An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 OCR AS Level Physics B (Advancing Physics) - H157 paper. Not affiliated with or reproduced from OCR.
卷一 甲部
Twenty multiple-choice questions, one mark each; spend a maximum of about 25 minutes; write the letter in the answer box.
20 题目 · 20 分
题目 1 · 選擇題
1 分
A 12 V supply of negligible internal resistance is connected across a fixed resistor of resistance \(6.0\ \text{k}\Omega\) in series with a second resistor of resistance \(2.0\ \text{k}\Omega\). The output is taken across the \(2.0\ \text{k}\Omega\) resistor. What is the output potential difference?
A.\(1.0\ \text{V}\)
B.\(3.0\ \text{V}\)
C.\(4.0\ \text{V}\)
D.\(9.0\ \text{V}\)
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解题
\(V_{out} = V_{in} \times \frac{R_2}{R_1+R_2} = 12 \times \frac{2.0}{6.0+2.0} = 3.0\ \text{V}\). Option D is the p.d. across the 6.0 kΩ resistor.
评分标准
Answer B (1 mark). Correct selection only. Distractors: D = p.d. across the wrong resistor; C = equal-ratio error; A = misuse of ratio.
题目 2 · 選擇題
1 分
A uniform metal wire has length \(2.0\ \text{m}\), diameter \(0.50\ \text{mm}\) and resistivity \(4.9\times10^{-7}\ \Omega\,\text{m}\). What is its resistance?
A.\(1.2\ \Omega\)
B.\(2.5\ \Omega\)
C.\(5.0\ \Omega\)
D.\(20\ \Omega\)
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解题
\(A = \pi (0.25\times10^{-3})^2 = 1.96\times10^{-7}\ \text{m}^2\). \(R = \rho L/A = 4.9\times10^{-7}\times2.0 / 1.96\times10^{-7} = 5.0\ \Omega\). Using the diameter as the radius gives 1.2 Ω.
评分标准
Answer C (1 mark). Method: area from radius (not diameter), then \(R=\rho L/A\). Distractor A arises from using diameter as radius.
题目 3 · 選擇題
1 分
A dynamics trolley of mass \(0.80\ \text{kg}\) moves at \(1.5\ \text{m s}^{-1}\) and collides with a stationary trolley of mass \(0.40\ \text{kg}\). The trolleys stick together. What is their common speed immediately after the collision?
A.\(0.50\ \text{m s}^{-1}\)
B.\(0.75\ \text{m s}^{-1}\)
C.\(1.0\ \text{m s}^{-1}\)
D.\(1.5\ \text{m s}^{-1}\)
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解题
Momentum is conserved: \(0.80\times1.5 = (0.80+0.40)v\), so \(v = 1.2/1.2 = 1.0\ \text{m s}^{-1}\).
评分标准
Answer C (1 mark). Conservation of momentum with combined mass. Distractor D ignores the added mass.
题目 4 · 選擇題
1 分
A resistance is found from \(R = V/I\). The potential difference is \(V = 6.0 \pm 0.1\ \text{V}\) and the current is \(I = 0.50 \pm 0.02\ \text{A}\). What is the best estimate of the absolute uncertainty in \(R\)?
A.\(0.1\ \Omega\)
B.\(0.7\ \Omega\)
C.\(1.2\ \Omega\)
D.\(2.4\ \Omega\)
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解题
\(R = 12\ \Omega\). Percentage uncertainties: V \(= 0.1/6.0 = 1.7\%\); I \(= 0.02/0.50 = 4.0\%\). Add for a quotient: 5.7%. \(0.057\times12 = 0.68 \approx 0.7\ \Omega\).
评分标准
Answer B (1 mark). Add percentage uncertainties for division, then convert to absolute. Distractor C: 12 × (0.1) misuse; D: doubling.
题目 5 · 選擇題
1 分
An audio signal has a maximum frequency of \(5.0\ \text{kHz}\). It is digitised at the minimum sampling rate that allows it to be reconstructed, using 12 bits per sample. What is the transmission rate?
A.\(6.0\times10^{4}\ \text{bit s}^{-1}\)
B.\(1.2\times10^{5}\ \text{bit s}^{-1}\)
C.\(1.2\times10^{4}\ \text{bit s}^{-1}\)
D.\(2.4\times10^{5}\ \text{bit s}^{-1}\)
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解题
Minimum sampling rate = \(2\times5.0\ \text{kHz} = 10\ \text{kHz}\). Transmission rate = samples per second × bits per sample = \(1.0\times10^4\times12 = 1.2\times10^5\ \text{bit s}^{-1}\).
评分标准
Answer B (1 mark). Distractor A uses the signal frequency without doubling; C is a power-of-ten slip.
题目 6 · 選擇題
1 分
A cell of e.m.f. \(1.5\ \text{V}\) and internal resistance \(0.50\ \Omega\) is connected across a \(2.5\ \Omega\) resistor. What is the power dissipated in the internal resistance of the cell?
