Teacher Eugene
Solutions & breakdown by Teacher Eugene

2025 O-Level Physics Paper 2 — Full Worked Solutions

GCE O-Level Physics (6091/02) · Paper 2 · October/November 2025 · 80 marks

Preview of the annotated PDF answer scheme, with Teacher Eugene's handwritten workings, highlighted keywords and concept notes Click to expand

Want the full blank question paper and answer key with annotations to help you ace your exams?

Annotations by Teacher Eugene

Section A Answer all questions · 70 marks

1 Kinematics & forces — velocity–time graph, parachutist 7 marks

Fig. 1.1 shows the velocity–time graph for a parachutist. The parachute opens at time t = 30 s and the parachutist lands on the ground at t = 60 s.
Two regions, P and Q, are labelled on the graph. Region P is between t = 0 and t = 20 s and region Q is between t = 20 s and t = 30 s.

Fig. 1.1: velocity–time graph of a parachutist. Velocity rises steeply to about 50 m/s over region P, stays at 50 m/s over region Q, drops sharply to about 5 m/s at 30 s when the parachute opens, then stays constant until 60 s
a Determine the distance travelled by the parachutist after the parachute opens. [1]
Show working
Concept: Distance = area under a velocity–time graph
After the parachute opens (t = 30 s to t = 60 s) the graph is a horizontal line at a velocity of 5.0 m/s.
Distance travelled = area under the graph from t = 30 s to t = 60 s = 5.0 × (60 − 30) = 150 m.
150 m
b(i) Describe how the magnitude of the acceleration of the parachutist varies in regions P and Q. [2]
Show working
Concept: Acceleration = gradient of a velocity–time graph
Region P: the parachutist experiences an initial constant acceleration before the acceleration decreases (the gradient gets less steep).
Region Q: the velocity is constant (flat line), so the parachutist experiences zero acceleration.
P: constant acceleration at first, then decreasing. Q: zero acceleration.
b(ii) Using ideas about forces, explain why the acceleration of the parachutist varies in region P. [3]
Show working
Concept: Resultant force: weight vs air resistance
Initially, the only force acting on the parachutist is the weight. Hence the parachutist experiences a constant acceleration equal to the gravitational field strength.
As the velocity increases, the air resistance acting on the parachutist increases as well. This causes the resultant force to decrease, since Fnet = weight − air resistance. Hence the acceleration decreases as well.
When the air resistance increases until it is equal to the weight, the resultant force is zero, causing the acceleration to be zero.
Weight only → constant acceleration; air resistance grows with speed → resultant force and acceleration decrease → air resistance = weight → resultant force zero.
c An identical parachute is used by a heavier person. The terminal velocity of the heavier parachutist after the parachute opens is 8 m/s.
Explain why the terminal velocity of the heavier parachutist is greater than that of the lighter parachutist. [1]
Show working
Concept: Terminal velocity: air resistance = weight
The heavier parachutist has a larger weight, and would take a longer time for the air resistance to be equal to its weight.
Hence the heavier parachutist experiences a greater terminal velocity than the lighter parachutist.
Larger weight → air resistance takes longer to equal the weight → greater terminal velocity.
2 Moments & centre of gravity — laptop lid 8 marks

Fig. 2.1 shows a top-down view of a laptop computer placed in two different positions on a table.

Fig. 2.1: top-down view of a laptop lying flat on a table and the same laptop standing on its side
a Using ideas about the centre of gravity, explain why the laptop is likely to fall over when placed on its side. [2]
Show working
Concept: Centre of gravity, line of action of the weight and stability
When the laptop is placed on its side and is displaced (or tilted) slightly, the line of action of the weight is more likely to lie outside its small / narrow base area.
This produces a net moment about the pivot that causes the laptop to fall over.
Small base → weight’s line of action falls outside the base → net moment about the pivot → laptop falls over.
b(i) Fig. 2.2 shows the lid of the laptop open at 30° to the horizontal. The lid stays open at this angle without any support. The smallest force applied at P that opens the lid further is 1.2 N.
State the principle of moments. [2]
Fig. 2.2: laptop lid open at 30 degrees to the horizontal, hinged at the bottom right, with a 1.2 N force applied at P perpendicular to the 25 cm lid
Show working
Concept: Principle of moments
When an object is in equilibrium, the sum of the clockwise moments about any pivot is equal to the sum of the anticlockwise moments about the same pivot.
In equilibrium: sum of clockwise moments = sum of anticlockwise moments about the same pivot.
b(ii) Calculate the moment of the 1.2 N force about the hinge of the laptop. [1]
Show working
Concept: Moment = force × perpendicular distance from the pivot
The 1.2 N force acts perpendicular to the lid, so the perpendicular distance from the hinge is the lid length, 25 cm = 0.25 m.
Moment = F × d = 1.2 × 0.25 = 0.30 N m.
0.30 N m
b(iii) Explain why a force smaller than 1.2 N can close the lid. [1]
Show working
Concept: Always consider ALL clockwise & anticlockwise moments
The weight of the lid acts downwards, producing an anticlockwise moment about the hinge, which helps to close the lid.
Thus, less moment and force are needed to close the lid than to open it.
The lid’s own weight gives an anticlockwise moment that helps close it, so a smaller force is needed.
b(iv) The lid is open at 30° to the horizontal. A student applies a vertical force F to open the lid, as shown in Fig. 2.3.
Calculate the minimum force F needed to open the lid further at this position. [2]
Fig. 2.3: laptop lid open at 30 degrees with a vertical force F applied at the top edge, 12.5 cm above the table, and a 25 cm lid length
Show working
Concept: Perpendicular distance from the line of action of F to the hinge
For a vertical force, the perpendicular distance from the hinge is the horizontal distance. Using tan 30° = opposite / adjacent: tan 30° = 12.5 / A, so A = 12.5 / tan 30° = 21.650 cm (recall TOA, CAH, SOH).
Minimum force: the moment of F must at least equal the 0.30 N m moment from part (ii). Taking moments about the pivot: F × (21.650 / 100) = 0.30.
F = 1.385 ≈ 1.4 N (2 s.f.).
Fig. 2.3 annotated: the horizontal distance from the hinge to the vertical force is marked as 21.650 cm
F = 1.4 N
3 Pressure — gas, manometer and gas particles 7 marks

