NEET Physics · NCERT Class 11

Mechanical Properties of Fluids notes

Pressure & Pascal's Law, Fluid Flow, Bernoulli's Principle, Viscosity & Stokes' Law and Surface Tension & Capillarity — explained topic by topic with key points, formulas, memory tricks and common mistakes.

  • 6 topics
  • 55 min read
  • 28 formulas
  • 23 PYQs
1

Chapter Overview

7 min

Mechanical Properties of Fluids studies liquids and gases at rest and in motion. Fluids cannot sustain shear stress at rest, so they flow and exert pressure normally on surfaces. The chapter begins with pressure in fluids, hydrostatic pressure and Pascal's law, which explains hydraulic lift and hydraulic press. Moving fluids are described using streamline flow, turbulent flow, continuity equation and Bernoulli's principle. Real fluids show viscosity, leading to drag, terminal velocity and Reynolds number. At the liquid surface, molecular forces produce surface tension, surface energy, capillary rise or fall and excess pressure in drops and soap bubbles. For NEET, this chapter is formula-rich, application-based and highly scoring.

Key points7

  • Pressure in a fluid at rest is the same in all directions at the same point.
  • Fluid pressure depends on depth, density and gravity, not on container shape.
  • In streamline flow through a narrow region, speed increases and pressure generally decreases.
  • Bernoulli's principle is valid for ideal, incompressible, non-viscous and steady flow.
  • Terminal velocity occurs when net force on a falling body in fluid becomes zero.
  • Surface tension makes liquid surfaces behave like stretched membranes.
  • NEET frequently asks hydraulic lift, continuity, Bernoulli applications, terminal velocity and capillarity formulas.

Memory tricks2

Continuity: A down means v up: smaller area gives larger speed.

Bernoulli: Fast fluid has low pressure at the same height.

Examples2

Daily Life Example

Hydraulic brakes use Pascal's law, perfume atomizers use Bernoulli's principle, and capillary action helps water rise in plants.

NEET Quick Check

If a pipe area becomes one-third, fluid speed becomes three times for steady incompressible flow.

Common mistakes3

Using Bernoulli Everywhere

Bernoulli's equation in this form applies only to ideal, incompressible, non-viscous, steady flow along a streamline.

Confusing Viscosity and Surface Tension

Viscosity is internal resistance to flow, while surface tension acts at the free surface of a liquid.

Ignoring Atmospheric Pressure

Absolute pressure includes atmospheric pressure, while gauge pressure excludes it.

Reference tables2

Quick revision8

Formulas5

Pressure
P=F/AP = F/A

Pressure is the normal force acting per unit area.

P
Pressure
F
Normal force
A
Area
Hydrostatic Pressure
P=P0+ρghP = P0 + ρgh

Pressure at depth h below the free surface of a liquid open to atmosphere.

P
Pressure at depth
P0
Pressure at free surface
ρ
Density of liquid
g
Acceleration due to gravity
h
Depth
Continuity Equation
A1v1=A2v2A1v1 = A2v2

For steady incompressible flow, volume flow rate remains constant.

A1, A2
Cross-sectional areas
v1, v2
Fluid speeds
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Diagrams3

2

Pressure & Pascal's Law

9 min

Pressure is the normal force acting per unit area. In a fluid at rest, pressure at a point acts equally in all directions and always normal to any surface. Hydrostatic pressure increases with depth because lower layers support the weight of fluid above them. At depth h below an open surface, pressure is P = P0 + ρgh, where P0 is atmospheric pressure. Pascal's law states that pressure applied to an enclosed incompressible fluid is transmitted equally and undiminished throughout the fluid. This explains hydraulic lift and hydraulic press, where a small force on a small piston produces a large force on a larger piston. The machine multiplies force, not energy.

Key points6

  • Pressure is scalar although force is vector.
  • Fluid pressure at rest depends on depth, fluid density and gravity.
  • Pressure does not depend on container shape or base area alone.
  • In hydraulic lift, the larger piston gives larger force because pressure is same but area is larger.
  • The smaller piston moves a greater distance, so ideal work input equals work output.
  • Pressure always acts normal to the surface of an object immersed in a fluid.

Memory tricks2

Hydrostatic Pressure: Deeper means more liquid above, so pressure increases.

Hydraulic Machines: Same pressure plus bigger area gives bigger force.

Examples2

Solved Example: Pressure at Depth

At 4 m depth in water, gauge pressure = ρgh = 1000 × 10 × 4 = 4 × 10^4 Pa.

Solved Example: Hydraulic Lift

If A2 = 25A1 and F1 = 80 N, then F2 = F1(A2/A1) = 80 × 25 = 2000 N.

