100+ Free Cameroon GCE A-Level Physics Practice Questions

Prepare for the Cameroon General Certificate of Education Advanced Level Physics (Subject Code 0780) exam with instant access — no signup required.

✓ No registration✓ No credit card✓ No hidden fees✓ Start practicing immediately

63.47% pass rate in the June 2026 session (Cameroon GCE Board, Performance by Subjects, results released 21 August 2026) Pass Rate

100+ Questions

100% Free

Loading practice questions...

Same family resources

Explore More Cameroon GCE Advanced Level

Continue into nearby exams from the same family. Each card keeps practice questions, study guides, flashcards, videos, and articles in one place.

2026 Statistics

Key Facts: Cameroon GCE A-Level Physics Exam

50 MCQs

Paper 1 Examination Format (1h 30m duration)

CGCEB Regulations & Syllabuses

33.3%

Paper 1 Percentage Weighting of Overall Subject Grade

Cameroon GCE Board Official Specification

Grade E

Minimum Pass Mark for University Admission Eligibility

Ministry of Secondary Education (MINESEC)

Cameroon GCE A-Level Physics (0780) evaluates Upper Sixth science candidates across mechanics, waves, electromagnetism, thermodynamics, and quantum physics through a 3-paper terminal assessment comprising a 50-question MCQ paper, a 3-hour structured theory paper, and a separately scheduled laboratory practical examination.

Sample Cameroon GCE A-Level Physics Practice Questions

Try these sample questions to test your Cameroon GCE A-Level Physics exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Which of the following represents the correct base SI units for Planck's constant, $h$?

A.kg m s^-1

B.kg m^2 s^-1

C.kg m^2 s^-2

D.kg m^-1 s^-2

Explanation: Planck's constant $h$ has units of Joule-seconds (J s). In base SI units, energy is expressed as Joules: $1\text{ J} = 1\text{ kg m}^2\text{ s}^{-2}$. Multiplying by seconds yields $[h] = \text{kg m}^2\text{ s}^{-2} \times \text{s} = \text{kg m}^2\text{ s}^{-1}$.

2A motorist travelling at a steady speed of $25\text{ m s}^{-1}$ sees an obstruction on the road. If the driver's reaction time is $0.60\text{ s}$ and the vehicle decelerates uniformly at $5.0\text{ m s}^{-2}$ after the brakes are applied, what is the total stopping distance?

A.62.5 m

B.77.5 m

C.15.0 m

D.92.5 m

Explanation: Total stopping distance is the sum of thinking distance and braking distance. Thinking distance $s_1 = u \times t_r = 25 \times 0.60 = 15.0\text{ m}$. Braking distance $s_2$ is found from $v^2 = u^2 - 2as \implies 0 = 25^2 - 2(5.0)s_2 \implies s_2 = 625 / 10 = 62.5\text{ m}$. Total distance $s = s_1 + s_2 = 15.0 + 62.5 = 77.5\text{ m}$.

3A projectile is launched from horizontal ground with initial speed $u$ at an angle $\theta$ to the horizontal. If the maximum height reached $H$ equals one-quarter of the horizontal range $R$ ($H = \frac{1}{4}R$), what is the launch angle $\theta$?

A.30°

B.45°

C.60°

D.75°

Explanation: Maximum height is $H = \frac{u^2 \sin^2\theta}{2g}$ and horizontal range is $R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin\theta \cos\theta}{g}$. Setting $H = \frac{1}{4}R$ gives $\frac{u^2 \sin^2\theta}{2g} = \frac{1}{4} \left(\frac{2u^2 \sin\theta \cos\theta}{g}\right) = \frac{u^2 \sin\theta \cos\theta}{2g}$. Dividing both sides by $\frac{u^2 \sin\theta}{2g}$ gives $\sin\theta = \cos\theta \implies \tan\theta = 1$, so $\theta = 45^\circ$.

4A ball is projected horizontally at $20\text{ m s}^{-1}$ from the top of a cliff of height $45\text{ m}$. Taking $g = 10\text{ m s}^{-2}$ and neglecting air resistance, what is the speed of the ball just before it strikes the ground?

A.30 m s^-1

B.36 m s^-1

C.50 m s^-1

D.25 m s^-1

Explanation: The horizontal velocity remains constant at $u_x = 20\text{ m s}^{-1}$. The vertical velocity upon landing is found from $v_y^2 = u_y^2 + 2gh = 0 + 2(10)(45) = 900 \implies v_y = 30\text{ m s}^{-1}$. The resultant speed is $v = \sqrt{u_x^2 + v_y^2} = \sqrt{20^2 + 30^2} = \sqrt{400 + 900} = \sqrt{1300} \approx 36.06\text{ m s}^{-1} \approx 36\text{ m s}^{-1}$.

5A block of mass $4.0\text{ kg}$ rests on a rough plane inclined at $30^\circ$ to the horizontal. If the coefficient of static friction between the block and the plane is $\mu = 0.70$ and $g = 9.8\text{ m s}^{-2}$, what is the magnitude of the frictional force acting on the block?

