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Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET PDF Download

Archimedes' Principle 

  • Archimedes made a significant discovery regarding how objects behave when immersed in a fluid, whether partially or completely. 
  •  He found that such objects experience a buoyant force equal to the weight of the fluid they displace. 
  •  This observation led to the formulation of Archimedes' principle, which is fundamentally rooted in the principles of fluid statics
  •  When an object is submerged in a fluid, the fluid exerts pressure on the object's surface, creating a force perpendicular to the surface at every point. 
  •  This force is determined by the pressure at that point multiplied by the area of the surface
  •  The overall effect of these forces results in a net upward force known as buoyancy or upthrust. 

Understanding the Forces at Play

To comprehend how buoyancy works, let's consider a scenario where an object is immersed in a fluid with a certain density. 

The forces acting on the vertical sides of the object will balance each other out. However, the top and bottom surfaces of the object experience different forces due to the varying pressure at different depths.

  • Force on the Top Surface: The top surface of the object experiences a downward force due to the pressure at that depth. This force can be calculated using the formula:

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

where A is the area, h1 is the depth of the top surface, and (𝜎𝑔+𝑃0) represents the pressure at that point.

  • Force on the Bottom Surface: The bottom surface of the object experiences an upward force due to the pressure at that greater depth. This force is given by:

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

where h2 is the depth of the bottom surface.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

  • Since the pressure increases with depth (h> h1), the upward force (𝑭𝑢𝑝) will be greater than the downward force (𝑭𝑑𝑜𝑤𝑛).
  • This difference in forces leads to a net upward force on the object, calculated as:
  • 𝑭𝑛𝑒𝑡 = 𝑭𝑢𝑝𝑭𝑑𝑜𝑤𝑛 = 𝑉ρ (g2g1)
  • This net force can also be expressed in terms of the volume of fluid displaced by the object.
  • If L is the height of the object, then:
  • 𝑭𝑛𝑒𝑡 = Vρ = gρ
  • where V is the volume of fluid displaced.
  • This force, representing the weight of the displaced fluid, is what we refer to as buoyancy or upthrust.
  • Buoyancy acts vertically upwards, opposing the weight of the object, and its point of action is through the center of gravity of the displaced fluid.
  • This principle explains why objects feel lighter in water and why some objects float while others sink.
  • Though we have derived this result for a body fully submerged in a fluid, it can be shown to hold good for partly submerged bodies or a body in more than one fluid also.
  • Upthrust is independent of all factors of the body such as its mass, size, density, etc. except the volume of the body inside the fluid.
  • Upthrust depends upon the nature of displaced fluid. This is why upthrust on a fully submerged body is more in seawater than in fresh water because its density is more than fresh water.
  • Apparent weight of the body of density ρ when immersed in a liquid of density σ.
    Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

  • If a body of volume V is immersed in a liquid of density σ, then its weight reduces.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

  • Relative density of a body (R.D.) is the ratio of the density of the body to the density of water. It can also be defined as:

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Floatation

(1) Translatory equilibrium

When a body with density ρ and volume V is immersed in a liquid with density σ, the forces acting on the body are:

  • The weight of the body: W = mg = V ρ g, which acts vertically downward through the center of gravity of the body.

  • The upthrust force: V σ g, which acts vertically upward through the center of buoyancy (the center of gravity of the displaced liquid).

(a) If density of body is greater than that of liquid ρ>σ:

  • Weight will be more than upthrust so the body will sink.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

(b) If density of body is equal to that of liquid ρ=σ:

  • Weight will be equal to upthrust so the body will float fully submerged in neutral equilibrium with its top surface in it just at the top of the liquid.
    Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

(c) If density of body is lesser than that of liquid ρ<σ:

  • Weight will be less than upthrust so the body will move upwards, and in equilibrium, will float partially immersed in the liquid. 

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

W = Vinρg

⇒V ρg = Vinσg

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

(2) Rotational Equilibrium

  •   When a floating body is slightly tilted from its equilibrium, the center of buoyancy B shifts. 
  •  The new vertical line through the shifted center of buoyancy B′ intersects the original vertical line at a point called the meta-center M
  •  If this meta-center is located above the center of gravity, the forces at G (weight of the body W) and B′ (upthrust) create a restoring couple that returns the body to its original position. 
  •  Therefore, for a floating body to maintain rotational equilibrium, the meta-center must always be higher than the center of gravity. 

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

  • If the meta-center falls below the center of gravity, the forces at G (weight) and B′ (upthrust) will create a couple that causes the floating body to topple.
  • That is why a wooden log cannot be made to float vertical in water or a boat is likely to capsize if the sitting passengers stand on it.
  • In these situations, CG becomes higher than MG and so the body will topple if slightly tilted.

