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Physics Part 2

Mechanical Properties of Fluids

Mechanical Properties of Fluids

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Study Action Plan (Sorted by Difficulty)
Atmospheric Pressure, Gauge Pressure and Manometer
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Bernoulli's Principle
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Capillary Action
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Dynamic Lift and Magnus Effect
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Equation of Continuity
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Hydraulic Machines
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Pascal's Law
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Pressure Variation with Depth
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Pressure, Density and Relative Density
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Stokes' Law
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Streamline, Laminar and Turbulent Flow
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Surface Tension
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Terminal Velocity
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Torricelli's Law
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Viscosity
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Learning Objectives

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Learning Objectives

  • Understand Bernoulli's Principle and its application in relating pressure, kinetic energy per unit volume, and potential energy per unit volume in streamline flow.
  • Calculate fluid dynamics using the Equation of Continuity, ensuring the product of cross-sectional area and velocity remains constant in incompressible fluid flow.
  • Apply Pascal's Law to determine pressure transmission in enclosed fluids and its applications in hydraulic systems.
  • Analyze the viscous drag force on spheres using Stokes' Law and its dependence on radius, velocity, and fluid viscosity.
  • Explore the concept of Surface Tension as a force per unit length at the liquid interface and its implications in various phenomena.
  • Investigate Capillary Action and the factors influencing the rise or fall of liquids in narrow tubes.
  • Derive the Pressure Variation with Depth formula and apply it to calculate pressure changes in fluids due to depth.
  • Examine the concept of Viscosity and its role in fluid resistance to deformation or flow.
  • Evaluate Dynamic Lift and the Magnus Effect in the context of lift forces on bodies moving through fluids.
  • Utilize Torricelli's Law to determine the speed of efflux of fluids under gravity from an orifice.
  • Define and calculate Pressure, Density, and Relative Density, and apply these concepts in numerical problems.
  • Differentiate between Atmospheric Pressure, Gauge Pressure, and use a Manometer for pressure-difference calculations.
  • Apply Pascal's Law in Hydraulic Machines to understand mechanical advantage and force transmission.
  • Distinguish between Streamline, Laminar, and Turbulent Flow, and understand the significance of critical speed.
  • Calculate Terminal Velocity using Stokes' Law and analyze the effects of buoyancy and density differences.

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Revision Notes & Summary

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Chapter Notes

Bernoulli's Principle

  • Definition: Relates pressure, kinetic energy per unit volume, and potential energy per unit volume in a streamline flow, stating their sum remains constant.
  • Equation: P+12ρv2+ρgh=constantP + \frac{1}{2} \rho v^2 + \rho gh = \text{constant}
  • Assumptions: Applies to incompressible, non-viscous fluids in steady flow.
  • Applications: Explains phenomena like lift on airplane wings and the functioning of carburetors.

Equation of Continuity

  • Definition: For incompressible fluid flow, the product of cross-sectional area and velocity remains constant along a streamline.
  • Equation: A1v1=A2v2A_1 v_1 = A_2 v_2
  • Conservation: Represents conservation of mass in fluid dynamics.

Pascal's Law

  • Statement: Pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel.
  • Applications: Basis for hydraulic lifts and hydraulic brakes.

Stokes' Law

  • Definition: Describes the viscous drag force on a sphere moving through a fluid.
  • Equation: F=6πηavF = 6 \pi \eta a v
  • Variables:
    • η\eta: Viscosity of the fluid
    • aa: Radius of the sphere
    • vv: Velocity of the sphere

Surface Tension

  • Definition: The force per unit length acting at the interface between a liquid and another medium.
  • Equation: Surface tension S=F2lS = \frac{F}{2l}
  • Phenomena: Explains capillary action and the formation of droplets.

