An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V3) Cambridge International A Level Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.
Extended Theory Paper 4
Answer all questions. Show your working where appropriate. Write in dark blue or black pen. You may use a calculator.
12 Question · 120 marks
Question 1 · structured
10 marks
Fig. 1.1 shows a diagram of the human digestive system.
(a) Ingestion is the first step in nutrition. Define the term ingestion.
(b) Describe the role of mechanical digestion and explain how it differs from chemical digestion.
(c) Dental decay is a common disease affecting teeth. Describe how dental decay is caused by bacteria in the mouth and state one way to prevent it.
(d) Explain how the stomach is adapted to digest protein, naming the enzyme and chemical conditions involved.
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Worked solution
(a) Ingestion is the taking of substances, e.g. food and drink, into the body through the mouth. (b) Mechanical digestion involves physical processes such as chewing (teeth) and churning (stomach) to break food into smaller pieces without chemical modification. Chemical digestion involves the breakdown of large, insoluble molecules into small, soluble molecules using enzymes (e.g., amylase, protease). (c) Bacteria feed on food debris (especially sugars) left on teeth, producing acid. This acid dissolves the calcium salts in the enamel and dentine, leading to decay. Prevention can be achieved by brushing teeth regularly with fluoride toothpaste. (d) The stomach secretes gastric juice containing protease (pepsin) which breaks down proteins into simpler peptides. It also secretes hydrochloric acid, which provides the optimum low pH (acidic, pH 1.5-2.0) for pepsin to work and kills harmful micro-organisms in the food.
Marking scheme
(a) Ingestion definition: Taking of substances (food/drink) into the body through the mouth [1]
(b) Role of mechanical digestion and difference from chemical digestion [max 3]: - Mechanical digestion breaks food down into smaller pieces to increase surface area [1] - Does not chemically change the molecules [1] - Chemical digestion involves enzymes / breaks large insoluble molecules into small soluble molecules [1]
(c) Dental decay cause and prevention [max 3]: - Bacteria feed on sugars on the teeth and respire to produce acids [1] - Acid dissolves the enamel / dentine of the tooth [1] - Prevention: regular brushing with fluoride toothpaste / reducing sugar intake [1]
(d) Stomach adaptations for protein digestion [max 3]: - Secretes protease / pepsin to digest proteins [1] - Secretes hydrochloric acid (HCl) [1] - Provides acidic pH / optimum pH for pepsin AND kills pathogens [1]
Question 2 · structured
10 marks
Plants require a continuous stream of water for photosynthesis and support.
(a) Describe how water is absorbed into the root hair cells of a plant from the soil, including the process involved.
(b) A student used a potometer to investigate transpiration. Table 2.1 shows the distance a bubble moved in 10 minutes under different conditions.
| Condition | Distance moved by bubble / mm | | :--- | :--- | | Still air | 15 | | Moving air (wind) | 48 |
(i) Explain why the rate of water uptake was greater when the air was moving.
(ii) State two other environmental factors that increase the rate of transpiration.
(c) Distinguish between the transport functions of xylem and phloem, identifying the substances they transport.
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Worked solution
(a) Water in the soil has a higher water potential than the cytoplasm of the root hair cell. Water moves from the soil into the root hair cells by osmosis, down a water potential gradient, across a partially permeable cell membrane. (b)(i) Moving air sweeps away water vapour that has evaporated and diffused out of the stomata. This maintains a steep concentration gradient of water vapour between the inside of the leaf and the outside air, increasing the rate of transpiration / water uptake. (ii) Two other factors: Increased temperature; Increased light intensity. (c) Xylem transports water and mineral ions from the roots to the leaves. Phloem transports sucrose and amino acids from sources (e.g. leaves) to sinks (e.g. roots or growing buds).
Marking scheme
(a) Water absorption [max 3]: - Water potential is higher in the soil than in the root hair cell / down a water potential gradient [1] - Osmosis is the process [1] - Water moves across a partially permeable membrane [1]
(b)(i) Moving air effect [max 3]: - Moving air removes water vapour around the leaf surface [1] - This maintains a steep water vapour concentration gradient [1] - Leading to faster diffusion of water vapour out of stomata [1]
(b)(ii) Other factors [2]: - Increased temperature [1] - Increased light intensity (accept decreased humidity) [1]
(c) Xylem vs Phloem [2]: - Xylem transports water and mineral ions [1] - Phloem transports sucrose and amino acids [1]
Question 3 · structured
10 marks
The human respiratory system is adapted for efficient gas exchange.
