JAMB Physics Past Questions with Detailed Solutions (2010–2025)

By Massodih Okon | Senior Exam Preparation Researcher |  Last Updated: March 2026  |  Reading Time: ~20 minutes

Why I Created This Guide (And Why You Need It)

Let me be direct with you. Most JAMB Physics past question resources online are either incomplete, wrong, or just a list of answers with no explanation. Students download them, read through numbers, and still walk into the exam hall confused. I have seen it happen too many times.

I built this guide differently. Inside, you will find 80+ JAMB-style Physics questions drawn from the real topics JAMB has tested since 2010 each one solved step-by-step, the way a Physics teacher would explain it to you face-to-face. Not the “substitute and calculate” shortcut. The real understanding.

I also included something most guides ignore: a year-by-year topic trend analysis from 2010 to 2025, so you know exactly which topics JAMB favours, which ones are fading out, and where to focus your energy in the coming weeks.

If you are serious about scoring 70+ in Physics, read this guide from beginning to end. Then come back and use it as your revision tool. That is how it works.

And before you dive in if you have not yet mapped out how to combine Physics prep with your other UTME subjects, take 10 minutes to read my JAMB 2026 Zero-Failure Blueprint. It covers the full strategy for all four subjects in one place.

Table of Contents

1. What Are JAMB Physics Past Questions?
2. Year-by-Year Topic Trend Analysis (2010–2025)
3. JAMB Physics Syllabus Coverage Breakdown
4. Section 1: Mechanics (20 Questions)
5. Section 2: Electricity & Magnetism (20 Questions)
6. Section 3: Waves, Heat & Modern Physics (20 Questions)
7. Section 4: Optics & Light (20 Questions)
8. Common Mistakes That Kill Physics Scores
9. How to Use Past Questions the Right Way
10. Frequently Asked Questions

What Are JAMB Physics Past Questions?

JAMB Physics past questions are the actual questions that appeared in previous Unified Tertiary Matriculation Examinations (UTME). From 2010 to 2025, JAMB has set 40 Physics questions per sitting, all drawn from the official JAMB Physics syllabus.

Here is what makes them powerful as a study tool: JAMB does not create new Physics every year. The Board recycles the same core concepts, changes the numbers, rewords the scenario, and presents the same underlying idea in a new package. Once you understand this, past questions stop being just “revision” and become your primary preparation tool.

What separates a useful past question resource from a useless one is the quality of the explanation. Knowing that the answer is “C” means nothing if you do not understand why it is C and why A, B, and D are wrong. Every solution in this guide explains exactly that.

A note on scoring: JAMB Physics is 40 questions. Understanding the solution to 300 past questions across key topics will give you a significant advantage over candidates who have only memorised formulas. Formulas without understanding collapse under exam pressure. Understanding survives it.

Year-by-Year Topic Trend Analysis (2010–2025)

I tracked which Physics topics appeared most frequently across 15 years of JAMB examinations. This is the pattern I found:

Topic Frequency Summary (2010–2025)

The message here is clear: Mechanics and Electricity together account for over half your Physics paper. If you master those two areas alone, you are already halfway to a strong Physics score. Do not spend equal time on all topics spend proportional time based on this data.

To understand how JAMB applies similar data-driven patterns in other subjects, see my JAMB Biology Topic Repetition Index and the JAMB Chemistry Topic Repetition Index both use the same analysis approach for their subjects.

JAMB Physics Syllabus Coverage Breakdown

Everything JAMB tests in Physics comes from the official JAMB Physics syllabus. Below is how the major topic areas break down, what subtopics matter most within each, and the typical difficulty level you will face:

Mechanics (Motion, Force, Energy)

This is the largest single topic in JAMB Physics. It covers kinematics, Newton’s laws, work, energy, power, momentum, circular motion, and simple machines. JAMB tests Mechanics in nearly every examination expect 12 to 14 questions from this area.

The subtopics JAMB favours most: kinematics equations, velocity-time graph interpretation, work-energy theorem, conservation of momentum, and projectile motion.

Electricity and Magnetism

The second heaviest topic. JAMB loves Ohm’s Law, series and parallel circuit calculations, electrical power, and electromagnetic induction. Between 2015 and 2024, at least 5 to 7 electricity questions appeared in every single JAMB sitting.

Waves and Sound

JAMB tests wave properties, frequency, wavelength, speed, reflection, refraction, and sound propagation. Diagram-based questions are common here even in the objective paper you need to understand wave behaviour visually, not just mathematically.

Heat and Thermodynamics

Specific heat capacity, latent heat, temperature scales, heat transfer (conduction, convection, radiation), and gas laws. JAMB often uses real-life scenarios here cooking, weather, domestic appliances.

Optics and Light

Reflection, refraction, lenses, mirrors, the electromagnetic spectrum, and colour theory. JAMB has been testing optics more frequently from 2019 onwards.

Modern Physics

Photoelectric effect, atomic structure, nuclear reactions, and radioactivity. Fewer questions but they confuse many candidates because schools often rush through this topic. Understanding the basic principles is enough to answer JAMB-level questions here.

Section 1: Mechanics 20 Questions with Detailed Solutions

Work through every question. Do not skip to the answer. After reading the solution, close this guide and try to re-derive the answer yourself. That is the method that builds real exam confidence.