Capillary Action

  • Definition: The rise or fall of a liquid in a narrow tube due to surface tension and adhesive forces.
  • Equation: h=2Scos⁡θρgah = \frac{2S \cos \theta}{\rho g a}
  • Variables:
    • hh: Height of the liquid column
    • SS: Surface tension
    • θ\theta: Contact angle
    • ρ\rho: Density of the liquid
    • aa: Radius of the tube

Pressure Variation with Depth

  • Equation: P=Pa+ρghP = P_a + \rho gh
  • Explanation: Pressure in a fluid increases with depth due to the weight of the fluid above.

Viscosity

  • Definition: A measure of a fluid's resistance to deformation or flow.
  • Equation: η=FAlv\eta = \frac{F}{A} \frac{l}{v}
  • Units: Poiseuille (Pl), N s m⁻², or Pa s

Dynamic Lift and Magnus Effect

  • Dynamic Lift: Force on a body moving through a fluid due to pressure differences.
  • Magnus Effect: Lift force on a spinning object due to differences in velocity and pressure.

Torricelli's Law

  • Definition: Describes the speed of efflux of a fluid under gravity from an orifice.
  • Equation: v=2ghv = \sqrt{2gh}

Pressure, Density and Relative Density

  • Pressure: P=FAP = \frac{F}{A}
  • Density: ρ=mV\rho = \frac{m}{V}
  • Relative Density: Ratio of the density of a substance to the density of a reference substance.

Atmospheric Pressure, Gauge Pressure and Manometer

  • Atmospheric Pressure: Pressure exerted by the weight of the atmosphere.
  • Gauge Pressure: Difference between absolute pressure and atmospheric pressure.
  • Manometer: Device for measuring pressure differences.

Hydraulic Machines

  • Principle: Based on Pascal's law.
  • Examples: Hydraulic lift and hydraulic brakes.

Streamline, Laminar and Turbulent Flow

  • Streamline Flow: Flow where each particle follows a smooth path.
  • Laminar Flow: Smooth, orderly fluid motion.
  • Turbulent Flow: Chaotic, irregular fluid motion.

Terminal Velocity

  • Definition: The constant velocity reached by a sphere falling through a viscous medium.
  • Equation: vt=2a2(ρ−σ)g9ηv_t = \frac{2a^2 (\rho - \sigma) g}{9\eta}
  • Variables:
    • vtv_t: Terminal velocity
    • aa: Radius of the sphere
    • ρ\rho: Density of the sphere
    • σ\sigma: Density of the fluid
    • gg: Acceleration due to gravity
    • η\eta: Viscosity of the fluid
This chapter covers the mechanical properties of fluids, focusing on principles such as Bernoulli's principle, Pascal's law, and the equation of continuity, which are fundamental to understanding fluid dynamics and applications in real-world scenarios.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

P1+12ρv12+ρgh1=P2+12ρv22+ρgh2P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2

B.

P1+ρv1+gh1=P2+ρv2+gh2P_1 + \rho v_1 + gh_1 = P_2 + \rho v_2 + gh_2

C.

P1+ρv12+12ρgh1=P2+ρv22+12ρgh2P_1 + \rho v_1^2 + \frac{1}{2}\rho gh_1 = P_2 + \rho v_2^2 + \frac{1}{2}\rho gh_2

D.

P1+ρgh1=P2+ρgh2P_1 + \rho gh_1 = P_2 + \rho gh_2
Correct Answer: A

Solution:

Bernoulli's equation for streamline flow is given by P+12ρv2+ρgh=constantP + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}, which applies to points 1 and 2.

Chapter Concept:

Bernoulli's Principle

A.

2 m/s

B.

4 m/s

C.

8 m/s

D.

16 m/s
Correct Answer: A

Solution:

According to the equation of continuity, A1v1=A2v2A_1v_1 = A_2v_2. If A1=12A2A_1 = \frac{1}{2}A_2, then v2=A1A2v1=12×4=2 m/sv_2 = \frac{A_1}{A_2}v_1 = \frac{1}{2} \times 4 = 2 \text{ m/s}.

Chapter Concept:

Bernoulli's Principle

A.

Gravitational forces

B.

Adhesive forces between the liquid and tube material

C.

Cohesive forces within the liquid

D.

