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    Physics (0625)

    0625/22 May/June 2025 Worked Answers (IGCSE Physics 0625 Extended)

    40 questions · 40 marks · 45 minutes

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    Worked answers for 40 questions
    1. An irregular solid has no simple formula for its volume. Lowering it into water in a measuring cylinder and reading the rise in water level gives the volume directly by displacement.
      Method:
      Select the measuring cylinder, which finds volume by water displacement.
      Examiner tips
      • Use displacement in a measuring cylinder for irregular solids.
    2. Question 2

      1 marksFree fall motion
      With no air resistance the only force on the ball is its weight. This gives a constant acceleration equal to the acceleration of free fall, about 9.8 m/s squared, so the ball speeds up at a steady rate.
      Method:
      With weight as the only force, the ball has constant acceleration of free fall.
      Examiner tips
      • No air resistance means constant acceleration due to gravity.
    3. Acceleration = change in velocity divided by time taken. The change in velocity is 100 - 20 = 80 m/s and the time is 40 s, so the acceleration = 80 / 40 = 2.0 m/s squared.
      Method:
      Compute change in velocity 80 m/s, divide by 40 s to get 2.0 m/s squared.
      Examiner tips
      • Acceleration is the gradient of a velocity-time graph: change in velocity over time.
    4. Squashing the foam only reduces its volume; no matter is added or removed, so the amount of material is the same. The mass therefore stays the same, and since weight = mass times gravitational field strength, the weight is unchanged too.
      Method:
      Compression changes volume only; mass is constant, so weight is constant.
      Examiner tips
      • Changing volume alone does not change the amount of matter (mass).
    5. Density = mass divided by volume. The mass is found with a balance. Because the block is a regular rectangular shape, its volume is found by measuring its length, width and height with a ruler and multiplying them.
      Method:
      Mass from a balance, volume from ruler measurements of the regular block.
      Examiner tips
      • A regular solid's volume comes from measured dimensions, not displacement.
    6. The spring force at release = spring constant times extension = 30 x 0.12 = 3.6 N upward. The weight = mass times g = 0.15 x 9.8 = 1.47 N downward. The resultant upward force = 3.6 - 1.47 = 2.13 N. The acceleration = resultant force divided by mass = 2.13 / 0.15 = about 14 m/s squared.
      Method:
      Spring force 3.6 N, weight 1.47 N, resultant 2.13 N, a = 2.13/0.15 = 14 m/s squared.
      Examiner tips
      • At release the resultant force is the spring force minus the weight.
    7. Question 7

      1 marksMomentum change and force
      The velocity reverses, so the change of momentum = m(v - (-v)) = 0.15 x (30 - (-30)) = 0.15 x 60 = 9.0 kg m/s. The average force = change of momentum divided by contact time = 9.0 / (1.0 x 10^-3) = 9000 N.
      Method:
      Momentum change = 0.15 x 60 = 9.0 kg m/s; force = 9.0 / 0.001 = 9000 N.
      Examiner tips
      • When an object bounces back, the change of velocity is twice the speed.
    8. Setting kinetic energy equal to gravitational potential energy: half times m times v squared = m times g times h. The mass cancels, giving h = v squared divided by (2 times g). Substituting the speed, h = 5.0 squared / (2 x 9.8).
      Method:
      Equate kinetic and potential energy, cancel mass, rearrange to h = 5.0 squared / (2 x 9.8).
      Examiner tips
      • Equating kinetic and potential energy gives h = v squared / (2 g).
    9. Wind is caused by the Sun heating the atmosphere unevenly, so wind energy comes from the Sun. Geothermal energy comes from heat inside the Earth, and nuclear fuel energy comes from the nuclei of atoms, so neither of these comes from the Sun.
      Method:
      Identify wind as Sun-driven; geothermal and nuclear are not.
      Examiner tips
      • Wind, waves and biomass trace back to the Sun; geothermal and nuclear do not.
    10. Question 10