A.19.6 N

B.23.8 N

C.33.9 N

D.39.2 N

Explanation: The component of the gravitational force pulling the block down the incline is $F_{\parallel} = mg \sin(30^\circ) = 4.0 \times 9.8 \times 0.5 = 19.6\text{ N}$. The maximum possible static friction is $f_{\max} = \mu R = \mu mg \cos(30^\circ) = 0.70 \times 4.0 \times 9.8 \times \frac{\sqrt{3}}{2} \approx 23.76\text{ N}$. Since $F_{\parallel} < f_{\max}$, the block remains in static equilibrium, so the actual frictional force equals the applied parallel force: $f = 19.6\text{ N}$.

6A rubber ball of mass $0.20\text{ kg}$ hits a vertical wall horizontally with a speed of $15\text{ m s}^{-1}$ and rebounds horizontally in the opposite direction at $10\text{ m s}^{-1}$. If the contact with the wall lasts for $0.050\text{ s}$, what is the average force exerted by the wall on the ball?

A.20 N

B.50 N

C.100 N

D.250 N

Explanation: Taking the initial direction as positive, initial velocity $u = +15\text{ m s}^{-1}$ and rebound velocity $v = -10\text{ m s}^{-1}$. The change in momentum is $\Delta p = m(v - u) = 0.20(-10 - 15) = 0.20(-25) = -5.0\text{ N s}$. The average force is $F = \frac{|\Delta p|}{\Delta t} = \frac{5.0}{0.050} = 100\text{ N}$.

7A trolley of mass $2.0\text{ kg}$ travelling at $6.0\text{ m s}^{-1}$ collides head-on with a stationary trolley of mass $4.0\text{ kg}$. If the two trolleys couple and move together after the collision, what is the loss in kinetic energy?

A.12 J

B.24 J

C.36 J

D.18 J

Explanation: By conservation of linear momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \implies (2.0)(6.0) + 0 = (2.0 + 4.0) v \implies 12 = 6.0 v \implies v = 2.0\text{ m s}^{-1}$. Initial kinetic energy $E_{k1} = \frac{1}{2}(2.0)(6.0)^2 = 36\text{ J}$. Final kinetic energy $E_{k2} = \frac{1}{2}(6.0)(2.0)^2 = 12\text{ J}$. Loss in kinetic energy $\Delta E_k = 36 - 12 = 24\text{ J}$.

8A force $F = (6x^2 + 4x)\text{ N}$ acts on a particle of mass $0.50\text{ kg}$ along the $x$-axis, moving it from $x = 0\text{ m}$ to $x = 2.0\text{ m}$. What is the total work done on the particle?

A.16 J

B.24 J

C.32 J

D.40 J

Explanation: Work done by a variable force is $W = \int_0^2 F\,dx = \int_0^2 (6x^2 + 4x)\,dx = \left[ 2x^3 + 2x^2 \right]_0^2 = 2(2^3) + 2(2^2) = 2(8) + 2(4) = 16 + 8 = 24\text{ J}$.

9A car of mass $1200\text{ kg}$ ascends a slope of $1\text{ in } 20$ (where $\sin\theta = \frac{1}{20}$) at a constant speed of $15\text{ m s}^{-1}$. If total frictional resistance opposing motion is $400\text{ N}$ and $g = 9.8\text{ m s}^{-2}$, what power must the engine develop?

A.8.82 kW

B.6.00 kW

C.14.82 kW

D.17.82 kW

Explanation: At constant velocity, the forward tractive force $F$ equals the sum of the parallel component of weight and resistive forces: $F = mg\sin\theta + F_{\text{resist}} = 1200 \times 9.8 \times \frac{1}{20} + 400 = 588 + 400 = 988\text{ N}$. Engine power $P = F v = 988\text{ N} \times 15\text{ m s}^{-1} = 14820\text{ W} = 14.82\text{ kW}$.

10A small bob of mass $m$ is suspended by a light inextensible string of length $L$ and rotates in a horizontal circle of radius $r$ at a constant angular speed $\omega$, forming a conical pendulum where the string makes an angle $\theta$ with the vertical. Which expression gives the period of oscillation $T$?

A.T = 2π √(L / g)

B.T = 2π √(L cos θ / g)

C.T = 2π √(L sin θ / g)

D.T = 2π √(L tan θ / g)

Explanation: Resolving vertically gives $T_{\text{tension}} \cos\theta = mg$, and horizontally $T_{\text{tension}} \sin\theta = m\omega^2 r$. Since $r = L\sin\theta$, we have $T_{\text{tension}} \sin\theta = m\omega^2 L\sin\theta \implies T_{\text{tension}} = mL\omega^2$. Substituting into the vertical equation: $mL\omega^2 \cos\theta = mg \implies \omega = \sqrt{\frac{g}{L\cos\theta}}$. Since $T = \frac{2\pi}{\omega}$, we find $T = 2\pi \sqrt{\frac{L\cos\theta}{g}}$.