Applications of floatation

(i) When a body floats, the weight of the body equals the upthrust. Therefore:

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Thus, the volume outside the liquid is:

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

This gives the fraction of volume outside the liquid:

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

(ii) For flotation:

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

(iii) If the same body floats in different liquids with densities σA and σB, respectively:

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Streamline, Laminar and Turbulent Flow

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET(i) Streamline Flow

 Streamline flow of a liquid occurs when each element of the liquid passing through a point follows the same path and has the same velocity as the preceding element at that point.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

  • A streamline is the path, straight or curved, where the tangent at any point indicates the direction of the flow of liquid at that point.
  • Two streamlines cannot intersect, and when streamlines are closer together, it indicates a higher velocity of the liquid particles at that location.
  • The path ABC in the figure represents a streamline, with v1, v2, and v3 being the velocities of the liquid at points A, B, and C, respectively.

(2) Laminar Flow

Laminar flow occurs when a liquid flows over a horizontal surface in steady layers with different velocities that do not mix with one another. This type of flow is referred to as laminar flow.

In laminar flow, the velocity of the liquid is always lower than its critical velocity. Laminar flow is often used interchangeably with streamlined flow.

(3) Turbulent Flow

 When a liquid flows at a speed higher than its critical velocity, the motion of its particles becomes chaotic or irregular. This type of flow is called turbulent flow.

  • In turbulent flow, both the path and velocity of liquid particles change continuously and unpredictably over time. 
  • Most of the external energy that maintains turbulent flow is used to create swirling motions (eddies) within the liquid, leaving only a small amount of energy for forward motion.
  • For example, eddies can be seen around the pillars of a bridge over a river.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Critical Velocity and Reynold's Number

The critical velocity is the speed up to which the flow of liquid remains streamlined, and above this velocity, the flow becomes turbulent.

Reynolds number is a dimensionless number that indicates the nature of the flow of a liquid through a pipe.

It is defined as the ratio of the inertial force per unit area to the viscous force per unit area for a fluid in motion.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

If a liquid with density ρ\rhoρ flows through a tube of radius rrr and cross-sectional area AAA, the mass of the liquid flowing per second is given by:

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

If the value of Reynold's number

(i) When the Reynolds number is between 0 and 2000, the liquid flow is streamlined or laminar.

(ii) When the Reynolds number is between 2000 and 3000, the liquid flow becomes unstable and transitions from streamlined to turbulent.

(iii) When the Reynolds number is above 3000, the liquid flow is definitely turbulent.

Equation of Continuity

The equation of continuity is derived from the principle of conservation of mass.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

  • This expression is known as the equation of continuity for the steady flow of an incompressible and non-viscous liquid.
  • The velocity of the flow does not depend on the type of liquid (assuming it is non-viscous).
  • The velocity of the flow increases when the cross-sectional area decreases and decreases when the cross-sectional area increases. This explains:
    (a) In hilly regions, where rivers are narrow and shallow (i.e., small cross-section), the water flows faster. In contrast, in plains, where rivers are wider and deeper (i.e., large cross-section), the water flows slower, making the deeper water appear still.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET 

      (b) When water falls from a tap, the velocity of the falling water increases with distance due to the action of gravity (i.e., v> v1). According to the equation of continuity, the cross-sectional area of the water stream decreases (i.e., A2 < A1), meaning the stream of falling water becomes narrower.

Energy of a Flowing Fluid

A flowing fluid in motion possesses the following three types of energy:

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

If you hang two balloons vertically with separate threads and blow out air with your mouth in between the gap of two balloons, to your surprise the balloons will come closer to each other rather than going far off. This is an illustration of Bernoulli's Equation, which we are going to study in detail in this document.

When the person blows air in between the balloons, the pressure drops and they come closer.When the person blows air in between the balloons, the pressure drops and they come closer.

Bernoulli's Principle

It states that: 

The sum of pressure energy, kinetic energy and potential energy per unit volume of an incompressible, non-viscous fluid in a streamlined irrotational flow remains constant along a streamline.

