October/November 2025 Paper 23 Worked Answers (IGCSE Physics 0625 Extended)

40 questions · 40 marks · 45 minutes

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1. Question 1

   1 marksPeriod of oscillation
   The pendulum passes the marker 15 more times after the start, so it completes 15 full oscillations. Total time = 60 s. Period = 60 / 15 = 4.0 s. Counting from the same direction each time ensures whole periods are being counted.
   Method:

   Identify the number of complete oscillations (15), divide total elapsed time (60 s) by 15 to get 4.0 s.
   Examiner tips

   • Only count passes in the same direction; each such pass represents one complete period.

2. Question 2

   1 marksAverage speed during deceleration
   Initial speed = deceleration x time = 5.0 x 10 = 50 m/s. Final speed = 0. For uniform deceleration, average speed = (initial + final) / 2 = (50 + 0) / 2 = 25 m/s.
   Method:

   Calculate initial speed = 5.0 x 10 = 50 m/s, then average speed = 50 / 2 = 25 m/s.
   Examiner tips

   • For uniform deceleration to rest, average speed = initial speed / 2.

3. Question 3

   1 marksTerminal velocity and motion description
   As the ball accelerates downward, air resistance increases until it equals the weight. At that point the resultant force is zero so acceleration is zero and velocity is constant. The velocity never decreases because the net force never becomes negative (upward). Acceleration drops from a positive value to zero.
   Method:

   Reason from forces: initially weight > air resistance so positive acceleration; eventually they balance so acceleration = 0 and velocity is constant.
   Examiner tips

   • Terminal velocity means constant velocity; the acceleration is zero, not negative.

4. Question 4

   1 marksMeasuring instruments
   A balance compares the unknown mass against known masses (or uses calibrated springs/strain gauges). A protractor measures angles, a stop-watch measures time, and a voltmeter measures potential difference. None of the other three instruments measure mass.
   Method:

   Match each instrument to the quantity it measures and select the one for mass.
   Examiner tips

   • Learn which instrument corresponds to each physical quantity: mass, time, angle, potential difference.

5. Question 5

   1 marksDensity and floating — layering of liquids
   Liquids that do not mix layer according to density: the least dense floats on top. J has the lowest density (1.2), so it is on top. K (1.8) is in the middle. L (2.4) is densest and sinks to the bottom. Order top to bottom: J, K, L.
   Method:

   Rank densities smallest to largest: 1.2, 1.8, 2.4. This ranking is also the top-to-bottom order.
   Examiner tips

   • Rank all liquids by density first; the ranking directly gives top-to-bottom order.

6. Question 6

   1 marksBalanced forces and constant speed
   The two forces are equal in magnitude and opposite in direction, so the resultant force is zero. By Newton's first law, a zero resultant force means no change in velocity, so the vehicle continues at constant speed in the same direction.
   Method:

   Calculate resultant force = 600 - 600 = 0 N, then apply Newton's first law to conclude constant speed.
   Examiner tips

   • Zero resultant force means zero acceleration, which means constant velocity (Newton's first law).

7. Question 7

   1 marksMoments and equilibrium
   Anticlockwise moment (left) = 4.0 N x 3.0 m = 12 N m. Clockwise moment (right) = 6.0 N x 2.0 m = 12 N m. The moments are equal so the principle of moments is satisfied and the beam is in equilibrium.
   Method:

   Compute both moments: left = 4.0 x 3.0 = 12 N m; right = 6.0 x 2.0 = 12 N m. Equal, so balanced.
   Examiner tips

   • Always compute moment = force x distance for EACH side before concluding balance.

8. Question 8

   1 marksImpulse and average acceleration
   Impulse = force x time = 160 x 0.050 = 8.0 N s. By the impulse-momentum theorem, impulse = mass x change in velocity, so mass = 8.0 / 20 = 0.40 kg. Average acceleration = force / mass = 160 / 0.40 = 400 m/s^2.
   Method:

   Step 1: impulse = 160 x 0.050 = 8.0 N s. Step 2: mass = 8.0 / 20 = 0.40 kg. Step 3: a = 160 / 0.40 = 400 m/s^2.
   Examiner tips

   • Impulse = F x t (not F x v). Use impulse-momentum to find mass before calculating acceleration.

9. Question 9

   1 marksEnergy transfers — chemical to kinetic
   A car engine burns chemical fuel (petrol or diesel) and converts that chemical energy into kinetic energy of the moving vehicle. An a.c. generator converts kinetic energy to electrical energy. A battery-powered torch converts chemical energy to light (and thermal) energy. A wind-up clock converts elastic (strain) energy to kinetic energy.
   Method:

   List input and output stores for each option; only the car engine has chemical in and kinetic out.
   Examiner tips

   • Trace the energy pathway: identify both the input store and output store for each device.