A syringe containing gas is connected to a manometer containing mercury, as shown in Fig. 3.1. Fig. 3.1 shows the apparatus when a force F is applied by the thumb to the piston.

Fig. 3.1: a syringe with a piston connected to a U-tube manometer containing mercury, with a ruler in cm beside the left limb
a The level of the mercury on each side of the manometer is shown on the ruler.
State the level of the mercury on each side when the gas inside the syringe is at atmospheric pressure. [1]
Show working
Concept: Manometer: equal pressure → equal levels
When the gas inside the syringe is at atmospheric pressure, both sides of the manometer are at the same pressure, so the mercury is at the same level on both sides.
In Fig. 3.1 the mercury is at about 1 cm on the left and 5 cm on the right (a difference of 4 cm); the same volume of mercury is displaced from one side to the other, so the levels return to the midpoint, 3 cm.
Left-hand side = 3 cm; right-hand side = 3 cm
b(i) Atmospheric pressure is 1.00 × 105 Pa. The density of mercury is 14 000 kg/m3. The gravitational field strength g is 10 N/kg.
Calculate the pressure of the gas in the syringe in Fig. 3.1.
Give your answer to 3 significant figures. [2]
Show working
Concept: Pressure in a liquid column: p = ρgh
The gas pressure is the atmospheric pressure plus the pressure from the column of mercury that is 4 cm higher on one side: pressure of gas = ρgh + patm.
= (14 000)(10)(4/100) + 1.00 × 105 = 105 600 Pa.
106 000 Pa (3 s.f.)
b(ii) The cross-sectional area of the piston is 1.2 × 10−4 m2.
Calculate the force F. [2]
Show working
Concept: Pressure = force ÷ area; balanced forces on the piston
The piston is in equilibrium. The gas pushes outwards on the inside of the piston with a force pgasA. On the outside, the atmosphere pushes inwards with patmA, and the thumb pushes inwards with F.
F + patmA = pgasA, so F = (pgas − patm)A = ρgh × A.
F = (105 600 − 1.00 × 105) × 1.2 × 10−4 = 5600 × 1.2 × 10−4 = 0.672 N ≈ 0.67 N (2 s.f.).
F = 0.67 N
c The force F is increased. The temperature of the gas does not change.
Explain in terms of the motion of the particles in the gas why the pressure of the gas increases. [2]
Show working
Concept: Gas pressure: particle collisions with the container walls
When force F is increased, the volume of the gas in the piston decreases. This increases the number of gas particles per unit volume as the particles come closer together.
Hence the gas particles collide with the walls of the syringe more often. Since the force exerted per unit area of the walls increases, the pressure of the gas increases.
Smaller volume → more particles per unit volume → more frequent collisions with the walls → greater force per unit area → pressure increases.
4 Light — image formation by a converging lens 5 marks

Fig. 4.1 is a scale diagram showing a lens with its principal focuses at F1 and at F2. An object O is shown near the lens.

Fig. 4.1: a converging lens with an upright object O to the left, principal focus F1 on the left and F2 on the right
a On Fig. 4.1:
• draw two rays from the tip of the object to show the formation of the image of the object
• draw the whole of the image and label it I. [2]
Show working
Concept: Ray diagram for a converging lens
Ray 1: from the tip of the object, parallel to the principal axis; after the lens it passes through the principal focus F2.
Ray 2: from the tip of the object, through the optical centre of the lens; it continues straight, undeviated.
The tip of the image is where the two rays meet (just beyond F2, below the principal axis). Draw the whole image as an arrow from the principal axis down to this point, and label it I.
Fig. 4.1 annotated: a ray parallel to the axis refracting through F2 and a ray through the optical centre meet below the axis just beyond F2, where the inverted image I is drawn
Image I is inverted, formed where the two rays cross.
b State whether the image is real or virtual. Explain your answer. [1]
Show working
Concept: Real vs virtual images
The image formed is real. The image is formed on the side of the lens that is opposite to where the object is placed at (the refracted rays actually meet).
Real — formed on the opposite side of the lens to the object.
c(i) State one change that occurs to the image when the object O is moved slightly further from the lens. [1]
Show working
Concept: Moving the object: the second ray passing through the optical centre
The first ray’s direction does not change. Imagine the second ray (through the optical centre) becoming slightly less steep: it now intersects the other ray at a position nearer to F2 and nearer to the principal axis.
So the image is formed closer to the axis and becomes smaller.
The image becomes smaller.
c(ii) State one change that occurs to the image when the object O is moved closer to the lens, so that O is between the lens and F1. [1]
Show working
Concept: Object inside the focal length → virtual image
The first ray’s direction does not change. Imagine the second ray becoming more steep: it can only intersect the other ray on the side of the object when the rays are extended backwards, which means the image is virtual.
Or just recall: if the distance from the object to the optical centre is less than the focal length, f, the image is virtual.
The image becomes virtual.
5 Electricity — electric fields and charging by induction 8 marks