About the Cameroon GCE A-Level Physics Exam

The Cameroon GCE Advanced Level Physics (Subject Code 0780) is the national terminal high school examination in physics conducted annually by the Cameroon GCE Board in Buea. Designed for students in Upper Sixth (Science stream), the syllabus covers foundational classical and modern physics, emphasizing dimensional rigor, mathematical derivation, physical modeling, experimental precision, and graphical interpretation. A strong performance in A-Level Physics is essential for admission into prestigious faculties of engineering (e.g., NAHPI Bambili, COT Buea, ENSP Yaoundé/Douala), medical schools (FHS Buea, CUSS Yaoundé), and higher teacher training colleges (ENS/ENSET) across Cameroon and the Commonwealth. Format note: this site's practice bank is 100 four-option multiple-choice questions covering the whole official syllabus. Paper 1 of the real examination is genuinely multiple choice (50 compulsory questions), so the format matches that paper, but the bank is a study aid only — it does not simulate the written theory/essay paper(s) or any practical examination, and its length does not describe the official exam.

Assessment

Official Advanced Level structure for subject code 0780 (Physics) per the Cameroon GCE Board June 2026 timetable (Form G6): Paper 1: 50 compulsory multiple-choice questions (1 hour 30 minutes); Paper 2: written theory/structured questions (3 hours); a separately scheduled practical examination (practical phase, 5–27 May 2026). Total written time is 4 hours 30 minutes. The Board does not publish per-paper mark weightings for individual subjects.

Time Limit

Paper 1: 1 hour 30 minutes; total written time 4 hours 30 minutes plus a separately scheduled practical examination.

Passing Score

Grade E or better (Cameroon GCE Advanced Level grades A, B, C, D and E are passes; O is a subsidiary pass and F is a fail)

Exam Fee

17,000 FCFA (Cameroon General Certificate of Education Board (CGCEB), Buea, South West Region, Cameroon)

Cameroon GCE A-Level Physics Exam Content Outline

20%

Curriculum Structure & Core Modules

Core learning objectives, topics, and problem solving in Curriculum Structure & Core Modules.

20%

Examination Papers & Weighting Breakdown

Core learning objectives, topics, and problem solving in Examination Papers & Weighting Breakdown.

20%

Grading Scale & University Admission Equivalences

Core learning objectives, topics, and problem solving in Grading Scale & University Admission Equivalences.

How to Pass the Cameroon GCE A-Level Physics Exam

What You Need to Know

Passing score: Grade E or better (Cameroon GCE Advanced Level grades A, B, C, D and E are passes; O is a subsidiary pass and F is a fail)
Assessment: Official Advanced Level structure for subject code 0780 (Physics) per the Cameroon GCE Board June 2026 timetable (Form G6): Paper 1: 50 compulsory multiple-choice questions (1 hour 30 minutes); Paper 2: written theory/structured questions (3 hours); a separately scheduled practical examination (practical phase, 5–27 May 2026). Total written time is 4 hours 30 minutes. The Board does not publish per-paper mark weightings for individual subjects.
Time limit: Paper 1: 1 hour 30 minutes; total written time 4 hours 30 minutes plus a separately scheduled practical examination.
Exam fee: 17,000 FCFA

Keys to Passing

Work through all 100 available questions
Review every answer and explanation
Track weak areas and revisit them
Use our AI tutor for tough concepts

Frequently Asked Questions

What is the format of Cameroon GCE A-Level Physics Paper 1?

Paper 1 consists of 50 compulsory four-option multiple-choice questions (MCQs) covering the entire syllabus. Candidates are given 1 hour and 30 minutes (90 minutes) to complete the paper,GCE Advanced Level Physics grade.

Are non-programmable electronic calculators permitted in the exam?

Yes, candidates are permitted and strongly encouraged to use standard, silent, non-programmable scientific calculators in all three papers (Paper 1, Paper 2, and Paper 3) along with standard mathematical tables and formula sheets provided by the CGCEB.

How should I allocate my time during Paper 1 MCQ?

With 50 questions in 90 minutes, candidates have an average of 1.8 minutes (1 minute 48 seconds) per question. It is recommended to answer straightforward conceptual and dimensional questions first, mark moderately complex numerical problems for a second pass, and reserve the final 10 minutes to verify bubble sheet responses and calculations.

What laboratory experiments are typically assessed in Paper 3?

Paper 3 assesses three experimental stations covering: (1) Mechanics (e.g., simple pendulum, compound pendulum, spiral spring oscillation, or friction/inclined plane), (2) Heat/Optics (e.g., focal length of lenses/mirrors, Snell's law using glass blocks, or cooling curves/specific heat), and (3) Electricity (e.g., potentiometer wire circuits, internal resistance using voltmeters/ammeters, or meter bridge resistance determination).

100+ Free Cameroon GCE A-Level Physics Practice Questions

Explore More Cameroon GCE Advanced Level

Cameroon GCE A-Level Accounting (0705)

Cameroon GCE A-Level Biology (0710)

Cameroon GCE A-Level Chemistry (0715)

Cameroon GCE A-Level Computer Science (0795)

Cameroon GCE A-Level Economics (0725)

Cameroon GCE A-Level English Language (0730)