Course Content
Class 11 Physics Chapter 1 Physical World
Section Name Topic Name 1 Physical World 1.1 What is physics? 1.2 Scope and excitement of physics 1.3 Physics, technology and society 1.4 Fundamental forces in nature 1.5 Nature of physical laws
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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
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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
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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
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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
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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
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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
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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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Class 11th Physics Online Class For 100% Result
About Lesson

Inertial and Gravitational Mass

Inertial Mass: – Inertial mass is defined as the mass of body by virtue of inertia of mass.

  • By Newton’s Law F=ma
  • m=F/a where m= inertial mass (as it is because of inertia of a body)

Gravitational Mass:-Gravitational mass is defined as the mass of the body by virtue of the gravitational force exerted by the earth.

  • By Gravitation Force of attraction –
    • F=GmM/r2
      • m=Fr2/GM where
  • m= mass of the object
  • F=force of attraction exerted by the earth
  • r=distance between object and earth
  • M=mass of the earth
  • Experimentally, Inertial mass= Gravitational mass

Problem: ( Class 11 Physics Gravitation )

Calculate the mass of the sun from the data given below:

Mean distance between Sun and Earth =1.5×1011m

Time taken by earth to complete one orbit around the sun = 1 year

Answer:

F= G MexMs/r2

Given: r= 1.5 x 1011m, v = 2π /r; T= 1 year =365 days =365x24x60x60sec

Fc= Mev2/r

F=Fc

GMeMs/r2 = Mev2/r = Me (2π r)/rT2

After calculation:-

Ms= 4n2r3/GT2

Ms=2×1030 kg.

Problem: ( Gravitation )

Let us assume that our galaxy consists of 2.5 × 1011 stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be 105 ly.

Answer

Mass of our galaxy Milky Way, M = 2.5 × 1011 solar mass

Solar mass = Mass of Sun = 2.0 × 1036 kg

Mass of our galaxy, M = 2.5 × 1011 × 2 × 1036 = 5 ×1041 kg

Diameter of Milky Way, d = 105 ly

Radius of Milky Way, r = 5 × 104 ly

1ly = 9.46 × 1015 m

 r =5 × 10× 9.46 × 1015

= 4.73 ×1020 m

Since a star revolves around the galactic centre of the Milky Way, its time period is

given by the relation:

T= (4 π2r3/G M)1/2

= ((4 x(3.14)2x (4.73) 3x1060)/6.67×10-11x5x1041)1/2

= (39.48×105.82×1030/33.35)1/2

=(125.27×1030)1/2=1.12×1016 s

1year = 365x324x60x60s

1s=1/365x324x60x60 years

Therefore,

1.12×1016/365x324x60x60

=3.35×108 years.

Problem: – ( Gravitation )

 Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 108  m. Show that the mass of Jupiter is about one-thousandth that of the sun.

Answer

Orbital period of Io =Tlo = 1.769 days =1.769x24x60x60s

Orbital radius of Io = RIo = 4.22 × 108 m

Satellite Iis revolving around the Jupiter

Mass of the latter is given by the relation:

Mj=4 π2Rlo3/G Tlo2

Where,

Mj = Mass of Jupiter

G = Universal gravitational constant

Orbital radius of the Earth,

Te = 365.25 days = 365.25x24x60x60s

Orbital radius of the Earth,

Re = 1AU = 1.496×1011m

Mass of the Sun is given as:

Ms= 4 π2Re3/G Te2

Ms/ M= (4 π2Re3/G Te2) x (G Tlo2/4 π2Rlo3)

 = (Re3/ Rlo3) x (Tlo2/ Te2)

= (1.769x24x60x60s/365.25x24x60x60s)2 x (1.496×1011m/4.22 × 108 m)

=1045.04

Ms/ Mj ~ 1000

Ms ~ 1000 Mj

Hence, it can be inferred that the mass of Jupiter is about one-thousandth that of the Sun.

Problem: –  ( Gravitation )

How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 × 108 km.

Answer:

Orbital radius of the Earth around the Sun, r = 1.5 × 1011 m

Time taken by the Earth to complete one revolution around the Sun,

T = 1 year = 365.25 days= 365.25 × 24 × 60 × 60 s

Universal gravitational constant, G = 6.67 × 10–11 Nm2 kg–2

Thus, mass of the Sun can be calculated using the relation

M=4 π2r3/GT2

= (4x (3.14)2x(1.5×1011)3)/(6.67×10-11 x(365.25 × 24 × 60 × 60))

=133.24×10/6.64×104

=2.0×1030Kg

Hence, the mass of the Sun is 2 × 1030 kg.

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