(a) Fig. 5.1 shows an isolated positive charge.

Fig. 5.1: an isolated positive charge, shown as a circle with a plus sign
a(i) On Fig. 5.1, draw at least four field lines to show the pattern of the electric field in the space around the charge.
Include the direction of the field lines. [1]
Show working
Concept: Electric field pattern around a point charge
Draw at least four straight field lines that start at the charge and spread out radially, evenly spaced.
Field lines point away from a positive charge, so put an arrowhead on each line pointing outwards.
Fig. 5.1 annotated: four red field lines with arrows pointing outwards from the positive charge, up, down, left and right
At least four evenly spaced radial lines, arrows pointing away from the positive charge.
a(ii) State what is shown by the direction of the electric field at any point. [1]
Show working
Concept: Definition of the direction of an electric field
The direction of the electric field shows the direction of the electric force acting on a positive test charge placed at that point.
The direction of the electric force on a positive test charge placed at that point.
b(i) A light metal ball is suspended by an insulating thread. The ball is initially uncharged and placed midway between two metal plates, as shown in Fig. 5.2. The plates are connected to a high voltage d.c. power supply and a sensitive ammeter.
On Fig. 5.2, draw the charges present on the metal surface of the ball. [1]
Fig. 5.2: an uncharged metal ball on an insulating thread between a positively charged left plate and a negatively charged right plate, connected to a power supply and ammeter
Show working
Concept: Charging by induction
The ball is uncharged overall, but the electric field between the plates makes the free electrons move.
Negative charges are induced on the left side of the ball (nearer the positive plate) and positive charges are left on the right side (nearer the negative plate).
Fig. 5.2 annotated: negative signs drawn on the left surface of the ball and positive signs on the right surface
Negative charges on the left surface, positive charges on the right surface.
b(ii)1 The ball is moved to the left so that it touches the left-hand plate, as shown in Fig. 5.3.
Describe what happens to the charges on the ball as it touches the plate. [2]
Fig. 5.3: the metal ball touching the positively charged left plate, with the thread tilted to the left
Show working
Concept: Neutralisation of induced charges
As the ball touches the plate, the negative charges that were induced on the left side of the ball move from the ball to the positive metal plate and are neutralised.
This leaves excess positive charges on the ball.
Negative charges move to the positive plate and are neutralised, leaving the ball positively charged.
b(ii)2 The ball is released.
Explain why the ball moves continuously to the right and back to the left. [2]
Show working
Concept: Like charges repel
When the ball touches the positive plate, it becomes positively charged. Since like charges repel, the ball moves to the right.
When the ball touches the negative plate, the plate’s negative charges move to the left side of the ball to neutralise the ball’s positive charges, and the ball becomes negatively charged. Since like charges repel, the ball moves back to the left.
Repelled by the plate it touches, so it swings across to the other plate, and back again.
b(ii)3 Describe the readings shown on the ammeter as the ball moves backwards and forwards. [1]
Show working
Concept: Moving charge = current
The readings on the ammeter fluctuate: it increases as the ball touches either plate, and decreases as the ball loses contact with either plate.
The ammeter reading fluctuates — it rises when the ball touches a plate and falls when it leaves.
6 Electricity — e.m.f., potential divider and the LDR 7 marks

Fig. 6.1 shows a circuit with a power supply of electromotive force (e.m.f.) 14 V, a resistor of resistance 600 Ω, a voltmeter and a light-dependent resistor (LDR).

