Full Syllabus Challenge 1 — Paper 4

Cambridge IGCSE Physics 0625 (Extended) — Topics 1–6: Motion, Forces & Energy · Thermal Physics · Waves · Electricity & Magnetism · Nuclear Physics · Space Physics

Duration 1 hour 15 minutes

Total Marks 80

Questions 7

Time Remaining 75:00

Syllabus 0625 / 0972

⚡ Cambridge Challenge Level — Full Syllabus

This paper covers all six topics at real Cambridge IGCSE difficulty. Scoring 50%+ is a solid B, 60%+ is an A, 70%+ is A*. Don't worry if this feels harder -- that's the point!

Instructions

Answer all questions in the spaces provided.
Show all working for calculations. Answers without working may not receive full marks.
You may use a calculator.
Take g = 10 N/kg unless a question states otherwise.
Speed of light c = 3.0 × 10⁸ m/s; speed of sound in air = 340 m/s
Specific heat capacity of water = 4200 J/(kg °C)
1 light-year = 9.5 × 10¹⁵ m; Hubble constant H₀ = 2.2 × 10⁻¹⁸ s⁻¹
Answer all questions first. When finished, click "Submit Exam" at the bottom of Question 7.
Your answers will be automatically graded when you submit the exam. The model answers will be shown for review.

Question Navigation

Question 1 -- The Burj Khalifa Express Lift

Total: 14 marks · Topic 1: Motion, Forces and Energy

The Burj Khalifa in Dubai has one of the fastest lifts (elevators) in the world. An express lift car has a mass of 1800 kg and carries 10 passengers of average mass 70 kg from the ground floor to an observation deck. Take g = 10 N/kg.

(a) [2 marks]

Calculate the total mass of the loaded lift and hence its total weight.

Mark Scheme -- Part (a)

Total mass = 1800 + (10 × 70) = 2500 kg

Weight W = mg = 2500 × 10 = 25 000 N

1 mark -- total mass 1800 plus 700 equals 2500 kg 1 mark -- weight mg 2500 times 10 equals 25000 N

⚠ If you missed marks here: Don't forget the passengers: total mass = 1800 + (10 × 70) = 2500 kg — using 1800 kg alone gives 18 000 N. And weight is a FORCE in newtons (W = mg = 25 000 N); writing "2500 kg" as the weight mixes up mass and weight.

(b) [4 marks]

At the start of the journey the lift accelerates upwards at 1.2 m/s².

(i) Calculate the resultant force on the loaded lift during this acceleration.

(ii) Hence calculate the tension in the supporting cable during the acceleration. Explain your reasoning.

Mark Scheme -- Part (b)

(i) F = ma stated or used

(i) F = 2500 × 1.2 = 3000 N (upwards)

(ii) The cable tension must both support the weight and provide the resultant force: T = W + F = 25 000 + 3000

(ii) T = 28 000 N

1 mark -- resultant force F equals ma used 1 mark -- F 2500 times 1.2 equals 3000 N 1 mark -- tension equals weight plus resultant force 25000 plus 3000 1 mark -- tension 28000 N

⚠ If you missed marks here: The trap in (ii): F = ma gives the RESULTANT force (3000 N), not the tension. Because the lift accelerates upwards, the cable must supply the weight PLUS the resultant: T = 25 000 + 3000 = 28 000 N — answering 3000 N or 25 000 N means the two forces were confused, and subtracting (22 000 N) has the direction backwards.

The complete journey has three stages:
• Stage 1: uniform acceleration from rest to 6.0 m/s in 5.0 s
• Stage 2: constant speed of 6.0 m/s for 60 s
• Stage 3: uniform deceleration from 6.0 m/s to rest in 4.0 s

Sketch the speed-time graph for the journey on paper, then use the graph (or otherwise) to show that the lift rises approximately 387 m. Show the distance for each stage separately.

Mark Scheme -- Part (c)

Stage 1: area of triangle = ½ × 5.0 × 6.0 = 15 m

Stage 2: area of rectangle = 6.0 × 60 = 360 m

Stage 3: area of triangle = ½ × 4.0 × 6.0 = 12 m

Total = 15 + 360 + 12 = 387 m (distance = area under speed-time graph)

1 mark -- stage 1 triangle half times 5 times 6 equals 15 m 1 mark -- stage 2 rectangle 6 times 60 equals 360 m 1 mark -- stage 3 triangle half times 4 times 6 equals 12 m 1 mark -- total area under graph 387 m

⚠ If you missed marks here: The classic slip is forgetting the ½ on the acceleration and deceleration triangles — 5 × 6 = 30 m and 4 × 6 = 24 m give a total of 414 m, not 387 m. This is a "show that" question, so each stage's area must be shown separately: 15 + 360 + 12 = 387 m; a bare final answer earns little.