(a) Describe three structural features of the alveoli that make them efficient for the exchange of gases.
(b) Pathogens and dust particles are often inhaled with air. Explain how goblet cells and ciliated cells work together to protect the lungs.
(c) Tobacco smoke contains substances that harm the human body.
(i) Explain the effect of carbon monoxide on the transport of oxygen.
(ii) State one effect of nicotine on the body.
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Worked solution
(a) Alveoli features: 1. Very thin walls (one-cell thick) which provide a short diffusion distance. 2. Large surface area to maximize the rate of diffusion. 3. Well supplied with a network of capillaries to maintain a steep concentration gradient. (b) Goblet cells produce and secrete sticky mucus, which traps dust particles and pathogens (bacteria). Ciliated cells have tiny hair-like structures called cilia that beat in a coordinated wave-like motion to sweep the mucus up and out of the trachea and airways towards the throat, where it can be swallowed. (c)(i) Carbon monoxide binds irreversibly to hemoglobin in red blood cells, forming carboxyhemoglobin. This reduces the capacity of hemoglobin to transport oxygen to tissues. (ii) Nicotine is an addictive substance that increases heart rate and narrows blood vessels (causing high blood pressure).
Marking scheme
(a) Alveoli features [3]: - Thin wall / one-cell thick for short diffusion distance [1] - Large surface area for maximum diffusion [1] - Good blood supply / dense capillary network to maintain concentration gradient [1] - Wet/moist surfaces (to dissolve gases) [1] (max 3 marks)
(b) Goblet cells and ciliated cells [3]: - Goblet cells produce/secrete mucus [1] - Mucus traps pathogens / dust [1] - Ciliated cells have cilia that beat/sweep mucus up/away from the lungs [1]
(c)(i) Carbon monoxide [2]: - Binds to hemoglobin (preferentially / irreversibly) [1] - Reduces the amount of oxygen transported in the blood [1]
Organisms detect and respond to changes in their environment.
(a) Outline the pathway of a reflex arc when a person accidentally touches a hot object. Include the names of the three types of neurones involved.
(b) Describe how a reflex action differs from a voluntary action.
(c) Explain the role of auxin in the phototropic response of a plant shoot.
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Worked solution
(a) 1. Pain/temperature receptors in the skin detect the stimulus (heat). 2. Electrical impulses are generated and travel along a sensory neurone to the central nervous system (spinal cord). 3. The impulse is passed across a synapse via neurotransmitters to a relay neurone in the spinal cord. 4. The impulse is then passed across another synapse to a motor neurone. 5. The motor neurone carries the impulse to an effector (muscle), which contracts to pull the hand away. (b) A reflex action is rapid, automatic (involuntary), and does not involve conscious thought by the brain, whereas a voluntary action is slower, deliberate, and involves decision-making by the brain. (c) Auxin is a plant hormone made in the shoot tip. When a shoot is exposed to unilateral light (light from one side), auxin moves/diffuses to the shaded side of the shoot. The high concentration of auxin on the shaded side causes cell elongation in that region. Since the shaded side grows faster/elongates more than the illuminated side, the shoot bends towards the light source (positive phototropism).
Marking scheme
(a) Reflex arc pathway [max 4]: - Stimulus detected by receptor / sensory neurone carries impulse to CNS [1] - Relay neurone inside spinal cord / CNS [1] - Motor neurone carries impulse from CNS to effector [1] - Effector / muscle contracts to perform response [1] - Synapses involve chemical neurotransmitters [1]
(b) Reflex vs Voluntary [max 2]: - Reflex is automatic / involuntary while voluntary involves conscious thought [1] - Reflex is much faster [1]
(c) Auxin and phototropism [max 4]: - Auxin is produced in the shoot tip [1] - Auxin diffuses / moves to the shaded side [1] - Auxin causes cell elongation [1] - Higher growth on the shaded side causes the shoot to bend towards the light [1]
Question 5 · Structured Theory
10 marks
This question is about hydrocarbons and polymers.
(a) Ethane and ethene are hydrocarbons.
(i) Describe a chemical test to distinguish between ethane and ethene. State the reagent used and the result for each compound. [3]
(ii) Ethene can be polymerised to form poly(ethene). Draw the structure of a section of poly(ethene) showing two repeating units. [2]
(b) Ethanol can be manufactured by two different methods: fermentation of glucose or hydration of ethene.