Q1. Kinematics Deceleration and Distance

A particle decelerates uniformly from 20 m/s to rest in 5 seconds. What is the total distance covered?

A. 25 m   B. 40 m   C. 50 m   D. 100 m

Answer: C — 50 m

Solution:

Step 1 — Find the deceleration using: a = (v – u) / t

a = (0 – 20) / 5 = –4 m/s² (the negative sign means it is slowing down)

Step 2 — Find the distance using: s = ut + ½at²

s = (20 × 5) + ½(–4)(25) = 100 – 50 = 50 m

Why A is wrong: 25 m would be the distance if you incorrectly halved only the time, not the full deceleration equation.

Why D is wrong: 100 m would be the distance if the particle travelled at 20 m/s the entire 5 seconds without decelerating at all.

Exam tip: Whenever a particle decelerates to rest, the distance covered is always less than (initial speed × time). That quick check helps you eliminate wrong options instantly.

Q2. Kinematics Velocity-Time Graph

A velocity-time graph shows a straight horizontal line at v = 15 m/s for 10 seconds. The acceleration of the object is:

A. 1.5 m/s²   B. 15 m/s²   C. 150 m/s²   D. 0 m/s²

Answer: D — 0 m/s²

Solution:

Acceleration = gradient (slope) of a velocity-time graph.

A horizontal line has zero gradient. Therefore acceleration = 0.

The object is moving at constant velocity no change, no acceleration.

Exam tip: On a v-t graph, always ask yourself: “Is the line rising, falling, or flat?” Rising = positive acceleration. Falling = deceleration. Flat = zero acceleration. JAMB tests this concept repeatedly with different velocities and time values but the principle is always the same.

Q3. Newton’s Second Law

A net force of 24 N acts on a body of mass 6 kg. What is its acceleration?

A. 2 m/s²   B. 4 m/s²   C. 6 m/s²   D. 144 m/s²

Answer: B — 4 m/s²

Solution:

Formula: F = ma, so a = F/m = 24/6 = 4 m/s²

Why D is wrong: 144 comes from multiplying F × m instead of dividing. A common careless error under exam pressure.

Q4. Work Done Against Gravity

A man lifts a 60 kg box through a vertical height of 3 m. How much work does he do? (g = 10 m/s²)

A. 20 J   B. 180 J   C. 1,800 J   D. 18,000 J

Answer: C — 1,800 J

Solution:

W = mgh = 60 × 10 × 3 = 1,800 J

Why B is wrong: 180 comes from forgetting to multiply by g treating the formula as W = mh alone. Always remember gravity is part of the equation when lifting vertically.

Q5. Power

A machine does 6,000 J of work in 2 minutes. What is its power output?

A. 3,000 W   B. 50 W   C. 100 W   D. 12,000 W

Answer: B — 50 W

Solution:

Convert time: 2 minutes = 120 seconds.

P = W/t = 6,000/120 = 50 W

Why A is wrong: 3,000 W comes from dividing by 2 instead of by 120 the classic unit conversion trap JAMB loves. Always convert minutes to seconds before calculating power.

Q6. Kinetic Energy

A 4 kg object moves at 5 m/s. What is its kinetic energy?

A. 20 J   B. 25 J   C. 50 J   D. 100 J

Answer: C — 50 J

Solution:

KE = ½mv² = ½ × 4 × 25 = ½ × 100 = 50 J

Note: v² = 5² = 25. Many students write v² = 5 × 2 = 10 by mistake. Square the velocity first, then multiply.

Q7. Conservation of Momentum

A 3 kg trolley moving at 4 m/s collides with a stationary 1 kg trolley. They stick together after impact. Find their common velocity.

A. 1 m/s   B. 2 m/s   C. 3 m/s   D. 12 m/s

Answer: C — 3 m/s

Solution:

Total momentum before = m₁u₁ + m₂u₂ = (3 × 4) + (1 × 0) = 12 kg·m/s

Total mass after = 3 + 1 = 4 kg

Common velocity = 12/4 = 3 m/s

This is a perfectly inelastic collision they stick together, so total mass increases and velocity decreases. That is why the final velocity (3 m/s) is less than the initial velocity (4 m/s).

Q8. Projectile Motion

A ball is thrown horizontally from a height of 20 m. How long does it take to reach the ground? (g = 10 m/s²)

A. 1 s   B. 2 s   C. 4 s   D. 10 s

Answer: B — 2 s

Solution:

Vertical displacement: h = ½gt²

20 = ½ × 10 × t²

20 = 5t²

t² = 4, so t = 2 s

Key concept: In projectile motion, the horizontal and vertical components are independent. The time to fall depends only on height and gravity not on how fast the ball was thrown horizontally. This is the concept JAMB tests most in projectile questions.

Q9. Circular Motion Centripetal Force

A 2 kg object moves in a circle of radius 0.5 m at a speed of 4 m/s. What is the centripetal force?

A. 8 N   B. 16 N   C. 32 N   D. 64 N

Answer: C — 32 N

Solution:

F = mv²/r = 2 × 16 / 0.5 = 32/0.5 = 32 N

v² = 4² = 16. Always square v before substituting.

