S6 Physics 2025 Past Paper: Full Worked Solutions with Explanations

If you are revising for your S6 Physics national examination, working through past papers with full solutions is the single most effective thing you can do. This page contains complete, step-by-step worked solutions for every question in the 2025 S6 Physics paper. Section A (70 marks) and Section B (30 marks). Study each solution carefully, not just the final answer.

If you find yourself struggling with any topic below, our guide on how to study effectively using proven techniques will help you build a better revision system before your exams.

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Section A : Attempt All Questions (70 Marks)

Question 1: Simple Harmonic Motion : True or False (4 marks)

a) In SHM, the restoring force is directly proportional to the displacement. TRUE

This is the defining condition of SHM: $F = - k x$ . The negative sign means the force always acts opposite to displacement towards the equilibrium position.

b) In SHM, displacement is maximum when velocity is maximum. FALSE

Displacement is maximum (equal to amplitude $A$ ) when velocity is zero. Maximum velocity occurs at the equilibrium position where displacement is zero. Energy is fully kinetic at equilibrium and fully potential at the amplitude.

c) The period of a simple pendulum depends on the mass of the bob. FALSE

The period is $T = 2 π \sqrt{\frac{L}{g}}$ . Mass does not appear only length $L$ and gravitational acceleration $g$ determine the period. This is why a heavy pendulum and a light pendulum of the same length swing at the same rate.

d) The total mechanical energy in SHM depends on the frequency of oscillation. TRUE

Total energy is $E = \frac{1}{2} m ω^{2} A^{2}$ where $ω = 2 π f$ . Since $E \propto ω^{2} \propto f^{2}$ , increasing frequency increases total energy for the same amplitude.

Question 2: Particle Physics: Multiple Choice (5 marks)

Sub	Answer	Explanation
I	c) All hadrons	Quarks are the building blocks of hadrons (baryons and mesons). Leptons such as electrons are NOT made of quarks.
II	b) Electron	Discovered by J.J. Thomson in 1897. The electron remains a fundamental elementary particle with no known substructure.
III	a) Three quarks	Baryons contain exactly three quarks. A proton = uud, a neutron = udd.
IV	c) Strong nuclear force	The strong force, mediated by gluons, binds protons and neutrons inside the nucleus, overcoming electromagnetic repulsion between protons.
V	d) W bosons	The W⁺, W⁻, and Z⁰ bosons are the exchange particles (mediators) of the weak nuclear force, responsible for radioactive beta decay.

Question 3: Electric and Gravitational Fields: Matching (4 marks)

Term	Description
a) Gravitational potential	Work done per unit mass to bring a test mass from infinity to that point
b) Electric field strength	Electric force per unit positive charge at a point: $E = F / q$
c) Electric potential	Work done per unit positive charge to bring a test charge from infinity to that point
d) Equipotential surface	A surface on which the electric (or gravitational) potential is the same at every point — no work is done moving along it

Key distinction: Gravitational potential uses unit mass; electric potential uses unit charge. Both are defined as zero at infinity.

Question 4: Wave Optics and Interference :Multiple Choice (5 marks)

Sub	Answer	Explanation
I	b) Compton Effect	The Compton Effect demonstrates particle behaviour of photons (momentum transfer). Diffraction, interference, and polarization are all wave phenomena.
II	b) Transverse waves	EM waves have electric and magnetic fields oscillating perpendicular to the direction of propagation — the definition of a transverse wave.
III	a) 0 rad	Constructive interference requires waves to be in phase: phase difference = 0, 2π, 4π... At 0 rad, crests perfectly align.
IV	c) (2n+1)λ/2	Destructive interference needs a half-wavelength path difference: λ/2, 3λ/2, 5λ/2... = $(2 n + 1) λ / 2$ for integer $n \geq 0$ .
V	c) Fringe width increases	Fringe width $w = \frac{λ D}{d}$ . Decreasing slit separation $d$ increases fringe width the bright fringes spread further apart.