Mathematically, it can be expressed as, Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Question for Mechanical Properties of Fluids: Part 2
Try yourself:
Which of the following best describes Bernoulli's principle?
View Solution

Proof of Bernoulli's Principle 

Assumptions:

  • The fluid is incompressible, non-viscous, non-rotational, and streamlined flow

Bernoulli`s TheoremBernoulli's Theorem

  • Mass of the fluid entering from side S
    dm1 = p A1 dx1 = p dV1
  • The work done in this displacement dx1 at point S is
    Wp1 = F1dx1 = P1A1dx1
    Wp1 = P1dV1    {∵ A1dx1 = dV1}


At the same time, the amount of fluid that moves out of the tube at point T is dm2 = pdV2

According to the equation of continuity  Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

The work done in the displacement of dm2 mass at point T

Wp2 = P2dV2

Now Apply the Work - Energy Theorem,
Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEETMechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET
Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET
Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEETMechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET
h = Gravitational head
Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET = volume head

Question for Mechanical Properties of Fluids: Part 2
Try yourself:
Which assumption is NOT required for Bernoulli's principle to hold?
View Solution

Bernoulli’s Equation at Constant Depth

When the fluid moves but its depth is constant—that is, 1=ℎ2.

Under that condition, Bernoulli’s equation becomes:

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Bernoulli’s Equation for Static Fluids

When the fluid is static, then v1 = v2 = 0, then Bernoulli’s equation is given as:

When v1 = v2 = 0P1 + ρgh1 = P2 + ρgh2
When h2 = 0P2 = P1 + ρgh1

Application of Bernoulli's principle 

1. Magnus effect

  • When a spinning ball is thrown, it deviates from its usual path in flight. This effect is called the Magnus effect and plays an important role in tennis, cricket, soccer, etc., as by applying appropriate spin the moving ball can be made to curve in any desired direction. 
  •  If a ball is moving from left to right and also spinning about a horizontal axis perpendicular to the direction of motion as shown in the figure, then relative to the ball air will be moving from right to left. 

(A)(A)            (B)(B)             (C)(C)

  • The resultant velocity of air above the ball will be V + rw while below it V - rw (shown in figure).
  • So in accordance with Bernoulli's principle pressure above the ball will be less than below it.
  • Due to this difference of pressure, an upward force will act on the ball and hence the ball will deviate from its usual path OA0 and will hit the ground at A1 following the path OA1 (figure shown).
  • i.e., if a ball is thrown with backspin, the pitch will curve less sharply prolonging the flight.
  • Similarly, if the spin is clockwise, i.e., the ball is thrown with topspin, the force due to pressure difference will act in the direction of gravity and so the pitch will curve more sharply shortening the flight.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Furthermore, if the ball is spinning about a vertical axis, the curving will be sideways as shown in the figure. producing the so-called out swing or in the swing.

2. Action of Atomiser

  •  The action of the aspirator, carburetor, paint gun, scent spray, or insect sprayer is based on Bernoulli's principle
  •  In all these, by means of motion of a piston P in a cylinder C, high-speed air is passed over a tube T dipped in liquid L to be sprayed. 
  •  High-speed air creates low pressure over the tube, due to which liquid (paint, scent, insecticide, or petrol) rises in it and is then blown off in very small droplets with expelled air. 

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

3. Working of Aeroplane

  • This is also based on Bernoulli's principle.
  • The wings of the aeroplane have tapering as shown in the figure.
  • Due to this specific shape of wings when the airplane runs, air passes at higher speed over it as compared to its lower surface.
  • This difference of air speeds above and below the wings, in accordance with Bernoulli's principle, creates a pressure difference, due to which an upward force called 'dynamic lift' ( = pressure difference × area of the wing) acts on the plane.
  • If this force becomes greater than the weight of the plane, the plane will rise up.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Question for Mechanical Properties of Fluids: Part 2
Try yourself:
Which principle explains the deviation of a spinning ball in flight?
View Solution

Example 1: If pressure and velocity at point A are P1 and V1 respectively & at point B is P2, V2 is the figure as shown. Comment on P1 and P2.  

Sol:  From the equation of continuity A1V1 = A2V2

here A1 > A2

⇒ V1 < V2 ....(1)

from Bernoulli's equation. We can write

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

 Torricelli's Law of Efflux (Velocity of efflux)

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

The cross-sectional area of the hole at A is greater than B If water comes in the tank with velocity vA and goes outside with velocity

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET
On Applying Bernoulli's theorem at A and B,
Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET 

Range (R) 

Let us find the range R on the ground.

Considering the vertical motion of the liquid.

(H -h) = Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET or Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Now, considering the horizontal motion,

R = vt or Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET or Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

From the expression of R, the following conclusions can be drawn,

(i) Rh = RH -h 

as Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET and Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

This can be shown in Figure.