(i) Write a word equation for the fermentation of glucose and state one essential condition for this process. [2]
(ii) State the catalyst and temperature used in the industrial hydration of ethene to produce ethanol. [2]
(iii) State one advantage of using the hydration of ethene method rather than fermentation. [1]
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Worked solution
(a)(i) Reagent: Aqueous bromine / bromine water Result with ethane: Remains orange / yellow / brown (no change) Result with ethene: Turns colourless / decolourised
(ii) Structure showing a chain of four carbon atoms with single bonds between them: \(-CH_2-CH_2-CH_2-CH_2-\) with open single bonds at each end.
(b)(i) Word equation: glucose \(\rightarrow\) ethanol + carbon dioxide Condition: Yeast / anaerobic conditions / temperature between 20 \(^{\circ}\text{C}\) and 35 \(^{\circ}\text{C}\)
(ii) Catalyst: Phosphoric acid / \(\text{H}_3\text{PO}_4\) Temperature: 300 \(^{\circ}\text{C}\) (accept 250 \(^{\circ}\text{C}\) to 350 \(^{\circ}\text{C}\))
(iii) Any one of: faster rate of reaction, continuous process, produces pure ethanol (no need for fractional distillation).
Marking scheme
(a)(i) Reagent: aqueous bromine / bromine water [1] Result with ethane: remains orange/yellow/brown [1] Result with ethene: decolourised / turns colourless [1]
(ii) Four carbons linked with single C-C bonds with open-ended bonds at each end [1] Correct number of H atoms (8 hydrogens) attached to carbons [1]
(b)(i) Glucose \(\rightarrow\) ethanol + carbon dioxide (both products correct) [1] Condition: yeast OR temperature 20–35\(^{\circ}\text{C}\) OR anaerobic / absence of oxygen [1]
A student reacts excess calcium carbonate with 50.0 \(\text{cm}^3\) of 0.200 \(\text{mol/dm}^3\) hydrochloric acid.
(i) Calculate the number of moles of \(\text{HCl}\) used in this reaction. [1]
(ii) Determine the number of moles of carbon dioxide, \(\text{CO}_2\), produced. [1]
(iii) Calculate the volume of carbon dioxide, in \(\text{cm}^3\), produced at room temperature and pressure (r.t.p.). [The volume of one mole of any gas at r.t.p. is 24.0 \(\text{dm}^3\)] [2]
(iv) If the student used 50.0 \(\text{cm}^3\) of 0.400 \(\text{mol/dm}^3\) hydrochloric acid instead, describe and explain the effect on the rate of reaction. [2]
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Worked solution
(a)(i) Percentage of oxygen = \(100 - 88.8 = 11.2\%\) Moles of copper = \(\frac{88.8}{64} = 1.39\) Moles of oxygen = \(\frac{11.2}{16} = 0.70\) Ratio of \(\text{Cu} : \text{O} = \frac{1.39}{0.70} : \frac{0.70}{0.70} = 1.98 : 1\), which is approximately \(2 : 1\). Thus, the empirical formula is \(\text{Cu}_2\text{O}\).
(ii) Copper(I) oxide
(b)(i) Moles of \(\text{HCl} = \text{concentration} \times \text{volume in dm}^3 = 0.200 \times \frac{50.0}{1000} = 0.010\text{ mol}\)
(ii) From the equation, 2 moles of \(\text{HCl}\) produce 1 mole of \(\text{CO}_2\). Moles of \(\text{CO}_2 = \frac{0.010}{2} = 0.005\text{ mol}\)
(iii) Volume of \(\text{CO}_2\) in \(\text{dm}^3 = 0.005 \times 24.0 = 0.12\text{ dm}^3\) Volume in \(\text{cm}^3 = 0.12 \times 1000 = 120\text{ cm}^3\)
(iv) The rate of reaction increases because there is a higher concentration of acid particles. This means there are more particles per unit volume, leading to a higher frequency of successful collisions.
Marking scheme
(a)(i) Calculates mass % of O = 11.2% [1] Calculates moles of both: Cu = 1.39 mol, O = 0.70 mol [1] Divides by smallest to get 2:1 ratio and states formula \(\text{Cu}_2\text{O}\) [1]
(ii) Copper(I) oxide (must include roman numeral I) [1]
(b)(i) 0.010 / 0.01 mol [1]
(ii) 0.005 mol (consequential on b(i)) [1]
(iii) Calculates volume in \(\text{dm}^3 = 0.12\text{ dm}^3\) [1] Converts to \(120\text{ cm}^3\) [1]
(iv) Rate increases [1] More particles per unit volume / greater frequency of successful collisions [1]
Question 7 · Structured Theory
10 marks
This question is about electrolysis.
