Mechanical Properties of Fluids — Class 11 Physics

1. Fluids and Pressure

A fluid is a substance that can flow and does not sustain a static shear stress. Liquids and gases are fluids. Pressure is normal force per unit area: P=F/A. Its SI unit is pascal (Pa).

In a fluid at rest, pressure at a point acts equally in all directions. Pressure is a scalar quantity.

2. Pressure Due to a Fluid Column

For a liquid of density ρ at depth h below a free surface open to atmospheric pressure, the pressure is P=P_atm+ρgh. The gauge pressure is P_g=ρgh.

Thus pressure increases with depth and depends on density and depth, not on the shape of the container.

3. Pascal’s Law and Hydraulic Machines

Pascal’s law states that an externally applied pressure change to an enclosed fluid is transmitted undiminished throughout the fluid. In a hydraulic lift, F₁/A₁=F₂/A₂, so a small force acting on a small piston can balance a larger force on a larger piston.

4. Atmospheric Pressure and Barometer

Atmospheric pressure is the pressure exerted by the atmosphere. A mercury barometer relates atmospheric pressure to the mercury column height through P_atm=ρgh, under the idealised assumptions of a standard barometer.

5. Buoyancy and Archimedes’ Principle

An immersed object experiences an upward buoyant force equal to the weight of the fluid displaced by it. Thus F_B=ρ_fluidV_displacedg.

An object floats when its average density is less than the fluid density and, in equilibrium, its weight equals the buoyant force.

6. Fluid Flow and Streamlines

A streamline is a curve whose tangent at every point gives the instantaneous direction of fluid velocity. For steady flow, streamlines do not cross because that would imply two velocity directions at the same point.

7. Equation of Continuity

For steady incompressible flow, conservation of mass gives A₁v₁=A₂v₂. Therefore fluid speed increases when the cross-sectional area of a pipe decreases.

8. Bernoulli’s Principle

For steady, incompressible, non-viscous flow along a streamline, P+½ρv²+ρgh=constant. The three terms represent pressure energy per unit volume, kinetic energy per unit volume and gravitational potential energy per unit volume.

Bernoulli’s equation helps explain flow through constrictions, Venturi meters and several idealised fluid-flow situations. It should not be applied blindly when viscosity, turbulence or strong unsteady effects are important.

9. Viscosity

Viscosity is internal friction between layers of a fluid. For a Newtonian fluid, shear stress is proportional to velocity gradient: shear stress=η(dv/dy), where η is the coefficient of viscosity. Its SI unit is Pa·s.

10. Stokes’ Law and Terminal Velocity

For a small sphere moving slowly through a viscous fluid, the viscous drag is F_v=6πηrv. This is Stokes’ law and applies under appropriate low-Reynolds-number conditions.

When a falling sphere reaches terminal velocity, the net force becomes zero. For a sphere of radius r and density ρ_s falling in a fluid of density ρ_f, the ideal terminal speed is v_t=2r²(ρ_s−ρ_f)g/(9η).

11. Surface Tension

Surface tension arises from cohesive molecular forces at a liquid surface. It can be defined as tangential force per unit length perpendicular to a line drawn on the surface. Its SI unit is N/m.

Surface tension tends to minimise surface area, which helps explain the nearly spherical shape of small liquid droplets.

12. Excess Pressure

For a liquid drop of radius R, excess pressure inside is ΔP=2T/R. For a soap bubble with two surfaces, ΔP=4T/R.

13. Capillarity

Capillary rise or fall occurs because of surface tension and the balance between adhesive and cohesive effects. For a narrow tube, the ideal capillary rise is h=2T cosθ/(ρgr), where θ is the contact angle and r is tube radius.

Worked Numerical 1 — Pressure at Depth

Question: Find the gauge pressure 5 m below the surface of water. Take ρ=1000 kg/m³ and g=9.8 m/s².

P_g=ρgh=1000×9.8×5=49,000 Pa.

Worked Numerical 2 — Hydraulic Lift

Question: A hydraulic lift has piston areas 0.01 m² and 0.5 m². What force on the smaller piston can balance a 10,000 N load on the larger piston?

F₁/A₁=F₂/A₂.

