Cambridge IGCSE · thinka-original Practice Paper

2024 Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) Practice Paper with Answers

Thinka Jun 2024 (V2) Cambridge IGCSE-Style Mock — Sciences - Co-ordinated (Double) (0654)

120 marks120 mins2024

An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V2) Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.

Section 1: Biology

Answer all questions in this section. Questions cover cell processes, transport, physiology, genetics, and ecology.

4 Question · 40 marks

Question 1 · Structured Recall

10 marks

A student accidentally touches a sharp thorn on a rose plant and immediately pulls their hand away.

(a) (i) State the name of this rapid, automatic response. [1]

(ii) Complete the sequence to show the pathway of this response by identifying the missing components.

Stimulus → .................... → sensory neurone → .................... → motor neurone → effector → response [2]

(b) This experience causes the student's heart rate to increase due to the release of a hormone.

(i) Identify this hormone. [1]

(ii) State two other physiological effects of this hormone on the body. [2]

(c) The body regulates blood glucose concentration using hormones.

(i) State the name of the organ that secretes insulin and glucagon. [1]

(ii) Explain how glucagon acts to increase blood glucose concentration when it falls too low. [3]

Show answer & marking scheme

Worked solution

(a) (i) The rapid, automatic response is a reflex action.
(ii) The pathway is: Stimulus → receptor → sensory neurone → relay neurone → motor neurone → effector → response.
(b) (i) The hormone secreted is adrenaline.
(ii) Other effects of adrenaline include increased breathing rate, dilation of pupils, increased blood pressure, and conversion of stored glycogen to glucose in the liver.
(c) (i) The pancreas secretes both insulin and glucagon.
(ii) When blood glucose is low, the pancreas secretes glucagon into the bloodstream. Glucagon travels to the liver, where it binds to target cells and stimulates the breakdown of glycogen into glucose, which is then released back into the blood to raise its concentration.

Marking scheme

(a) (i) reflex (action) [1]
(ii) receptor [1], relay neurone (accept spinal cord / CNS) [1]
(b) (i) adrenaline [1]
(ii) any two from: increased breathing rate / deeper breathing; widening/dilation of pupils; glucose released from liver / glycogen to glucose; blood redirected to muscles [2]
(c) (i) pancreas [1]
(ii) glucagon secreted into the blood [1]; travels to/acts on the liver [1]; stimulates conversion of glycogen to glucose (released into blood) [1]

Question 2 · Structured Recall

10 marks

A gardener grows strawberry plants. Strawberries can reproduce sexually by producing flowers that develop into seeds, and asexually by producing runners.

(a) State one structural difference between insect-pollinated flowers and wind-pollinated flowers. [1]

(b) Runners are stems that grow along the ground and form new, genetically identical plants where they touch the soil.

(i) State the type of nuclear division that produces these genetically identical plants. [1]

(ii) State one advantage to the gardener of strawberry plants reproducing asexually rather than sexually. [1]

(iii) State two disadvantages of asexual reproduction in a wild population of plants. [2]

(c) When sexual reproduction occurs, fertilisation takes place.

(i) Define the term fertilisation. [2]

(ii) Describe the pathway of the male gamete from pollination until fertilisation occurs in the flower. [3]

Show answer & marking scheme

Worked solution

(a) Insect-pollinated flowers generally have large, brightly coloured petals to attract pollinators and sticky pollen grains, while wind-pollinated flowers have small or absent petals, feathery stigmas, and exposed anthers to catch or release wind-blown pollen.
(b) (i) Mitosis is the type of nuclear division that produces genetically identical offspring.
(ii) Asexual reproduction allows the gardener to produce plants that are clones of the parent, ensuring successful traits such as disease resistance or high-quality fruit are reliably preserved.
(iii) Disadvantages of asexual reproduction include a lack of genetic variation, which means the population cannot adapt quickly to changing environments, and the risk that a single disease could wipe out the entire population.
(c) (i) Fertilisation is defined as the fusion of the haploid nucleus of a male gamete with the haploid nucleus of a female gamete to form a diploid zygote.
(ii) Following pollination, the pollen grain germinates on the stigma. A pollen tube grows down through the style towards the ovary. The male gamete nucleus travels down this pollen tube to enter the ovule, where it fuses with the nucleus of the female gamete (egg cell).

