Cambridge IGCSE Physics Revision

Diameter = 10/20 = 0.50 mm; radius = 0.25 mm. Measuring a longer total width makes the ruler’s fixed resolution a smaller fraction of the measured total before dividing by the turn count.

Magnitude = √(3²+4²) = 5 m/s; its direction is north of east (tan θ = 4/3).

Total distance = 250 m; total time = 50 s; average speed = 250/50 = 5.0 m/s.

With no drag the velocity–time graph is a straight line of constant gradient g and the acceleration–time graph is horizontal at g. With drag the velocity–time curve progressively flattens as its gradient (acceleration) falls; at terminal velocity it is horizontal, acceleration is zero and drag balances weight.

W=mg=0.60×9.8=5.88 N (about 5.9 N).

Its mass is unchanged, but the Moon’s gravitational field strength is lower, so W=mg is smaller.

Volume = 20 cm³; ρ=54/20=2.7 g/cm³.

X floats on Y because its density is lower.

Taking right as positive, Fᵣ=9−4=5 N right.

k=F/x=5.0/0.025=200 N/m. a=F/m=6/2=3 m/s² in the resultant-force direction.

M=Fd=15×0.20=3.0 N m.

Clockwise total=7 N m. The 2 N force gives 3 N m, so a further 4 N m anticlockwise is needed.

Its centre of gravity is low and the support base is wide, so the weight line remains within the base through a larger tilt.

Δp=0−(0.50×8)=−4 kg m/s; F=Δp/Δt=−4/0.20=−20 N, opposite its initial motion.

30 J is transferred to other stores/surroundings; 100 J total remains accounted for.

Ek=½×2×3²=9 J; ΔEp=2×10×1.0=20 J.

W=Fd=20×3=60 J.

Tidal output is predictable and renewable, but only suitable sites exist and output follows the tide. Wind is renewable and has no fuel or operational emissions, but the wind supply at a site varies, so output is intermittent.

η=350/500×100%=70%.

P=ΔE/t=1200/20=60 W.

p=F/A=240/0.08=3000 Pa.

Δp=ρgΔh=1000×10×0.5=5000 Pa.

Condensation.

Rapid air molecules collide with it randomly and unevenly, continually changing its direction and speed.

Gas particles are much farther apart, leaving large empty spaces; particles in a solid are already close together.

T=27+273=300 K.

p₁V₁=p₂V₂; p₂=(100×0.30)/0.20=150 kPa.

Rails expand when heated; gaps provide room for expansion and reduce buckling.

A gas generally expands most; its widely separated particles can increase their mean spacing substantially.

It increases.

ΔE=mcΔθ=0.50×900×4=1800 J.

The most energetic surface particles escape, leaving a lower average kinetic energy in the liquid and therefore a lower temperature.

Warmth gives more particles enough energy to escape; moving air removes vapour above the surface and maintains evaporation.

Heat one end of each equally, attach identical wax pins at equal distances, and compare the time/order they fall while controlling rod dimensions and heat input.

Metals have mobile electrons that carry thermal energy efficiently; wood lacks this carrier mechanism and is a much poorer conductor.

Heating expands the air and lowers its density relative to surrounding cooler air, so it rises and is replaced by cooler air.

Dull black is the better emitter.

For emission, fill a Leslie cube with hot water and compare each face with an infrared detector held at a fixed distance and angle; keep cube temperature and detector position fixed and record readings. Dull black gives the largest reading, shiny metal the smallest. For absorption, put equal volumes of water at the same initial temperature into identical black and shiny cans; place them equally far from an infrared lamp and record probe temperatures at fixed intervals. Keep area, volume, starting temperature and exposure constant. The black can has the greater temperature-rise rate.

It transfers energy from the heater through the base to the contents efficiently.

Hot coolant transfers energy to the radiator by convection; metal conducts it to the outer surface, and moving air carries energy away by convection (with radiation also possible).

A wavefront joins points in phase; wavelength is spacing between adjacent points in phase; frequency is cycles per second; crest and trough are maximum positive and negative displacements; amplitude is maximum displacement from equilibrium. v=fλ=5×0.12=0.60 m/s.

No. Spreading is small when the gap is much larger than the wavelength.

28° to the normal.

6 cm behind the mirror; the image is the same perpendicular distance from the mirror.

Send a narrow ray-box beam through the curved face of a semicircular transparent block toward its centre. Vary incidence at the flat face and observe the refracted ray travel along the boundary at the critical angle; above it, no ray exits and the ray reflects inside. Optical fibres/endoscopes guide light by repeated total internal reflection at the glass–cladding boundary.

n=1/sin30°=2.0.

