"Properties of Fluids"

About

Fluids — liquids and gases — surround us and obey a different set of mechanical rules than solids. This chapter introduces fluid statics (pressure, Pascal's law, buoyancy) and fluid dynamics (equation of continuity, Bernoulli's theorem). You will also learn about surface tension, viscosity, and their real-world applications.

Key Concepts

9.1 Pressure

Pressure is the force exerted per unit area normal to the surface:

$P=\frac{F}{A}$

SI unit: pascal (Pa) = N/m²

Other units: 1 atm = $1.01\times 10^5$ Pa, 1 bar = $10^5$ Pa, 1 torr = 1 mm of Hg

Pressure is a scalar quantity.

Since $P\propto \frac{1}{A}$, for the same force:

• Small area → High pressure (needle pierces skin)
• Large area → Low pressure (skis prevent sinking in snow)

9.2 Pressure in a Liquid

At a depth $h$ below the surface of a liquid of density $\rho$:

$P=P_0+\rho gh$

Where $P_0$ is atmospheric pressure at the surface.

• Pressure in a liquid increases linearly with depth
• Pressure is the same at all points at the same horizontal level (Pascal's principle)

9.3 Pascal's Law

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

Hydraulic lift: Uses Pascal's law to multiply force.

$\frac{F_1}{A_1}=\frac{F_2}{A_2}⟹F_2=F_1\times \frac{A_2}{A_1}$

A small force on a small piston can balance a much larger weight on a larger piston — provided the pressures are equal.

9.4 Archimedes' Principle

When a body is wholly or partially immersed in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced by it.

$F_B={\rho }_{\text{fluid}}\cdot V_{\text{displaced}}\cdot g$

Law of floatation: A body floats when its weight equals the weight of the fluid displaced.

9.5 Flow of Fluids

Streamline flow: Fluid particles follow smooth, non-intersecting paths. Velocity at each point is constant in time.

Turbulent flow: Irregular, chaotic motion with eddies and vortices.

Equation of continuity (for incompressible fluids): $A_1{v}_1=A_2{v}_2$

The product $Av$ (volume flow rate) remains constant.

9.6 Bernoulli's Theorem

For an ideal fluid (incompressible, non-viscous) in streamline flow:

$P+\frac{1}{2}\rho v^2+\rho gh=\text{constant}$

Applications:

• Aeroplane lift: Faster airflow over curved wing top → lower pressure → upward lift
• Venturi meter: Measures fluid flow rate
• Atomiser/Spray gun: Fast air over nozzle creates low pressure, drawing liquid up

9.7 Surface Tension

Surface tension is the property of a liquid that makes its free surface behave like a stretched elastic membrane.

$S=\frac{F}{l}$

SI unit: N/m

Surface energy: $E=S\times \Delta A$ (work done to increase surface area)

Applications:

• Water droplets are spherical (minimum surface area)
• Insects walk on water
• Capillary rise/tube action

Capillary rise: $h=\frac{2S\cos \theta }{r\rho g}$

9.8 Viscosity

Viscosity is the internal friction of a fluid — its resistance to flow.

Newton's law of viscosity: $F=\eta A\frac{dv}{dx}$

Where $\eta$ is the coefficient of viscosity. SI unit: Pa⋅s or N⋅s/m² or poiseuille.

Stokes' law: For a spherical body moving through a viscous fluid with terminal velocity $v_t$:

$F=6\pi \eta r{v}_t$

Terminal velocity: $v_t=\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta }$

INTEXT QUESTIONS 9.1

Q1. Why are the shoes used for skiing on snow made big in size?

Ans: Skiing shoes are made big in size to reduce pressure on the snow surface. According to $P=F/A$, when the contact area is larger, the skier's weight is distributed over a greater area, resulting in lower pressure per unit area. This prevents the skier from sinking deep into the snow and allows gliding smoothly. Normal-sized shoes would create high pressure and cause the skier to sink.

Q2. Calculate the pressure at the bottom of an ocean at a depth of 1500 m. Take the density of sea water 1.024 × 10³ kg m⁻³, atmospheric pressure = 1.01 × 10⁵ Pa and g = 9.80 m s⁻².

Ans:

• $h=1500$ m, $\rho =1.024\times 10^3$ kg/m³, $P_0=1.01\times 10^5$ Pa, $g=9.80$ m/s²

$P=P_0+\rho gh$

$P=1.01\times 10^5+(1.024\times 10^3)(9.80)(1500)$

$P=1.01\times 10^5+1.505\times 10^7$

$P=0.101\times 10^7+1.505\times 10^7=1.606\times 10^7\text{ Pa}$

Q3. An elephant of weight 5000 kgf is standing on the bigger piston of area 10 m² of a hydraulic lift. Can a boy of 25 kg wt standing on the smaller piston of area 0.05 m² balance or lift the elephant?

Ans:

• $W_1=5000$ kgf, $A_1=10$ m²
• $W_2=25$ kgf, $A_2=0.05$ m²

Pressure on bigger piston: $P_1=\frac{5000}{10}=500$ kgf/m²

Pressure on smaller piston: $P_2=\frac{25}{0.05}=500$ kgf/m²

Since both pressures are equal, the boy can balance the elephant but cannot lift it. To lift, the boy would need to apply slightly more force to create higher pressure on the smaller piston.

