Cambridge IGCSE · Thinka-original Practice Paper

2025 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) Practice Paper with Answers

Thinka Jun 2025 (V2) Cambridge International A Level-Style Mock — Sciences - Co-ordinated (Double) (0654)

120 marks120 mins2025

An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V2) Cambridge International A Level Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.

Biology Section

Answer all questions. Questions 1 to 4 cover various biological systems, cell processes, and ecological dynamics.

4 Question · 40 marks

Question 1 · structured

10 marks

An image of a plant palisade mesophyll cell and a human liver cell are studied by a student.

(a) Name two cellular structures that are present in a typical palisade mesophyll cell but absent from a typical liver cell. [2]

(b) State the function of the following organelles:
(i) ribosomes [1]
(ii) mitochondria [1]

(c) A student uses a microscope to observe a cell. The length of the cell in the micrograph (image size) is measured as 4.8 cm. The actual length of the cell is 0.08 mm. Calculate the magnification of the microscope. Show your working. [2]

(d) Red blood cells are highly specialized to transport oxygen throughout the human body.
(i) Describe two structural features of a red blood cell. [2]
(ii) Explain how these structural features adapt the red blood cell for its function of oxygen transport. [2]

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Worked solution

(a) Palisade mesophyll cells (plant cells) contain a cell wall, chloroplasts, and a large permanent vacuole, all of which are absent from liver cells (animal cells).

(b) (i) Ribosomes are the sites of protein synthesis.
(ii) Mitochondria are the sites of aerobic respiration, releasing energy for cellular processes.

(c) First, convert units so they are the same: 4.8 cm = 48 mm.
Use the formula: Magnification = Image size / Actual size
Magnification = 48 mm / 0.08 mm = 600.
Therefore, the magnification is x600.

(d) (i) Structural features of a red blood cell include: biconcave disc shape, no nucleus, and the presence of haemoglobin.
(ii) The biconcave shape increases the surface area to volume ratio of the cell, which allows for faster diffusion of oxygen into and out of the cell. Alternatively, having no nucleus provides more space inside the cell to pack in more haemoglobin, allowing more oxygen to be transported.

Marking scheme

(a) 1 mark for each correct structure (max 2): cell wall, chloroplast, permanent vacuole. Reject: nucleus, cell membrane.

(b) (i) 1 mark for protein synthesis / making proteins.
(ii) 1 mark for aerobic respiration / release of energy. Reject: "producing energy".

(c) 1 mark for correct unit conversion (e.g., 48 mm or 48000 micrometres) OR correct substitution of values.
1 mark for correct calculation: 600 (or x600).

(d) (i) 1 mark for each described feature (max 2): biconcave disc shape, absence of nucleus, contains haemoglobin, flexible.
(ii) 1 mark for linking a feature to its function, and 1 mark for the explanation: e.g., biconcave shape increases surface area to volume ratio [1] which increases the rate of oxygen diffusion [1]; OR lack of nucleus leaves more room [1] for more haemoglobin to transport more oxygen [1].

Question 2 · structured

10 marks

Gas exchange in humans occurs across the alveolar membrane in the lungs.

(a) Identify three structural features of alveoli that make them efficient for gas exchange. [3]

(b) Describe the pathway taken by an oxygen molecule as it travels from the trachea to the alveoli. [2]

(c) The composition of inspired and expired air is different.
(i) Describe how the percentage of oxygen and carbon dioxide changes between inspired air and expired air. [2]
(ii) Describe how limewater can be used to compare the carbon dioxide concentration in inspired and expired air, including the expected results. [3]

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Worked solution

(a) Alveoli have several adaptations: 1. They are very thin (only one cell thick), which provides a short diffusion distance. 2. They have a massive total surface area, maximizing the rate of diffusion. 3. They have an excellent blood supply from a network of capillaries, which maintains a steep concentration gradient. 4. They are ventilated to maintain a concentration gradient.

(b) The oxygen molecule travels down the trachea, then enters one of the bronchi, then travels through the smaller bronchioles, and finally reaches the alveolus.

