An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V1) Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.
Paper 21 Multiple Choice
Answer all forty multiple choice questions. Choose the single best answer A, B, C or D.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
During human reproduction, where does fertilisation usually occur, and what is the correct term for the resulting fertilised egg cell?
A.in the oviduct, forming a zygote
B.in the oviduct, forming an embryo
C.in the uterus, forming a zygote
D.in the uterus, forming an embryo
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Worked solution
Fertilisation typically occurs in the oviduct (also known as the Fallopian tube). The fusion of the male and female gametes produces a fertilised egg cell called a zygote.
Marking scheme
1 mark for correct option. Correct answer is A.
Question 2 · multiple-choice
1 marks
An enzyme-catalysed reaction is investigated at different temperatures. At \(60^\circ\text{C}\), the rate of reaction is found to be zero. Which statement explains this observation?
A.The kinetic energy of the enzyme molecules has decreased to zero.
B.The activation energy of the reaction has increased.
C.The shape of the active site has changed permanently.
D.The enzyme and substrate molecules are colliding too rapidly.
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Worked solution
At high temperatures, the thermal energy disrupts the bonds maintaining the three-dimensional structure of the enzyme. This permanently alters the shape of its active site, causing denaturation. As a result, the substrate can no longer bind.
Marking scheme
1 mark for correct option. Correct answer is C.
Question 3 · multiple-choice
1 marks
Which statement about the hydrocarbons ethane and ethene is correct?
A.Both ethane and ethene decolourise aqueous bromine.
B.Ethane is saturated whereas ethene is unsaturated.
C.Ethene can be polymerised to form poly(ethane).
D.Ethane contains a carbon-carbon double bond.
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Worked solution
Ethane is an alkane and contains only single carbon-carbon bonds, meaning it is saturated. Ethene is an alkene and contains a carbon-carbon double bond, meaning it is unsaturated.
Marking scheme
1 mark for correct option. Correct answer is B.
Question 4 · multiple-choice
1 marks
An acid can be defined as a proton donor and a base as a proton acceptor. Which substance acts as a base when dissolved in water?
A.\(\text{HCl}\)
B.\(\text{HNO}_3\)
C.\(\text{NH}_3\)
D.\(\text{H}_2\text{SO}_4\)
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Worked solution
Ammonia (\(\text{NH}_3\)) dissolves in water and accepts a proton (\(\text{H}^+\)) from water molecules to form ammonium ions (\(\text{NH}_4^+\)) and hydroxide ions (\(\text{OH}^-\)). It therefore acts as a proton acceptor (base).
Marking scheme
1 mark for correct option. Correct answer is C.
Question 5 · multiple-choice
1 marks
A toy car of mass \(2.0\text{ kg}\) accelerates from rest to a speed of \(6.0\text{ m/s}\). What is the kinetic energy of the toy car at this speed?
A.\(6.0\text{ J}\)
B.\(12\text{ J}\)
C.\(18\text{ J}\)
D.\(36\text{ J}\)
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Worked solution
The formula for kinetic energy is \(E_k = \frac{1}{2}mv^2\). Substituting the given values: \(E_k = 0.5 \times 2.0\text{ kg} \times (6.0\text{ m/s})^2 = 36\text{ J}\).
Marking scheme
1 mark for correct option. Correct answer is D.
Question 6 · multiple-choice
1 marks
Two resistors, with resistances of \(3.0\ \Omega\) and \(6.0\ \Omega\), are connected in parallel. What is the combined resistance of this combination?
A.\(2.0\ \Omega\)
B.\(4.5\ \Omega\)
C.\(9.0\ \Omega\)
D.\(18\ \Omega\)
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Worked solution
For resistors in parallel, the formula is \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\). Here, \(\frac{1}{R_p} = \frac{1}{3.0} + \frac{1}{6.0} = \frac{2}{6.0} + \frac{1}{6.0} = \frac{3}{6.0} = \frac{1}{2.0}\). Taking the reciprocal gives \(R_p = 2.0\ \Omega\).
Marking scheme
1 mark for correct option. Correct answer is A.
Question 7 · multiple-choice
1 marks
A ray of light in air enters a rectangular glass block of refractive index \(1.5\) at an angle of incidence of \(45^\circ\). What is the angle of refraction in the glass?
A.\(28^\circ\)
B.\(30^\circ\)
C.\(45^\circ\)
D.\(60^\circ\)
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Worked solution
According to Snell's law, \(n = \frac{\sin(i)}{\sin(r)}\). Rearranging to solve for \(\sin(r)\): \(\sin(r) = \frac{\sin(45^\circ)}{1.5} = \frac{0.7071}{1.5} = 0.4714\). Calculating the inverse sine gives \(r \approx 28^\circ\).
Marking scheme
1 mark for correct option. Correct answer is A.
Question 8 · multiple-choice
1 marks
During the electrolysis of concentrated aqueous sodium chloride using inert electrodes, which products are formed at the anode and the cathode?
A.chlorine at the anode, hydrogen at the cathode
B.oxygen at the anode, sodium at the cathode
C.chlorine at the anode, sodium at the cathode
D.oxygen at the anode, hydrogen at the cathode
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Worked solution
At the anode, chloride ions (\(\text{Cl}^-\)) are discharged preferentially because they are in high concentration, forming chlorine gas. At the cathode, hydrogen ions (\(\text{H}^+\)) are discharged preferentially over sodium ions because hydrogen is less reactive, producing hydrogen gas.
Marking scheme
1 mark for correct option. Correct answer is A.
Question 9 · multiple-choice
1 marks
Which statement correctly describes the placenta?
A.It allows direct mixing of maternal and fetal blood to facilitate rapid diffusion of gases.
B.It acts as a complete barrier that prevents all pathogens and drugs from reaching the fetus.
C.It provides a large surface area for the exchange of oxygen, nutrients, and urea.
D.It secretes progesterone to stimulate the breakdown of the uterine lining during pregnancy.
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Worked solution
The placenta has villi that provide a large surface area for the diffusion of oxygen, nutrients (like glucose and amino acids), and urea. Direct mixing of maternal and fetal blood does not occur to protect the fetus from high pressure and certain pathogens. Some harmful substances, like nicotine, alcohol, and the rubella virus, can cross the placenta. Progesterone maintains the uterine lining rather than breaking it down.
Marking scheme
1 mark for the correct option C.
Question 10 · multiple-choice
1 marks
An enzyme-catalysed reaction is investigated at different pH values. The rate of reaction is highest at pH 2 and decreases rapidly as the pH increases above pH 4. Which statement explains why the rate of reaction decreases above pH 4?
A.The kinetic energy of the enzyme and substrate molecules decreases, reducing the rate of effective collisions.
B.The active site of the enzyme changes shape, so the substrate can no longer fit.
C.The activation energy of the reaction is lowered, preventing the reaction from taking place.
D.The peptide bonds within the enzyme are broken, hydrolysing the enzyme into free amino acids.
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Worked solution
As pH increases above the optimum value, the active site of the enzyme is denatured, changing its shape so the substrate can no longer fit and bind to it. pH changes do not affect the kinetic energy of molecules, which is determined by temperature. pH changes do not lower the activation energy in a way that prevents the reaction. Denaturation changes the three-dimensional conformation of the protein but does not hydrolyse it into free amino acids.
Marking scheme
1 mark for the correct option B.
Question 11 · multiple-choice
1 marks
Aqueous copper(II) sulfate is electrolysed using copper electrodes. Which statement describes what occurs during this electrolysis?
A.At the anode, copper atoms are oxidised to form copper(II) ions, and the anode decreases in mass.
B.At the cathode, copper(II) ions are oxidised to form copper atoms, and the cathode increases in mass.
C.Hydroxide ions are discharged at the anode to produce oxygen gas.
D.Hydrogen ions are discharged at the cathode to produce hydrogen gas.
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Worked solution
When copper electrodes are used, the copper anode dissolves by losing electrons (oxidation): Cu(s) -> Cu2+(aq) + 2e-. This causes the anode to decrease in mass. At the cathode, copper(II) ions gain electrons (reduction): Cu2+(aq) + 2e- -> Cu(s), depositing copper metal. Neither hydroxide nor hydrogen ions are discharged because copper atoms at the anode are more easily oxidised, and copper(II) ions at the cathode are more easily reduced.
Marking scheme
1 mark for the correct option A.
Question 12 · multiple-choice
1 marks
Consider the following reversible reaction in aqueous solution: NH4+ + H2O <=> NH3 + H3O+. Which species act as the proton donor and the proton acceptor in the forward reaction?
A.Proton donor: NH4+ ; Proton acceptor: H2O
B.Proton donor: H2O ; Proton acceptor: NH4+
C.Proton donor: NH3 ; Proton acceptor: H3O+
D.Proton donor: NH4+ ; Proton acceptor: NH3
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Worked solution
An acid is a proton donor and a base is a proton acceptor. In the forward reaction, NH4+ loses a hydrogen ion (proton) to form NH3, so it acts as the proton donor. H2O gains a proton to form H3O+, so it acts as the proton acceptor.
Marking scheme
1 mark for the correct option A.
