CLASS 11 MECHANICAL PROPERTIES OF FLUID

by Admin
11-Dec-2025

MECHANICAL PROPERTIES OF FLUID

✤ FLUIDS :- It is defined as the thrust per unit area

P = Thrust/AREA P= F/A unit is Nm-²or Pascal (Pa)

NOTE: (1)P is a scalar Quantity.

② For small area P is maximum.

✤ Pascal's law: In the absence of Gravity (Neg. gravity) & liquid in rest every particle has same force (Pressure) at same horizontal in a plane!

It states that when pressure is applied on an ideal fluid then pressure is distributing uniformly in all possible direction.

✤ Application:

Hydraulic Lift: It is leased on Pascal's law & we can produce large force by applying small force

Pressure at smaller end

P₁= F / a

Pressure at bigger end

P₂=F/A

So F= f x (A/a)

 F>>f we have A>>a

✤ HYDRAULIC BREAK

Pressure Exerted by liquid Column:

Let a liquid of density filled in a container

Mass of liquid = vol x density

= πr²h x p

= Ahp

Weight of liquid = mg

= Ahpg

We know Pressure= Thrust/Area = Ahpg/A

Note: PXR P does not depend upon the vol. of fluid & it depends upon height.

If Vol. is same then

P₁>P₂>P₃

This phenomenon is known as Hydraulic Paradox

Hydrostatic

✤ AIR PRESSURE :- < Pressure exerted by Air >

(Atmospheric Pressure)

P= Pressure exerted by Hg

Column height 76 cm

P = p.g.h

= 13.6x9.8x76x10-2

P = 1.01x105 Pascal

* Question Solving tip:-

To find how much piston move out

L₁A₁ = L₂A₂ i.e. Volume=equal

1 torr = 1 mm Hg

1 atm = 101.3 KPa = 1.013 bar = 760 torr

1 atm = 1.013x10⁶ dyne/cm² = 1.013x10⁵ Pa

1 bar = 10⁶ dyne cm² = 10⁵ N/m²

1 millibar = 10³ bar = 10² dyne/cm² = 10² N/m²

FOR YOUR KNOWLEDGE :-

✤ The mean pressure on the walls of a vessel containing liquid upto height is hpg/2.

✤ Overpressure is seen in inflated tyres of the human circulatory the absolute pressure is greater than atmospheric pressure therefore Gauge pressure.

✤ Gauge pressure is negative when absolute pressure is less than atmospheric pressure.

✤ A diver in water at depth 10 m is under twice the atmospheric pressure.

✤ Systolic pressure is 120 mm Hg

✤ Diastolic pressure is 80 mm Hg

✤ When depth of sea is 100km, the increase in pressure 100 atm

✤ The pressure at the centre of the earth is estimated to be 3-billion atmosphere

✤ The atmospheric pressure is nearly 100 k Pa. The tyres of a car are usually inflated to a pressure 200 k Pa

✤ A drop in the atmospheric pressure by 10 mm of Hg osmosis a sighing an approaching storm.

✤ Density: It is defined as mass per unit volume. d = m/v Sl unit is kg/m³

✤ Relative Density: It is defined as the ratio of density of a given Substance & density of water. It is also known as specific gravity

Rd = ds/dH₂O

Note:-

Pressure on the object = P₀ + pgh

Gauge pressure: It is the difference b/w total pressure & atmosphere pressure.

Gauge Pressure = P - P₀

= P₀ + pgh - P₀

= pgh

✤ Height of atmosphere: It is the vertical distance from surface at where atmosphere is present.

We know,

P= Pgh

h = 1.01x10⁵ / Pg = 1.01x10⁵ / 1.3x9.8 = 101x10⁵ / 12.74 = 10100x10³ / 1294 = 8x10³ km

✤ Surface tension: It is defined as the ability of a fluid by which it Gain minimum surface area.

Surface tension is always equal to the force per unit length of tangent

On the free surface

S=P Sunit & N/m

Dim : MT-2

Application:

Needles supported on water surface.

Endless wet thread on a soap film

Rain drops are generally spherical in shape

Small mercury droplets are spherical and larger ones tend to flattened

The hair of a painting brush cling together when taken out of water

A leg floats on water due to surface tension

Oil spreads on cold water but remains a drop on hot water.

Surface Energy: →

This defined as the amount of work done against surface tension to increase surface area.