Use 1/f=1/u+1/v and m=−v/u. (a) 1/v=1/40−1/120=1/60, so v=+60 cm=1.5f and m=−0.50: real, inverted, half-height. Draw the parallel ray through the far focus and central ray straight; they meet beyond 2F. (b) 1/v=1/40−3/80=−1/80, so v=−80 cm=−2f and m=+3: virtual, upright, three times as tall. The outgoing rays diverge; dashed backward extensions meet on the object side.

Using 1/f=1/u+1/v gives 1/v=1/40−3/80=−1/80, so v=−80 cm=−2f. Magnification m=−v/u=+3: upright, virtual and three times taller on the object side. A diverging lens corrects short-sightedness by spreading rays before they enter the eye.

Violet.

One frequency (and therefore one wavelength in a fixed medium).

They can transmit through the atmosphere and are suitable for directed satellite communication; the syllabus also distinguishes low-orbit and geostationary applications.

Repeaters can reconstruct the intended discrete values accurately, limiting accumulated noise; analogue amplification also amplifies noise.

The pulse travels to the wall and back: d=vt/2=340×0.40/2=68 m.

Depth=vt/2=1500×0.10/2=75 m.

A compass uses a permanent magnet because its needle must retain a field without a power supply. A scrapyard crane uses an electromagnet because current switches the lifting field on and off to pick up and release steel.

A stronger magnetic field.

Negative charge, because electrons carry negative charge.

The force direction on a positive charge at that point.

In series with the part of the circuit whose current is measured.

I=Q/t=12/4=3 A.

E=W/Q=6/2=3 V.

V=W/Q=18/3=6 V.

R=V/I=4/0.5=8 Ω.

Resistance doubles because R is directly proportional to length.

E=2×3=6 kW h; cost=6×0.25=1.50 currency units.

The gallery shows one-cell and two battery-of-cells alternatives, a plain two-terminal power supply, a d.c. supply with fixed polarity, and an a.c. supply with alternating polarity. In the circuit, put cell, switch, lamp and ammeter in series and the voltmeter in parallel across the lamp.

The triangle/anode toward the bar/cathode is forward-biased, so current can flow and the LED emits light. Reversing it reverse-biases the LED; it blocks current in the simple model and stays off.

Rtotal=4+6=10 Ω.

1/R=1/6+1/3=3/6=1/2, so R=2 Ω.

Its p.d. doubles because V=IR at constant I.

V2/V1=R2/R1=6/2=3; V2=9 V.

When opened it disconnects the appliance from the live supply, making the appliance safer to handle.

Move the magnet faster, use a stronger magnet or increase coil turns, changing linked magnetic flux more rapidly.

Fleming’s right-hand rule gives +z: out of the page toward the viewer (dot). If motion reverses to −x while B stays +y, current reverses to −z: into the page away from the viewer (cross).

The alternating e.m.f. crosses zero and changes sign.

Concentric circles centred on the wire.

The wire field is weaker farther away. A solenoid field is strong and nearly uniform in its central interior but weaker outside and near its ends. Increasing current or turns per unit length strengthens it; reversing the current reverses direction.

The force direction reverses.

For the wire, first finger east (+x), middle finger north (+y), so the left-hand thumb gives force into the page, −z (cross, away from the viewer). For the particle, +x×+y gives +z (dot, out toward the viewer) for positive charge; the electron’s force reverses to −z (cross, into the page away from the viewer).

Increase number of turns, current or magnetic-field strength.

To reverse the coil forces at the correct time so its rotation continues in the same direction.

Vs=Vp×Ns/Np=12×1000/200=60 V; it is step-up.

Current halves; loss becomes (1/2)²=1/4 as large.

It becomes a 2+ ion.

Most alpha particles pass straight through the thin metal foil with little/no deflection.

N=A−Z=23−11=12.

It splits into smaller nuclei, usually with energy release; total nucleon number and charge balance across the equation.

120/2=60 counts/min.

45−8=37 counts/min.

Gamma radiation.

An alpha particle has kinetic energy and a +2 charge, so it interacts strongly with atomic electrons and transfers energy rapidly along its path, removing many electrons. Gamma has kinetic energy but no charge, so it does not ionise directly; it can transfer energy to charged secondary electrons, which then ionise atoms.

No. Decay is spontaneous and random: the exact decay time of an individual nucleus cannot be predicted. Only the probability of decay in an interval, or the statistical behaviour of a large population, can be described.

The daughter is ²³⁴₉₀Th; nucleon and proton totals balance.