Question 5: Lenses and Prisms : Multiple Choice (4 marks)

Sub	Answer	Explanation
I	b) Its first focus	A ray through the front focal point refracts and emerges parallel to the principal axis.
II	d) Between focal point and lens	Object closer than focal length → converging lens acts as a magnifying glass → virtual, upright, magnified image.
III	b) Virtual, smaller, upright	A diverging lens always produces a virtual, diminished, upright image regardless of object position.
IV	c) Angle of incidence = angle of emergence	At minimum deviation, the ray passes symmetrically through the prism: $i_{1} = i_{2}$ .

Question 6: Thermodynamics and Energy: Multiple Choice (4 marks)

Sub	Answer	Explanation
I	b) Energy into less useful form, usually heat	Energy degradation = converting high-grade energy (electricity) into low-grade heat that cannot be fully recovered for useful work.
II	d) Heat engine	Absorbs heat from a hot reservoir, converts some to work, rejects remainder to a cold reservoir — operating in a cycle.
III	b) Kinetic energy of fluid → mechanical energy	A turbine's blades are turned by steam or water, converting the fluid's kinetic energy into rotational mechanical energy.
IV	c) Fossil fuels	Thermal power plants burn coal, oil, or natural gas to produce steam that drives turbines connected to generators.

Question 7: Communications Technology: Matching (5 marks)

Term	Match
a) Base Transceiver Station (BTS)	Relays signals between mobile phones and the network
b) Frequency Modulation (FM)	Frequency of carrier signal altered at rate of audio frequency; amplitude constant
c) Satellite Communication	Transmits radio signals for long-distance communication
d) Mobile Switching Center (MSC)	Routes calls and connects mobile devices to the wider network
e) Phase Modulation	Amplitude and frequency of carrier remain constant; only phase changes

Question 8: Cosmology: Fill in the Blanks (4 marks)

a) The Milky Way is a large barred spiral galaxy containing our solar system.

b) The redshift of light from most galaxies is evidence that the universe is expanding, supporting the Big Bang theory. The Doppler redshift shows galaxies moving away from us. Hubble's Law: $v = H_{0} d$ quantifies this.

c) The reciprocal of Hubble's constant gives the approximate age of the Universe: Age $\approx \frac{1}{H_{0}} \approx 13.8$ billion years.

Question 9: Black Body Radiation and Climate: True or False (4 marks)

a) Energy first increases then decreases as wavelength increases. TRUE

Black body emission curves show a peak wavelength governed by Wien's Displacement Law: $λ_{m a x} \propto \frac{1}{T}$ . Emission rises to the peak then falls — hence the characteristic "hump" shape.

b) All objects emit radiation if temperature > 0 K. TRUE

Only at absolute zero (0 K) would thermal radiation cease. Every object above 0 K emits electromagnetic radiation — humans at body temperature radiate mostly in the infrared.

c) Climate change refers only to rising surface temperature. FALSE

Climate change encompasses changes in precipitation patterns, sea-level rise, ocean acidification, extreme weather events, and shifting seasons not just temperature.

d) Reduced ice extent is a positive feedback because ice has greater albedo than ocean. FALSE

The reasoning is backwards. Ice does have higher albedo than ocean water , it reflects more radiation. When ice melts, darker ocean is exposed, absorbing more radiation and causing more warming. The feedback is positive, but not because of ice's albedo rather because its loss reduces overall surface reflectivity.

Question 10: Gravitation: Multiple Choice (5 marks)

I) Distance doubled → gravitational force:

F = \frac{G m_{1} m_{2}}{r^{2}}

If $r \to 2 r$ : $F^{'} = \frac{G m_{1} m_{2}}{(2 r)^{2}} = \frac{F}{4}$ . d) Reduced to a quarter.

II) Period of satellite independent of: b) Mass of the satellite.

From $T = 2 π \sqrt{\frac{r^{3}}{G M_{E}}}$ , the satellite's own mass cancels. Period depends only on orbital radius and the mass of the central body.

III) Satellite moving from X to Y (toward Sun):

Gravity does positive work as the satellite moves closer, so kinetic energy (and speed) increases. Angular momentum $L = m v r$ is conserved (no external torque).

c) Speed increases, angular momentum remains constant.