(c) (i) Inspired air contains approximately 21% oxygen and 0.04% carbon dioxide. Expired air contains approximately 16% oxygen (a decrease) and 4% carbon dioxide (an increase).
(ii) Set up two test tubes with equal volumes of clear limewater. Bubble inspired air through one tube and expired air through the other for the same number of breaths or rate. The limewater in the expired air tube will turn cloudy (milky) much faster than the inspired air tube, showing expired air has a higher concentration of CO2.

Marking scheme

(a) 1 mark for each correct feature (max 3): thin walls / one cell thick, large surface area, good blood supply / capillary network, moist surface, well ventilated.

(b) 1 mark for mentioning bronchus and bronchiole. 1 mark for correct order (trachea -> bronchus -> bronchiole -> alveolus).

(c) (i) 1 mark for describing the decrease in oxygen percentage. 1 mark for describing the increase in carbon dioxide percentage.
(ii) 1 mark for bubbling/passing inspired and expired air through separate tubes of limewater. 1 mark for stating that expired air turns limewater cloudy/milky. 1 mark for comparing that inspired air remains clear or takes much longer to change, showing expired air has more carbon dioxide.

Question 3 · structured

10 marks

Humans possess a complex immune system to defend against pathogens.

(a) State what is meant by the term pathogen. [1]

(b) The body uses different types of barriers and cells to defend against pathogens.
(i) State one physical barrier and one chemical barrier that prevent pathogens from entering body tissues. [2]

(ii) Describe the role of phagocytes in protecting the body against pathogens. [2]

(c) Vaccines are used to induce active immunity.
(i) Explain how a vaccine results in active immunity against a specific disease. [3]

(ii) Explain why active immunity provides long-term protection against a disease, whereas passive immunity does not. [2]

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Worked solution

(a) A pathogen is defined as a disease-causing organism.

(b) (i) A physical barrier is a structural block, such as the skin or hairs in the nose. A chemical barrier is a chemical substance that destroys or traps pathogens, such as stomach acid (hydrochloric acid), tears (containing lysozyme), or mucus.
(ii) Phagocytes detect pathogens, engulf (ingest) them in a process called phagocytosis, and then use digestive enzymes inside the cell to destroy them.

(c) (i) A vaccine contains a dead, weakened, or harmless form of a pathogen (or its antigens). When injected, the lymphocytes recognize these foreign antigens and produce specific antibodies to target them. Importantly, they also produce memory cells.
(ii) Active immunity results in the production of memory cells. If the real pathogen enters the body in the future, these memory cells quickly recognize it and produce large amounts of antibodies very rapidly. Passive immunity is the temporary acquisition of antibodies (e.g., from breast milk or injection) which eventually break down; since no memory cells are produced, it offers no long-term protection.

Marking scheme

(a) 1 mark for: disease-causing organism.

(b) (i) 1 mark for physical barrier (skin, hairs in nose). 1 mark for chemical barrier (stomach acid, mucus, tears).
(ii) 1 mark for: engulfing / ingesting / phagocytosis of pathogens. 1 mark for: digesting / breaking down / killing the pathogen using enzymes.

(c) (i) 1 mark for stating that vaccine contains dead/weakened/harmless pathogen or antigens. 1 mark for stating that lymphocytes produce specific antibodies in response. 1 mark for stating that memory cells are produced.
(ii) 1 mark for stating active immunity produces memory cells that remain in the body (ready for rapid antibody production). 1 mark for stating passive immunity only provides temporary antibodies / antibodies are broken down / no memory cells are made.

Question 4 · structured

10 marks

Ecosystems consist of communities of organisms interacting with each other and their environment.

(a) Define the following terms:
(i) population [1]
(ii) community [1]

(b) Consider the following forest food web:
• Oak tree → Caterpillar → Blue tit → Hawk
• Oak tree → Aphid → Ladybird → Blue tit
(i) Identify one producer and one secondary consumer from this food web. [2]
(ii) Explain why food chains rarely contain more than five trophic levels. [3]

(c) The carbon cycle involves the movement of carbon through the atmosphere, biosphere, and geosphere.
State three processes in the carbon cycle that release carbon dioxide into the atmosphere. [3]

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Worked solution

(a) (i) A population is a group of organisms of one species, living in the same area, at the same time.
(ii) A community is all of the populations of different species in an ecosystem.