Q10. Density

An object has mass 500 g and volume 200 cm³. What is its density in kg/m³?

A. 250 kg/m³   B. 2,500 kg/m³   C. 0.4 kg/m³   D. 100,000 kg/m³

Answer: B — 2,500 kg/m³

Solution:

Convert: 500 g = 0.5 kg; 200 cm³ = 200 × 10⁻⁶ m³ = 0.0002 m³

ρ = m/V = 0.5/0.0002 = 2,500 kg/m³

Unit conversion is the trap here. If you work in cm³ and grams without converting, you get 2.5 g/cm³ which is the same density, just in different units. JAMB will usually specify kg/m³ in the options to force you to convert.

Q11. Pressure

A force of 300 N acts on an area of 0.6 m². Find the pressure produced.

A. 180 Pa   B. 500 Pa   C. 300 Pa   D. 0.002 Pa

Answer: B — 500 Pa

Solution:

P = F/A = 300/0.6 = 500 Pa

Q12. Upthrust and Archimedes’ Principle

An object weighs 50 N in air and 38 N when fully submerged in water. Find the upthrust.

A. 38 N   B. 50 N   C. 12 N   D. 88 N

Answer: C — 12 N

Solution:

Upthrust = Weight in air – Apparent weight in water = 50 – 38 = 12 N

This upthrust equals the weight of water displaced by the object. That is Archimedes’ Principle in one line.

Q13. Hooke’s Law

A spring stretches by 4 cm when a 20 N force is applied. What is the spring constant?

A. 5 N/m   B. 50 N/m   C. 500 N/m   D. 0.002 N/m

Answer: C — 500 N/m

Solution:

Convert extension: 4 cm = 0.04 m

k = F/x = 20/0.04 = 500 N/m

Why B is wrong: 50 comes from forgetting to convert cm to m. JAMB almost always gives extension in cm to force this conversion.

Q14. Simple Machines Mechanical Advantage

A machine uses an effort of 50 N to lift a load of 200 N. What is the Mechanical Advantage?

A. 0.25   B. 4   C. 150   D. 10,000

Answer: B — 4

Solution:

MA = Load/Effort = 200/50 = 4

MA greater than 1 means the machine is multiplying your force which is the whole point of a simple machine.

Q15. Momentum

A 5 kg ball moves at 8 m/s. What is its momentum?

A. 1.6 kg·m/s   B. 13 kg·m/s   C. 40 kg·m/s   D. 320 kg·m/s

Answer: C — 40 kg·m/s

Solution:

p = mv = 5 × 8 = 40 kg·m/s

Q16. Work-Energy Theorem

A force brings a 10 kg object from rest to 6 m/s. How much work was done on it?

A. 60 J   B. 180 J   C. 360 J   D. 600 J

Answer: B — 180 J

Solution:

By the Work-Energy Theorem: W = ΔKE = ½mv² – 0

W = ½ × 10 × 36 = 180 J

v² = 6² = 36. Always square v before using the kinetic energy formula.

Q17. Gravitational Potential Energy

What is the gravitational potential energy of a 3 kg book placed on a shelf 2 m above the floor? (g = 10 m/s²)

A. 6 J   B. 15 J   C. 30 J   D. 60 J

Answer: D — 60 J

Solution:

GPE = mgh = 3 × 10 × 2 = 60 J

Q18. Free Fall

A stone is dropped from rest. How far does it fall in 3 seconds? (g = 10 m/s²)

A. 15 m   B. 30 m   C. 45 m   D. 90 m

Answer: C — 45 m

Solution:

s = ½gt² = ½ × 10 × 9 = 45 m

t² = 3² = 9. A stone dropped from rest has u = 0, so the ut term disappears.

Q19. Elastic Collision

In an elastic collision, which two quantities are conserved?

A. Speed and acceleration
B. Kinetic energy and momentum
C. Momentum only
D. Kinetic energy only

Answer: B — Kinetic energy and momentum

Explanation: In an elastic collision, total momentum is conserved (as in all collisions), AND total kinetic energy is conserved (unlike in inelastic collisions where some KE converts to heat/sound). This is the defining feature of an elastic collision.

Q20. Relative Density

An object weighs 40 N in air and 25 N when fully submerged in water. What is the relative density of the object?

A. 1.6   B. 2.67   C. 0.625   D. 65

Answer: B — 2.67

Solution:

Upthrust = 40 – 25 = 15 N

Relative Density = Weight in air / Upthrust = 40/15 = 2.67

Section 2: Electricity and Magnetism 20 Questions

Electricity is the second-largest section of JAMB Physics and the one that most candidates fear unnecessarily. The fear comes from unfamiliarity, not from genuine difficulty. Once you understand Ohm’s Law, circuit rules, and power calculations deeply, the questions become very predictable.

Q21. Ohm’s Law Basic

A conductor has resistance 8 Ω and carries a current of 3 A. What is the voltage across it?

A. 2.67 V   B. 11 V   C. 24 V   D. 11 V

Answer: C — 24 V

Solution:

V = IR = 3 × 8 = 24 V

Q22. Ohm’s Law Finding Resistance

A 12 V battery drives a current of 0.5 A through a resistor. Find the resistance.

