Class 11 Physics: Mechanical Properties of Fluids | Physics Q&A

CBSE Class 11 › Unit VII: Properties of Bulk Matter

Mechanical Properties of Fluids

NCERT Chapter 9 • Pressure, Bernoulli’s Principle, Viscosity & Surface Tension

NCERT 2025–26 Unit VII • ~10 Marks JEE Main • 1-2 Questions

Illustration of fluid flow demonstrating Bernoulli's Principle. — Fluid Dynamics: The physics of liquids and gases in motion.

1. Pressure and Pascal’s Law

Pressure ( $P$ ): The normal force acting per unit area. It is a scalar quantity.

P = \dfrac{F}{A}

Unit: Pascal (Pa) or N/m². 1 atm $\approx 1.013 \times 10^5$ Pa.

1.1 Solids vs Fluids & Density

Solids resist shear stress and maintain shape. Fluids (liquids and gases) offer negligible resistance to shear stress and flow under its action.

Liquids: Incompressible, fixed volume, free surface.
Gases: Highly compressible, fill entire container.

Density: Mass per unit volume, $\rho = m/V$ . SI unit: kg/m³.

Relative Density: Ratio of a substance’s density to that of water at 4°C (where $\rho_{\text{water}} = 1000$ kg/m³). It is dimensionless.

1.2 Variation of Pressure with Depth

In a fluid at rest, pressure increases with depth due to the weight of the fluid column above.

P = P_a + \rho g h

Where $P_a$ is atmospheric pressure, $\rho$ is density, and $h$ is depth.

1.3 Hydrostatic Paradox

The pressure at a given depth in a fluid is independent of the shape or volume of the container. This is why water levels equalize in connected vessels of different shapes.

Three differently shaped vessels with same water level. — Hydrostatic paradox: Pressure depends only on depth, not container shape.

1.4 Atmospheric Pressure & Gauge Pressure

At sea level, atmospheric pressure is measured using a mercury barometer: $P_a = \rho_{\text{Hg}} g h$ . With $h \approx 76$ cm, $P_a \approx 1.013 \times 10^5$ Pa.

Gauge Pressure: Pressure relative to atmosphere: $P_{\text{gauge}} = P - P_a$ .

1.5 Pascal’s Law

Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel.

Hydraulic lift schematic showing force multiplication using Pascal's Law. — Pascal’s Law allows us to lift heavy loads with small forces (Hydraulic Lift).

Application (Hydraulic Lift):
Since $P_1 = P_2$ , we have $\dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}$ .
Therefore, $F_2 = F_1 \left( \dfrac{A_2}{A_1} \right)$ . A large force $F_2$ is generated by a small force $F_1$ if $A_2 \gg A_1$ .

Problem 1: In a car lift, compressed air exerts a force $F_1$ on a small piston of radius 5 cm. This pressure is transmitted to a second piston of radius 15 cm. If the mass of the car is 1350 kg, find $F_1$ . ( $g=9.8$ m/s²)

Show Answer

Area ratio $\dfrac{A_1}{A_2} = \dfrac{\pi r_1^2}{\pi r_2^2} = \left( \dfrac{5}{15} \right)^2 = \dfrac{1}{9}$ .
Load force $F_2 = mg = 1350 \times 9.8 = 13230$ N.
$F_1 = F_2 \times \dfrac{A_1}{A_2} = 13230 \times \dfrac{1}{9} = 1470$ N.

2. Streamline Flow

Streamline (Laminar) Flow: Flow in which every particle passing a point follows the same path as the preceding particles. It is steady and orderly.

Turbulent Flow: Irregular and chaotic flow occurring above a critical speed.

Comparison of orderly streamline flow and chaotic turbulent flow. — Flow transitions from orderly (streamline) to chaotic (turbulent) above critical velocity.

Equation of Continuity

For an incompressible fluid in streamline flow, the product of area and velocity is constant (Conservation of Mass).

A_1 v_1 = A_2 v_2 = \text{Constant}

This implies that velocity is higher where the tube is narrower ( $v \propto 1/A$ ).

3. Bernoulli’s Principle

For streamline flow of an ideal fluid, the sum of pressure energy, kinetic energy, and potential energy per unit volume is constant.

P + \dfrac{1}{2}\rho v^2 + \rho g h = \text{Constant}

Assumptions:

Steady, incompressible, non-viscous flow.

Key Insight:

Where speed is high, pressure is low (and vice-versa). This explains dynamic lift in airplanes.

