Physics Cycle

PHYSICS: VISCOSITY — CALCULATION QUESTIONS & ANSWERS
(WAEC | UTME/JAMB | AP Physics | IGCSE Syllabus)
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1. DEFINITIONS & KEY CONCEPTS
1.1 What is Viscosity?
Viscosity (η) is the measure of a fluid's internal resistance to flow. It is caused by the frictional forces between adjacent layers of fluid moving at different velocities. The higher the viscosity, the "thicker" the fluid.

SI Unit: Pascal-second (Pa·s) or kg·m⁻¹·s⁻¹ Other Unit: Poise (P); 1 Pa·s = 10 P

1.2 Types of Viscosity
Table
Type Symbol Definition
Dynamic viscosity η Resistance to shear flow
Kinematic viscosity ν ν = η / ρ (where ρ = density)

1.3 Factors Affecting Viscosity
• Temperature: For liquids, viscosity decreases as temperature increases. For gases, viscosity increases as temperature increases.
• Pressure: Viscosity of liquids increases slightly with pressure.
• Nature of fluid: Molecular structure and intermolecular forces.
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2. STOKES' LAW
2.1 Statement
When a small spherical object moves slowly through a viscous fluid, the viscous drag force (F) opposing its motion is given by:
F = 6π η r v
Where:
• F = Viscous drag force (N)
• η = Coefficient of viscosity of the fluid (Pa·s)
• r = Radius of the sphere (m)
• v = Velocity of the sphere (m/s)

2.2 Conditions for Stokes' Law to Apply
1. The sphere must be small and smooth.
2. The velocity must be low (laminar flow, Re < 1).
3. The fluid must be homogeneous and infinite in extent (no wall effects).
4. The sphere must be rigid and spherical.
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3. TERMINAL VELOCITY
3.1 Definition
Terminal velocity (vₜ) is the constant maximum velocity attained by an object falling through a viscous fluid when the net force on it becomes zero.

3.2 Derivation of Terminal Velocity
Consider a sphere of radius r and density ρₛ falling through a fluid of density ρf and viscosity η.
Forces acting on the sphere:
1. Weight (W) downward: W = (4/3)π r³ ρₛ g
2. Upthrust/Buoyancy (U) upward: U = (4/3)π r³ ρf g
3. Viscous drag (F) upward: F = 6π η r v
At terminal velocity (v = vₜ): Net force = 0 W = U + F
(4/3)π r³ ρₛ g = (4/3)π r³ ρf g + 6π η r vₜ
Solving for vₜ:
vₜ = [2 r² g (ρₛ - ρf)] / (9 η)
Alternative form using diameter (d = 2r):
vₜ = [d² g (ρₛ - ρf)] / (18 η)
3.3 Key Relationships
• vₜ ∝ r² (terminal velocity is proportional to the square of the radius)
• vₜ ∝ (ρₛ - ρf) (terminal velocity increases with density difference)
• vₜ ∝ 1/η (terminal velocity is inversely proportional to viscosity)

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4. CALCULATION QUESTIONS & SOLUTIONS
QUESTION 1 (WAEC/UTME Style)
A small sphere of radius 2.0 mm falls through a viscous liquid of viscosity 0.5 Pa·s. Calculate the viscous drag force on the sphere when it moves at a velocity of 0.04 m/s.
Solution: Given:
• r = 2.0 mm = 2.0 × 10⁻³ m
• η = 0.5 Pa·s
• v = 0.04 m/s
Using Stokes' Law: F = 6π η r v F = 6 × π × 0.5 × (2.0 × 10⁻³) × 0.04 F = 6 × 3.142 × 0.5 × 2.0 × 10⁻³ × 0.04 F = 7.54 × 10⁻⁴ N
Answer: F = 7.5 × 10⁻⁴ N (to 2 s.f.)

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QUESTION 2 (WAEC/UTME Style)
A metal ball of weight W falls through a column of glycerine of viscosity η. If the ball experiences an upthrust U and terminal velocity is attained, which equation correctly relates these quantities?
Solution: At terminal velocity: Weight = Upthrust + Viscous drag W = U + F
From Stokes' Law, F = 6π η r vₜ Therefore: W = U + 6π η r vₜ

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QUESTION 3 (IGCSE/AP Style)
A steel ball bearing of radius 5.0 mm falls at a constant speed of 0.030 m/s through oil which has a viscosity of 0.3 Pa·s and density 900 kg/m³. Determine the viscous drag acting on the ball bearing.
Solution: Given:
• r = 5.0 mm = 5.0 × 10⁻³ m
• v = 0.030 m/s
• η = 0.3 Pa·s
Using Stokes' Law: F = 6π η r v F = 6 × π × 0.3 × (5.0 × 10⁻³) × 0.030 F = 8.48 × 10⁻⁴ N
Answer: F = 8.5 × 10⁻⁴ N