800→400→200→100 is three half-lives; T½=12/3=4 h.

Some beta passes through thin paper, and detected intensity decreases with thickness; alpha is stopped too easily and gamma passes through too much.

The work must be authorised and directly teacher-supervised; the student must not repeat it alone. Minimise exposure time and handling, keep the source away from people with suitable forceps/tongs, use the prescribed shielding, then promptly return it to secure, labelled, shielded storage.

Minimise exposure time, increase distance using tongs and add suitable shielding.

Earth’s axis is tilted. At that part of the orbit the Northern Hemisphere tilts toward the Sun, receiving higher-angle sunlight and longer days, while the Southern Hemisphere tilts away, receiving lower-angle sunlight and shorter days.

v=2πr/T≈2π×1.5×10¹¹/3.0×10⁷≈3.1×10⁴ m/s.

t=d/c=1.5×10¹¹/(3.0×10⁸)=500 s (about 8.3 min).

As it falls closer, gravitational potential energy decreases and kinetic energy/speed increases; total mechanical energy is conserved.

Hydrogen and helium.

Fusion of hydrogen into helium releases energy.

Distance: how far light travels in one year.

When most of the hydrogen in the central region has fused into helium, the core contracts and outer layers expand and cool. A supernova nebula contains hydrogen from the star’s outer layers together with newly formed heavier elements.

Their light wavelength is stretched; this is evidence that galaxies recede as the Universe expands.

H₀=v/d=2.2×10⁵/1.0×10²²=2.2×10⁻¹⁷ s⁻¹.

Papers 2, 4 and 6; the Extended route is eligible for A*–G. Yes, 0625 also has a March exam series in India (as well as June and November).

AO3, experimental skills and investigations.

Write 7635 without a space and 12 000 with a space after the first three digits. A sketch shows key features in correct relative positions/proportions and labels important parts, but omits unnecessary detail.

Independent: wire length. Dependent: resistance (from V/I). Control: wire material/cross-sectional area and temperature.

12.5 mm, the nearest half-division.

Yes. The assumed limit is ±10%, and 9% lies within it (subject to the question’s wording and relevant uncertainty).

A force meter; check its range/scale (and zero) before measuring.

The normal, perpendicular to the surface at the point of incidence.

Put the ammeter in series with the resistor and the voltmeter in parallel across it.

Precision is high; accuracy is low (a systematic error may be present).

m/s. Use a triangle whose hypotenuse covers at least half the best-fit straight line, then calculate Δy/Δx.

Yes. Paper 5 is the laboratory Practical Test.

No. It uses the same experimental contexts and skill requirements, assessed without requiring experiments during the paper.

Frequency f: Hz; wavelength λ: m; focal length f: m (distinguish by context); angle θ: degrees; potential difference V: volts; current I: amperes; e.m.f. E: volts; resistance R: ohms; count rate: counts/s or counts/min; half-life t½: units of time such as s, min, h, days, weeks or years.

The same Core entries still apply: current I in A (or mA), resistance R in Ω and half-life t½ in time units; they are not Supplement-only.

V=πr²h=π(0.050)²(0.20)=1.57×10⁻³ m³. Use radius rather than diameter and express radius and length in the same unit before substitution.

sin θ=3/5=0.6, so θ=sin⁻¹(0.6)≈36.9°.

Diameter = 10/20 = 0.50 mm; radius = 0.25 mm. Measuring a longer total width makes the ruler’s fixed resolution a smaller fraction of the measured total before dividing by the turn count.

Magnitude = √(3²+4²) = 5 m/s; its direction is north of east (tan θ = 4/3).

Total distance = 250 m; total time = 50 s; average speed = 250/50 = 5.0 m/s.

With no drag the velocity–time graph is a straight line of constant gradient g and the acceleration–time graph is horizontal at g. With drag the velocity–time curve progressively flattens as its gradient (acceleration) falls; at terminal velocity it is horizontal, acceleration is zero and drag balances weight.

W=mg=0.60×9.8=5.88 N (about 5.9 N).

Its mass is unchanged, but the Moon’s gravitational field strength is lower, so W=mg is smaller.

Volume = 20 cm³; ρ=54/20=2.7 g/cm³.

X floats on Y because its density is lower.

Taking right as positive, Fᵣ=9−4=5 N right.

k=F/x=5.0/0.025=200 N/m. a=F/m=6/2=3 m/s² in the resultant-force direction.

M=Fd=15×0.20=3.0 N m.

Clockwise total=7 N m. The 2 N force gives 3 N m, so a further 4 N m anticlockwise is needed.