10. Question 10

    1 marksEnergy resources — solar origin
    Wind is caused by unequal heating of the Earth's surface by the Sun, which creates convection currents in the atmosphere. Geothermal energy comes from heat inside the Earth (radioactive decay and residual formation heat). Nuclear energy comes from fission of uranium nuclei. Tidal energy comes from gravitational interactions dominated by the Moon, not the Sun.
    Method:

    Trace the origin of each resource back to its primary energy source; only wind traces directly to the Sun.
    Examiner tips

    • Solar-derived resources include wind, waves, hydroelectric and biomass. Geothermal and nuclear are not solar.

11. Question 11

    1 marksGas pressure and particle motion
    Lowering temperature reduces the average kinetic energy of gas particles; they move more slowly. Slower particles travel less distance per second, so they hit the container walls less frequently. Fewer collisions per second means a smaller force on the walls and therefore lower pressure.
    Method:

    Lower temperature reduces kinetic energy and speed of particles, which reduces collision frequency with walls, which reduces pressure.
    Examiner tips

    • Link temperature to kinetic energy, kinetic energy to speed, and speed to collision frequency with the walls.

12. Question 12

    1 marksBoyle's law calculation
    Using Boyle's law at constant temperature: $p_1 V_1 = p_2 V_2$. So $p_2 = 150 \times 20 / 12 = 250 \text{ kPa}$. Change in pressure = 250 - 150 = 100 kPa.
    Method:

    Calculate $p_2 = 150 \times 20 / 12 = 250 \text{ kPa}$. Change = 250 - 150 = 100 kPa.
    Examiner tips

    • Use $p_1 V_1 = p_2 V_2$ and then subtract the original pressure to find the change.

13. Question 13

    1 marksSpecific heat capacity calculation
    Using $E = mc\Delta T$: $E = 2000 \times 0.200 \times 5 = 2000 \text{ J}$. Note that mass must be converted from grams to kilograms: 200 g = 0.200 kg.
    Method:

    Convert 200 g to 0.200 kg, then E = 2000 x 0.200 x 5 = 2000 J.
    Examiner tips

    • Always convert mass to kg when using $E = mc\Delta T$; specific heat capacity is in J/(kg °C).

14. Question 14

    1 marksChanges of state and energy input
    Changes that go from a lower-energy state to a higher-energy state require energy input. Melting (J) breaks bonds in a solid, requiring energy. Vaporisation (K) breaks bonds in a liquid, requiring energy. Condensation (L) and freezing (M) both release energy as particles form stronger bonds.
    Method:

    Identify which changes move up the energy scale (solid < liquid < gas); those changes require energy input.
    Examiner tips

    • Solid to liquid and liquid to gas: energy in. Gas to liquid and liquid to solid: energy out.

15. Question 15

    1 marksThermal conduction in metals
    Metals are good conductors because they contain free (delocalised) electrons. These electrons gain kinetic energy at the hot end and move through the lattice, transferring energy to cooler regions. The positive metal ions are held in a fixed lattice; they vibrate but do not move from place to place.
    Method:

    Recall that metallic conductors have mobile free electrons that transfer thermal energy; ions vibrate but stay in the lattice.
    Examiner tips

    • The key to metal conduction is free electrons: they move, while ions stay fixed and only vibrate.

16. Question 16

    1 marksConvection in air
    When air is heated it expands, so the same mass occupies a larger volume: its density decreases. This less-dense warm air experiences a net upward buoyancy force and rises to the top of the room. Cooler, denser air sinks to the bottom. This convection current means warmer air accumulates near the ceiling.
    Method:

    State that heating reduces air density; less dense warm air rises to the top of the cave.
    Examiner tips

    • Heating a gas increases particle spacing (expands it), reducing density, causing it to rise by buoyancy.

17. Question 17

    1 marksThermal radiation and surface temperature
    A star radiates power from its entire surface. If two stars emit the same total power but star P has a larger surface area, then P radiates less power per unit area. The power radiated per unit area increases with temperature (Stefan-Boltzmann law), so the star with lower power per unit area must have a lower surface temperature. Therefore P is cooler than Q.
    Method:

    Reason that power per unit area = total power / surface area; larger area gives smaller power per unit area; therefore lower surface temperature.
    Examiner tips

    • Power per unit area determines surface temperature; the same total power over a bigger surface gives lower temperature.