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QUESTION 4 (WAEC/UTME Style)
Calculate the terminal velocity of a steel ball of radius 3.0 mm falling through glycerol of viscosity 1.5 Pa·s. (Density of steel = 7800 kg/m³, density of glycerol = 1300 kg/m³, g = 10 m/s²)
Solution: Given:
• r = 3.0 mm = 3.0 × 10⁻³ m
• η = 1.5 Pa·s
• ρₛ = 7800 kg/m³
• ρf = 1300 kg/m³
• g = 10 m/s²
Using terminal velocity formula: vₜ = [2 r² g (ρₛ - ρf)] / (9 η) vₜ = [2 × (3.0 × 10⁻³)² × 10 × (7800 - 1300)] / (9 × 1.5) vₜ = [2 × 9.0 × 10⁻⁶ × 10 × 6500] / 13.5 vₜ = 1.17 / 13.5 vₜ = 0.0867 m/s
Answer: vₜ = 8.7 × 10⁻² m/s (or 0.087 m/s)

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QUESTION 5 (AP/IGCSE Style)
A student drops a steel sphere with radius 4.0 mm into a cylinder of glycerol. The sphere reaches terminal velocity and takes 3.9 s to fall 0.50 m. Calculate the viscosity of glycerol. (Density of steel = 7800 kg/m³, density of glycerol = 1300 kg/m³)
Solution: Given:
• r = 4.0 mm = 4.0 × 10⁻³ m
• Distance = 0.50 m
• Time = 3.9 s
• ρₛ = 7800 kg/m³
• ρf = 1300 kg/m³
First, find terminal velocity: vₜ = distance / time = 0.50 / 3.9 = 0.128 m/s
Rearrange terminal velocity formula to find η: η = [2 r² g (ρₛ - ρf)] / (9 vₜ) η = [2 × (4.0 × 10⁻³)² × 9.81 × (7800 - 1300)] / (9 × 0.128) η = [2 × 1.6 × 10⁻⁵ × 9.81 × 6500] / 1.152 η = 2.04 / 1.152 η = 1.77 Pa·s
Answer: η ≈ 1.8 Pa·s (to 2 s.f.)

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QUESTION 6 (WAEC/UTME Style)
A raindrop with diameter 1.0 mm descends through air of viscosity 1.8 × 10⁻⁵ Pa·s. Find the viscous force acting on the drop when its speed is 2.0 m/s.
Solution: Given:
• d = 1.0 mm, so r = 0.5 mm = 5.0 × 10⁻⁴ m
• η = 1.8 × 10⁻⁵ Pa·s
• v = 2.0 m/s
Using Stokes' Law: F = 6π η r v F = 6 × π × (1.8 × 10⁻⁵) × (5.0 × 10⁻⁴) × 2.0 F = 3.39 × 10⁻⁷ N
Answer: F = 3.4 × 10⁻⁷ N

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QUESTION 7 (IGCSE/AP Style — Deduction)
A spherical clay particle has radius 2.5 × 10⁻⁷ m. Deduce whether this particle takes more than 6 months to fall 4 m in still water. (Viscosity of water = 1.0 × 10⁻³ Pa·s, density of water = 1000 kg/m³, density of clay = 2650 kg/m³)
Solution: Given:
• r = 2.5 × 10⁻⁷ m
• η = 1.0 × 10⁻³ Pa·s
• ρₛ = 2650 kg/m³
• ρf = 1000 kg/m³
• g = 9.81 m/s²
Calculate terminal velocity: vₜ = [2 × (2.5 × 10⁻⁷)² × 9.81 × (2650 - 1000)] / (9 × 1.0 × 10⁻³) vₜ = [2 × 6.25 × 10⁻¹⁴ × 9.81 × 1650] / (9.0 × 10⁻³) vₜ = 2.02 × 10⁻⁹ / 9.0 × 10⁻³ vₜ = 2.24 × 10⁻⁷ m/s
Time to fall 4 m: t = distance / vₜ = 4 / (2.24 × 10⁻⁷) = 1.79 × 10⁷ s
Convert to months: 1 month ≈ 30 days = 30 × 24 × 3600 = 2.59 × 10⁶ s t = 1.79 × 10⁷ / 2.59 × 10⁶ = 6.9 months
Conclusion: Yes, the particle takes more than 6 months to fall 4 m.