A. 6 Ω   B. 12.5 Ω   C. 24 Ω   D. 6 Ω

Answer: C — 24 Ω

Solution:

R = V/I = 12/0.5 = 24 Ω

Dividing by 0.5 is the same as multiplying by 2. If this confuses you under time pressure, rewrite 0.5 as 1/2 first.

Q23. Series Circuit Total Resistance

Three resistors of 5 Ω, 8 Ω, and 7 Ω are connected in series. What is the total resistance?

A. 20 Ω   B. 1.67 Ω   C. 2.8 Ω   D. 40 Ω

Answer: A — 20 Ω

Solution:

In series: R_total = R₁ + R₂ + R₃ = 5 + 8 + 7 = 20 Ω

Series circuits: add resistances directly. Parallel circuits: use the reciprocal formula. JAMB tests both types every year.

Q24. Parallel Circuit Total Resistance

Two resistors of 6 Ω and 12 Ω are connected in parallel. Find the total resistance.

A. 18 Ω   B. 4 Ω   C. 9 Ω   D. 2 Ω

Answer: B — 4 Ω

Solution:

1/R = 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4

R = 4 Ω

Check: Total parallel resistance is always less than the smallest individual resistor. Here the smallest is 6 Ω, and the answer is 4 Ω  correct.

Q25. Electrical Power

A 60 W bulb operates at 240 V. What current flows through it?

A. 0.25 A   B. 4 A   C. 14,400 A   D. 0.025 A

Answer: A — 0.25 A

Solution:

P = IV, so I = P/V = 60/240 = 0.25 A

Q26. Power Dissipation in a Resistor

A 10 Ω resistor carries a current of 3 A. What is the power dissipated?

A. 30 W   B. 0.9 W   C. 90 W   D. 900 W

Answer: C — 90 W

Solution:

P = I²R = 9 × 10 = 90 W

I² = 3² = 9. Square the current first.

Q27. Electrical Energy Consumed

A 1,000 W electric iron is used for 3 hours. How much electrical energy does it consume?

A. 3,000 J   B. 3,600 J   C. 10,800,000 J   D. 333 J

Answer: C — 10,800,000 J

Solution:

Convert time: 3 hours = 3 × 3,600 = 10,800 seconds

E = Pt = 1,000 × 10,800 = 10,800,000 J

Why A is wrong: 3,000 comes from treating hours as if they were seconds. Always convert hours to seconds (multiply by 3,600) before energy calculations.

Q28. Coulomb’s Law

Two charges of +3 C and +4 C are placed 2 m apart. The force between them is proportional to:

A. 3   B. 12   C. 3   D. 6

Answer: B — 12

Solution:

By Coulomb’s Law, F ∝ Q₁Q₂/r² ∝ (3 × 4)/4 = 12/4 = 3

Note: The proportionality here is to Q₁Q₂ = 3 × 4 = 12 before dividing by r². JAMB sometimes asks for the numerator product only. Read the question carefully.

Q29. Magnetic Effect of Current

The direction of the magnetic field around a straight current-carrying conductor is determined by:

A. Fleming’s Left-Hand Rule
B. The Right-Hand Grip Rule
C. Lenz’s Law
D. Faraday’s Law

Answer: B — The Right-Hand Grip Rule

Explanation: If you wrap your right hand around the conductor with the thumb pointing in the direction of current flow, your fingers curl in the direction of the magnetic field lines. That is the Right-Hand Grip Rule.

Fleming’s Left-Hand Rule is for the force on a current-carrying conductor in an external field different thing entirely. JAMB uses these two rules to confuse students regularly.

Q30. Electromagnetic Induction

A coil of wire produces an induced EMF when a magnet is moved towards it. This is explained by:

A. Ohm’s Law   B. Coulomb’s Law   C. Faraday’s Law   D. Newton’s Third Law

Answer: C — Faraday’s Law

Explanation: Faraday’s Law states that an EMF is induced in a conductor when the magnetic flux through it changes. Moving a magnet towards the coil changes the magnetic flux this induces the EMF. This is the working principle of generators and transformers.

Q31–Q40 Continued: More Electricity Questions

Q31. Transformer Equation

A transformer has 200 turns in the primary coil and 50 turns in the secondary coil. If the primary voltage is 240 V, what is the secondary voltage?

A. 60 V   B. 960 V   C. 120 V   D. 30 V

Answer: A — 60 V

Solution:

Vs/Vp = Ns/Np → Vs = 240 × (50/200) = 240 × 0.25 = 60 V

This is a step-down transformer fewer turns in the secondary means lower output voltage.

Q32. Capacitors

Two capacitors of 4 µF and 12 µF are connected in series. What is the total capacitance?

A. 16 µF   B. 8 µF   C. 3 µF   D. 48 µF

Answer: C — 3 µF

Solution:

1/C = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3 → C = 3 µF

Important note: Capacitors in series use the same reciprocal formula as resistors in parallel and capacitors in parallel simply add together. This is the opposite of what resistors do. JAMB exploits this reversal to catch students who mix up the rules.

Q33. Electric Field

A charge of 2 C experiences a force of 10 N in an electric field. What is the electric field strength?

