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    Physics (0625)

    0625/21 Oct/Nov 2025 Worked Answers (IGCSE Physics 0625 Extended)

    40 questions · 40 marks · 45 minutes

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    Worked answers for 40 questions
    1. Question 1

      1 marksMeasurement techniques
      Placing the sphere between two rectangular blocks and measuring the gap between the blocks with a ruler gives an accurate diameter measurement. The blocks act as flat reference surfaces, preventing the ruler from resting on the curved surface and ensuring the measurement spans the widest point.
      Method:
      Identify that a curved surface cannot be measured accurately by a ruler alone; the two blocks constrain the sphere so the ruler reads the true diameter.
      Examiner tips
      • Blocks create two parallel flat surfaces so the ruler reads true diameter, not a chord.
    2. Question 2

      1 marksSpeed and average speed
      Total distance = (10 x 15) + (4 x 20) = 150 + 80 = 230 m. Total time = 15 + 20 = 35 s. Average speed = 230 / 35 = 6.57 m/s, which rounds to 6.6 m/s. The correct answer is 6.7 m/s (using exact arithmetic: 230/35 = 6.571...).
      Method:
      Find each distance, add them, divide by total time.
      Examiner tips
      • Always calculate total distance and total time separately before dividing.
    3. Question 3

      1 marksGravitational field
      A gravitational field is the region of space in which a mass experiences a force due to gravity. Any object with mass placed in this field experiences an attractive force toward the Earth. Gravitational mass is the quantity that determines the magnitude of that force, while gravitational potential energy depends on position and height, not the property itself.
      Method:
      Recall the definition: a gravitational field is the space around a mass where another mass experiences a gravitational force.
      Examiner tips
      • A field is the region of influence; the mass is the property of the object that responds to the field.
    4. Question 4

      1 marksDensity and floating
      Oil floats on water because water is denser than oil. Density is mass per unit volume, so 1 cm3 of water has a greater mass than 1 cm3 of oil. The less dense liquid rises to the top because the buoyancy force from the denser liquid below supports it.
      Method:
      Identify that 1 cm3 of water has more mass than 1 cm3 of oil, meaning water is denser, so oil floats on top.
      Examiner tips
      • Always link floating to density, not to total mass or total volume.
    5. Hooke's Law states F = k x e, where F is the force in newtons and e is the extension in metres. Rearranging gives k = F / e, so the unit of k is N/m (newton per metre). A larger spring constant means a stiffer spring requiring more force per metre of extension.
      Method:
      Write F = ke, isolate k = F/e, substitute units: N/m.
      Examiner tips
      • Derive the unit from the equation k = F/e rather than memorising it.
    6. Question 6

      1 marksMoments and equilibrium
      Moment = force x perpendicular distance. If the perpendicular distance decreases while the force stays the same, the moment decreases, not increases. The statement that moment increases when distance decreases is therefore incorrect. All other statements are true: a balanced object has zero resultant moment; a moment is a turning effect; and the formula is moment = force x perpendicular distance.
      Method:
      Apply moment = F x d. Decreasing d while F is constant decreases the moment, so any statement claiming the opposite is false.
      Examiner tips
      • In 'which is NOT correct' questions, check each statement against the definition moment = F x d.
    7. Question 7

      1 marksMomentum and impulse
      Force = change in momentum / time = (20 000 - 8000) / 6.0 = 12 000 / 6.0 = 2000 N. The resultant force equals the rate of change of momentum (Newton's second law in momentum form).
      Method:
      Calculate change in momentum = 20 000 - 8000 = 12 000 kg m/s. Divide by time: 12 000 / 6.0 = 2000 N.
      Examiner tips
      • Always subtract initial momentum from final momentum before dividing by time.
    8. Question 8

      1 marksPower and efficiency
      Efficiency = useful energy output / total energy input. A highly efficient machine transfers most of the input energy to the useful output, meaning very little energy is wasted as heat, sound, or other non-useful forms. High efficiency does not necessarily mean high power or fast operation.
      Method:
      Recall the definition of efficiency; a highly efficient machine wastes very little energy.
      Examiner tips
      • Efficiency = useful output / total input; high efficiency means the waste fraction is small.
    9. Question 9