Its centre of gravity is low and the support base is wide, so the weight line remains within the base through a larger tilt.

Δp=0−(0.50×8)=−4 kg m/s; F=Δp/Δt=−4/0.20=−20 N, opposite its initial motion.

30 J is transferred to other stores/surroundings; 100 J total remains accounted for.

Ek=½×2×3²=9 J; ΔEp=2×10×1.0=20 J.

W=Fd=20×3=60 J.

Tidal output is predictable and renewable, but only suitable sites exist and output follows the tide. Wind is renewable and has no fuel or operational emissions, but the wind supply at a site varies, so output is intermittent.

η=350/500×100%=70%.

P=ΔE/t=1200/20=60 W.

p=F/A=240/0.08=3000 Pa.

Δp=ρgΔh=1000×10×0.5=5000 Pa.

Condensation.

Rapid air molecules collide with it randomly and unevenly, continually changing its direction and speed.

Gas particles are much farther apart, leaving large empty spaces; particles in a solid are already close together.

T=27+273=300 K.

p₁V₁=p₂V₂; p₂=(100×0.30)/0.20=150 kPa.

Rails expand when heated; gaps provide room for expansion and reduce buckling.

A gas generally expands most; its widely separated particles can increase their mean spacing substantially.

It increases.

ΔE=mcΔθ=0.50×900×4=1800 J.

The most energetic surface particles escape, leaving a lower average kinetic energy in the liquid and therefore a lower temperature.

Warmth gives more particles enough energy to escape; moving air removes vapour above the surface and maintains evaporation.

Heat one end of each equally, attach identical wax pins at equal distances, and compare the time/order they fall while controlling rod dimensions and heat input.

Metals have mobile electrons that carry thermal energy efficiently; wood lacks this carrier mechanism and is a much poorer conductor.

Heating expands the air and lowers its density relative to surrounding cooler air, so it rises and is replaced by cooler air.

Dull black is the better emitter.

For emission, fill a Leslie cube with hot water and compare each face with an infrared detector held at a fixed distance and angle; keep cube temperature and detector position fixed and record readings. Dull black gives the largest reading, shiny metal the smallest. For absorption, put equal volumes of water at the same initial temperature into identical black and shiny cans; place them equally far from an infrared lamp and record probe temperatures at fixed intervals. Keep area, volume, starting temperature and exposure constant. The black can has the greater temperature-rise rate.

It transfers energy from the heater through the base to the contents efficiently.

Hot coolant transfers energy to the radiator by convection; metal conducts it to the outer surface, and moving air carries energy away by convection (with radiation also possible).

A wavefront joins points in phase; wavelength is spacing between adjacent points in phase; frequency is cycles per second; crest and trough are maximum positive and negative displacements; amplitude is maximum displacement from equilibrium. v=fλ=5×0.12=0.60 m/s.

No. Spreading is small when the gap is much larger than the wavelength.

28° to the normal.

6 cm behind the mirror; the image is the same perpendicular distance from the mirror.

Send a narrow ray-box beam through the curved face of a semicircular transparent block toward its centre. Vary incidence at the flat face and observe the refracted ray travel along the boundary at the critical angle; above it, no ray exits and the ray reflects inside. Optical fibres/endoscopes guide light by repeated total internal reflection at the glass–cladding boundary.

n=1/sin30°=2.0.

Use 1/f=1/u+1/v and m=−v/u. (a) 1/v=1/40−1/120=1/60, so v=+60 cm=1.5f and m=−0.50: real, inverted, half-height. Draw the parallel ray through the far focus and central ray straight; they meet beyond 2F. (b) 1/v=1/40−3/80=−1/80, so v=−80 cm=−2f and m=+3: virtual, upright, three times as tall. The outgoing rays diverge; dashed backward extensions meet on the object side.

Using 1/f=1/u+1/v gives 1/v=1/40−3/80=−1/80, so v=−80 cm=−2f. Magnification m=−v/u=+3: upright, virtual and three times taller on the object side. A diverging lens corrects short-sightedness by spreading rays before they enter the eye.

Violet.

One frequency (and therefore one wavelength in a fixed medium).

They can transmit through the atmosphere and are suitable for directed satellite communication; the syllabus also distinguishes low-orbit and geostationary applications.

Repeaters can reconstruct the intended discrete values accurately, limiting accumulated noise; analogue amplification also amplifies noise.

The pulse travels to the wall and back: d=vt/2=340×0.40/2=68 m.

Depth=vt/2=1500×0.10/2=75 m.