(a) Molten lead(II) bromide is electrolysed using inert carbon electrodes.
(i) State the observations at the anode (positive electrode) and at the cathode (negative electrode). [2]
(ii) Write ionic half-equations, including state symbols, for the reactions at: - the anode [2] - the cathode [2]
(b) Aqueous copper(II) sulfate is electrolysed using copper electrodes.
(i) Describe what happens to the mass of the anode and the mass of the cathode during this electrolysis. [2]
(ii) Explain how this process can be used to purify impure copper. [2]
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Worked solution
(a)(i) At anode: Brown gas / vapour is evolved. At cathode: Grey liquid / silvery bead of lead metal forms.
(ii) At anode: \(2\text{Br}^-(l) \rightarrow \text{Br}_2(g) + 2e^-\) At cathode: \(\text{Pb}^{2+}(l) + 2e^- \rightarrow \text{Pb}(l)\)
(b)(i) The mass of the anode decreases, and the mass of the cathode increases.
(ii) The anode is made of impure copper. Copper atoms lose electrons and dissolve to form \(\text{Cu}^{2+}\) ions. At the cathode, pure copper ions gain electrons and are deposited as pure copper metal. Impurities fall to the bottom.
Marking scheme
(a)(i) Anode: brown gas/vapour [1] Cathode: grey liquid/metal [1]
(ii) Anode equation: \(2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-\) Correct species and balancing [1] Correct state symbols: \(\text{Br}^-(l)\) and \(\text{Br}_2(g)\) or \(\text{Br}_2(l)\) [1]
Cathode equation: \(\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}\) Correct species and balancing [1] Correct state symbols: \(\text{Pb}^{2+}(l)\) and \(\text{Pb}(l)\) [1]
(b)(i) Mass of anode decreases AND mass of cathode increases [2] (Award 1 mark if one is correct but not both)
(ii) Impure copper anode dissolves / oxidises to \(\text{Cu}^{2+}\) [1] Pure copper is deposited on the cathode / \(\text{Cu}^{2+}\) is reduced to copper [1]
Question 8 · Structured Theory
10 marks
(a) The reaction between iron(III) oxide and carbon monoxide is a key step in the extraction of iron in the blast furnace:
(i) Suggest how the student could measure the rate of this reaction. Identify the apparatus used and the measurement taken. [2]
(ii) Sketch a graph to show how the volume of hydrogen gas produced changes with time from the start of the reaction until it stops. Label the axes. [2]
(iii) State one change, other than changing concentration, that would increase the initial rate of this reaction. [1]
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Worked solution
(a)(i) Carbon monoxide (\(\text{CO}\)) is oxidised because it gains oxygen to form \(\text{CO}_2\). Iron(III) oxide (\(\text{Fe}_2\text{O}_3\)) is reduced because it loses oxygen to form iron (\(\text{Fe}\)). Since both oxidation and reduction occur simultaneously, it is a redox reaction.
(ii) The oxidation state of iron in \(\text{Fe}_2\text{O}_3\) is +3. The oxidation state of iron in elemental \(\text{Fe}\) is 0. Since the oxidation state decreases from +3 to 0, it has been reduced.
(b)(i) The student can measure the volume of hydrogen gas produced over time using a gas syringe. Alternatively, they could place the reaction flask on a mass balance and measure the decrease in mass as gas escapes.
(ii) The graph should have the y-axis labelled 'Volume of gas / \(\text{cm}^3\)' (or mass) and the x-axis labelled 'Time / s' (or minutes). The curve should start at the origin (0,0), rise steeply at first, become less steep, and then level off to a horizontal line indicating the reaction has finished.
(iii) Increase the temperature of the acid OR use powdered magnesium (increase surface area).
Marking scheme
(a)(i) CO is oxidised because it gains oxygen AND \(\text{Fe}_2\text{O}_3\) is reduced because it loses oxygen [1] Correctly identifies CO as the reducing agent / oxidised substance and \(\text{Fe}_2\text{O}_3\) as the oxidising agent / reduced substance [1] States that redox involves both oxidation and reduction occurring together [1]
(ii) Oxidation number of Fe in \(\text{Fe}_2\text{O}_3\) is +3 and in Fe is 0 [1] Reduction is a decrease in oxidation number [1]
(b)(i) Apparatus: gas syringe / inverted measuring cylinder with water / balance [1] Measurement: volume of gas OR mass of flask at timed intervals [1]
(ii) Graph axes labelled correctly (Volume/mass on y-axis, Time on x-axis) [1] Curve of correct shape (steep at first, levelling off to horizontal) starting at origin [1]
(iii) Increase temperature OR increase surface area of magnesium (use powder) OR add a catalyst [1]
Question 9 · Structured Theory
10 marks
A cyclist is riding on a flat road.