IV) Gravitational PE vs gravitational potential:

$G P E = m V_{g}$ — it depends on both the object's mass $m$ and the gravitational potential $V_{g} = - \frac{G M}{r}$ . The potential itself depends only on the source mass and distance.

a) GPE = mV_g.

V) Force of Earth on satellite vs force of satellite on Earth:

Newton's Third Law — action-reaction pairs are always equal in magnitude and opposite in direction. b) F₁ = F₂.

Question 11: Total Internal Reflection and Optical Fibre (5 marks)

I-A) Critical angle = 30°, find refractive index:

n = \frac{1}{\sin θ_{c}} = \frac{1}{\sin 30 °} = \frac{1}{0.5} = 2

d) n = 2

I-B) Speed of light in the medium:

v = \frac{c}{n} = \frac{3 \times 10^{8}}{2} = 1.5 \times 10^{8} m/s

a) 1.5 × 10⁸ m/s

II) Critical angle at core-cladding interface ( $n_{1} = 1.45, n_{2} = 1.42$ ):

\sin θ_{c} = \frac{n_{2}}{n_{1}} = \frac{1.42}{1.45} = 0.9793

θ_{c} = \arcsin (0.9793) = 78.3 °

c) 78.3°

III) Best fibre for long-distance, minimal distortion:

b) Single mode fibre. Single mode fibre has a very narrow core (~9 µm) that allows only one propagation path, eliminating modal dispersion the main cause of signal distortion over long distances.

Question 12: Work and Energy (5 marks)

I-a) Work is a vector quantity. FALSE

Work is a scalar. $W = F d \cos θ$ has magnitude only, no direction. It can be positive, negative, or zero, but that sign is not a direction in space.

I-b) Work done by all forces = KE of body. FALSE

The work-energy theorem states that net work done equals the change in kinetic energy: $W_{n e t} = Δ K E$ , not the KE itself.

II) Diver: mass = 77 kg, height = 10 m, g = 9.81 m/s²

A) Potential energy before dropping:

P E = m g h = 77 \times 9.81 \times 10 = 7553.7 J

d) 7553.7 J

B) Kinetic energy at water surface:

By conservation of energy (neglecting air resistance): all PE converts to KE.

K E = P E = 7553.7 J

c) 7553.7 J

C) Velocity at water surface:

v = \sqrt{\frac{2 K E}{m}} = \sqrt{\frac{2 \times 7553.7}{77}} = \sqrt{196.2} = 14.0 m/s

d) 14.0 m/s

Question 13: Electric Field and Gauss's Law (5 marks)

Given: $Q = 4 \times 10^{- 7}$ C, $r = 0.09$ m, $k = 9 \times 10^{9}$ Nm²/C²

I-A) Electric potential at point P:

V = \frac{k Q}{r} = \frac{9 \times 10^{9} \times 4 \times 10^{- 7}}{0.09} = \frac{3600}{0.09} = 4 \times 10^{4} V

c) 4×10⁴ V

I-B) Electric field at point P:

E = \frac{k Q}{r^{2}} = \frac{9 \times 10^{9} \times 4 \times 10^{- 7}}{(0.09)^{2}} = \frac{3600}{0.0081} = 4.4 \times 10^{5} N/C

d) 4.4×10⁵ N/C

II-A) Increasing Gaussian surface radius → electric flux:

b) Remains the same. By Gauss's Law: $Φ = \frac{Q_{e n c}}{ε_{0}}$ . Flux depends only on enclosed charge, not the surface size.

II-B) Electric charge enclosed (Φ = 6×10³ Nm²/C):

Q = ε_{0} Φ = 8.854 \times 10^{- 12} \times 6 \times 10^{3} = 5.31 \times 10^{- 8} C

c) 5.31×10⁻⁸ C

Question 14: Natural Disasters : True or False (5 marks)

a) Rainfall, slope steepness, earthquakes, and rock layer arrangement cause landslides. TRUE

b) Epicenter is the point inside Earth where an earthquake begins. FALSE

The focus (or hypocenter) is inside the Earth. The epicenter is the point on the surface directly above the focus.

c) Elevated or floating houses are creative solutions for flood-prone areas. TRUE

d) Early warning systems and evacuation routes are key for tsunami-prone cities. TRUE

e) Underground bunkers in every home are more beneficial than public alert systems for cyclone-prone cities. FALSE

Public alert systems serve far more people and are practically achievable. Underground bunkers are unsuitable for cyclone protection, which requires evacuation not shelter-in-place.