(d) [4 marks]

During Stage 2 the lift rises at a constant 6.0 m/s.

(i) Calculate the useful power output of the motor during Stage 2 (the rate at which work is done against gravity on the 2500 kg lift).

(ii) The electrical input power to the motor during Stage 2 is 200 kW. Calculate the efficiency of the lift system and state what happens to the wasted energy.

Mark Scheme -- Part (d)

(i) P = work done / time = Fv = weight × speed = 25 000 × 6.0

(i) P = 150 000 W = 150 kW

(ii) Efficiency = useful power / input power = 150 / 200 = 0.75 = 75%

(ii) Wasted energy is transferred to thermal energy (heat) in the motor, cables and pulleys due to friction / electrical resistance (and some sound)

1 mark -- power equals force times speed 25000 times 6 1 mark -- useful power 150000 W or 150 kW 1 mark -- efficiency 150 divided 200 equals 75 percent 1 mark -- wasted energy thermal heat friction motor resistance sound

⚠ If you missed marks here: Power against gravity = WEIGHT × speed = 25 000 × 6.0 = 150 kW — using the mass (2500 × 6 = 15 000) forgets g and is 10× too small. For the wasted energy, "it's lost" scores nothing: name the store and cause — thermal energy from friction/electrical resistance in the motor, cables and pulleys.

Question 2 -- Engine Cooling in Chennai Traffic

Total: 10 marks · Topic 2: Thermal Physics

A taxi idling in slow Chennai traffic relies on its cooling system to stop the engine overheating. The engine transfers thermal energy to water-based coolant, which is pumped through a radiator at the front of the car. Specific heat capacity of the coolant = 4200 J/(kg °C).

(a) [3 marks]

The engine transfers 42 000 J of thermal energy to the coolant every second. Coolant flows through the engine at a rate of 1.2 kg per second.

Calculate the rise in temperature of the coolant as it passes through the engine. Show your working.

Mark Scheme -- Part (a)

E = mcΔT rearranged: ΔT = E / (mc)

ΔT = 42 000 / (1.2 × 4200) = 42 000 / 5040

ΔT = 8.3 °C (accept 8.3–8.4 °C)

1 mark -- equation E equals mc delta T rearranged 1 mark -- substitution 42000 divided 1.2 times 4200 or 5040 1 mark -- temperature rise 8.3 or 8.4 degrees

⚠ If you missed marks here: Everything here is "per second", so the energy (42 000 J) pairs with the mass that flows in that second (1.2 kg) — no time conversion needed. The usual slips: rearranging wrongly (ΔT = E/(mc), so divide by BOTH 1.2 and 4200 — dividing by 4200 alone gives 10 °C) or multiplying instead of dividing.

(b) [3 marks]

The radiator is made of thin metal tubes with many closely spaced metal fins. The fins are painted matt black, and a fan blows air across them.

Explain how the design of the radiator uses all three methods of thermal energy transfer to cool the coolant. Name each method.

Mark Scheme -- Part (b)

Conduction: thermal energy passes through the thin metal tube walls and fins because metal is a good conductor (free electrons transfer energy); thin walls and large fin area speed this up

Convection: air warmed between the fins expands, becomes less dense and rises (and the fan forces fresh cool air across), carrying energy away in convection currents

Radiation: the matt black surface is a good emitter of infrared radiation, so the fins radiate energy away faster than a shiny surface would

1 mark -- conduction metal good conductor free electrons thin walls fins 1 mark -- convection warm air less dense rises fan cool air currents 1 mark -- radiation matt black good emitter infrared

⚠ If you missed marks here: Each mark needs the METHOD NAMED and tied to a design feature: conduction through the thin metal walls/fins, convection as warmed air rises (say "hot AIR rises", never "heat rises") helped by the fan, and radiation from the matt black surface — which here is a good EMITTER, not absorber. Missing any one of the three loses that mark.

The cooling system includes an expansion tank, which gives the coolant somewhere to go as it warms up.

Explain, in terms of molecules, why the coolant needs extra space when its temperature rises.