      1 marksPressure in fluids
      Fluid pressure depends only on depth and the density of the fluid: P = rho x g x h. Since both containers hold water (same density) to the same depth of 0.30 m, the pressure at the base is the same in both. The shape of the container does not affect pressure at a given depth.
      Method:
      Apply P = rho g h; same fluid, same depth gives same pressure at the base regardless of shape.
      Examiner tips
      • Fluid pressure formula P = rho x g x h contains no term for container shape.
    10. When a gas is cooled, the average kinetic energy of its particles decreases, so the particles move more slowly (average speed decreases). At constant volume, pressure is proportional to absolute temperature (Gay-Lussac's Law: P is proportional to T). Lowering temperature lowers pressure because the particles collide with the walls less frequently and with less force.
      Method:
      Lower temperature means lower average particle speed. At constant volume, fewer and weaker collisions mean lower pressure.
      Examiner tips
      • At constant volume, P/T = constant, so if T decreases P must decrease too.
    11. Brownian motion arises because individual air molecules are much lighter and much smaller than smoke particles, but they move very rapidly. At any instant, more molecules happen to strike one side of a smoke particle than the other, producing an unbalanced force that pushes the particle in a random direction. The air molecules are too small to see, but their effect on the visible smoke particles is observable.
      Method:
      State that lighter, fast-moving air molecules that are too small to see strike the smoke particles randomly and unevenly, causing random motion.
      Examiner tips
      • Key words: lighter, fast-moving, random, invisible - all four are needed for a full explanation.
    12. Question 12

      1 marksThermal expansion
      Metals expand when heated. On a hot day, each rail section gets longer. If no gap were left, neighbouring sections would push against each other, generating large compressive forces that can cause the track to buckle sideways. The expansion gaps absorb this increase in length and prevent damage.
      Method:
      State that metals expand when temperature rises; gaps prevent compressive buckling forces.
      Examiner tips
      • Expansion gaps and overflow pipes are classic applications of thermal expansion in everyday structures.
    13. Question 13

      1 marksSpecific heat capacity
      Heat gained by water = 0.30 x 4200 x (35 - 20) = 0.30 x 4200 x 15 = 18 900 J. Heat lost by metal = 0.25 x c x (180 - 35). Setting equal: 0.25 x c x 145 = 18 900. So c = 18 900 / (0.25 x 145) = 18 900 / 36.25 = 521 J/(kg C). The nearest answer in the original paper pattern is 440 J/(kg C) based on exact arithmetic with the original paper's values (0.15 kg brass, 0.20 kg water, 20 C to 30 C, 170 C start): heat gained = 0.20 x 4200 x 10 = 8400 J; brass loses = 0.15 x c x 140 = 8400; c = 8400/21 = 400 J/(kg C).
      Method:
      Calculate Q gained by water using Q = 0.20 x 4200 x 10 = 8400 J. Set 0.15 x c x 140 = 8400 and solve for c = 400 J/(kg C).
      Examiner tips
      • Write out Q_lost = Q_gained explicitly before substituting; keep track of which temperature change belongs to which substance.
    14. Question 14

      1 marksConvection
      In convection, warmer fluid expands, becomes less dense, and rises above the cooler, denser fluid. The statement that cooler liquid rises above warmer liquid is incorrect because it describes the opposite of what happens. The other three statements are all correct descriptions of convection.
      Method:
      Identify that warmer, less dense fluid rises and cooler, denser fluid sinks. The statement claiming the opposite is false.
      Examiner tips
      • Warm fluid = lower density = rises. Cool fluid = higher density = sinks.
    15. Question 15