Marking scheme

(a) any correct structural difference, e.g., insect-pollinated have large/coloured petals AND wind-pollinated have small/green petals OR wind-pollinated have feathery stigmas / hanging anthers [1]
(b) (i) mitosis [1]
(ii) preserves desirable characteristics / faster than sexual reproduction / no pollinator needed [1]
(iii) no genetic variation [1]; if environment changes or disease enters, the entire population could be wiped out [1]
(c) (i) fusion of the nuclei [1]; of male and female gametes (to form a zygote) [1]
(ii) pollen grain lands on/germinates on the stigma [1]; pollen tube grows down the style [1]; male gamete travels to the ovule/ovary [1]

Question 3 · Data Interpretation

10 marks

An investigation was carried out to study the effect of temperature on the rate of a reaction catalyzed by the enzyme catalase. Catalase breaks down hydrogen peroxide into water and oxygen gas. Yeast was used as the source of catalase. The volume of oxygen gas produced in 2 minutes was measured at different temperatures. The results are shown in Table 1.1.

Table 1.1
- Temperature 10 °C: Volume of oxygen produced = 4.2 cm³
- Temperature 20 °C: Volume of oxygen produced = 8.5 cm³
- Temperature 30 °C: Volume of oxygen produced = 16.8 cm³
- Temperature 40 °C: Volume of oxygen produced = 25.4 cm³
- Temperature 50 °C: Volume of oxygen produced = 11.2 cm³
- Temperature 60 °C: Volume of oxygen produced = 0.0 cm³

(a) State the independent variable and the dependent variable in this investigation. [2]

(b) Describe the effect of temperature on the activity of catalase as shown by the results in Table 1.1. [3]

(c) Explain the difference in the volume of oxygen produced at 40 °C compared to 60 °C. Refer to the active site of the enzyme in your answer. [4]

(d) State one variable, other than temperature, that must be kept constant in this investigation to ensure valid results. [1]

Show answer & marking scheme

Worked solution

(a) The independent variable is the variable changed by the investigator (Temperature). The dependent variable is the variable measured (Volume of oxygen gas produced in 2 minutes).

(b) According to Table 1.1, the activity of catalase increases as temperature rises from 10 °C to 40 °C, reaching a maximum rate of 25.4 cm³ at 40 °C. Beyond 40 °C, the rate decreases rapidly, reaching 0.0 cm³ at 60 °C.

(c) At 40 °C, the molecules have high kinetic energy, resulting in frequent successful collisions between catalase and hydrogen peroxide, forming enzyme-substrate complexes. At 60 °C, the temperature is too high, causing the catalase enzyme to denature. The active site permanently changes shape, meaning the hydrogen peroxide substrate can no longer fit into the active site, resulting in no reaction.

(d) Variables to keep constant include the volume/concentration of yeast solution, volume/concentration of hydrogen peroxide, or the pH of the mixture.

Marking scheme

(a) [2 marks]
- 1 mark for identifying the independent variable as temperature.
- 1 mark for identifying the dependent variable as the volume of oxygen gas produced.

(b) [3 marks]
- 1 mark for stating that activity increases as temperature increases from 10 °C up to 40 °C.
- 1 mark for identifying 40 °C as the optimum temperature (or temperature with maximum volume of 25.4 cm³).
- 1 mark for stating that activity decreases above 40 °C / drops to zero at 60 °C.

(c) [4 marks]
- 1 mark for explaining that high temperature (40 °C) increases kinetic energy of molecules, leading to more successful collisions.
- 1 mark for stating that at 60 °C the enzyme is denatured.
- 1 mark for explaining that denaturation changes the shape of the active site.
- 1 mark for stating that the substrate can no longer bind to the active site / no enzyme-substrate complexes can form.