18. Question 18

    1 marksDiffraction of waves through a gap
    Diffraction (spreading) is most pronounced when the gap width is comparable to or smaller than the wavelength, and least when the gap is much wider than the wavelength. The wavelength here is 2 cm. A 10 cm gap is five times the wavelength, so the waves pass through almost as a straight beam with the least spreading. The 1 cm and 2 cm gaps, being smaller than or equal to the wavelength, produce the most diffraction.
    Method:

    Compare each gap with the 2 cm wavelength; the widest gap, many times the wavelength, gives the least diffraction.
    Examiner tips

    • Maximum diffraction: gap width equals wavelength. Least diffraction: gap width much greater than wavelength.

19. Question 19

    1 marksReflection of light — angle relationships
    By the law of reflection, the angle of reflection equals the angle of incidence, both measured from the normal to the mirror. Angle of reflection = 35°. The angle between the incident ray and the mirror surface is the complement of the angle of incidence: 90° - 35° = 55°.
    Method:

    Apply the law of reflection: reflection angle = 35°. Surface angle = 90° - 35° = 55°.
    Examiner tips

    • Always measure reflection angles from the normal, not the mirror surface.

20. Question 20

    1 marksRefraction — angle of refraction calculation
    Refractive index $n = c / v = 3.0 \times 10^8 / 2.25 \times 10^8 = 1.333$. Snell's law: $n_1 \sin\theta_1 = n_2 \sin\theta_2$, so $\sin\theta_2 = \sin 40° / 1.333 = 0.6428 / 1.333 = 0.482$. Therefore $\theta_2 = \arcsin(0.482) = 28.8° \approx 28°$. The ray bends towards the normal when entering a denser medium.
    Method:

    Find n = 3.0/2.25 = 1.333. The sine of the refraction angle = (the sine of 40 degrees) / 1.333 = 0.482, so the refraction angle is 28 degrees.
    Examiner tips

    • When entering a denser medium (lower speed), the ray bends towards the normal, so the refraction angle must be smaller than the incidence angle.

21. Question 21

    1 marksLenses — short sight correction
    In a short-sighted eye the lens is too powerful (or the eyeball too long), so parallel rays from a distant object converge to a focus before reaching the retina. A diverging (concave) lens pre-diverges the rays so that the eye then focuses them correctly onto the retina.
    Method:

    Identify short sight as over-convergence; choose the lens type that reduces convergence (diverging/concave).
    Examiner tips

    • Short-sighted: image in front of retina, corrected by diverging lens. Long-sighted: image behind retina, corrected by converging lens.

22. Question 22

    1 marksElectromagnetic spectrum — properties
    All electromagnetic waves travel at the same speed in a vacuum ($3.0 \times 10^8 \text{ m/s}$). Gamma rays have the shortest wavelength (and highest frequency) in the EM spectrum. Radio waves have the longest wavelength and lowest frequency. Visible light is in between.
    Method:

    Confirm all EM waves have the same speed, then recall the wavelength order of the spectrum to identify gamma rays as shortest.
    Examiner tips

    • All EM waves travel at c = 3.0 x 10^8 m/s in a vacuum. Rank from longest to shortest wavelength: radio, microwave, IR, visible, UV, X-ray, gamma.

23. Question 23

    1 marksSound waves — amplitude and frequency
    Amplitude controls loudness: larger amplitude means louder sound. Frequency controls pitch: lower frequency means lower pitch. When amplitude increases, the sound gets louder. When frequency decreases, the pitch drops. The result is a louder sound with a lower pitch.
    Method:

    Treat amplitude and frequency separately: amplitude up means louder; frequency down means lower pitch.
    Examiner tips

    • Loudness depends on amplitude; pitch depends on frequency. These are independent properties.

24. Question 24

    1 marksMagnetic field of a solenoid
    Inside a solenoid the magnetic field lines are parallel and closely spaced, indicating a strong, uniform field. Outside the ends the field lines diverge (spread out), so the field is weaker. The field at the centre (Q) is therefore greater than at the outside positions P and R, which are approximately equal to each other.
    Method:

    Recall the solenoid field pattern: uniform strong field inside, weaker diverging field outside ends.
    Examiner tips

    • Field line density represents field strength: dense parallel lines inside solenoid means strong field; sparse lines outside means weak field.