Question 15: de Broglie Wavelength of an Electron (5 marks)

Given: $V = 200$ V, $e = 1.6 \times 10^{- 19}$ C, $m_{e} = 9.1 \times 10^{- 31}$ kg, $h = 6.63 \times 10^{- 34}$ J·s

I) Kinetic energy:

E_{k} = e V = 1.6 \times 10^{- 19} \times 200 = 3.2 \times 10^{- 17} J = 32 \times 10^{- 18} J

c) 32×10⁻¹⁸ J

II) Momentum:

p = \sqrt{2 m_{e} E_{k}} = \sqrt{2 \times 9.1 \times 10^{- 31} \times 3.2 \times 10^{- 17}} = \sqrt{5.824 \times 10^{- 47}} = 7.63 \times 10^{- 24} kg·m/s

b) 7.63×10⁻²⁴ kg·m/s

III) de Broglie wavelength:

λ = \frac{h}{p} = \frac{6.63 \times 10^{- 34}}{7.63 \times 10^{- 24}} = 8.69 \times 10^{- 11} m

c) 8.69×10⁻¹¹ m

⚠️ Struggling with Section A? Many S6 students find the multiple-choice questions deceptively hard because they test conceptual understanding, not just formulas. Read our guide on how to score above 80% in A-Level sciences in Rwanda ,the same strategies apply across all science subjects.

Section B: Worked Solutions (30 Marks, Any 3 Questions)

Question 16: X-Ray Tube (10 marks)

I-a) How to increase X-ray intensity:

Increase the filament heating current (or filament voltage $V_{h}$ ). This produces more thermionic electrons per second by thermionic emission, increasing the number of X-ray photons emitted per second , hence higher intensity.

I-b) How to increase X-ray penetrating power:

Increase the anode accelerating voltage $V$ . Higher voltage gives electrons more kinetic energy. When they decelerate on hitting the tungsten target, they produce X-rays of higher frequency and shorter wavelength (Bremsstrahlung radiation) with greater penetrating power.

II-a) Electric charge transferred to anode per second ( $I = 15.0$ mA):

Q = I \times t = 15.0 \times 10^{- 3} \times 1 = 0.015 C = 15 mC

II-b) Number of electrons hitting anode per second:

N = \frac{Q}{e} = \frac{0.015}{1.6 \times 10^{- 19}} = 9.375 \times 10^{16} electrons/s

III) Velocity of electrons striking the target ( $V_{a c c} = 10$ kV):

Using energy conservation $e V = \frac{1}{2} m_{e} v^{2}$ :

v = \sqrt{\frac{2 e V}{m_{e}}} = \sqrt{\frac{2 \times 1.6 \times 10^{- 19} \times 10 \times 10^{3}}{9.1 \times 10^{- 31}}} = \sqrt{3.516 \times 10^{15}} = 5.93 \times 10^{7} m/s

IV) Two real-life applications of X-rays:

Medical radiography — imaging broken bones, chest infections, and dental cavities
Airport security scanners — detecting concealed weapons or prohibited items in luggage

Question 17: Thermodynamic Cycle WXYZ (10 marks)

a) Isochoric (isovolumetric) processes occur on vertical lines of a P-V diagram (constant volume):

W → X: constant volume at $V = 1.42 \times 10^{- 3}$ m³
Z → Y: constant volume at $V = 0.44 \times 10^{- 3}$ m³

b) Temperature at point Y ( $T_{X} = 310$ K):

At X: $P_{X} = 1.0 \times 10^{5}$ Pa, $V_{X} = 1.42 \times 10^{- 3}$ m³. At Y: $P_{Y} = 5.2 \times 10^{5}$ Pa, $V_{Y} = 0.44 \times 10^{- 3}$ m³.

