An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V3) Cambridge IGCSE Sciences - Co-ordinated (Double) (0654) paper. Not affiliated with or reproduced from Cambridge.
卷二 選擇題 (Extended)
There are forty questions on this paper. Answer all questions. Choose the one you consider correct.
40 题目 · 40 分
题目 1 · 選擇題
1 分
A paramecium (a unicellular organism) moves away from a drop of dilute acid placed on a microscope slide. Which characteristics of living organisms are demonstrated by this behavior?
A.excretion and movement
B.nutrition and sensitivity
C.sensitivity and movement
D.respiration and excretion
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解题
Sensitivity is the ability to detect and respond to stimuli in the environment. Movement is an action by an organism causing a change of position or place. By detecting the acid and swimming away, the paramecium demonstrates sensitivity and movement.
评分标准
1 mark for the correct option C.
题目 2 · 選擇題
1 分
The rate of an enzyme-controlled reaction increases as the temperature is raised from \(20^\circ\text{C}\) to \(40^\circ\text{C}\). However, above \(50^\circ\text{C}\), the rate decreases rapidly to zero. Which statement explains why the rate decreases above \(50^\circ\text{C}\)?
A.The activation energy of the reaction has increased.
B.The kinetic energy of the enzyme molecules has decreased.
C.The shape of the active site has changed, preventing substrate binding.
D.The substrate molecules have been denatured by the high temperature.
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解题
At high temperatures (above the optimum temperature), enzymes are denatured. This means the specific three-dimensional shape of their active site is permanently altered, so the substrate can no longer fit into the active site.
评分标准
1 mark for the correct option C.
题目 3 · 選擇題
1 分
The reaction between hydrogen and chlorine to form hydrogen chloride is exothermic. Which statement about the energy changes in this reaction is correct?
A.More energy is absorbed to break bonds in the reactants than is released when new bonds are formed.
B.More energy is released when new bonds are formed than is absorbed to break bonds in the reactants.
C.The activation energy is equal to the total energy released.
D.The products have a higher energy content than the reactants.
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解题
In any exothermic reaction, the total energy released when new bonds are formed (bond making) is greater than the total energy absorbed to break existing bonds (bond breaking).
评分标准
1 mark for the correct option B.
题目 4 · 選擇題
1 分
A student uses paper chromatography to analyze a dye. The solvent front travels \(8.0\text{ cm}\) from the baseline. A component of the dye travels \(6.4\text{ cm}\) from the baseline. What is the \(R_f\) value of this component?
A.0.80
B.1.25
C.1.40
D.0.16
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解题
The retention factor is calculated as: \(R_f = \frac{\text{distance travelled by substance}}{\text{distance travelled by solvent}} = \frac{6.4\text{ cm}}{8.0\text{ cm}} = 0.80\).
评分标准
1 mark for the correct option A.
题目 5 · 選擇題
1 分
Object X has a mass of \(m\) and moves with a speed of \(v\). Object Y has a mass of \(2m\) and moves with a speed of \(2v\). What is the ratio of the kinetic energy of Object Y to the kinetic energy of Object X?
A.2
B.4
C.8
D.16
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解题
The kinetic energy is given by \(E_k = \frac{1}{2} m v^2\). For Object X: \(E_{kX} = \frac{1}{2} m v^2\). For Object Y: \(E_{kY} = \frac{1}{2} (2m) (2v)^2 = \frac{1}{2} (2m) (4v^2) = 8 \left(\frac{1}{2} m v^2\right) = 8 E_{kX}\). The ratio of Object Y to Object X is therefore 8.
评分标准
1 mark for the correct option C.
题目 6 · 選擇題
1 分
A \(4.0\ \Omega\) resistor and a \(12.0\ \Omega\) resistor are connected in parallel. What is the combined resistance of this combination?
A.3.0 \(\Omega\)
B.8.0 \(\Omega\)
C.16.0 \(\Omega\)
D.48.0 \(\Omega\)
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解题
For resistors in parallel, \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}\). Substituting the values: \(\frac{1}{R_p} = \frac{1}{4.0} + \frac{1}{12.0} = \frac{3}{12.0} + \frac{1}{12.0} = \frac{4}{12.0} = \frac{1}{3.0}\). Therefore, \(R_p = 3.0\ \Omega\).
评分标准
1 mark for the correct option A.
题目 7 · 選擇題
1 分
In humans, the allele for brown eyes (\(B\)) is dominant to the allele for blue eyes (\(b\)). Two parents who both have brown eyes have a child with blue eyes. What is the probability that their next child will also have blue eyes?
A.0\%
B.25\%
C.50\%
D.75\%
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解题
Because both brown-eyed parents have a blue-eyed child (genotype \(bb\)), each parent must be a carrier of the recessive allele, making their genotypes \(Bb\) and \(Bb\). A cross of \(Bb \times Bb\) yields a offspring genotype ratio of \(1\ BB : 2\ Bb : 1\ bb\). The probability of producing a blue-eyed child (\(bb\)) is \(\frac{1}{4}\), which is \(25\%\).
评分标准
1 mark for the correct option B.
题目 8 · 選擇題
1 分
Which row correctly describes the relative wavelength, frequency, and speed in a vacuum of ultraviolet waves compared to radio waves?
A.Wavelength: longer, Frequency: higher, Speed in a vacuum: same
B.Wavelength: shorter, Frequency: higher, Speed in a vacuum: same
C.Wavelength: shorter, Frequency: lower, Speed in a vacuum: faster
D.Wavelength: longer, Frequency: lower, Speed in a vacuum: slower
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解题
Ultraviolet waves have a shorter wavelength and a higher frequency than radio waves. All electromagnetic waves travel at the same speed in a vacuum (approximately \(3 \times 10^8\text{ m/s}\)).
评分标准
1 mark for the correct option B.
题目 9 · 選擇題
1 分
An enzyme-catalysed reaction is carried out at different temperatures. The rate of reaction increases from 10 degrees C to 40 degrees C, but then drops rapidly to zero by 60 degrees C. Which statement explains the rapid decrease in rate above 40 degrees C?
A.The kinetic energy of the substrate molecules decreases.
B.The enzyme molecules are denatured, changing the shape of the active site.
C.The activation energy of the reaction increases.
D.The substrate molecules are broken down by the heat.
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解题
Above the optimum temperature, the thermal energy breaks bonds holding the enzyme's three-dimensional structure together. This denatures the enzyme, permanently changing the shape of the active site so that the substrate can no longer bind.
评分标准
Award 1 mark for the correct option B.
题目 10 · 選擇題
1 分
A toy car of mass 0.40 kg is traveling at a constant speed of 5.0 m/s. It then accelerates to a constant speed of 10.0 m/s. What is the increase in the kinetic energy of the car?
A.5.0 J
B.10.0 J
C.15.0 J
D.20.0 J
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解题
Initial kinetic energy = 0.5 * 0.40 * 5.0^2 = 5.0 J. Final kinetic energy = 0.5 * 0.40 * 10.0^2 = 20.0 J. Increase in kinetic energy = 20.0 J - 5.0 J = 15.0 J.
评分标准
Award 1 mark for the correct option C.
题目 11 · 選擇題
1 分
Two resistors, one of 12.0 ohms and one of 6.0 ohms, are connected in parallel. This parallel combination is connected in series with a 4.0 ohm resistor and a 12.0 V battery. What is the current flowing through the 4.0 ohm resistor?
A.1.0 A
B.1.5 A
C.2.0 A
D.3.0 A
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解题
First, find the combined resistance of the parallel pair: 1/Rp = 1/12.0 + 1/6.0 = 3/12.0, so Rp = 4.0 ohms. The total resistance of the circuit is Rp + 4.0 = 4.0 + 4.0 = 8.0 ohms. Using Ohm's law, total current = V / R = 12.0 V / 8.0 ohms = 1.5 A. Since the 4.0 ohm resistor is in series with the combination, the total current of 1.5 A flows through it.
评分标准
Award 1 mark for the correct option B.
题目 12 · 選擇題
1 分
Two substances, X and Y, are tested for their physical properties. Substance X has a high melting point and conducts electricity only when molten or dissolved in water. Substance Y has a low melting point and does not conduct electricity under any conditions. Which row correctly identifies the bonding type in X and Y?
A.X is covalent, Y is ionic
B.X is ionic, Y is covalent
C.X is metallic, Y is ionic
D.X is ionic, Y is metallic
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解题
Substance X is ionic because ionic compounds have high melting points due to strong electrostatic attractions between oppositely charged ions, and they conduct electricity only when molten or aqueous because the ions are free to move. Substance Y is simple covalent because it has a low melting point due to weak intermolecular forces and contains no free ions or electrons to conduct electricity.
评分标准
Award 1 mark for the correct option B.
题目 13 · 選擇題
1 分
In a certain plant species, the allele for red flowers (R) is dominant to the allele for white flowers (r). A heterozygous red-flowered plant is crossed with a white-flowered plant. What is the expected ratio of flower colours in the offspring?
A.all red
B.3 red : 1 white
C.1 red : 1 white
D.1 red : 3 white
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解题
The cross is Rr (heterozygous red) x rr (homozygous recessive white). The gametes produced by the red parent are R and r, and by the white parent are all r. The offspring genotypes will be 50% Rr (red phenotype) and 50% rr (white phenotype), giving a 1:1 ratio.
评分标准
Award 1 mark for the correct option C.