25. Question 25

    1 marksElectric charge — charging by friction
    When two materials are rubbed together, electrons (not protons) are transferred. Protons are locked in atomic nuclei and do not move. The glass rod becomes positively charged because it loses electrons to the silk cloth, leaving the rod with fewer electrons than protons.
    Method:

    Positive charge means electron deficit; state that the rod loses electrons to the cloth.
    Examiner tips

    • Charging by friction always involves electron transfer, never proton transfer.

26. Question 26

    1 marksConventional current and electron flow
    By convention, conventional current flows from the positive terminal through the external circuit to the negative terminal. In reality, electrons (negative charge) flow in the opposite direction: from the negative terminal to the positive terminal through the external circuit.
    Method:

    Recall that conventional current flows from positive terminal; electrons flow opposite, from negative terminal.
    Examiner tips

    • Conventional current and electron flow are always in opposite directions in the external circuit.

27. Question 27

    1 marksElectromotive force — definition
    Electromotive force (e.m.f.) is defined as the energy supplied by a source (battery or generator) per unit charge that passes through it in driving charge around the complete circuit. Current is the charge per unit time; potential difference is the energy transferred per unit charge between two points; power is the energy per unit time.
    Method:

    Match the definition 'energy per unit charge, whole circuit, from source' to e.m.f.
    Examiner tips

    • E.m.f. is the energy per charge from the source around the whole circuit; p.d. is energy per charge between two points.

28. Question 28

    1 marksElectrical energy transferred
    Energy transferred $E = V \times I \times t$. Convert: current 50 mA = 0.050 A; time 2.0 min = 120 s. $E = 6.0 \times 0.050 \times 120 = 36 \text{ J}$.
    Method:

    Convert 50 mA to 0.050 A and 2.0 min to 120 s. Then E = 6.0 x 0.050 x 120 = 36 J.
    Examiner tips

    • Always convert mA to A and minutes to seconds before using E = VIt.

29. Question 29

    1 marksCircuit diagrams — lamp lighting
    A lamp only lights when a complete, unbroken conducting path exists from one terminal of the battery, through the lamp, and back to the other terminal. Any break in the circuit (open switch, disconnected wire) prevents current from flowing and the lamp does not light.
    Method:

    Identify the diagram where there is a complete, unbroken loop including both battery terminals and the lamp.
    Examiner tips

    • Trace the path from one battery terminal through every component and back to the other terminal; if the loop is unbroken the lamp lights.

30. Question 30

    1 marksKirchhoff's current law — series and parallel
    By Kirchhoff's first law, the total current entering a junction equals the total current leaving it. Total current entering the parallel section = 6.0 A. Branch 1 carries 4.0 A. Therefore branch 2 carries 6.0 - 4.0 = 2.0 A.
    Method:

    Total current = branch 1 + branch 2. So branch 2 = 6.0 - 4.0 = 2.0 A.
    Examiner tips

    • Apply Kirchhoff's first law at every junction: currents in = currents out.

31. Question 31

    1 marksPotential dividers — variable p.d.
    When the rheostat resistance is at maximum, almost all the supply voltage drops across the rheostat and nearly 0 V appears across the lamp. When the rheostat is at 0 ohms, the full 9.0 V appears across the lamp. So the lamp p.d. can vary from 0 V to 9.0 V by adjusting the rheostat.
    Method:

    Consider the two extremes of rheostat setting to find minimum and maximum p.d. across the lamp: 0 V to 9.0 V.
    Examiner tips

    • A rheostat in series acts as a potential divider: minimum resistance gives maximum p.d. across the lamp; maximum resistance gives minimum p.d. across the lamp.

32. Question 32

    1 marksElectromagnetic induction — force on induced current
    By Lenz's law, the force on the induced current acts in a direction that opposes the change causing the induction. The wire is moving downward, so the force on the induced current acts upward, opposing the downward motion. This is consistent with conservation of energy.
    Method:

    State Lenz's law: the force on the induced current opposes the downward motion, so the force is upward.
    Examiner tips

    • Apply Lenz's law: the induced force always opposes the motion that caused the induction.

33. Question 33

    1 marksRadioactivity — corrected count rate
    Mean background count = (160 + 170 + 150) / 3 = 480 / 3 = 160 counts per 10 minutes = 16 counts/min. Corrected count rate = measured rate - background rate = 500 - 16 = 484 counts/min.
    Method:

    Find mean background: (160+170+150)/3 = 160 per 10 min = 16 per min. Corrected = 500 - 16 = 484 counts/min.
    Examiner tips

    • Always convert background count to per-minute rate before subtracting from the measured count rate.