Common mistakes3

Thinking Hydraulic Lift Creates Energy

A hydraulic lift multiplies force, but the larger piston moves a smaller distance, so energy is conserved ideally.

Using Total Liquid Volume

Pressure at depth depends on height of liquid column, not total volume of liquid.

Confusing Gauge and Absolute Pressure

Gauge pressure is ρgh, while absolute pressure is P0 + ρgh.

Reference tables2

Quick revision8

Formulas5

Pressure
P=F/AP = F/A

Pressure is normal force per unit area.

P
Pressure
F
Normal force
A
Area
Hydrostatic Pressure
P=P0+ρghP = P0 + ρgh

Absolute pressure at depth h in a liquid open to atmosphere.

P0
Atmospheric pressure or pressure at free surface
ρ
Density of liquid
g
Acceleration due to gravity
h
Depth below free surface
Gauge Pressure
Pg=ρghPg = ρgh

Excess pressure over atmospheric pressure at depth h.

Pg
Gauge pressure
ρgh
Pressure due to liquid column
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Diagrams4

3

Fluid Flow

8 min

Fluid flow describes the motion of liquids and gases. In streamline flow, each fluid particle follows a smooth path, and the velocity of fluid at a point remains constant with time. In turbulent flow, the motion is irregular, chaotic and contains eddies. The equation of continuity is based on conservation of mass. For steady incompressible flow, the volume of fluid crossing every cross-section per second is the same, so A1v1 = A2v2. Thus, fluid moves faster through narrow sections and slower through wider sections. The volume flow rate is Q = Av. Ideal fluid assumptions include incompressible, non-viscous, steady and irrotational flow, which are important before applying Bernoulli's principle.

Key points6

  • Continuity equation comes from conservation of mass.
  • For incompressible fluid, density remains constant.
  • If streamlines crossed, a fluid particle at one point would have two velocities, which is impossible.
  • Flow rate has SI unit m³ s^-1.
  • Turbulence is more likely at high speed, large pipe diameter or low viscosity.
  • Continuity is often used together with Bernoulli equation in NEET.

Memory tricks2

Continuity Shortcut: Area × velocity is constant: small area means large velocity.

Streamlines: Streamlines do not cross because one point cannot have two flow directions.

Examples2

Numerical Problem

Water flows through area 8 cm² at 3 m/s and then through area 2 cm². Using A1v1 = A2v2, v2 = 8×3/2 = 12 m/s.

Real-Life Example

When you partially close the mouth of a water hose, water comes out faster because the exit area decreases.

Common mistakes3

Using A1v1 = A2v2 for Compressible Gas Blindly

This simple form assumes incompressible flow with constant density.

Thinking Flow Rate Depends Only on Speed

Flow rate depends on both area and speed: Q = Av.

Confusing Turbulent Flow with Fast Streamline Flow

High speed may cause turbulence, but streamline flow can still be fast under suitable conditions.

Reference tables3

Quick revision8

Formulas3

Volume Flow Rate
Q=AvQ = Av

Volume of fluid crossing a cross-section per unit time.

Q
Volume flow rate
A
Area of cross-section
v
Fluid speed
Equation of Continuity
A1v1=A2v2A1v1 = A2v2

For incompressible steady flow, flow rate remains constant along the pipe.

A1, A2
Cross-sectional areas
v1, v2
Speeds of fluid at those sections
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Diagrams3

4

Bernoulli's Principle

10 min

Bernoulli's principle is the conservation of mechanical energy for ideal fluid flow. For steady, incompressible, non-viscous flow along a streamline, the sum P + 1/2ρv² + ρgh remains constant. These terms represent pressure energy, kinetic energy and potential energy per unit volume. At the same height, when fluid speed increases, pressure decreases. This pressure-velocity relationship explains many applications: a venturimeter measures flow speed using pressure difference, airplane wings produce lift due to pressure difference, and atomizers and carburetors use fast air to reduce pressure and draw liquid into the air stream. NEET frequently asks direct formulas and conceptual applications.

Key points7

  • Bernoulli's equation is applied along a streamline.
  • Pressure energy per unit volume is P.
  • Kinetic energy per unit volume is 1/2ρv².
  • Potential energy per unit volume is ρgh.
  • Continuity equation is often needed before applying Bernoulli.
  • Bernoulli does not account for viscous energy loss in real fluids.
  • A pressure difference can create lift or suction-like effects.

Memory tricks2

Bernoulli Shortcut: Same height: speed up means pressure down.

Venturimeter: Narrow throat is the fast-low-pressure zone.