Mark Scheme -- Part (c)

As temperature rises the molecules gain kinetic energy and move faster, so on average they push slightly further apart

The same mass of liquid therefore occupies a larger volume (the liquid expands), so without the expansion tank the pressure would rise and could burst the system

1 mark -- molecules gain kinetic energy move faster further apart 1 mark -- same mass larger volume liquid expands pressure would rise

⚠ If you missed marks here: The classic wrong answer: "the molecules expand / get bigger" — molecules stay the same size; they move FASTER and their average SEPARATION increases. The second mark is the consequence: same mass now occupies a larger volume, so the liquid needs the extra space of the expansion tank.

(d) [2 marks]

The air trapped inside the sealed expansion tank is warmed from 30 °C to 90 °C. The volume of the trapped air stays approximately constant.

Explain, using the kinetic particle model, why the pressure of the trapped air increases.

Mark Scheme -- Part (d)

At the higher temperature the air molecules have more kinetic energy and move faster

They collide with the container walls more frequently and with greater force per collision, so the average force per unit area (pressure) on the walls increases

1 mark -- higher temperature molecules more kinetic energy faster 1 mark -- collide walls more often greater force per collision pressure increases

⚠ If you missed marks here: "The air expands" contradicts the question — the volume is FIXED here; it's the temperature that changes. The particle story: hotter molecules move faster, so they hit the walls MORE OFTEN and EACH collision pushes harder — both effects raise the force per unit area. Compare with expansion questions where speed is constant and only collision frequency changes.

Question 3 -- Light, Prisms and Radio Waves

Total: 12 marks · Topic 3: Waves

A photography student in Tokyo studies how her DSLR camera works. Inside the viewfinder, light is redirected by a glass pentaprism using total internal reflection. The glass of the prism has a refractive index of 1.55.

(a) [3 marks]

Inside the prism, a ray of light strikes a glass–air surface at an angle of incidence of 45°.

Calculate the critical angle of the glass, and determine whether total internal reflection occurs at this surface. Justify your answer.

Mark Scheme -- Part (a)

sin c = 1/n = 1/1.55 = 0.645

c = 40° (accept 40.2°)

45° is greater than the critical angle and the light is travelling in the denser medium (glass) towards air, so total internal reflection does occur

1 mark -- sin critical angle equals 1 divided n equals 0.645 1 mark -- critical angle 40 degrees 1 mark -- 45 exceeds critical angle denser medium glass total internal reflection occurs

⚠ If you missed marks here: sin c = 1/1.55 = 0.645, then take sin⁻¹ to get c = 40° — stopping at 0.645 (not an angle) is the usual slip. The last mark needs the explicit comparison AND conditions: 45° > 40° and the light is inside the DENSER glass heading to air, so TIR occurs — a bare "yes" doesn't score.

(b) [3 marks]

The green focusing light used by the camera has a wavelength of 5.46 × 10⁻⁷ m in air.

(i) Calculate the frequency of this light in air. (c = 3.0 × 10⁸ m/s)

(ii) Calculate the speed of this light inside the glass (n = 1.55), and state what happens to its frequency as it enters the glass.

Mark Scheme -- Part (b)

(i) f = v/λ = 3.0 × 10⁸ / 5.46 × 10⁻⁷ = 5.5 × 10¹⁴ Hz

(ii) speed in glass = c/n = 3.0 × 10⁸ / 1.55 = 1.9 × 10⁸ m/s

(ii) The frequency stays the same (it is fixed by the source); it is the speed and wavelength that decrease in the glass

1 mark -- frequency v divided wavelength 5.5 times 10 to 14 Hz 1 mark -- speed glass c divided n 1.9 times 10 to 8 m/s 1 mark -- frequency unchanged fixed source speed wavelength decrease

⚠ If you missed marks here: Watch the powers of ten in f = v/λ = 3.0 × 10⁸ / 5.46 × 10⁻⁷ ≈ 5.5 × 10¹⁴ Hz — a wrong exponent (10¹⁵, 10¹³) is the top error. In (ii), speed in glass = c/n = 1.9 × 10⁸ m/s (divide by n, don't multiply), and the FREQUENCY is unchanged — saying it decreases is the classic mistake.

The student's grandparents live in a valley behind a large hill. They can receive long-wave radio broadcasts (wavelength 1500 m) clearly, but FM radio (wavelength 3 m) is very weak in the valley.

Explain this difference using the idea of diffraction.