Fig. 6.1: a 14 V supply in series with a 600 ohm resistor and an LDR, with a voltmeter connected across the LDR
a Define electromotive force. [1]
Show working
Concept: Definition: e.m.f.
Electromotive force is the work done by the source in driving a unit charge around a complete circuit.
E.m.f. = work done by the source in driving a unit charge around a complete circuit.
b(i) Bright light is incident on the LDR. The reading on the voltmeter is 2.8 V.
Show that the resistance of the LDR is 150 Ω. [2]
Show working
Concept: Potential divider
The voltmeter is across the LDR, and the LDR is in series with the 600 Ω resistor, so the 14 V is shared in proportion to resistance: VLDR = (RLDR / Rtotal) × Vtotal.
2.8 = RLDR / (RLDR + 600) × 14, so 2.8(RLDR + 600) = 14RLDR and 1680 = 11.2RLDR.
RLDR = 150 Ω (shown)
b(ii) The LDR is covered so that no light is incident on the LDR.
Explain why the reading on the voltmeter is nearly 14 V. [1]
Show working
Concept: Potential divider — LDR in the dark
When no light is incident on the LDR, the resistance of the LDR increases to its maximum value, such that it is a large proportion of the whole circuit’s effective resistance.
Since the potential difference across the LDR is directly proportional to the ratio of the LDR’s resistance to the whole circuit’s effective resistance, it increases to nearly 14 V.
LDR resistance becomes very large → takes almost all of the 14 V.
c(i) A student attempts to use the circuit to switch on a filament lamp when it becomes dark in the room. The lamp is first tested on its own and found to glow very dimly when connected to a potential difference (p.d.) of 3.0 V and to glow brightly when the p.d. is 14 V. The lamp is then connected in parallel with the LDR, as shown in Fig. 6.2. [2]
Fig. 6.2: the circuit of Fig. 6.1 with a filament lamp connected in parallel with the LDR instead of the voltmeter
Show working
Concept: Resistors in parallel
The LDR is initially in bright light and the lamp does not glow. This is expected by the student. The resistance of the filament lamp in the circuit is 50 Ω and the resistance of the LDR is 150 Ω. Calculate the combined resistance of the filament lamp and the LDR.
1 / R = 1/150 + 1/50, so R = (1/150 + 1/50)−1 = 37.5 Ω.
Resistance = 37.5 Ω
c(ii) The LDR is then covered. The student expects the p.d. across the lamp to be nearly 14 V. The student expects the lamp to glow brightly but the lamp does not glow.
Explain why. You may assume that the resistance of the lamp does not change. [1]
Show working
Concept: Parallel branch with a very large resistance
When no light is incident on the LDR, its resistance increases to its maximum value. As such, the current in the circuit flows through the path that carries the lamp only.
Hence the potential difference across the lamp = 50 / 650 × 14 = 1.08 V, which is lower than 3.0 V (the p.d. at which the lamp only glows very dimly).
The p.d. across the lamp is only 1.08 V, lower than 3.0 V, so the lamp does not glow.
7 Electromagnetic induction — the a.c. generator 8 marks

Fig. 7.1 shows a simple alternating current (a.c.) generator.

Fig. 7.1: a simple a.c. generator with a rotating coil between the N and S poles of a magnet, and components X and Y connected to the output voltage terminals
a State the names of the components labelled X and Y. [2]
Show working
Concept: Parts of an a.c. generator
X is attached to the rotating coil: a slip ring.
Y presses against the slip ring and carries the current out: a carbon brush.
X: slip ring; Y: carbon brush
b Explain why a voltage is induced in the coil when it is rotated. [1]
Show working
Concept: Faraday’s law of electromagnetic induction
When the coil rotates, the coil cuts the magnetic field of the magnet, causing a change in magnetic flux.
By Faraday’s Law, an electromotive force is thus induced in the coil.
The coil cuts magnetic field lines → changing magnetic flux → e.m.f. induced (Faraday’s Law).
c(i) Fig. 7.2 shows the variation of the output voltage with time.
Fig. 7.1 shows the coil in a horizontal position.
On Fig. 7.2, mark two points on the time axis at which the coil is horizontal. Label your points X. [1]
Fig. 7.2: a sinusoidal output voltage graph with time in ms, peaking at 1 ms and 5 ms, crossing zero at 2 ms and 4 ms
Show working
Concept: Coil position vs induced voltage
In the horizontal position shown in Fig. 7.1 the coil is moving across the magnetic field lines at the greatest rate, so the induced voltage is at its maximum.
Mark the two points X on the time axis directly beneath two peaks of the graph, for example t = 1 ms and t = 5 ms (the orange marks show one full period of 4 ms).
Fig. 7.2 annotated: crosses on the time axis directly below the peaks at 1 ms and 5 ms
Two Xs on the time axis directly below two peaks (e.g. 1 ms and 5 ms).
c(ii) Calculate the frequency of the alternating voltage output. [2]
Show working
Concept: Frequency = 1 ÷ period
From the graph, one complete cycle takes 4 ms: period = 4 × 10−3 s.
f = 1 / T = 1 / (4 × 10−3) = 250 Hz.
250 Hz
c(iii) Changes are made to the a.c. generator and the output voltage is shown by curve B on Fig. 7.3. The original output voltage is shown by curve A.
Suggest two changes to the a.c. generator that together change the output voltage from A to B. [2]
Fig. 7.3: curve A (dashed) with large amplitude and a long period, and curve B (solid) with smaller amplitude and a shorter period
Show working
Concept: Reading a voltage–time graph: amplitude and period
Observe: curve B has a smaller amplitude and a shorter period than curve A.
1. The coil of the a.c. generator was rotated at a faster speed (shorter period = faster speed). Note: this by itself actually increases the peak output voltage (the amplitude), but the second change can reduce it significantly, to the extent that it becomes lower than curve A, thereby matching curve B.
2. The number of turns of the coil decreases. OR the magnets were replaced with weaker ones. OR the magnets were placed further apart from each other.
Faster rotation + a smaller induced voltage (fewer turns / weaker magnets / magnets further apart).
8 Waves — infrared, microwaves, optical fibres and fuses 9 marks