(b) (i) The producer is the Oak tree (which photosynthesizes to make its own organic nutrients). The secondary consumers are the Blue tit (when eating the caterpillar) and the Ladybird (when eating the aphid).
(ii) Energy is lost at each trophic level. Only about 10% of the energy is successfully transferred to the next level. Energy is lost through respiration, metabolic heat, movement, excretion, or because parts of the organism are not eaten. Consequently, after 4 or 5 trophic levels, the amount of energy remaining is too small to sustain another trophic level.

(c) Carbon dioxide is returned to the atmosphere through: 1. Respiration (by plants, animals, and decomposers). 2. Combustion (burning of fossil fuels, wood, or biomass). 3. Decomposition (breakdown of organic material by decomposers, which release carbon dioxide via respiration).

Marking scheme

(a) (i) 1 mark for: group of organisms of one species, living in same area at same time.
(ii) 1 mark for: all populations of different species in an ecosystem (at same time).

(b) (i) 1 mark for producer: Oak tree. 1 mark for secondary consumer: Blue tit OR Ladybird.
(ii) 1 mark for stating that energy is lost at each level / only about 10% of energy is transferred. 1 mark for listing at least two ways energy is lost (respiration, heat, movement, excretion, uneaten parts). 1 mark for explaining that after 4 or 5 levels, there is insufficient energy left to support another trophic level.

(c) 1 mark for each correct process (max 3): respiration (by plants/animals/decomposers), combustion / burning (of wood/fossil fuels), decomposition / decay.

Chemistry Section

Answer all questions. Questions 5 to 8 cover bonding, reaction rates, metals, and organic polymerisation.

4 Question · 40 marks

Question 1 · structured

10 marks

This question is about chemical bonding and structures. (a) Describe the bonding in ammonia, \(NH_3\). Explain in terms of electron sharing or transfer and electrostatic forces. [3 marks] (b) Describe the arrangement of outer-shell electrons in a molecule of ammonia. [2 marks] (c) Explain why magnesium fluoride, \(MgF_2\), has a very high melting point, whereas ammonia is a gas at room temperature. [3 marks] (d) Describe the arrangement and movement of particles in gaseous ammonia. [2 marks]

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Worked solution

(a) Ammonia contains covalent bonds. There is sharing of electrons between the nitrogen and hydrogen atoms. The bond is the electrostatic attraction between the shared pairs of electrons and the positive nuclei of both atoms. (b) In an ammonia molecule, the nitrogen atom shares three of its outer-shell electrons with three hydrogen atoms (forming three single covalent bonds). The nitrogen atom also has one lone pair of non-bonding outer-shell electrons. (c) Magnesium fluoride is an ionic compound with a giant ionic lattice. There are strong electrostatic forces of attraction between oppositely charged magnesium ions, \(Mg^{2+}\), and fluoride ions, \(F^-\), which require a large amount of energy to break. Ammonia consists of simple covalent molecules. The intermolecular forces between these molecules are very weak and require very little thermal energy to overcome. (d) In the gaseous state, ammonia particles are randomly arranged with large spaces between them, and they move rapidly and randomly in all directions.

Marking scheme

(a) [3 marks]: 1 mark for covalent / shared pairs of electrons; 1 mark for electrostatic attraction; 1 mark for attraction being between shared electrons and positive nuclei. | (b) [2 marks]: 1 mark for three single covalent bonds / three shared pairs; 1 mark for one lone pair of electrons on the nitrogen. | (c) [3 marks]: 1 mark for magnesium fluoride has giant ionic lattice / strong electrostatic forces between oppositely charged ions; 1 mark for ammonia has weak intermolecular forces; 1 mark for stating that much less energy is needed to overcome weak intermolecular forces than strong ionic bonds. | (d) [2 marks]: 1 mark for random / irregular arrangement / far apart; 1 mark for rapid / random movement.