Let surface tension of the membrane is 's'

Then in δx displacement of PQ

Workdone is w, w= f x δx

we have s= f/l

⇒ f = s x l

⇒ w = s x l x δx

= sC x δA)

w= sxδA

The work is stored in the form of energy

E = s x δA SI unit : Joules

Excess Pressure: It is the difference inside pressure & outside pressure of a drop or lobule

✤ Excess Pressure = P₁-Po

Excess pressure inside a liquid drop: -

Let due to excess pressure 'P' radius drop increased to R+δR

:. work done by the pressure

w = F x δR

= P x A x δR

= P x 4πR² δR

Now change in surface area = 4π(R+δR)² - 4πR²

= 4π [R² + δR² + 2RδR - R²]

= 8πRδR [ Neg. δR²]

If surface tension is s then,

Surface Energy = s x 8πRδR — (1)

Surface Energy Occur due to work done

:. P x 4πR² δR = s x 8πR x δR

:. P.R= 2S

P = 2S/R

Excess Pressure inside a soap bubble:

In drop there are two free surfaces v,

Surface Energy = s x 16πR δR

P x 4πR² δR = s x 16πR δR

P = 4S/R

CAPILLARY :- It is a very fine test tube of Glass – Chair like.

The phenomenon of rise of liquid like Water is called Capillarity.

Main Principle of capillary wise is pressure difference.

Ascent formula:

Here Cos θ = r / R, r = R cos θ Notes:-

form (1) h = 2s cos θ/rpg -1 ≤ Cos θ ≤ 1

This is always smaller equal to 0.

Notes :- (1) h ∝ 1/r

h₁/h₂ = r2 / r1

(2) h ∝ S

h₁ / h₂ = S1 / S2

Height of liquid in insufficient tube :-

We have, h ∝ Cosθ In this case meniscus become flattened

& -1 ≤ Cosθ ≤ 1

So, h ≤ 1

Cohesive force :- It is the force of attraction which act b/w the atoms / molecules of same element

Example: Mercury (Hg)

Adhesive Force: - It is the force attraction b/w the atoms / molecules of two different element.

Example: H₂O & Glass

Angle of Contact:-

Adhesive > Cohesive Cohesive > Adhesive Cohesive = Adhesive

θ < 90° θ > 90° θ = 90°

Angle made by tangent to the meniscus with the wall of container

Hydrodynamics :-

(liquid in motion)

Viscous force : It is a self-adjusting backward dragging force which act b/w the layers of fluids. This property is called Viscosity.

Factors on which Viscous force depends:-

-F ∝ KA - (1)

-F ∝ p / dx - (2)

σ = W/2l mg/2l

radius is given

m = σ/mg = σ/pid

It two liquid drop on soap bubble are coalesce under isothermal conditions to form a single bubble, then the radius eq the bubble

formed

r² = r₁²+ r₂²

➤ If Glass plate is immersed in water then,

Apparent weight = Force acting vertically downward (Weight)+ Force due to Surface tension [σx2(l+t)] – Upthrust

(Volume of liquid displaced cxlxb/2xt ) xρxg

t stands for thickness

➤ If downward force is acting then P = 2(σ/t)xσ

If plate is placed vertical to the surface P = σx 2(l+t)

* For Your knowledge:-

✤ Tiny fog droplets that they are rigid enough to behave like solids & resist fairly large deforming forces.

✤ When an air bubble of radius R lies at a depth h below the free surface of a liquid of density ρ & surface tension σ, the excess pressure inside the bubble.

P = 2σ/R + hρg

✤ Pressure required to blow the bubble = Pressure Outside the bubble + Excess Pressure.

Excess pressure a dome can survive = Px 4πr²

Load = 4σ/R x 4πr²

✤ Amount of liquid flowing per second through a pipe :- Amount of liquid flowing per second is directly proportional to the 4 power of radius & inversely proportional to the coefficient of Viscosity & length

V ∝ r⁴ - (i)

V ∝ r⁴ - (ii)

V ∝ 1/l - (iii)

V ∝ 1/η - (iv)

from (i), (ii), (iii) & (iv)

V ∝ Pπr⁴/8ηl

✤ Buoyancy :-

* It is the phenomenon by which any fluid exert upward force against weight of the body which is immersed in liquid.

* It depend upon the density of fluid.

✤ STOKE'S FORCE :-

When any spherical body fall freely in a fluid then a force act onit. This force is F = 6π ηrv

❈ Importance of stoke's force (Uses) :-

→ It is used in cloud formation.

→ Manufacturing of Parachute.

✤ TERMINAL VELOCITY :-

It is the uniform Velocity of a lead falling freely in a fluid when sum of buoyant & stoke's force become equal to the weight of the ball

i.e. Fb + Fs = W → (i)

We know, w=mg

w=4/3πr³ρxg FB=6πηrv

FB=4/3πr³σg

From (1) 4/3πr³ρg +6πηrv=4/3πr³ρg.