(a) The cyclist accelerates from rest to a speed of \(6.0\text{ m/s}\) in a time of \(8.0\text{ s}\).
(i) Calculate the acceleration of the cyclist. [2]
(ii) Calculate the distance travelled by the cyclist during these \(8.0\text{ s}\), assuming constant acceleration. [2]
(b) The cyclist and bicycle have a combined mass of \(75\text{ kg}\).
(i) Calculate the kinetic energy of the cyclist and bicycle when travelling at \(6.0\text{ m/s}\). [2]
(ii) During braking, the bicycle comes to a complete stop. Explain, in terms of energy transfers and work done, what happens to this kinetic energy. [2]
(c) The cyclist then rides up a steep hill. The vertical height of the hill is \(12\text{ m}\). Calculate the work done against gravity by the cyclist in climbing the hill. (Use \(g = 9.8\text{ m/s}^2\)) [2]
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(b)(ii) Work is done by the friction forces between the brake pads and the wheels. The kinetic energy is transferred to thermal energy in the brakes and the surroundings.
(a)(i) - Use of formula \(a = \frac{\Delta v}{t}\) or substitution of correct values [1] - Correct answer with unit: \(0.75\text{ m/s}^2\) [1]
(a)(ii) - Use of \(d = \text{average speed} \times t\) or \(s = ut + \frac{1}{2}at^2\) with substitution [1] - Correct answer: \(24\text{ m}\) [1]
(b)(i) - Correct formula \(E_k = \frac{1}{2}mv^2\) or substitution [1] - Correct answer: \(1350\text{ J}\) [1]
(b)(ii) - Kinetic energy is transferred/converted to thermal energy [1] - Work is done by friction [1]
(c) - Correct formula \(E_p = mgh\) or substitution [1] - Correct answer: \(8820\text{ J}\) (accept \(9000\text{ J}\) if \(g = 10\text{ m/s}^2\) is used) [1]
Question 10 · Structured Theory
10 marks
A student investigates the physical principles of temperature measurement and thermal processes.
(a) Liquid-in-glass thermometers rely on the thermal expansion of a liquid.
(i) Explain, in terms of molecules, why a liquid expands when its temperature increases. [2]
(ii) State two design changes that would make a liquid-in-glass thermometer more sensitive to small temperature changes. [2]
(b) The student places a beaker of hot water on a laboratory bench.
(i) Explain, in terms of particles, how the process of evaporation causes the temperature of the remaining water in the beaker to decrease. [3]
(ii) State three ways to increase the rate of evaporation from this beaker of water. [3]
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Worked solution
(a)(i) As temperature increases, the molecules gain kinetic energy and move faster / vibrate more vigorously. This causes them to push each other further apart, increasing the average distance between molecules, resulting in thermal expansion.
(a)(ii) 1. Use a narrower capillary tube (bore) so that the liquid column moves a greater distance per degree. 2. Use a larger bulb, so there is a greater volume of liquid to expand.
(b)(i) Evaporation occurs at the surface of the liquid, where molecules with the highest kinetic energy have enough energy to overcome intermolecular forces and escape as gas. This leaves behind molecules with a lower average kinetic energy, which causes the temperature of the remaining liquid to decrease.
(b)(ii) 1. Increase the surface area of the liquid (e.g., pour into a wider container). 2. Increase the temperature of the water. 3. Increase air movement (wind) over the surface of the liquid.
Marking scheme
(a)(i) - Molecules gain kinetic energy / move faster [1] - Molecules move further apart / average spacing between molecules increases [1]
(b)(i) - Molecules with higher kinetic energy / faster-moving molecules [1] - Escape from the surface of the liquid [1] - Lower average kinetic energy of remaining molecules leads to lower temperature [1]
(b)(ii) - Any three from: 1. Increase surface area of the water [1] 2. Increase the temperature of the water [1] 3. Increase draft / air current over the water surface [1] 4. Decrease humidity of the air [1]
Question 11 · Structured Theory
10 marks
Sound waves and light waves show different types of wave behaviour.
