Class 12 Physics Fluid statics Notes

Fluid Statics (Hydrostatic)

Fluid

The substance which can flow from one point to another point is called fluid. Example ⇒ liquid & gas.

Hydrostatics (Fluid Statics)

The study of fluid at rest is called hydrostatics.

Density

Mass per unit volume is called density. Its unit is kg m⁻³.

Relative Density (Specific Gravity)

Relative density of a substance is the ratio of its density to the density of water at 4°C.

Relative density (S_r) = ρ / ρ_w

where, ρ = density of substance

ρ_w = density of water = 1 g/cm³ = 1000 kg/m³

Pressure

Force per unit area is called pressure. Its unit is N/m² or pascal (Pa).

In fluid, pressure (P) = hρg

Derivation of Pressure in Liquid

Let us consider a liquid of density ‘ρ’ contained in a vessel of cross-sectional area ‘A’ up to height ‘h’ as shown in figure.

Liquid in a vessel of cross-sectional area A and height h

Now, total mass of liquid:

m = volume (V) × density (ρ)

= Area (A) × height (h) × density (ρ)

= Ahρ

Weight of liquid is given by:

W = mg

= Ahρg

So, the normal force on the bottom of the vessel is:

F = W

= Ahρg

Finally, the pressure exerted by liquid at the bottom of vessel is:

P = F/A

= Ahρg / A

P = hρg

This is the required relation for pressure due to liquid and this relation shows that pressure due to liquid is directly proportional to:

density of liquid,
height of liquid in vessel,
acceleration due to gravity at that place.

Principle of Flotation

Principle of flotation states, “The weight of a floating body is equal to the weight of liquid displaced.”

OR

“A body is said to be floating in liquid if upthrust acting on the body is greater or equal to the weight of the body.”

When a body is immersed in a liquid there are three cases:

i) Weight of body is greater than upthrust (W > U)

W > U

Vρg > Vρ₁g

ρ > ρ₁

In this case, body sinks and set up at bottom.

ii) Weight of body is equal to upthrust

W = U

Vρg = Vρ₁g

ρ = ρ₁

In this case, the body lies inside liquid with upper surface on the surface of liquid.

iii) Weight of body is less than upthrust

W < U

Vρg < Vρ₁g

ρ < ρ₁

In this case, a part of body lies inside the liquid and remaining part outside the liquid surface.

Upthrust, Archimedes’ Principle, Pascal’s Law & Centre of Buoyancy

Upthrust

The upward force exerted by a fluid on an object which is completely or partially immersed in fluid is called upthrust or buoyancy.

Archimedes’ Principle

It states, “When a body fully or partially immersed in a fluid, it experiences an upthrust which is equal to the weight of fluid displaced by the body.”

Upthrust = Weight of fluid displaced

Pascal’s Law

It states that, “When a pressure is applied to an enclosed liquid then the pressure is equally transmitted to every portion of it.”

Centre of Buoyancy

The centre of gravity of the displaced liquid is called centre of buoyancy.

Centre of gravity and centre of buoyancy on the same vertical line

A floating body is said to be in equilibrium if centre of gravity and centre of buoyancy of the body lie in same vertical line.

Surface Tension

Cohesive and Adhesive Force

Cohesive force: The force of attraction between similar molecules is called cohesive force. e.g. force between water molecules.

Adhesive force: The force of attraction between dissimilar molecules is called adhesive force. e.g. force between water & glass molecules.

Surface Tension

The property by virtue of which liquid surface behaves as stretched membrane and tends to occupy the minimum surface area is called surface tension.

Mathematically: Surface tension is the force per unit length of an imaginary line drawn in the plane of liquid surface acting at right angle of this line.

If ‘F’ be the force acting on the imaginary line of length ‘l’, then:

T = F/l

Note: Surface-tension expressions for different objects

Case	Source condition	Expression
Object	Object of length l on surface of liquid	T = F/l
Object	Object of length l inside the liquid	T = F/(2l)
Ring	Ring of perimeter 2πr on surface	T = F/(2πr)
Ring	Ring of perimeter 2πr inside liquid	T = F/(4πr)
Rectangle	Rectangle of perimeter 2(l+b) on surface	T = F/[2(l+b)]
Rectangle	Rectangle of perimeter 2(l+b) inside liquid	T = F/[4(l+b)]

Representative surface-tension cases from the scanned notes

Surface Energy

The amount of workdone against surface tension per unit change in surface area is called surface energy.