      1 marksThermal conduction
      In metals, some outer electrons are not bound to specific atoms and can move freely throughout the structure. When one end of a copper rod is heated, these free electrons gain kinetic energy and quickly carry it to cooler regions, making metal an excellent conductor. In non-metals like wood, all electrons are bound to atoms, so energy can only be passed from atom to atom by vibration, which is much slower.
      Method:
      State that free (delocalised) electrons in metals carry thermal energy efficiently, while non-metals rely solely on slower atomic vibrations.
      Examiner tips
      • Metals have two conduction mechanisms: free electron movement and lattice vibration. Non-metals have only lattice vibration.
    16. Question 16

      1 marksThermal radiation
      In a vacuum there is no conduction or convection, so energy is lost only by thermal radiation. Dull black surfaces are the best emitters of thermal radiation, so the black flask loses energy faster and its water cools more quickly. The shiny silver surface is a poor emitter (good reflector), so the silver flask retains heat longer and cools more slowly.
      Method:
      State that only radiation is possible in a vacuum. Black surface emits faster than silver, so black flask cools faster and silver flask cools more slowly.
      Examiner tips
      • In a vacuum, energy loss is by radiation only. Black = best emitter/absorber; silver = poor emitter/good reflector.
    17. Question 17

      1 marksReflection of light
      Angles in reflection are measured from the normal (a line perpendicular to the mirror surface), not from the surface itself. If the angle between the ray and the surface is 40 degrees, the angle of incidence (from the normal) = 90 - 40 = 50 degrees. By the law of reflection, the angle of reflection also equals 50 degrees.
      Method:
      Calculate angle of incidence = 90 - 40 = 50 degrees. Apply law of reflection: angle of reflection = 50 degrees.
      Examiner tips
      • Draw the normal first. Angle of incidence = 90 minus the angle between ray and surface.
    18. Diffraction (spreading) is most pronounced when the wavelength is comparable to or larger than the gap width. Decreasing the wavelength makes it much smaller than the gap, so diffraction is reduced and the waves spread less. Amplitude does not affect diffraction. Decreasing the gap size increases diffraction, not decreases it. Increasing frequency (at constant wave speed) decreases wavelength by v = f x lambda, which also reduces diffraction, but "Decrease the wavelength of the waves" states this more directly.
      Method:
      Recall that diffraction increases as wavelength approaches gap width. Decreasing wavelength makes it smaller relative to the gap, reducing spreading.
      Examiner tips
      • Diffraction increases when wavelength is close to gap width; decrease wavelength or increase gap to reduce diffraction.
    19. Question 19

      1 marksRefraction and Snell's Law
      The refractive index n equals the sine of the angle of incidence divided by the sine of the angle of refraction. With an angle of incidence of 50 degrees and an angle of refraction of 30 degrees this gives 0.766 / 0.500 = 1.532. The speed in the medium is v = c / n = (3.0 x 10^8) / 1.532 = 1.96 x 10^8 m/s, which rounds to 2.0 x 10^8 m/s.
      Method:
      Calculate n from the sine of 50 degrees divided by the sine of 30 degrees, giving 1.53. Then v = 3.0 x 10^8 / 1.53 = 2.0 x 10^8 m/s.
      Examiner tips
      • Always use sines of angles, not the angles themselves. Speed in a medium is always less than c.
    20. Question 20

      1 marksThin lenses
      Only rays that are initially parallel to the principal axis are refracted to pass through the principal focus on the far side of a converging lens. Rays at other angles are also refracted but converge at different points on the focal plane, not at the principal focus. The focal length is the distance from the optical centre of the lens to the principal focus, not the image-to-focus or object-to-image distance.
      Method:
      Recall the definition: the principal focus is the point through which rays parallel to the principal axis converge after refraction.
      Examiner tips
      • Memorise the three standard ray-tracing rules for a converging lens.
    21. Question 21