F₁=10,000×0.01/0.5=200 N.

Worked Numerical 3 — Buoyant Force

Question: A body displaces 0.002 m³ of water. Find the buoyant force. Take ρ=1000 kg/m³ and g=9.8 m/s².

F_B=ρVg=1000×0.002×9.8=19.6 N.

Worked Numerical 4 — Continuity Equation

Question: Water flows through a pipe of area 4×10⁻³ m² at 2 m/s. It enters a section of area 1×10⁻³ m². Find the speed there.

A₁v₁=A₂v₂, so v₂=(4×10⁻³×2)/(1×10⁻³)=8 m/s.

Worked Numerical 5 — Bernoulli Principle

Question: Water flows horizontally from a wide pipe at 2 m/s into a narrow section at 6 m/s. Take ρ=1000 kg/m³. Find P₁−P₂ for ideal steady flow at the same height.

Bernoulli gives P₁+½ρv₁²=P₂+½ρv₂².

P₁−P₂=½ρ(v₂²−v₁²)=½(1000)(36−4)=16,000 Pa.

Worked Numerical 6 — Stokes’ Law

Question: A small sphere of radius 1 mm moves through a fluid with viscosity 0.1 Pa·s at 0.02 m/s. Find the viscous drag.

F=6πηrv=6π(0.1)(1×10⁻³)(0.02)≈3.77×10⁻⁵ N.

Worked Numerical 7 — Excess Pressure in a Drop

Question: A water drop has radius 1 mm and surface tension 0.072 N/m. Find the excess pressure inside it.

ΔP=2T/R=2(0.072)/(1×10⁻³)=144 Pa.

Worked Numerical 8 — Capillary Rise

Question: A liquid has surface tension 0.072 N/m, density 1000 kg/m³ and contact angle 0°. In a capillary of radius 0.5 mm, take g=9.8 m/s². Find the rise.

h=2T cosθ/(ρgr)=2(0.072)(1)/[1000(9.8)(0.5×10⁻³)]≈0.0294 m = 2.94 cm.

Explained MCQs

Pressure at the same depth in a connected liquid at rest is:
Answer: The same when the liquid is the same and the points are at the same level. Hydrostatic pressure depends on depth and density, not container shape.
When a pipe becomes narrower in steady incompressible flow, fluid speed:
Answer: Increases. Continuity gives Av=constant.
According to Bernoulli’s equation, when speed increases at the same height in ideal flow, pressure:
Answer: Decreases. The kinetic-energy-per-volume term increases, so pressure must decrease if the total remains constant.
What is the SI unit of coefficient of viscosity?
Answer: Pa·s.
Which has greater excess pressure for the same radius and surface tension: a liquid drop or soap bubble?
Answer: Soap bubble. A bubble has two surfaces, giving ΔP=4T/R rather than 2T/R.
Why does water rise in a clean narrow glass capillary?
Answer: Surface-tension effects and wetting produce an upward capillary force. For a wetting liquid, the contact angle is less than 90° and cosθ is positive.

Competency Question

Water flows through a pipe that gradually narrows. Predict how its speed and ideal pressure change as the pipe becomes narrower, and explain your answer using both the continuity equation and Bernoulli’s principle.

HOTS

A small steel ball falls through a viscous liquid. Initially it accelerates, but after some time its speed becomes constant. Explain why the speed becomes constant and identify the force balance at terminal velocity.

Common Mistakes

Confusing absolute pressure with gauge pressure.
Using Bernoulli’s equation without checking its assumptions.
Forgetting that continuity A v=constant applies to steady incompressible flow.
Using the liquid-drop formula for a soap bubble.
Ignoring the contact angle in capillary-rise problems.
Applying Stokes’ law outside its range of validity.

Quick Revision

P=F/A
P=P_atm+ρgh
F₁/A₁=F₂/A₂
F_B=ρVg
A₁v₁=A₂v₂
P+½ρv²+ρgh=constant
F_v=6πηrv
v_t=2r²(ρ_s−ρ_f)g/(9η)
ΔP(drop)=2T/R
ΔP(soap bubble)=4T/R
h=2T cosθ/(ρgr)

Class 11 Physics – Mechanical Properties of Fluids