题目 14 · 選擇題
1 分
An organic compound reacts quickly with aqueous bromine in the dark to turn the orange-brown solution colourless. Which type of compound is this, and what type of reaction occurs?
A.alkane, addition reaction
B.alkene, addition reaction
C.alkane, substitution reaction
D.alkene, substitution reaction
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解题
Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond, which allows them to undergo addition reactions with bromine water, decolourising it rapidly. Alkanes are saturated and require UV light to undergo substitution reactions with bromine.
评分标准
Award 1 mark for the correct option B.
题目 15 · 選擇題
1 分
Which statement correctly compares infrared waves with ultraviolet waves?
A.Infrared waves have a higher frequency and a longer wavelength than ultraviolet waves.
B.Infrared waves have a lower frequency and a longer wavelength than ultraviolet waves.
C.Infrared waves have a higher frequency and a shorter wavelength than ultraviolet waves.
D.Infrared waves have a lower frequency and a shorter wavelength than ultraviolet waves.
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解题
In the electromagnetic spectrum, infrared radiation has longer wavelengths and lower frequencies than visible light, whereas ultraviolet radiation has shorter wavelengths and higher frequencies than visible light. Therefore, infrared has a lower frequency and a longer wavelength than ultraviolet.
评分标准
Award 1 mark for the correct option B.
题目 16 · 選擇題
1 分
What is the percentage by mass of nitrogen in ammonium sulfate, (NH4)2SO4? [Ar values: N = 14, H = 1, S = 32, O = 16]
A.10.6%
B.21.2%
C.28.0%
D.42.4%
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解题
First, calculate the relative formula mass (Mr) of (NH4)2SO4: Mr = 2 * [14 + (4 * 1)] + 32 + (4 * 16) = 2 * 18 + 32 + 64 = 132. The total mass of nitrogen in one formula unit is 2 * 14 = 28. The percentage by mass of nitrogen = (28 / 132) * 100 = 21.2%.
评分标准
Award 1 mark for the correct option B.
题目 17 · 選擇題
1 分
The muscular wall of the left ventricle of the human heart is thicker than that of the right ventricle. What is the reason for this?
A.The left ventricle pumps blood around the body, which requires a higher pressure than pumping blood to the lungs.
B.The left ventricle pumps blood to the lungs, which are further from the heart than other organs.
C.The left ventricle contains deoxygenated blood, which is more viscous than oxygenated blood.
D.The left ventricle has to pump a larger volume of blood with each contraction than the right ventricle.
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解题
The left ventricle is responsible for pumping oxygenated blood to all the tissues and organs in the body (systemic circulation). This requires a much higher pressure to overcome the resistance of the systemic blood vessels compared to the right ventricle, which only pumps blood a short distance to the lungs (pulmonary circulation). Therefore, the left ventricle has a thicker muscular wall to generate this higher pressure.
评分标准
1 mark: Identify that the left ventricle pumps blood around the body at higher pressure compared to the right ventricle pumping to the lungs.
题目 18 · 選擇題
1 分
An enzyme-controlled reaction is heated from 15 °C to 30 °C. Which statement correctly explains why the rate of reaction increases?
A.The kinetic energy of the molecules increases, resulting in more frequent collisions between enzymes and substrates.
B.The activation energy of the reaction decreases as the temperature rises.
C.The enzyme molecules change shape to allow more different types of substrates to bind.
D.The rate of denaturation of the enzyme increases, exposing more active sites.
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解题
Increasing the temperature increases the kinetic energy of both the enzyme and substrate molecules. This makes them move faster, resulting in more frequent collisions and a higher probability of successful collisions per unit time, thereby increasing the rate of reaction.
评分标准
1 mark: Relate the temperature increase to increased kinetic energy and higher collision frequency.
题目 19 · 選擇題
1 分
A section of an addition polymer is shown: \(-CH_2-CHCl-CH_2-CHCl-CH_2-CHCl-\). What is the formula of the monomer used to make this polymer?
A.\(CH_2=CHCl\)
B.\(CH_3-CH_2Cl\)
C.\(CHCl=CHCl\)
D.\(CH_2=CH_2\)
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解题
The repeating unit of the polymer is \(-CH_2-CHCl-\). To find the monomer, we replace the single carbon-carbon bond in the repeating unit with a double bond. This gives \(CH_2=CHCl\), which is chloroethene.
评分标准
1 mark: Deduce the correct monomer formula by identifying the repeating unit and introducing a double bond.
题目 20 · 選擇題
1 分
Which statement correctly explains why graphite conducts electricity but diamond does not, even though both are made entirely of carbon atoms?
A.In graphite, each carbon atom is bonded to three others, leaving one outer electron free to move, whereas in diamond all outer electrons are localized in bonds.
B.Graphite has a simple molecular structure with weak forces, whereas diamond has a giant covalent lattice with strong bonds.
C.Graphite contains mobile positive ions that conduct charge, whereas diamond contains only neutral atoms.
D.In graphite, carbon atoms are held together by ionic bonds, whereas in diamond they are held by covalent bonds.
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解题
In graphite, each carbon atom forms covalent bonds with three other carbon atoms in a hexagonal layer structure. This leaves one of the four valency electrons delocalised and free to move along the layers, conducting electricity. In diamond, each carbon atom is tetrahedrally bonded to four other carbon atoms, so all valency electrons are localized in covalent bonds, and none are free to move.
评分标准
1 mark: Identify the bonding difference in carbon atoms that leads to free/delocalised electrons in graphite but not in diamond.
题目 21 · 選擇題
1 分
A potential divider circuit is constructed using a fixed resistor and a thermistor connected in series with a constant d.c. power supply. The temperature of the room decreases. How do the resistance of the thermistor and the potential difference (p.d.) across the fixed resistor change?
A.The resistance of the thermistor increases, and the p.d. across the fixed resistor decreases.
B.The resistance of the thermistor increases, and the p.d. across the fixed resistor increases.
C.The resistance of the thermistor decreases, and the p.d. across the fixed resistor decreases.
D.The resistance of the thermistor decreases, and the p.d. across the fixed resistor increases.
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解题
For a standard NTC thermistor, a decrease in temperature causes its resistance to increase. Since the thermistor is in series with the fixed resistor, the total resistance of the circuit increases, which decreases the current flowing through the circuit. Since \(V = I \times R\), the potential difference across the fixed resistor (whose resistance \(R\) is constant) must decrease.
评分标准
1 mark: Correctly identify that thermistor resistance increases and the p.d. across the fixed resistor decreases as a result of the reduced current.
题目 22 · 選擇題
1 分
A trolley of mass 2.0 kg moving at a velocity of 4.0 m/s collides with a stationary trolley of mass 3.0 kg. The two trolleys stick together and move off with a common velocity \(v\). Assuming no external forces act on the system, what is the value of \(v\)?
A.1.6 m/s
B.2.0 m/s
C.2.7 m/s
D.4.0 m/s
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解题
By the law of conservation of momentum, total momentum before collision equals total momentum after collision. \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \implies (2.0 \times 4.0) + (3.0 \times 0) = (2.0 + 3.0) v \implies 8.0 = 5.0 v \implies v = 1.6\text{ m/s}\).
评分标准
1 mark: Use conservation of momentum equation to calculate the correct common velocity.
题目 23 · 選擇題
1 分
A stable star has a mass much greater than that of the Sun. Which sequence correctly shows the life cycle of this star after the stable period?
A.red supergiant \(\rightarrow\) supernova \(\rightarrow\) neutron star or black hole
B.red giant \(\rightarrow\) planetary nebula \(\rightarrow\) white dwarf
C.red giant \(\rightarrow\) supernova \(\rightarrow\) white dwarf
D.red supergiant \(\rightarrow\) planetary nebula \(\rightarrow\) black hole
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解题
A massive star (mass much greater than the Sun) expands to form a red supergiant after its stable hydrogen-fusing period. It then explodes in a supernova, leaving behind either a neutron star or a black hole, depending on the remaining core mass.
评分标准
1 mark: Recall the correct life cycle stages of a high-mass star.
题目 24 · 選擇題
1 分
The refractive index of a transparent plastic block is 1.50. What is the critical angle for light travelling from this plastic block into air?
A.41.8°
B.48.2°
C.30.0°
D.22.5°
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解题
The relationship between refractive index \(n\) and critical angle \(c\) is given by \(\sin c = \frac{1}{n}\). Substituting \(n = 1.50\) gives \(\sin c = \frac{1}{1.50} \approx 0.6667\). Thus, \(c = \arcsin(0.6667) \approx 41.8^\circ\).
评分标准
1 mark: Apply the equation \(\sin c = 1/n\) and calculate the correct critical angle.
题目 25 · 選擇題
1 分
An enzyme-controlled reaction is carried out at various temperatures. Which statement correctly explains why the rate of reaction decreases above the optimum temperature?
A.The kinetic energy of the substrate molecules decreases, leading to fewer successful collisions.
B.The enzyme molecules denature because the active site changes shape, preventing the substrate from binding.
C.The activation energy of the reaction increases, making it harder for the substrate to react.
D.The substrate molecules denature, so they can no longer fit into the active site of the enzyme.
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解题
Above the optimum temperature, the increased thermal energy causes the enzyme molecules to vibrate excessively. This disrupts the hydrogen bonds and other interactions holding the enzyme's three-dimensional structure together, changing the shape of the active site (denaturation). Consequently, the substrate can no longer fit into the active site, and the rate of reaction decreases.