Infrared radiation and microwaves are types of wave which both travel at a speed of 3.0 × 108 m/s in a vacuum.

a State one other property that is the same for both infrared radiation and microwaves. [1]
Show working
Concept: Properties of electromagnetic waves
Both infrared radiation and microwaves are electromagnetic waves, so they both do not require a medium to travel.
They both do not require a medium to travel.
b(i) Fig. 8.1 shows two points, A and B, on the surface of the Earth. Two different methods are used to send a message from A to B: an infrared signal is sent through an optical fibre across the surface of the Earth, and a microwave signal is sent via a satellite in orbit. Fig. 8.2 shows the path of the infrared signal in the optical fibre.
Explain why the infrared stays inside the optical fibre as it travels from A to B. [2]
Fig. 8.1 (not to scale): points A and B on the surface of the Earth joined by an optical fibre along the surface, and a dashed path from A up to a satellite and down to B
Fig. 8.2 (not to scale): the zigzag path of an infrared signal reflecting off the inside walls of an optical fibre from A to B
Show working
Concept: Total internal reflection
The infrared is travelling from an optically denser medium to an optically less dense medium, and it is also travelling at an angle of incidence greater than the critical angle of the optical fibre.
Hence the infrared undergoes total internal reflection that makes it stay inside the optical fibre.
Denser → less dense medium, angle of incidence > critical angle → total internal reflection.
b(ii) The distance travelled by the infrared signal is 3600 km and the refractive index of the material in the optical fibre is 1.5.
Calculate the time taken for the infrared signal to travel from A to B. [2]
Show working
Concept: Refractive index = c ÷ v; time = distance ÷ speed
Refractive index n = c / v, so 1.5 = (3.0 × 108) / v and v = 2.0 × 108 m/s.
Time taken = distance / speed = (3600 × 103) / (2.0 × 108) = 0.018 s.
0.018 s
b(iii) The time taken for the microwave signal to travel from A to B is much larger than the time taken for the infrared signal to travel from A to B.
Suggest one reason for this. [1]
Show working
Concept: Time = distance ÷ speed
The microwave signal takes a longer time to travel from A to B because it travels a much longer path from A to B in space (up to the satellite and back down), while the infrared signal travels a much shorter route along the Earth’s surface.
The microwave signal travels a much longer path (via the satellite in space).
c One type of oven contains a microwave source and a grill that emits infrared. Both the microwave source and the infrared grill can be used at the same time. The source of microwaves in the oven has a power input of 800 W. The grill is a red-hot metal plate at the top of the oven with a power input of 1100 W. The input voltage to the oven is 230 V.
Suggest, with reasons and a calculation, whether an 8.0 A fuse is suitable to connect the oven to the mains supply. [3]
Show working
Concept: Electrical power: P = IV
Both can be used at the same time, so the fuse must carry the total current.
Microwave source: I = P / V = 800 / 230 = 3.478 A. Grill: I = 1100 / 230 = 4.782 A.
Total current = 3.478 + 4.782 = 8.26 A ≈ 8.3 A.
Since the total current used by the oven and grill is 8.3 A, which exceeds the 8.0 A rating on the fuse, an 8.0 A fuse would not be suitable (it would melt whenever both are used together).
Not suitable: total current 8.3 A is greater than the 8.0 A rating.
9 Nuclear physics — decay of potassium-40 and half-life 11 marks

Potassium-40 is an isotope that decays into either calcium-40 or argon-40.
The percentage of potassium-40 and argon-40 atoms in a sample of rock can be used to determine the age of the rock.
(a) Table 9.1 shows the number of protons and neutrons in a nucleus of the three isotopes involved and the number of electrons in a neutral atom of each.