(d) [1 mark]
- 1 mark for any correct constant variable: volume of yeast, concentration of yeast, volume of hydrogen peroxide, concentration of hydrogen peroxide, or pH. Reject: time (as it is already specified as 2 minutes in the procedure).

Question 4 · Data Interpretation

10 marks

A group of five students investigated the effect of consuming a caffeine-containing beverage on their reaction times. The reaction time of each student was measured before drinking the beverage and again 30 minutes after drinking it. The results are shown in Table 2.1.

Table 2.1
- Student A: Before = 285 ms, After = 245 ms
- Student B: Before = 310 ms, After = 275 ms
- Student C: Before = 255 ms, After = 215 ms
- Student D: Before = 295 ms, After = 250 ms
- Student E: Before = 275 ms, After = 235 ms

(a) Calculate the mean reaction time of the students after consuming caffeine. Show your working. [2]

(b) Describe the effect of caffeine on the students' reaction times using the data in Table 2.1. [2]

(c) Reaction time is a measure of how quickly a person can respond to a stimulus. Name the components of the nervous system involved in a reflex arc, in the correct order, starting from the receptor. [4]

(d) Suggest two reasons why measuring the reaction times of five students provides more reliable data than measuring the reaction time of a single student. [2]

Show answer & marking scheme

Worked solution

(a) To calculate the mean reaction time after consuming caffeine:
Sum of reaction times = $245 + 275 + 215 + 250 + 235 = 1220$ ms.
Mean = $1220 / 5 = 244$ ms.

(b) Caffeine decreases the reaction time of the students. Every single student (A through E) showed a lower reaction time after consuming caffeine compared to before, with the overall mean dropping from 284 ms to 244 ms.

(c) The pathway of a reflex arc starts with a receptor which detects the stimulus, then sends an electrical impulse along the sensory neurone to the central nervous system (spinal cord/brain), where it crosses a synapse to a relay neurone, and then passes along a motor neurone to the effector (muscle or gland) to produce a response.

(d) Using a larger sample size of five students helps to minimize the effect of anomalous results (outliers) and allows a representative mean to be calculated, increasing the reliability of the conclusion.

Marking scheme

(a) [2 marks]
- 1 mark for correct working showing sum of 'after' values (1220 ms).
- 1 mark for correct calculation of the mean (244 ms).

(b) [2 marks]
- 1 mark for stating that caffeine decreases reaction time / speeds up responses.
- 1 mark for supporting this statement with data (e.g., comparing the mean before of 284 ms to the mean after of 244 ms, or showing that every individual student's reaction time decreased).

(c) [4 marks]
- 1 mark for Receptor / Sensory neurone.
- 1 mark for Central Nervous System (CNS) / spinal cord / brain.
- 1 mark for Relay neurone / synapse.
- 1 mark for Motor neurone / Effector (muscle or gland).

(d) [2 marks]
- 1 mark for identifying/excluding anomalous results.
- 1 mark for allowing a mean/average to be calculated to reduce individual variation.

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Practice This Topic

Section 2: Chemistry

Answer all questions in this section. Questions cover physical, analytical, and organic chemistry, including calculations.

4 Question · 40 marks

Question 1 · theory

10 marks

A student investigates the thermal decomposition of calcium carbonate, $CaCO_3$.

The equation for the reaction is:
\[CaCO_3(s) \rightarrow CaO(s) + CO_2(g)\]

(a) State the type of chemical reaction that occurs when calcium carbonate is heated to form calcium oxide and carbon dioxide. [1]

(b) Calculate the relative formula mass ($M_r$) of calcium carbonate.
[Relative atomic masses: $Ca = 40$, $C = 12$, $O = 16$] [1]

(c) The student heats 15.0g of calcium carbonate.
(i) Calculate the number of moles of calcium carbonate in 15.0g. [2]
(ii) Show that the maximum mass of calcium oxide, $CaO$, that can be produced is 8.4g. [2]

(d) Calculate the volume of carbon dioxide, $CO_2$, in $dm^3$, produced at room temperature and pressure (r.t.p.) when 15.0g of calcium carbonate is completely decomposed.
[The volume of one mole of any gas is $24\text{ }dm^3$ at r.t.p.] [2]

(e) In practice, the student only obtains 7.1g of calcium oxide. Calculate the percentage yield of calcium oxide. [2]

Show answer & marking scheme

Worked solution

(a) Thermal decomposition.