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QUESTION 8 (UTME/WAEC Style)
A powder comprising particles of various sizes is stirred up in a vessel filled to a height of 0.50 m with water. Find the radius of the largest particle that will remain in suspension after 1 hour. (Density of powder = 2500 kg/m³, viscosity of water = 0.01 poise = 1.0 × 10⁻³ Pa·s)
Solution: Given:
• h = 0.50 m
• t = 1 hour = 3600 s
• ρₛ = 2500 kg/m³
• ρf = 1000 kg/m³
• η = 1.0 × 10⁻³ Pa·s
Terminal velocity of largest particle: vₜ = h / t = 0.50 / 3600 = 1.39 × 10⁻⁴ m/s
Rearrange to find r: vₜ = [2 r² g (ρₛ - ρf)] / (9 η) r² = (9 η vₜ) / [2 g (ρₛ - ρf)] r² = (9 × 1.0 × 10⁻³ × 1.39 × 10⁻⁴) / [2 × 9.81 × (2500 - 1000)] r² = 1.25 × 10⁻⁶ / 29430 r² = 4.25 × 10⁻¹¹ r = 6.52 × 10⁻⁶ m = 6.52 μm
Answer: r ≈ 6.5 μm

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QUESTION 9 (AP/IGCSE Style — Graphical Analysis)
A student investigates the relationship between terminal velocity v and diameter d of ball bearings falling through a viscous fluid. Explain why a graph of v on the y-axis and d² on the x-axis should be a straight line through the origin.
Solution: From the terminal velocity equation: v = [g (ρₛ - ρf) / (18 η)] × d²
This is in the form y = mx, where:
• y = v
• x = d²
• m = g(ρₛ - ρf)/(18η) = constant
Since all quantities in m are constants, the graph of v against d² is a straight line passing through the origin (when d = 0, v = 0).

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QUESTION 10 (UTME/WAEC Style — Conceptual)
Explain the motion of a bubble of air rising through a vertical column of water.
Solution: When a bubble is released at the bottom:
1. Initially: Upthrust > Weight + Viscous drag, so the bubble accelerates upward.
2. As velocity increases: Viscous drag increases (F ∝ v).
3. At terminal velocity: Upthrust = Weight + Viscous drag, so the bubble moves at constant velocity.
For a bubble, weight is negligible compared to upthrust, so approximately: Upthrust ≈ Viscous drag at terminal velocity.

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5. EXPERIMENTAL DETERMINATION OF VISCOSITY
5.1 Falling-Ball Method (Stokes' Method)
Apparatus: Tall glass cylinder, viscous liquid, steel ball bearings, stopwatch, metre rule, micrometer screw gauge, thermometer.
Procedure:
1. Measure the diameter of the ball bearing using a micrometer screw gauge.
2. Fill the cylinder with the liquid and mark two points A and B a known distance apart.
3. Gently drop the ball bearing at the center of the liquid surface.
4. Start the stopwatch when the ball passes mark A (after reaching terminal velocity).
5. Stop the stopwatch when the ball passes mark B.
6. Repeat for different ball bearings and calculate the mean time.
7. Measure the temperature of the liquid.
Calculations:
• vₜ = distance AB / mean time
• Plot vₜ against r² or d²
• Gradient = 2g(ρₛ - ρf)/(9η)
• Calculate η from the gradient
5.2 Precautions
• Ensure the ball bearing falls along the axis of the cylinder (to avoid wall effects).
• Allow the ball to reach terminal velocity before timing.
• Use a wide cylinder (diameter >> ball diameter).
• Maintain constant temperature.
• Repeat measurements to obtain mean values.

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6. COMPARISON TABLE: VISCOSITY VS. FRICTION
Table
Property Viscosity Friction
Medium Between layers of fluid Between solid surfaces
Dependence on area Directly proportional to area Independent of area
Dependence on velocity Proportional to relative velocity Independent of velocity
Dependence on normal reaction Independent Directly proportional
Cause Molecular attraction/cohesion Surface irregularities
Nature Opposes relative motion between fluid layers Opposes relative motion between solids
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7. APPLICATIONS OF VISCOSITY
1. Lubrication: Selection of oils based on viscosity for different temperatures.
2. Shock absorbers: High viscosity liquids dampen motion.
3. Blood circulation: Viscosity affects blood flow through arteries and veins.
4. Sedimentation: Used to determine particle sizes in suspensions.
5. Weather: Viscosity of air affects raindrop terminal velocity.
6. Food industry: Controlling flow of honey, syrups, and oils.

