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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

Bernoulli’s Principle

  • For a streamline fluid flow, the sum of the pressure (P), the kinetic energy per unit volume (ρv2/2) and the potential energy per unit volume (ρgh) remain constant.
  • Mathematically:- P+ ρv2/2 + ρgh = constant
    • where P= pressure ,
    • E./ Volume=1/2mv2/V = 1/2v2(m/V) = 1/2ρv2
    • E./Volume = mgh/V = (m/V)gh = ρgh

Derive: Bernoulli’s equation


  1. Fluid flow through a pipe of varying width.
  2. Pipe is located at changing heights.
  3. Fluid is incompressible.
  4. Flow is laminar.
  5. No energy is lost due to friction:applicable only to non-viscous fluids.
  • Mathematically: –
  • Consider the fluid initially lying between B and D. In an infinitesimal timeinterval Δt, this fluid would have moved.
    • Suppose v1= speed at B and v2= speedat D, initial distance moved by fluid from to C=v1Δt.
    • In the same interval Δtfluid distance moved by D to E = v2Δt.
    • P1= Pressureat A1, P2=Pressure at A2.
    • Work done on the fluid atleft end (BC) W1 = P1A1(v1Δt).
    • Work done by the fluid at the other end (DE)W2 = P2A2(v2Δt)
  • Net work done on the fluid is W1 – W= (P1A1v1Δt− P2A2v2Δt)
  • By the Equation of continuity Av=constant.
    • P1Av1Δt – P2A2v2Δt where A1v1Δt =P1ΔV and A2v2Δt = P2ΔV.
  • Therefore Work done = (P1− P2) ΔVequation (a)
    • Part of this work goes in changing Kinetic energy, ΔK = (½)m (v22 – v12) and part in gravitational potential energy,ΔU =mg (h2 − h1).
  • The total change in energy ΔE= ΔK +ΔU = (½) m (v22 – v12) + mg (h2 − h1). (i)
  • Density of the fluid ρ =m/V or m=ρV
  • Therefore in small interval of time Δt, small change in mass Δm
    • Δm=ρΔV (ii)
  • Putting the value from equation (ii) to (i)
  • ΔE = 1/2 ρΔV (v22 – v12) + ρgΔV (h2 − h1)  equation(b)
  • By using work-energy theorem: W = ΔE
    • From (a) and (b)
    • (P1-P2) ΔV =(1/2) ρΔV (v22 – v12) + ρgΔV (h2 − h1)
    • P1-P2 = 1/2ρv22 – 1/2ρv12+ρgh2 -ρgh1(By cancelling ΔV from both the sides).
  • After rearranging we get,P1 + (1/2) ρ v12 + ρg h1 = (1/2) ρ v22 + ρg h2
  • P+(1/2) ρv2+ρg h = constant.
  • This is the Bernoulli’s equation.
Mechanical Properties of Fluids
Mechanical Properties of Fluids

Bernoulli’s equation: Special Cases

  1. When a fluid is at rest. This means v1=v2=0.
  • From Bernoulli’s equation P1 + (1/2) ρ v12 + ρg h1 = (1/2) ρ v22 + ρg h2
  • By puttingv1=v2=0 in the above equation changes to
    • P1-P2= ρg(h2-h1). This equation is same as when the fluids are at rest.
  1. When the pipe is horizontal.h1=h2.This means there is no Potential energy by the virtue of height.
  • Therefore from Bernoulli’s equation(P1 + (1/2) ρ v12 + ρg h1 = (1/2) ρ v22 + ρg h2)
  • By simplifying,P+(1/2) ρ v2 = constant.


Water flows through a horizontal pipeline of varying cross-section.If the pressure of waterequals 6cm of mercury at a point where the velocity of flow is 30cm/s, what is the pressure at the another point where the velocity of flow is 50m/s?