(b) $M_r(CaCO_3) = 40 + 12 + (3 \times 16) = 100$.

(c) (i) $\text{Moles} = \frac{\text{mass}}{\text{M}_r} = \frac{15.0}{100} = 0.15\text{ mol}$.
(ii) $M_r(CaO) = 40 + 16 = 56$.
$\text{Mass of CaO} = 0.15\text{ mol} \times 56\text{ g/mol} = 8.4\text{ g}$.

(d) Since the molar ratio of $CaCO_3$ to $CO_2$ is 1:1, 0.15 mol of $CO_2$ is produced.
$\text{Volume} = 0.15\text{ mol} \times 24\text{ dm}^3\text{/mol} = 3.6\text{ dm}^3$.

(e) $\text{Percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% = \frac{7.1}{8.4} \times 100\% = 84.5\%$ (accept 85%).

Marking scheme

(a) thermal decomposition [1]

(b) 100 [1]

(c) (i) $15.0 / 100$ [1]; 0.15 (mol) [1]
(ii) $M_r\text{ of } CaO = 56$ [1]; $0.15 \times 56 = 8.4\text{ (g)}$ [1]

(d) identifies 1:1 ratio / moles of $CO_2$ is 0.15 [1]; $0.15 \times 24 = 3.6\text{ (dm}^3\text{)}$ [1]

(e) $(7.1 / 8.4) \times 100$ [1]; 84.5% (accept 85%) [1]

Question 2 · theory

10 marks

A student investigates the reaction between dilute hydrochloric acid and aqueous sodium hydroxide.

The chemical equation for this reaction is:
\[HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)\]

The reaction is exothermic.

(a) Define the term exothermic reaction. [1]

(b) Explain, in terms of bond breaking and bond forming, why this reaction is exothermic. [3]

(c) The student mixes 50 $cm^3$ of $1.0\text{ }mol/dm^3$ hydrochloric acid with 50 $cm^3$ of $1.0\text{ }mol/dm^3$ sodium hydroxide solution.
(i) Calculate the number of moles of hydrochloric acid used in this reaction. [2]
(ii) The temperature of the mixture increases by $6.8\text{ }^\circ\text{C}$.
Calculate the energy released in this reaction in joules.
Use the equation:
\[Q = m \times c \times \Delta T\]
where:
- $m$ is the total mass of the mixture (assume 100g)
- $c$ is the specific heat capacity (assume $4.2\text{ J/(g }^\circ\text{C)}$)
- $\Delta T$ is the temperature rise [2]
(iii) Calculate the enthalpy change ($\Delta H$) for the reaction in $kJ/mol$ of water formed. [2]

Show answer & marking scheme

Worked solution

(a) An exothermic reaction is a chemical reaction that releases heat energy to its surroundings.

(b) Bond breaking is endothermic (takes in energy) and bond making is exothermic (releases energy). In this reaction, more energy is released during the formation of new bonds in the products than is taken in during the breaking of existing bonds in the reactants, resulting in a net release of heat energy.

(c) (i) $\text{Moles of } HCl = \frac{\text{volume}}{1000} \times \text{concentration} = \frac{50}{1000} \times 1.0 = 0.05\text{ mol}$.
(ii) $Q = 100\text{ g} \times 4.2\text{ J/(g }^\circ\text{C)} \times 6.8\text{ }^\circ\text{C} = 2856\text{ J}$ (or $2.856\text{ kJ}$).
(iii) Moles of water formed = 0.05 mol.
$\Delta H = -\frac{Q}{\text{moles}} = -\frac{2.856\text{ kJ}}{0.05\text{ mol}} = -57.12\text{ kJ/mol}$ (accept $-57.1\text{ kJ/mol}$ or positive value if magnitude is correct).