Question 13 · multiple-choice
1 marks
Which compound can undergo addition polymerisation to form a polymer with the repeating unit -[CH2-CHCl]n- ?
A.chloroethane
B.chloroethene
C.1,2-dichloroethane
D.1,2-dichloroethene
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Worked solution
The polymer shown is poly(chloroethene), commonly known as PVC. It is formed by the addition polymerisation of the unsaturated monomer chloroethene (vinyl chloride), which contains a carbon-carbon double bond (CH2=CHCl). Chloroethane and 1,2-dichloroethane are saturated and cannot polymerise. 1,2-dichloroethene would polymerise to form repeating units of -[CHCl-CHCl]n-.
Marking scheme
1 mark for the correct option B.
Question 14 · multiple-choice
1 marks
An electric motor is used to lift a load of mass 20 kg vertically upwards through a height of 6.0 m in 8.0 s. The motor has an efficiency of 60%. What is the electrical power input to the motor? (Use g = 10 m/s^2)
A.90 W
B.150 W
C.250 W
D.400 W
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Worked solution
First, calculate the useful work done: W = mgh = 20 kg * 10 m/s^2 * 6.0 m = 1200 J. Next, calculate the useful power output: P_out = W / t = 1200 J / 8.0 s = 150 W. Finally, use the efficiency formula to find the electrical power input: Efficiency = (P_out / P_in) * 100% => 60% = 150 W / P_in => P_in = 150 / 0.60 = 250 W.
Marking scheme
1 mark for the correct option C.
Question 15 · multiple-choice
1 marks
A circuit consists of three resistors of resistance R1 = 4.0 Ohm, R2 = 6.0 Ohm, and R3 = 12.0 Ohm connected in parallel across a 12.0 V battery of negligible internal resistance. What is the total current leaving the battery, and the current passing through the 12.0 Ohm resistor?
A.Total current = 6.0 A; Current through 12.0 Ohm resistor = 1.0 A
B.Total current = 6.0 A; Current through 12.0 Ohm resistor = 6.0 A
C.Total current = 1.5 A; Current through 12.0 Ohm resistor = 1.0 A
D.Total current = 1.5 A; Current through 12.0 Ohm resistor = 0.5 A
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Worked solution
In a parallel circuit, the voltage across each resistor is equal to the battery voltage (12.0 V). The current through the 12.0 Ohm resistor is I3 = V / R3 = 12.0 V / 12.0 Ohm = 1.0 A. The currents through the other branches are I1 = 12.0 V / 4.0 Ohm = 3.0 A, and I2 = 12.0 V / 6.0 Ohm = 2.0 A. The total current leaving the battery is the sum of these branch currents: I_total = 3.0 A + 2.0 A + 1.0 A = 6.0 A.
Marking scheme
1 mark for the correct option A.
Question 16 · multiple-choice
1 marks
A ray of light travels from air into a transparent plastic block. The angle of incidence is 45 degrees and the angle of refraction is 28 degrees. The speed of light in a vacuum is 3.0 * 10^8 m/s. What is the speed of light in the plastic block?
A.1.5 * 10^8 m/s
B.2.0 * 10^8 m/s
C.2.3 * 10^8 m/s
D.4.5 * 10^8 m/s
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Worked solution
First, calculate the refractive index n of the plastic: n = sin(i) / sin(r) = sin(45) / sin(28) = 0.7071 / 0.4695 = 1.51. Next, use the relationship between refractive index and wave speed: n = c / v => v = c / n = (3.0 * 10^8 m/s) / 1.51 = 1.99 * 10^8 m/s, which rounds to 2.0 * 10^8 m/s.
Marking scheme
1 mark for the correct option B.
Question 17 · MCQ
1 marks
During the human menstrual cycle, which hormone is responsible for both stimulating the repair and growth of the uterus lining and, at high levels, triggering the release of luteinising hormone (LH) from the pituitary gland?
A.follicle-stimulating hormone (FSH)
B.estrogen
C.progesterone
D.testosterone
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Worked solution
Estrogen is secreted by the developing follicles in the ovary. It stimulates the repair and thickening of the uterine lining (endometrium). As estrogen levels peak, they exert a positive feedback effect on the pituitary gland, triggering a surge in luteinising hormone (LH), which induces ovulation.
Marking scheme
Award 1 mark for selecting B. Rejects other options as FSH stimulates follicle development, progesterone maintains the uterus lining, and testosterone is the male sex hormone.
Question 18 · MCQ
1 marks
With reference to the collision theory, which statement explains why the rate of an enzyme-controlled reaction decreases above its optimum temperature?
A.The kinetic energy of the substrate molecules decreases, reducing the frequency of effective collisions.
B.The enzyme molecules gain kinetic energy and move too quickly to successfully bind to substrate molecules.
C.Thermal energy alters the specific three-dimensional shape of the active site, so the substrate is no longer complementary.
D.The activation energy of the reaction increases, making it harder for the substrate to bind to the active site.
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Worked solution
At temperatures above the optimum, the excessive thermal energy disrupts the weak bonds (like hydrogen bonds) that maintain the enzyme's specific three-dimensional tertiary structure. This causes the enzyme to denature, altering the shape of its active site so that the substrate is no longer complementary and can no longer bind.
Marking scheme
Award 1 mark for identifying that denaturation involves alteration of the active site shape such that it is no longer complementary to the substrate. Reject options suggesting substrate kinetic energy decreases or activation energy increases.
Question 19 · MCQ
1 marks
In the reaction between hydrogen chloride gas and ammonia gas to form ammonium chloride:
Which statement describes the behavior of ammonia, $\text{NH}_3$?
A.It acts as an acid because it accepts a proton.
B.It acts as an acid because it donates a proton.
C.It acts as a base because it accepts a proton.
D.It acts as a base because it donates a proton.
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Worked solution
According to the Brønsted-Lowry theory, an acid is a proton ($H^+$) donor and a base is a proton ($H^+$) acceptor. In this reaction, the $\text{HCl}$ molecule donates a proton to the ammonia molecule ($\text{NH}_3$), converting it to an ammonium ion ($\text{NH}_4^+$). Therefore, ammonia acts as a base because it accepts a proton.
Marking scheme
Award 1 mark for selecting C. Correctly identifies base behaviour as proton acceptance.
Question 20 · MCQ
1 marks
Ethene reacts with steam in the presence of a catalyst to produce ethanol. What are the correct conditions for this industrial hydration process?
A.temperature of $100^\circ\text{C}$, pressure of $1\text{ atm}$, yeast catalyst
B.temperature of $300^\circ\text{C}$, pressure of $60\text{ atm}$, phosphoric acid catalyst
C.temperature of $450^\circ\text{C}$, pressure of $200\text{ atm}$, iron catalyst
D.temperature of $200^\circ\text{C}$, pressure of $5\text{ atm}$, nickel catalyst
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Worked solution
The industrial manufacture of ethanol by the catalytic hydration of ethene requires a temperature of $300^\circ\text{C}$, a pressure of $60\text{ atm}$ (or $6000\text{ kPa}$), and a concentrated phosphoric acid catalyst ($\text{H}_3\text{PO}_4$).
Marking scheme
Award 1 mark for identifying the correct industrial conditions (B). Reject options involving yeast fermentation, nickel catalysts (used for hydrogenation), or iron catalysts (used in the Haber process).
Question 21 · MCQ
1 marks
Concentrated aqueous sodium chloride (brine) is electrolysed using inert carbon electrodes. Which ionic half-equation represents the reaction occurring at the anode?
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Worked solution
In the electrolysis of concentrated aqueous $\text{NaCl}$ (brine), the ions present are $\text{Na}^+$, $\text{H}^+$, $\text{Cl}^-$, and $\text{OH}^-$. At the anode (positive electrode), negative ions are attracted. Because it is a concentrated solution, the chloride ions ($\text{Cl}^-$) are preferentially discharged over hydroxide ions ($\text{OH}^-$) to form chlorine gas: $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-$.
Marking scheme
Award 1 mark for selecting A. Reject B and C since they represent cathode reactions (reduction). Reject D because it is the reaction for dilute sodium chloride or sulfate solutions where oxygen is released.
Question 22 · MCQ
1 marks
A tennis ball of mass $0.06\text{ kg}$ travels horizontally at a speed of $25\text{ m/s}$. It hits a vertical wall and rebounds in the opposite direction at a speed of $15\text{ m/s}$. What is the magnitude of the change in momentum (impulse) of the ball?
A.$0.6\text{ kg}\cdot\text{m/s}$
B.$1.0\text{ kg}\cdot\text{m/s}$
C.$1.5\text{ kg}\cdot\text{m/s}$
D.$2.4\text{ kg}\cdot\text{m/s}$
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Worked solution
Taking the initial direction of motion as positive: Initial velocity, $v_i = 25\text{ m/s}$ Final velocity, $v_f = -15\text{ m/s}$ (opposite direction)
Using $\Delta p = m \cdot \Delta v = m(v_f - v_i)$: $\Delta p = 0.06 \cdot (-15 - 25) = 0.06 \cdot (-40) = -2.4\text{ kg}\cdot\text{m/s}$
The magnitude of the change in momentum is $2.4\text{ kg}\cdot\text{m/s}$.