Surface energy (σ) = Workdone in increasing surface area / Change in surface area

Fig: Surface energy

Let us consider a rectangular frame ABCD of wire in which BC can be slid horizontally. Suppose it is dipped in a soap solution such that a thin layer is deposited over an area. Due to surface tension (T), the layer tends to minimize surface area. Suppose force ‘F’ is applied perpendicularly to BC towards outwards. So, the membrane is extended by distance ‘x’ horizontally.

F = T × 2l

[Since the soap film has two surfaces in contact with air, so total length of wire = 2l]

So, workdone by this force F:

W = F × d

= F × x

W = T × 2l × x

Hence,

σ = Workdone / Change in area

= (T × 2l × x) / (2 × l × x)

σ = T

This surface energy is equal to surface tension.

Shape of Liquid Meniscus

Water forms a concave meniscus; mercury forms a convex meniscus

i) Convex Meniscus

In convex meniscus, the cohesive force between mercury molecules is greater than the adhesive force between mercury molecules and glass molecules.

F_A < F_C

ii) Concave Meniscus

In concave meniscus, the adhesive force between mercury molecule and water molecule is greater than the cohesive force between water molecules.

F_A > F_C

Angle of Contact

The angle between tangent to the liquid meniscus at point of contact with the surface of solid inside the liquid surface is called angle of contact. For example, angle of contact for water glass is acute and angle of contact for mercury glass is obtuse.

Angle of contact in water is acute; in mercury it is obtuse

Capillarity

The phenomena of rise or fall of the liquid in a capillary tube when dipped partially in it is called capillarity.

For example:

When a capillary tube with narrow bore is dipped in water, the level of water rises in capillary tube.
When same capillary tube is dipped in mercury, the level of liquid falls in capillary tube.

Capillary rise in water and depression in mercury

Measurement of Surface Tension by Capillary Rise Tube

Fig: Capillary rise

Let us consider a capillary tube of radius ‘r’ is dipped in a liquid of density ‘ρ’. Let ‘h’ be the height above which liquid rise in the capillary tube. The surface tension acting tangentially at concave meniscus is equal and opposite to the relation. If ‘θ’ be angle of contact, then T can be resolved into two components—T sinθ along horizontal and T cosθ along vertical. The horizontal components cancel each other and vertical components add up together which pull liquid upwards.

The total vertical force acting upward throughout the circumference is given by:

F = T cosθ × 2πr — (i)

This force pulls the liquid upwards until weight of rise liquid equals to upward force.

Now, volume of liquid in the tube above free surface of liquid is given by:

V = [Volume of cylinder of radius r & height h]

+ [Volume of cylinder of radius r & height r]

− [Volume of hemisphere of radius r]

V = πr²h + πr³ − (2/3)πr³

V = πr²h + (1/3)πr³

V = πr²(h + r/3)

So, weight of liquid rise:

W = mass of liquid × g

= Vρg

= πr²(h + r/3)ρg — (ii)

At equilibrium:

Total upward force = Weight of liquid rise

T cosθ × 2πr = πr²(h + r/3)ρg

T cosθ = (h + r/3)ρrg / 2

T = [(h + r/3)ρrg] / (2 cosθ) — (iii)

Thus, knowing r, h, θ, ρ and g value, T can be calculated.

For narrow tube: h ≫ r/3, so h + r/3 ≈ h.

T = ρrgh / (2 cosθ)

h = 2T cosθ / (ρrg)

Numericals — Surface Tension & Capillarity

Q.1 — Capillary rise

A capillary tube of 0.3 m diameter is placed vertically inside a liquid of density 800 kgm⁻³, surface tension 5×10⁻⁴ Nm⁻¹ and angle of contact 30°. Calculate the height to which the liquid rises in the capillary tube.

diameter (d) = 0.3 m

radius (r) = 0.3/2 = 0.15 m

density of liquid (ρ) = 800 kg/m³

surface tension (T) = 5×10⁻⁴ N/m

angle of contact (θ) = 30°

h = 2T cosθ / (ρrg)

h = [2(5×10⁻⁴) cos30°] / [800×0.15×10]

h = 7.2×10⁻⁷ m

Q.2 — Rectangular plate and surface tension

A rectangular plate of dimensions 6 cm by 4 cm and thickness 2 mm is placed with its largest face flat on the surface of water. Calculate the downward force on the plate due to surface tension assuming zero angle of contact. What is the downward force if the plate is placed vertical so that its longest side just touches the water?

