Course Content
Class 11 Physics Chapter 4 Motion In A Plane
4 Motion in a plane 4.1 Introduction 4.2 Scalars and vectors 4.3 Multiplication of vectors by real numbers 4.4 Addition and subtraction of vectors – graphical method 4.5 Resolution of vectors 4.6 Vector addition – analytical method 4.7 Motion in a plane 4.8 Motion in a plane with constant acceleration 4.9 Relative velocity in two dimensions 4.10 Projectile motion 4.11 Uniform circular motion
Class 11 Physics Chapter 5 Laws of motion
Section Name Topic Name 5 Laws of motion 5.1 Introduction 5.2 Aristotle’s fallacy 5.3 The law of inertia 5.4 Newton’s first law of motion 5.5 Newton’s second law of motion 5.6 Newton’s third law of motion 5.7 Conservation of momentum 5.8 Equilibrium of a particle 5.9 Common forces in mechanics 5.10 Circular motion 5.11 Solving problems in mechanics
Class 11 Physics Chapter 6 Work Energy and Power
Section Name Topic Name 6 Work Energy and power 6.1 Introduction 6.2 Notions of work and kinetic energy : The work-energy theorem 6.3 Work 6.4 Kinetic energy 6.5 Work done by a variable force 6.6 The work-energy theorem for a variable force 6.7 The concept of potential energy 6.8 The conservation of mechanical energy 6.9 The potential energy of a spring 6.10 Various forms of energy : the law of conservation of energy 6.11 Power 6.12 Collisions
Class 11 Physics Chapter 7 Rotation motion
Topics Introduction Centre of mass Motion of COM Linear Momentum of System of Particles Vector Product Angular velocity Torque & Angular Momentum Conservation of Angular Momentum Equilibrium of Rigid Body Centre of Gravity Moment of Inertia Theorem of perpendicular axis Theorem of parallel axis Moment of Inertia of Objects Kinematics of Rotational Motion about a Fixed Axis Dynamics of Rotational Motion about a Fixed Axis Angular Momentum In Case of Rotation about a Fixed Axis Rolling motion
Class 11 Physics Chapter 9 mechanics properties of solid
Section Name Topic Name 9 Mechanical Properties Of Solids 9.1 Introduction 9.2 Elastic behaviour of solids 9.3 Stress and strain 9.4 Hooke’s law 9.5 Stress-strain curve 9.6 Elastic moduli 9.7 Applications of elastic behaviour of materials
Class 11 Physics Chapter 11 Thermal Properties of matter
Section Name Topic Name 11 Thermal Properties of matter 11.1 Introduction 11.2 Temperature and heat 11.3 Measurement of temperature 11.4 Ideal-gas equation and absolute temperature 11.5 Thermal expansion 11.6 Specific heat capacity 11.7 Calorimetry 11.8 Change of state 11.9 Heat transfer 11.10 Newton’s law of cooling
Class 11 Physics Chapter 14 Oscillations
Section Name Topic Name 14 Oscillations 14.1 Introduction 14.2 Periodic and oscilatory motions 14.3 Simple harmonic motion 14.4 Simple harmonic motion and uniform circular motion 14.5 Velocity and acceleration in simple harmonic motion 14.6 Force law for simple harmonic motion 14.7 Energy in simple harmonic motion 14.8 Some systems executing Simple Harmonic Motion 14.9 Damped simple harmonic motion 14.10 Forced oscillations and resonance
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About Lesson


  • Viscosity is the property of a fluid that resists the force tending to cause the fluid to flow.
  • It is analogous to friction in solids.
  • Example:-
    • Consider 2 glasses one filled with water and the other filled with honey.
    • Water will flow down the glass very rapidly whereas honey won’t.This is because honey is more viscous than water.
    • Therefore in order to make honey flow we need to apply greater amount of force. Because honey has the property to resist the motion.
  • Viscosity comes into play when there is relative motion between the layers of the fluid.The different layers are not moving at the same pace.