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8. QUICK REFERENCE FORMULA SHEET
Table
Quantity Formula Variables
Stokes' Law F = 6π η r v F=drag, η=viscosity, r=radius, v=velocity
Terminal Velocity vₜ = 2r²g(ρₛ-ρf)/(9η) ρₛ=sphere density, ρf=fluid density
Terminal Velocity (diameter) vₜ = d²g(ρₛ-ρf)/(18η) d=diameter
Viscosity from experiment η = 2r²g(ρₛ-ρf)/(9vₜ) Rearranged from above
Upthrust U = (4/3)π r³ ρf g Archimedes' principle
Weight of sphere W = (4/3)π r³ ρₛ g Mass = volume × density

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9. EXAM TIPS & COMMON MISTAKES
Tips:
• Always convert mm to m and cm to m before calculations.
• Remember that diameter = 2 × radius.
• For terminal velocity problems, always check if the object has reached terminal velocity.
• When plotting graphs, ensure axes are labeled with quantities and units.
Common Mistakes:
• Confusing radius with diameter in formulas.
• Forgetting to subtract upthrust from weight in force balance.
• Using wrong units (e.g., leaving radius in mm).
• Assuming Stokes' Law applies at high velocities (turbulent flow).

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10. SYLLABUS ALIGNMENT
Table
Exam Board Topics Covered
WAEC Viscosity, terminal velocity, Stokes' Law, coefficient of viscosity
UTME/JAMB Qualitative treatment of viscosity, terminal velocity, Stokes' Law
AP Physics 1/2 Fluid properties, viscosity, drag forces, terminal velocity
IGCSE Viscous drag, Stokes' Law, terminal velocity, experimental methods
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End of Notes
These notes are comprehensive and cover all calculation types typically examined in WAEC, UTME, AP, and IGCSE physics. Practice the derivations and ensure you understand the force balance at terminal velocity, as this is the foundation for all viscosity calculations.

How viscous are you?