评分标准
Award 1 mark for the correct option B. - Reject option A because kinetic energy increases with temperature. - Reject option C because enzymes do not change the thermodynamic activation energy threshold of reactions in this manner when they denature. - Reject option D because substrate molecules do not typically denature in this context.
题目 26 · 選擇題
1 分
Iron is extracted from iron(III) oxide, \( \text{Fe}_2\text{O}_3 \), using carbon monoxide. The equation for the reaction is shown.
What is the maximum mass of iron that can be obtained from 80 g of iron(III) oxide? [Relative atomic masses, \( A_r \): \( \text{O} = 16 \), \( \text{Fe} = 56 \)]
A.28 g
B.56 g
C.112 g
D.160 g
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解题
Step 1: Calculate the relative formula mass (\( M_r \)) of \( \text{Fe}_2\text{O}_3 \): \( M_r = (2 \times 56) + (3 \times 16) = 112 + 48 = 160 \).
Step 2: Calculate the number of moles of \( \text{Fe}_2\text{O}_3 \): \( \text{moles} = \frac{80\text{ g}}{160\text{ g/mol}} = 0.5\text{ mol} \).
Step 3: Determine the moles of iron (\( \text{Fe} \)) produced from the stoichiometry: From the balanced equation, 1 mole of \( \text{Fe}_2\text{O}_3 \) produces 2 moles of \( \text{Fe} \). \( \text{moles of Fe} = 0.5 \times 2 = 1.0\text{ mol} \).
Step 4: Calculate the mass of iron: \( \text{mass} = 1.0\text{ mol} \times 56\text{ g/mol} = 56\text{ g} \).
评分标准
Award 1 mark for the correct option B. - Option A incorrectly assumes a 1:1 molar ratio. - Option C is the mass if the stoichiometry is inverted or reactant mass is used directly. - Option D is the molar mass of the oxide.
题目 27 · 選擇題
1 分
A trolley of mass 4.0 kg moving at 3.0 m/s collides with a stationary trolley of mass 2.0 kg. The two trolleys stick together and move off with a common velocity \( v \). What is the value of \( v \)?
A.1.5 m/s
B.2.0 m/s
C.3.0 m/s
D.6.0 m/s
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解题
According to the law of conservation of momentum, total momentum before collision equals total momentum after collision.
After collision, they stick together, so the combined mass is: \( m_{\text{total}} = 4.0\text{ kg} + 2.0\text{ kg} = 6.0\text{ kg} \)
\( p_{\text{final}} = m_{\text{total}} \times v \) \( 12 = 6.0 \times v \) \( v = 2.0\text{ m/s} \)
评分标准
Award 1 mark for the correct option B. - Option A is obtained by dividing momentum by the wrong combination of masses. - Option C is the initial speed of the first trolley. - Option D is incorrect conservation step.
题目 28 · 選擇題
1 分
Which row correctly identifies the type of blood carried by the pulmonary artery, and the chamber of the heart from which it exits?
A.oxygenated blood, exiting from the left ventricle
B.oxygenated blood, exiting from the right ventricle
C.deoxygenated blood, exiting from the left ventricle
D.deoxygenated blood, exiting from the right ventricle
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解题
The pulmonary artery carries deoxygenated blood away from the heart to the lungs so that it can be oxygenated. This vessel exits from the right ventricle of the heart during ventricular systole.
评分标准
Award 1 mark for the correct option D.
题目 29 · 選擇題
1 分
Which statement correctly describes a reaction of alkenes?
A.They react with bromine water to turn it from blue to colorless.
B.They undergo substitution reactions with chlorine in the presence of ultraviolet light.
C.They undergo addition reactions with steam in the presence of an acid catalyst to produce alcohols.
D.They are saturated hydrocarbons that burn completely to produce carbon monoxide and hydrogen.
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解题
Alkenes are unsaturated hydrocarbons containing a carbon-carbon double bond, allowing them to undergo addition reactions. They react with steam in the presence of an acid catalyst (such as phosphoric acid) to produce alcohols. They also react with bromine water to turn it from orange/brown to colorless, not blue to colorless.
评分标准
Award 1 mark for the correct option C. - Option A incorrectly states the initial color of bromine water as blue. - Option B describes a substitution reaction, which is typical for alkanes, not alkenes. - Option D incorrectly states that alkenes are saturated hydrocarbons.
题目 30 · 選擇題
1 分
A ray of light in glass is incident on a boundary with air. The critical angle for the glass-air boundary is \( 42^\circ \). Which path does the ray take if its angle of incidence is \( 45^\circ \)?
A.It is partially refracted into the air and partially reflected back into the glass.
B.It is completely refracted into the air along the boundary.
C.It is totally internally reflected back into the glass.
D.It passes straight through the boundary without changing direction.
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解题
When light travels from a more optically dense medium (glass) to a less optically dense medium (air), it can undergo total internal reflection. Total internal reflection occurs if the angle of incidence is greater than the critical angle. Since the angle of incidence (\( 45^\circ \)) is greater than the critical angle (\( 42^\circ \)), the light is totally internally reflected back into the glass.
评分标准
Award 1 mark for the correct option C. - Option A describes behavior when the angle of incidence is less than the critical angle. - Option B describes behavior when the angle of incidence is exactly equal to the critical angle. - Option D describes behavior when light is incident along the normal.
题目 31 · 選擇題
1 分
During sexual reproduction in a flowering plant, what is the correct path taken by the male gametes to reach the female gamete?
A.down the style inside a pollen tube, entering the ovary, then entering an ovule
B.down the filament inside a pollen tube, entering the anther, then entering the ovary
C.through the air from the anther directly to the ovule inside the ovary
D.down the style by active transport through the phloem sieve tubes to the ovary
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解题
After pollination, pollen grains land on the stigma. The pollen grain germinates and grows a pollen tube down the style, which enters the ovary and finally enters an ovule through the micropyle, delivering the male gametes to fertilize the female gamete (egg cell).
评分标准
Award 1 mark for the correct option A.
题目 32 · 選擇題
1 分
Silicon(IV) oxide, \( \text{SiO}_2 \), has a macromolecular structure similar to diamond. Which properties does silicon(IV) oxide have?
A.low melting point and conducts electricity when molten
B.high melting point and conducts electricity when solid
C.low melting point and does not conduct electricity when solid
D.high melting point and does not conduct electricity when solid
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解题
Silicon(IV) oxide has a giant covalent macromolecular structure. All atoms are linked by strong covalent bonds, requiring a large amount of thermal energy to break, resulting in a high melting point. Since it does not contain free-moving ions or delocalized electrons, it does not conduct electricity in either solid or molten states.
评分标准
Award 1 mark for the correct option D.
题目 33 · 選擇題
1 分
An experiment investigates the rate of an enzyme-controlled reaction. At \(37^\circ\text{C}\), the reaction produces \(15\text{ cm}^3\) of gas in 2 minutes. When the temperature is increased to \(65^\circ\text{C}\), no gas is produced. Which statement explains this result?
A.The kinetic energy of the enzyme molecules decreased at the higher temperature.
B.The enzyme was denatured at \(65^\circ\text{C}\), changing the shape of its active site.
C.The activation energy of the reaction was lowered too much at \(65^\circ\text{C}\).
D.The substrate concentration became the limiting factor at \(65^\circ\text{C}\).
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解题
Increasing the temperature to \(65^\circ\text{C}\) provides too much thermal energy, causing the weak bonds holding the enzyme's 3D structure together to break. This permanently changes the shape of the active site (denaturation), meaning the substrate can no longer bind to it.
评分标准
Award 1 mark for selecting B. Award 0 marks for any other option.
题目 34 · 選擇題
1 分
What is the volume of carbon dioxide gas, measured at r.t.p., produced when \(5.0\text{ g}\) of calcium carbonate reacts completely with excess dilute hydrochloric acid?
[Relative atomic masses: \(\text{C} = 12\), \(\text{O} = 16\), \(\text{Ca} = 40\). One mole of any gas occupies \(24\text{ dm}^3\) at r.t.p.]
A.1.2 dm³
B.2.4 dm³
C.12 dm³
D.24 dm³
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解题
First, calculate the relative formula mass (\(M_r\)) of \(\text{CaCO}_3\): \(M_r(\text{CaCO}_3) = 40 + 12 + (3 \times 16) = 100\).
Next, find the moles of \(\text{CaCO}_3\): \(\text{moles} = 5.0\text{ g} / 100\text{ g/mol} = 0.05\text{ mol}\).
Since the molar ratio of \(\text{CaCO}_3\) to \(\text{CO}_2\) in the reaction \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O}\) is 1:1, the moles of \(\text{CO}_2\) produced is \(0.05\text{ mol}\).
Finally, calculate the volume of \(\text{CO}_2\) at r.t.p.: \(\text{Volume} = 0.05\text{ mol} \times 24\text{ dm}^3\text{/mol} = 1.2\text{ dm}^3\).
评分标准
Award 1 mark for selecting A. Award 0 marks for any other option.
题目 35 · 選擇題
1 分
A student connects two \(6.0\ \Omega\) resistors in parallel. This parallel combination is then connected in series with a \(3.0\ \Omega\) resistor and a \(12\text{ V}\) power supply. What is the total current flowing from the power supply?
A.1.3 A
B.2.0 A
C.3.0 A
D.4.0 A
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解题
1. Find the equivalent resistance of the two \(6.0\ \Omega\) resistors connected in parallel: \(R_p = \frac{6.0 \times 6.0}{6.0 + 6.0} = 3.0\ \Omega\).