Marking scheme
Award 1 mark for the correct calculation showing the opposite signs for velocity resulting in a change of $2.4\text{ kg}\cdot\text{m/s}$ (D).
Question 23 · MCQ
1 marks
An ideal step-down transformer has a primary coil with $1200$ turns and a secondary coil with $150$ turns. The primary voltage is $240\text{ V}$. If the current in the secondary circuit is $4.0\text{ A}$, what is the current in the primary circuit?
A.$0.5\text{ A}$
B.$1.0\text{ A}$
C.$8.0\text{ A}$
D.$32\text{ A}$
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Worked solution
For an ideal transformer, the power input equals the power output ($V_p \cdot I_p = V_s \cdot I_s$). Also, the turns ratio is related to voltage and current by: $\frac{I_p}{I_s} = \frac{N_s}{N_p}$ Substituting the known values: $\frac{I_p}{4.0} = \frac{150}{1200} = \frac{1}{8}$ $I_p = 4.0 \cdot \frac{1}{8} = 0.5\text{ A}$.
Marking scheme
Award 1 mark for the correct application of the transformer equation to find $0.5\text{ A}$ (A).
Question 24 · MCQ
1 marks
A nucleus of bismuth-212 ($^{212}_{83}\text{Bi}$) decays by emitting a $\beta$-particle (electron) to form a nucleus of polonium-212 ($^{212}_{84}\text{Po}$). Which statement about this radioactive decay process is correct?
A.A proton in the bismuth nucleus has changed into a neutron.
B.A neutron in the bismuth nucleus has changed into a proton.
C.The mass number of the nucleus has increased by one.
D.The proton number of the nucleus has decreased by one.
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Worked solution
In $\beta$-decay, a neutron inside the parent nucleus decays into a proton and an electron (the $\beta$-particle). The proton stays in the nucleus, increasing the atomic number by 1 (from 83 to 84), while the total nucleon number (mass number) remains unchanged at 212 because the total number of protons plus neutrons does not change.
Marking scheme
Award 1 mark for B. Correctly identifies the nuclear transition of a neutron into a proton during beta decay.
Question 25 · multiple-choice
1 marks
Which hormone triggers ovulation during the human menstrual cycle, and from which gland is it secreted?
A.estrogen from the ovary
B.LH from the pituitary gland
C.progesterone from the corpus luteum
D.FSH from the ovary
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Worked solution
Luteinising hormone (LH) is secreted by the pituitary gland. A sudden surge in LH levels triggers ovulation, which is the release of a mature egg from the ovary. Estrogen and progesterone are secreted by the ovaries, while follicle-stimulating hormone (FSH) stimulates follicle development.
Marking scheme
B is correct. 1 mark for identifying both the hormone and its source gland.
Question 26 · multiple-choice
1 marks
How does an enzyme increase the rate of a metabolic reaction?
A.by increasing the kinetic energy of the substrate molecules
B.by lowering the activation energy of the reaction
C.by increasing the temperature of the cellular environment
D.by altering the overall energy change of the reaction
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Worked solution
Enzymes act as biological catalysts by lowering the activation energy required for a reaction to occur. They do not increase the temperature of the cell, alter the overall enthalpy change (energy change) of the reaction, or increase the kinetic energy of substrate molecules.
Marking scheme
B is correct. 1 mark for identifying the lowering of activation energy.
Question 27 · multiple-choice
1 marks
A tennis ball of mass 0.15 kg is moving at 20 m/s towards a wall. It hits the wall and rebounds in the opposite direction at 15 m/s. What is the magnitude of the impulse exerted by the wall on the ball?
A.0.75 N s
B.2.25 N s
C.3.00 N s
D.5.25 N s
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Worked solution
Impulse is equal to the change in momentum: Impulse = m * v_f - m * v_i. Taking the direction towards the wall as positive, the initial velocity is +20 m/s and the final velocity is -15 m/s. Change in momentum = 0.15 * (-15 - 20) = 0.15 * (-35) = -5.25 kg m/s. The magnitude of this impulse is 5.25 N s.
Marking scheme
D is correct. 1 mark for the correct calculation of impulse magnitude using opposite signs for the velocities.
Question 28 · multiple-choice
1 marks
Which ester and other product are formed when propanoic acid reacts with ethanol in the presence of an acid catalyst?
A.ethyl propanoate and water
B.propyl ethanoate and water
C.ethyl propanoate and hydrogen
D.propyl ethanoate and hydrogen
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Worked solution
A carboxylic acid and an alcohol react to form an ester and water. Ethanol reacts with propanoic acid to form ethyl propanoate and water. The alcohol provides the alkyl prefix (ethyl) and the carboxylic acid provides the carboxylate suffix (propanoate).
Marking scheme
A is correct. 1 mark for identifying ethyl propanoate and water as the products.
Question 29 · multiple-choice
1 marks
According to the Brønsted-Lowry theory of acids and bases, in which of these equations does water act as a base?
A.H2O + NH3 ⇌ OH- + NH4+
B.H2O + CO3^2- ⇌ OH- + HCO3-
C.HCl + H2O ⇌ H3O+ + Cl-
D.H2O + NH2^- ⇌ OH- + NH3
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Worked solution
According to Brønsted-Lowry theory, a base is a proton (H+) acceptor. In reaction C, H2O accepts a proton from HCl to form H3O+, therefore acting as a base. In all other options, H2O donates a proton to form OH-, meaning it acts as an acid.
Marking scheme
C is correct. 1 mark for identifying the reaction in which water accepts a proton.
Question 30 · multiple-choice
1 marks
Two 6.0 Ω resistors are connected in parallel with each other. This parallel combination is then connected in series with a 3.0 Ω resistor and a battery. What is the total combined resistance of this circuit network?
A.1.5 Ω
B.3.0 Ω
C.6.0 Ω
D.15.0 Ω
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Worked solution
For the parallel part, 1/Rp = 1/6.0 + 1/6.0 = 2/6.0 = 1/3.0, so Rp = 3.0 Ω. For the series connection, R_total = Rp + 3.0 = 3.0 + 3.0 = 6.0 Ω.
Marking scheme
C is correct. 1 mark for calculating the parallel resistance first and then adding the series resistance.
Question 31 · multiple-choice
1 marks
During strenuous exercise, human muscle cells perform anaerobic respiration. Which statement correctly describes the products and consequences of this process?
A.Lactic acid is produced, and an oxygen debt is built up which is repaid by rapid breathing after exercise.
B.Carbon dioxide and ethanol are produced, which build up in the muscles and cause fatigue.
C.Lactic acid is produced, which is immediately transported to the lungs to be exhaled.
D.Carbon dioxide and water are produced, releasing more energy per gram of glucose than aerobic respiration.
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Worked solution
Anaerobic respiration in human muscle cells produces lactic acid, which causes muscle fatigue and builds up an oxygen debt. This oxygen debt is repaid after exercise by deep, rapid breathing, which provides the oxygen needed to break down the lactic acid in the liver.
Marking scheme
A is correct. 1 mark for identifying lactic acid and the subsequent repayment of oxygen debt.
Question 32 · multiple-choice
1 marks
Which products are formed at the anode and the cathode during the electrolysis of concentrated aqueous sodium chloride using inert electrodes?
A.hydrogen at the anode and chlorine at the cathode
B.chlorine at the anode and hydrogen at the cathode
C.oxygen at the anode and sodium at the cathode
D.chlorine at the anode and sodium at the cathode
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Worked solution
Concentrated aqueous sodium chloride contains Na+, Cl-, H+, and OH- ions. At the cathode (negative electrode), H+ ions are discharged in preference to Na+ ions because hydrogen is less reactive, forming hydrogen gas. At the anode (positive electrode), Cl- ions are discharged in preference to OH- because chloride is in a high concentration, forming chlorine gas.
Marking scheme
B is correct. 1 mark for correctly matching the anode product (chlorine) and cathode product (hydrogen).
Question 33 · multiple-choice
1 marks
A tissue sample from a plant, with a cell water potential of \(-500\text{ kPa}\), is placed in a beaker containing a concentrated salt solution with a water potential of \(-900\text{ kPa}\).
Which statement describes the movement of water and the state of the plant cells after some time?
A.Water moves out of the cells, and the cells undergo plasmolysis.
B.Water moves out of the cells, and the cells become turgid.
C.Water moves into the cells, and the cells undergo plasmolysis.
D.Water moves into the cells, and the cells become turgid.
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Worked solution
Water moves from a region of higher water potential (\(-500\text{ kPa}\)) to a region of lower water potential (\(-900\text{ kPa}\)) by osmosis. As water leaves the plant cells, the vacuole shrinks and the cytoplasm pulls away from the cell wall, causing the cells to become plasmolysed (undergo plasmolysis).
Marking scheme
A is correct because water moves down a water potential gradient from \(-500\text{ kPa}\) to \(-900\text{ kPa}\) by osmosis, resulting in plasmolysis. B, C, and D are incorrect as they describe incorrect directions of movement or incorrect cellular states.
Question 34 · multiple-choice
1 marks
Which row correctly identifies the genetic state of the nuclei in the male gamete, the female gamete, and the zygote during sexual reproduction in plants?