Mark Scheme -- Part (c)

Waves diffract (spread) around an obstacle significantly only when their wavelength is comparable to (or larger than) the size of the obstacle

The 1500 m long-wave signal has a wavelength similar to the size of the hill, so it diffracts strongly over/around the hill and reaches the valley

The 3 m FM signal has a wavelength much smaller than the hill, so it diffracts very little and the valley lies in a radio "shadow"

1 mark -- diffraction significant when wavelength comparable size obstacle 1 mark -- 1500 m long wave similar size hill diffracts strongly reaches valley 1 mark -- 3 m FM much smaller little diffraction shadow

⚠ If you missed marks here: "Long-wave is stronger / travels further" scores nothing — the whole answer hangs on comparing WAVELENGTH to the SIZE of the hill: 1500 m is comparable to the hill so it diffracts strongly over it; 3 m is far smaller so it barely bends and the valley sits in a radio shadow. State the general rule (diffraction is significant when λ ≈ obstacle size) for the first mark.

(d) [3 marks]

The student designs an experiment to measure the speed of sound directly. She places two microphones 2.5 m apart, both connected to a fast electronic timer. A wooden clapper is struck beyond the first microphone: the sound starting the timer at microphone 1 and stopping it at microphone 2. The timer reads 7.4 ms (0.0074 s).

(i) Calculate the speed of sound given by this experiment.

(ii) Suggest one improvement to the experiment that would reduce the effect of timing errors, and explain why it helps.

Mark Scheme -- Part (d)

(i) v = d/t = 2.5 / 0.0074

(i) v = 338 m/s (accept 337–340 m/s)

(ii) e.g. increase the separation of the microphones (longer time, so the fixed timing uncertainty is a smaller fraction of the reading) / repeat several times and take an average to reduce random error

1 mark -- speed distance divided time 2.5 divided 0.0074 1 mark -- 338 m/s or about 340 1 mark -- increase separation longer time smaller percentage uncertainty or repeat average random error

⚠ If you missed marks here: In (i) mind the milliseconds: 7.4 ms = 0.0074 s, so v = 2.5/0.0074 = 338 m/s — dividing by 7.4 gives 0.34 m/s, a thousand times off. In (ii), no echo here, so NO factor of 2 — and the improvement mark needs a reason: a bigger separation makes the measured time longer, so the fixed timing uncertainty becomes a smaller PERCENTAGE of it ("repeat it" alone needs "...and average to reduce random error").

Question 4 -- Greenhouse Electronics in Pune

Total: 14 marks · Topic 4: Electricity

A horticulturist near Pune grows orchids in a greenhouse. A sensing circuit monitors the temperature, and an electric heater keeps seedlings warm on cold nights.

(a) [4 marks]

The sensing circuit is a 12 V battery connected in series with a thermistor and a 400 Ω fixed resistor. At 20 °C the thermistor has a resistance of 800 Ω.

Calculate:
(i) the total resistance of the circuit
(ii) the current in the circuit
(iii) the potential difference across the fixed resistor and across the thermistor at 20 °C.

Mark Scheme -- Part (a)

(i) Total R = 800 + 400 = 1200 Ω

(ii) I = V/R = 12 / 1200 = 0.010 A (10 mA)

(iii) V across fixed resistor = IR = 0.010 × 400 = 4.0 V

(iii) V across thermistor = 0.010 × 800 = 8.0 V (the two p.d.s add up to the 12 V supply)

1 mark -- total resistance 800 plus 400 equals 1200 ohm 1 mark -- current 12 divided 1200 equals 0.010 A 1 mark -- pd fixed resistor 0.010 times 400 equals 4.0 V 1 mark -- pd thermistor 8.0 V sum equals 12 supply

⚠ If you missed marks here: In series the SAME current (12/1200 = 0.010 A) flows through both components — the classic error is using 12 V across each component separately (12/400 and 12/800). Check your two p.d.s with the built-in test: 4.0 V + 8.0 V must add up to the 12 V supply.

(b) [3 marks]

On a hot afternoon the thermistor's resistance falls to 200 Ω.

(i) Calculate the new potential difference across the 400 Ω fixed resistor.

(ii) Explain how this circuit could be used to switch on a cooling fan automatically when the greenhouse gets hot.