34. Question 34

    1 marksDeflection of radiation in a magnetic field
    A magnetic field exerts a force on moving charged particles ($F = Bqv$). Gamma rays carry no charge, so the field exerts no force and they travel in a straight line. Alpha particles (charge +2e) and beta particles (charge -e) are oppositely charged, so the magnetic force acts in opposite directions, deflecting them in opposite directions. Alpha particles (greater mass and charge) follow a less curved path than the lighter beta particles.
    Method:

    Identify charges: gamma neutral (straight path), alpha positive and beta negative (opposite curves).
    Examiner tips

    • No charge means no magnetic force; opposite charges give opposite deflections.

35. Question 35

    1 marksSpontaneous and random nature of radioactive decay
    All radioactive decay, whether alpha or beta, is both spontaneous (it occurs without any external trigger such as temperature, pressure, or chemical state) and random (it is impossible to predict exactly when a particular nucleus will decay or which nucleus in a sample will be next). These properties apply equally to alpha and beta decay.
    Method:

    Recall that all radioactive decay is both spontaneous and random; apply this to both alpha and beta.
    Examiner tips

    • Spontaneous and random applies to ALL types of radioactive decay without exception.

36. Question 36

    1 marksBeta decay — nuclear change
    In beta-minus decay, a neutron inside the nucleus is transformed into a proton. The electron produced in this process is immediately emitted as the beta particle. This increases the proton number (atomic number) by 1, while the nucleon number stays the same because one neutron has become one proton.
    Method:

    Recall beta-minus process: neutron converts to proton; emitted particle is electron (beta minus).
    Examiner tips

    • Beta-minus decay: neutron to proton inside nucleus, electron emitted. Proton number increases by 1; nucleon number unchanged.

37. Question 37

    1 marksRadiation safety — shielding materials
    Alpha particles are stopped by paper or a few centimetres of air. Beta particles are stopped by a few millimetres of aluminium. Gamma rays are the most penetrating and require several centimetres of lead or metres of concrete to be adequately absorbed. Since the container must stop all three, it must be made of thick lead, which stops even gamma rays (and therefore also stops alpha and beta).
    Method:

    Identify the most penetrating radiation (gamma); select the material that stops gamma (lead), which will also stop alpha and beta.
    Examiner tips

    • Penetrating power: alpha (paper) < beta (aluminium) < gamma (lead). To stop all three, choose lead.

38. Question 38

    1 marksOrbital period of a satellite
    The satellite travels one complete orbit (a circle of circumference $2\pi r$) at constant orbital speed $v$. Period $T = \frac{2\pi r}{v} = \frac{2\pi \times 8.0 \times 10^6}{4000} = \frac{5.027 \times 10^7}{4000} = 12 566 \text{ s}$. Converting: $12566 / 60 = 209.4 \text{ min} \approx 3 \text{ h } 29 \text{ min}$.
    Method:

    Calculate $T = 2\pi \times 8.0 \times 10^6 / 4000 = 12 566 \text{ s} \approx 3 \text{ h } 29 \text{ min}$.
    Examiner tips

    • Orbital period = $2\pi r / v$. Remember the full circumference formula and convert seconds to hours and minutes carefully.

39. Question 39

    1 marksThe Sun — stellar composition and emission
    The Sun consists mainly of hydrogen (about 73%) and helium (about 25%), with trace amounts of heavier elements. At its surface temperature of about 5800 K, Wien's law places the peak of its emission spectrum at about 500 nm, which is in the green part of the visible spectrum. The overall emission covers visible light and near-infrared/near-ultraviolet. X-rays are emitted mainly by the corona (very high temperature), not the photosphere.
    Method:

    Recall Sun's composition (H and He) and surface temperature placing peak emission in visible/near-IR.
    Examiner tips

    • The Sun: mainly hydrogen and helium; surface ~5800 K; peak emission in visible light (with near-IR and near-UV).

40. Question 40

    1 marksSupernova remnants
    When a massive star explodes as a supernova, it ejects its outer layers into space, forming an expanding cloud of gas called a nebula (or supernova remnant). The collapsed core that remains becomes a neutron star (if the mass is below about 3 solar masses) or a black hole for more massive cores. White dwarfs are the remnants of low-mass stars that do not explode as supernovae. Red supergiants are the pre-supernova stage, not a product of the supernova.
    Method:

    Recall that a supernova simultaneously ejects gas (nebula) and compresses the core into a neutron star.
    Examiner tips

    • Supernova of massive star produces: expanding nebula (ejected outer layers) + neutron star (collapsed core). White dwarf is for low-mass stars.