2. Find the total resistance by adding the series resistor: \(R_{\text{total}} = R_p + 3.0 = 3.0 + 3.0 = 6.0\ \Omega\).
3. Use Ohm's Law to calculate the total current: \(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{6.0\ \Omega} = 2.0\text{ A}\).
评分标准
Award 1 mark for selecting B. Award 0 marks for any other option.
题目 36 · 選擇題
1 分
Which row correctly describes the structural features of a vein?
D.wall one cell thick, narrow lumen, valves absent
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解题
Veins carry blood under low pressure back to the heart. Therefore, they have a thin muscular wall and a wide lumen to reduce resistance to blood flow, and they contain valves to prevent the backflow of blood.
评分标准
Award 1 mark for selecting B. Award 0 marks for any other option.
题目 37 · 選擇題
1 分
Propene reacts with steam in the presence of an acid catalyst at high temperature and pressure. What is the main product of this reaction?
A.propane
B.propanol
C.propyl propanoate
D.poly(propene)
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解题
The reaction of an alkene (propene) with steam is an hydration reaction (a type of addition reaction) that yields an alcohol (propanol).
评分标准
Award 1 mark for selecting B. Award 0 marks for any other option.
题目 38 · 選擇題
1 分
A ray of light in air strikes the flat surface of a glass block. The angle of incidence is \(45^\circ\) and the angle of refraction inside the glass is \(28^\circ\). What is the refractive index of the glass?
Use Snell's Law: \(n = \frac{\sin i}{\sin r}\) \(n = \frac{\sin 45^\circ}{\sin 28^\circ} = \frac{0.71}{0.47} \approx 1.51\). Thus, the refractive index is approximately 1.5.
评分标准
Award 1 mark for selecting C. Award 0 marks for any other option.
题目 39 · 選擇題
1 分
Which sequence represents the correct order of the stages in the life cycle of a star with a mass much larger than the Sun?
A.nebula \(\rightarrow\) stable star \(\rightarrow\) red giant \(\rightarrow\) white dwarf
B.nebula \(\rightarrow\) stable star \(\rightarrow\) red supergiant \(\rightarrow\) supernova \(\rightarrow\) neutron star
C.stable star \(\rightarrow\) nebula \(\rightarrow\) supernova \(\rightarrow\) black hole
D.stable star \(\rightarrow\) red supergiant \(\rightarrow\) nebula \(\rightarrow\) white dwarf
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解题
A massive star starts as a nebula, becomes a stable massive star, expands into a red supergiant, explodes in a supernova, and leaves behind a neutron star (or a black hole).
评分标准
Award 1 mark for selecting B. Award 0 marks for any other option.
题目 40 · 選擇題
1 分
A leafy shoot is placed in a potometer to measure the rate of transpiration. Which set of environmental conditions would result in the highest rate of transpiration?
A.high humidity, high temperature, wind
B.low humidity, high temperature, wind
C.low humidity, low temperature, still air
D.high humidity, low temperature, still air
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解题
Transpiration rate increases when: 1. Humidity is low (creating a steeper concentration gradient of water vapor between the leaf and air). 2. Temperature is high (increasing the kinetic energy of water molecules and rate of evaporation). 3. Wind is present (removing water vapor from near the leaf surface, maintaining a steep concentration gradient).
评分标准
Award 1 mark for selecting B. Award 0 marks for any other option.
Answer all questions. Show your working and use appropriate units.
12 题目 · 120 分
题目 1 · Structured
10 分
A metallic wire has a resistance of 15 ohms. (a) Calculate the resistance of a wire made of the same metal that has twice the length and half the cross-sectional area. (b) Explain how the movement of free electrons in a metal wire constitutes an electric current, describing the direction of their movement relative to the potential difference. (c) Two resistors of 4.0 ohms and 6.0 ohms are connected in parallel. Calculate the combined resistance of the resistors. (d) Define the term electromotive force (e.m.f.) of a cell.
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解题
Part (a): Resistance \( R = \rho \frac{l}{A} \). If length \( l \) is doubled (\( \times 2 \)) and area \( A \) is halved (\( \times \frac{1}{2} \)), the resistance is multiplied by \( 4 \). Thus, new resistance = \( 15 \times 4 = 60\,\Omega \). Part (b): Metals contain a lattice of positive ions and delocalised electrons. When a potential difference is applied, the free electrons drift towards the positive potential. Current is defined as the rate of flow of charge. Part (c): \( \frac{1}{R_p} = \frac{1}{4.0} + \frac{1}{6.0} = \frac{3+2}{12} = \frac{5}{12} \Rightarrow R_p = 2.4\,\Omega \). Part (d): E.m.f. is the work done per unit charge in driving charge around a complete circuit.
评分标准
(a) [3 marks] Formula \( R \propto \frac{l}{A} \) or similar [1 mark]; recognition that resistance increases by a factor of 4 [1 mark]; correct final answer of 60 ohms [1 mark]. (b) [3 marks] Free/delocalised electrons [1 mark]; move/drift from negative to positive potential [1 mark]; flow of charge is current [1 mark]. (c) [2 marks] Correct formula used \( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \) [1 mark]; correct answer 2.4 (ohms) [1 mark]. (d) [2 marks] Energy transferred/work done per unit charge [1 mark]; in driving charge around a complete circuit [1 mark].
题目 2 · Structured
10 分
A block of granite has a volume of 45 cubic centimetres. The density of granite is 2.7 grams per cubic centimetre. (a) Calculate the mass of the granite block. (b) Determine the weight of this block on Earth where the gravitational field strength \( g = 9.8\,\text{N/kg} \). (c) A cyclist accelerates from rest to a speed of 6.0 metres per second in 8.0 seconds. Calculate the acceleration of the cyclist. (d) The mass of the cyclist and bicycle is 85 kilograms. Calculate the kinetic energy of the cyclist and bicycle when travelling at 6.0 metres per second. (e) Identify which two quantities from the following list are vectors: energy, acceleration, time, force, distance.
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解题
Part (a): Mass \( m = \text{density} \times \text{volume} = 2.7\,\text{g/cm}^3 \times 45\,\text{cm}^3 = 121.5\,\text{g} \). Part (b): Mass in kg = \( 0.1215\,\text{kg} \). Weight \( W = m \times g = 0.1215\,\text{kg} \times 9.8\,\text{N/kg} = 1.1907\,\text{N} \approx 1.2\,\text{N} \). Part (c): Acceleration \( a = \frac{v - u}{t} = \frac{6.0 - 0}{8.0} = 0.75\,\text{m/s}^2 \). Part (d): Kinetic energy \( E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 85\,\text{kg} \times (6.0\,\text{m/s})^2 = 1530\,\text{J} \). Part (e): Acceleration and force are vectors because they have both magnitude and direction.
评分标准
(a) [2 marks] \( m = d \times V \) or \( 2.7 \times 45 \) [1 mark]; mass = 121.5 (g) [1 mark]. (b) [2 marks] Convert mass to kg (\( 0.1215\,\text{kg} \)) [1 mark]; calculate weight \( 0.1215 \times 9.8 = 1.19\,\text{N} \) (accept 1.2 N) [1 mark]. (c) [2 marks] \( a = \frac{\Delta v}{t} \) or \( \frac{6.0}{8.0} \) [1 mark]; acceleration = 0.75 (m/s^2) [1 mark]. (d) [2 marks] \( E_k = \frac{1}{2} m v^2 \) or \( \frac{1}{2} \times 85 \times 6.0^2 \) [1 mark]; kinetic energy = 1530 (J) [1 mark]. (e) [2 marks] 1 mark for each correct vector: acceleration [1 mark], force [1 mark].
题目 3 · Structured
10 分
(a) Humans have a double circulation of blood. Explain the advantages of a double circulation compared to the single circulation of a fish. (b) Describe the function of capillaries and explain how their structure is adapted to their function. (c) Platelets are cellular fragments in blood. (i) State the function of platelets in the body. (ii) Suggest why a low platelet count is dangerous if a person gets a cut.
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解题
Part (a): In double circulation, blood is pumped to the lungs first (low pressure) and then returned to the heart to be pumped to the body at high pressure. This increases the rate of blood flow and oxygen delivery to tissues. Part (b): Capillaries have thin walls (one cell thick) to facilitate fast diffusion. They also have narrow lumens to slow down blood flow for efficient exchange. Part (c)(i): Platelets plug damaged blood vessels and release chemicals to initiate clotting. (c)(ii): Insufficient platelets mean slower clot formation, resulting in prolonged bleeding and an open pathway for infections.
评分标准
(a) [3 marks] Separation of oxygenated and deoxygenated blood [1 mark]; higher blood pressure around body/lower pressure to lungs [1 mark]; faster/more efficient delivery of oxygen/nutrients [1 mark]. (b) [3 marks] Exchange of substances between blood and tissues [1 mark]; walls are one cell thick [1 mark]; short diffusion distance [1 mark]. (c) (i) [2 marks] Blood clotting [1 mark]; scab/clot formation to prevent bleeding [1 mark]. (ii) [2 marks] Slower/less clotting [1 mark]; leading to increased blood loss or entry of pathogens [1 mark].
题目 4 · Structured
10 分
(a) Explain, using collision theory, why increasing the temperature of a reaction mixture increases the rate of reaction. (b) Catalysts are used to increase the rate of reactions. State how a catalyst increases the rate of a reaction. (c) Calcium carbonate reacts with dilute hydrochloric acid to produce calcium chloride, water, and carbon dioxide. (i) Write the balanced chemical equation with state symbols for this reaction. (ii) Describe a test for carbon dioxide gas and state the observation for a positive result.
