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

Atmospheric Pressure

  • Pressure exerted by the weight of the atmosphere.
  • Atmosphere is a mixture of different gases. All these gas molecules together constitute some weight. By virtue of this weight there is some pressure exerted by the atmosphere on all the objects.
  • This pressure is known as atmospheric pressure.
  • Value of atmospheric pressure at sea level is 1.01*105
  • 1atm = 1.01*105Pa

Problem:- What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10–1 N m–1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.

Answer:-

Radius of the mercury drop, r = 3.00 mm = 3 × 10–3 m

Surface tension of mercury, S = 4.65 × 10–1 N m–1

Atmospheric pressure, P0 = 1.01 × 105 Pa

Total pressure inside the mercury drop

= Excess pressure inside mercury + Atmospheric pressure

=2S/r + P0

= (2×4.65×10-1)/(3×10-3)

= 1.0131 × 105 = 1.01 ×105 Pa

Excess pressure = 2S/r = (2×4.65×10-1)/3×10-3

= 310 Pa

How to measure atmospheric pressure?

  • Atmospheric pressure is measured by Mercury Barometer.
  • Mercury barometer consists of trough filled with mercury(Hg).There is a tube which also contains mercury and it is invertedinside the trough.
  • The one end of tube is closed and other end of the tube is placed inverted inside the trough.
  • The inverted tube which also contains mercury up to a certain level and the space above mercury in the tubeis occupied by the vapours of mercury. The pressure can be considered as 0 at this place.
  • The atmosphere will exert some atmospheric pressure on the mercury level as a result the level of mercury decreases in the trough andit increases in the tube.
  • This increase in level will determine how much pressure was exerted by the atmosphere.
  • The pressure exerted is directly ∝to the increase in the mercury column of the tube.
  • We can say that pressure at point A is same as pressure at point B.
  • Patm=hρg.
  • It is measured in terms of how many mm of Hg rose in the column.
  • Greater the height greater is the atmospheric pressure.
  • When the height in this column becomes 76cm Hg we can say that the pressure applied is equal to 1atm.

Units of Pressure:-

1.SI unit: Pascal (Pa)

  • Pressure is always measured by taking sea level as the reference level. At sea level P=1.01*105 Pa.
  1. Atmosphere (atm)
    • Reference level is at sea level.
    • Pressure equivalent of 76cm of Hg column
    • 1atm=76cm of Hg column
    • 1atm=1.01*105 Pa
    •  
  2. Torr
  • Pressure equivalent of 1mm of Hg column.
  • 1torr =133 Pa
  1. Bar
  • 1bar = 10Pa

Problem:- A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10–2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?

Answer

The weight that the soap film supports, W = 1.5 × 10–2 N

Length of the slider, l = 30 cm = 0.3 m

A soap film has two free surfaces.

Total length = 2l = 2 × 0.3 = 0.6 m

Surface tension, S = (Force or Weight)/2l

=(1.5×10-2)/0.6 = 2.5×10-2N/m

Therefore, the surface tension of the film is 2.5 × 10–2 N m–1

Gauge Pressure

  • Pressure difference between the system and the atmosphere.
  • From relation P=Pa+ ρgh where P= pressure at any point, Pa = atmospheric pressure.
  • We can say that Pressure at any point is always greater than the atmospheric pressure by the amount ρgh.
  • P-Pa=ρgh where
  • P =pressure of the system, Pa=atmospheric pressure,
  • (P-Pa) = pressure difference between the system and atmosphere.
  • hρg = Gauge pressure.

How to measure Gauge pressure

  • Gauge pressure is measured by Open Tube Manometer.
  • Open Tube Manometer is a U-shaped tube which is partially filled with mercury(Hg).
  • One end is open and other end is connected to some device where pressure is to be determined.This means it is like a system.
  • The height to which the mercury column will rise depends on the atmospheric pressure. Similarly depending on the pressure of the system the height of mercury in another tube rises.
  • The pressure difference between these two heights is the difference between the atmospheric pressure and system.
  • This difference in pressure is the gauge pressure.
  • Consider if the level of mercury column is same in both the U-tubes.
  • Patm=P, therefore the difference between the atmospheric pressure and the pressure of the system is 0.
  • Gauge Pressure is 0.
  • Patm = 760torr.