VISCOSITY
1. Introduction to Viscosity
Viscosity is an important property of fluids (liquids and gases). It explains why some fluids flow easily while others flow slowly.
Examples:
• Water flows easily → low viscosity
• Honey flows slowly → high viscosity
• Engine oil is more viscous than water
• Air also possesses viscosity, although smaller than most liquids.
Viscosity plays an important role in fluid flow, lubrication, transportation of liquids, blood circulation, and industrial processes.
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2. Meaning of Viscosity
Viscosity is the internal friction or resistance between layers of a fluid that opposes their relative motion.
OR
Viscosity is the property of a fluid that resists flow.
Molecular Explanation
Fluid molecules attract one another.
When adjacent layers move with different speeds:
• Molecular attraction causes friction
• This internal friction resists motion
• Energy is lost as heat.
Hence, viscosity is caused by intermolecular forces.
Example:
• Honey has stronger molecular attraction than water, so it flows more slowly.
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3. Types of Viscosity
A. Dynamic (Absolute) Viscosity
This measures resistance to fluid motion.
Symbol:
η (eta)
Most WAEC, UTME and IGCSE problems involve dynamic viscosity.
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B. Kinematic Viscosity
Defined as:
Kinematic viscosity = Dynamic viscosity / Density
Formula:
ν = η/ρ
Where:
• ν = kinematic viscosity
• η = dynamic viscosity
• ρ = density
SI unit:
m²s⁻¹
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4. Coefficient of Viscosity
The coefficient of viscosity is the force required to move a unit area of one fluid layer over another at unit velocity gradient.
Symbol:
η (eta)
It measures how viscous a fluid is.
Large η:
• High viscosity
• Slow flow
Small η:
• Low viscosity
• Fast flow
Examples:
Fluid Relative Viscosity
Water Low
Petrol Very low
Oil Moderate
Honey High
Glycerine Very high
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5. Newton’s Law of Viscosity
Statement
The viscous force between layers of a fluid is directly proportional to:
• Area of contact
• Velocity gradient
and acts opposite to motion.
Mathematically:
F=\eta A\frac{dv}{dx}
Where:
• F = viscous force (N)
• η = coefficient of viscosity
• A = area (m²)
• dv/dx = velocity gradient
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6. Explanation of Velocity Gradient
Velocity gradient is the rate of change of velocity with distance between layers.
Formula:
dv/dx
Meaning:
• dv = difference in velocity
• dx = separation between layers
Greater velocity gradient:
• Greater viscous force
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7. Proof/Derivation of Newton’s Law of Viscosity
Consider:
• Two fluid layers
• Area A
• Velocity difference dv
• Distance dx
Experiment shows:
1. Force is proportional to area
F ∝ A
2. Force proportional to velocity difference
F ∝ dv
3. Force inversely proportional to separation
F ∝ 1/dx
Combining:
F ∝ A(dv/dx)
Introducing constant η:
F=\eta A\frac{dv}{dx}
This is Newton’s Law of Viscosity.
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8. SI Unit of Viscosity
From:
F = ηA(dv/dx)
Making η subject:
η = Fdx/Adv
SI units:
• F = N
• dx = m
• A = m²
• dv = ms⁻¹
Therefore:
η = Ns/m²
OR
kgm⁻¹s⁻¹
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CGS Unit
Poise (P)
Relationship:
1 Pa·s = 10 P
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9. Dimensional Formula of Viscosity
Using:
η = Fdx/Adv
Dimensions:
[F] = MLT⁻²
Therefore:
[\eta]=ML^{-1}T^{-1}
Dimension:
ML⁻¹T⁻¹
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10. Terminal Velocity
Meaning
Terminal velocity is the constant maximum velocity attained by a body falling through a fluid when resultant force becomes zero.
At this stage:
• Weight = Upthrust + Viscous force
Acceleration becomes zero.
Body moves with constant speed.
Example:
• Raindrops
• Parachute motion
• Steel ball in oil
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11. Forces Acting on Falling Steel Ball
Three forces act:
A. Weight (W)
Downward
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B. Upthrust (Buoyant Force)
Upward
Due to displaced liquid.
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C. Viscous Drag (Fv)
Upward
Opposes motion.
Initially:
• Weight > opposing forces
• Ball accelerates
Later:
• Drag increases
Eventually:
Weight = Upthrust + Drag
Then:
Resultant force = 0
Terminal velocity reached.
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12. Stokes’ Law
Statement
The viscous force acting on a sphere moving through a fluid is directly proportional to:
• Radius of sphere
• Velocity
• Fluid viscosity.
Formula:
F=6\pi\eta rv
Where:
• F = viscous drag
• η = viscosity
• r = radius
• v = velocity
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Assumptions of Stokes’ Law
Applicable when:
• Sphere is smooth
• Fluid is uniform
• Flow is laminar
• Velocity is small
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13. Derivation of Terminal Velocity Formula
At terminal velocity:
Weight = Upthrust + Drag
Weight:
\frac{4}{3}\pi r^3\rho_s g
Upthrust:
\frac{4}{3}\pi r^3\rho_f g
Drag:
6\pi\eta rv
Substituting:
(4/3)πr³ρsg = (4/3)πr³ρfg + 6πηrv
Solving:
Terminal velocity:
v=\frac{2r^2g(\rho_s-\rho_f)}{9\eta}
Where:
• v = terminal velocity
• r = radius
• ρs = density of sphere
• ρf = density of fluid
• η = viscosity
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14. Experiment to Determine Terminal Velocity of Steel Ball in Fluid
Aim
To determine terminal velocity of steel ball falling through fluid.
Apparatus
• Tall glass tube
• Glycerine/oil
• Steel ball
• Stopwatch
• Metre rule
• Marker
Procedure
1. Fill tube with glycerine.
2. Mark two points.
3. Release steel ball.
4. Observe motion.
5. Measure time between marks.
6. Repeat and average.
Observation
Ball:
• Accelerates first
• Then moves uniformly.
Uniform speed = terminal velocity.
Calculation
Terminal velocity:
v=\frac{d}{t}
Where:
• d = distance
• t = time
Precautions
• Use transparent tube
• Avoid air bubbles
• Release gently
• Repeat readings
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15. Factors Affecting Viscosity
1. Temperature
Liquids:
• Higher temperature → lower viscosity
Gases:
• Higher temperature → higher viscosity
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2. Nature of Fluid
Stronger intermolecular force:
• Greater viscosity
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3. Pressure
Usually:
• Increased pressure increases viscosity slightly.
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4. Presence of Impurities
Impurities often increase viscosity.
Example:
• Dirty oil
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5. Molecular Size
Large molecules:
• Greater resistance
• Higher viscosity
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16. Effects of Viscosity
Positive Effects
• Lubrication
• Controlled flow
• Damping motion
Negative Effects
• Energy loss
• Reduced machine efficiency
• Heat production
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17. Applications of Viscosity
Engineering
• Lubricating engines
• Hydraulic systems
Medicine
• Blood flow studies
• Syringe design
Aviation
• Aircraft design
• Air resistance studies
Food Industry
• Syrup
• Paint
• Milk processing
Petroleum Industry
• Oil transport
• Refining
Meteorology
• Rainfall and cloud motion
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18. Advantages and Disadvantages
Advantages
• Prevents wear
• Enables lubrication
• Stabilizes motion
Disadvantages
• Causes friction
• Wastes energy
• Slows machines
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19. Worked Examples
Example 1
A fluid has:
• η = 0.5 Ns/m²
• A = 2 m²
• dv/dx = 4 s⁻¹
Find force.
Solution:
F = ηA(dv/dx)
F = 0.5 × 2 × 4
F = 4 N
Answer:
4 N
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Example 2
A ball travels:
• 0.5 m
• in 2 s
Find terminal velocity.
Solution:
v = d/t
v = 0.5/2
v = 0.25 ms⁻¹
Answer:
0.25 ms⁻¹
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20. WAEC/UTME/AP/IGCSE Exam Facts and Tips
1. Viscosity means internal friction.
2. η is viscosity symbol.
3. Unit:
• Pa·s
• Ns/m²
4. Dimension:
• ML⁻¹T⁻¹
5. Terminal velocity occurs when resultant force is zero.
6. High viscosity means slow flow.
7. Stokes’ law applies to small spheres in laminar flow.
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21. Common Examination Mistakes
Avoid:
• Confusing viscosity with density
• Forgetting SI units
• Mixing drag and upthrust
• Ignoring terminal condition
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22. Revision Summary
• Viscosity = resistance to fluid flow
• Caused by intermolecular friction
• Newton’s law:
F = ηA(dv/dx)
• SI unit:
Pa·s or Ns/m²
• Dimension:
ML⁻¹T⁻¹
• Terminal velocity occurs at zero acceleration
• Stokes’ law:
F = 6πηrv
• Viscosity affects machines, medicine, transport and industries.