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解题
Part (a): Temperature increase raises kinetic energy, particles move faster, more frequent collisions, more particles have the required activation energy, leading to more successful collisions. Part (b): Lower activation energy means more collisions are successful. Part (c)(i): State symbols are (s) for calcium carbonate, (aq) for hydrochloric acid and calcium chloride, (l) for water, and (g) for carbon dioxide. Part (c)(ii): Carbon dioxide gas reacts with calcium hydroxide (limewater) to form insoluble calcium carbonate, turning it milky.
评分标准
(a) [3 marks] Particles have more kinetic energy/move faster [1 mark]; more frequent collisions [1 mark]; greater proportion of collisions have energy >= activation energy / more successful collisions per unit time [1 mark]. (b) [2 marks] Provides alternative pathway [1 mark]; with lower activation energy [1 mark]. (c) (i) [3 marks] Correct reactants and products [1 mark]; correctly balanced [1 mark]; correct state symbols [1 mark]. (ii) [2 marks] Bubble through limewater [1 mark]; turns milky/cloudy [1 mark].
题目 5 · Structured
10 分
(a) State how the Big Bang Theory describes the beginning and evolution of the Universe. (b) Complete the stages in the life cycle of a star with a mass much larger than the Sun, starting from a stable star: very large mass stable star -> [Stage 1] -> [Stage 2] -> [Stage 3] (c) The Sun has planets orbiting it. (i) Describe the relationship between the orbital speed and the orbital radius of the planets as the orbital radius increases. (ii) Explain why the planets remain in orbit around the Sun.
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解题
Part (a): Big Bang posits cosmic expansion from a singularity. Part (b): Massive stars expand into red super giants, explode in supernovae, leaving behind a neutron star or black hole. Part (c)(i): Speed is inversely proportional to the square root of radius. Part (c)(ii): Gravity acts as the centripetal force.
评分标准
(a) [3 marks] Began as a single hot/dense point/singularity [1 mark]; rapid expansion [1 mark]; continues to expand and cool [1 mark]. (b) [3 marks] 1 mark for each stage: red super giant [1 mark]; supernova [1 mark]; black hole (or neutron star) [1 mark]. (c) (i) [2 marks] Speed decreases [1 mark]; as orbital radius increases [1 mark]. (ii) [2 marks] Gravitational force/pull of the Sun [1 mark]; provides centripetal force/acts towards the centre [1 mark].
题目 6 · Structured
10 分
A water wave has a wavelength of 0.12 metres and a frequency of 4.0 hertz. (a) (i) Calculate the speed of the wave. (ii) Describe what is meant by the frequency of a wave. (b) Explain what is meant by: (i) the amplitude. (ii) the wavelength. (c) Light travels from air into a rectangular glass block. (i) State what happens to the speed of light as it enters the glass block. (ii) State the direction of bending of the light ray relative to the normal as it enters the glass block at an angle of incidence of 45 degrees. (d) An object is placed at a distance of less than the focal length from a thin converging lens. State two characteristics of the image formed.
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解题
Part (a)(i): \( v = f \lambda = 4.0 \times 0.12 = 0.48\,\text{m/s} \). Part (a)(ii): definition of frequency. Part (b)(i), (ii): definitions of wave properties. Part (c)(i): Glass has a higher refractive index than air, so light slows down. Part (c)(ii): Bending is towards the normal since the speed decreases. Part (d): The lens acts as a magnifying glass, creating a virtual, upright, and magnified image.
评分标准
(a) (i) [2 marks] \( v = f \lambda \) or \( 4.0 \times 0.12 \) [1 mark]; speed = 0.48 (m/s) [1 mark]. (ii) [1 mark] Number of waves/oscillations passing a point per second [1 mark]. (b) (i) [1 mark] Maximum displacement from undisturbed/equilibrium position [1 mark]. (ii) [1 mark] Distance between two adjacent/consecutive crests/troughs [1 mark]. (c) (i) [1 mark] Decreases [1 mark]. (ii) [1 mark] Bends towards the normal [1 mark]. (d) [2 marks] Any two from: upright [1 mark], magnified [1 mark], virtual [1 mark].
题目 7 · Structured
10 分
Copper and magnesium are metals. (a) Copper is a transition element. State three characteristic properties of transition elements that are not shown by Group I elements. (b) Magnesium is in Group II and period 3 of the Periodic Table. (i) Determine the electronic configuration of a magnesium atom. (ii) Deduce the chemical formula of magnesium oxide. (c) Magnesium exists as three stable isotopes: \( ^{24}\text{Mg} \), \( ^{25}\text{Mg} \), and \( ^{26}\text{Mg} \). (i) Describe the similarity and the difference between these three isotopes. (ii) Explain why isotopes of the same element have the same chemical properties. (d) Brass is an alloy of copper and zinc. Explain, in terms of structure, why brass is harder than pure copper.
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解题
Part (a): Transition metal properties are unique compared to Group I/II. Part (b)(i): Magnesium has 12 electrons, so 2, 8, 2. Part (b)(ii): Mg has charge 2+, O has 2-, so MgO. Part (c)(i): isotopes differ only in neutrons. Part (c)(ii): chemical reactions depend on outer electrons. Part (d): Alloys are stronger due to lattice distortion.
评分标准
(a) [3 marks] Any three transition metal properties not shown by Group I: high density [1 mark]; high melting point [1 mark]; form coloured compounds [1 mark]; act as catalysts [1 mark]. (b) (i) [1 mark] 2.8.2 [1 mark]. (ii) [1 mark] MgO [1 mark]. (c) (i) [2 marks] Similarity: same number of protons/electrons (accept atomic number) [1 mark]; Difference: different number of neutrons (accept mass number) [1 mark]. (ii) [1 mark] Same electronic configuration / same number of valence electrons [1 mark]. (d) [2 marks] Zinc atoms are different size to copper atoms [1 mark]; this disrupts the regular layers / prevents layers sliding [1 mark].
题目 8 · Structured
10 分
(a) Calculate the relative molecular mass (\( M_r \)) of magnesium sulfate, \( \text{MgSO}_4 \). [Relative atomic masses: \( A_r(\text{Mg}) = 24 \), \( A_r(\text{S}) = 32 \), \( A_r(\text{O}) = 16 \)] (b) Determine the amount, in moles, of magnesium sulfate in 6.1 grams of magnesium sulfate. (c) A student reacts 5.6 grams of iron, \( \text{Fe} \), with 0.15 moles of hydrochloric acid, \( \text{HCl} \). The equation for the reaction is: \( \text{Fe(s)} + 2\text{HCl(aq)} \rightarrow \text{FeCl}_2\text{(aq)} + \text{H}_2\text{(g)} \) (i) Calculate the amount, in moles, of iron used. [Relative atomic mass: \( A_r(\text{Fe}) = 56 \)] (ii) Show, by calculation, which reactant is the limiting reactant.
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解题
Part (a): \( M_r(\text{MgSO}_4) = 24 + 32 + (4 \times 16) = 120 \). Part (b): \( \text{moles} = 6.1 / 120 = 0.0508 \approx 0.051\,\text{mol} \). Part (c)(i): \( \text{moles of Fe} = 5.6 / 56 = 0.10\,\text{mol} \). Part (c)(ii): stoichiometric ratio Fe : HCl is 1 : 2. 0.10 mol of Fe requires 0.20 mol of HCl. Since we only have 0.15 mol of HCl, HCl is the limiting reactant.
评分标准
(a) [2 marks] Correct calculation \( 24 + 32 + 64 \) [1 mark]; relative molecular mass = 120 [1 mark]. (b) [2 marks] \( \text{moles} = \frac{6.1}{120} \) [1 mark]; moles = 0.051 (mol) (accept 0.05) [1 mark]. (c) (i) [2 marks] \( \text{moles of Fe} = \frac{5.6}{56} \) [1 mark]; moles = 0.10 (mol) [1 mark]. (ii) [4 marks] Moles of HCl required for 0.10 mol of Fe is \( 2 \times 0.10 = 0.20\,\text{mol} \) [1 mark]; moles of HCl available is 0.15 mol [1 mark]; comparison showing \( 0.15 < 0.20 \) [1 mark]; correct conclusion that HCl is the limiting reactant [1 mark].
题目 9 · structured
10 分
An electric winch is used to lift a cargo crate of mass \(80\text{ kg}\) vertically upwards.
(a) Define the term velocity. [2]
(b) (i) Calculate the weight of the \(80\text{ kg}\) crate. Take the acceleration of free fall, \(g = 9.8\text{ m/s}^2\). [1]
(ii) Calculate the work done in lifting this crate vertically upwards through a height of \(12\text{ m}\). [2]
(iii) The lifting process takes \(8.0\text{ s}\). Calculate the useful power output of the winch. [2]
(c) The electrical power input supplied to the winch motor is \(1.5\text{ kW}\). Calculate the efficiency of the winch system. [3]
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解题
(a) Velocity is defined as the speed of an object in a specified direction, or the rate of change of displacement.
(b) (i) Weight is calculated as: \(W = m \times g = 80\text{ kg} \times 9.8\text{ m/s}^2 = 784\text{ N}\).
(ii) Work done is calculated as: \(W = F \times d = 784\text{ N} \times 12\text{ m} = 9408\text{ J}\).