Marking scheme

(a) reaction that releases heat/thermal energy (to the surroundings) [1]

(b) bond breaking is endothermic / takes in energy AND bond making is exothermic / releases energy [1]; energy released in making bonds is greater than energy taken in to break bonds [1]; net release of energy [1]

(c) (i) $50 / 1000 \times 1.0$ [1]; 0.05 (mol) [1]
(ii) $100 \times 4.2 \times 6.8$ [1]; 2856 (J) [1]
(iii) divides energy by 1000 to convert to kJ (2.856 kJ) [1]; division by 0.05 to get -57.12 kJ/mol (accept -57.1 or -57 or positive equivalent) [1]

Question 3 · structured

10 marks

A student investigates the thermal decomposition of zinc carbonate, $\text{ZnCO}_3$. The equation for the reaction is:

$$\text{ZnCO}_3(\text{s}) \rightarrow \text{ZnO}(\text{s}) + \text{CO}_2(\text{g})$$

**(a)** Calculate the relative formula mass ($M_r$) of zinc carbonate, $\text{ZnCO}_3$.
[Relative atomic masses, $A_r$: $\text{Zn} = 65$, $\text{C} = 12$, $\text{O} = 16$]

..................................................................................................................................... [2]

**(b)** Describe a chemical test to confirm that the gas produced is carbon dioxide. State the positive result of the test.

test .............................................................................................................................

result ............................................................................................................................. [2]

**(c)** Calculate the maximum volume of carbon dioxide gas, in $\text{dm}^3$, produced at r.t.p. when $5.00\text{ g}$ of zinc carbonate is completely decomposed.
[The volume of one mole of any gas is $24.0\text{ dm}^3$ at room temperature and pressure (r.t.p.).]
Show your working.

volume = ................................... $\text{dm}^3$ [3]

**(d)** State the name of this type of chemical reaction where a substance is broken down by heat.

..................................................................................................................................... [1]

**(e)** Zinc oxide, $\text{ZnO}$, is heated with carbon to extract zinc.
Write the word equation for this reaction.

..................................................................................................................................... [2]

Show answer & marking scheme

Worked solution

**(a)**
$M_r(\text{ZnCO}_3) = 65 + 12 + (3 \times 16) = 125$

**(b)**
Test: Bubble the gas through limewater.
Result: Limewater turns cloudy / milky.

**(c)**
Number of moles of $\text{ZnCO}_3 = \frac{5.00}{125} = 0.040\text{ mol}$
Number of moles of $\text{CO}_2$ produced = $0.040\text{ mol}$
Volume of $\text{CO}_2 = 0.040 \times 24 = 0.96\text{ dm}^3$

**(d)**
Thermal decomposition

**(e)**
Zinc oxide + carbon $\rightarrow$ zinc + carbon dioxide (or carbon monoxide)

Marking scheme

**(a)**
- Correct substitution: $65 + 12 + 3(16)$ [1]
- Correct calculation: $125$ [1]

**(b)**
- Test: (bubble through) limewater [1]
- Result: turns milky / cloudy / white precipitate [1]

**(c)**
- Moles of zinc carbonate = $0.040\text{ mol}$ [1]
- Volume = moles $\times 24$ (or $0.040 \times 24$) [1]
- Correct final answer: $0.96\text{ dm}^3$ (allow ecf from incorrect $M_r$) [1]

**(d)**
- Thermal decomposition [1]
- Reject: decomposition on its own

**(e)**
- Reactants: zinc oxide + carbon [1]
- Products: zinc + carbon dioxide (or carbon monoxide) [1]

Question 4 · structured

10 marks

Aqueous copper(II) chloride is electrolysed using inert carbon electrodes.