      1 marksStates of matter and shape
      A solid keeps its own fixed shape. Liquids and gases can flow and take the shape of their container. So a sample that takes the shape of its container must be either a liquid or a gas.
      Method:
      A container-shaped sample flows, so it must be a liquid or a gas.
      Examiner tips
      • Only solids keep a fixed shape; liquids and gases flow to fit the container.
    11. In the collision the fast molecule transfers some momentum to the smoke particle, so the smoke particle starts to move. The light molecule bounces off (rebounds) with a changed direction. This is the basis of Brownian motion.
      Method:
      The molecule transfers momentum, so the smoke particle moves and the molecule rebounds.
      Examiner tips
      • Brownian motion comes from molecules transferring momentum to larger particles.
    12. At constant temperature, pressure multiplied by volume is constant: p1 V1 = p2 V2. So 360 x 2.0 = 120 x V2, giving 720 = 120 V2, so V2 = 720 / 120 = 6.0 cubic centimetres.
      Method:
      Apply p1 V1 = p2 V2 to get V2 = 720 / 120 = 6.0 cubic centimetres.
      Examiner tips
      • For a gas at constant temperature, p x V stays constant.
    13. Question 13

      1 marksThermal expansion of a gas
      Gases expand far more than solids or liquids for the same temperature rise. Warming the trapped air makes it expand much more than the glass or the mercury, so the expanding air pushes the mercury thread up the tube.
      Method:
      The trapped air expands much more than glass or mercury, pushing the thread up.
      Examiner tips
      • Gases expand much more than solids or liquids for the same temperature change.
    14. The energy supplied by the heater is power multiplied by time, P times t. This equals m times c times theta. Rearranging for c gives c = P t divided by (m times theta).
      Method:
      Set P t = m c theta and rearrange to c = P t / (m theta).
      Examiner tips
      • Heater energy P x t equals m x c x temperature change; rearrange for c.
    15. Question 15

      1 marksBest thermal conductor
      Thermal energy travels along each rod by conduction. The rod that conducts heat fastest reaches the wax soonest, melting it first. Since the silver ball falls first, silver must be the best thermal conductor of the four metals.
      Method:
      The first ball to fall is on the best conductor, which is silver.
      Examiner tips
      • Faster conduction means the wax melts and the ball drops sooner.
    16. Emission increases with a darker surface and a larger surface area. X is large and black, so it emits the most. Comparing the two white spheres, the large one Z emits more than the small one Y because it has a greater surface area. So the order is X, then Z, then Y.
      Method:
      Rank by darkness and size: X (large black) highest, then Z (large white), then Y (small white).
      Examiner tips
      • Compare both surface colour and surface area when ranking emission.
    17. Question 17

      1 marksDiffraction and wavelength
      Wavelength = speed divided by frequency. "speed 350 m/s, frequency 280 Hz" gives 350 / 280 = 1.25 m. "speed 300 m/s, frequency 400 Hz" gives 300 / 400 = 0.75 m. "speed 300 m/s, frequency 280 Hz" gives 300 / 280 = 1.07 m. "speed 350 m/s, frequency 400 Hz" gives 350 / 400 = 0.88 m. The longest wavelength is 1.25 m, so the wave with speed 350 m/s and frequency 280 Hz diffracts the most.
      Method:
      Calculate each wavelength; 350/280 = 1.25 m is longest, so it diffracts the most.
    18. Question 18

      1 marksAmplitude of a wave
      The amplitude of a wave is the maximum displacement of a point from its undisturbed (rest) position. It is measured from the middle line to a crest, or from the middle line to a trough.
      Method:
      Recall the definition of amplitude as the maximum displacement from rest.
      Examiner tips
      • Amplitude is from the rest position to a peak, not crest-to-trough.
    19. Question 19

      1 marksSeismic S-waves
      Seismic S-waves (secondary waves) are transverse waves. In a transverse wave the particles vibrate at right angles (perpendicular) to the direction in which the wave travels.
      Method:
      Identify the S-wave as transverse with perpendicular particle vibration.
      Examiner tips
      • S-waves are transverse; P-waves are longitudinal.
    20. Question 20