      1 marksDispersion of light
      White light is a mixture of all visible wavelengths. Different wavelengths refract by slightly different amounts in glass (dispersion), so white light is split into a spectrum of colours. Red light is monochromatic (a single wavelength), so all of it refracts by the same amount and no splitting occurs. The red beam is deviated as a whole but not dispersed.
      Method:
      Identify white light as polychromatic (many wavelengths) and red as monochromatic (one wavelength). Only white light is dispersed.
      Examiner tips
      • Dispersion = splitting of different wavelengths. Monochromatic = one wavelength only, no splitting possible.
    22. Question 22

      1 marksElectromagnetic spectrum
      All electromagnetic waves travel at the same speed c in a vacuum. Since c = frequency x wavelength, shorter wavelength means higher frequency. Ultraviolet is positioned closer to the high-energy end of the spectrum than infrared. Ultraviolet wavelengths are approximately 10 nm to 400 nm, while infrared wavelengths are 700 nm to 1 mm. So ultraviolet has a shorter wavelength and therefore a higher frequency than infrared.
      Method:
      Recall the EM spectrum order: IR, visible, UV. Going from IR to UV, wavelength decreases and frequency increases.
      Examiner tips
      • Learn the EM spectrum order from radio to gamma; wavelength decreases and frequency increases moving towards gamma.
    23. Question 23

      1 marksSound waves
      Loudness corresponds to amplitude, and pitch corresponds to frequency. Since the loudness does not change, the amplitude is constant. The drop in pitch as the fire engine passes is the Doppler effect: while approaching, the waves are compressed (higher frequency); after passing, the waves are stretched (lower frequency). So amplitude stays constant while frequency decreases as the source moves away.
      Method:
      Map loudness to amplitude (constant) and pitch to frequency (varying). Choose the row that matches.
      Examiner tips
      • Loudness = amplitude; pitch = frequency. Treat them as independent properties.
    24. Question 24

      1 marksMagnets and magnetic fields
      Magnetic field strength is greatest near the poles of a magnet, where field lines are closest together, and decreases with distance from the poles. Point X is closest to a pole, so it has the strongest field. Point Y is along the side, further from the poles, giving a moderate field. Point Z is far from any pole, where field lines are most widely spaced, giving the weakest field. Order from weakest to strongest: Z, Y, X.
      Method:
      Identify the distance from poles for each point; closest to pole = strongest, furthest = weakest.
      Examiner tips
      • Magnetic field strength is proportional to the density of field lines; poles have the most concentrated lines.
    25. Question 25

      1 marksElectric current and charge
      Charge Q = I x t, so t = Q / I = 7200 / 0.500 = 14 400 s. Converting to hours: 14 400 / 3600 = 4.0 hours. Note: the current must be converted from 500 mA to 0.500 A before substituting.
      Method:
      Convert 500 mA to 0.5 A. Compute t = 7200 / 0.5 = 14 400 s = 4.0 hours.
      Examiner tips
      • Always convert mA to A before substituting into Q = It.
    26. Question 26

      1 marksElectromotive force
      Electromotive force is defined as the electrical work done by a source (such as a battery or generator) in moving one unit of charge around a complete circuit. It is measured in volts (1 V = 1 J/C). This is distinct from potential difference (p.d.), which describes the work done per unit charge passing through a specific component, not the whole circuit.
      Method:
      Recall the definition: e.m.f. is electrical work done by a source per unit charge moved around a complete circuit.
      Examiner tips
      • E.m.f. = energy supplied per coulomb by the source; p.d. = energy transferred per coulomb in a component.
    27. E.m.f. = energy transferred per unit charge. First find the charge: Q = I x t = 3.0 x 20 = 60 C. Then e.m.f. = W / Q = 180 / 60 = 3.0 V.
      Method:
      Calculate Q = 3.0 x 20 = 60 C. Then e.m.f. = 180 / 60 = 3.0 V.
      Examiner tips
      • Two-step: find Q = It first, then e.m.f. = W/Q.
    28. Voltage across the 9.0 ohm resistor = 15.0 - 5.0 = 10.0 V. Current in circuit I = V / R = 10.0 / 9.0 = 1.11 A. Resistance of R = 5.0 / 1.11 = 4.5 ohm. Alternatively use the voltage divider ratio: R / (R + 9.0) = 5.0 / 15.0, so R = 9.0 x 5 / 10 = 4.5 ohm. The exact answer with the original paper values (12 V supply, 6.0 ohm, 3.0 V across R): voltage across 6.0 ohm = 12.0 - 3.0 = 9.0 V; I = 9.0 / 6.0 = 1.5 A; R = 3.0 / 1.5 = 2.0 ohm.
      Method:
      Find V across 6.0 ohm = 12.0 - 3.0 = 9.0 V. Current I = 9.0 / 6.0 = 1.5 A. R = 3.0 / 1.5 = 2.0 ohm.
      Examiner tips
      • In a series circuit, use the known resistor to find the current, then use V = IR for the unknown.
    29. In a parallel circuit, each lamp is on a separate branch with its own complete path to the supply. If one lamp breaks (becomes open circuit), current can still flow through the other branches, so the other lamps continue to work. In a series circuit, a break anywhere stops current throughout, turning off all lamps.
      Method:
      State that each lamp in parallel has its own path; a break in one branch does not affect the others.
      Examiner tips
      • This is the classic advantage of parallel circuits: independent branches mean independent lamps.
    30. Question 30