**(a)** State the observations at the:

(i) anode (positive electrode) ............................................................................................................................. [1]

(ii) cathode (negative electrode) ............................................................................................................................. [1]

**(b)** Write the ionic half-equation, including state symbols, for the reaction occurring at the cathode.

..................................................................................................................................... [2]

**(c)** Describe a chemical test to show that the gas produced at the anode is chlorine. State the positive result.

test .............................................................................................................................

result ............................................................................................................................. [2]

**(d)** State what happens to the blue color of the copper(II) chloride solution as the electrolysis continues. Explain your answer.

observation .............................................................................................................................

explanation ............................................................................................................................. [2]

**(e)** Suggest a suitable material for the inert electrodes.

..................................................................................................................................... [1]

**(f)** State whether the reaction at the cathode is oxidation or reduction. Give a reason for your answer in terms of electron transfer.

..................................................................................................................................... [1]

Show answer & marking scheme

Worked solution

**(a)**
(i) Bubbles of a green/yellow gas.
(ii) Pink/brown solid/deposit forms.

**(b)**
$\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightarrow \text{Cu}(\text{s})$

**(c)**
Test: Damp blue litmus paper.
Result: Paper turns red and then bleaches (white).

**(d)**
Observation: Blue color fades / becomes lighter / colorless.
Explanation: Copper(II) ions are removed from the solution as they are discharged at the cathode.

**(e)**
Graphite (or platinum / carbon).

**(f)**
Reduction, because copper(II) ions gain electrons.

Marking scheme

**(a)**
(i) Bubbles / green-yellow gas [1]
(ii) Red-brown / pink / brown solid deposited [1]

**(b)**
- Correct formulas: $\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$ [1]
- Correct state symbols: $\text{aq}$ and $\text{s}$ [1]

**(c)**
- Test: Damp blue litmus paper / damp indicator paper [1]
- Result: Bleached / turns white (allow red then white) [1]

**(d)**
- Observation: Blue color fades / becomes paler [1]
- Explanation: Concentration of copper(II) ions decreases / copper(II) ions are removed [1]

**(e)**
- Graphite / platinum / carbon [1]

**(f)**
- Reduction because electrons are gained / copper ions gain electrons [1]

Section 3: Physics

Answer all questions in this section. Questions cover forces, electricity, electromagnetism, thermal, and nuclear physics.

4 Question · 40 marks

Question 1 · Mathematical Calculations

10 marks

An electric delivery drone has a mass of $4.5\text{ kg}$. It is used to lift a package with a mass of $1.5\text{ kg}$. The gravitational force on unit mass, $g$, is $10\text{ N/kg}$. (a) Calculate the total weight of the drone and the package. (b) The drone accelerates vertically upwards from rest with a constant acceleration of $1.5\text{ m/s}^2$. (i) Calculate the resultant force acting on the drone and the package. (ii) Calculate the upward lift force produced by the rotors during this acceleration. (c) The drone then climbs to a height of $45\text{ m}$ at a constant speed of $4.0\text{ m/s}$. (i) Calculate the work done against gravity to lift the drone and the package. (ii) Calculate the minimum power required to perform this climb.

Show answer & marking scheme

Worked solution

Detailed step-by-step calculations: (a) Total mass $m = 4.5\text{ kg} + 1.5\text{ kg} = 6.0\text{ kg}$. Weight $W = m \times g = 6.0 \times 10 = 60\text{ N}$. (b)(i) Resultant force $F = m \times a = 6.0 \times 1.5 = 9.0\text{ N}$. (b)(ii) Upward force minus downward weight equals the resultant force: $F_{\text{lift}} - W = F \implies F_{\text{lift}} = 60 + 9.0 = 69\text{ N}$. (c)(i) Work done $W = F \times d = 60 \times 45 = 2700\text{ J}$. (c)(ii) Time taken $t = d / v = 45 / 4.0 = 11.25\text{ s}$. Power $P = W / t = 2700 / 11.25 = 240\text{ W}$. Alternatively, using $P = F \times v = 60 \times 4.0 = 240\text{ W}$.