INTRODUCTION
Electromagnetic induction is the production of an electromotive force (emf) across an electrical conductor in a changing magnetic field. This topic extends that principle to mutual induction and its most important application—the transformer.

📌 1.1 TRANSFORMER
🔹 Definition
A transformer is an electrical device used to increase (step-up) or decrease (step-down) alternating voltage using electromagnetic induction.
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📌 1.2 STRUCTURE OF A TRANSFORMER
A transformer consists of:
Primary coil (Np) → connected to input AC source
Secondary coil (Ns) → connected to load
Soft iron core → laminated to reduce energy losses
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📌 1.3 WORKING PRINCIPLE
Based on mutual induction:
Mutual inductance is the property of two coils whereby an induced emf is generated in one coil due to a change in current in the other coil.

When two coils are placed close together, a changing current in the primary coil produces a changing magnetic flux that links with the secondary coil, inducing an emf in it.
Alternating current in the primary coil produces a changing magnetic field
This induces an emf in the secondary coil

Mutual Inductance: Formula, Derivation, Units, Dimensions, and Factors
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Unit of Mutual Inductance
"Unit of " M="henry (H)"
1" " H=1" " (V⋅s)/A

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Dimensions of Mutual Inductance
[M]=ML^2 T^(-2) A^(-2)

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Factors Affecting Mutual Inductance
Number of turns in both coils
Cross-sectional area of the coils
Distance (proximity) between the coils
Permeability of the core material (μ)
Orientation (alignment) of the coils

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1.3 THE TRANSFORMER
Definition
A transformer is a static electrical device that transfers electrical energy from one circuit to another through mutual induction, without any change in frequency, while changing the voltage and current levels.

Principle of Operation
A transformer works on the principle of mutual induction between two coils linked by a common magnetic flux. When an alternating voltage is applied to the primary coil, it produces an alternating magnetic flux in the core. This changing flux links with the secondary coil and induces an alternating emf in it.
Key Point: Transformers work only with AC, not DC, because a changing magnetic flux is required for induction.

Structure of a Transformer
Component Description Function
Primary Coil Winding connected to the AC source Receives electrical energy
Secondary Coil Winding connected to the load Delivers electrical energy
Laminated Soft Iron Core Thin sheets of soft iron, insulated and stacked Provides low-reluctance path for magnetic flux; reduces eddy currents
Insulation Varnish or paper between laminations Prevents electrical contact between laminations
Frame/Bobbin Mechanical support Holds the assembly together

📌 1.4 TRANSFORMER EQUATION
V_s/V_p =N_s/N_p
Where:
V_s = Secondary voltage (V)
V_p = Primary voltage (V)
N_s = Number of turns in secondary
N_p = Number of turns in primary
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Types of Transformers
Type Turns Ratio Voltage Current Application

Step-Up Transformer N_s>N_p Increases Decreases Power transmission, X-ray machines