Mark Scheme -- Part (b)

(i) New total R = 600 Ω, so I = 12/600 = 0.020 A

(i) V across fixed resistor = 0.020 × 400 = 8.0 V

(ii) As temperature rises the thermistor resistance falls, so the fixed resistor takes a larger share of the supply p.d.; when this voltage passes a set level it can trigger a switching circuit / relay that turns the fan on

1 mark -- new total 600 ohm current 0.020 A 1 mark -- pd fixed resistor now 8.0 V 1 mark -- temperature rises thermistor resistance falls voltage share rises triggers relay switch fan

⚠ If you missed marks here: The resistance changed, so the current changes too — recalculate it first: I = 12/600 = 0.020 A, then V = 0.020 × 400 = 8.0 V; keeping the old 0.010 A gives 4.0 V and loses both marks. In (ii), remember a thermistor's resistance FALLS as it gets hotter (the reverse is a common error), so the fixed resistor's share of the 12 V RISES and can trigger a relay/switch.

The greenhouse heater is marked 230 V, 460 W.

Calculate:
(i) the current in the heater when operating normally
(ii) the resistance of the heating element
(iii) the energy transferred, in kWh, when the heater runs for 8.0 hours overnight
(iv) the cost of running the heater overnight, if electricity costs ₹8.00 per kWh.

Mark Scheme -- Part (c)

(i) I = P/V = 460 / 230 = 2.0 A

(ii) R = V/I = 230 / 2.0 = 115 Ω

(iii) E = 0.46 kW × 8.0 h = 3.68 kWh (accept 3.7 kWh)

(iv) Cost = 3.68 × 8.00 = ₹29.44 (accept ₹29–30)

1 mark -- current P divided V 460 divided 230 equals 2.0 A 1 mark -- resistance 230 divided 2 equals 115 ohm 1 mark -- energy 0.46 kW times 8 equals 3.68 kWh 1 mark -- cost 3.68 times 8 equals 29.44 rupees

⚠ If you missed marks here: For (iii) the kWh shortcut is power in kW × time in HOURS: 0.46 × 8.0 = 3.68 kWh — students who compute joules (460 × 28 800 s) then forget to convert lose the mark, and 460 × 8 = 3680 "kWh" forgets the W→kW step. In (i)-(ii), I = P/V = 2.0 A first, THEN R = V/I = 115 Ω.

(d) [3 marks]

The heater has a metal case, an earth wire, and a fuse.

One night a loose live wire inside the heater touches the metal case. Describe, step by step, the sequence of events that protects anyone who later touches the case.

Mark Scheme -- Part (d)

The case becomes live, but the earth wire connects the case to earth and provides a very low resistance path, so a very large current flows from live through the case to earth

This large current melts (blows) the fuse, which is fitted in the live wire

The circuit is broken and the heater is disconnected from the live supply, so the case is no longer live and cannot give an electric shock

1 mark -- earth wire low resistance path large current flows case to earth 1 mark -- large current melts blows fuse fitted live wire 1 mark -- circuit broken heater disconnected case no longer live no shock

⚠ If you missed marks here: "The earth wire carries the electricity safely away" on its own misses the sequence the question asks for: LOW-RESISTANCE earth path → very LARGE current flows → that current melts the fuse (which must be in the LIVE wire) → circuit broken, case no longer live. The fuse and earth wire work TOGETHER — answers crediting only one usually lose marks.

Question 5 -- Induction in the Laboratory and the Phone Charger

Total: 10 marks · Topic 4: Electromagnetic Effects

A student at a school in Singapore investigates electromagnetic induction, then applies the same physics to the mains charger of her phone.

(a) [3 marks]

The student drops a bar magnet, north pole first, straight through a vertical coil of wire connected to a data logger that records the induced e.m.f. The trace shows a small pulse followed by a larger pulse in the opposite direction.

Explain these observations: why is an e.m.f. induced at all, why are the two pulses in opposite directions, and why is the second pulse larger?

Mark Scheme -- Part (a)

An e.m.f. is induced because the moving magnet causes the magnetic field through the coil to change (field lines are cut by the coil)

The pulses are in opposite directions because the field change reverses: first the field through the coil is increasing (magnet entering), then decreasing (magnet leaving)

The second pulse is larger (and narrower) because the magnet has accelerated under gravity, so it is moving faster and the field changes at a greater rate

1 mark -- emf induced magnetic field through coil changing field lines cut 1 mark -- opposite directions field increasing entering decreasing leaving reversed 1 mark -- second pulse larger magnet accelerated faster greater rate change

⚠ If you missed marks here: Each of the three observations has its own mark, and each needs "change": an e.m.f. is induced because the field through the coil is CHANGING (cutting field lines) — not just "because there's a magnet". Opposite pulses: field increasing as the magnet enters, decreasing as it leaves. Bigger second pulse: gravity has made the magnet FASTER, so the field changes at a greater rate.