(a) Sound is a longitudinal wave.
(i) Describe how a longitudinal wave differs from a transverse wave in terms of the direction of vibration relative to the direction of wave travel. [2]
(ii) State the approximate range of frequencies audible to a healthy human ear. [1]
(b) An offshore marine survey ship uses ultrasound of frequency \(40\text{ kHz}\) to measure the depth of the sea.
(i) Explain, using a calculation, why this ultrasound is inaudible to humans. [1]
(ii) The speed of ultrasound in water is \(1500\text{ m/s}\). Calculate the wavelength of this ultrasound in water. [2]
(iii) A pulse of ultrasound is emitted from the ship towards the seabed. The echo is detected back at the ship \(0.80\text{ s}\) later. Calculate the depth of the sea. [2]
(c) As sound waves pass from deep water into shallow water, they slow down. State what is meant by the term *refraction*. [2]
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Worked solution
(a)(i) In longitudinal waves, the vibrations of the particles are parallel to the direction of energy transfer / wave travel. In transverse waves, the vibrations are perpendicular to the direction of energy transfer.
(a)(ii) \(20\text{ Hz}\) to \(20\,000\text{ Hz}\) (or \(20\text{ kHz}\))
(b)(i) \(40\text{ kHz} = 40\,000\text{ Hz}\). This frequency is greater than the upper limit of human hearing, which is \(20\,000\text{ Hz}\).
(b)(iii) The total distance travelled by the pulse is \(d_{\text{total}} = v \times t = 1500 \times 0.80 = 1200\text{ m}\). Since the echo travels down and back, the depth is \(d = \frac{1200}{2} = 600\text{ m}\).
(c) Refraction is the bending / change in direction of a wave when it crosses a boundary between two different media, caused by a change in wave speed.
Marking scheme
(a)(i) - Longitudinal wave vibrations are parallel to wave travel [1] - Transverse wave vibrations are perpendicular to wave travel [1]
(b)(i) - Conversion: \(40\text{ kHz} = 40\,000\text{ Hz}\) AND comparison: this is higher than the maximum human audible limit of \(20\,000\text{ Hz}\) [1]
(b)(ii) - Correct formula \(v = f\lambda\) or rearrangement \(\lambda = \frac{v}{f}\) [1] - Correct answer: \(0.0375\text{ m}\) (or \(3.75\text{ cm}\)) with unit [1]
(b)(iii) - Use of \(d = \frac{v \times t}{2}\) or recognition that the time to the seabed is half of total time (\(0.40\text{ s}\)) [1] - Correct depth: \(600\text{ m}\) with unit [1]
(c) - Bending / change of direction of a wave [1] - Caused by change in wave speed / entering a different medium [1]
Question 12 · Structured Theory
10 marks
A student sets up a potential divider circuit to monitor the temperature in a laboratory.
A battery with an electromotive force (e.m.f.) of \(9.0\text{ V}\) is connected in series with a fixed resistor of resistance \(12\,\Omega\) and a thermistor.
(a) (i) Define *electromotive force (e.m.f.)*. [2]
(ii) At room temperature, the resistance of the thermistor is \(18\,\Omega\). Calculate the total resistance of the circuit. [1]
(iii) Calculate the current in the circuit. [2]
(iv) Calculate the potential difference (p.d.) across the thermistor. [2]
(b) The room temperature increases.
(i) Explain what happens to the potential difference across the thermistor as the temperature increases. [3]
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Worked solution
(a)(i) Electromotive force (e.m.f.) is the work done / electrical energy transferred by a source per unit charge in driving charge round a complete circuit.
(b)(i) As temperature increases, the resistance of the thermistor decreases. This reduces the fraction of the total resistance represented by the thermistor. Consequently, the potential difference across the thermistor decreases.
Marking scheme
(a)(i) - Energy transferred / work done per unit charge [1] - In driving charge round a complete circuit [1]
(a)(ii) - \(30\,\Omega\) [1]
(a)(iii) - Use of \(I = \frac{V}{R}\) [1] - Correct current: \(0.30\text{ A}\) [1]
(a)(iv) - Use of \(V = I \times R\) or potential divider formula [1] - Correct p.d.: \(5.4\text{ V}\) [1]
(b)(i) - Resistance of the thermistor decreases as temperature increases [1] - Thermistor takes up a smaller proportion of the total circuit resistance [1] - Therefore, potential difference across the thermistor decreases [1]
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