Examples2

Numerical Example

Water speed changes from 2 m/s to 6 m/s at same height. Pressure drop = 1/2ρ(36 - 4) = 1/2×1000×32 = 16000 Pa.

Application Example

In a perfume atomizer, fast air over the tube lowers pressure and pulls liquid perfume upward into the air stream.

Common mistakes3

Applying Bernoulli Between Any Two Random Points

Use Bernoulli along the same streamline in ideal steady flow.

Ignoring Height Term

If two points are at different heights, include ρgh.

Assuming Bernoulli Explains All Lift Alone

For NEET, Bernoulli pressure difference is enough, but real aerodynamic lift also involves flow direction and Newton's laws.

Reference tables3

Quick revision8

Formulas5

Bernoulli Equation
P+1/2ρv2+ρgh= constantP + 1/2ρv² + ρgh =\text{ constant}

Total mechanical energy per unit volume remains constant along a streamline for ideal flow.

P
Pressure
ρ
Fluid density
v
Fluid speed
h
Height
Bernoulli at Same Height
P1+1/2ρv12=P2+1/2ρv22P1 + 1/2ρv1² = P2 + 1/2ρv2²

Simplified Bernoulli equation when height difference is negligible.

P1, P2
Pressures at two points
v1, v2
Speeds at two points
Pressure Difference from Speed Difference
P1−P2=1/2ρ(v22−v12)P1 - P2 = 1/2ρ(v2² - v1²)

At the same height, pressure falls where speed rises.

P1 - P2
Pressure difference
v2² - v1²
Difference of squared speeds
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Diagrams4

5

Viscosity & Stokes' Law

10 min

Viscosity is the internal friction of a fluid that opposes relative motion between its layers. A more viscous fluid, such as honey, flows slowly compared with water. The coefficient of viscosity, η, measures the viscous nature of a fluid. When a small spherical body moves slowly through a viscous fluid, it experiences viscous drag given by Stokes' law: F = 6πηrv. As a sphere falls through a fluid, its speed increases until weight is balanced by buoyant force and viscous drag. Then acceleration becomes zero and it moves with terminal velocity. Reynolds number predicts whether fluid flow is streamline or turbulent. These ideas are important in raindrops, sedimentation, blood flow and oil flow.

Key points7

  • Viscosity arises due to relative motion between fluid layers.
  • Liquids generally become less viscous when temperature increases.
  • Gases generally become more viscous when temperature increases.
  • Stokes' law is valid for small spherical bodies moving slowly in viscous fluid.
  • Terminal velocity is proportional to r² for a small sphere.
  • Higher viscosity lowers terminal velocity.
  • Reynolds number is dimensionless.

Memory tricks3

Viscosity: Viscosity is fluid friction: honey has more, water has less.

Terminal Velocity: Terminal means final constant speed: weight is balanced by upward forces.

Stokes' Law: Remember 6πηrv: six-pi-eta-radius-velocity.

Examples2

Numerical Example

If radius of a small sphere is doubled, terminal velocity becomes four times, assuming all other factors remain constant.

Application Example

Small raindrops fall slowly because viscous drag becomes comparable to their weight and they reach terminal velocity.

Common mistakes3

Applying Stokes' Law to Any Shape

Stokes' law in this form is for small spherical bodies moving slowly through viscous fluid.

Thinking Terminal Velocity Means No Forces

Forces still act, but net force is zero.

Forgetting Radius Squared in Terminal Velocity

Terminal velocity of a sphere is proportional to r², not r.

Reference tables3

Quick revision8

Formulas4

Newton's Law of Viscosity
F=ηA(dv/dx)F = ηA(dv/dx)

Viscous force between fluid layers is proportional to area and velocity gradient.

F
Viscous force
η
Coefficient of viscosity
A
Area of fluid layer
dv/dx
Velocity gradient perpendicular to flow
Stokes' Law
F=6πηrvF = 6πηrv

Viscous drag on a small sphere moving slowly through a viscous fluid.

F
Viscous drag force
η
Coefficient of viscosity
r
Radius of sphere
v
Speed of sphere
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Diagrams4

6

Surface Tension & Capillarity

11 min

Surface tension is the property of a liquid surface by which it behaves like a stretched elastic membrane. It arises because molecules at the surface experience unbalanced cohesive forces. Surface tension is defined as force per unit length or surface energy per unit area. It explains spherical drops, floating needles, soap bubbles and insects walking on water. The angle of contact decides whether a liquid wets a solid surface. In a narrow capillary tube, liquid may rise or fall depending on adhesive and cohesive forces. Water rises in glass because angle of contact is acute, while mercury falls because its angle is obtuse. Curved liquid surfaces create excess pressure inside drops and bubbles, which is very important for NEET.