Surface tension of water, T = 7.0×10⁻² N m⁻¹

l = 6 cm = 6×10⁻² m

b = 4 cm = 4×10⁻² m

t = 2 mm = 2×10⁻³ m

Downward force on the plate due to surface tension:

F₁ = T × 2(l+b)

= 7×10⁻² × 2(6+4)×10⁻²

F₁ = 1.4×10⁻² N

Again, downward force when the plate is placed vertical and its largest side just touches:

F₂ = T × 2(l+t)

= 7×10⁻² × 2(6×10⁻² + 2×10⁻³)

F₂ = 8.68×10⁻³ N

Q.3 — Breaking one drop into eight drops

Find the workdone required to break up a drop of water of radius 5×10⁻³ m into eight drops of water assuming isothermal condition.

Surface tension of water, T = 7.2×10⁻³ N/m

Radius of big drop (R) = 5×10⁻³ m

Number of drops (n) = 8

Weight/volume of big drop = weight/volume of small 8 drops

(4/3)πR³ρg = 8(4/3)πr³ρg

R³ = 8r³

R = 2r

r = R/2 = 2.5×10⁻³ m

Workdone = T × change in area

W = T[8×4πr² − 4πR²]

= 7.2×10⁻³[4π(8r² − R²)]

W = 2.26×10⁻⁶ J

Q.4 — Breaking one drop into one million droplets

Calculate the work done in breaking a drop of water of 2 mm diameter into million droplets of same size. T of water = 72×10⁻³ N/m.

D = 2 mm, R = 10⁻³ m

T = 72×10⁻³ N/m

Volume of big drop = volume of million drops

R³ = 10⁶r³

R = 10²r

r = R/100 = 10⁻⁵ m

Again, workdone is:

W = T × change in area of drop

= 72×10⁻³[10⁶×4πr² − 4πR²]

= 72×10⁻³×4(3.14)[10⁶r² − R²]

W = 8.95×10⁻⁵ J

Viscosity

The property by virtue of which a fluid layer opposes the relative motion of its different layer is called viscosity.

Generally: Viscosity means the frictional force acting between the layers of fluid that oppose the relative motion between them.

Newton’s Formula for Viscosity

Fig: Different layers of liquid flowing over the solid surface

Let us consider a liquid flowing over the fixed solid horizontal surface. The liquid flows in different layers parallel to the fixed surface. The layer in contact with the fixed surface is at rest while the velocity of other layer increases uniformly upward.

Let ‘A’ be the area of layers in contact and v & v+dv are the velocity of layer at distance x & x+dx from fixed surface respectively, then dv/dx is velocity gradient.

According to Newton’s law of viscosity, viscous force ‘F’ acting between two layers of liquid is:

i) directly proportional to area of contact:

F ∝ A — (i)

ii) directly proportional to velocity gradient:

F ∝ dv/dx — (ii)

Combining eqn (i) and (ii), then:

F ∝ A(dv/dx)

F = −ηA(dv/dx)

where ‘η’ is proportionality constant called coefficient of viscosity. The negative sign indicates that viscous force decreases the relative motion/velocity between two layers.

η = −F / [A(dv/dx)]

If A = 1 m² and dv/dx = 1 s⁻¹, then η = −F.

Thus, coefficient of viscosity is defined as the viscous force acting per unit area having unit velocity gradient.

Dimensional formula of coefficient of viscosity

η = F / [A(dv/dx)]

= [MLT⁻²] / ([L²][T⁻¹])

[η] = [ML⁻¹T⁻¹]

Poiseuille’s Formula

Fig: Streamline flow of liquid in capillary tube

Poiseuille concluded that the volume ‘V’ of the liquid flowing per second through a capillary tube is:

directly proportional to the pressure difference between two ends: V ∝ P,
directly proportional to the fourth power of radius of tube: V ∝ r⁴,
inversely proportional to the coefficient of viscosity of liquid: V ∝ 1/η,
inversely proportional to the length of capillary tube: V ∝ 1/l.

Combining equations (i), (ii), (iii) & (iv), we get:

V ∝ Pr⁴/(ηl)

V = kPr⁴/(ηl), where k = π/8

V = πPr⁴/(8ηl)

This is called Poiseuille’s formula.

Derivation of Poiseuille’s Formula by Dimensional Method

Consider a liquid through a capillary tube of radius r with length ‘l’ such that pressure difference between two end of tube is P = P₁ − P₂.