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Worked solution
Gametes (pollen nucleus and egg cell nucleus) are haploid (contain one set of chromosomes, \(n\)). During fertilisation, the haploid male and female gamete nuclei fuse to form a diploid zygote nucleus (contains two sets of chromosomes, \(2n\)).
Marking scheme
A is correct because gametes are haploid and the zygote is diploid. B, C, and D represent incorrect combinations of haploid/diploid states.
Question 35 · multiple-choice
1 marks
Concentrated aqueous sodium chloride is electrolysed using inert carbon electrodes.
Which gases are liberated at the cathode and at the anode?
A.cathode: hydrogen; anode: chlorine
B.cathode: hydrogen; anode: oxygen
C.cathode: sodium; anode: chlorine
D.cathode: sodium; anode: oxygen
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Worked solution
During the electrolysis of concentrated aqueous sodium chloride: - At the cathode, hydrogen ions (\(\text{H}^+\)) are preferentially reduced over sodium ions (\(\text{Na}^+\)), producing hydrogen gas (\(\text{H}_2\)). - At the anode, chloride ions (\(\text{Cl}^-\)), being in high concentration, are preferentially oxidised over hydroxide ions (\(\text{OH}^-\)), producing chlorine gas (\(\text{Cl}_2\)).
Marking scheme
A is correct because hydrogen gas is produced at the negative electrode (cathode) and chlorine gas is produced at the positive electrode (anode). B, C, and D are incorrect because they name incorrect products resulting from a misunderstanding of preferential discharge rules.
Question 36 · multiple-choice
1 marks
An addition polymer has the repeating structure shown:
What is the name of the monomer used to make this polymer?
A.ethene
B.propene
C.but-1-ene
D.propane
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Worked solution
The repeating unit has a backbone of two carbon atoms with a methyl group (\(\text{---CH}_3\)) on one of them. This corresponds to the monomer propene, \(\text{CH}_2\text{=CH---CH}_3\), which undergoes addition polymerisation to form poly(propene).
Marking scheme
B is correct because propene is the monomer that polymerises to form this repeating unit. A is incorrect (ethene forms poly(ethene), with no methyl groups). C is incorrect (but-1-ene would have an ethyl side chain). D is incorrect because propane is an alkane and cannot undergo addition polymerisation.
Question 37 · multiple-choice
1 marks
Barium sulfate is an insoluble salt.
Which method is most suitable for preparing a pure, dry sample of barium sulfate?
A.React solid barium carbonate with dilute sulfuric acid, then evaporate the mixture to dryness.
B.Mix aqueous barium chloride with dilute sulfuric acid, filter the mixture, wash the residue with water, and dry it.
C.Titrate aqueous barium hydroxide with dilute sulfuric acid using an indicator, then crystallise the salt.
D.Heat solid barium oxide with dilute sulfuric acid, filter the mixture, and crystallise the filtrate.
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Worked solution
Barium sulfate is an insoluble salt and is best prepared by precipitation. Mixing soluble barium chloride (\(\text{BaCl}_2\text{(aq)}\)) with dilute sulfuric acid (\(\text{H}_2\text{SO}_4\text{(aq)}\)) produces a precipitate of barium sulfate:
The insoluble barium sulfate is separated by filtration, washed with distilled water to remove soluble hydrochloric acid, and then dried.
Marking scheme
B is correct because precipitation of an insoluble salt is achieved by mixing two soluble reactants, followed by filtration, washing, and drying. A is incorrect because barium sulfate is insoluble and coats the barium carbonate, stopping the reaction. C is incorrect because titration is typically used for preparing soluble salts. D is incorrect as it is not a standard or viable laboratory method.
Question 38 · multiple-choice
1 marks
An electric car of mass \(1200\text{ kg}\) accelerates uniformly along a straight, horizontal road.
Its speed increases from \(10\text{ m/s}\) to \(25\text{ m/s}\) in a time interval of \(5.0\text{ s}\).
What is the resultant force acting on the car during this time?
A.1200 N
B.2400 N
C.3600 N
D.6000 N
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Worked solution
First, calculate the acceleration (\(a\)) of the car: \[a = \frac{v - u}{t} = \frac{25\text{ m/s} - 10\text{ m/s}}{5.0\text{ s}} = 3.0\text{ m/s}^2\]
Next, use Newton's second law (\(F = m \times a\)) to find the resultant force: \[F = 1200\text{ kg} \times 3.0\text{ m/s}^2 = 3600\text{ N}\]
Marking scheme
C is correct. First find acceleration using \(a = \Delta v / t = 15 / 5 = 3\text{ m/s}^2\), then force using \(F = ma = 1200 \times 3 = 3600\text{ N}\). A is incorrect (uses \(a = v/t = 25/5 = 5\) or incorrect manipulation). B is incorrect (uses change in speed of \(10\text{ m/s}\) instead of \(15\text{ m/s}\)). D is incorrect (uses \(F = m \times v / t = 1200 \times 25 / 5 = 6000\text{ N}\)).
Question 39 · multiple-choice
1 marks
A resistor of resistance \(4.0\ \Omega\) and a resistor of resistance \(12.0\ \Omega\) are connected in parallel.
This parallel combination is connected in series with a \(3.0\ \Omega\) resistor and a \(12.0\text{ V}\) d.c. power supply.
What is the current drawn from the power supply?
A.1.0 A
B.2.0 A
C.3.0 A
D.4.0 A
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Worked solution
First, calculate the equivalent resistance of the two parallel resistors, \(R_p\): \[\frac{1}{R_p} = \frac{1}{4.0\ \Omega} + \frac{1}{12.0\ \Omega} = \frac{3 + 1}{12.0} = \frac{4}{12.0}\] \[R_p = 3.0\ \Omega\]
Next, find the total resistance of the circuit, \(R_{total}\), by adding the series resistor: \[R_{total} = R_p + 3.0\ \Omega = 3.0\ \Omega + 3.0\ \Omega = 6.0\ \Omega\]
Finally, use Ohm's law to calculate the total current, \(I\): \[I = \frac{V}{R_{total}} = \frac{12.0\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\]
Marking scheme
B is correct because the parallel equivalent resistance is \(3.0\ \Omega\), the total resistance is \(6.0\ \Omega\), and the current is \(12.0\text{ V} / 6.0\ \Omega = 2.0\text{ A}\). A is incorrect (uses simple sum of all three: \(19\ \Omega\) giving approx. \(0.6\text{ A}\) or other calculation error). C is incorrect. D is incorrect.
Question 40 · multiple-choice
1 marks
A ray of light travelling in air strikes the flat boundary of a transparent plastic block at an angle of incidence of \(45^\circ\).
The refractive index of the plastic is \(1.5\).
What is the angle of refraction inside the plastic block, to the nearest degree?
A.28°
B.30°
C.45°
D.60°
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Worked solution
Using Snell's Law: \[n = \frac{\sin i}{\sin r}\]
Rearrange to solve for \(\sin r\): \[\sin r = \frac{\sin i}{n} = \frac{\sin 45^\circ}{1.5} \approx \frac{0.7071}{1.5} \approx 0.4714\]
Now find the angle of refraction \(r\): \[r = \arcsin(0.4714) \approx 28.1^\circ\]
To the nearest degree, the angle of refraction is \(28^\circ\).
Marking scheme
A is correct because substituting \(i = 45^\circ\) and \(n = 1.5\) into \(n = \sin i / \sin r\) yields \(r \approx 28^\circ\). B is incorrect (close approximation but not mathematically correct). C is incorrect because the angle must decrease when light enters a more optically dense medium. D is incorrect.
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Answer all twelve structured theory questions. Write answers in the spaces provided.
12 Question · 120 marks
Question 1 · structured
10 marks
Fig. 1.1 shows a diagram of a lily flower during pollination.
(a) Define the term pollination. [2]
(b) Describe the pathway of a male gamete from pollination to fertilisation in a flowering plant. [3]
(c) State two advantages and two disadvantages of asexual reproduction compared to sexual reproduction in plants in the wild. [4]
(d) State the cell structure where genetic material is stored in the male gamete of a plant. [1]
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Worked solution
(a) Pollination is the transfer of pollen grains from the anther to the stigma.
(b) After pollination, the pollen grain germinates on the stigma. A pollen tube grows down through the style into the ovary. The male gamete travels down this tube and enters the ovule through the micropyle to fuse with the female gamete (fertilisation).
(c) Advantages: 1. Fast / rapid reproduction method. 2. Requires only one parent (no pollinators needed).
Disadvantages: 1. No genetic variation among offspring (susceptible to the same diseases). 2. Increased competition for resources as offspring grow close to the parent.
(d) Nucleus.
Marking scheme
(a) transfer of pollen grains [1]; from anther to stigma [1]
(b) pollen grain germinates / grows pollen tube [1]; down the style into the ovary [1]; male gamete passes down tube and enters ovule / fuses with female gamete [1]
(c) two advantages: e.g., faster / only one parent needed / preserves successful traits [2]; two disadvantages: e.g., no genetic variation / population susceptible to same disease / crowding / competition [2]
(d) nucleus [1]
Question 2 · structured
10 marks
A student investigates the effect of pH on the activity of the enzyme catalase, which breaks down hydrogen peroxide into water and oxygen.