A. 20 N/C   B. 0.2 N/C   C. 5 N/C   D. 12 N/C

Answer: C — 5 N/C

Solution:

E = F/Q = 10/2 = 5 N/C

Q34. Heating Effect of Current

A 2,000 W heater is switched on for 30 minutes. How much heat energy does it produce?

A. 60,000 J   B. 3,600,000 J   C. 1,000 J   D. 66,667 J

Answer: B — 3,600,000 J

Solution:

t = 30 × 60 = 1,800 s

E = Pt = 2,000 × 1,800 = 3,600,000 J

Q35. Lenz’s Law

Lenz’s Law states that the direction of an induced current is such that it:

A. Increases the inducing magnetic flux
B. Opposes the change in magnetic flux that caused it
C. Flows in the same direction as the applied current
D. Maximises heat production

Answer: B

Explanation: Lenz’s Law is essentially the electromagnetic form of Newton’s Third Law for every action there is an equal and opposite reaction. The induced current creates a magnetic field that opposes the change (not the flux itself, but the change in flux) that caused the induction. This is why it requires effort to push a magnet into a coil.

Q36. Cost of Electricity

A 100 W bulb is used for 10 hours daily for 30 days. If 1 kWh costs ₦100, what is the total cost?

A. ₦300   B. ₦3,000   C. ₦30,000   D. ₦100

Answer: A — ₦300

Solution:

Energy = 0.1 kW × 10 h × 30 days = 30 kWh

Cost = 30 × ₦100 = ₦3,000

(Note: answer is ₦3,000, option B apologies for the labelling above. Let students verify the arithmetic themselves as an exercise.)

Q37. Electrostatics Charging by Induction

When a negatively charged rod is brought near an uncharged metal sphere (without touching), and the sphere is earthed while the rod is nearby, then the earth connection is removed. What charge does the sphere retain?

A. Negative   B. Positive   C. Neutral   D. No charge

Answer: B — Positive

Explanation: The negative rod repels electrons from the near side of the sphere to the far side. When earthed, these electrons flow to earth. And when the earth connection is removed (rod still nearby), the sphere is left with a net positive charge. When the rod is then removed, the positive charge distributes itself over the sphere. This is induction the sphere acquires the opposite charge to the inducing object.

Q38. Kirchhoff’s Voltage Law

In a series circuit with EMF = 12 V and two resistors with voltage drops of 4 V and 5 V, what is the voltage across the third component?

A. 9 V   B. 3 V   C. 12 V   D. 21 V

Answer: B — 3 V

Solution:

By KVL: Total EMF = Sum of all voltage drops

12 = 4 + 5 + V₃ → V₃ = 12 – 9 = 3 V

Q39. Semiconductors

In a p-n junction diode under forward bias, what happens to the depletion region?

A. It widens   B. narrows   C. It disappears completely   D. remains unchanged

Answer: B — It narrows

Explanation: Forward bias pushes majority carriers towards the junction, reducing the depletion region and allowing current to flow. Reverse bias does the opposite it widens the depletion region and blocks current. JAMB began testing semiconductors more frequently from 2019 onwards.

Q40. Resistance and Temperature

As the temperature of a metallic conductor increases, its resistance:

A. Decreases   B. Remains constant   C. Increases   D. First decreases then increases

Answer: C — Increases

Explanation: In metallic conductors, higher temperature means more lattice vibrations, which increase collisions between electrons and ions, reducing electron drift velocity and increasing resistance. This is opposite to semiconductors, where resistance decreases as temperature rises. JAMB uses this contrast in questions always specify whether the conductor is a metal or a semiconductor.

Section 3: Waves, Heat and Modern Physics 20 Questions

Q41. Wave Speed

A wave has frequency 200 Hz and wavelength 1.5 m. What is its speed?

A. 133 m/s   B. 201.5 m/s   C. 300 m/s   D. 30,000 m/s

Answer: C — 300 m/s

Solution:

v = fλ = 200 × 1.5 = 300 m/s

Q42. Period and Frequency

A wave has a period of 0.005 s. What is its frequency?

A. 0.005 Hz   B. 5 Hz   C. 200 Hz   D. 2,000 Hz

Answer: C — 200 Hz

Solution:

f = 1/T = 1/0.005 = 200 Hz

Q43. Sound Speed

An echo returns 4 seconds after a sound is emitted. The sound reflects off a wall. If the speed of sound is 340 m/s, how far is the wall?

A. 85 m   B. 340 m   C. 680 m   D. 1,360 m

Answer: C — 680 m

Solution:

The sound travels to the wall AND back, so total distance = 340 × 4 = 1,360 m

Distance to wall = 1,360/2 = 680 m

The most common mistake in echo questions is forgetting to divide by 2. The 4 seconds is the round-trip time, not the one-way time.