(iii) Power is calculated as: \(P = \frac{W}{t} = \frac{9408\text{ J}}{8.0\text{ s}} = 1176\text{ W}\) (or \(1.18\text{ kW}\)).
(c) The useful power output is \(1176\text{ W}\). The total electrical power input is \(1.5\text{ kW} = 1500\text{ W}\). The efficiency is calculated as: \(\text{Efficiency} = \frac{\text{useful power output}}{\text{total power input}} \times 100\% = \frac{1176\text{ W}}{1500\text{ W}} \times 100\% = 78.4\%\) (or \(0.784\)).
评分标准
(a) speed; [1] in a given direction / rate of change of displacement; [1]
(b) (i) 784 N (allow 780 N); [1] (ii) formula used W = F * d or weight * distance; [1] 9408 J (or 9400 J); [1] (iii) formula used P = W / t; [1] 1176 W (accept 1180 W or 1.18 kW); [1]
(c) converting power input to W (1500 W); [1] use of efficiency formula: useful power / power input; [1] 78.4% or 0.784 (accept 78% or 0.78); [1]
题目 10 · structured
10 分
A student investigates the reduction of iron(III) oxide by carbon monoxide to extract iron. The balanced chemical equation for the reaction is: \(\text{Fe}_2\text{O}_3(\text{s}) + 3\text{CO}(\text{g}) \rightarrow 2\text{Fe}(\text{s}) + 3\text{CO}_2(\text{g})\)
(a) Explain, in terms of oxygen transfer, why iron(III) oxide is reduced in this reaction. [1]
(b) Determine the relative formula mass (\(M_r\)) of iron(III) oxide, \(\text{Fe}_2\text{O}_3\). [\(A_r\): \(\text{Fe} = 56\); \(\text{O} = 16\)] [2]
(c) A mixture of \(40\text{ g}\) of iron(III) oxide is reacted with \(25.2\text{ g}\) of carbon monoxide.
(i) Calculate the number of moles of iron(III) oxide in \(40\text{ g}\). [1]
(ii) Calculate the number of moles of carbon monoxide in \(25.2\text{ g}\). [\(M_r\) of \(\text{CO} = 28\)] [1]
(iii) Show by calculation which of the reactants is the limiting reactant. [3]
(d) Calculate the maximum mass of iron (\(\text{Fe}\)) that can be produced in this reaction. [2]
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解题
(a) Reduction is defined as the loss/removal of oxygen. Iron(III) oxide loses oxygen to form iron metal.
(b) The relative formula mass of \(\text{Fe}_2\text{O}_3\) is: \(M_r = (2 \times 56) + (3 \times 16) = 112 + 48 = 160\).
(ii) Moles of \(\text{CO}\): \(\text{moles} = \frac{\text{mass}}{M_r} = \frac{25.2\text{ g}}{28\text{ g/mol}} = 0.90\text{ mol}\).
(iii) According to the balanced equation, \(1\text{ mol}\) of \(\text{Fe}_2\text{O}_3\) reacts with \(3\text{ mol}\) of \(\text{CO}\). Therefore, \(0.25\text{ mol}\) of \(\text{Fe}_2\text{O}_3\) requires: \(0.25 \times 3 = 0.75\text{ mol}\) of \(\text{CO}\). Since \(0.90\text{ mol}\) of \(\text{CO}\) is present, \(\text{CO}\) is in excess, which makes iron(III) oxide, \(\text{Fe}_2\text{O}_3\), the limiting reactant.
(d) From the equation, \(1\text{ mol}\) of \(\text{Fe}_2\text{O}_3\) produces \(2\text{ mol}\) of \(\text{Fe}\). Therefore, \(0.25\text{ mol}\) of \(\text{Fe}_2\text{O}_3\) produces: \(0.25 \times 2 = 0.50\text{ mol}\) of \(\text{Fe}\).
The maximum mass of iron produced is: \(\text{mass} = \text{moles} \times A_r = 0.50\text{ mol} \times 56\text{ g/mol} = 28\text{ g}\).
评分标准
(a) loss / removal of oxygen (from iron(III) oxide); [1]
(c) (i) 0.25 (mol); [1] (ii) 0.90 (mol); [1] (iii) calculation of CO moles required for 0.25 mol of Fe2O3 (0.75 mol) OR calculation of Fe2O3 moles required for 0.90 mol of CO (0.30 mol); [1] comparison showing CO is in excess / required CO is less than available (0.75 < 0.90) OR required Fe2O3 is more than available (0.30 > 0.25); [1] correct deduction that iron(III) oxide is the limiting reactant; [1]
(d) mol ratio used to calculate moles of Fe produced = 0.50 (mol); [1] 28 g; [1]
题目 11 · structured
10 分
The mammalian heart is adapted to pump blood efficiently around the body.
(a) (i) State the name of the blood vessel that supplies the heart muscle itself with oxygenated blood. [1]
(ii) Name the valve located between the left atrium and the left ventricle. [1]
(b) Explain why the wall of the left ventricle is much thicker than the wall of the right ventricle. [2]
(c) Coronary heart disease (CHD) occurs when the blood supply to the heart muscle is restricted.
(i) Explain how diet and lifestyle choices can increase the risk of developing CHD. [3]
(ii) Describe how CHD can be treated medically or surgically. [3]
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解题
(a) (i) The coronary arteries branch off the aorta and supply the heart cardiac muscle tissue with oxygen and nutrients. (ii) The bicuspid valve (also called the mitral valve) is the left atrioventricular valve.
(b) The left ventricle must pump blood through the systemic circulation to the entire body, which is a much longer distance and requires generating a much higher pressure. The right ventricle only pumps blood to the lungs through the pulmonary circulation, which is a short distance and requires lower pressure to avoid damaging lung capillaries.
(c) (i) A diet rich in saturated fats and cholesterol raises blood cholesterol levels, leading to the formation of atheromas/plaques (fatty deposits) in coronary arteries. Smoking introduces nicotine and carbon monoxide, which damage artery walls, promote blood clot formation, and raise blood pressure. Lack of exercise and sedentary lifestyles can lead to obesity and high blood pressure, increasing cardiac workload. (ii) CHD can be treated medically using statins (to lower blood cholesterol) or aspirin (to prevent clots). Surgical treatments include angioplasty, where a tiny balloon is inflated to widen the narrowed artery and a stent is inserted to keep it open, or coronary artery bypass graft (CABG) surgery, where a vessel from another part of the body is grafted to bypass the blockage.
评分标准
(a) (i) coronary artery; [1] (ii) bicuspid / mitral / left atrioventricular valve; [1]
(b) left ventricle pumps blood to the rest of the body / further distance AND right ventricle pumps blood to the lungs / shorter distance; [1] left ventricle must pump at a higher pressure; [1]
(c) (i) Any three from: diet high in saturated fats/cholesterol increases blood cholesterol levels / forms fatty plaques/atheroma in coronary arteries; smoking damages coronary artery walls / increases blood clotting / raises blood pressure; lack of physical exercise increases risk of high blood pressure / obesity; high stress levels raise blood pressure / stress hormones damage arteries; [3]
(ii) Any three from: statins / drugs to lower cholesterol levels; aspirin / blood thinners to reduce clotting; angioplasty / inserting a balloon to widen arteries; stents to keep arteries open; coronary bypass surgery / grafting another vessel to bypass the blockage; [3]
题目 12 · structured
10 分
Waves can be classified into different types depending on their properties.
(a) Sound waves are longitudinal waves and light waves are transverse waves.
(i) Explain, in terms of particle vibration and wave propagation, the difference between a longitudinal wave and a transverse wave. [2]
(ii) State the name of the region of the electromagnetic spectrum that has the longest wavelength. [1]
(b) A ray of light travels from air into a glass block. The angle of incidence in air is \(45^\circ\) and the refractive index of the glass is \(1.5\).
(i) Calculate the angle of refraction inside the glass. [3]
(ii) State what happens to the speed and frequency of the light wave as it enters the glass from air. [2]
(c) Calculate the wavelength of a sound wave with a frequency of \(4.0\text{ kHz}\) travelling through water at a speed of \(1500\text{ m/s}\). [2]
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解题
(a) (i) In a longitudinal wave, the particles of the medium vibrate/oscillate parallel (in the same direction) to the direction of wave travel / energy propagation. In a transverse wave, the particles vibrate/oscillate perpendicular (at right angles) to the direction of wave travel / energy propagation. (ii) Radio waves have the longest wavelength in the electromagnetic spectrum.
(ii) Upon entering a optically denser medium like glass, the speed of the light wave decreases. However, the frequency of the wave depends only on the source and remains constant.
(c) Using the wave equation: \(v = f \lambda \implies \lambda = \frac{v}{f}\) First, convert frequency to Hz: \(f = 4.0\text{ kHz} = 4000\text{ Hz}\). Then calculate wavelength: \(\lambda = \frac{1500\text{ m/s}}{4000\text{ Hz}} = 0.375\text{ m}\).
评分标准
(a) (i) longitudinal: oscillations/vibrations of particles are parallel to wave travel / energy transfer; [1] transverse: oscillations/vibrations of particles are perpendicular to wave travel / energy transfer; [1] (ii) radio waves; [1]
(b) (i) recall/use of formula: n = sin i / sin r; [1] substitution/rearrangement: sin r = sin 45 / 1.5; [1] 28.1 degrees or 28 degrees; [1] (ii) speed: decreases; [1] frequency: stays constant / remains unchanged; [1]
(c) conversion of frequency to Hz: 4.0 kHz = 4000 Hz; [1] correct calculation of wavelength: 1500 / 4000 = 0.375 m (accept 0.38 m); [1]
Paper 6 Alternative to Practical
Answer all questions. Write your answers in the spaces provided.