Table 2.1 shows the volume of oxygen gas produced in 2.0 minutes at different pH values.
Table 2.1 pH | Volume of oxygen gas produced / cm³ 4.0 | 5.0 5.0 | 12.0 6.0 | 28.0 7.0 | 45.0 8.0 | 24.0 9.0 | 8.0
(a) State the optimum pH for catalase in this investigation. Explain your answer with reference to Table 2.1. [2]
(b) Explain, in terms of collisions and active sites, why enzyme activity is lower at pH 4.0 than at the optimum pH. [3]
(c) Calculate the rate of oxygen production at the optimum pH in cm³/s. Show your working. [2]
(d) Describe the chemical test for oxygen gas and its positive result. [2]
(e) State the chemical elements present in catalase. [1]
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Worked solution
(a) The optimum pH is 7.0. This is because pH 7.0 shows the maximum volume of oxygen gas produced (45.0 cm³), which indicates the highest enzyme activity.
(b) At pH 4.0, the enzyme is denatured. This changes the shape of the active site so that the substrate (hydrogen peroxide) can no longer fit into it. As a result, successful collisions between the substrate and active site are reduced, leading to lower enzyme activity.
(c) Volume of oxygen at pH 7.0 = 45.0 cm³ Time = 2.0 minutes = 120 seconds Rate = 45.0 / 120 = 0.375 cm³/s (or 0.38 cm³/s).
(d) Test: Insert a glowing splint into a test tube containing the gas. Result: The splint relights.
(e) Carbon, hydrogen, oxygen, nitrogen (and sometimes sulfur).
Marking scheme
(a) pH 7.0 [1]; has the highest volume of oxygen produced / maximum activity [1]
(b) enzyme / active site is denatured / changes shape [1]; substrate can no longer fit into the active site / is not complementary [1]; fewer successful collisions / no enzyme-substrate complexes formed [1]
(c) 120 seconds seen / conversion of 2 mins to 120 s [1]; 0.375 (cm³/s) [1]
(b) Molar mass of \(CaCO_3 = 40 + 12 + (3 \times 16) = 100\text{ }g/mol\). Number of moles of \(CaCO_3 = 5.0 / 100 = 0.05\text{ }mol\). From the balanced equation, 1 mole of \(CaCO_3\) produces 1 mole of \(CO_2\). So, moles of \(CO_2 = 0.05\text{ }mol\). Volume of \(CO_2 = 0.05 \times 24 = 1.2\text{ }dm^3\).
(c) An acid is a proton donor.
(d) Calcium nitrate.
(e) Red (or pink).
Marking scheme
(a) correct formulae of reactants and products: \(CaCO_3\), \(HCl\), \(CaCl_2\), \(H_2O\), \(CO_2\) [1]; balancing correct: 2 in front of \(HCl\) [1]; all state symbols correct: (s), (aq), (aq), (l), (g) [1]
(b) \(M_r\) of \(CaCO_3 = 100\) [1]; moles of \(CaCO_3 = 0.05\text{ }mol\) [1]; 1:1 mole ratio of \(CaCO_3\) to \(CO_2\) used / moles of \(CO_2 = 0.05\text{ }mol\) [1]; volume of \(CO_2 = 1.2\text{ }dm^3\) [1]
(c) proton donor [1]
(d) calcium nitrate [1]
(e) red / pink [1]
Question 4 · structured
10 marks
Fractional distillation of petroleum produces several fractions, including naphtha. Naphtha contains alkanes that can be cracked.
(a) Define the term hydrocarbon. [1]
(b) (i) Decane (C₁₀H₂₂) can be cracked to form octane (C₈H₁₈) and one other molecule. Write a balanced symbol equation for this reaction. [2]
(ii) State the name of the other molecule produced in this cracking reaction. [1]
(c) Describe a chemical test to distinguish between a saturated hydrocarbon and an unsaturated hydrocarbon, and state the result for each. [3]
(d) Draw the displayed formula of ethene. [2]
(e) State one major environmental consequence of burning fossil fuels like octane. [1]
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Worked solution
(a) A hydrocarbon is a compound containing only carbon and hydrogen atoms.
(b) (i) \(C_{10}H_{22} \rightarrow C_8H_{18} + C_2H_4\) (ii) Ethene
(c) Test: Add aqueous bromine (bromine water) to the hydrocarbons. Result for saturated: Stays orange / yellow / brown (no change). Result for unsaturated: Turns colourless (decolourises).
(d) The displayed formula of ethene has two carbon atoms joined by a double bond, with each carbon atom bonded to two hydrogen atoms: H-C=C-H with H's on each C (double bond shown: C=C).
(e) Release of carbon dioxide, which is a greenhouse gas that contributes to global warming.
Marking scheme
(a) compound containing carbon and hydrogen only [1] (reject: contains carbon and hydrogen, must say ONLY)
(b) (i) formula of reactants and products: \(C_{10}H_{22}\) and \(C_8H_{18} + C_2H_4\) [1]; fully balanced equation [1] (ii) ethene [1]
(d) C=C double bond shown [1]; 4 C-H single bonds shown correctly [1]
(e) produces carbon dioxide / greenhouse gas / contributes to climate change or global warming [1]
Question 5 · structured
10 marks
A toy car of mass 0.50 kg starts from rest and accelerates uniformly to a speed of 6.0 m/s in 4.0 seconds.
(a) Calculate the acceleration of the toy car. Show your working and state the unit. [3]
(b) Calculate the kinetic energy of the toy car when it is travelling at 6.0 m/s. Show your working. [2]
(c) The car then travels at a constant speed of 6.0 m/s for a distance of 18 m. Calculate the time taken to travel this distance. [2]
(d) A resistive force of 0.80 N acts against the motion of the car. Calculate the work done by the motor to overcome this resistive force over the distance of 18 m. [3]
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A 12 V battery is connected in series with a switch, a 4.0 Ω resistor, and a parallel combination of two lamps, A and B. Lamp A has a resistance of 6.0 Ω and Lamp B has a resistance of 3.0 Ω.
(a) Calculate the combined resistance of Lamp A and Lamp B connected in parallel. Show your working. [2]
(b) Calculate the total resistance of the entire circuit when the switch is closed. [1]
(c) Show that the total current in the circuit when the switch is closed is 2.0 A. [2]
(d) Calculate the potential difference across the parallel combination of lamps. Show your working. [2]
(e) Calculate the power dissipated in the 4.0 Ω resistor. Show your working. [2]
(f) State what happens to the current in the 4.0 Ω resistor if Lamp A breaks. Explain your answer. [1]
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(c) Total current \(I = \frac{V}{R_{\text{total}}} = \frac{12}{6.0} = 2.0\text{ }A\).
(d) Potential difference across parallel lamps \(V_p = I \times R_p = 2.0 \times 2.0 = 4.0\text{ }V\).
(e) Power dissipated in 4.0 Ω resistor \(P = I^2 \times R = 2.0^2 \times 4.0 = 4.0 \times 4.0 = 16.0\text{ }W\).
(f) The current decreases. If Lamp A breaks, the parallel branch resistance increases from 2.0 Ω to 3.0 Ω, which increases the total circuit resistance to 7.0 Ω, thus decreasing the total current.
(b) The critical angle is the angle of incidence in the more optically dense medium that results in an angle of refraction of 90° in the less dense medium.
(d) In a transverse wave, particles vibrate perpendicular (at right angles) to the direction of wave travel. In a longitudinal wave, particles vibrate parallel to the direction of wave travel.
Marking scheme
(a) (i) \(n = \frac{\sin i}{\sin r}\) [1] (ii) \(\sin r = \frac{\sin 45^\circ}{1.50}\) or 0.4714 [1]; 28.1 (°) (accept 28) [1]
(b) angle of incidence (in dense medium) [1]; that produces angle of refraction of 90° (in less dense medium) [1]
(c) \(\sin c = 1/n\) [1]; \(\sin c = 1/1.50 = 0.6667\) [1]; 41.8 (°) (accept 42) [1]
(d) transverse: vibrations are perpendicular to wave direction [1]; longitudinal: vibrations are parallel to wave direction [1]
Question 8 · structured
10 marks
Molten zinc chloride (ZnCl₂) is electrolysed using inert carbon electrodes.
(a) State the name of the product formed at: (i) the cathode [1] (ii) the anode [1]
(b) Construct the ionic half-equation, including state symbols, for the reaction that occurs at: (i) the cathode [2] (ii) the anode [2]
(c) Explain, in terms of particles and movement, why solid zinc chloride does not conduct electricity but molten zinc chloride does. [2]
(d) Describe a safe chemical test to confirm the identity of the gas produced at the anode, and state the positive result of the test. [2]
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Worked solution
(a) (i) Cathode product: Zinc (ii) Anode product: Chlorine
(b) (i) At the cathode: \(Zn^{2+}(l) + 2e^- \rightarrow Zn(l)\) (or \(Zn(s)\)) (ii) At the anode: \(2Cl^-(l) \rightarrow Cl_2(g) + 2e^-\)
(c) In solid zinc chloride, the ions are held in fixed positions in a giant lattice and cannot move. In molten zinc chloride, the giant lattice is broken and the ions are free to move and carry the charge.