(b) [3 marks]

Her phone charger contains a step-down transformer. The primary coil has 4600 turns and is connected to the 230 V a.c. mains. The output is 5.0 V a.c.

(i) Calculate the number of turns on the secondary coil.

(ii) The charger delivers 3.0 A at 5.0 V to the phone. Assuming the transformer is 100% efficient, calculate the current drawn from the mains.

Mark Scheme -- Part (b)

(i) Ns/Np = Vs/Vp so Ns = 4600 × 5.0/230

(i) Ns = 100 turns

(ii) Output power = 5.0 × 3.0 = 15 W; Ip = P/Vp = 15/230 = 0.065 A (65 mA)

1 mark -- turns ratio equation Ns equals 4600 times 5 divided 230 1 mark -- secondary 100 turns 1 mark -- power 15 W primary current 15 divided 230 equals 0.065 A

⚠ If you missed marks here: For (i) keep the ratio the right way up: Ns = 4600 × 5.0/230 = 100 turns — step-DOWN means fewer secondary turns, so 211 600 turns should ring alarm bells. For (ii) use power conservation (Vp Ip = Vs Is): Ip = 15/230 = 0.065 A — the mains current is SMALLER than 3.0 A because the voltage is higher; applying the turns ratio the wrong way gives 138 A, clearly absurd.

Explain how the transformer transfers energy from the primary coil to the secondary coil, and why it cannot work on d.c.

Mark Scheme -- Part (c)

The alternating current in the primary produces a continually changing magnetic field in the soft iron core, which links the secondary coil

This changing field induces an alternating e.m.f. in the secondary; with d.c. the field would be constant, no change means no induced e.m.f., so a transformer cannot work on d.c.

1 mark -- alternating current primary changing magnetic field iron core links secondary 1 mark -- changing field induces emf secondary dc constant field no induction

⚠ If you missed marks here: The key word the examiner scans for is CHANGING: the a.c. makes a continually changing magnetic field in the core, and only a changing field induces an e.m.f. in the secondary. For d.c., say WHY it fails — steady current → constant field → no change → no induced e.m.f.; "d.c. doesn't work in transformers" restates the question.

(d) [2 marks]

In a second experiment, a light aluminium rod rests on two horizontal metal rails inside a magnetic field that points vertically downwards. When the student switches on a current through the rod, the rod rolls along the rails.

(i) Explain why the rod experiences a force, and name the rule used to predict its direction.

(ii) State what happens to the rod if the current direction is reversed.

Mark Scheme -- Part (d)

(i) A current-carrying conductor in a magnetic field experiences a force (the motor effect); the force is perpendicular to both the current and the field, and its direction is given by Fleming's left-hand rule

(ii) Reversing the current reverses the direction of the force, so the rod rolls the opposite way along the rails

1 mark -- current carrying conductor magnetic field force motor effect Fleming left hand rule 1 mark -- reversing current reverses force rod rolls opposite way

⚠ If you missed marks here: Name the rule precisely: Fleming's LEFT-hand rule (the right-hand rule is for generators — mixing them up is the classic slip), and state the cause: a current-carrying conductor in a magnetic field experiences a force perpendicular to both the current and the field. In (ii) one word matters: reversed current → reversed force → the rod rolls the OPPOSITE way.

Question 6 -- Caesium-137 and Food Irradiation

Total: 10 marks · Topic 5: Nuclear Physics

A spice-processing plant in Gujarat uses a sealed caesium-137 source to irradiate packaged spices before export, killing bacteria and insect pests. Caesium-137 (¹³⁷₅₅Cs) emits beta particles and gamma rays, and has a half-life of 30 years.

(a) [3 marks]

Caesium-137 decays by beta (β⁻) emission to an isotope of barium (Ba), which then emits a gamma ray.

(i) Write the balanced nuclear equation for the beta decay, giving the nucleon number and proton number of the barium nuclide and of the beta particle.

(ii) Explain why the gamma emission does not change the nucleon number or the proton number of the barium nucleus.