Question 2 · structured

10 marks

A student investigates the rate of reaction between magnesium ribbon and dilute hydrochloric acid: \(Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)\). (a) State two ways to increase the rate of this reaction, other than increasing the temperature. [2 marks] (b) Using the collision theory, explain how increasing the temperature increases the rate of this reaction. Refer to activation energy in your answer. [4 marks] (c) Describe how the rate of reaction changes as the reaction proceeds, and explain why this change occurs. [3 marks] (d) Name one piece of laboratory apparatus that can be used to collect and measure the volume of hydrogen gas produced. [1 mark]

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Worked solution

(a) The rate of reaction can be increased by: 1. Increasing the concentration of the hydrochloric acid. 2. Increasing the surface area of the magnesium (e.g., using magnesium powder instead of ribbon). (b) When the temperature is increased, the reacting particles gain kinetic energy and move faster. This leads to more frequent collisions between particles. Crucially, a much larger proportion of colliding particles now have energy equal to or greater than the activation energy. This increases the frequency of successful collisions (more successful collisions per unit time). (c) As the reaction proceeds, the rate of reaction decreases (the reaction slows down) until it eventually stops. This is because the reactant particles (magnesium and acid) are being used up, causing their concentration/amount to decrease. Consequently, the frequency of collisions between reactant particles decreases. (d) A gas syringe can be used to collect and measure the volume of the gas.

Marking scheme

(a) [2 marks]: 1 mark for each valid method (e.g., increase acid concentration, use powdered magnesium, add a catalyst). | (b) [4 marks]: 1 mark for particles having more kinetic energy / moving faster; 1 mark for more frequent collisions; 1 mark for more particles having energy greater than or equal to the activation energy; 1 mark for more successful collisions per unit time / higher frequency of successful collisions. | (c) [3 marks]: 1 mark for rate decreases / slows down; 1 mark for reactants used up / concentration decreases; 1 mark for collision frequency decreases. | (d) [1 mark]: 1 mark for gas syringe (accept: graduated gas cylinder inverted over water).

Question 3 · structured

10 marks

Iron is extracted from its ore, hematite, in a blast furnace. (a) Write the word equation and the balanced chemical equation for the reduction of iron(III) oxide by carbon monoxide. [3 marks] (b) Explain the role of limestone, \(CaCO_3\), in the blast furnace. Include at least one balanced chemical equation in your answer. [3 marks] (c) Pure iron is relatively soft, but it can be alloyed with carbon to produce steel. Explain, in terms of structure, why steel is harder and stronger than pure iron. [3 marks] (d) State one condition required for the rusting of iron, other than oxygen. [1 mark]

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Worked solution

(a) Word equation: iron(III) oxide + carbon monoxide \rightarrow iron + carbon dioxide. Chemical equation: \(Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2\). (b) Limestone (calcium carbonate) is added to remove acidic impurities (silica/silicon dioxide). It undergoes thermal decomposition to form calcium oxide and carbon dioxide: \(CaCO_3 \rightarrow CaO + CO_2\). The calcium oxide (a basic oxide) then reacts with the silicon dioxide (an acidic oxide) to form slag (liquid calcium silicate): \(CaO + SiO_2 \rightarrow CaSiO_3\). (c) Pure iron has a giant metallic lattice structure with regular layers of identical iron atoms. These layers can easily slide over each other when a force is applied, making pure iron malleable. Steel is an alloy of iron and carbon. Carbon atoms are of a different size compared to iron atoms. Their presence disrupts the regular arrangement of the layers, preventing the layers from sliding over each other easily. (d) Water (or moisture) is required for rusting, in addition to oxygen.

Marking scheme

(a) [3 marks]: 1 mark for correct word equation; 1 mark for correct chemical formulas in equation; 1 mark for correct balancing of the equation: \(Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2\). | (b) [3 marks]: 1 mark for stating thermal decomposition of limestone / \(CaCO_3 \rightarrow CaO + CO_2\); 1 mark for calcium oxide reacting with silicon dioxide / silica to form slag; 1 mark for correct slag formation equation: \(CaO + SiO_2 \rightarrow CaSiO_3\). | (c) [3 marks]: 1 mark for pure iron having regular layers of atoms that slide over each other easily; 1 mark for carbon atoms being of a different size; 1 mark for carbon atoms disrupting the regular arrangement and preventing the layers from sliding. | (d) [1 mark]: 1 mark for water / moisture / water vapour.