(d) Test: Insert damp blue litmus paper (or starch-iodide paper) into the gas. Result: The litmus paper turns red and then bleaches white.
Marking scheme
(a) i. zinc [1] ii. chlorine [1]
(b) i. \(Zn^{2+} + 2e^- \rightarrow Zn\) [1]; state symbols correct: (l) for reactants, (l) or (s) for products [1] ii. \(2Cl^- \rightarrow Cl_2 + 2e^-\)[1]; state symbols correct: (l) for reactants, (g) for products [1]
(c) solid: ions are in fixed positions / lattice [1]; molten: ions are free to move / flow [1] (reject: free electrons)
(d) test: damp blue litmus paper [1]; result: bleaches / turns white [1]
Question 9 · structured
10 marks
The menstrual cycle in human females is coordinated by several hormones that interact via feedback loops. Figure 1.1 represents the changes in the thickness of the uterus lining and the blood concentrations of two ovarian hormones, hormone X and hormone Y, over a typical 28-day cycle.
[Figure 1.1: A graph showing thickness of the uterus lining rising from day 5 to day 28. Hormone X rises first, peaking around day 12. Hormone Y rises after ovulation on day 14 and peaks around day 21.]
(a) (i) Identify the two ovarian hormones represented in the cycle:
(ii) State the name of the endocrine structure in the ovary that secretes hormone X before day 14, and the endocrine structure that secretes hormone Y after ovulation.
(b) FSH and LH are pituitary hormones that play essential roles in the cycle. Explain the roles of FSH and LH in initiating and progressing the events of the menstrual cycle leading up to ovulation. [3]
(c) During the second half of the cycle (days 15 to 28), high levels of progesterone (hormone Y) are present.
Describe the feedback effect of progesterone on the secretion of FSH and LH from the pituitary gland, and explain the biological significance of this feedback if fertilisation does not occur. [3]
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(ii) Structure secreting X: ovarian follicle / Graafian follicle Structure secreting Y: corpus luteum
(b) FSH is released by the pituitary gland and stimulates the development and maturation of follicles in the ovary. It also stimulates these follicle cells to produce and secrete estrogen. LH is released in a surge near the middle of the cycle, triggering ovulation (the release of the mature egg) and subsequently promoting the development of the remaining follicle tissue into the corpus luteum.
(c) Progesterone exerts negative feedback on the pituitary gland and hypothalamus, which inhibits the secretion of both FSH and LH. The biological significance of this inhibition is that it prevents the development and maturation of any new follicles or further ovulation during this phase of the cycle. If fertilisation does not occur, the corpus luteum degenerates, progesterone levels fall, removing the negative feedback and allowing FSH levels to rise again to start the next cycle.
Marking scheme
(a) (i) - Estrogen [1] - Progesterone [1]
(ii) - (Ovarian / Graafian) follicle (accept ovary) [1] - Corpus luteum [1]
(b) - FSH stimulates follicle growth / maturation in the ovary [1] - FSH stimulates follicle cells to secrete estrogen [1] - LH triggers ovulation / egg release [1] - LH stimulates development of the corpus luteum [1] (Max 3 marks)
(c) - Progesterone inhibits / decreases secretion of FSH and LH (negative feedback) [1] - This prevents further follicle development / ovulation [1] - If fertilisation fails, falling progesterone removes the inhibition, allowing FSH to rise and initiate a new cycle [1]
Question 10 · structured
10 marks
Esters are organic compounds with characteristic sweet smells, and polyesters such as Terylene are polymers constructed from ester linkages.
(a) (i) Draw the fully displayed structure of the ester ethyl propanoate, showing all atoms and all bonds. [2]
(ii) Name the alcohol and the carboxylic acid required to synthesise ethyl propanoate, and state the reaction conditions needed for this reaction to occur.
(i) State the type of polymerisation reaction used to form Terylene, and explain how this type of polymerisation differs from addition polymerisation. [2]
(ii) Complete the description to show the repeating unit of Terylene. Use block diagrams to represent the carbon-containing groups in the monomers. Show all the atoms and bonds in the ester linkage. [3]
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Worked solution
(a) (i) Fully displayed structure of ethyl propanoate: Propanoate part: 3 carbons, ethyl part: 2 carbons. \(\text{H}-\text{C}(\text{H})(\text{H})-\text{C}(\text{H})(\text{H})-\text{C}(=\text{O})-\text{O}-\text{C}(\text{H})(\text{H})-\text{C}(\text{H})(\text{H})-\text{H}\) All bonds including \(\text{C}-\text{H}\), \(\text{C}-\text{C}\), \(\text{C}=\text{O}\), and \(\text{C}-\text{O}\) must be drawn out.
(ii) Alcohol: ethanol Carboxylic acid: propanoic acid Conditions: heat / warm with a concentrated sulfuric acid catalyst
(b) (i) Condensation polymerisation. In addition polymerisation, monomers join together without the loss of any other substances, while in condensation polymerisation, a small molecule (such as water or hydrogen chloride) is eliminated as each linkage is formed.
(ii) The repeating unit has the structure: \(\text{[-O-}\square\text{-O-CO-}\diamond\text{-CO-]_n}\) showing continuation bonds at each end and the ester linkage clearly as \(\text{-O-C(=O)-}\) or \(\text{-CO-O-}\).
Marking scheme
(a) (i) - Correct ester group structure \(\text{-C(=O)-O-}\) with all bonds shown [1] - Correct ethyl group (2-carbon chain) and propanoate group (3-carbon chain) with all hydrogens and bonds fully displayed [1]
(b) (i) - Condensation polymerisation [1] - Elimination / loss of a small molecule / water (which does not happen in addition polymerisation) [1]
(ii) - Correct ester linkage drawn out with all atoms and bonds: \(\text{-O-C(=O)-}\) or \(\text{-C(=O)-O-}\) [1] - Alternating blocks sequence showing diol block and dicarboxylic acid block [1] - Correct continuation bonds on both open ends of the repeating unit [1]
Question 11 · structured
10 marks
A tennis player hits a tennis ball of mass \(0.058\text{ kg}\) that is initially travelling horizontally towards the racket at a speed of \(32\text{ m/s}\). After being struck, the ball travels horizontally in the opposite direction at a speed of \(42\text{ m/s}\).
(a) (i) Calculate the change in momentum of the tennis ball during the collision. Show your working and state the direction of this change relative to the ball's initial motion.
change in momentum = ............................................................ \(\text{kg m/s}\)
(ii) The ball is in contact with the racket strings for a time of \(4.5\text{ ms}\).
Calculate the average force exerted by the racket on the ball.
average force = ............................................................ \(\text{N}\) [2]
(b) (i) Define the term *impulse* and state how it relates to momentum. [2]
(ii) When a car undergoes a collision, crumple zones are designed to deform.
Explain, in terms of momentum, impulse, and force, how crumple zones protect the passengers in a car during a high-speed impact. [3]
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Worked solution
(a) (i) Let the initial direction of the ball be negative, and the opposite direction after collision be positive. Initial momentum, \(p_i = m \times u = 0.058\text{ kg} \times (-32\text{ m/s}) = -1.856\text{ kg m/s}\) Final momentum, \(p_f = m \times v = 0.058\text{ kg} \times 42\text{ m/s} = 2.436\text{ kg m/s}\) Change in momentum, \(\Delta p = p_f - p_i = 2.436 - (-1.856) = 4.292\text{ kg m/s}\) (or \(4.3\text{ kg m/s}\)) Direction: opposite to the initial direction of motion / in the direction of the final motion.
(ii) Force, \(F = \frac{\Delta p}{\Delta t}\) \(\Delta t = 4.5\text{ ms} = 0.0045\text{ s}\) \(F = \frac{4.292\text{ kg m/s}}{0.0045\text{ s}} = 953.8\text{ N}\) (or \(954\text{ N}\), accept \(950\text{ N}\) using \(4.3\text{ kg m/s}\))
(b) (i) Impulse is defined as the force acting on an object multiplied by the time interval over which it acts (\(F \times t\)). It is equal to the change in momentum of the object (\(\Delta p\)).
(ii) During a collision, the car must lose all its momentum to come to a stop, requiring a specific impulse. Crumple zones undergo plastic deformation, which increases the time duration of the impact (\(\Delta t\)). Since the force is given by \(F = \frac{\Delta p}{\Delta t}\), increasing the time for the same change in momentum dramatically reduces the average force experienced by the passengers, reducing severe injuries.
