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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
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

Variation of pressure with depth

  • Consider a cylindrical object inside a fluid;consider 2 different positions for this object.
  • Fluid is at rest therefore the force along the horizontal direction is 0.
  • Forces along the vertical direction:-
    • Consider two positions 1 and 2.
    • Force at position 1 is perpendicular to cross sectional area A, F1= P1
    • Similarly F2=P2
  • Total force Fnet= -F1+F2 as F1 is along negative y axis therefore it is –ive. And F2 is along +ive y -axis.
  • Fnet =(P2-P1)A
  • This net force will be balanced by the weight of the cylinder(m).
  • Therefore under equilibrium condition
    • Fnet=mg=weight of the cylinder = weight of the fluid displaced.
    • =ρ Vg where ρ=density=volume of the fluid
    • =ρhAg   where V=hA(h=height and A= area)
  • Therefore (P2-P1) A=ρhAg
  • P2-P= ρhg,Therefore the difference in the pressure is dependent on height of the cylinder.
  • Consider the top of the cylinder exposed to air therefore P1=Pa(where Pa= P1 is equal to atmospheric pressure.)
  • Then P2=Pa+ ρhg
  • Conclusion: The pressure P, at depth below the surface of a liquid open to the atmosphere is greater than atmospheric pressure by an amount ρhg.
  • The pressure is independent of the cross sectional or base area or the shape of the container.
Mechanical Properties of Fluids
Mechanical Properties of Fluids

Problem:- What is the pressure on aswimmer 10 m below the surface of a lake?


 Here, h = 10 m and ρ = 1000 kg m-3. Take g = 10 m s–2

P = Pa + ρgh

= 1.01 × 105 Pa + 1000 kg m–3 × 10 m s–2 × 10 m

= 2.01 × 105 Pa

≈ 2 atm

This is a 100% increase in pressure fromsurface level. At a depth of 1 km the increase in

Pressure is 100 atm! Submarines are designedto withstand such enormous pressures.

Problem:- A vertical off-shore structure is built to withstand a maximum stress of 109 Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.

Answer:- Yes

The maximum allowable stress for the structure, P = 109 Pa

Depth of the ocean, d = 3 km = 3 × 103 m

Density of water, ρ = 103 kg/m3

Acceleration due to gravity, g = 9.8 m/s2

The pressure exerted because of the sea water at depth, d = ρdg

= 3 × 103 × 103 × 9.8 = 2.94 × 107 Pa

The maximum allowable stress for the structure (109 Pa) is greater than the pressure of the sea water (2.94 × 107 Pa). The pressure exerted by the ocean is less than the pressure that the structure can withstand. Hence, the structure is suitable for putting up on top of an oil well in the ocean.

Hydrostatic Paradox

  • Hydrostatic Paradox means: – hydro = water, static =at rest

            Paradox meansthat something taking place surprisingly.

  • Consider 3 vessels of very different shapes (like thin rectangular shape, triangular and some filter shape) and we have a source from which water enters into these 3 vessels.
  • Water enters through the horizontal base which is the base of these 3 vessels we observe that the level of water in all the 3 vessels is same irrespective of their different shapes.
  • This is because pressure at some point at the base of these 3 vessels is same.
  • The water will rise in all these 3 vessels till the pressure at the top is same as the pressure at the bottom.
  • As pressure is dependent only on height therefore in all the 3 vessels the height reached by the water is same irrespective of difference in their shapes.
  • This experiment is known as Hydrostatic Paradox.
Mechanical Properties of Fluids
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