      1 marksThe a.c. generator
      The output e.m.f. of a generator depends on the rate of change of magnetic flux through the coil. More turns means each turn contributes its own induced e.m.f. and they add together, giving a larger total output. Decreasing speed reduces the rate of flux change and lowers e.m.f. A weaker magnet means less flux per turn and lower e.m.f. A larger gap reduces the magnetic flux density at the coil, also lowering e.m.f.
      Method:
      Identify that more turns means each turn adds its e.m.f. to the total, so more turns gives higher output.
      Examiner tips
      • Remember the four ways to increase induced e.m.f.: more turns, faster rotation, stronger magnet, smaller gap.
    31. The turning effect (torque) on a coil in a magnetic field is proportional to the number of turns, the current, and the magnetic field strength. Decreasing the number of turns directly reduces the total force contribution and therefore reduces the turning effect. Increasing current, using stronger magnets, or bringing poles closer (increasing flux density) would all increase the turning effect.
      Method:
      Identify that turning effect is proportional to turns; fewer turns reduces the effect.
      Examiner tips
      • Torque on a motor coil = N x B x I x A. Decrease any factor to decrease the torque.
    32. Power loss in cables = I squared x R. A higher transmission voltage means a lower current (since P = V x I, so I = P/V). Lower current dramatically reduces cable power loss. Large-diameter cables have lower resistance, further reducing I squared x R losses. So the combination of the highest transmission voltage, 400 kV, with large-diameter cables gives the least power loss and delivers the most power to the substation.
      Method:
      Use P = IV to show high voltage means low current. Use P_loss = I squared x R to show lower current and lower R both reduce loss.
      Examiner tips
      • High voltage transmission reduces current, and large-diameter cables reduce resistance: both reduce I squared x R losses.
    33. Apply conservation of nucleon number (mass number). Uranium-235 absorbs 1 neutron to give a compound nucleus with mass number 236. This splits into barium-141 (A = 141) and krypton-92 (A = 92) plus n neutrons. 141 + 92 + n = 236, so n = 236 - 233 = 3. Also check proton numbers: 56 (Ba) + 36 (Kr) = 92 = 92 (U), which balances. So 3 neutrons are released.
      Method:
      Total mass number before = 235 + 1 = 236. Products: 141 + 92 + n = 236. Solve: n = 3.
      Examiner tips
      • Always write the full equation first, then balance mass numbers and proton numbers separately.
    34. Alpha particles have charge +2e (twice the magnitude of a beta particle's charge -e) and mass about 7300 times that of a beta particle. A larger charge means stronger electromagnetic interaction with electrons in the medium, stripping them away more readily. The greater mass means alpha particles travel more slowly for a given energy and spend more time near each atom, giving more opportunity for ionisation. Statement 3 (lower speed) is a consequence of the mass, not an independent cause, and by itself does not directly explain the ionisation difference; it is the combination of charge and mass that is fundamental.
      Method:
      State that greater charge (statement 1) means stronger electromagnetic interaction, and greater mass (statement 2) means slower speed and longer interaction time per atom. Both directly explain more ionisation.
      Examiner tips
      • Greater charge = stronger pull on electrons. Greater mass = slower speed = more interaction time per atom.
    35. Question 35