Applications

Venturi Meter: Measures flow speed.
Torricelli’s Law: Speed of efflux from a tank hole ( $v = \sqrt{2gh}$ ).
Dynamic Lift: Curved wings (aerofoils) create pressure difference.

Problem 2: Water flows through a horizontal pipe. At a point where cross-section is 10 cm², speed is 1 m/s and pressure is 2000 Pa. At another point, cross-section is 5 cm². Find pressure at this point.

Show Answer

1. Continuity: $A_1 v_1 = A_2 v_2 \Rightarrow 10(1) = 5(v_2) \Rightarrow v_2 = 2$ m/s.
2. Bernoulli (Horizontal $h_1=h_2$ ): $P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2$ .
$2000 + 0.5(1000)(1)^2 = P_2 + 0.5(1000)(2)^2$ .
$2000 + 500 = P_2 + 2000 \Rightarrow P_2 = 500$ Pa.

4. Viscosity

Viscosity is the internal friction between fluid layers. It opposes relative motion.

F = \eta A \dfrac{dv}{dx}

Where $\eta$ is the Coefficient of Viscosity. Unit: Poiseuille (Pl) or Pa-s.

Stokes’ Law

The viscous drag force on a sphere of radius $r$ moving with velocity $v$ is:

F = 6\pi \eta r v

Terminal Velocity ( $v_t$ )

The constant maximum speed attained by a falling body when Viscous Force + Buoyancy = Weight.

v_t = \dfrac{2 r^2 (\rho - \sigma) g}{9 \eta}

Problem 3 (Terminal Velocity): A copper ball of radius 2 mm falls through oil. Given $\rho_{\text{copper}} = 8900$ kg/m³, $\rho_{\text{oil}} = 900$ kg/m³, $\eta = 0.1$ Pa·s, find $v_t$ .

Show Answer

$v_t = \dfrac{2 r^2 (\rho - \sigma) g}{9 \eta} = \dfrac{2 (0.002)^2 (8900 - 900) \times 10}{9 \times 0.1}$
$= \dfrac{2 \times 4 \times 10^{-6} \times 8000 \times 10}{0.9} = \dfrac{0.64}{0.9} \approx 0.71$ m/s.

5. Surface Tension

The property of a liquid surface to behave like a stretched elastic membrane. It arises due to cohesive forces.

S = \dfrac{F}{L} \quad \text{or} \quad S = \dfrac{\text{Surface Energy}}{\text{Area}}

5.1 Angle of Contact ( $\theta$ )

Acute angle ( $\theta < 90^\circ$ ): Liquid wets the solid (e.g., water on glass).
Obtuse angle ( $\theta > 90^\circ$ ): Liquid does not wet the solid (e.g., mercury on glass).

5.2 Excess Pressure

Due to surface tension, pressure inside a curved surface is higher than outside.

Liquid Drop: $P_{excess} = \dfrac{2S}{R}$
Soap Bubble: $P_{excess} = \dfrac{4S}{R}$ (Two surfaces)

5.3 Capillarity

The rise or fall of liquid in a narrow tube.

Capillary rise and fall showing concave and convex menisci. — Surface tension causes liquids to rise or fall in narrow tubes.

h = \dfrac{2S \cos \theta}{r \rho g}

Problem 4 (Capillary Rise): Water rises to a height of 5 cm in a capillary tube of radius 0.5 mm. Find the surface tension of water. ( $\rho = 1000$ kg/m³, $g = 10$ m/s², $\theta = 0^\circ$ )

Show Answer

$h = \dfrac{2S \cos\theta}{r \rho g} \Rightarrow S = \dfrac{h r \rho g}{2 \cos\theta}$
$S = \dfrac{0.05 \times 0.5 \times 10^{-3} \times 1000 \times 10}{2 \times 1} = 0.125$ N/m.

← Previous Chapter Next Chapter: Thermal Props →

1. Pressure and Pascal’s Law

1.1 Solids vs Fluids & Density

1.2 Variation of Pressure with Depth

1.3 Hydrostatic Paradox

1.4 Atmospheric Pressure & Gauge Pressure

1.5 Pascal’s Law

2. Streamline Flow

Equation of Continuity

3. Bernoulli’s Principle

Applications

4. Viscosity

Stokes’ Law

Terminal Velocity ()

5. Surface Tension

5.1 Angle of Contact ()

5.2 Excess Pressure

5.3 Capillarity

Terminal Velocity ( $v_t$ )

5.1 Angle of Contact ( $\theta$ )