Mark Scheme -- Part (a)

(i) ¹³⁷₅₅Cs → ¹³⁷₅₆Ba + β : barium is 137/56 (nucleon number unchanged, proton number up by 1)

(i) The beta particle is ⁰₋₁e (an electron emitted when a neutron changes into a proton)

(ii) A gamma ray is an electromagnetic wave / pure energy with no mass and no charge, so emitting it leaves both the nucleon number and the proton number unchanged

1 mark -- barium 137 56 nucleon unchanged proton up one 1 mark -- beta particle 0 minus 1 electron neutron changes proton 1 mark -- gamma electromagnetic wave no mass no charge numbers unchanged

⚠ If you missed marks here: In beta decay the proton number goes UP by one (55 → 56) while the nucleon number stays 137 — writing 136 or dropping the proton number to 54 are the classic errors; the numbers must balance across the arrow with the beta particle written as ⁰₋₁e. For (ii): gamma is an electromagnetic wave with NO mass and NO charge, so nothing in the nucleus count changes.

(b) [3 marks]

A monitoring detector beside a small test source of caesium-137 records a corrected count rate of 4800 counts/s when the source is new. Company records predict the corrected count rate will be 300 counts/s after 120 years.

Use these figures to show that the half-life of caesium-137 is 30 years. Set out your reasoning clearly.

Mark Scheme -- Part (b)

4800 → 2400 → 1200 → 600 → 300: the count rate has halved 4 times

So 120 years corresponds to 4 half-lives: half-life = 120 / 4

Half-life = 30 years, as required

1 mark -- 4800 2400 1200 600 300 halved 4 times 1 mark -- 120 years equals 4 half lives 120 divided 4 1 mark -- half life 30 years

⚠ If you missed marks here: The trap is the number of halvings: 4800 → 2400 → 1200 → 600 → 300 is FOUR half-lives, not five (count the arrows, not the numbers) — miscounting gives 24 years. It's a "show that" question, so write out the halving chain and 120/4 = 30 years explicitly; quoting 30 years without working scores nothing.

Explain two reasons why gamma radiation is suitable for sterilising sealed packets of spices.

Mark Scheme -- Part (c)

Gamma rays are highly penetrating, so they pass right through the sealed packaging and the spices, killing bacteria/pests throughout without opening the packets

The radiation kills microorganisms but no radioactive material touches the food, so the spices do not become radioactive and are safe to eat

1 mark -- gamma highly penetrating passes through packaging kills bacteria throughout 1 mark -- food does not become radioactive safe no radioactive material transferred

⚠ If you missed marks here: "It kills bacteria" restates the question — the physics marks are for WHY gamma suits SEALED packets: it is highly PENETRATING, so it passes right through the packaging and sterilises the contents without opening them; and the food itself does not become radioactive because no radioactive material touches it. Alpha or beta would be stopped by the packaging — that comparison is what makes gamma the right choice.

(d) [2 marks]

A customer worries that irradiated spices are "contaminated with radiation".

Explain the difference between irradiation and contamination, and use it to explain why the customer's worry is unfounded.

Mark Scheme -- Part (d)

Irradiation means the object is exposed to radiation from outside; contamination means radioactive atoms are transferred onto or into the object and keep decaying there

The spices were only irradiated -- the sealed source never touches them, no radioactive atoms are transferred, so once removed from the beam the spices emit no radiation

1 mark -- irradiation exposed radiation outside contamination radioactive atoms transferred keep decaying 1 mark -- spices only irradiated source never touches emit no radiation afterwards

⚠ If you missed marks here: The definitions must be crisp: IRRADIATION = exposed to radiation from outside; CONTAMINATION = radioactive ATOMS transferred onto/into the object, which keep decaying there. The second mark applies it: the sealed source never touches the spices, no radioactive atoms transfer, so once out of the beam the spices emit nothing — "the radiation wears off" is wrong physics and scores zero.

Question 7 -- Satellites, Redshift and Stellar Distances

Total: 10 marks · Topic 6: Space Physics

ISRO operates the INSAT series of communications satellites in geostationary orbit. Meanwhile, astronomers at the Indian Institute of Astrophysics in Bengaluru analyse the light from distant galaxies and nearby stars.

(a) [3 marks]

An INSAT satellite in geostationary orbit has an orbital radius of 4.22 × 10⁷ m (measured from the centre of the Earth) and an orbital period of 24 hours (86 400 s).

(i) Calculate the orbital speed of the satellite using v = 2πr / T.

(ii) Explain why a geostationary orbit is useful for a television broadcast satellite.