      1 marksImage in a plane mirror
      A plane mirror forms a virtual image because the rays only appear to come from behind the mirror. The image is the same distance behind the mirror as the object is in front of it, and it is the same size as the object.
      Method:
      Apply the rules for a plane mirror: virtual image, same distance behind as the object in front.
      Examiner tips
      • Plane-mirror images are virtual, upright, same size and at equal distance.
    21. Angles in the refractive index formula are measured from the normal, not from the boundary. The angle of incidence from the normal is 90 - 50 = 40 degrees and the angle of refraction from the normal is 90 - 64 = 26 degrees. So the refractive index is the sine of 40 degrees divided by the sine of 26 degrees.
      Method:
      Convert 50 and 64 degrees from the boundary to 40 and 26 degrees from the normal; n = sine 40 / sine 26.
      Examiner tips
      • Always measure angles of incidence and refraction from the normal.
    22. An enlarged, upright image on the same side as the object is a virtual image. A converging lens produces this only when the object is placed inside the focal length, that is closer to the lens than its principal focus.
      Method:
      Recognise the magnifying-glass case: the object is closer than the principal focus.
      Examiner tips
      • Virtual, upright, enlarged image means the object is inside the focal length.
    23. Question 23

      1 marksSpeed of light
      Light, like all electromagnetic waves, travels through air (and a vacuum) at approximately 3.0 x 10^8 metres per second. This is about 300 000 km/s.
      Method:
      Recall the standard value: 3.0 x 10^8 m/s.
      Examiner tips
      • Remember the speed of light as 3.0 x 10^8 m/s.
    24. Question 24

      1 marksUltrasound depth sounding
      The pulse travels to the seabed and back, so the total distance is speed times time = 1500 x 4.6 = 6900 m. The depth is half of this because the pulse covers the depth twice: 6900 / 2 = 3450 m, which is about 3500 m.
      Method:
      Total distance 6900 m, depth = 6900 / 2 = about 3500 m.
      Examiner tips
      • For echo methods, halve the total distance to get the one-way distance.
    25. Question 25

      1 marksAudible frequency ranges
      Both can hear a frequency only if it lies inside both ranges. The overlap starts at the higher lower-limit (150 Hz, set by the dolphin) and ends at the lower upper-limit (20 kHz, set by the human). So both can hear from 150 Hz to 20 kHz.
      Method:
      Take the dolphin's 150 Hz minimum and the human's 20 kHz maximum to get 150 Hz to 20 kHz.
      Examiner tips
      • The shared range is the overlap: highest minimum to lowest maximum.
    26. Question 26

      1 marksCharging by induction
      The positive charge on the balloon attracts the free electrons in the wall towards the surface nearest the balloon. This leaves the wall surface negatively charged close to the balloon, so the opposite charges attract and the balloon sticks. This is charging by induction.
      Method:
      The positive balloon attracts wall electrons, inducing an opposite charge that attracts the balloon.
      Examiner tips
      • Charging by induction moves electrons in the wall, creating an opposite surface charge.
    27. An ammeter measures the current through the lamp, so it must be in series with the lamp. A voltmeter measures the potential difference across the lamp, so it must be connected in parallel with the lamp. The resistance is then voltage divided by current.
      Method:
      Connect the ammeter in series and the voltmeter in parallel with the lamp.
      Examiner tips
      • Ammeter in series, voltmeter in parallel: always.
    28. Resistance is proportional to length, so a shorter wire has a smaller resistance. Resistance is inversely proportional to cross-sectional area, so a thinner wire (smaller diameter) has a greater resistance.
      Method:
      Shorter wire has smaller resistance; thinner wire has greater resistance.
      Examiner tips
      • Resistance rises with length and falls with cross-sectional area.
    29. Question 29

      1 marksPotential divider equation
      In a series circuit the same current flows through both resistors, and the supply voltage is shared in proportion to the resistances. The fraction across R1 is R1 divided by the total resistance, so V1 = V multiplied by R1 divided by (R1 + R2).
      Method:
      Apply the potential divider rule: V1 = V x R1 / (R1 + R2).
      Examiner tips
      • The potential divider shares voltage in the ratio of the resistances.
    30. Question 30