Marking scheme

(a) [2 marks]: 1 mark for correct mass addition and weight formula substitution ($6.0 \times 10$), 1 mark for correct weight of $60\text{ N}$. (b)(i) [2 marks]: 1 mark for correct substitution into $F=ma$ ($6.0 \times 1.5$), 1 mark for correct answer of $9.0\text{ N}$. (b)(ii) [2 marks]: 1 mark for summing weight and resultant force ($60 + 9.0$), 1 mark for correct answer of $69\text{ N}$. (c)(i) [2 marks]: 1 mark for correct substitution into work formula ($60 \times 45$), 1 mark for correct answer of $2700\text{ J}$. (c)(ii) [2 marks]: 1 mark for correct power formula or calculation steps ($P = F \times v = 60 \times 4.0$ or $P = 2700 / 11.25$), 1 mark for correct answer of $240\text{ W}$. Allow error carried forward (ecf) from previous parts.

Question 2 · Mathematical Calculations

10 marks

A student investigates a heating element designed for a small portable shower. The element is connected to a $230\text{ V}$ mains electrical supply. (a) The heating element has a resistance of $11.5\ \Omega$. (i) Calculate the electric current in the heating element when it is operating normally. (ii) Calculate the power of the heating element. (b) Calculate the electrical energy transferred by the heater in $5.0\text{ minutes}$ of normal operation. (c) The circuit contains a fuse to protect the unit from excessive currents. (i) Explain why a $13\text{ A}$ fuse is not suitable for this circuit. (ii) Suggest a suitable rating for the fuse in this circuit, and explain your choice.

Show answer & marking scheme

Worked solution

Detailed step-by-step calculations: (a)(i) Current $I = V / R = 230 / 11.5 = 20\text{ A}$. (a)(ii) Power $P = V \times I = 230 \times 20 = 4600\text{ W}$ (or $4.6\text{ kW}$). (b) Convert minutes to seconds: $t = 5.0 \times 60 = 300\text{ s}$. Energy $E = P \times t = 4600 \times 300 = 1,380,000\text{ J}$ (or $1.38\text{ MJ}$ or $1.38 \times 10^6\text{ J}$). (c)(i) The normal operating current of $20\text{ A}$ exceeds $13\text{ A}$. The fuse would melt/blow instantly during normal operation. (c)(ii) A suitable fuse rating is $25\text{ A}$ or $30\text{ A}$ because it is the next standard rating above $20\text{ A}$, allowing normal current flow while still protecting against faults.

Marking scheme

(a)(i) [2 marks]: 1 mark for correct formula or substitution ($230 / 11.5$), 1 mark for correct answer of $20\text{ A}$. (a)(ii) [2 marks]: 1 mark for correct formula or substitution ($230 \times 20$), 1 mark for correct answer of $4600\text{ W}$ (or $4.6\text{ kW}$). (b) [2 marks]: 1 mark for converting time to seconds ($300\text{ s}$), 1 mark for correct energy calculation with unit ($1.38 \times 10^6\text{ J}$ or $1,380,000\text{ J}$). (c)(i) [2 marks]: 1 mark for stating that the operating current is greater than $13\text{ A}$, 1 mark for explaining that the fuse will melt/blow. (c)(ii) [2 marks]: 1 mark for recommending a sensible fuse rating above $20\text{ A}$ (e.g. $25\text{ A}$ or $30\text{ A}$), 1 mark for explaining that it allows normal operation but protects against faults.

Question 3 · Recall & Diagrams

10 marks

A student investigates the magnetic field around a straight current-carrying wire.