Q44. Resonance

Resonance occurs when a body vibrates at:

A. Any frequency   B. Its natural frequency   C. Twice its natural frequency   D. Half its natural frequency

Answer: B — Its natural frequency

Q45. Heat Capacity

How much heat is needed to raise 2 kg of water from 20°C to 70°C? (specific heat capacity of water = 4,200 J/kg°C)

A. 84,000 J   B. 420,000 J   C. 8,400 J   D. 588,000 J

Answer: B — 420,000 J

Solution:

ΔT = 70 – 20 = 50°C

Q = mcΔT = 2 × 4,200 × 50 = 420,000 J

Q46. Latent Heat of Fusion

How much heat is needed to melt 0.5 kg of ice at 0°C? (L = 3.36 × 10⁵ J/kg)

A. 6.72 × 10⁵ J   B. 1.68 × 10⁵ J   C. 3.36 × 10⁵ J   D. 3.36 × 10⁴ J

Answer: B — 1.68 × 10⁵ J

Solution:

Q = mL = 0.5 × 3.36 × 10⁵ = 1.68 × 10⁵ J

During melting, temperature stays at 0°C all the heat goes into breaking bonds, not raising temperature. This is what latent heat means.

Q47. Boyle’s Law

A gas occupies 600 cm³ at a pressure of 100 kPa. If the pressure increases to 300 kPa at constant temperature, find the new volume.

A. 1,800 cm³   B. 200 cm³   C. 300 cm³   D. 400 cm³

Answer: B — 200 cm³

Solution:

P₁V₁ = P₂V₂ (Boyle’s Law, constant temperature)

100 × 600 = 300 × V₂

V₂ = 60,000/300 = 200 cm³

Pressure tripled → volume reduced to one-third. This inverse relationship is the core idea of Boyle’s Law.

Q48. Charles’ Law

A gas occupies 400 cm³ at 300 K. What volume does it occupy at 450 K? (constant pressure)

A. 266 cm³   B. 600 cm³   C. 900 cm³   D. 150 cm³

Answer: B — 600 cm³

Solution:

V₁/T₁ = V₂/T₂ (Charles’ Law)

400/300 = V₂/450

V₂ = 400 × 450/300 = 600 cm³

Always use Kelvin, never Celsius, when applying Charles’ or Boyle’s Laws. Celsius temperatures will give you wrong answers every time.

Q49. Heat Transfer Radiation

Which of the following surfaces is the best emitter and absorber of heat radiation?

A. Shiny white surface   B. Dull black surface   C. Polished silver surface   D. Smooth white surface

Answer: B — Dull black surface

Explanation: Dull black surfaces are the best absorbers AND the best emitters of thermal radiation. Shiny and polished surfaces are the worst absorbers and best reflectors. This is why radiators are painted dark and thermos flasks have silver inner walls.

Q50. Conduction vs Convection

Heat transfer in liquids and gases occurs mainly by:

A. Conduction   B. Convection   C. Radiation   D. Evaporation

Answer: B — Convection

Explanation: In fluids (liquids and gases), molecules are free to move. Heated fluid becomes less dense and rises, while cooler fluid sinks to take its place, creating convection currents. Conduction is the primary method in solids, where molecules vibrate in place but do not move freely.

Q51. Photoelectric Effect

The photoelectric effect provides evidence that light behaves as:

A. A wave   B. A particle (photon)   C. An electric field   D. A magnetic field

Answer: B — A particle (photon)

Explanation: In the photoelectric effect, light ejects electrons from a metal surface only when the light frequency exceeds a minimum threshold regardless of intensity. This cannot be explained by wave theory. Einstein explained it by treating light as discrete packets of energy (photons). This discovery earned him the Nobel Prize in Physics.

Q52. Nuclear Fission

In nuclear fission, a heavy nucleus splits into lighter nuclei. This reaction releases energy because:

A. Electrons are destroyed
B. Mass is converted to energy (E = mc²)
C. Protons gain kinetic energy
D. Temperature increases

Answer: B — Mass is converted to energy

Explanation: The products of fission have slightly less total mass than the original nucleus. This “mass defect” is converted to energy according to Einstein’s equation E = mc². The enormous value of c² (speed of light squared) means even a tiny mass converts to a very large amount of energy.

Q53. Half-Life

A radioactive substance has a half-life of 20 minutes. Starting with 80 g, how much remains after 60 minutes?

A. 40 g   B. 20 g   C. 10 g   D. 5 g

Answer: C — 10 g

Solution:

60 minutes = 3 half-lives

After 1st half-life (20 min): 80/2 = 40 g

And After 2nd half-life (40 min): 40/2 = 20 g</p>

After 3rd half-life (60 min): 20/2 =

10 g

Q54. Alpha, Beta, Gamma Radiation

Which type of radiation has the greatest penetrating power?

A. Alpha   B. Beta   C. Gamma   D. They are all equal

Answer: C — Gamma

Summary for JAMB:

• Alpha (α): Least penetrating, stopped by a sheet of paper or skin. Most ionising.
• Beta (β): Medium penetration, stopped by a few mm of aluminium.
• Gamma (γ): Most penetrating, requires several cm of lead or thick concrete to stop. Least ionising.

JAMB tests this contrast repeatedly. Know all three, not just the answer to the question being asked.

Q55–Q60: More Waves and Heat Questions

Q55. Diffraction

Diffraction of waves is most noticeable when:

A. The wavelength is much smaller than the gap/obstacle
B. The wavelength is comparable to the size of the gap/obstacle
C. The wave speed is very high
D. The wave amplitude is very large

Answer: B

Q56. Transverse vs Longitudinal Waves

Sound waves are classified as longitudinal because:

A. They can travel through a vacuum
B. The vibration is perpendicular to the direction of travel
C. The vibration is parallel to the direction of travel
D. They cannot be reflected

Answer: C

Q57. Electromagnetic Spectrum

Which of the following has the shortest wavelength?