      1 marksConverting a.c. to d.c.
      A diode only allows current to flow in one direction. When connected to an alternating supply it blocks the reverse half of each cycle, so the output current is in one direction only. This process is rectification, which converts a.c. to d.c.
      Method:
      Identify the diode as the one-way component that rectifies a.c. to d.c.
      Examiner tips
      • A diode conducts one way only, so it rectifies alternating current.
    31. The two resistors are in parallel, so they have the same potential difference. Across the 4.0 ohm resistor this is V = I x R = 1.0 x 4.0 = 4.0 V. The current in the 2.0 ohm resistor is then I = V / R = 4.0 / 2.0 = 2.0 A. The total current from the cell is the sum of the branch currents: 1.0 + 2.0 = 3.0 A.
      Method:
      Shared p.d. 4.0 V, 2.0 ohm branch carries 2.0 A, total = 1.0 + 2.0 = 3.0 A.
      Examiner tips
      • Parallel resistors share the same voltage; total current is the sum of the branch currents.
    32. The magnetic field around a straight wire forms concentric circles. Using the right-hand grip rule with the current pointing into the page, the field lines run clockwise when viewed from the front (looking in the direction of the current).
      Method:
      Apply the right-hand grip rule for current into the page: clockwise circular field.
      Examiner tips
      • Right-hand grip rule gives the circular field direction around a wire.
    33. Question 33

      1 marksTransformer primary current
      For a 100% efficient transformer the input power equals the output power. The output power = secondary voltage times secondary current = 5.0 x 2.0 = 10 W. The input power = primary voltage times primary current = 3.0 x primary current. Setting these equal: primary current = 10 / 3.0 = about 3.3 A.
      Method:
      Output power 10 W equals input power; primary current = 10 / 3.0 = about 3.3 A.
      Examiner tips
      • For an ideal transformer, primary power equals secondary power.
    34. Large-angle deflections happen when an alpha particle passes very close to a tiny, dense, positively charged nucleus and is strongly repelled. This shows that the positive charge and most of the mass of the atom are concentrated in a very small central nucleus.
      Method:
      Large deflections come from a close approach to a tiny, dense, positive nucleus.
      Examiner tips
      • Large-angle scattering shows a small, dense, positive nucleus.
    35. Question 35

      1 marksCorrected count rate
      Some of the measured count rate is due to background radiation that is always present. The corrected count rate for the source is found by measuring the background count rate separately and subtracting it from the measured count rate.
      Method:
      Subtract the separately measured background count rate from the detector reading.
      Examiner tips
      • Always subtract the background count rate to get the corrected count rate.
    36. Beta particles are far less massive than alpha particles, so for a given field they are deflected much more than alpha particles. The statement that alpha particles are deflected the most is therefore incorrect. The other three statements are all correct.
      Method:
      Beta is deflected more than alpha, so the alpha-most statement is the incorrect one.
      Examiner tips
      • Lighter beta particles are deflected more than heavier alpha particles.
    37. Question 37

      1 marksHalf-life fractions
      60 minutes is 60 divided by 20 = 3 half-lives. After each half-life the amount remaining halves: 1 then one half then one quarter then one eighth. So after 3 half-lives one eighth of the original sample remains.
      Method:
      60 / 20 = 3 half-lives, so the fraction left is one eighth.
      Examiner tips
      • Count whole half-lives and halve the amount each time.
    38. Question 38

      1 marksMotion in the Solar System
      The Moon takes about one month to complete one orbit of the Earth. The Earth orbits the Sun once per year (about 365 days) and rotates once every 24 hours; the Sun does not orbit the Earth.
      Method:
      Identify the correct period: the Moon orbits the Earth in about a month.
      Examiner tips
      • Day = Earth's spin, month = Moon's orbit, year = Earth's orbit of the Sun.
    39. Question 39

      1 marksNuclear reaction in stars
      A stable star is powered by nuclear fusion, in which hydrogen nuclei join together to form helium, releasing a very large amount of energy. This outward energy flow balances the inward pull of gravity.
      Method:
      Identify fusion of hydrogen into helium as the stellar energy source.
      Examiner tips
      • Stable stars fuse hydrogen into helium.
    40. First convert the speed: 3000 km/s = 3.0 x 10^6 m/s. The distance = speed divided by the Hubble constant = (3.0 x 10^6) / (2.2 x 10^-18) = about 1.4 x 10^24 m.
      Method:
      Convert 3000 km/s to 3.0 x 10^6 m/s, divide by 2.2 x 10^-18 to get about 1.4 x 10^24 m.
      Examiner tips
      • Watch unit conversions and the direction of the division in Hubble's law.

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