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 2 Unit and Measurements
Unit and Measurements
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Class 11 Physics Chapter 3 Motion In A Straight Line
Section Name Topic Name 3 Motion in a Straight Line 3.1 Introduction 3.2 Position, path length and displacement 3.3 Average velocity and average speed 3.4 Instantaneous velocity and speed 3.5 Acceleration 3.6 Kinematic equations for uniformly accelerated motion 3.7 Relative velocity
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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 &amp; 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 8 Gravitation
Section Name Topic Name 8 Gravitation 8.1 Introduction 8.2 Kepler’s laws 8.3 Universal law of gravitation 8.4 The gravitational constant 8.5 Acceleration due to gravity of the earth 8.6 Acceleration due to gravity below and above the surface of earth 8.7 Gravitational potential energy 8.8 Escape speed 8.9 Earth satellite 8.10 Energy of an orbiting satellite 8.11 Geostationary and polar satellites 8.12 Weightlessness
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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 10 Mechanical Properties of Fluids
Section Name Topic Name 10 Mechanical Properties Of Fluids 10.1 Introduction 10.2 Pressure 10.3 Streamline flow 10.4 Bernoulli’s principle 10.5 Viscosity 10.6 Reynolds number 10.7 Surface tension
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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 12 Thermodynamics
Section Name Topic Name 12 Thermodynamics 12.1 Introduction 12.2 Thermal equilibrium 12.3 Zeroth law of thermodynamics 12.4 Heat, internal energy and work 12.5 First law of thermodynamics 12.6 Specific heat capacity 12.7 Thermodynamic state variables and equation of state 12.8 Thermodynamic processes 12.9 Heat engines 12.10 Refrigerators and heat pumps 12.11 Second law of thermodynamics 12.12 Reversible and irreversible processes 12.13 Carnot engine
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Class 11 Physics Chapter 13 Kinetic Theory
Section Name Topic Name 13 Kinetic Theory 13.1 Introduction 13.2 Molecular nature of matter 13.3 Behaviour of gases 13.4 Kinetic theory of an ideal gas 13.5 Law of equipartition of energy 13.6 Specific heat capacity 13.7 Mean free path
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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 11 Physics Chapter 15 Waves
Section Name Topic Name 15 Waves 15.1 Introduction 15.2 Transverse and longitudinal waves 15.3 Displacement relation in a progressive wave 15.4 The speed of a travelling wave 15.5 The principle of superposition of waves 15.6 Reflection of waves 15.7 Beats 15.8 Doppler effect
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Class 11th Physics Online Class For 100% Result

Expression for Acceleration due to gravity

• Consider any object of mass ‘m’ at a point A on the surface of the earth.
• The force of gravity between the body and earth can be calculated as
• F =G m M/Re2 (1) where
• m=mass of the body
• Me = mass of the Earth
• Re= distance between the body and the earthis same as the radius of the earth
• Newton’s Second law states that
• F=ma (2)

Comparing the equations (1) and (2)

• F=m (G mMe/Re2)
• (G mMe/Re2) is same as g(acceleration due to gravity)
• Therefore, the expression forAcceleration due to gravity.
• g=G Me/Re2

Problem: Calculate the acceleration due to gravity on the (a) earth’s surface (b) at a height of 1.5×10m from the earth surface .Given: Re=6.4×106m; M= 5.98×1024 kg?

(a) g= GMe/Re2

= 6.67×10-11×5.98×1024/(6.4 x106)2

= 9.74m/s2

(b)  h= 1.5×105m (given)

g (h)= GMe/(Re+h)2

= 6.67×10-11×5.98×1024/ (6.4 x106+ 1.5×105)

=9.30m/s2

Problem:- Calculate the mass of the moon if the free fall acceleration near its surface is known to be 1.62m/s2.Radius of the moon is 1738km?

g=1. 62m/s2

Rm = 1738km =1738 x 103 m

G=6.67×10-11Nm2/kg2

g= GMm/Rm2

Mm=gRm2/G

=1.62x (1738×103)2/6.67×10-11

=7.34×1022 kg

Acceleration due to gravity below the surface of earth

• To calculate acceleration due to gravity below the surface of the earth (between the surface and centre of the earth).
• Density of the earth is constant throughout. Therefore,
• ρ = M/ (4/3π Re3) equation(1)
• where
• M= mass of the earth
• Volume of sphere= 4/3π Re3
• Re = radius of the earth.
• As entire mass is concentrated at the centre of the earth.
• Therefore density can be written as

ρ = M/ (4/3π Rs3)   equation (2)

• Comparing equation (1) and (2)
• Me/Ms = Re3/Rs3 where Rs = (Re-d) 3
• d= distance of the body form the centre to the surface of the earth.
• Therefore,
• Me/Ms = Re3/(Re-d)3
• Ms = Me(Re-d)3/Re3 from equation(3)
• To calculate Gravitational force (F) between earth and point mass m at a depth d below the surface of the earth.

Above figure shows the value of g at a depth d. In this case only the smaller sphere of radius (Re–d) contributes to g.

• F= G m Ms/(Re-d)2
• g=F/m where g= acceleration due to gravity at point d below the surface of the earth.
• g= GMs/(Re-d)2
• Putting the value of Ms from equation (3)

= GM(Re-d)3/Re3(Re-d)2

=GM(Re-d)/Re3

• We know g=GMe/Re2   equation (4)
• g(d) = GMe/ Re(1-d/Re)
• From equation (4)
• g (d) =  g(1-d/Re)

Acceleration due to gravity above the surface of earth

• To calculate the value of acceleration due to gravity of a point mass m at a height h above thesurface of the earth.

Above figure shows the value ofacceleration due to gravity g at a height h above the surface of the earth.

• Force of gravitation between the object and the earth will be
• F= G mMe/(Re+h)2  where
• m = mass of the object, Re = radius of the earth
• g(h) = F/m = GMe/(Re+h)2 = GMe/[Re2(1+h/Re)2]
• h << Re (as radius of the earth is very large)
• By calculating we will get,

g(h) = g(1-2h/Re)

• Conclusion: –The value of acceleration due to gravity varies on the surface, above the surface and below the surface of the earth.

Problem:-  ( Gravitation )

Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface?

Weight of a body of mass m at the Earth’s surface, W = mg = 250 N

Body of mass m is located at depth, d=1/2 Re

Where,

Acceleration due to gravity at depth g (d) is given by the relation:

g’= (1-d/Re)g

= (1- Re/2x Re) g= ½ g

Weight of the body depth d,

W’=mg’

=m 1/2g =1/2mg=1/2W

=1/2 x250=125N

Problem:- ( Gravitation )

A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

Answer:  Weight of the body, W = 63 N

Acceleration due to gravity at height h from the Earth’s surface is given by the relation:

g’=g/(1+h/Re)2

Where,

g = Acceleration due to gravity on the Earth’s surface

Re = Radius of the Earth

For h= Re /2

g’ = g/(1+ Re/2x Re)2 = g(1+1/2)2 = 4/9g

Weight of a body of mass m at height h is given as:

W’=mg’

=mx4/9g=4/9mg

=4/9W

=4/9×63=28N