Table 9.1
number of protonsnumber of neutronsnumber of electrons
potassium-40192119
calcium-40202020
argon-40182218
a(i) Explain why the number of electrons is equal to the number of protons in a neutral atom. [2]
Show working
Concept: Charge balance in a neutral atom
An electron is negatively charged and a proton is positively charged to the same extent.
In a neutral atom, the overall charge of the atom is zero.
Equal and opposite charges on electrons and protons → overall charge zero.
a(ii) When a nucleus of potassium-40 decays to calcium-40, a beta particle (β) is emitted.
Using Table 9.1, complete the nuclide equation for this decay. [2]
Show working
Concept: Nuclide equations: conserve nucleon number and proton number
From Table 9.1, calcium-40 has 20 protons (and a nucleon number of 40). Proton number: 19 = 20 + (−1), so the beta particle has a proton number of −1.
Nucleon number: 40 = 40 + 0, so the beta particle has a nucleon number of 0.
4019K → 4020Ca +   0−1β
a(iii) The decay of potassium-40 to argon-40 occurs when an electron from the electron shells in a potassium-40 atom is absorbed by the nucleus.
Using data from Table 9.1, deduce what happens inside the nucleus when the electron is absorbed. [2]
Show working
Concept: Good reasoning skills required: deduce from the data
We deduce the answer from these pieces of info: (1) the number of protons (+ve charge) from potassium-40 to argon-40 decreases by 1 (19 → 18); (2) the number of neutrons (no charge) increases by 1 (21 → 22); (3) the number of electrons (−ve charge) decreases by 1 (19 → 18).
Hence, when the electron is absorbed into the nucleus of potassium-40, it causes a proton in the atom to turn into a neutron. The total mass number of the atom remains the same (40).
A proton in the nucleus turns into a neutron (e− + p+ → n); the mass number stays at 40.
b(i) Calcium-40 and argon-40 are both stable. Argon-40 formed in liquid rock quickly escapes into the air. After the liquid rock solidifies, argon-40 is trapped inside. A liquid sample of the rock solidifies at time t = 0. At t = 0, there is no argon-40 in the rock. Table 9.2 shows how the number of atoms of potassium-40 and argon-40 in a sample of rock vary with t. Also shown is the number of calcium-40 atoms formed after t = 0. There are initially 1000 potassium-40 atoms in the sample.
Using data from Table 9.2, determine the number of potassium-40 atoms present at t = 400 million years. [1]
Table 9.2
t / million yearsnumber of potassium-40 atomsnumber of argon-40 atomsnumber of calcium-40 atoms formed after t = 0
0100000
2008951194
40021178
60071730253
80064237321
100057445381
120051451435
Show working
Concept: Conservation of atoms: every K-40 atom decays to Ar-40 or Ca-40
Each potassium-40 atom that decays becomes either an argon-40 atom or a calcium-40 atom, so atoms remaining = initial atoms − argon-40 atoms − calcium-40 atoms.
At t = 400 million years: 1000 − 21 − 178 = 801.
801
b(ii)1 The number of argon-40 atoms in the sample at any time is 10.5% of the number of potassium-40 atoms that have decayed. The half-life of potassium-40 is 1300 million years.
Calculate the number of potassium-40 atoms in the sample at t = 3900 million years. [1]
Show working
Concept: Half-life
Number of half-lives = 3900 / 1300 = 3.
1000 → (1 half-life) 500 → (2) 250 → (3) 125. OR 1000 × (½)3 = 125.
125
b(ii)2 Calculate the number of argon-40 atoms in the sample at t = 3900 million years. [1]
Show working
Concept: Percentage of the atoms that decayed
Number of potassium-40 atoms decayed = 1000 − 125 = 875.
Number of argon-40 atoms = (10.5 / 100) × 875 = 91.875 ≈ 92. The number of atoms must be an integer (whole number)!
92
b(iii)1 The age of a rock is the time since it solidified. A sample of rock from a meteorite is tested. The ratio of the number of argon-40 atoms to the remaining number of potassium-40 atoms in the meteorite is 0.078.
Using data from Table 9.2, determine the age of the meteorite. [1]
Show working
Concept: Using a ratio from the table
From Table 9.2, find the row where (argon-40 atoms) / (potassium-40 atoms) = 0.078.
At 1000 million years, the ratio = 45 / 574 = 0.0784 ≈ 0.078.
1000 million years
b(iii)2 Suggest why this method of dating is not used to find the age of rocks that solidified only a few years ago. [1]
Show working
Concept: Half-life and measurable decay
The half-life of potassium-40 is very long. For a few years, there will not be much decay of the potassium-40 into argon-40 and calcium-40.
Hence it will not be possible to have an accurate measurement of the age of the rock.
Very long half-life → too little decay in a few years to measure the age accurately.

Section B Answer one question · 10 marks · both options published

10 Energy resources, thermal conduction, transformers & efficiency 10 marks

Fig. 10.1 represents a power station used to generate electricity.