Key points7

  • Surface tension decreases with increase in temperature.
  • Soap solution has lower surface tension than pure water.
  • A liquid drop tends to be spherical to minimize surface area.
  • Surface energy increases when surface area increases.
  • Capillary action depends on surface tension, angle of contact, density and tube radius.
  • A soap bubble has two surfaces, so its excess pressure is twice that of a liquid drop of same radius.
  • For capillary fall, cosθ is negative because angle of contact is greater than 90 degrees.

Memory tricks3

Drop vs Bubble: Drop has one surface: 2S/R. Soap bubble has two surfaces: 4S/R.

Capillary Formula: Rise is high when tube is thin: h is inversely proportional to r.

Wetting: Water wets glass and rises; mercury avoids glass and falls.

Examples3

Solved Example: Capillary Rise

If capillary radius is reduced to half, h = 2S cosθ/(ρgr) becomes double.

Solved Example: Bubble Pressure

For a soap bubble of radius R and surface tension S, excess pressure inside is 4S/R because it has two surfaces.

Everyday Application

Detergents reduce surface tension of water, helping it spread and enter small spaces in cloth for better cleaning.

Common mistakes3

Using Soap Bubble Formula for Liquid Drop

Liquid drop has one surface, so excess pressure is 2S/R, not 4S/R.

Forgetting cosθ in Capillary Rise

The sign of cosθ decides rise or fall.

Thinking Surface Tension Increases with Temperature

Surface tension generally decreases when temperature increases.

Reference tables3

Quick revision8

Formulas6

Surface Tension
S=F/lS = F/l

Surface tension is force acting per unit length along the liquid surface.

S
Surface tension
F
Surface force
l
Length of line over which force acts
Surface Energy
Surface energy =S×ΔA\text{Surface energy }= S × ΔA

Work required to increase surface area of a liquid by ΔA.

S
Surface tension
ΔA
Increase in surface area
Capillary Rise or Fall
h=2Scosθ/(ρgr)h = 2S cosθ/(ρgr)

Height of rise or fall of liquid in a capillary tube.

h
Capillary rise or fall
S
Surface tension
θ
Angle of contact
ρ
Density of liquid
r
Radius of capillary tube
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Diagrams4

Videos for Mechanical Properties of Fluids

  • MECHANICAL PROPERTIES OF FLUIDS 01 | Pressure in Fluids | Physics | Class 11th/NEET/JEEPW English Medium · English
  • MECHANICAL PROPERTIES OF FLUIDS 02 | Pascal’s Law & Archimedes Principle | Physics | Class 11/NEET/PW English Medium · English
  • MECHANICAL PROPERTIES OF FLUIDS 03 | Law Of Flotation | Physics | Class 11th/NEET/JEEPW English Medium · English
  • MECHANICAL PROPERTIES OF FLUIDS 04 | Bernoulli’s Principle | Physics | Class 11th/NEET/JEEPW English Medium · English
  • MECHANICAL PROPERTIES OF FLUIDS 05 | Velocity of Efflux | Physics | Class 11th/NEET/JEEPW English Medium · English
  • Mechanical Properties of Fluids || AnimationDigital Guruji · English
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NEET PYQs from Mechanical Properties of Fluids

The latest 3 of 23 previous-year questions from this chapter.

  1. NEET 2026Set 50MediumQ1

    In the measurement of viscosity of liquids using terminal velocity experiment, spherical balls of same radius but having different densities are used. The variation of the terminal velocity ($v$) with the ratio of density of spherical ball ($\sigma$) to density of the liquid ($\rho$), is best represented by:

    • AGraph A
    • BGraph B
    • CGraph C
    • DGraph D
  2. NEET 2026Set 50MediumQ2

    Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between $P$ and $Q$ is $15\ \mathrm{Nm^{-2}}$. The area of cross-section at $P$ and $Q$ are $40\ \mathrm{cm^2}$ and $20\ \mathrm{cm^2}$, respectively. The rate of flow of water through the pipe, in $\mathrm{cm^3s^{-1}}$, is: [Take density of water = $1000\ \mathrm{kg\ m^{-3}}$]

    • A$100$
    • B$200$
    • C$300$
    • D$400$
  3. NEET 2026Set 11EasyQ3

    A submarine is designed to withstand an absolute pressure of 100 atm. How deep can it go below the water surface? (Consider density of water = 1000 kg m⁻³, 1 atm = 1 × 10⁵ Pa and gravitational acceleration g = 10 m/s²)

    • A990 m
    • B9900 m
    • C99 m
    • D9000 m

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