Using the ideal gas law $\frac{P_{X} V_{X}}{T_{X}} = \frac{P_{Y} V_{Y}}{T_{Y}}$ :

T_{Y} = 310 \times \frac{5.2 \times 10^{5} \times 0.44 \times 10^{- 3}}{1.0 \times 10^{5} \times 1.42 \times 10^{- 3}} = 310 \times \frac{228.8}{142} = 310 \times 1.611 \approx 499 K

c-i) What does the shaded area WXYZ represent?

The enclosed area on a P-V diagram represents the net work done by the gas in one complete cycle. This follows from $W = \int P d V$ . The gas does positive work during expansion and negative work during compression; the net is the area of the loop.

c-ii) Efficiency of cycle WXYZ:

Heat input: $Q_{i n} = W_{n e t} + Q_{o u t} = 610 + 1300 = 1910$ J

η = \frac{W_{n e t}}{Q_{i n}} = \frac{610}{1910} \approx 0.319 = 31.9\%

d) Change in internal energy during adiabatic compression X→Y:

For an adiabatic process, $Q = 0$ . By the First Law of Thermodynamics $Δ U = Q - W_{b y g a s}$ :

Δ U = 0 - (- 210) = + 210 J

Internal energy increases by 210 J — all the work done on the gas goes into raising its internal energy.

Question 18: Sound Waves (10 marks)

I-A) From the waveform diagram:

Amplitude = the maximum displacement from equilibrium position
Wavelength = the distance spanning one complete cycle of the wave

I-B) Effect of increasing loudness at constant pitch:

Amplitude: Increases — louder sound carries more energy → greater displacement of particles → larger amplitude
Wavelength: Unchanged — pitch = frequency = constant. Since $v = f λ$ and both $v$ and $f$ are unchanged, $λ$ stays the same

II-A) Wavelength in closed pipe ( $L = 2.1$ m), 7th harmonic:

L = \frac{7 λ}{4} \Rightarrow λ = \frac{4 L}{7} = \frac{4 \times 2.1}{7} = \frac{8.4}{7} = 1.2 m

c) 1.2 m

II-B) Fundamental frequency:

7th harmonic frequency: $f_{7} = \frac{v}{λ} = \frac{340}{1.2} = 283.3$ Hz

f_{1} = \frac{f_{7}}{7} = \frac{283.3}{7} = 40.48 Hz

b) 40.476 Hz

III) Experiment to compare speed of sound in steel, water, and air:

Aim: To compare the speed of sound through three states of matter.

Materials: Signal generator, two identical microphones, oscilloscope, 2 m steel rod, water-filled tube, measuring tape.

Method:

Place one microphone at each end of the steel rod connected to an oscilloscope. Strike one end sharply. Measure time delay $Δ t$ between signals. Speed = length/Δt.
Repeat with a water-filled tube of the same length using waterproof microphones.
Repeat in air over the same distance.

Why these materials? Steel (solid) has highest molecular density and elasticity; water (liquid) intermediate; air (gas) lowest. This demonstrates that sound travels fastest in solids and slowest in gases, governed by molecular spacing and intermolecular forces.

Question 19: Electric Circuits (10 marks)

I) Current in ammeter — circuits A, B, C:

d) The same in all cases. The same EMF source drives the same resistance $R$ with an ammeter in series. Rearranging series components does not change total current.

II) Heat produced per second ( $V = 12$ V source):

Circuit A: Single 2Ω → $P = V^{2} / R = 144 / 2 = 72$ W
Circuit B: Two 2Ω in series → $R_{t o t a l} = 4 Ω$ → $P = 144 / 4 = 36$ W
Circuit C: Two 2Ω in parallel → $R_{t o t a l} = 1 Ω$ → $P = 144 / 1 = 144$ W

c) Maximum in Circuit C — parallel combination has lowest resistance → highest current → maximum power.

III) Factors affecting resistance of a metallic conductor:

D) All of the above. From $R = \frac{ρ L}{A}$ : material resistivity $ρ$ , length $L$ , and cross-sectional area $A$ all determine resistance.