According to Poiseuille’s, the volume per second ‘V’ depends upon:

the pressure gradient (P/l),
radius of capillary tube (r),
coefficient of viscosity (η).

In dimensional form:

V ∝ (P/l)^a r^b η^c

V = k(P/l)^a r^b η^c — (i)

where k = π/8 called proportionality constant.

The dimensional equation of eqn (i) is:

[M⁰L³T⁻¹] = [ML⁻²T⁻²]^a[L]^b[ML⁻¹T⁻¹]^c

Equating dimensions:

a + c = 0

−2a + b − c = 3

−2a − c = −1

a = 1, c = −1, b = 4

Putting a = 1, b = 4, c = −1 in eqn (i):

V = (π/8)(P/l)r⁴η⁻¹

V = πPr⁴/(8lη)

Which is Poiseuille’s formula.

Stoke’s Law

When a spherical body falls through liquid, the upthrust and viscous force act in upward direction and weight of the body acts in downward direction. With increase in velocity of the body, the viscous force also increases and at certain time when total upward forces equal to downward force, the body falls with constant velocity called terminal velocity.

Stoke’s law found that the viscous force (F) acting on a spherical body of radius ‘r’ moving with terminal velocity ‘v’ in a fluid of coefficient of viscosity ‘η’ is:

F = 6πηrv

Fig: Body falling downward in liquid

Derivation of Stoke’s Law by Dimensional Method

Consider a spherical body of radius ‘r’ falling through a liquid of coefficient of viscosity (η) with terminal velocity ‘v’. According to Stoke’s, viscous force (F) depends upon:

coefficient of viscosity of liquid (η),
terminal velocity of the body (v),
radius of spherical body (r).

In dimensional form:

F ∝ η^av^br^c

F = kη^av^br^c

where k is proportionality constant = 6π.

∴ F = 6πη^av^br^c — (i)

Dimensional equation of eqn (i):

[MLT⁻²] = [ML⁻¹T⁻¹]^a[LT⁻¹]^b[L]^c

[MLT⁻²] = [M^aL^−a+b+cT^−a−b]

Equating dimensions of both sides:

a = 1

−a − b = −2 ⇒ b = 1

−a + b + c = 1 ⇒ c = 1

Putting a = 1, b = 1, c = 1 in eqn (i):

F = 6πη¹v¹r¹

F = 6πηrv

This is Stoke’s law.

Measurement of Coefficient of Viscosity by Using Stoke’s Law

Let us consider a sphere of radius ‘r’ and density ‘ρ’ falling through a liquid of density ‘σ’ with terminal velocity ‘v’. In this case force acting on a body are:

Fig: Sphere falling in a liquid

i) Weight of sphere (W) = mg:

W = Vρg

= (4/3)πr³ρg [in downward direction]

ii) Upthrust (U) = weight of liquid displaced:

U = (4/3)πr³σg

iii) Viscous force:

F = 6πηrv [in upward direction]

When speed of the body in downward direction increases, the viscous force is also increased and at certain time the upward force acting on a body equals to downward force and body attains equilibrium.

At equilibrium:

F + U = W

6πηrv + (4/3)πr³σg = (4/3)πr³ρg

6πηrv = (4/3)πr³(ρ − σ)g

η = 2r²(ρ − σ)g / (9v)

This is required expression for coefficient of viscosity.

Bernoulli’s Theorem

Bernoulli’s theorem states, “For stream-line flow of an ideal liquid (non-viscous and incompressible), the total energy (the sum of P.E., K.E. & pressure energy) per unit mass remains constant at every cross-section throughout the flow.”

v²/2 + gh + P/ρ = K (constant)

Fig: Bernoulli’s theorem

Let us consider a pipe AB in which non-viscous and incompressible fluid is flowing. Let a₁, P₁, v₁, h₁ and a₂, P₂, v₂, h₂ be the area of cross-section, pressure, velocity and height at end A & B respectively. Also let in time Δt the fluid at end A reach to A′ and at end B′ reach to B.