(ii) R is maximum at H/2 and R max = H.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Proof R= 4 (H h – h2)
For R to be maximum, Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

or H - 2h = 0 or h = H/2

That is, R is maximum at h=H/2

and Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET 

Example 2: A cylindrical dark 1 m in radius rests on a platform 5 m high. Initially, the tank is filled with water up to a height of 5 m. A plug whose area is 10–4 m2 is removed from an orifice on the side of the tank at the bottom Calculate 

(a) the initial speed with which the water flows from the orifice 

(b) the initial speed with which the water strikes the ground and 

(c) time taken to empty the tank to half its original volume 

(d) Does the time to empty the tank depend upon the height of the stand?

Sol: The situation is shown in the figure:

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET
(a) As the speed of flow is given by

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET
Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET
(b) As the initial vertical velocity of water is zero, so is its vertical velocity when it hits the ground
Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

So the initial speed with which water strikes the ground.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

(c) When the height of the water level above the hole is y, the velocity of flow will be Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET and so the rate of flow

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET  Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Which on integration improper limits gives

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET 

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET
 = 9.2 × 103s – ~ 2.5 h

(d) No, as an expression of t is independent of the height of the stand.

 4. Venturimeter

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

The figure shows a venturi meter used to measure flow speed in a pipe of non-uniform cross-section. 

We apply Bernoulli's equation to the wide (point 1) and narrow (point 2) parts of the pipe, with h1 = h2

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET
From the continuity equation

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET
Substituting and rearranging, we get
Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Because A1 is greater than A2, v2 is greater than v1 and hence the pressure P2 is less than P1

A net force to the right acceleration the fluid as it enters the narrow part of the tube (called the throat) and a net force to the left slows as it leaves. 

The pressure difference is also equal to ρgh, where h is the difference in liquid level in the two tubes. Substituting in Eq. (1), we get

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

5. Pivot Tube

  • It is a device used to measure the flow velocity of fluid
  • It is a U-shaped tube that can be inserted in a tube or in the fluid-flowing space as shown in the figure shown. 
  •  In the U tube, a liquid that is immiscible with the fluid is filled up to a level C and the short opening M is placed in the fluid flowing space against the flow. 
  • This allows a few of the fluid particles to enter into the tube and exert a pressure on the liquid in limb A of U tube. 
  •  Due to this, the liquid level changes as shown in the figure shown. 

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

  • At end B fluid is freely flowing, which exerts approximately negligible pressure on this liquid. The pressure difference at ends A and B can be given by measuring the liquid level difference h as it is a gas, then  PA - PB = ρhg
  • It is the liquid of density, then PA - PB = h(ρ - ρg)g
  • Now if we apply Bernoulli's equation at the ends of A and B we'll have

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Now by using equations, we can evaluate the velocity v, with which the fluid is flowing.

Note: A pivot tube is also used to measure the velocity of airplanes with respect to the wind. It can be mounted at the top surface of the plain and hence the velocity of wind can be measured with respect to the plane.

6. Siphon

It is a pipe used to drain liquid at a lower height but the pipe initially rises and then comes down 

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

let velocity of outflow be v and the pipe is of uniform cross-section A. Applying Bernoulli's Equation between P (top of the tank) and R (Opening of pipe) we get 

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

here velocity is considered zero at P since the area of the tank is very large compared to the area of the pipe

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Naturally for siphon to work 

h> 0

Now as the area of the pipe is constant by the equation of continuity

as Av = constant the velocity of flow inside the siphon is also constant between Q and R

(P + pgh + 1/2 pv2)Q = (P + pgh + 1/2 pv2)R ⇒ PQ + pgh2 = P0 - pgh1 (v is same)

⇒ PQ = P0 - Pg (h1 + h2) as PQ = 0    ⇒ P0 > pg (h1 + h2) means (h1 + h2) should not be more than P/pg for siphon to work.

Viscosity

  • The property of a fluid by virtue of which an internal frictional force acts between its different layers which opposes their relative motion is called viscosity.
  • These internal frictional force is called viscous force.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

  • Viscous forces are inter-molecular forces acting between the molecules of different layers of liquid moving with different velocities.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

where, (dv/dx) = rate of change of velocity with distance called velocity gradient, A = area of cross-section and = coefficient of viscosity.

  • SI unit of η is Nsm-2 or pascal-second or decapoise. Its dimensional formula is [ML-1T-1].
  • The knowledge of the coefficient of viscosity of different oils and its variation with temperature helps us to select a suitable lubricant for a given machine.
  • Viscosity is due to transport of momentum. The value of viscosity (and compressibility) for ideal liquid is zero.
  • The viscosity of air and of some liquids are utilised for damping the given parts of some instruments.
  • The knowledge of viscosity of some organic liquids is used in determining the molecular weight and shape of large organic moleculars like proteins and cellulose.