Marking scheme
(a) (i) - Correct substitution into \(\Delta p = m(v - u)\) accounting for opposite directions: \(0.058 \times (42 - (-32))\) [1] - Correct magnitude of \(4.3\text{ kg m/s}\) (or \(4.29\text{ kg m/s}\)) [1] - Direction stated as opposite to the initial velocity / in direction of hit [1]
(ii) - Uses \(F = \frac{\Delta p}{\Delta t}\) with time converted to seconds (\(0.0045\text{ s}\)) [1] - Correct calculation of force: \(950\text{ N}\) or \(954\text{ N}\) (allow ecf from (a)(i)) [1]
(b) (i) - Impulse is Force \(\times\) Time [1] - Impulse is equal to change in momentum [1]
(ii) - Crumple zone increases the duration / time of the impact (\(\Delta t\)) [1] - For the same change in momentum / same impulse required [1] - This reduces the average force on the passenger compartment/passengers (as \(F \propto 1/\Delta t\)) [1]
Question 12 · structured
10 marks
The electrolysis of concentrated aqueous sodium chloride (brine) is an important industrial process carried out in a chlor-alkali cell.
(a) (i) Name the product released at the anode (positive electrode) and the product released at the cathode (negative electrode) during this electrolysis.
(ii) Write the ionic half-equations, including state symbols, for the reactions occurring at:
the anode: ............................................................
the cathode: ............................................................ [4]
(b) (i) As the electrolysis of brine continues, the remaining electrolyte solution around the cathode becomes increasingly alkaline. Explain why this happens, and identify the alkaline chemical produced. [2]
(ii) Both gases produced in the process are collected and used in other chemical industries. State one major industrial use for each of the gases produced.
Anode gas use: ............................................................
Hydorogen gas use: ............................................................ [2]
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Worked solution
(a) (i) Anode product: chlorine gas Cathode product: hydrogen gas
(b) (i) Water molecules slightly dissociate to form hydrogen (\(\text{H}^+\)) and hydroxide (\(\text{OH}^-\)) ions. The \(\text{H}^+\dots\) ions are discharged at the cathode as hydrogen gas, leaving an excess of \(\text{OH}^-\dots\) ions in the solution around the cathode. These hydroxide ions associate with the remaining sodium ions (\(\text{Na}^+\dots\)) to produce sodium hydroxide, which is a strong alkali.
(ii) Anode gas (chlorine) use: manufacture of bleach / sterilising drinking water / water in swimming pools / making PVC. Cathode gas (hydrogen) use: manufacture of ammonia (Haber process) / manufacture of margarine / rocket fuel.
(ii) - Anode equation: balanced reactants and products [1], correct state symbols [1] - Cathode equation: balanced reactants and products [1], correct state symbols [1]
(b) (i) - Hydrogen ions (\(\text{H}^+\)) are discharged / reduced, leaving a high concentration of hydroxide ions (\(\text{OH}^-\)) [1] - Alkaline chemical produced: sodium hydroxide (\(\text{NaOH}\)) [1]
(ii) - Chlorine use: water purification / PVC manufacture / bleaching [1] - Hydrogen use: making ammonia / margarine manufacture (hydrogenation) [1]
Paper 61 Alternative to Practical
Answer all experimental design and analysis questions. Include units where necessary.
6 Question · 60 marks
Question 1 · Practical
10 marks
A student investigates the effect of pH on the activity of the enzyme amylase.
Amylase digests starch into maltose. The student mixes amylase and starch solutions at different pH values. At 30-second intervals, a sample of the mixture is added to a drop of iodine solution on a spotting tile.
(a) State the color of iodine solution: (i) in the presence of starch, (ii) when all the starch has been digested. [2]
(b) Table 1.1 shows the time taken for the starch to be completely digested at different pH values.
**Table 1.1** | pH | Time taken for starch to be digested / s | | :---: | :---: | | 4.0 | 360 | | 5.0 | 180 | | 6.0 | 60 | | 7.0 | 120 | | 8.0 | 300 |
(i) State the optimum pH for this amylase enzyme. Explain your answer with reference to the data in Table 1.1. [2] (ii) Suggest how the student could modify the method to find a more precise value for the optimum pH. [2]
(c) State two variables that must be kept constant to ensure a fair test in this investigation. [2]
(d) Explain why repeating the experiment at each pH and calculating an average improves the quality of the data. [2]
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Worked solution
(a) (i) Iodine solution turns blue-black in the presence of starch. (ii) In the absence of starch (when fully digested), the iodine solution remains its original orange-brown/yellow-brown color.
(b) (i) The optimum pH is 6.0 because the reaction rate is fastest, indicated by the shortest time taken for digestion (60 s). (ii) The student can test additional pH values at smaller intervals around pH 6.0, such as 5.5, 6.0, and 6.5, to pin down the exact optimum more precisely.
(c) Any two from: temperature of the water bath, volume of starch solution, volume of amylase solution, concentration of starch/amylase.
(d) Repeating allows the identification of anomalous results and enables calculation of a more reliable average, minimizing the impact of random errors.
Marking scheme
*(a) (i)* blue-black / black [1] *(a) (ii)* orange-brown / yellow-brown / brown / yellow [1] (reject: colorless / blue) *(b) (i)* pH 6.0 [1] and because it has the shortest time / fastest rate of reaction [1] *(b) (ii)* test at smaller pH intervals [1] between pH 5.0 and pH 7.0 / around pH 6.0 [1] *(c)* any two from: temperature, volume of starch, volume of amylase, concentration of starch/amylase [2] *(d)* allows identification of anomalous results / reduces effect of random error / increases reliability [2] (reject: increases accuracy / makes it a fair test)
Question 2 · Practical
10 marks
A student investigates the rate of reaction between calcium carbonate (marble chips) and dilute hydrochloric acid.
(a) Draw a labelled diagram of a suitable apparatus setup to react marble chips with acid and collect and measure the volume of carbon dioxide gas produced. [3]
(b) Table 2.1 shows the volume of gas collected at 30-second intervals.
(i) State why the volume of gas remains constant after 180 seconds. [1] (ii) Calculate the average rate of reaction during the first 60 seconds. Include the unit. [2]
(c) Sketch a graph to show how the volume of gas collected changes over time for this experiment. On the same graph, draw a line to represent the results if the experiment was repeated using hydrochloric acid of a lower concentration, keeping all other variables constant. Label the two lines clearly. [2]
(d) Describe the chemical test used to confirm that the gas produced is carbon dioxide. State the positive result. [2]
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Worked solution
(a) A correct diagram should show a reaction vessel (e.g., conical flask) containing acid and marble chips, fitted with a stopper and delivery tube. The tube must lead to a gas syringe or to an inverted measuring cylinder filled with water in a trough. All main parts must be labelled.
b) (i) The reaction has stopped because the limiting reactant (hydrochloric acid) has been fully consumed. (ii) Rate = \(\frac{38.0\text{ cm}^3}{60\text{ s}} = 0.633\text{ cm}^3/\text{s}\).
(c) The sketch should show time on the x-axis and gas volume on the y-axis. The curve for the lower concentration should have a shallower gradient (less steep) at the beginning, indicating a slower rate, and should eventually level off at either a lower volume (if acid was limiting) or the same volume but at a later time.
(d) Bubble the gas through limewater. If carbon dioxide is present, the limewater turns cloudy/milky/chalky.
Marking scheme
*(a)* conical flask/reaction vessel containing marble chips and acid, with a stopper [1]; delivery tube connected to a gas syringe OR inverted measuring cylinder in a trough of water [1]; correct labels of flask, delivery tube, and gas measuring apparatus [1] *(b) (i)* reaction finished / reactant(s) used up [1] (accept: acid fully reacted) *(b) (ii)* 38.0 / 60 [1] = 0.63 [1] (accept 0.633) unit: cm3/s [1] *(c)* both curves start at origin, first curve rises steeply and levels off [1]; second curve is less steep / below the first curve [1] *(d)* test: bubble through limewater [1]; result: turns cloudy / milky / white precipitate [1]
Question 3 · Practical
10 marks
A student investigates the effectiveness of two different materials, Bubble Wrap and Felt, as thermal insulators for a beaker of hot water.
Beaker A is wrapped in Bubble Wrap. Beaker B is wrapped in Felt. The temperature of the water in each beaker is recorded every minute for 5 minutes.
**Table 3.1** | Time / min | Temperature of Beaker A / \(^\circ\text{C}\) | Temperature of Beaker B / \(^\circ\text{C}\) | | :---: | :---: | :---: | | 0 | 85.0 | 85.0 | | 1 | 81.5 | 79.5 | | 2 | 78.5 | 75.0 | | 3 | 76.0 | 71.5 | | 4 | 74.0 | 68.5 | | 5 | 72.5 | 66.0 |
(a) (i) Calculate the total temperature drop for the water in both Beaker A and Beaker B over the 5-minute period. [2] (ii) State which material is the better thermal insulator. Explain your answer with reference to the results in Table 3.1. [2]
(b) State two variables that must be kept constant to ensure a fair comparison. [2]
(c) Suggest one modification to the beakers to reduce heat loss from the top. [1]
(d) The student wants to extend the investigation to find out how the thickness of the insulation affects the rate of cooling. Describe how the student could adapt this experiment to investigate this. [3]
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Worked solution
(a) (i) Temperature drop in Beaker A = \(85.0 - 72.5 = 12.5\,^\circ\text{C}\). Temperature drop in Beaker B = \(85.0 - 66.0 = 19.0\,^\circ\text{C}\). (ii) Bubble Wrap (Beaker A) is the better insulator because the temperature of the water fell by a smaller amount (\(12.5\,^\circ\text{C}\)) than in the beaker wrapped in Felt (\(19.0\,^\circ\text{C}\)).
