in a narrow pipe?

▶ Show Solution

By continuity A v is constant, so a smaller area means a larger speed.

Answer: Because A v is constant (continuity).

Example 8

Q: A body weighs 50 N in air and 30 N in water. Find the buoyant force.

▶ Show Solution

Buoyant force = loss of weight = 50 minus 30.

= 20 N.

Answer: 20 N.

Example 9

Q: State where the pressure is lowest in a pipe carrying flowing water.

▶ Show Solution

By Bernoulli, pressure is lowest where the speed is highest, in the narrow part.

Answer: In the narrowest, fastest part.

Example 10

Q: Find the speed in a 0.002 m² pipe if water enters a 0.008 m² pipe at 1 m/s.

▶ Show Solution

0.008 × 1 = 0.002 × v2.

v2 = 4 m/s.

Answer: 4 m/s.

9. Practice Sets A to D

Set A – Multiple Choice (Basic)

1. Pressure at depth h in a fluid is: (a) hρ (b) hρg (c) ρg (d) hg

2. Pascal’s law concerns pressure that is: (a) lost with depth (b) transmitted equally (c) zero (d) one-directional

3. Archimedes’ principle gives the: (a) weight (b) buoyant force (c) pressure (d) density

4. The equation of continuity is: (a) A1 = A2 (b) v1 = v2 (c) A1 v1 = A2 v2 (d) P1 = P2

5. By Bernoulli, where speed is high the pressure is: (a) high (b) low (c) zero (d) unchanged

▶ Reveal Answers

1. (b) hρg.

2. (b) transmitted equally.

3. (b) buoyant force.

4. (c) A1 v1 = A2 v2.

5. (b) low.

Set B – Short Answer (Understanding)

1. Write the formula for pressure at a depth in a fluid.

2. State Pascal’s law.

3. State Archimedes’ principle.

4. State the equation of continuity.

5. State Bernoulli’s principle in words.

▶ Reveal Answers

1. P = h ρ g.

2. Pressure applied to an enclosed fluid is transmitted equally and undiminished throughout.

3. An immersed body feels an upward force equal to the weight of fluid displaced.

4. For an incompressible fluid, A1 v1 = A2 v2; area times speed is constant.

5. Where a fluid moves faster its pressure is lower, and the reverse.

Set C – Application and Reasoning

1. Find the pressure at 8 m depth (ρ = 1000, g = 10).

2. Find the buoyant force when 0.003 m³ of water is displaced (ρ = 1000, g = 10).

3. Water at 3 m/s in 0.012 m² enters a 0.004 m² pipe. Find the new speed.

4. Why is a dam thicker at the bottom?

5. Explain how a hydraulic press multiplies force.

▶ Reveal Answers

1. P = 8 × 1000 × 10 = 80000 Pa.

2. Buoyant force = 0.003 × 1000 × 10 = 30 N.

3. 0.012 × 3 = 0.004 × v2, so v2 = 9 m/s.

4. Because pressure increases with depth, so the wall must be stronger lower down.

5. Pressure is the same throughout, so a small force on a small area gives a large force on a large area.

Set D – Higher Order (Challenge)

1. A hydraulic press has pistons of area 0.01 m² and 0.05 m². A 100 N force on the small piston gives what force on the large one?

2. Explain, using Bernoulli, how an aeroplane wing produces lift.

3. A ship floats by displacing water. Explain why it can be made of steel.

4. Water enters a pipe at 2 m/s and area 0.02 m² and leaves at 0.005 m². Find the exit speed.

5. Why does the pressure fall in the narrow part of a pipe?

▶ Reveal Answers

1. Pressure = 100 ÷ 0.01 = 10000 Pa; force on large piston = 10000 × 0.05 = 500 N.

2. The wing makes air move faster over the top, lowering the pressure there; the higher pressure below gives an upward lift.

3. Although steel is dense, the ship’s shape displaces a large volume of water whose weight equals the ship’s, so it floats.

4. 0.02 × 2 = 0.005 × v2, so v2 = 8 m/s.

5. Because the fluid moves faster there, and by Bernoulli faster flow means lower pressure.

Chapter Summary

Fluid

A liquid or gas that flows and cannot sustain shear at rest.

Pressure

P = F ÷ A; in a fluid at rest P = hρg, growing with depth.

Pascal’s Law

Applied pressure is transmitted equally throughout an enclosed fluid.

Archimedes

Buoyant force equals the weight of fluid displaced.

Continuity

A1 v1 = A2 v2 for an incompressible fluid.

Bernoulli

Faster flow means lower pressure.

Quantity

Unit

Symbol

Pressure

pascal

Pressure with depth

hρg

Continuity

A1 v1 = A2 v2

–

8-Point Exam Quick-Check

Pressure = force divided by area, in pascals.

In a fluid at rest, P = h rho g, increasing with depth.

Pascal: applied pressure is transmitted equally throughout an enclosed fluid.

Archimedes: buoyant force = weight of fluid displaced.

Continuity: A1 v1 = A2 v2 for an incompressible fluid.

A fluid speeds up where the pipe narrows.

Bernoulli: faster flow has lower pressure.

Hydraulic presses use Pascal’s law to multiply force.

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Class 11 Physics Chapter 9: Mechanical Properties of Fluids, Complete Notes and Practice

This revision guide follows the current NCERT Class 11 Physics syllabus and develops fluids at rest and in motion, covering pressure and its increase with depth, Pascal’s law and the hydraulic press, Archimedes’ principle and buoyancy, the equation of continuity, and Bernoulli’s principle, with two diagrams, ten worked examples and graded practice. Visit SchoolRevise.com to revise, practise and excel.