Question 4 · structured

10 marks

This question is about synthetic polymers. (a) Ethene undergoes addition polymerisation to form poly(ethene). (i) Describe the structure of an ethene monomer, including its bonding. [1 mark] (ii) Describe the structure of the repeat unit of poly(ethene). [2 marks] (iii) State one key difference between addition polymerisation and condensation polymerisation. [1 mark] (b) Terylene is a synthetic polyester. (i) State the names of the two different functional groups present in the monomers used to make this polyester. [2 marks] (ii) Describe the linkage formed when these monomers react, including the bonds involved. [2 marks] (c) State one environmental problem caused by the disposal of non-biodegradable synthetic polymers in landfill sites, and suggest one way to reduce this problem. [2 marks]

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Worked solution

(a)(i) An ethene monomer contains two carbon atoms joined by a double covalent bond, with each carbon atom also bonded to two hydrogen atoms by single covalent bonds (formula \(C_2H_4\)). (ii) The repeat unit of poly(ethene) is represented as \(-[CH_2-CH_2]-\). It features a single covalent bond between the two carbon atoms, each of which is bonded to two hydrogen atoms, with open-ended single bonds extending through the brackets to indicate continuation. (iii) In addition polymerisation, the polymer is the only product. In condensation polymerisation, two different monomers typically react to produce the polymer plus a small molecule, such as water or hydrogen chloride. (b)(i) The monomers used are a dicarboxylic acid (containing carboxylic acid groups, \(-COOH\)) and a diol (containing alcohol/hydroxyl groups, \(-OH\)). (ii) The linkage is an ester linkage. It consists of a carbonyl group (\(C=O\)) from the carboxylic acid monomer bonded to an oxygen atom (\(-O-\)) from the alcohol monomer, represented as \(-C(=O)-O-\). (c) Problem: Synthetic polymers are non-biodegradable and persist in landfills for hundreds of years, causing waste accumulation and potential harm to wildlife. Solution to reduce this problem: recycling the plastics, incinerating them to produce energy (with scrubbers to remove toxic gases), or developing and using biodegradable alternative polymers.

Marking scheme

(a)(i) [1 mark]: 1 mark for describing a carbon-carbon double bond and four single C-H bonds. | (a)(ii) [2 marks]: 1 mark for single carbon-carbon bond; 1 mark for four single carbon-hydrogen bonds (two on each carbon) and open bonds at each end. | (a)(iii) [1 mark]: 1 mark for stating that addition polymerisation produces only the polymer, while condensation polymerisation also produces a small molecule (or that addition requires C=C double bonds while condensation requires two functional groups). | (b)(i) [2 marks]: 1 mark for carboxylic acid (or carboxyl); 1 mark for alcohol (or hydroxyl). | (b)(ii) [2 marks]: 1 mark for naming the linkage as an ester linkage; 1 mark for describing it as a carbon-oxygen double bond (C=O) bonded to a carbon-oxygen single bond (C-O). | (c) [2 marks]: 1 mark for identifying a valid environmental problem (e.g. non-biodegradable / persists in landfill / danger to wildlife); 1 mark for a valid solution (e.g. recycling / incinerating for energy / using biodegradable polymers).

Physics Section

Answer all questions. Questions 9 to 12 cover kinematics, electricity, waves, nuclear physics, and space physics.

4 Question · 40 marks

Question 1 · structured

10 marks

A model rocket is launched vertically upwards from rest.

(a) The rocket accelerates from rest with a constant acceleration of \(5.0\text{ m/s}^2\) for \(6.0\text{ s}\).

(i) Calculate the velocity of the rocket at \(t = 6.0\text{ s}\).

(ii) Calculate the distance travelled by the rocket during the first \(6.0\text{ s}\).

(b) At \(t = 6.0\text{ s}\), the rocket engine turns off, and the rocket continues to move upwards until it reaches its maximum height. Air resistance is negligible. Take the acceleration due to gravity, \(g\), to be \(9.8\text{ m/s}^2\).

