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

Compressibility

  • Compressibility is the measure of compression of a substance.
  • Reciprocal of bulk modulus is termed as ‘Compressibility’.
  • Mathematically:
  • k=1/B = – (1/p) (ΔV/V)
  • It is denoted by ‘k’.
  • k(solids)<k(liquids)<k(gases)

Problem:-The average depth of Indian Ocean is about 3000 m.

 Calculate the fractional compression, ΔV/V, of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 × 109 N m–2. (Take g = 10 m s–2)

Answer: – The pressure exerted by a 3000 m column of water on the bottom layer

p = hρ g = 3000 m × 1000 kg m–3 × 10 m s–2

= 3 × 107 kg m–1 s-2

= 3 × 107 N m–2

Fractional compression ΔV/V, is

ΔV/V = stress/B = (3 × 107 N m-2)/ (2.2 × 109 N m–2)

= 1.36 × 10-2 or 1.36 %

Problem: The bulk modulus for water is 2.1GPa.Calculate the contraction in volume of 200ml of water is subjected to a pressure of 2MPa.

Answer:-

B=2.1GPa = 2.1 x109Pa.

V=200ml = 2×10-6ml.

P=2MPa = 200×106 Pa.

B=- (1/p) (ΔV/V) = ΔV = pV/B

            = (2×106x200x10-6)/ = 2.1 x109

B=0.19ml

Problem: – Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Answer:

Initial volume, V1 = 100.0l = 100.0 × 10–3 m3

Final volume, V2 = 100.5 l = 100.5 ×10–3 m3

Increase in volume, ΔV = V– V1 = 0.5 × 10–3 m3

Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 105 Pa

Bulk Modulus = Δp / ΔV/ V1 = Δp x V1/ ΔV

= (100×1.013 × 105x100x10-3)/0.5×10-3

=2.026×106 Pa

Bulk modulus of air= 1.0×105Pa

Therefore,

Bulk modulus of water/ Bulk modulus of air

=2.026×106/1.0×105 =2.026×104

This ratio is very high because air is more compressible than water.

Problem:-

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 103 kg m–3?

Answer:-

Let the given depth be h.

Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 105 Pa

Density of water at the surface, ρ1 = 1.03 × 103 kg m–3

Let ρ2 be the density of water at the depth h.

Let V1 be the volume of water of mass m at the surface.

Let V2 be the volume of water of mass m at the depth h.

Let ΔV be the change in volume.

ΔV = V1 – V2

=m (1/ ρ1 – 1/ ρ2)

Therefore, Volumetric strain = ΔV/ V1

=m (1/ ρ1 – 1/ ρ2) x ρ1/m

Therefore, ΔV/ V1 = (1- ρ1/ ρ2)     … (i)

Bulk modulus, B = p V1/ ΔV

ΔV/ V= p/B

Compressibility of water =1/B =45.8×10-11 Pa-1

Therefore, ΔV/ V1= 80×1.013×105x45.8×10-11 = 3.71×10-3    … (ii)

For equations (i) and (ii), we get:

1- ρ/ ρ2 = 3.71 x10-3

ρ2 = 1.03×103/(1-(3.71×10-3)

=1.034×103kgm-3

Therefore, the density of water at the given depth (h) is 1.034 × 103 kg m–3.

Problem: – Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Answer:-

 Hydraulic pressure exerted on the glass slab, p = 10 atm = 10 × 1.013 × 105 Pa

 Bulk modulus of glass, B = 37 × 109 Nm–2

Bulk modulus, B= p/ (ΔV/V)

Where,

ΔV/V = Fractional change in volume

Therefore,

ΔV/ V = p/B

=10×1.013×105

=2.73×10-5

Hence, the fractional change in the volume of the glass slab is 2.73 × 10–5.

Problem: – How much should the pressure on a litre of water is changed to compress it by 0.10%?

Answer:

Volume of water, V = 1 L

It is given that water is to be compressed by 0.10%.

Fractional change = ΔV/V =0.1/100×1 = 10-3

Bulk modulus, B= ρ/ ΔV/V

p=B x ΔV/V

Bulk Modulus of water, B = 2.2×109Nm-2

p=2.2×109x10-3

=2.2x106Nm-2

Therefore, the pressure on water should be 2.2 ×10Nm–2.

Problem: – The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression, ΔV/V, of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 × 109 N m–2. (Take g = 10 m s–2)

Answer:- The pressure exerted by a 3000 m column of water on the bottom layer

p = hρ g = 3000 m × 1000 kg m–3 × 10 m s–2

= 3 × 107 kg m–1 s-2

= 3 × 107 N m–2

Fractional compression ΔV/V, is

ΔV/V = stress/B = (3 × 107 N m-2)/ (2.2 × 109 N m–2)

= 1.36 × 10-2 or 1.36 %

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