Force acting on the fluid at end A:

F₁ = P₁a₁

W₁ = (P₁a₁)(v₁Δt)

Similarly, workdone by fluid at end B′ to B:

W₂ = P₂a₂v₂Δt

Thus, net workdone on the fluid from moving A to B:

W = W₁ − W₂

W = P₁a₁v₁Δt − P₂a₂v₂Δt — (i)

From equation of continuity:

a₁v₁ = a₂v₂ = av (say)

a₁v₁Δt = a₂v₂Δt = avΔt = V

W = avΔt(P₁ − P₂)

W = (P₁ − P₂)V — (ii)

Let ‘m’ be the mass of liquid flowing through the pipe in Δt. Then change in K.E.:

ΔK.E. = ½mv₂² − ½mv₁² — (iii)

Similarly, change in P.E.:

ΔP.E. = mgh₂ − mgh₁ — (iv)

Total workdone = Total change in energy:

(P₁ − P₂)V = ½mv₂² − ½mv₁² + mgh₂ − mgh₁

P₁ − P₂ = ½ρv₂² − ½ρv₁² + ρgh₂ − ρgh₁

P₁/ρ − P₂/ρ = v₂²/2 − v₁²/2 + gh₂ − gh₁

P₁/ρ + v₁²/2 + gh₁ = P₂/ρ + v₂²/2 + gh₂

P/ρ + gh + v²/2 = constant

Equation of Continuity

Fig: Steady flow of liquid

Let us consider a liquid flowing through pipe of cross-sectional area ‘a₁’ and ‘a₂’ at left & right end respectively. Also let v₁, ρ₁ and v₂, ρ₂ are velocities and density of liquid at respective end.

Volume per second of liquid entering into the pipe at left end:

V₁ = a₁v₁

Again, mass of liquid entering per second at left end:

m₁ = ρ₁V₁

m₁ = a₁v₁ρ₁

Similarly, mass of liquid per second at right end:

m₂ = a₂v₂ρ₂

If there is no loss of liquid in tube, then:

m₁ = m₂

a₁v₁ρ₁ = a₂v₂ρ₂

Also, if liquid is incompressible, then ρ₁ = ρ₂:

a₁v₁ = a₂v₂

av = constant

This is called equation of continuity. This equation states that if the area of cross-section of the tube becomes larger then liquid’s speed becomes smaller and vice-versa.

Numericals — Viscosity, Stoke’s Law, Bernoulli & Continuity

Q.5 — Mass of an aeroplane

Calculate mass of an aeroplane with the wings of area 55 m² flying horizontally. The velocity of air above & below the wings is 155 m/s and 140 m/s respectively.

Source note: the question states wing area 55 m², but the handwritten solution uses A = 50 m². The calculation below preserves the solution exactly as written.

Area of plane (A) = 50 m²

Speed of air over lower wing (v₁) = 140 m/s

Speed of air over upper wing (v₂) = 155 m/s

ρ_air = 1.293 kg/m³

Pressure of air over lower & upper wing are P₁ & P₂ respectively.

Then, by Bernoulli’s equation:

(P₁ − P₂)/ρ = v₂²/2 − v₁²/2 [h₁ = h₂]

(P₁ − P₂)A / 1.29 = 50[(155)² − (140)²]/2 = 110625

F = 110625 × 1.29 = 142706.25 N

mg = 142706.25

m = 142706.25/9.8 = 1.456×10⁴ kg

Q.6 — Terminal velocity of an air bubble

Calculate the magnitude and direction of the terminal velocity of an 1 mm radius air bubble rising in an oil of viscosity 0.20 N m⁻²s and specific gravity of 0.9 & density of air 1.29 kg/m³.

Source note: the handwritten substitution uses η = 0.02 N s m⁻² (0.2 poise), while the question line appears to show 0.20 N m⁻²s. The calculation below follows the handwritten solution.

r = 1 mm = 10⁻³ m

ρ_oil = specific gravity × density of water = 0.9×10³ kg/m³

ρ_air = 1.29 kg/m³

η = 0.02 N s m⁻² = 0.2 poise

For air bubble, from Stoke’s law:

v_t = 2r²(ρ_air − ρ_oil)g/(9η)

= 2(10⁻³)²(1.29 − 900)×10 / [9(0.02)]

= −9.99×10⁻³ m/s

Magnitude = 0.0099 m/s; direction = upward

Q.7 — Eight raindrops coalesce

Eight/three/two spherical raindrops of equal size are falling vertically through air with terminal velocity 0.15 m/s. If they coalesce what would be the terminal velocity?