(i) Describe the motion of the rocket after the engine turns off.

(ii) Calculate the time taken from when the engine turns off to when the rocket reaches its maximum height.

(c) Define the term velocity and state how it differs from speed.

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Worked solution

(a)(i)
Using the formula for acceleration:
\(v = u + at\)
Since the rocket starts from rest, \(u = 0\text{ m/s}\):
\(v = 0 + (5.0\text{ m/s}^2 \times 6.0\text{ s}) = 30\text{ m/s}\)

(a)(ii)
Using the formula for distance:
\(s = ut + \frac{1}{2}at^2\)
\(s = 0 + \frac{1}{2} \times 5.0\text{ m/s}^2 \times (6.0\text{ s})^2\)
\(s = 2.5 \times 36 = 90\text{ m}\)

(b)(i)
Once the engine turns off, the only force acting on the rocket is its weight (gravity). It undergoes uniform (constant) deceleration at a rate of \(9.8\text{ m/s}^2\) until its velocity becomes zero at the highest point.

(b)(ii)
Using \(v = u + at\) where \(u = 30\text{ m/s}\), \(v = 0\text{ m/s}\) at maximum height, and \(a = -g = -9.8\text{ m/s}^2\):
\(0 = 30 - 9.8t\)
\(9.8t = 30\)
\(t = \frac{30}{9.8} \approx 3.1\text{ s}\) (or \(3.06\text{ s}\))

(c)
Velocity is defined as the rate of change of displacement (or speed in a given direction). It is a vector quantity (has both magnitude and direction), whereas speed is a scalar quantity (has magnitude only).

Marking scheme

(a)(i) [2 marks total]
- \(v = u + at\) or correct substitution shown [1 mark]
- \(30\text{ m/s}\) (ignore units unless incorrect) [1 mark]

(a)(ii) [2 marks total]
- \(s = \frac{1}{2}at^2\) or correct substitution shown [1 mark]
- \(90\text{ m}\) [1 mark]

(b)(i) [2 marks total]
- Decelerating / slowing down [1 mark]
- Constantly / uniformly / at a rate of \(g\) or \(9.8\text{ m/s}^2\) [1 mark]

(b)(ii) [2 marks total]
- \(t = \frac{v}{g}\) or \(\frac{30}{9.8}\) [1 mark]
- \(3.1\text{ s}\) (accept \(3.06\text{ s}\) or \(3.0\text{ s}\) if \(g=10\text{ m/s}^2\) is used) [1 mark]

(c) [2 marks total]
- Velocity is rate of change of displacement / speed in a specified direction [1 mark]
- Velocity is a vector (has direction) whereas speed is a scalar (no direction) [1 mark]

Question 2 · structured

10 marks

A student designs a potential divider circuit to monitor temperature changes in a room.

(a) The potential divider circuit consists of a \(6.0\text{ V}\) d.c. power supply, a fixed resistor of \(1200\ \Omega\), and a thermistor connected in series. An analog voltmeter is connected across the fixed resistor.

(i) Draw a circuit diagram representing this arrangement.

(ii) At room temperature, the resistance of the thermistor is \(800\ \Omega\). Calculate the reading on the voltmeter.

(iii) The temperature of the room increases. State and explain what happens to the voltmeter reading.

(b) State the formula relating charge, current, and time. Calculate the charge passing through the circuit in \(5.0\text{ minutes}\) when the current in the circuit is \(3.0\text{ mA}\).

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Worked solution

(a)(i)
The circuit diagram must show:
- A d.c. voltage source (cell or battery or d.c. supply symbol)
- A fixed resistor (rectangular box) connected in series with a thermistor (rectangular box with a diagonal line starting flat at the bottom and going up to the right)
- A voltmeter connected in parallel specifically across the fixed resistor.

