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  As the particle is moving in the same way the projections are also moving.

• When the particle is moving in the upper part of circle then the projections start moving towards left.
• When the particle is moving in the lower part of the circle then the projections are moving towards right.
• We can conclude that the particle is swinging from left to right and again from right to left.
• This to and fro motion is SHM.

Answer: – (a) Time period, t = 2 s

Amplitude, A = 3 cm

At time, t = 0, the radius vector OP makes an angle π/2 with the positive x-axis, phase angle Φ = +π/2

Therefore, the equation of simple harmonic motion for the x-projection of OP, at the time t, is given by the displacement equation:

A= cos [(2 πt/T) + Φ]

=3cos (2 πt/2 + π/2) = -3sin (2πt/2)

=-3sinπt cm.

(b) Time Period, t = 4 s

Amplitude, a = 2 m

At time t = 0, OP makes an angle π with the x-axis, in the anticlockwise direction, Hence, phase angle Φ = +π

Therefore, the equation of simple harmonic motion for the x-projection of OP, at the time t, is given as:

=a cos [(2 πt/T) + Φ]

=2 cos [(2 πt/T) + π]

x=-2 cos (π/2 t) m

Velocity in Simple Harmonic Motion

• Uniform Circular motion can be defined as motion of an object in a circle at a constant speed.
• Consider a particle moving in circular path
• The velocity at any point P at any time t will be tangential to the point P.
• Consider θ = ωt+ φ where
• θ = angular position
• ω = angular velocity of the particle

ap = -ω2Awhere

• A = radius of the circle
• (-ive sign shows it is pointing towards the centre of the circle.)
• Consider the acceleration of the projection of the particle P’ on the x-axis.
• Accelerationwill be given as
• a(t) = -acosθ
• a(t) = – ap cos(ωt + φ)
• 2A cos (ωt + φ)
• a(t) = – ω2x(t) (Using x(t)= A cos (ωt + φ))

To verify expression for acceleration when calculated directly from SHM –

• Displacement in SHM is = A cos (ωt + φ)
• Velocity v(t) in SHM is = -Aw sin (ωt + φ)

Therefore,

• a(t) = dv/dt
• = – ω2A cos (ωt + φ)
• a(t) = – ω2x(t) (Using x(t)= A cos (ωt + φ))

Equation of acceleration of the particle which executesSHM:-

a(t) = – ω2x(t)

We can conclude that:-

1. a is proportional to displacement
2. acceleration is always directed towards the centre(in circular motion centre is mean position of the SHM)

From above we can say that

• SHM is the projection of the uniform circular motion such that centre of uniform circular motion becomes the mean position of the SHM and the radius of the circular motion is the amplitude of the SHM.

Problem: -A body oscillates with SHM according to the equation (in SI units),   x = 5 cos [2π t + π/4]

At t = 1.5 s, calculate the (a) displacement,(b) speed and (c) acceleration of the body?

The angular frequency ω of the body

= 2π s–1

and its time period T = 1 s.

At t = 1.5 s

(a) Displacement = (5.0 m) cos [(2π s–1)1.5 s + π/4]

= (5.0 m) cos [(3π + π/4)]

= –5.0 x 0.707 m

= –3.535 m

(b) The speed of the body

= – (5.0 m)(2π s–1) sin [(2π s–1) 1.5 s+ π/4]

= – (5.0 m) (2π s–1) sin [(3π + π/4)]

= 10π (0.707) m s–1

= 22 m s–1

(c) The acceleration of thebody

= – (2π s–1)2displacement

= – (2π s-1)2 (–3.535 m)

= 140 m s–2

F1 = -kx (force exerted by the spring onthe left side, trying to

pull the mass towards the mean position)

F2 = -kx   (force exerted by the spring onthe right side, trying to pull the mass towards the mean position)

The net force, F, acting on the mass is thengiven by,

F = –2kx

Therefore, the force acting on the mass is proportional to the displacement and is directedtowards the mean position; therefore, the motion executed by the mass is simple harmonic. The time period of oscillations is given as:-

T=2π√m/2k

Problem:-The acceleration due to gravity on the surface of moon is 1.7 ms–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 ms–2)

Answer: -Acceleration due to gravity on the surface of moon,g’ = 1.7 m s–2

Acceleration due to gravity on the surface of earth, g = 9.8 m s–2

Time period of a simple pendulum on earth, T = 3.5 s

T=2π√l/g

where l =length of the pendulum

l= T2/ (2π) 2x g

= (3.5)2/ (4x (3.14)2) x 9.8 m

The length of the pendulum remains constant,

On moon’s surface, time period, T’= 2π√l/g’

=2π ((3.5)2/4x(3.14)2 x 9.8)/1.7

=8.4s

Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.

Problem:-

(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle:

T = 2π√ (m/k). A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than 2π√ (l/g)

Think of a qualitative argument to appreciate this result.

(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?

(d) What is the frequency of oscillation of a simple pendulum mounted in a cabinthat is freely falling under gravity?

(a) For a simple pendulum, force constant or spring factor k is proportional to mass m; therefore, m cancels out in denominator as well as in numerator. That is why the time period of simple pendulum is independent of the mass of the bob.

(b) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:

F = –mg sinθ

where,

F = Restoring force

m = Mass of the bob

g = Acceleration due to gravity

θ = Angle of displacement

For small θ, sinθ≈θ

For large θ, sinθ is greater than θ.

This decreases the effective value of g.

Hence, the time period increases as:

T = 2π√ (l/g)

(c) Yes, because the working of the wrist watch depends on spring action and it has nothing to do with gravity.

(d) Gravity disappears for a man under free fall, so frequency is zero