Fig. 10.1 (not to scale): a geothermal power station with water pumped down through cracks in hot dry rock and steam rising to a turbine, condenser and generator connected to a pylon
a(i) State the name of the renewable energy resource used in this type of power station. [1]
Show working
Concept: Renewable energy resources
Water is pumped down to hot, dry rock deep underground and returns as steam, so the energy comes from the heat inside the Earth.
Geothermal energy
a(ii) State the name of one other renewable energy resource. [1]
Show working
Concept: Renewable energy resources
Any one other renewable resource is accepted.
Wind / solar / biomass (any one)
b Cold water pumped into the ground is warmed as it passes through cracks in the rock. Thermal energy from deeper in the Earth is conducted through the rock.
Describe, using ideas about molecules, how thermal conduction occurs in the rock. [2]
Show working
Concept: Thermal conduction: vibrating particles collide and pass on energy
The rock particles that are hotter vibrate more vigorously and collide with their neighbouring, less energetic particles.
Thermal energy is transferred to those neighbours, causing them to gain thermal energy and vibrate faster as well. At the cracks, the same process passes the energy on to the water particles.
Hotter particles vibrate and collide with less energetic neighbours, passing on thermal energy.
c(i) The generator produces a.c. (alternating current) at 25 kV and this voltage is increased to 250 kV.
State the name of the device used to increase the voltage at the power station. [1]
Show working
Concept: Transformers
A device that increases an a.c. voltage is a step-up transformer.
(Step-up) transformer
c(ii)1 High-voltage a.c. is used for most transmission of electricity across long distances.
Explain why a high voltage is used. [1]
Show working
Concept: Recall: power loss = I2R
A high voltage will cause the current to be lower for the same power transmitted (P = IV).
Hence the power loss due to the resistance of the transmission wires will be reduced.
High voltage → lower current for the same power → less power lost as heat (I2R) in the wires.
c(ii)2 Explain why a.c. is used instead of d.c. [1]
Show working
Concept: Transformers need a changing magnetic field
An a.c. is used to produce an alternating magnetic field at the primary coil of the transformer, so that an electromotive force can be induced at the secondary coil of the transformer.
Only a.c. gives a changing magnetic field in the primary coil, which induces an e.m.f. in the secondary coil.
d A gas-fired power station has an efficiency of 55%. The useful output power is 220 MW.
Calculate the energy wasted in each minute in this power station. [3]
Show working
Concept: Efficiency = useful ÷ total × 100%
Efficiency = (useful power / total power) × 100%: 55% = (220 × 106) / total power × 100%, so total power = 4 × 108 W.
Power wasted = 4 × 108 − 220 × 106 = 1.80 × 108 W.
Energy wasted per minute = power × time = 1.80 × 108 × 60 = 1.08 × 1010 J.
1.08 × 1010 J
11 Thermal physics — states of matter, latent heat & specific latent heat 10 marks

A very cold solid sample of substance X is placed in a container and heated.
Fig. 11.1 shows how the temperature of the sample changes with time.

Fig. 11.1: temperature against time for substance X. A flat section at −110 °C, a rising section from −110 to 78 °C, a flat section at 78 °C, then a rise again
a Compare the arrangement and the motion of the particles in a solid with their arrangement and motion in a liquid. [3]
Show working
Concept: Particle model of solids and liquids
Solid: the particles are very closely packed in an orderly manner (a regular arrangement). The particles are also vibrating in their fixed positions.
Liquid: the particles are slightly further apart but still closely packed in a disorderly manner. The particles in a liquid are able to slide past one another.
Solid: close, orderly, vibrating about fixed positions. Liquid: close, disorderly, sliding past one another.
b Explain, in terms of the particles, why latent heat is needed to change a liquid into a gas. [1]
Show working
Concept: Latent heat and intermolecular forces
Latent heat is taken in by the liquid particles to overcome the intermolecular forces of attraction between the liquid particles and to expand against atmospheric pressure, to turn into the gaseous state.
Energy is needed to overcome intermolecular forces and to expand against atmospheric pressure.
c(i) On Fig. 11.1, mark:
• one point where boiling occurs and label it B
• one point where only evaporation occurs and label it E. [2]
Show working
Concept: Reading a heating curve
Boiling happens at the boiling point, the flat section at 78 °C: mark B anywhere on that plateau.
The sloping section between the two plateaux is the liquid warming up: it is not yet boiling, so there is only evaporation at the surface. Mark E on that slope.
Fig. 11.1 annotated: B marked on the 78 degree plateau and E marked on the rising section
B on the 78 °C plateau; E on the sloping section between −110 °C and 78 °C.
c(ii) State one way in which evaporation differs from boiling. [1]
Show working
Concept: Evaporation vs boiling
Evaporation occurs only at the surface of the liquid, while boiling occurs throughout the whole liquid.
Evaporation occurs only at the surface; boiling occurs throughout the liquid.
d A liquid sample of X is initially at 48 °C and absorbs a total of 87 000 J of energy by heating. This energy is just sufficient to raise the temperature of the liquid to its boiling point and change the liquid completely into gas at the boiling point. The volume of the liquid in the sample is 120 cm3. The density of the liquid in the sample is 0.79 g/cm3. The specific heat capacity of the liquid is 2.5 J/(g °C).
Using this data and information from Fig. 11.1, calculate the specific latent heat of vaporisation of the liquid. [3]
Show working
Concept: Energy absorbed = heating the liquid + changing its state
Mass of liquid = ρ × V = 0.79 × 120 = 94.8 g.
The boiling point from Fig. 11.1 is 78 °C. Heat absorbed by the liquid = heat used to raise the temperature + heat used to change state to gas: Q = mcΔθ + mℓv.
87 000 = [94.8 × 2.5 × (78 − 48)] + [94.8 × ℓv]
ℓv = 842.7 ≈ 843 J/g.
843 J/g

Meet Your Tutor

The Tutor Behind These Solutions

Teacher Eugene
Teacher Eugene
Lower Sec Science · Combined Science · Physics · 2 years of teaching

I enjoy providing relevant learning materials that dissect complex concepts, and I customise my teaching style and pace to each tutee — for a visual learner, that might mean drawing models and mind maps to explain the concepts.

When I teach Math and Science, I help my tutees understand the concepts behind the formulae, so that even when questions are tweaked, they can still solve them.

More info

Superpower: I can prove formulae to you, and I bring fun stuff to class to demonstrate Science concepts (balloons, anyone?)