(a)(ii)
Total resistance of the series circuit:
\(R_{\text{total}} = R_{\text{fixed}} + R_{\text{thermistor}} = 1200\ \Omega + 800\ \Omega = 2000\ \Omega\)

Using the potential divider formula for the voltage across the fixed resistor:
\(V_{\text{out}} = V_{\text{supply}} \times \frac{R_{\text{fixed}}}{R_{\text{total}}}\)
\(V_{\text{out}} = 6.0\text{ V} \times \frac{1200\ \Omega}{2000\ \Omega} = 3.6\text{ V}\)

Alternatively, calculate the current:
\(I = \frac{V}{R} = \frac{6.0\text{ V}}{2000\ \Omega} = 0.0030\text{ A}\)
Then the voltage across the fixed resistor:
\(V = I \times R_{\text{fixed}} = 0.0030\text{ A} \times 1200\ \Omega = 3.6\text{ V}\)

(a)(iii)
When the temperature increases, the resistance of the thermistor decreases.
Since the thermistor resistance decreases, the total resistance of the circuit decreases, causing the current to increase. Because the current increases, the potential difference across the fixed resistor (\(V = I \times R\)) increases. (Or: the fixed resistor now takes up a larger fraction of the total resistance, so it takes a larger share of the input voltage). The voltmeter reading increases.

(b)
Formula: \(Q = I \times t\)
Convert time to seconds:
\(t = 5.0\text{ minutes} = 5.0 \times 60 = 300\text{ s}\)
Convert current to amperes:
\(I = 3.0\text{ mA} = 0.0030\text{ A}\)
Calculate charge:
\(Q = 0.0030\text{ A} \times 300\text{ s} = 0.90\text{ C}\)

Marking scheme

(a)(i) [3 marks total]
- Correct symbols used for d.c. supply/cell, fixed resistor, and thermistor [1 mark]
- Fixed resistor and thermistor connected in series with the power supply [1 mark]
- Voltmeter connected in parallel across the fixed resistor [1 mark]

(a)(ii) [3 marks total]
- Calculation of total resistance: \(1200 + 800 = 2000\ \Omega\) [1 mark]
- Correct substitution into potential divider or current equation [1 mark]
- Correct final voltmeter reading of \(3.6\text{ V}\) [1 mark]

(a)(iii) [2 marks total]
- Correctly states that resistance of the thermistor decreases [1 mark]
- Correctly explains that voltmeter reading increases because the fixed resistor gets a larger proportion of the supply voltage (or circuit current increases) [1 mark]

(b) [2 marks total]
- Correct formula stated: \(Q = I \times t\) or equivalent [1 mark]
- Correct conversion of units and calculation: \(0.90\text{ C}\) (accept \(0.9\text{ C}\); reject \(0.015\text{ C}\) from not converting minutes/mA) [1 mark]

Question 3 · structured

10 marks

(a) Sound is a longitudinal wave.

(i) Explain what is meant by a longitudinal wave. Your answer should include reference to how the particles of the medium move and the terms compressions and rarefactions.

(ii) A sound wave traveling through water has a speed of \(1500\text{ m/s}\) and a frequency of \(2.5\text{ kHz}\). Calculate the wavelength of this sound wave.

(b) Electromagnetic waves travel through space.

(i) State two physical properties that are shared by all electromagnetic waves.

(ii) Light waves can undergo reflection and refraction. Contrast these two processes in terms of what happens to the speed and direction of the light wave.

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Worked solution

(a)(i)
In a longitudinal wave, the particles of the medium oscillate (or vibrate) back and forth parallel to the direction of wave travel (and energy transfer).
- A compression is a region in the longitudinal wave where the particles are closest together (causing high pressure).
- A rarefaction is a region where the particles are furthest apart (causing low pressure).

(a)(ii)
Using the wave equation:
\(v = f \lambda\)
Rearranging for wavelength (\(\lambda\)):
\(\lambda = \frac{v}{f}\)
Given:
\(v = 1500\text{ m/s}\)
\(f = 2.5\text{ kHz} = 2500\text{ Hz}\)
Calculation:
\(\lambda = \frac{1500}{2500} = 0.60\text{ m}\)

(b)(i)
Any two properties from:
- They all travel at the same speed in a vacuum (approximately \(3.0 \times 10^8\text{ m/s}\)).
- They are all transverse waves.
- They can all travel through a vacuum.
- They all transfer energy.