At R1:- v1 = 30cm/s =0.3m/s

P1=ρg h=6×10-2x13600x9.8=7997N/m2

At R2:- v2=50cm/s=0.5m/s

From Bernoulli’s equation: – P+ (1/2) ρ v2+ρg h=constant

P1+ (1/2) ρ v12 = P2(1/2) ρ v22

7997+1/2x 1000x (0.3)2 = P2+1/2x 1000x (0.5)2


=ρg h2 = h2x13600x9.8

h2 = 5.9cmHg.

Torricelli’s law

  • Torricelli law states that the speed of flow of fluid from an orifice is equal to the speed that it would attain if falling freely for a distance equal to the height of the free surface of the liquid above the orifice.
  • Consider any vessel which has an orifice (slit)filled with some fluid.
  • The fluid will start flowing through the slit and according to Torricelli law the speed with which the fluid will flow is equal to the speed with which a freely falling bodyattains such that the height from which the body falls is equal to the height of the slit from the free surface of the fluid.
  • Let the distance between the free surface and the slit = h
  • Velocity with which the fluid flows is equal to the velocity with which a freely falling body attains if it is falling from a height h.

Derivation of the Law:-

  • Let A1= area of the slit (it is very small), v1= Velocity with which fluid is flowing out.
    • A2=Area of the free surface of the fluid,v2=velocity of the fluid at the free surface.
  • From Equation of Continuity, Av=constant.Therefore A1v1 = A2v2.
    • From the figure, A2>>>A1, This implies v2<<v1(This meansfluid is at rest on the free surface), Therefore v2~ 0.
  • Using Bernoulli’s equation,
    • P+ (1/2) ρ v2+ρgh = constant.
  • Applying Bernoulli’s equation at the slit:
    • Pa+(1/2) ρ v12+ρgy1(Equation 1) where Pa=atmospheric pressure,y1=height of the slit from the base.
  • Applying Bernoulli’s equation at the surface:
  • P+ρgy2(Equation 2) where as v2=0 therefore (1/2) ρ v12=0, y2=height of the free surface from the base.
  • By equating(1) and (2),
  • Pa+ (1/2) ρ v12+ ρgy1= P+ρgy2
  • (1/2) ρ v12 = (P-Pa) + ρg(y2-y1)
  • =(P-Pa) ρgh (where h=(y2-y1))
  • v12=2/ρ [(P-Pa) + ρgh]
  • Therefore v1=√2/ρ [(P-Pa) + ρgh].This is the velocity by which the fluid will come out of the small slit.
  • v1 is known as Speed of Efflux. This means the speed of the fluid outflow.
Mechanical Properties of Fluids
Mechanical Properties of Fluids

Case1:- The vessel is not closed it is open to atmosphere that means P=Pa.

  • Therefore v1=√2gh.This is the speed of a freely falling body.
  • This is accordance to Torricelli’s law which states that the speed by which the fluid is flowing out of a small slit of a container is same as the velocity of a freely falling body.
  • Case2:-Tank is not open to atmosphere but P>>Pa.
    • Therefore 2gh is ignored as it is very very large, hence v1= √2P/ρ.
    • The velocity with which the fluid will come out of the container is determined by the Pressure at the free surface of the fluid alone.

Problem: –Calculate the velocity of emergence of a liquid from a hole in the side of a wide cell 15cm below the liquid surface?


By using Torricelli’slaw v1=√2gh

=√2×9.8x15x10-2 m/s


Problem:- Torricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984kgm–3. Determine the height of the wine column for normal atmospheric pressure.


Density of mercury, ρ1 = 13.6 × 103kg/m3

Height of the mercury column, h1 = 0.76 m

Density of French wine, ρ2 = 984 kg/m3

Height of the French wine column = h2

Acceleration due to gravity, g = 9.8 m/s2

The pressure in both the columns is equal, i.e.

Pressure in the mercury column= Pressure in the French wine column

ρ1h1g = ρ2h2g

h2= ρ1h1/ ρ2

= (13.6×103x0.76)/984

= 10.5 m

Hence, the height of the French wine column for normal atmospheric pressure is 10.5 m.

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