A compass uses a permanent magnet because its needle must retain a field without a power supply. A scrapyard crane uses an electromagnet because current switches the lifting field on and off to pick up and release steel.

A stronger magnetic field.

Negative charge, because electrons carry negative charge.

The force direction on a positive charge at that point.

In series with the part of the circuit whose current is measured.

I=Q/t=12/4=3 A.

E=W/Q=6/2=3 V.

V=W/Q=18/3=6 V.

R=V/I=4/0.5=8 Ω.

Resistance doubles because R is directly proportional to length.

E=2×3=6 kW h; cost=6×0.25=1.50 currency units.

The gallery shows one-cell and two battery-of-cells alternatives, a plain two-terminal power supply, a d.c. supply with fixed polarity, and an a.c. supply with alternating polarity. In the circuit, put cell, switch, lamp and ammeter in series and the voltmeter in parallel across the lamp.

The triangle/anode toward the bar/cathode is forward-biased, so current can flow and the LED emits light. Reversing it reverse-biases the LED; it blocks current in the simple model and stays off.

Rtotal=4+6=10 Ω.

1/R=1/6+1/3=3/6=1/2, so R=2 Ω.

Its p.d. doubles because V=IR at constant I.

V2/V1=R2/R1=6/2=3; V2=9 V.

When opened it disconnects the appliance from the live supply, making the appliance safer to handle.

Move the magnet faster, use a stronger magnet or increase coil turns, changing linked magnetic flux more rapidly.

Fleming’s right-hand rule gives +z: out of the page toward the viewer (dot). If motion reverses to −x while B stays +y, current reverses to −z: into the page away from the viewer (cross).

The alternating e.m.f. crosses zero and changes sign.

Concentric circles centred on the wire.

The wire field is weaker farther away. A solenoid field is strong and nearly uniform in its central interior but weaker outside and near its ends. Increasing current or turns per unit length strengthens it; reversing the current reverses direction.

The force direction reverses.

For the wire, first finger east (+x), middle finger north (+y), so the left-hand thumb gives force into the page, −z (cross, away from the viewer). For the particle, +x×+y gives +z (dot, out toward the viewer) for positive charge; the electron’s force reverses to −z (cross, into the page away from the viewer).

Increase number of turns, current or magnetic-field strength.

To reverse the coil forces at the correct time so its rotation continues in the same direction.

Vs=Vp×Ns/Np=12×1000/200=60 V; it is step-up.

Current halves; loss becomes (1/2)²=1/4 as large.

It becomes a 2+ ion.

Most alpha particles pass straight through the thin metal foil with little/no deflection.

N=A−Z=23−11=12.

It splits into smaller nuclei, usually with energy release; total nucleon number and charge balance across the equation.

120/2=60 counts/min.

45−8=37 counts/min.

Gamma radiation.

An alpha particle has kinetic energy and a +2 charge, so it interacts strongly with atomic electrons and transfers energy rapidly along its path, removing many electrons. Gamma has kinetic energy but no charge, so it does not ionise directly; it can transfer energy to charged secondary electrons, which then ionise atoms.

No. Decay is spontaneous and random: the exact decay time of an individual nucleus cannot be predicted. Only the probability of decay in an interval, or the statistical behaviour of a large population, can be described.

The daughter is ²³⁴₉₀Th; nucleon and proton totals balance.

800→400→200→100 is three half-lives; T½=12/3=4 h.

Some beta passes through thin paper, and detected intensity decreases with thickness; alpha is stopped too easily and gamma passes through too much.

The work must be authorised and directly teacher-supervised; the student must not repeat it alone. Minimise exposure time and handling, keep the source away from people with suitable forceps/tongs, use the prescribed shielding, then promptly return it to secure, labelled, shielded storage.

Minimise exposure time, increase distance using tongs and add suitable shielding.

Earth’s axis is tilted. At that part of the orbit the Northern Hemisphere tilts toward the Sun, receiving higher-angle sunlight and longer days, while the Southern Hemisphere tilts away, receiving lower-angle sunlight and shorter days.

v=2πr/T≈2π×1.5×10¹¹/3.0×10⁷≈3.1×10⁴ m/s.

t=d/c=1.5×10¹¹/(3.0×10⁸)=500 s (about 8.3 min).

As it falls closer, gravitational potential energy decreases and kinetic energy/speed increases; total mechanical energy is conserved.

Hydrogen and helium.

Fusion of hydrogen into helium releases energy.

Distance: how far light travels in one year.

When most of the hydrogen in the central region has fused into helium, the core contracts and outer layers expand and cool. A supernova nebula contains hydrogen from the star’s outer layers together with newly formed heavier elements.