External pressure applied on the liquid
Correct Answer: B

Solution:

Capillary action is caused by adhesive forces between the liquid and the tube material, along with surface tension.

Chapter Concept:

Capillary Action

A.

The meniscus will be concave and the liquid will rise

B.

The meniscus will be concave and the liquid will fall

C.

The meniscus will be convex and the liquid will rise

D.

The meniscus will be convex and the liquid will fall
Correct Answer: A

Solution:

When a liquid wets the tube, the meniscus is concave, and due to surface tension, the liquid rises in the capillary tube.

Chapter Concept:

Equation of Continuity

A.

1 m/s1 \ m/s

B.

2 m/s2 \ m/s

C.

4 m/s4 \ m/s

D.

8 m/s8 \ m/s
Correct Answer: C

Solution:

Using the equation of continuity, A1v1=A2v2A_1v_1 = A_2v_2, where A1=0.1 m2A_1 = 0.1 \ m^2, v1=2 m/sv_1 = 2 \ m/s, A2=0.05 m2A_2 = 0.05 \ m^2. Solving for v2v_2, we get v2=A1v1A2=0.1 m2×2 m/s0.05 m2=4 m/sv_2 = \frac{A_1v_1}{A_2} = \frac{0.1 \ m^2 \times 2 \ m/s}{0.05 \ m^2} = 4 \ m/s.

Chapter Concept:

Equation of Continuity

A.

1.01×105 Pa1.01 \times 10^5 \text{ Pa}

B.

1.03×105 Pa1.03 \times 10^5 \text{ Pa}

C.

1.28×105 Pa1.28 \times 10^5 \text{ Pa}

D.

1.33×105 Pa1.33 \times 10^5 \text{ Pa}
Correct Answer: C

Solution:

The pressure difference due to mercury is ΔP=ρgh=13.6×103×9.8×0.2=2.67×104 Pa\Delta P = \rho gh = 13.6 \times 10^3 \times 9.8 \times 0.2 = 2.67 \times 10^4 \text{ Pa}. Therefore, the absolute pressure of the gas is P=Pa+ΔP=1.01×105+2.67×104=1.28×105 PaP = P_a + \Delta P = 1.01 \times 10^5 + 2.67 \times 10^4 = 1.28 \times 10^5 \text{ Pa}.

Chapter Concept:

Atmospheric Pressure, Gauge Pressure and Manometer

A.

h=2Scos⁡θρgah = \frac{2S \cos \theta}{\rho g a}

B.

h=Scos⁡θρgah = \frac{S \cos \theta}{\rho g a}

C.

h=2Sρgah = \frac{2S}{\rho g a}

D.

h=Sρgah = \frac{S}{\rho g a}
Correct Answer: A

Solution:

The capillary rise hh is given by the expression h=2Scos⁡θρgah = \frac{2S \cos \theta}{\rho g a}, where SS is the surface tension, ρ\rho is the density of the liquid, gg is the acceleration due to gravity, and aa is the radius of the tube.

Chapter Concept:

Pressure, Density and Relative Density

A.

vt=2r2(ρs−ρm)g9ηv_t = \frac{2r^2(\rho_s - \rho_m)g}{9\eta}

B.

vt=9η2r2(ρs−ρm)gv_t = \frac{9\eta}{2r^2(\rho_s - \rho_m)g}

C.

vt=9r2(ρs+ρm)g2ηv_t = \frac{9r^2(\rho_s + \rho_m)g}{2\eta}

D.

vt=2η9r2(ρs−ρm)gv_t = \frac{2\eta}{9r^2(\rho_s - \rho_m)g}
Correct Answer: A

Solution:

According to Stokes' Law, the terminal velocity vtv_t of a sphere falling through a viscous medium is given by vt=2r2(ρs−ρm)g9ηv_t = \frac{2r^2(\rho_s - \rho_m)g}{9\eta}. This expression accounts for the balance of gravitational force, buoyant force, and viscous drag force acting on the sphere.

Chapter Concept:

Terminal Velocity

A.