Step-Down Transformer N_sV_p, N_s>N_p
Step-down transformer → V_s

WORK DONE IN A FORCE FIELD & ENERGY CONVERSION
1. Meaning of Work in Physics
Work is said to be done when a force causes a body to move through a distance in the direction of the force.
Formula for Work Done
W=Fscos⁡θ
Where:
W= work done
F= applied force
s= displacement
θ= angle between force and displacement
SI Unit of Work
Joule (J)
Dimension of Work
[ML^2 T^(-2) ]

Important Facts
Work is a scalar quantity.
Work is positive if force and displacement are in the same direction.
Work is negative if force opposes motion.
Work is zero if:
displacement is zero
force is perpendicular to displacement
Examples
Pulling a box along a floor
Lifting a bucket from a well
A crane raising a load
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2. WORK DONE IN A FORCE FIELD
A force field is a region where a body experiences force without physical contact.
Examples:
Gravitational field
Electric field
Magnetic field
In WAEC, UTME, AP and IGCSE Physics, emphasis is mainly on the gravitational field.
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3. WORK DONE IN A GRAVITATIONAL FIELD
When a body is lifted vertically upward, work is done against gravity.
Formula
W=mgh
Where:
m= mass (kg)
g= acceleration due to gravity (9.8" " m/s^2 or 10" " m/s^2)
h= height (m)
Derivation
Since:
Force=Weight=mg

And:
Work=Force×distance

Therefore:
W=mg×h

Hence:
W=mgh

SI Unit
Joule (J)
Dimension
[ML^2 T^(-2) ]

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4. POTENTIAL ENERGY (P.E.)
Potential energy is the energy possessed by a body due to its position or configuration.
For gravitational potential energy:
PE=mgh
m
h
PE=mgh=299.88
Gravity is fixed at g = 9.8 in this visualization.
h
Important Facts
P.E. increases with height.
A raised object stores gravitational energy.
P.E. is maximum at the highest point.
Applications
Water stored in dams
Roller coasters
Hydroelectric power stations
Pendulums
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5. KINETIC ENERGY (K.E.)
Kinetic energy is the energy possessed by a moving body.
Formula
KE=1/2 mv^2
m_1
m_2
v
m1m2
Where:
m= mass
v= velocity
Derivation
From equation of motion:
v^2-u^2=2as

Using:
F=ma

Then:
a=(v^2-u^2)/2s

Substitute into force equation:
F=m((v^2-u^2)/2s)

Multiply both sides by s:
Fs=1/2 m(v^2-u^2)

Since Fs=Work:
W=1/2 mv^2-1/2 mu^2

If u=0:
KE=1/2 mv^2

SI Unit
Joule (J)
Dimension
[ML^2 T^(-2) ]

Applications
Moving vehicles
Bullets
Flowing water
Wind turbines
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6. WORK DONE IN LIFTING A BODY
When lifting a body upward:
Work is done against gravity.
Chemical energy in muscles converts to gravitational potential energy.
Formula
W=mgh

Important Notes
The heavier the body, the more work required.
Greater height means greater work done.
Example
A 5" " kgobject is lifted through 10" " m.
Using:
W=mgh
W=5×10×10
W=500" " J

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7. FALLING BODIES
When a body falls freely under gravity:
Potential energy decreases.
Kinetic energy increases.
Total mechanical energy remains constant if air resistance is ignored.
Energy Transformation During Falling
At the top:
Maximum P.E.
Minimum K.E.
During fall:
P.E. converts to K.E.
At the ground:
Minimum P.E.
Maximum K.E.
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8. CONSERVATION OF MECHANICAL ENERGY
The total mechanical energy of a system remains constant provided no energy is lost through friction or air resistance.
Formula
PE+KE="constant"
Or:
mgh+1/2 mv^2="constant"

Example
A 2" " kgstone falls from 20" " m.
At top:
PE=2×10×20=400J

At ground:
KE=400J

Hence:
PE_top=KE_bottom

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9. TRANSFORMATION OF ENERGY
Energy transformation means conversion from one form of energy to another.
According to the law of conservation of energy:
Energy cannot be created or destroyed but can be transformed from one form to another.
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10. COMMON ENERGY CONVERSIONS
Initial Energy Converted Energy Example
Chemical Heat Burning fuel
Chemical Mechanical Car engine
Electrical Light Electric bulb
Electrical Heat Electric iron
Electrical Sound Speaker
Light Electrical Solar panel
Mechanical Electrical Generator
Potential Kinetic Falling object
Nuclear Heat/Electrical Nuclear plant
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11. LAW OF CONSERVATION OF ENERGY
The total energy of an isolated system remains constant.
Key Principle
Energy" " before=Energy" " after