(b) Two control variables are: the volume of hot water in each beaker, the starting temperature of the water, the thickness of the insulation layers used, or the room temperature.
(c) Place a lid (made of cardboard or polystyrene) on top of each beaker.
(d) To investigate thickness, the student should use a single insulating material (e.g., Bubble Wrap) and wrap the beakers with different numbers of layers (e.g., 1, 2, 3, and 4 layers), keeping the volume of water, starting temperature, and duration of cooling the same, then compare the temperature drops.
Marking scheme
*(a) (i)* Beaker A: 12.5 (°C) [1]; Beaker B: 19.0 (°C) [1] *(a) (ii)* Bubble Wrap / Beaker A [1] and because it has a smaller temperature drop / water stays hotter [1] *(b)* any two from: volume of water, starting temperature, same surface area covered, room temperature [2] *(c)* add a lid / cover the top [1] *(d)* use only one type of insulation material [1]; vary the number of layers / thickness (at least 3 different thicknesses) [1]; keep water volume / starting temperature constant [1]
Question 4 · Practical
10 marks
A student wants to investigate the effect of wind speed on the rate of transpiration in a leafy shoot.
(a) Draw a labelled diagram of a simple potometer setup that can be used to measure water uptake by a leafy shoot. Include labels for the capillary tube and the reservoir. [3]
(b) Plan an experiment to investigate how wind speed affects the rate of transpiration in a leafy shoot.
You are provided with: - a leafy shoot in a potometer - an electric fan with three speed settings (Low, Medium, High) - a ruler and a stop-watch
In your plan, include: - a brief description of the method, - how you will vary the independent variable, - what you will measure and how you will ensure accuracy, - the key control variables and how they are kept constant, - how you will process your results to draw a conclusion. [7]
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Worked solution
(a) The diagram of the potometer must show: 1. A leafy shoot sealed tightly into a glass tube filled with water. 2. A horizontal capillary tube with a graduated scale (ruler) underneath and a visible air bubble. 3. A water reservoir with a tap/syringe to return the bubble to the start position.
(b) Experimental Plan: - **Method**: Set up the potometer with the leafy shoot in a room with stable conditions. Position the electric fan at a fixed distance (e.g., 50 cm) from the shoot to create the wind. - **Independent Variable**: Vary the wind speed using the fan settings: off (no wind), Low, Medium, and High. - **Dependent Variable & Accuracy**: For each setting, measure the distance moved by the air bubble in the capillary tube over a fixed time of 5 minutes using the ruler scale. To ensure accuracy, ensure the system is completely airtight using petroleum jelly, and let the plant adapt to each new environment for 2 minutes before taking readings. - **Control Variables**: Keep constant the distance of the fan from the shoot, the room temperature, the light intensity (keep room lights constant), and use the same leafy shoot throughout. - **Repetition**: Repeat the measurement at each wind speed setting three times and calculate the average distance moved. - **Conclusion**: Plot a bar chart of average distance moved per 5 minutes (representing transpiration rate) against the fan speed setting to draw a conclusion.
Marking scheme
*(a)* leafy shoot sealed into the tube [1]; horizontal capillary tube with scale and bubble shown [1]; water reservoir with tap correctly connected [1] *(b) Plan (7 marks total)*: - **independent variable**: vary wind speed using fan settings (off, low, medium, high) [1] - **dependent variable**: measure distance moved by air bubble [1] in a fixed time interval / using a stop-watch [1] - **accuracy/procedure**: ensure airtight seal / leave time for plant to adjust before reading / reset bubble using reservoir [1] - **control variables**: keep fan distance constant OR keep room temperature / light intensity constant [1] - **repetition**: repeat each setting at least twice and calculate average [1] - **processing**: plot a bar chart / graph of distance/rate against wind speed OR state how the trend would show the relationship [1]
Question 5 · Practical
10 marks
A student investigates the temperature change during the neutralisation reaction between dilute hydrochloric acid (HCl) and aqueous sodium hydroxide (NaOH).
In each experiment, the student mixes different volumes of the two solutions in a cup. The total volume of the mixture is always kept constant at 50 \(\text{cm}^3\). The initial temperature of both solutions is 20.0 \(^\circ\text{C}\).
(a) State the material of the cup that the student should use to conduct these experiments to minimise heat loss to the surroundings. [1]
(b) Explain why the temperature change, \(\Delta T\), is greatest in Experiment 3. [2]
(c) Suggest why it is important to keep the total volume of the mixture constant at 50 \(\text{cm}^3\) in all five experiments. [2]
(d) (i) Describe the trend shown in the temperature change, \(\Delta T\), as the volume of HCl increases from 10.0 \(\text{cm}^3\) to 40.0 \(\text{cm}^3\). [2] (ii) State one source of experimental error in this neutralisation experiment and describe how it can be reduced. [3]
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Worked solution
(a) A polystyrene cup (or plastic foam cup) is used because it is a good thermal insulator and minimizes heat loss to the surroundings.
(b) In Experiment 3, equal volumes of HCl and NaOH are mixed. Assuming equal concentrations, this represents the exact stoichiometric ratio where the maximum amount of acid reacts with the maximum amount of alkali, producing the maximum heat energy.
(c) Keeping the total volume constant ensures that the same mass of water is being heated by the reaction in each experiment. Since \(Q = mc\Delta T\), a constant volume (mass, \(m\)) allows the temperature changes (\(\Delta T\)) to be directly proportional to the heat released (\(Q\)), making the experiments directly comparable.
(d) (i) The temperature change first increases as volume of HCl increases from 10.0 \(\text{cm}^3\) to 25.0 \(\text{cm}^3\), reaches a peak at 25.0 \(\text{cm}^3\), and then decreases as the volume of HCl increases further to 40.0 \(\text{cm}^3\). (ii) **Source of error**: Heat is lost from the cup to the surroundings (which leads to lower recorded maximum temperatures). **Reduction**: Put a lid on the cup, use a nested double polystyrene cup, or insulate the sides of the cup with cotton wool.
Marking scheme
*(a)* polystyrene / styrofoam / plastic foam (cup) [1] (reject: glass / beaker) *(b)* complete neutralisation occurs / maximum amount of reaction takes place / largest amount of water formed [1]; maximum heat is released [1] *(c)* same mass/volume of liquid is heated [1]; to make temperature changes directly comparable / keep heat capacity constant [1] *(d) (i)* increases to a maximum (at 25 cm3) [1]; and then decreases [1] *(d) (ii)* heat loss to surroundings / evaporation [1]; use a lid [1]; wrap cup in insulation / use a double cup [1]
Question 6 · Practical
10 marks
A student investigates how the length of a resistance wire affects its electrical resistance.
The student sets up a circuit containing a cell, an ammeter, a voltmeter, and a resistance wire. Crocodile clips are used to connect different lengths of the wire into the circuit.
(a) Draw a circuit diagram showing how the cell, ammeter, voltmeter, and resistance wire should be connected. Voltmeter and ammeter symbols must be correct. [3]
(b) (i) Use the equation \(R = \frac{V}{I}\) to calculate the value of resistance, **X**, for the 60.0 cm length of wire in Table 6.1. Show your working and give your answer to two decimal places. [2] (ii) State the relationship between the length of the wire, \(L\), and its resistance, \(R\). [1]
(c) Suggest one reason why the switch in the circuit should be opened between taking readings. [2]
(d) State one safety hazard associated with this experiment and explain how it can be minimised. [2]
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Worked solution
(a) The circuit diagram should show: - A cell/battery in series with a switch, an ammeter, and the test resistance wire. - A voltmeter connected in parallel across the two ends of the test wire. - Correct symbols for cell, switch, ammeter (circle with A), voltmeter (circle with V), and wire.
(b) (i) \(R = \frac{1.60\text{ V}}{0.27\text{ A}} = 5.9259...\,\Omega\). Rounding to two decimal places gives **5.93** \(\Omega\). (ii) As the length of the wire, \(L\), increases, the resistance, \(R\), increases proportionally (it is directly proportional because doubling length from 20 to 40 cm doubles resistance from 2 to 4 \(\Omega\)).
(c) Leaving the current on causes the wire to heat up. An increase in temperature increases the resistance of the metal wire, which would make the test unfair. Opening the switch prevents heating and also prevents the battery/cell from running down.
(d) **Hazard**: The wire can become very hot and cause skin burns if touched. **Minimisation**: Do not touch the wire while the circuit is closed, or use a switch to only pass current briefly, or keep the current low.
Marking scheme
*(a)* ammeter connected in series with the wire [1]; voltmeter connected in parallel across the wire [1]; all standard symbols correct (cell, ammeter, voltmeter, variable/test wire) [1] *(b) (i)* 1.60 / 0.27 [1] = 5.93 [1] (must be 2 d.p., unit symbol Ι is in header, accept if written) *(b) (ii)* resistance is directly proportional to length [1] (accept: resistance increases as length increases) *(c)* to prevent the wire from heating up [1]; because temperature increase changes/increases resistance [1] *(d)* hot wire / burns [1]; do not touch wire when current is flowing / switch off between readings [1]
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