IV) Kirchhoff's Laws — Figure 9:

Circuit: 3.0 V (left branch, $I_{1}$ ), 1.5 V (right branch, $I_{3}$ ), 10Ω (top), 20Ω (middle, $I_{2}$ )

KCL at node b: $I_{1} = I_{2} + I_{3}$ ...(1)

KVL Loop 1 (left, clockwise):

3.0 = 10 I_{1} + 20 I_{2} ...(2)

KVL Loop 2 (right, clockwise):

20 I_{2} = 1.5 \Rightarrow I_{2} = 0.075 A ...(3)

From (2): $10 I_{1} = 3.0 - 20 (0.075) = 1.5 \Rightarrow I_{1} = 0.15$ A

From (1): $I_{3} = 0.15 - 0.075 = 0.075$ A

I_{1} = 0.15 A, I_{2} = 0.075 A, I_{3} = 0.075 A

Question 20: Atomic Physics and Quantum Mechanics (10 marks)

Sub	Answer	Explanation
I	b) Z=13; A=27	Z = number of protons = 13. A = protons + neutrons = 13 + 14 = 27.
II	c) Nucleus is positive with most of atom's mass	Rutherford's gold foil experiment: most alpha particles pass through; a few deflect at large angles → small, dense, positive nucleus.
III	b) Balmer series	Transitions ending at n=2 are the Balmer series (visible light). n=4→n=2 falls here.
IV	d) Discrete spectrum of radiation	Electrons occupy specific energy levels → emit only specific photon energies → line spectrum, not continuous.

B) Hydrogen atom calculations:

a) Energy at n = 6:

E_{n} = \frac{E_{1}}{n^{2}} = \frac{- 2.18 \times 10^{- 18}}{36} = - 6.06 \times 10^{- 20} J

b) Wavelength of photon emitted (n=4 → n=3):

$E_{4} = - 1.36 \times 10^{- 19}$ J, $E_{3} = - 2.42 \times 10^{- 19}$ J

Δ E = E_{4} - E_{3} = 1.06 \times 10^{- 19} J

λ = \frac{h c}{Δ E} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{1.06 \times 10^{- 19}} = 1.876 \times 10^{- 6} m \approx 1876 nm (infrared — Paschen series)

c) Radius of electron orbit at n = 6 (Bohr model):

r_{n} = n^{2} a_{0} = 36 \times 5.29 \times 10^{- 11} = 1.904 \times 10^{- 9} m \approx 1.90 nm

How to Use This Paper for Maximum Exam Gain

Reading worked solutions is only half the work. For each question you got wrong, go back to the topic and rebuild from first principles — do not just memorise the answer. Past papers reward students who understand why each step works, not those who memorise patterns.

If you consistently struggle with particular topics — SHM, circuits, optics — that is a signal you need structured support before the exam, not just more past papers. Our guide on why Rwandan students struggle in science subjects and how to fix it explains the root causes and what actually helps.

You might also recognise signs that a student around you needs more targeted help see our piece on 7 signs a child is struggling academically. And if anxiety around exams is affecting your performance, read about math and science anxiety and how to overcome it.

Need personalised help with S6 Physics or any other Subject ?

Working through past papers alone only takes you so far. At Mathrone Academy, our vetted S6 Physics tutors can identify exactly where you lose marks and build a targeted revision plan before your national exams. Find an S6 Physics tutor → OR send us a Whatsapp message on +250786684285

For students who prefer self-paced revision, our A-Level Physics course covers the full REB syllabus with worked examples, practice questions, and video explanations — structured unit by unit so nothing is missed before exam day.

For exam strategy beyond physics, our article on Rwanda national exams 2026 study guide and timetable will help you plan your final weeks across all subjects. And if you are thinking about what comes after S6, explore our guides on S4 and S6 subject streams and how to find the best private tutor in Kigali.

⚠️Important Disclaimer

The solutions on this page are prepared by the Mathrone Academy team for revision and learning purposes only. This is not an official NESA marking scheme or REB-approved answer guide. While every effort has been made to ensure accuracy, answers and explanations may differ from the official examiners' marking guide. Always refer to your school teacher or the official NESA publications for authoritative marking guidance. Mathrone Academy accepts no responsibility for any discrepancies between these solutions and official results. if you want to download 2025 Physics II Question Paper Click HERE Or visit NESA WEBSITE under resources category.

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