6πηrv=4/3πr³g (ρ-σ)

v=2/9 r²g (ρ-σ)/η

Note:- ① If ρ>σ (2) if ρ =σ

⇒ρ-σ>0 ρ- =0

So v>0 i.e body will sink v=0 i.e. body will float inside the

③ If ρ<σ liquid

ρ-σ<0

i.e body will float upper layer of liquid

* NATURE OF FLOW OF LIQUID :-

① Streamline : In this type of flow of liquid every particle has same velocity at point as that of preceding one. Ex: Flowing water in pipes

② Laminar Flow: In this type of liquid flows in form of layers

Ex: Natural fountain.

③ Turbulent Flow: His the Zig-Zag flow.

Ex: Fountain

✤ Reynald’s & Number: There are the pure number to describe the nature of blew of liquid.

NR=DPη Here, D-diameter

ρ= density

v= Velocity

n= Coeffeicient of Viscosity.

1) If 0

2) If 2000< N< 3000 then flows laminar car turbulent.

3) If NR>2000, then flow is turbulent.

EQUATION OF CONTINUITY :-

It states that amount of liquid onto r per second in a pipe W_equal E to the amount of liquid ejected per second amount of liquid per second = πa₂xρ

amount of liquid ejected = v₂axρ

By Conservation of mass

v₁a₁ρ=v₂a₂ρ

v₁a₁=v₂a₂

Va = Constant

If Q is given, what's the nature of flow Re = 4PQ/pidn

✤ PE in fluid:

It is defined as the energy gain by fluid due to its height from some horizontal

1) PE = mgh

2) PE per unit Volume

= mgh/v

= Pgh

3) PE per unit mass = mgh/m

= gh

KE in fluid: w.k.d

This the energy possessed by a fluid due to its motion

1) KE = 1/2 mv2

2) KE per unit Volume

V = Volume = mv2/2v

3) KE per unit mass = 1/2 mv2/m = 1/2 v2

Pressure Energy in fluid:

Amount of work done to p liquid without imparting all.

By applying external force for work done

dw = f x dx and dx=xdx/a

= P A dx

=P(A dx)

=P (AXdx)

= PdV

The work is stored in the form of pressure energy.

PE = ∫pdW

PE = PV

Pressure per unit mass = PV/m = (v/m = 1/ρ)

Pressure per unit volume = P/ ρ

✤ Bernoulli theorem: It states that in streamline flows ideal liquid, sum of all energies at every cross section is always constant per unit mass or per unit volume.

P+ρgh+ 1/2ρv2 = Constant

Proof:

Let a liquid (ideal) is entering with v1 & exit from another end with v2, so by equation of Continuity

a1v1 = a2v2 = C Vol. per second (1)

Work done on liquid = P1V1

by the liquid = P2V2

Network done = P1V1-P2V2

Change in K.E. = 1/2 mv2^2- 1/2 mv1^2

Change in P.E. = mgh2-mgh1

By work - Energy Principle

work done = change in energies

P1V1 - P2V2 = mgh2 - mgh1 + 1/2 mv22 - 1/2 mv12

P₁v+ mgh₁ + ½ mv² = P₂ V+mgh₂ + ½ mv²

Dividing by V on both side

P₁V + mgh₁ + ½ mv² = P₂V+mgh₂ + ½ mv²

P₁ + Pgh₁ + ½ pv² = P₂ + Pgh₂ + ½ pv²

P+pgh+ ½ pv² = Constant

✤ LIMITATIONS OF BERNOULLI'S EQUATION:-

① Bernoulli's equation ideally applies to fluids with zero viscosity or non-viscous fluids. In case of viscous fluids, one need to take into

account the work done against viscous drag.

② Bernoulli's equation is appreciable only to streamline flow of liquid & not when flow is turbulent.

③ Bernoulli's equation is applicable only to incompressible fluids because it does not take into account the elastic energy of the fluids.

* VELOCITY HEAD:-

① KE per unit weight = KE / mg

= ½ mv²/mg

= ½ v²/ g

It is also called Velocity head

② Gravitational head, PE per unit weight

= PE = mgh/mg

= h

③ Pressure heads PE per pound weight = Pv/mg = P/Pg

❈ APPLICATION:-

① Lift on aeroplane:-

Speed of air = v₂

Pressure = P₂

Speed of air = v₁

Pressure = P₁

If the airplane has speed = 0,

So speed of air above the airplane is more than the speed of air below the plane

i.e P₁ > P₂

The difference in pressure gives extra lift.

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