Q4. If a pointed needle is pressed against your skin, you are hurt but if the same force is applied by a rod on your skin nothing may happen. Why?

Ans: This happens due to the difference in pressure. For the same applied force:

• Needle: Contact area is very small → $P=F/\text{small area}$ = high pressure → pierces skin
• Rod: Contact area is much larger → $P=F/\text{large area}$ = low pressure → distributed over a larger area, causing no harm

Pressure is inversely proportional to area for the same force.

Q5. A body of 50 kgf is put on the smaller piston of area 0.1 m² of a big hydraulic lift. Calculate the maximum weight that can be balanced on the bigger piston of area 10 m² of this hydraulic lift.

Ans:

• $F_1=50$ kgf, $A_1=0.1$ m², $A_2=10$ m²

Using Pascal's law: $\frac{F_1}{A_1}=\frac{F_2}{A_2}$

$F_2=F_1\times \frac{A_2}{A_1}=50\times \frac{10}{0.1}=50\times 100=5000\text{ kgf}$

Maximum weight that can be balanced = 5000 kgf

Terminal Exercise

1. Define pressure. Explain why: (a) a sharp knife cuts better than a blunt one, (b) camels have broad feet.

2. State and explain Pascal's law. Describe the working of a hydraulic lift.

3. Derive the expression for pressure at a point inside a liquid: $P=P_0+\rho gh$.

4. State Archimedes' principle. A body weighs 50 N in air and 40 N when fully immersed in water. Find: (a) buoyant force, (b) volume of the body, (c) density of the body.

5. State and explain the equation of continuity. Water flows through a pipe of varying cross-section. If the diameter at one section is 4 cm and velocity is 2 m/s, find the velocity where the diameter narrows to 2 cm.

6. State Bernoulli's theorem. Explain how an aeroplane gets lift using this principle.

7. Define surface tension. Derive the expression for capillary rise: $h=\frac{2S\cos \theta }{r\rho g}$.

8. Define coefficient of viscosity. State Newton's law of viscosity and Stokes' law.

9. A spherical ball of radius 2 mm and density $8\times 10^3$ kg/m³ is dropped in a liquid of density $1\times 10^3$ kg/m³. Find the terminal velocity. ($\eta =0.8$ N⋅s/m², g = 10 m/s²)

10. Explain why: (a) oil rises in the wick of a lamp, (b) detergents reduce surface tension, (c) a steel needle can float on water.

11. Water is flowing through a horizontal pipe with a pressure of $2\times 10^5$ Pa at a point where velocity is 2 m/s. At another point, the velocity increases to 4 m/s. Find the pressure at the second point. ($\rho =1000$ kg/m³)

12. A U-tube contains water and oil separated by mercury. The density of oil is 0.8 g/cm³ and water is 1 g/cm³. If the oil column is 10 cm high, find the height of the water column that balances it.

Worked Examples

Example 1: Pressure at Depth

Problem: Find the pressure 50 m below the surface of a lake. ($\rho =1000$ kg/m³, $P_0=1.01\times 10^5$ Pa, g = 10 m/s²)

Solution: $P=P_0+\rho gh=1.01\times 10^5+1000\times 10\times 50$ $=1.01\times 10^5+5.0\times 10^5=6.01\times 10^5\text{ Pa}$

Example 2: Hydraulic Lift

Problem: A hydraulic press has a small piston of radius 5 cm and a large piston of radius 25 cm. What force on the small piston will support a 1000 kg load? (g = 10 m/s²)

Solution: $\frac{F_1}{A_1}=\frac{F_2}{A_2}⟹F_1=F_2\times \frac{A_1}{A_2}=F_2\times \frac{r_1^2}{r_2^2}$

$F_1=10000\times \frac{25}{625}=10000\times \frac{1}{25}=400\text{ N}$

Example 3: Bernoulli's Theorem

Problem: Water flows through a horizontal pipe. At a wide section (area 0.1 m²), the pressure is $3\times 10^5$ Pa and velocity 2 m/s. At a narrow section (area 0.02 m²), find the velocity and pressure. ($\rho =1000$ kg/m³)

Solution: By continuity: $A_1{v}_1=A_2{v}_2⟹v_2=\frac{0.1\times 2}{0.02}=10$ m/s

By Bernoulli (horizontal pipe, $h_1=h_2$): $P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$ $3\times 10^5+\frac{1}{2}(1000)(4)=P_2+\frac{1}{2}(1000)(100)$ $3\times 10^5+2000=P_2+50000$ $P_2=3.02\times 10^5-0.5\times 10^5=2.52\times 10^5\text{ Pa}$

Common Mistakes

1. Forgetting atmospheric pressure when calculating total pressure: $P_{\text{total}}=P_0+\rho gh$.
2. Confusing force multiplication with work multiplication: Hydraulic lifts multiply FORCE, not work. Work is conserved.
3. Applying Bernoulli's theorem to viscous fluids: It's valid only for ideal (non-viscous, incompressible) fluids.
4. Forgetting that buoyant force depends on displaced volume, not depth: Buoyant force is the same at all depths as long as the body is fully submerged.
5. Mixing up surface tension and viscosity: Surface tension is a surface property; viscosity is a bulk property.

Quick Revision