Coefficient of Viscosity

  • Coefficient of viscosity is the measure of degree to which a fluid resists flow under an applied force.
  • This means how much resistance does a fluid have to its motion.
  • Ratio of shearing stress to the strain rate.
  • It is denoted by ‘η’.
  • Mathematically
    • Δt=time , displacement =Δx
    • Therefore,
      • shearing stress =Δx/l where l= length
      • Strain rate=Δx/lΔt
    • η=shearing stress/strain rate
      • (F/A)/(Δx/lΔt) = (Fl)/vA where Δx/t=v
    • Therefore η=(Fl)/ vA
    • I. Unit:- Poiseiulle (PI)/Pa/Nsm-2
    • Dimensional Formula: [ML-1T-1]
Mechanical Properties of Fluids
Mechanical Properties of Fluids


The metal block moves to the rightbecause of the tension in the string. The tension

T is equal in magnitude to the weight of thesuspended mass m. Thus the shear force F is

F = T = mg = 0.010 kg × 9.8 ms–2 = 9.8 × 10-2 N

Shear stress on the fluid = F/A =9.8 x10–2/0.10

Strain rate =v/l =0.085/0.030

η= stress/strain rate

= (9.8×10-2x 0.30 x10-3m)/ (0.085ms-1x0.10m2)

= 3.45 × 10-3 Pa s

Stokes Law

  • The force that retards a sphere moving through a viscous fluid is directly ∝to the velocity and the radius of the sphere, and the viscosity of the fluid.
  • Mathematically:-F =6πηrv where
    • Let retarding force F∝v where v =velocity of the sphere
      • F ∝ r where r=radius of the sphere
      • F∝η where η=coefficient of viscosity
      • 6π=constant
    • Stokes law is applicable only to laminar flow of liquids.It is not applicable to turbulent law.
    • Example:-Falling raindrops
      • Consider a single rain drop, when rain drop is falling it is passing through air.
      • The air has some viscosity; there will be some force which will try to stop the motion of the rain drop.
      • Initially the rain drop accelerates but after some time it falls with constant velocity.
      • As the velocity increases the retarding force also increases.
      • There will be viscous force Fv and bind force Fbacting in the upward direction.There will also be Fggravitational force acting downwards.
      • After some time Fg = Fr (Fv+Fb)
      • Net Force is 0. If force is 0 as a result acceleration also becomes 0.
Mechanical Properties of Fluids
Mechanical Properties of Fluids

Terminal Velocity

  • Terminal velocity is the maximum velocity of a body moving through a viscous fluid.
  • It is attained when force of resistance of the medium is equal and opposite to the force of gravity.
  • As the velocity is increasing the retarding force will also increase and a stage will come when the force of gravity becomes equal to resistance force.
  • After that point velocity won’t increase and this velocity is known as terminal velocity.
  • It is denoted by ‘vt’.Wheret=terminal.
  • Mathematically:-
    • Terminal velocity is attained when Force of resistance = force due to gravitational attraction.
    • 6πηrv =mg
    • 6πηrv = densityxVg (Because density=m/V), density=ρ – σ where ρ and σ are the densities of the sphere and the viscous medium resp.
    • 6πηrv = (ρ – σ)x4/3πr3g where Volume of the sphere(V) =4/3πr3
    • By simplifying
      • =(ρ – σ)gx4/3r2x1/(6η)
    • vt =2r2(ρ – σ)g/9 η .This is the terminal velocity. Where(v=vt)
Mechanical Properties of Fluids
Mechanical Properties of Fluids

Problem: The terminal velocity of acopper ball of radius 2.0 mm fallingthrough a tank of oil at 20oC is 6.5 cm s-1.Compute the viscosity of the oil at 20oC.Density of oil is 1.5 × 103 kg m-3, density of

copper is 8.9 × 103 kg m-3.


Given: – vt = 6.5 × 10-2 ms-1, a = 2 × 10-3 m,g = 9.8 ms-2, ρ = 8.9 × 103 kg m-3,

σ =1.5 × 103 kg m-3. From Equation: -vt =2r(ρ – σ) g/9 η

=2/9(2 x10-3 m2 x 9.8ms-2/ 6.5×10-2ms-1) x 7.4 103 kgm-3

=9.9 x 10-1 kgm-1s-1

Problem: Calculate the terminal velocity in air of an oil drop of radius 2×105 m from the following data g=9.8m/s2; coefficient of viscosity of air =1.8×10-5Pas; density of oil=900kg/m3.The up thrust of air may be neglected?


Radius r =2×10-5 m



ρ =900kg/m3

v=vtwhen 6πηrv = mg

6πηrv =ρx4/3r3g

Simplifying: – vt =2r2gρ/9π = (2 x (2×10-5)2x9.8×900)/9×1.8×10-5


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