6 题目 · 60 分
题目 1 · Practical Structured
10 分
A student investigates the effect of temperature on the rate of oxygen production by the enzyme catalase in a yeast suspension.
Catalase breaks down hydrogen peroxide into water and oxygen gas according to the word equation: $$\text{hydrogen peroxide} \rightarrow \text{water} + \text{oxygen}$$
The volume of oxygen gas produced in 2 minutes is measured at different temperatures. The results are shown in Table 1.1.
(a) Plot a graph of the volume of oxygen produced in 2 minutes (y-axis) against temperature (x-axis). Draw a smooth curve of best fit. [4]
(b) Using your graph, state the optimum temperature for the catalase enzyme in this yeast suspension. [1]
(c) With reference to the structure of enzymes, explain why the volume of oxygen produced decreases significantly at \(60\,^{\circ}\text{C}\) compared to \(40\,^{\circ}\text{C}\). [2]
(d) Identify one variable, other than temperature, that must be kept constant in this investigation to ensure a fair test, and state how it would be controlled. [2]
(e) Suggest a more accurate piece of apparatus to measure the volume of oxygen gas produced than a standard measuring cylinder inverted over water. [1]
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解题
(a) Graph plotting requirements: - Axes labelled with quantity and unit (Temperature / \(^{\circ}\text{C}\) on x-axis, Volume of oxygen produced / \(\text{cm}^3\) on y-axis). - Suitable linear scale chosen so that points occupy at least half the grid. - All 5 points correctly plotted to within half a small square. - A smooth, single-line curve of best fit drawn through the points.
(b) The peak of the curve occurs at \(40\,^{\circ}\text{C}\).
(c) High thermal energy disrupts the weak bonds in the enzyme structure, permanently altering the three-dimensional shape of its active site (denaturation). Consequently, the substrate molecule no longer fits into the active site.
(d) Controlled variables include: - Concentration of hydrogen peroxide solution: controlled by using the same stock bottle. - Volume of yeast suspension used: controlled by measuring with a graduated pipette or syringe. - pH of the reaction mixture: controlled using a buffer solution.
(e) A gas syringe is much more accurate as it has finer graduations and avoids errors due to gas dissolving in water.
评分标准
(a) [4 marks total]: - 1 mark for correct axes labels with units. - 1 mark for sensible linear scales covering at least half of the grid. - 1 mark for all points plotted correctly (allow \(\pm 0.5\) small square). - 1 mark for a smooth curve of best fit going through the peak.
(b) [1 mark]: - Correct optimum temperature read from peak of graph (typically \(40\,^{\circ}\text{C}\)).
(c) [2 marks total]: - 1 mark for stating the enzyme is denatured. - 1 mark for explaining that the shape of the active site changes / substrate can no longer fit.
(d) [2 marks total]: - 1 mark for identifying a valid variable (e.g., concentration/volume of hydrogen peroxide, volume/concentration of yeast suspension, pH). - 1 mark for stating how it is controlled (e.g., use a syringe to measure exact volume, use a buffer).
(e) [1 mark]: - Accept: Gas syringe. - Reject: Burette.
题目 2 · Practical Structured
10 分
A student wants to investigate the effect of different concentrations of sodium chloride solution on the mass of potato tissue cylinders to determine the internal concentration of potato cells.
(a) Plan an investigation to find the concentration of sodium chloride solution that is isotonic to the potato tissue (where there is no net movement of water into or out of the cells). [7]
Your plan should include: - the apparatus and materials you will need - a brief description of the method, including how you will vary the independent variable - the measurements you will make and how you will ensure they are as valid as possible - the variables you will control - how you will process your results to find the isotonic concentration.
(b) Explain why it is important to blot the potato cylinders dry with a paper towel before weighing them. [1]
(c) During a trial run, a potato cylinder placed in a high-concentration sodium chloride solution showed a percentage mass change of \(-12.4\%\).
State the direction of water movement and explain this change in terms of water potential. [2]
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解题
(a) Experimental Plan Guidelines: 1. **Apparatus & Materials**: Potato, cork borer, scalpel, ruler, paper towels, electronic balance, test tubes/beakers, sodium chloride solutions of different concentrations, distilled water. 2. **Independent Variable**: Use a range of at least 5 different salt concentrations (e.g., 0.0, 0.2, 0.4, 0.6, 0.8 \(\text{mol/dm}^3\)). 3. **Dependent Variable**: Measure the mass of each potato cylinder before and after immersion using an electronic balance. 4. **Controlled Variables**: Keep cylinder diameter constant (use same cork borer), same initial length, same volume of solution, same temperature, and same immersion time (e.g., 1 hour). 5. **Method**: Cut potato cylinders, blot dry, record initial mass. Place one cylinder in each solution. Leave for 1 hour. Remove, blot dry, record final mass. 6. **Processing**: Calculate percentage change in mass: \(\frac{\text{final mass} - \text{initial mass}}{\text{initial mass}} \times 100\). Plot a graph of percentage mass change (y-axis) against NaCl concentration (x-axis). The concentration where the line crosses the x-axis (0% change) is the isotonic point.
(b) Surface liquid is not part of the tissue and varies in volume; blotting ensures only the tissue mass is measured.
(c) Since the external solution has a lower water potential (hypertonic) than the potato cytoplasm, water leaves the vacuole of the cells by osmosis down a water potential gradient, causing a decrease in mass.
评分标准
(a) [7 marks total] - 1 mark for each of the following points: - Identifies at least 5 different concentrations of sodium chloride solution. - Explains how potato cylinders are prepared to a uniform size (e.g., using a cork borer and cutting to same length). - States that initial mass of each cylinder is recorded. - Identifies the key variable of immersion time and states it must be kept constant (e.g., 30-60 minutes). - States that cylinders must be blotted dry before reweighing. - Explains that percentage change in mass is calculated to allow comparison. - Explains how to use the results (e.g., plot graph of concentration vs % mass change and locate where the curve crosses the x-axis / 0% mass change line).
(b) [1 mark]: - To remove surface liquid/water that would inaccurately increase the final mass.
(c) [2 marks total]: - 1 mark for stating water moves out of the cells. - 1 mark for stating that water moves from high water potential (inside cells) to low water potential (salt solution) / by osmosis.
题目 3 · Practical Structured
10 分
A student is given a solid hydrated salt, compound **Y**, which is green. They dissolve the salt in distilled water to prepare solution **Y** and carry out several qualitative tests. The tests and observations are recorded in Table 3.1.
### Table 3.1 | Test number | Test | Observation | | :--- | :--- | :--- | | **1** | Add a few drops of aqueous sodium hydroxide to solution **Y**, then add in excess. | Green precipitate formed, insoluble in excess. | | **2** | Add a few drops of aqueous ammonia to solution **Y**, then add in excess. | Green precipitate formed, insoluble in excess. | | **3** | Add dilute nitric acid followed by aqueous barium nitrate to solution **Y**. | Thick white precipitate formed. | | **4** | Add dilute nitric acid followed by aqueous silver nitrate to solution **Y**. | No change / solution remains clear. |
(a) Identify the ions present in compound **Y**: - (i) the cation [1] - (ii) the anion [1]
(b) State the chemical formula of the anhydrous salt present in **Y**. [1]
(c) Explain why dilute nitric acid is added before the barium nitrate solution in Test 3. [2]
(d) Describe how the student would carry out a flame test on the solid sample of compound **Y** to confirm the absence of any Group I metal contaminants. Include the expected observation if sodium ions were present. [3]
(e) Identify which test reagent from Table 3.1 should be replaced, and what it should be replaced with, if the student wanted to test for chloride ions instead of sulfate ions. [2]
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解题
(a) (i) A green precipitate with both sodium hydroxide and ammonia, which is insoluble in excess, indicates the presence of iron(II) ions, \(\text{Fe}^{2+}\). (ii) The thick white precipitate with barium nitrate in acidic conditions confirms the presence of sulfate ions, \(\text{SO}_4^{2-}\).
(b) The formula of iron(II) sulfate is \(\text{FeSO}_4\).
(c) Dilute nitric acid decomposes any carbonate impurities (producing carbon dioxide gas). If carbonate ions were present, they would react with barium ions to form barium carbonate, which is also a white precipitate, leading to a false positive for sulfate.
(d) Steps for a flame test: 1. Use a clean nichrome or platinum wire. 2. Clean the wire by dipping in concentrated hydrochloric acid and holding in a hot Bunsen flame until no colour is observed. 3. Dip the wire into the solid sample and place it in the outer edge of a blue (non-luminous) Bunsen flame. - Observation for sodium: a bright, persistent yellow flame.
(e) To test for chloride, use Test 4 reagents (dilute nitric acid and aqueous silver nitrate), which yield a white precipitate of silver chloride if chloride is present.
评分标准
(a) [2 marks total]: - (i) 1 mark for iron(II) / \(\text{Fe}^{2+}\) (reject iron / \(\text{Fe}^{3+}\)) - (ii) 1 mark for sulfate / \(\text{SO}_4^{2-}\)
(c) [2 marks total]: - 1 mark for stating it removes/reacts with carbonate ions. - 1 mark for explaining this prevents the formation of barium carbonate white precipitate / false positive.