5.15×106 Pa5.15 \times 10^6 \text{ Pa}

B.

6.16×106 Pa6.16 \times 10^6 \text{ Pa}

C.

5.16×106 Pa5.16 \times 10^6 \text{ Pa}

D.

6.15×106 Pa6.15 \times 10^6 \text{ Pa}
Correct Answer: C

Solution:

The absolute pressure is given by P=Pa+ρghP = P_a + \rho gh. Substituting Pa=1.01×105 PaP_a = 1.01 \times 10^5 \text{ Pa}, ρ=1.03×103 kg/m3\rho = 1.03 \times 10^3 \text{ kg/m}^3, g=9.8 m/s2g = 9.8 \text{ m/s}^2, and h=500 mh = 500 \text{ m}, we get P=1.01×105+1.03×103×9.8×500=5.16×106 PaP = 1.01 \times 10^5 + 1.03 \times 10^3 \times 9.8 \times 500 = 5.16 \times 10^6 \text{ Pa}.

Chapter Concept:

Pressure Variation with Depth

A.

The pressure at point A is higher than at point B

B.

The pressure at point A is lower than at point B

C.

The pressure at point A is equal to the pressure at point B

D.

The pressure at point A is zero
Correct Answer: B

Solution:

In a horizontal flow, according to Bernoulli's principle, an increase in velocity results in a decrease in pressure. Therefore, the pressure at point A is lower than at point B.

Chapter Concept:

Bernoulli's Principle

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True or False

Correct Answer: True

Solution:

Pressure is indeed a scalar quantity. Although it is defined as force per unit area, the 'force' in this context refers to the component of force normal to the surface, not a vector quantity.

Chapter Concept :

Pressure, Density and Relative Density

Correct Answer: True

Solution:

Surface tension is defined as the force per unit length acting at the interface between a liquid and another medium, related to the extra energy of surface molecules.

Chapter Concept :

Surface Tension

Correct Answer: True

Solution:

The pressure in a fluid indeed increases with depth because of the weight of the fluid above, as described by the equation P=Pa+ρghP = P_a + \rho gh, where PaP_a is the atmospheric pressure, ρ\rho is the fluid density, gg is the acceleration due to gravity, and hh is the depth.

Chapter Concept :

Pressure Variation with Depth

Correct Answer: False

Solution:

Bernoulli's equation is applicable to steady, incompressible flows with negligible viscosity. It does not hold for turbulent flows where velocity and pressure fluctuate.

Chapter Concept :

Bernoulli's Principle

Correct Answer: False

Solution:

The pressure inside a spherical drop is more than the pressure outside due to surface tension, as described by the equation (Pi−Po)=2Sr(P_i - P_o) = \frac{2S}{r}, where SS is the surface tension and rr is the radius of the drop.

Chapter Concept :

Atmospheric Pressure, Gauge Pressure and Manometer

Correct Answer: True

Solution:

Stokes' law states that the terminal velocity VtV_t of a sphere falling through a viscous medium is given by Vt=2a2(p−σ)g9ηV_t = \frac{2a^2 (p-\sigma)g}{9\eta}, where aa is the radius of the sphere. This shows that VtV_t is directly proportional to a2a^2.

Chapter Concept :

Terminal Velocity

Correct Answer: True

Solution:

Pascal's Law indeed states that a change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of the containing vessel.

Chapter Concept :

Pascal's Law

Correct Answer: True

Solution:

Torricelli's Law states that the speed of efflux, v1v_1, from an orifice is given by v1=2ghv_1 = \sqrt{2gh}, which is the same as the speed of a freely falling body from height hh.

Chapter Concept :

Torricelli's Law

Correct Answer: False

Solution:

Surface tension is defined as the force per unit length acting at the interface between a liquid and another medium, not per unit area.

Chapter Concept :

Surface Tension

Correct Answer: True

Solution:

Torricelli's Law states that the speed of efflux of a fluid from an orifice is the same as the speed of a freely falling body from the same height.

Chapter Concept :

Torricelli's Law

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