Variation of Viscosity

  • The viscosity of liquids decreases with increase in temperature

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

where, η0 and ηt is are coefficient of viscosities at 0°C and t°C, α and β are constants.

  • The viscosity of gases increases with increase in temperatures as η ∝ √T
  • The viscosity of liquids increases with increase in pressure but the viscosity of water decreases with increase in pressure.
  • The viscosity of gases do not changes with pressure.

Poiseuille’s Formula

The rate of flow (v) of liquid through a horizontal pipe for steady flow is given by

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

where, p = pressure difference across the two ends of the tube. r = radius of the tube, n = coefficient of viscosity.

This is known as Poiseuille's equation.

The Rate of Flow of Liquid

Rate of flow of liquid through a tube is given by v = (P/R)

where, R = (8 ηl/πr4), called liquid resistance and p = liquid pressure.

  • Series combination of Tubes
    Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

  •  Parallel combination of tubes

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

Stoke's Law

When a sphere moves through a thick liquid, like honey, the resistance it faces depends on how fast it's moving, how big the sphere is, and how thick the liquid is. This resistance is called Stokes' drag. 

Imagine stirring honey with a spoon—the harder and faster you stir (velocity), the bigger the spoon (radius), and the thicker the honey (viscosity), the more resistance you feel. 

Illustration of Stoke`s LawIllustration of Stoke's Law

This resistance force is calculated using a formula that takes into account the viscosity of the liquid, the size of the sphere, and how fast it's moving.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

  • where F is the frictional force — known as Stokes' drag — acting on the interface between the fluid and the particle
  • η is the viscosity
  • R is the radius of the spherical object
  • V is the flow velocity

Tips and Tricks to Learn:

  • When a liquid is in equilibrium, the force on its surface is perpendicular everywhere.
  • In a liquid, the pressure is uniform at the same horizontal level.
  • Pressure is the same in all directions at any point within a liquid.
  • The pressure is perpendicular to the surface of the fluid.
  • The pressure at any point in a liquid depends on its depth (hhh), density, and acceleration due to gravity.
  • Pressure is independent of the shape of the container or the total mass of the liquid.
  • Force is a vector quantity, while pressure is a tensor quantity.
  • Pressure and density in fluids play the same role as force and mass do in solids.
  • Bar and millibar are units used for measuring pressure in meteorology.
  • A sudden drop in atmospheric pressure indicates a storm.
  • The first scientific weather forecast was made using a water barometer constructed in the 17th century by Von Guericke, who predicted a storm after a sudden fall in the water column height.

Mechanical Properties of Fluids: Part 2 | Physics Class 11 - NEET

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FAQs on Mechanical Properties of Fluids: Part 2 - Physics Class 11 - NEET

1. What is Archimedes' Principle and how does it relate to buoyancy?
Ans. Archimedes' Principle states that an object submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced by the object. This principle explains why objects float or sink in fluids based on their density compared to the fluid. If the object's density is less than that of the fluid, it will float; if it is greater, it will sink.
2. What are the differences between streamline, laminar, and turbulent flow in fluids?
Ans. Streamline flow, also known as laminar flow, occurs when fluid particles move in parallel layers with minimal disruption between them, resulting in smooth and orderly motion. Turbulent flow, on the other hand, is characterized by chaotic changes in pressure and flow velocity, causing eddies and vortices. The transition between laminar and turbulent flow depends on the Reynolds number, which relates to fluid velocity, density, viscosity, and characteristic length.
3. How does the Equation of Continuity apply to fluid flow?
Ans. The Equation of Continuity states that for an incompressible fluid, the mass flow rate must remain constant from one cross-section of a pipe to another. Mathematically, it is expressed as A1V1 = A2V2, where A is the cross-sectional area and V is the fluid velocity. This means that if the area of the pipe decreases, the velocity of the fluid must increase to maintain a constant flow rate.
4. What is Bernoulli's Principle and what are its applications in fluid mechanics?
Ans. Bernoulli's Principle states that in a flowing fluid, an increase in velocity occurs simultaneously with a decrease in pressure or potential energy. This principle is applied in various situations, such as explaining how airplane wings generate lift, the operation of carburetors, and the functioning of venturi meters. It highlights the conservation of energy in fluid flow.
5. How can Bernoulli's Equation be used for static fluids?
Ans. Bernoulli's Equation can be modified for static fluids by considering the pressure exerted by the fluid column. In this case, the equation simplifies to P + ρgh = constant, where P is the pressure at a point, ρ is the fluid density, g is the acceleration due to gravity, and h is the height of the fluid above a reference level. This form helps in calculating pressures at different depths in a fluid and is essential in hydrostatics.
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