Fun fact: My upper sec Physics teacher (Mr Emerson Foo) created the EM spectrum song that got millions of views on YouTube — and I still know it by heart!

Student Testimonials
Testimonial for Teacher Eugene Testimonial for Teacher Eugene
Eugene's Class Schedule
MonS3 Combined Science Chemistry & Physics · 5:00–7:00 PMJurong East
MonS4 Combined Science Chemistry & Physics · 7:30–9:30 PMJurong East
TueS1 Science · 5:00–7:00 PMJurong East
ThuS3 Pure Physics · 5:00–7:00 PMJurong East
ThuS4 Pure Physics · 7:30–9:30 PMJurong East
SatS1 Science · 1:00–3:00 PMTan Kah Kee
SatS3 Pure Physics · 3:00–5:00 PMTan Kah Kee
SatS4 Pure Physics · 5:30–7:30 PMTan Kah Kee
SunS1 Science · 2:00–4:00 PMNovena
SunS3 Pure Physics · 4:00–6:00 PMNovena

Other Physics Tutors

Teacher Jarryl
Jarryl
Chemistry · Physics
Teacher Pamela
Pamela
Physics · Combined Science · H2 Physics
Teacher Zoey
Zoey
Lower Sec Math · Physics · IP Physics · H2 Physics

All Sec 3 & Sec 4 Pure Physics Class Slots

Subject
Location
Day
Start End
Tutor
S3 Pure PhysicsKovanMon4:30 PM6:30 PM
S3 Pure PhysicsKovanTue7:30 PM9:30 PMJay
S3 Pure PhysicsBishanWed4:30 PM6:30 PM
S3 Pure PhysicsMarine ParadeWed7:30 PM9:30 PMJay
S3 Pure PhysicsTampinesThu4:30 PM6:30 PM
S3 Pure PhysicsJurong EastThu5:00 PM7:00 PM
S3 Pure PhysicsTan Kah KeeSat3:00 PM5:00 PM
S3 Pure PhysicsNovenaSun4:00 PM6:00 PM
S4 Pure PhysicsKovanMon6:30 PM8:30 PM
S4 Pure PhysicsBishanTue5:00 PM7:00 PM
S4 Pure PhysicsTampinesTue6:30 PM8:30 PM
S4 Pure PhysicsNovenaTue7:15 PM9:15 PM
S4 Pure PhysicsBishanWed6:30 PM8:30 PM
S4 Pure PhysicsKovanThu4:30 PM6:30 PMJay
S4 Pure PhysicsTampinesThu6:30 PM8:30 PM
S4 Pure PhysicsJurong EastThu7:30 PM9:30 PM
S4 Pure PhysicsKovanFri4:30 PM6:30 PM
S4 Pure PhysicsKovanSat9:30 AM11:30 AM
S4 Pure PhysicsPunggolSat11:00 AM1:00 PM
S4 Pure PhysicsMarine ParadeSat3:00 PM5:00 PMJay
S4 Pure PhysicsTan Kah KeeSat5:30 PM7:30 PM
S4 Pure PhysicsBishanSun4:00 PM6:00 PMJay
Secondary Sch Tuition → JC Tuition →
Testimonial

FAQ

Common questions about this paper

What topics are tested in the 2025 O-Level Physics Paper 2 (6091/02)?

Paper 2 tests velocity–time graphs and the forces on a falling parachutist, moments and the centre of gravity, pressure and manometers, gas pressure in terms of particles, image formation by a converging lens, electric fields and charging by induction, potential dividers and the LDR, electromagnetic induction and the a.c. generator, electromagnetic waves and total internal reflection in optical fibres, fuse ratings, radioactive decay and half-life, geothermal energy, transformers and efficiency, and states of matter and latent heat.

How do you find the distance travelled from a velocity–time graph?

The distance travelled is the area under the velocity–time graph. For a section where the velocity is constant, the area is a rectangle: distance = velocity × time. In Question 1(a), after the parachute opens the parachutist moves at a constant 5.0 m/s for 30 s, so the distance is 5.0 × 30 = 150 m.

Why does a falling parachutist reach a terminal velocity?

At first the only force is the weight, so the acceleration is constant. As the speed rises, air resistance rises too, so the resultant force (weight − air resistance) and the acceleration decrease. When the air resistance equals the weight, the resultant force is zero, the acceleration is zero and the parachutist falls at a constant terminal velocity.

Why does infrared stay inside an optical fibre?

The infrared travels from an optically denser medium (the fibre) towards an optically less dense medium (the surroundings) at an angle of incidence greater than the critical angle. It therefore undergoes total internal reflection at every boundary and stays inside the fibre all the way from A to B.

How do you calculate the specific latent heat of vaporisation when the liquid is first heated and then boiled?

Split the energy into two parts: the energy to raise the liquid to its boiling point (mcΔθ) and the energy to change it into gas at the boiling point (mℓv). Set the total energy absorbed equal to the sum, then solve for ℓv. In Question 11(d), 87 000 = 94.8 × 2.5 × (78 − 48) + 94.8 × ℓv, which gives about 843 J/g.