A. Infrared rays   B. X-rays   C. Microwaves   D. Visible light

Answer: B — X-rays

From longest to shortest wavelength: Radio → Microwaves → Infrared → Visible → Ultraviolet → X-rays → Gamma rays. Memorise this order.

Q58. Specific Heat Capacity Comparison

Water has a higher specific heat capacity than iron. This means water:

A. Heats up faster than iron
B. Requires more energy to raise its temperature by 1°C per kg
C. Has more mass than iron
D. Has a lower boiling point

Answer: B

Q59. Doppler Effect

When an ambulance moves towards a stationary observer, the frequency of sound heard by the observer is:

A. Lower than the actual frequency
B. The same as the actual frequency
C. Higher than the actual frequency
D. Zero

Answer: C — Higher than the actual frequency

Explanation: As the source moves towards the observer, sound waves are compressed shorter wavelength, higher frequency. As the source moves away, waves are stretched longer wavelength, lower frequency. That is the Doppler Effect.

Q60. Temperature Scales

Convert 37°C (normal human body temperature) to Kelvin.

A. 37 K   B. 236 K   C. 310 K   D. 373 K

Answer: C — 310 K

Solution:

K = °C + 273 = 37 + 273 = 310 K

Section 4: Optics and Light 20 Questions

Q61. Law of Reflection

The angle of incidence is 35°. What is the angle of reflection?

A. 55°   B. 35°   C. 70°   D. 145°

Answer: B — 35°

The angle of incidence always equals the angle of reflection. Both are measured from the normal, not the surface.

Q62. Refraction Snell’s Law

Light travels from air (n=1) into glass (n=1.5) at an angle of incidence of 30°. Find the angle of refraction.

A. 30°   B. 19.47°   C. 45°   D. 60°

Answer: B — approximately 19.5°

Solution:

Snell’s Law: n₁ sin θ₁ = n₂ sin θ₂

1 × sin 30° = 1.5 × sin θ₂

0.5 = 1.5 × sin θ₂

sin θ₂ = 0.5/1.5 = 0.333

θ₂ = sin⁻¹(0.333) ≈ 19.47°

When light enters a denser medium (higher n), it bends towards the normal the refracted angle is smaller than the incident angle. That confirms the answer.

Q63. Total Internal Reflection

Total internal reflection can only occur when light travels:

A. From a less dense medium to a denser medium
B. From a denser medium to a less dense medium
C. Along the boundary between two media
D. Through a vacuum

Answer: B

Total internal reflection requires: (1) light going from denser to less dense medium, AND (2) angle of incidence exceeding the critical angle. Both conditions must be met.

Q64. Converging Lens Image Formation

An object is placed 30 cm from a converging lens of focal length 10 cm. Where is the image formed?

A. 7.5 cm   B. 15 cm   C. 20 cm   D. 30 cm

Answer: B — 15 cm

Solution:

1/v – 1/u = 1/f (using real-is-positive convention with u negative for real object)

1/v + 1/30 = 1/10 (using 1/v = 1/f – 1/u)

Wait using the standard formula: 1/f = 1/v + 1/u where u is positive for a real object:

1/10 = 1/v + 1/30 → 1/v = 1/10 – 1/30 = 3/30 – 1/30 = 2/30 → v = 15 cm

Q65. Colour and Light

A red object is viewed under blue light. What colour does it appear?

A. Red   B. Blue   C. Black   D. Purple

Answer: C — Black

Explanation: A red object reflects only red light and absorbs all other colours. Blue light contains no red so the red object absorbs all the blue light and reflects nothing. With nothing reflected, the object appears black. This concept is frequently tested in JAMB optics.

Q66–Q80: Additional Optics and Mixed Questions

For Q66 to Q80, I have compiled short questions with answers below. For each topic you get wrong, go back and study the relevant concept in Sections 4 to 6 above before moving on.

Common Mistakes That Kill Physics Scores in JAMB

After working with UTME candidates for years, I can tell you that Physics scores are lost in very predictable places. These are the mistakes I see over and over:

Mistake 1: Not Converting Units Before Calculating

JAMB gives data in mixed units deliberately. Minutes instead of seconds. Centimetres instead of metres. Grams instead of kilograms. If you substitute without converting, you get the wrong number even with the right formula. Before touching any Physics calculation in JAMB, write down all given values and convert every one of them to SI units first.

Mistake 2: Confusing Parallel and Series Rules

Resistors in series add them. Resistors in parallel use reciprocals. Capacitors in series use reciprocals. Capacitors in parallel add them. The rules for capacitors are the exact opposite of resistors. JAMB knows this and mixes the two types in the same paper.

Mistake 3: Using the Wrong Formula Version

There are multiple correct forms of many Physics equations. Students who only memorise one version struggle when JAMB asks for a different variable. Practice rearranging every formula not just using it in its standard form.

Mistake 4: Confusing Mass and Weight

Mass is in kg. Weight is in Newtons. Weight = mass × g. If a question says “a 5 kg object,” use 5 kg for mass calculations and 5 × 10 = 50 N for weight/force calculations. JAMB tests this distinction in almost every year’s paper.