Problem:- What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10–2 N m–1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × 105 Pa).

Answer:

Excess pressure inside the soap bubble is 20 Pa;

Pressure inside the air bubble is

Soap bubble is of radius, r = 5.00 mm = 5 × 10–3 m

Surface tension of the soap solution, S = 2.50 × 10–2 Nm–1

Relative density of the soap solution = 1.20

Density of the soap solution, ρ = 1.2 × 103 kg/m3

Air bubble formed at a depth, h = 40 cm = 0.4 m

Radius of the air bubble, r = 5 mm = 5 × 10–3 m

1 atmospheric pressure = 1.01 × 105 Pa

Acceleration due to gravity, g = 9.8 m/s2

Hence, the excess pressure inside the soap bubble is given by the relation:

P=4S/r

=(4×2.5×10-2)/5×10-3

=20Pa

Therefore, the excess pressure inside the soap bubble is 20 Pa.

The excess pressure inside the air bubble is given by the relation:

P’=2S/r

=(2×2.5×10-2)/5×10-3

=10Pa

Therefore, the excess pressure inside the air bubble is 10 Pa.

At a depth of 0.4 m, the total pressure inside the air bubble

= Atmospheric pressure + hρg + P’

=1.01×105+0.4×1.2×103x9.8+10

=1.057×105Pa

=1.06×105Pa

Therefore, the pressure inside the air bubble is 1.06×105Pa.

Absolute Pressure

  • Absolute pressure is defined as the pressure above the zero value of pressure.
  • It is the actual pressure which a substance has.
  • It is measured against the vacuum.
  • Absolute pressure is measured relative to absolute zero pressure.
  • It is sum of atmospheric pressure and gauge pressure.
  • P =Pa+hρg where P = pressure at any point, Pa = atmospheric pressure and hρg= gauge pressure.
  • Therefore P =Pa + Gauge Pressure. Where P = absolute pressure.
  • It is measured with the help of barometer.

Problem:  The density of theatmosphere at sea level is 1.29 kg/m3.Assume that it does not change withaltitude. Then how high would theatmosphere extend?

Answer:

From equation: – P=Pa+ρgh

ρgh = 1.29 kg m–3 × 9.8 m s2 × h m = 1.01 × 105 Pa

∴ h = 7989 m ≈ 8 km

In reality the density of air decreases withheight. So does the value of g. The atmospheric

cover extends with decreasing pressure over100 km. We should also note that the sea level

atmospheric pressure is not always 760 mm ofHg. A drop in the Hg level by 10 mm or more isa sign of an approaching storm.

Problem:- At a depth of 1000 m in anocean (a) what is the absolute pressure?(b) What is the gauge pressure? (c) Findthe force acting on the window of 20 cm × 20 cm of a submarine at this

depth, the interior of which is maintainedat sea-level atmospheric pressure. (Thedensity of sea water is 1.03 × 103 kg m-3,g = 10m s–2.)

Answer:

Here h = 1000 m and ρ = 1.03 × 103 kg m-3.

(a) From Eq. P2 − P1= ρgh, absolute pressure

P = Pa + ρgh

= 1.01 × 105 Pa+ 1.03 × 103 kg m–3 × 10 m s–2 × 1000 m

= 104.01 × 105 Pa

≈ 104 atm

(b) Gauge pressure is P − Pa = ρgh = Pg

Pg = 1.03 × 103 kg m–3 × 10 ms2 × 1000 m

= 103 × 105 Pa

≈ 103 atm

(c) The pressure outside the submarine isP = Pa + ρgh and the pressure inside it isPa. Hence, the net pressure acting on thewindow is gauge pressure, Pg = ρgh. Sincethe area of the window is A = 0.04 m2, theforce acting on it is

F = Pg A = (103 × 105 Pa) × 0.04 m2 = 4.12 × 105 N

 

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