(d) [3 marks total]: - 1 mark for describing cleaning the wire using acid and heating. - 1 mark for placing the wire with sample in a blue/non-luminous flame. - 1 mark for stating that a yellow flame indicates sodium.
(e) [2 marks total]: - 1 mark for using silver nitrate (with nitric acid). - 1 mark for noting it produces a white precipitate with chloride ions.
题目 4 · Practical Structured
10 分
An experiment is carried out to investigate the rate of reaction between excess calcium carbonate (marble chips) and dilute hydrochloric acid: $$\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}$$
A student monitors the reaction by measuring the total mass of the flask and its contents over time. The results are shown in Table 4.1.
(a) Explain why the mass of the flask and its contents decreases during the reaction. [1]
(b) State the purpose of placing a loose plug of cotton wool in the neck of the flask during the reaction. [1]
(c) On a grid, plot a graph of the total loss in mass / \(\text{g}\) (y-axis) against time / \(\text{s}\) (x-axis). Draw a smooth curve of best fit. [4]
(d) Use your graph to: - (i) determine the time at which the reaction was complete. [1] - (ii) explain how the shape of the graph shows that the rate of the reaction decreases over time. [2]
(e) State what change could be made to the calcium carbonate to increase the rate of this reaction without changing its mass. [1]
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解题
(a) The carbon dioxide gas evolved escapes into the atmosphere since the flask is open, causing a measurable reduction in the mass of the system.
(b) The cotton wool plug acts as a physical barrier to stop aerosol droplets of acid spray from being carried out by the escaping gas, which would artificially increase the apparent mass loss.
(c) Graph plotted with time on the x-axis (0 to 300 s) and total loss in mass on the y-axis (0.00 to 1.40 g). All points plotted accurately. A smooth curve leveling off at 1.30 g is drawn.
(d) (i) The reaction is complete when the mass no longer changes, which corresponds to the point where the curve becomes flat (horizontal) at \(240\,\text{s}\). (ii) Rate is represented by the gradient (slope) of the curve. The curve is steepest at the start (highest rate) and progressively gets flatter, showing that the rate of reaction is decreasing as the acid is used up.
(e) Crushing the marble chips into smaller pieces increases the total surface area available for collisions with hydrochloric acid molecules, speeding up the reaction.
评分标准
(a) [1 mark]: - Gas / carbon dioxide escapes from the flask.
(b) [1 mark]: - Prevents loss of acid spray / liquid splashes, but allows gas to escape.
(c) [4 marks total]: - 1 mark for correct axes labels with units (Time / \(\text{s}\) on x-axis and Total loss in mass / \(\text{g}\) on y-axis). - 1 mark for a linear scale with plotted points spanning at least 50% of both axes. - 1 mark for all 8 points correctly plotted (within half a small square). - 1 mark for a smooth curve of best fit that starts at the origin and levels off horizontally.
(d) [3 marks total]: - (i) 1 mark for stating \(240\,\text{s}\) (accept range \(220\text{--}240\,\text{s}\)). - (ii) 1 mark for mentioning the gradient/slope decreases. - 1 mark for linking the decreasing slope to a decreasing rate of reaction.
(e) [1 mark]: - Use smaller chips / powder / increase the surface area of the calcium carbonate.
题目 5 · Practical Structured
10 分
A student investigates how the electrical resistance of a constantan wire varies with its length. They use a d.c. power supply, an ammeter, a voltmeter, a switch, and a metre rule to measure different lengths of wire.
(a) Draw a circuit diagram to show how this apparatus is connected to measure the resistance of a test wire of length \(L\). [2]
(b) The student records the current \(I\) and potential difference \(V\) for five different lengths \(L\) of the wire. The results are shown in Table 5.1.
Calculate the resistance \(R\) of the wire for each length using the equation: $$R = \frac{V}{I}$$
Record the values of \(R\) to two significant figures in Table 5.1. [2]
(c) Describe the relationship between the length of the wire \(L\) and its electrical resistance \(R\). [1]
(d) State one variable regarding the physical properties of the wire that must be kept constant during this experiment. [1]
(e) During the experiment, the constantan wire becomes warm. - (i) Explain why allowing the wire to heat up significantly affects the accuracy of the resistance measurements. [2] - (ii) Suggest a practical step the student should take during the procedure to prevent the wire from becoming too warm. [2]
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解题
(a) The circuit diagram should show: - Power source, switch, ammeter, and the portion of the constantan wire connected in a single series loop. - Voltmeter connected in parallel across the specific length \(L\) of constantan wire under test.
(b) Calculations of resistance: - At \(20\,\text{cm}\): \(R = 0.52 / 0.40 = 1.3\,\Omega\) - At \(40\,\text{cm}\): \(R = 1.04 / 0.40 = 2.6\,\Omega\) - At \(60\,\text{cm}\): \(R = 1.56 / 0.40 = 3.9\,\Omega\) - At \(80\,\text{cm}\): \(R = 2.08 / 0.40 = 5.2\,\Omega\) - At \(100\,\text{cm}\): \(R = 2.60 / 0.40 = 6.5\,\Omega\)
(c) Since \(R/L\) remains constant (\(1.3 / 20 = 0.065\,\Omega/\text{cm}\)), resistance is directly proportional to the length of the wire.
(d) The wire must have uniform thickness (constant diameter/cross-sectional area) and be made of the same material throughout.
(e) (i) As metal wire heats up, the lattice ions vibrate more vigorously, causing more collisions with conducting electrons. This increases the resistance. Temperature thus acts as an uncontrolled variable. (ii) Turn off the current (by opening the switch) immediately after recording each pair of \(V\) and \(I\) values, allowing the wire to cool down before the next length is measured.
评分标准
(a) [2 marks total]: - 1 mark for correct symbols of power supply, switch, ammeter, and voltmeter. - 1 mark for correct connections: ammeter in series, voltmeter in parallel across the test wire.
(b) [2 marks total]: - 1 mark for at least three correct calculations of resistance. - 1 mark for all values correct to 2 significant figures (1.3, 2.6, 3.9, 5.2, 6.5).
(c) [1 mark]: - Resistance is directly proportional to length / resistance increases as length increases linearly.
(e) [4 marks total]: - (i) 1 mark for stating that resistance increases as temperature increases. - 1 mark for stating that temperature acts as an uncontrolled variable (making it an unfair test). - (ii) 1 mark for turning off the current / opening the switch between readings. - 1 mark for explaining this allows the wire to cool down / prevents heat buildup.
题目 6 · Practical Structured
10 分
A student determines the focal length \(f\) of a thin converging lens. They place an illuminated object (the letter 'E') at a distance \(u\) from the lens and adjust the position of a screen on the other side of the lens until a sharp image of the object is formed.
The distance from the lens to the screen is the image distance \(v\).
(a) Suggest one precaution the student should take when setting up the apparatus to ensure that the measurements of \(u\) and \(v\) are as accurate as possible. [1]
(b) Describe how the student can ensure that they have located the exact position for a sharp image on the screen. [1]
(c) In one trial, the student sets the object distance \(u = 24.0\,\text{cm}\). The sharp image is formed at a screen distance \(v = 48.0\,\text{cm}\).
Calculate the focal length \(f\) of the lens using the formula: $$f = \frac{u \times v}{u + v}$$
Show your working. [2]
(d) The magnification \(m\) of the image in this trial is calculated using: $$m = \frac{v}{u}$$
- (i) Calculate the magnification \(m\). [1] - (ii) Describe the appearance of the image formed on the screen in this trial compared to the original object, with reference to its size and orientation. [2]
(e) The student repeats the experiment for several larger values of \(u\). They find that the image on the screen becomes very small and dim, making it difficult to find the exact focus.
Suggest two changes to the setup or environment that would make it easier to see and locate this faint image. [2]
(f) State what happens to the image distance \(v\) as the object distance \(u\) is increased. [1]
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解题
(a) Precautionary steps: - Keep the ruler parallel to the optical bench to avoid parallax errors. - Align the center of the illuminated object, the center of the lens, and the center of the screen on the same horizontal axis.
(b) The screen should be moved past the sharp point in both directions so the student can judge the region of best focus, then settle in the middle of this narrow range.
(d) (i) \(m = 48.0 / 24.0 = 2.0\) (no unit). (ii) Since \(m > 1\), the image is magnified (larger). A real image formed on a screen by a single converging lens is always inverted.
(e) To locate a faint image easily: - Darken the room (turn off ambient lights) to increase contrast. - Increase the brightness of the light source behind the object.
(f) According to the lens equation, as \(u\) increases, \(v\) decreases and approaches the focal length \(f\).
评分标准
(a) [1 mark]: - Align centers of object, lens, and screen horizontally / place ruler parallel to optical bench / perform in a dark room.
(b) [1 mark]: - Move the screen back and forth to find the sharpest focus (or find the midpoint of the range of clarity).
(c) [2 marks total]: - 1 mark for correct substitution of numbers: \((24.0 \times 48.0) / (24.0 + 48.0)\). - 1 mark for correct final value with unit: \(16.0\,\text{cm}\) (accept \(16\,\text{cm}\)).
(d) [3 marks total]: - (i) 1 mark for correct calculation of magnification: \(2.0\) (accept \(2\)). - (ii) 1 mark for stating the image is magnified / larger. - 1 mark for stating the image is inverted / upside down.
(e) [2 marks total]: - 1 mark for darkening the room / using a black shield around the screen. - 1 mark for using a brighter lamp / light source.
(f) [1 mark]: - Image distance decreases.
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