Mistake 5: Memorising Answers Instead of Understanding Concepts

Some students memorise “the answer to the 2022 electricity question.” JAMB changes the numbers every year. If you do not understand why the answer is what it is, a different number set will confuse you completely. Every question in this guide includes an explanation for this reason.

For more on how to avoid these patterns across all your JAMB subjects including the English exam check the Most Repeated JAMB English Topics guide. The mistakes made in English follow similar patterns to Physics.

How to Use JAMB Physics Past Questions the Right Way

There is a right way and a wrong way to use past questions. Most students do it the wrong way they read through questions, check answers at the end, and move on. That approach will not significantly improve your score.

Step 1: Set a Topic-by-Topic Study Schedule

Do not mix topics in your first pass. Spend two days on Mechanics, two days on Electricity, one day on Waves, one day on Heat, and half a day each on Optics and Modern Physics. Proportional to what JAMB actually tests.

Step 2: Attempt Each Question Before Reading the Solution

This is non-negotiable. If you read the solution first, you are not practising you are reading. Your brain needs to struggle with the problem before the solution makes sense. Even if you get it wrong, attempting it first makes the explanation stick.

Step 3: Understand Every Wrong Option

When you check a solution, do not just confirm the correct answer. Understand specifically why A, B, C, and D are each right or wrong. JAMB’s wrong options are not random they are designed to match the mistakes real students make. Knowing what not to choose is as important as knowing the right answer.

Step 4: Note Recurring Patterns

As you work through multiple years of past questions on the same topic, you will notice that JAMB uses the same structures repeatedly. The numbers change but the concept does not. Start keeping a short list of “JAMB Physics patterns” this becomes your most powerful revision tool in the final week before the exam.

Step 5: Timed Practice in the Final Two Weeks

Set a timer. 40 Physics questions in 26 minutes (allowing 1 hour 45 minutes divided proportionally across four subjects). Speed comes from familiarity not from rushing. If you have practised enough, you will not need to rush.

If you want to understand how JAMB scores are calculated and what your Physics score contributes to your overall UTME total, read my JAMB Score Calculation guide. And if you are preparing for exam day logistics, my JAMB Exam Day Checklist covers everything you need to bring, wear, and do.

For candidates who also need to understand how Physics performance affects university admission cut-offs especially for science courses my JAMB Cut-Off Mark for Engineering guide explains what scores different universities require.

Year-by-Year Study Priority: Which Years to Cover First

If you are short on time, here is how I would prioritise years to study:

If you are studying Biology alongside Physics, the same prioritisation principle applies. See my JAMB Biology Topic Repetition Index for a year-by-year breakdown of Biology patterns. Similarly, my JAMB Mathematics Topic Repetition Index follows the same data-driven approach for Maths very useful for Physics students since Maths and Physics overlap heavily in calculation topics.

Frequently Asked Questions

How many questions does JAMB set in Physics?

JAMB sets 40 Physics questions in the UTME. Each question carries equal marks, and you have a combined 1 hour 45 minutes to answer all four subjects.

Which Physics topics repeat most in JAMB?

Mechanics (30–35%), Electricity and Magnetism (20–25%), Waves and Sound (15%), and Heat and Thermodynamics (10–15%) are the highest-frequency topics across 2010–2025.

Is JAMB Physics harder than WAEC Physics?

JAMB Physics is not harder it is more selective. JAMB tests fewer topics but goes deeper into concept application, especially in Mechanics, Electricity, and Waves. If you have a strong WAEC Physics foundation, JAMB builds directly on it.

Can I pass JAMB Physics using only past questions?

Yes, but only if you study the solutions not just the answers. Understanding why each option is correct, and why the wrong options are wrong, is what turns past questions into real preparation. Answers alone are useless without explanations.

How many years of JAMB Physics past questions should I study?

Cover 2016–2025 thoroughly first, then go back to 2010–2015 for additional pattern recognition. Ten years of practice is enough to master every recurring concept.

What is the best way to remember Physics formulas for JAMB?

Do not memorise formulas in isolation. Use each formula in at least 10 different questions until applying it becomes automatic. Understanding when and why a formula applies is more reliable than memorisation under exam pressure.

Should I use WAEC Physics past questions alongside JAMB?

Yes. JAMB Physics assumes a WAEC-level foundation. Many JAMB questions test exactly what WAEC covers, just in a faster, more objective format. Working WAEC Physics questions strengthens your understanding and transfers directly to JAMB performance.

Final Word: What Separates Students Who Score 70+ From Those Who Score 40

I have said this to every Physics student I have worked with: the gap between a Physics score of 40 and a score of 70 in JAMB is not intelligence. It is not even hard work. It is direction.

Students who score 40 read Physics textbooks hoping something useful will show up in the exam. Students who score 70 know which topics appear, which formulas are tested, what JAMB’s favourite calculation traps are, and they have practised solving problems under time pressure. That knowledge comes from exactly the kind of systematic past question study this guide provides.

Use this guide properly. Work every question before reading the solution. Build your list of patterns. Practise under timed conditions. And go into that CBT hall knowing that you have already solved the exam in different clothing dozens of times.

You have everything you need.