Importance
Basis of modern engineering
Used in machine design
Used in electricity generation
Used in thermodynamics
Applications
Hydroelectric stations
Solar panels
Electric generators
Wind turbines
Internal combustion engines
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12. POWER
Power is the rate of doing work.
Formula
P=W/t
Where:
P= power
W= work done
t= time
SI Unit
Watt (W)
Dimension
[ML^2 T^(-3) ]

Other Forms
P=Fv
P=VI

Applications
Electric motors
Engines
Generators
Electrical appliances
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13. WORLD ENERGY RESOURCES
Energy resources are natural sources from which useful energy is obtained.
WAEC, UTME, AP and IGCSE classify them into:
Renewable resources
Non-renewable resources
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14. RENEWABLE ENERGY RESOURCES
These can be replenished naturally.
Examples
(a) Solar Energy
Energy from the Sun.
Applications:
Solar panels
Solar cookers
Solar dryers
Advantages:
Clean
Unlimited
Environmentally friendly
Disadvantages:
Expensive installation
Depends on sunlight
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(b) Wind Energy
Produced using wind turbines.
Applications:
Electricity generation
Water pumping
Advantages:
Renewable
No pollution
Disadvantages:
Wind speed varies
Noisy turbines
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(c) Hydroelectric Energy
Obtained from falling water.
Applications:
Electricity generation
Advantages:
Cheap after installation
Renewable
Disadvantages:
Expensive dams
Flooding of land
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(d) Biomass Energy
Energy from plant and animal waste.
Applications:
Cooking gas
Biofuel
Advantages:
Cheap
Reduces waste
Disadvantages:
Produces smoke
Deforestation risk
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(e) Tidal and Ocean Wave Energy
Energy from sea tides and waves.
Advantages:
Renewable
Predictable
Disadvantages:
Expensive technology
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15. NON-RENEWABLE ENERGY RESOURCES
These are exhaustible.
Examples
(a) Coal
Advantages:
Cheap
High energy output
Disadvantages:
Air pollution
Greenhouse gases
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(b) Petroleum (Crude Oil)
Uses:
Fuel
Lubricants
Plastics
Advantages:
Easy transport
High energy value
Disadvantages:
Oil spills
Pollution
________________________________________
(c) Natural Gas
Advantages:
Cleaner than coal
Efficient
Disadvantages:
Fire hazard
Limited supply
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(d) Nuclear Energy
Obtained from radioactive materials like uranium.
Advantages:
Very high energy output
Disadvantages:
Radiation danger
Disposal problems
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16. ENERGY CRISIS
Energy crisis occurs when energy demand exceeds supply.
Causes
Population growth
Overdependence on fossil fuels
Poor management
Environmental issues
Solutions
Renewable energy use
Energy conservation
Diversification
Efficient appliances
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17. ENERGY EFFICIENCY
Efficiency measures useful energy output compared to total input.
Formula
η=Useful Energy Output/Total Energy Input×100%
Applications
Machines
Engines
Electrical appliances
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18. IMPORTANT WAEC/UTME/IGCSE/AP EXAM TIPS
Frequently Tested Areas
Derivation of K.E. and P.E.
Energy conversion diagrams
Conservation of energy calculations
Falling body problems
Renewable vs non-renewable energy
Efficiency calculations
Power equations
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19. COMMON EXAM TRAPS
Confusing work with force
Forgetting SI units
Using wrong value of g
Ignoring energy losses
Mixing power and energy
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20. QUICK FORMULA SUMMARY
Quantity Formula
Work Done W=Fscos⁡θ
Potential Energy PE=mgh
Kinetic Energy KE=1/2 mv^2
Power P=W/t
Efficiency η=Output/Input×100%
Mechanical Energy ME=PE+KE
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21. REAL-LIFE APPLICATIONS
Concept Application
Potential Energy Dams
Kinetic Energy Moving vehicles
Conservation of Energy Roller coasters
Solar Energy Solar panels
Wind Energy Turbines
Hydroelectric Energy Power stations
Nuclear Energy Electricity generation
Power Engines and motors
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22. SUMMARY
Work is done when force causes displacement.
Energy is the ability to do work.
Potential energy depends on height.
Kinetic energy depends on motion.
Mechanical energy is conserved in the absence of friction.
Energy changes from one form to another.
Renewable energy sources are sustainable.
Non-renewable energy sources are exhaustible.
Efficient energy use reduces waste and environmental damage.
The topics “Work Done in a Force Field”, “Conservation of Energy”, “Energy Conversion”, and “World Energy Resources” are major parts of the WAEC, UTME, AP and IGCSE Physics syllabuses.