Mark Scheme -- Part (a)

(i) v = 2π × 4.22 × 10⁷ / 86 400 = 2.65 × 10⁸ / 86 400

(i) v = 3070 m/s (accept 3.1 × 10³ m/s)

(ii) Its period equals the Earth's rotation period, so it stays above the same point on the equator; receiving dishes can point at a fixed position in the sky and never need to track the satellite

1 mark -- v equals 2 pi times 4.22 times 10 to 7 divided 86400 1 mark -- orbital speed 3070 or 3100 m/s 1 mark -- period equals Earth rotation stays above same point dishes fixed never track

⚠ If you missed marks here: In (i) use the FULL circumference: v = 2πr/T = 2π × 4.22 × 10⁷ / 86 400 ≈ 3070 m/s — forgetting the 2π (488 m/s) or using 24 instead of 86 400 s are the usual slips. In (ii), "it stays still" is imprecise: it ORBITS with a 24-hour period matching Earth's rotation, so it stays above the SAME point on the equator and dishes never need re-aiming.

(b) [3 marks]

In the laboratory, a particular hydrogen spectral line has a wavelength of 656 nm. In the light from a distant galaxy, the same line is observed at 672 nm.

(i) Name this effect and state what it tells us about the motion of the galaxy.

(ii) Explain how observations like this, made for many galaxies at different distances, provide evidence that the Universe is expanding.

Mark Scheme -- Part (b)

(i) The wavelength has increased (shifted towards the red end of the spectrum): this is redshift

(i) It shows the galaxy is moving away (receding) from Earth

(ii) Almost all galaxies show redshift, and the more distant a galaxy is, the greater its redshift / recession speed; this is exactly what is expected if space itself is expanding, carrying galaxies apart

1 mark -- wavelength increased shifted red end redshift 1 mark -- galaxy moving away receding from Earth 1 mark -- more distant greater redshift recession speed universe expanding

⚠ If you missed marks here: The wavelength got LONGER (656 → 672 nm), so this is REDSHIFT and the galaxy is moving AWAY — "blueshift" or "moving towards us" is the reversal to watch for. In (ii), one galaxy isn't evidence: the mark is for the PATTERN — nearly all galaxies are redshifted AND the more distant ones show greater redshift, which is what an expanding Universe predicts.

A galaxy is measured to be receding at 5.5 × 10⁶ m/s.

Use the Hubble equation v = H₀d, with H₀ = 2.2 × 10⁻¹⁸ s⁻¹, to calculate the distance to this galaxy in metres.

Mark Scheme -- Part (c)

d = v / H₀ = 5.5 × 10⁶ / 2.2 × 10⁻¹⁸

d = 2.5 × 10²⁴ m

1 mark -- d equals v divided H0 5.5 times 10 to 6 divided 2.2 times 10 to minus 18 1 mark -- distance 2.5 times 10 to 24 m

⚠ If you missed marks here: Rearrange first: d = v/H₀ (dividing by H₀, not multiplying — multiplying gives a ridiculous 10⁻¹¹ m). The power-of-ten step is where marks go: 10⁶ ÷ 10⁻¹⁸ = 10^6−(−18) = 10²⁴, so d = 2.5 × 10²⁴ m — getting 10¹² means the negative exponent was mishandled.

(d) [2 marks]

Sirius, the brightest star in the night sky, is 8.6 light-years from Earth. (1 light-year = 9.5 × 10¹⁵ m)

(i) Calculate the distance to Sirius in metres.

(ii) Explain why an astronomer observing Sirius tonight is seeing the star as it was in the past.

Mark Scheme -- Part (d)

(i) d = 8.6 × 9.5 × 10¹⁵ = 8.2 × 10¹⁶ m

(ii) The light now arriving left Sirius 8.6 years ago (light takes 8.6 years to cross that distance), so we see the star as it was 8.6 years in the past

1 mark -- distance 8.6 times 9.5 times 10 to 15 equals 8.2 times 10 to 16 m 1 mark -- light left 8.6 years ago see star as it was in past

⚠ If you missed marks here: A light-year is a DISTANCE, not a time — so (i) is just 8.6 × 9.5 × 10¹⁵ = 8.2 × 10¹⁶ m. In (ii), "because it's far away" isn't enough: say that light takes 8.6 YEARS to cross that distance, so the light arriving tonight left Sirius 8.6 years ago — we see the star as it WAS then.