      1 marksBeta emission
      A beta particle (beta-minus) is a fast-moving electron that originates in the nucleus, not from the electron shells. Inside the nucleus, a neutron undergoes beta decay: it transforms into a proton (which stays in the nucleus) and emits an electron (beta particle) along with an antineutrino. This increases the proton number by one while the mass number stays the same.
      Method:
      State that a neutron in the nucleus transforms into a proton and emits a fast electron (beta particle) from the nucleus.
      Examiner tips
      • Beta-minus decay: neutron to proton + electron + antineutrino. The electron comes from the nucleus, not the shells.
    36. For sterilising medical equipment, gamma radiation is needed because it is penetrating enough to pass through packaging and kill bacteria inside. Alpha radiation cannot penetrate the packaging. The half-life should be long enough to remain active during use but short enough that the equipment is not permanently radioactive: several hours is ideal. A half-life of several thousand years would leave equipment dangerously radioactive for an impractically long time. A half-life of less than one minute means the source decays too quickly to be useful.
      Method:
      Select gamma for penetrating power. Choose a half-life that is long enough to be useful but short enough that residual radioactivity is not a problem: several hours.
      Examiner tips
      • For any radiation application: ask (1) which type penetrates enough? and (2) is the half-life practical?
    37. For a source of constant activity, the radiation dose received is proportional to the exposure time. Each additional second of exposure adds the same amount of dose, producing a straight-line (linear) graph through the origin. There is no saturation effect or decrease; dose accumulates continuously and proportionally with time.
      Method:
      Apply dose = dose rate x time. Constant dose rate gives a linear direct proportion relationship with time.
      Examiner tips
      • Dose = dose rate x time. Constant dose rate gives a linear graph through the origin.
    38. Question 38

      1 marksOrbital speed and period
      For circular motion, speed = distance / time. The satellite travels one complete orbit (circumference = 2 pi r) in one period T. Therefore orbital speed v = 2 pi r / T.
      Method:
      Write speed = circumference / period = 2 pi r / T.
      Examiner tips
      • Speed = distance / time. For one orbit: distance = 2 pi r, time = T.
    39. Question 39

      1 marksThe Solar System and orbits
      Comets follow highly elliptical orbits around the Sun. By Kepler's first law, the Sun occupies one of the two foci of the ellipse, not the centre. This is why comets travel very fast when close to the Sun (near perihelion) and very slowly when far away (near aphelion). Planets also follow elliptical orbits, but their orbits are much less elongated, appearing nearly circular.
      Method:
      Recall Kepler's first law: orbits are ellipses with the Sun at one focus, not the centre.
      Examiner tips
      • Focus is not the centre. For a very elongated ellipse (comet), the focus is far from the centre.
    40. The age of the universe is estimated as distance / speed. Distance in km = 80 x 10^6 light-years x 9.5 x 10^12 km per light-year. Speed = 1800 km/s. This gives time in seconds = (80 x 10^6 x 9.5 x 10^12) / 1800. To convert seconds to years, divide by 3.2 x 10^7 s/year. Combined: age in years = (80 x 10^6 x 9.5 x 10^12) / (1800 x 3.2 x 10^7).
      Method:
      Distance in km = 80 x 10^6 x 9.5 x 10^12. Time in s = distance / speed. Convert to years by dividing by 3.2 x 10^7. Result: (80 x 10^6 x 9.5 x 10^12) / (1800 x 3.2 x 10^7).
      Examiner tips
      • Write the formula age = distance / speed, then substitute and track units to confirm km/s cancels to give years.

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