(b)(ii)
- Reflection: The light wave bounces off a boundary. Its direction changes, but its speed (and wavelength) remains completely constant as it stays in the same medium.
- Refraction: The light wave enters a new medium. Its speed changes (it slows down in a denser medium, or speeds up in a less dense medium), and its direction also changes (unless it enters along the normal).

Marking scheme

(a)(i) [3 marks total]
- Particles vibrate/oscillate parallel to the direction of wave travel / energy transfer [1 mark]
- Compressions described as regions of high density/pressure / where particles are close together [1 mark]
- Rarefactions described as regions of low density/pressure / where particles are spread apart [1 mark]

(a)(ii) [2 marks total]
- Use of \(v = f\lambda\) with frequency converted to \(2500\text{ Hz}\) [1 mark]
- Wavelength \(= 0.60\text{ m}\) (or \(0.6\text{ m}\)) [1 mark]

(b)(i) [2 marks total]
- Any two of: travel at \(3.0 \times 10^8\text{ m/s}\) in vacuum / are transverse / can travel through vacuum / transfer energy [2 marks, 1 mark for each correct property]

(b)(ii) [3 marks total]
- Reflection: direction changes but speed remains constant [1 mark]
- Refraction: speed changes as the wave enters a new medium [1 mark]
- Refraction: direction changes (unless at normal) [1 mark]

Question 4 · structured

10 marks

(a) The Sun is a stable star in the main sequence.

(i) Explain how the forces acting on the Sun keep it in stable equilibrium.

(ii) State the main nuclear reaction that provides energy in stars like the Sun, and describe what happens to the nuclei during this reaction.

(iii) Describe the life cycle of a star with a mass similar to the Sun, starting from when it leaves the main sequence stage.

(b) Light from distant galaxies is observed to show redshift.

(i) State what is meant by redshift and explain what this observation tells us about the motion of these galaxies.

(ii) State the name of the cosmological theory that redshift provides evidence for.

Show answer & marking scheme

Worked solution

(a)(i)
The Sun is kept stable by a balance between two forces:
- The inward force of gravitational attraction, which pulls the matter of the star inwards.
- The outward force due to thermal expansion/radiation pressure, which is produced by the energy released from nuclear fusion reactions in the core.
These two opposing forces are equal in magnitude, so the star is in a stable, balanced state (hydrostatic equilibrium).

(a)(ii)
- The nuclear reaction is nuclear fusion.
- During this reaction, lighter hydrogen nuclei (or protons) fuse together under extreme temperature and pressure to form heavier helium nuclei, releasing a large amount of energy in the process.

(a)(iii)
When a low-mass star like the Sun leaves the main sequence:
1. It expands and cools at the surface to become a red giant.
2. When fusion stops in the core, the outer layers of dust and gas are ejected, forming a planetary nebula.
3. The remaining core collapses under gravity to form a hot, dense white dwarf.

(b)(i)
- Redshift is the increase in the observed wavelength (or decrease in frequency) of light emitted from distant stars/galaxies as it travels to Earth.
- This observation tells us that the galaxies are moving away from the Earth / us (receding), which means the Universe is expanding.

(b)(ii)
The Big Bang theory.

Marking scheme

(a)(i) [2 marks total]
- Inward force of gravity / gravitational attraction [1 mark]
- Balanced by outward thermal pressure / radiation pressure from fusion [1 mark]

(a)(ii) [2 marks total]
- Identify nuclear fusion [1 mark]
- Description: hydrogen nuclei fuse to form helium nuclei [1 mark]

(a)(iii) [3 marks total]
- Stage 1: expands to become a red giant [1 mark]
- Stage 2: ejects outer layers / planetary nebula formed [1 mark]
- Stage 3: leaves behind a hot, dense core / white dwarf [1 mark]

(b)(i) [2 marks total]
- Redshift: wavelength of observed light increases / is shifted towards the red end of the spectrum [1 mark]
- Interpretation: galaxies are moving away from us / the Universe is expanding [1 mark]

(b)(ii) [1 mark total]
- Big Bang theory [1 mark]

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