For 8 small drops of radius r and one big drop of radius R:

8 × (4/3)πr³ = (4/3)πR³

R³ = (2r)³

R = 2r

By Stoke’s relation v ∝ r²:

v₁/v₂ = r²/R²

v₂ = v₁R²/r²

= 0.15(2r)²/r²

v₂ = 0.6 m/s

Q.8 — Steel ball in castor oil

Castor oil at 20°C has a coefficient of viscosity 2.42 N m⁻²s and density 940 kg/m³. Calculate the terminal velocity of the steel ball of radius 2.00 mm falling under gravity in the oil, taking density of steel as 7800 kgm⁻³.

T = 20°C

η = 2.42 N s m⁻²

σ = 940 kg/m³

ρ = 7800 kg/m³

r = 2 mm = 2×10⁻³ m

v_t = 2r²(ρ−σ)g/(9η)

= 2(2×10⁻³)²(7800−940)×10 / [9(2.42)]

v_t = 0.025 m/s

Q.9 — Water through non-uniform pipe

Water flows steadily through a horizontal pipe of non-uniform cross-section. If the pressure of the water is 4×10⁴ N m⁻² at a point where the velocity of flow is 2 m s⁻¹ & cross-section is 0.02 m², what is the pressure at the point where cross section reduces to 0.01 m²?

Non-uniform horizontal pipe

P₁ = 4×10⁴ N/m²

v₁ = 2 m/s

A₁ = 0.02 m²

A₂ = 0.01 m²

ρ_water = 1000 kg/m³

From equation of continuity:

A₁v₁ = A₂v₂

(0.02)2 = (0.01)v₂

v₂ = 4 m/s

Again, from Bernoulli’s principle (where h₁ = h₂):

P₁/ρ + v₁²/2 = P₂/ρ + v₂²/2

4×10⁴/1000 + (2)²/2 = P₂/1000 + (4)²/2

P₂ = 34000 N/m²

Q.10 — Glass ball falling through glycerol

What is the terminal velocity of the glass ball falling through a tall jar containing glycerol? The densities of the glass ball and glycerol are 8.5 g/cc and 1.32 g/cc respectively & viscosity of the glycerol is 0.85 poise & radius of the ball is 2 mm.

ρ = 8.5 g/cc = 8.5×10³ kg/m³

σ = 1.32 g/cc = 1.32×10³ kg/m³

η = 0.85 poise = 0.085 N s m⁻²

r = 2 mm = 2×10⁻³ m

v_t = 2r²(ρ−σ)g/(9η)

= 2(2×10⁻³)²(8.5−1.32)×10³×10 / [9(0.085)]

v_t = 0.75 m/s

The last handwritten sentence calls it a “steel ball”, although the question asks for a glass ball. The numerical value above is preserved from the source.

Fluid Statics (Hydrostatic)

Fluid

Hydrostatics (Fluid Statics)

Density

Relative Density (Specific Gravity)

Pressure

Derivation of Pressure in Liquid

Principle of Flotation

i) Weight of body is greater than upthrust (W > U)

ii) Weight of body is equal to upthrust

iii) Weight of body is less than upthrust

Upthrust, Archimedes’ Principle, Pascal’s Law & Centre of Buoyancy

Upthrust

Archimedes’ Principle

Pascal’s Law

Centre of Buoyancy

Surface Tension

Cohesive and Adhesive Force

Surface Tension

Note: Surface-tension expressions for different objects

Surface Energy

Shape of Liquid Meniscus

i) Convex Meniscus

ii) Concave Meniscus

Angle of Contact

Capillarity

Measurement of Surface Tension by Capillary Rise Tube

Numericals — Surface Tension & Capillarity

Q.1 — Capillary rise

Q.2 — Rectangular plate and surface tension

Q.3 — Breaking one drop into eight drops

Q.4 — Breaking one drop into one million droplets

Viscosity

Newton’s Formula for Viscosity

Dimensional formula of coefficient of viscosity

Poiseuille’s Formula

Derivation of Poiseuille’s Formula by Dimensional Method

Stoke’s Law

Derivation of Stoke’s Law by Dimensional Method

Measurement of Coefficient of Viscosity by Using Stoke’s Law

Bernoulli’s Theorem

Equation of Continuity

Numericals — Viscosity, Stoke’s Law, Bernoulli & Continuity

Q.5 — Mass of an aeroplane

Q.6 — Terminal velocity of an air bubble

Q.7 — Eight raindrops coalesce

Q.8 — Steel ball in castor oil

Q.9 — Water through non-uniform pipe

Q.10 — Glass ball falling through glycerol

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