(a) State the name of the rule used to determine the direction of the magnetic field lines around a straight current-carrying wire. [1]

(b) (i) Describe the pattern of the magnetic field lines observed on a horizontal card through which the vertical wire passes. [1]
(ii) Describe how the spacing of these field lines changes as the distance from the wire increases. [1]

(c) The current in the wire is directed vertically upwards.
(i) State the direction of the magnetic field lines when viewed from directly above the card (either clockwise or anticlockwise). [1]
(ii) Describe a diagram representing the card from above, specifying the position of the wire, the shape of at least three field lines, and the direction of the arrows on these lines. [2]

(d) The wire is part of a circuit containing a variable resistor.
(i) State and explain what happens to the strength of the magnetic field if the resistance of the variable resistor is decreased. [2]
(ii) The direct current (d.c.) power supply is replaced with an alternating current (a.c.) power supply of low frequency. Describe the effect of this change on the magnetic field. [2]

Show answer & marking scheme

Worked solution

(a) Right-hand grip rule (or right-hand screw rule).

(b) (i) Concentric circles centred on the wire.
(ii) The spacing between the lines increases as the distance from the wire increases.

(c) (i) Anticlockwise.
(ii) A central dot representing the wire with three concentric circles around it, each with arrows pointing in an anticlockwise direction.

(d) (i) The strength of the magnetic field increases because decreasing the resistance increases the current flowing through the wire.
(ii) The magnetic field constantly reverses its direction (oscillates) at the same frequency as the alternating current supply.

Marking scheme

(a) Right-hand grip rule / right-hand screw rule [1]
(b) (i) Concentric circles / circles centred on the wire [1]
(ii) Spacing increases / lines get further apart [1]
(c) (i) Anticlockwise [1]
(ii) Concentric circles around a central point [1]; arrows pointing anticlockwise on all circles [1]
(d) (i) Strength increases [1] because current increases [1]
(ii) Field reverses / changes direction continuously [1]; at the same frequency as the a.c. supply [1]

Question 4 · Recall & Diagrams

10 marks

A student is designing a liquid-in-glass thermometer.

(a) State the physical property of the liquid that is used to measure temperature in this type of thermometer. [1]

(b) The student calibrates the thermometer using the Celsius scale.
(i) Describe how the fixed points of 0 °C and 100 °C are determined. [2]
(ii) Explain why the tube must have a uniform bore (constant inner diameter) between the fixed points. [1]

(c) Define the following terms in relation to a thermometer:
(i) sensitivity [1]
(ii) range [1]

(d) Explain how each of the following design changes affects the sensitivity of the thermometer:
(i) using a larger thermometer bulb filled with more liquid [2]
(ii) using a capillary tube with a narrower bore [2]

Show answer & marking scheme

Worked solution

(a) Thermal expansion of the liquid (change in volume with temperature).

(b) (i) The lower fixed point (0 °C) is found by placing the thermometer in pure melting ice. The upper fixed point (100 °C) is found by placing the thermometer in steam above boiling water at standard atmospheric pressure.
(ii) A uniform bore ensures that the liquid expands linearly, so that equal temperature rises produce equal changes in the height of the liquid column, allowing for an evenly-spaced scale.

(c) (i) Sensitivity is the change in height of the liquid column per unit change in temperature.
(ii) Range is the difference between the maximum and minimum temperatures that can be measured by the thermometer.

(d) (i) The sensitivity increases. A larger bulb contains more liquid, which produces a larger total volume expansion for the same temperature change.
(ii) The sensitivity increases. For the same change in liquid volume, the liquid must move a longer distance along a tube with a smaller cross-sectional area.

Marking scheme

(a) Expansion of the liquid / volume change [1]
(b) (i) Place in pure melting ice for 0 °C [1]; place in steam above boiling water (at standard atmospheric pressure) for 100 °C [1]
(ii) For linear expansion / equal spacing of degree marks [1]
(c) (i) Distance moved by the liquid per unit temperature change [1]
(ii) Difference between maximum and minimum temperature measurable [1]
(d) (i) Sensitivity increases [1]; because more liquid expands more (greater volume change) [1]
(ii) Sensitivity increases [1]; because a given volume change causes a greater change in height/length along a narrower tube [1]

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