A.\(0.063\ \text{W}\)
B.\(0.125\ \text{W}\)
C.\(0.25\ \text{W}\)
D.\(0.63\ \text{W}\)
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解题
\(I = \varepsilon/(R+r) = 1.5/3.0 = 0.50\ \text{A}\). \(P = I^2 r = 0.25\times0.50 = 0.125\ \text{W}\). Option D is the power in the 2.5 Ω load.
评分标准
Answer B (1 mark). Current from total resistance, then \(I^2r\). Distractor D: load power; C: \(Ir\) without squaring.
题目 7 · 選擇題
1 分
A thin converging lens has focal length \(10\ \text{cm}\). A small object is placed \(30\ \text{cm}\) from the lens. Which row gives the image distance and the magnitude of the linear magnification?
Answer C (1 mark). Distractor A is \(h/\lambda\) (momentum); B and D are prefix (power-of-ten) errors.
题目 9 · 選擇題
1 分
A copper wire has a resistance of \(2.5\ \Omega\). A second wire is made of the same copper. It is twice as long and has half the diameter of the first wire. What is the resistance of the second wire?
A.\(5.0\ \Omega\)
B.\(20\ \Omega\)
C.\(10\ \Omega\)
D.\(40\ \Omega\)
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解题
\(R = \rho L / A\) and \(A \propto d^2\). Doubling L doubles R. Halving d reduces A to one quarter, so R is multiplied by 4. Overall factor = 2 x 4 = 8. \(R = 8 \times 2.5 = 20\ \Omega\).
评分标准
B1 for B (20 ohm). Distractors: A = only length doubled and area halved (factor 2); C = factor 4 (area dependence treated as linear in d); D = factor 16 (R proportional to L squared and d^-2 slip). No partial marks for multiple choice.
题目 10 · 選擇題
1 分
A potential divider is made from a \(12\ \text{V}\) supply of negligible internal resistance, a fixed resistor of \(4.0\ \text{k}\Omega\) and a light-dependent resistor (LDR) connected in series. The LDR is the lower component and has a resistance of \(2.0\ \text{k}\Omega\) in bright light. A voltmeter of very high resistance is connected across the LDR. What is the voltmeter reading in bright light?
B1 for D (4.0 V). Distractors: A = voltage across the fixed resistor; B = half of supply (equal resistances assumed); C = 12 / 6 slip (current-like value).
题目 11 · 選擇題
1 分
A cell of e.m.f. \(6.0\ \text{V}\) and internal resistance \(0.80\ \Omega\) is connected across a resistor of resistance \(2.2\ \Omega\). What is the power dissipated in the internal resistance of the cell?
B1 for A (3.2 W). Distractors: B = I^2 r / 2 slip; C = total power from cell (EI); D = power in the 2.2 ohm load resistor.
题目 12 · 選擇題
1 分
A trolley of mass \(0.80\ \text{kg}\) moves at \(1.5\ \text{m s}^{-1}\) along a level track and collides with a stationary trolley of mass \(0.40\ \text{kg}\). The trolleys stick together. Frictional forces are negligible. What is the speed of the trolleys immediately after the collision?
A.\(0.50\ \text{m s}^{-1}\)
B.\(1.0\ \text{m s}^{-1}\)
C.\(0.75\ \text{m s}^{-1}\)
D.\(1.5\ \text{m s}^{-1}\)
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解题
Momentum is conserved: \(0.80 \times 1.5 = (0.80+0.40)v\), so \(v = 1.20/1.20 = 1.0\ \text{m s}^{-1}\).
评分标准
B1 for B (1.0 m/s). Distractors: A = divided by wrong total mass (2.4); C = half the initial speed (equal masses assumed); D = speed unchanged.
题目 13 · 選擇題
1 分
The density of a metal cube is found from its mass and the length of one side. The side is \((2.00 \pm 0.02)\ \text{cm}\) and the mass is \((21.6 \pm 0.2)\ \text{g}\). What is the percentage uncertainty in the calculated density?
A.\(3.9\%\)
B.\(1.9\%\)
C.\(3.0\%\)
D.\(5.8\%\)
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解题
Percentage uncertainty in side = 1.0%, so in volume (side cubed) = 3.0%. Percentage uncertainty in mass = 0.2/21.6 = 0.93%. Density = mass/volume, so add percentages: 3.0 + 0.93 = 3.93%, i.e. 3.9%.
评分标准
B1 for A (3.9%). Distractors: B = forgot to multiply the length uncertainty by 3; C = volume uncertainty only; D = 3 x (1.0 + 0.93) slip.
题目 14 · 選擇題
1 分
An audio signal with a maximum frequency of \(5.0\ \text{kHz}\) is sampled at \(12\ \text{kHz}\) using 8 bits per sample. How much data is produced in \(30\ \text{s}\)?
A.\(0.36\ \text{Mbit}\)
B.\(5.8\ \text{Mbit}\)
C.\(2.9\ \text{Mbit}\)
D.\(23\ \text{Mbit}\)
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解题
Samples per second = 12000. Bits per second = 12000 x 8 = 96000. Over 30 s: 96000 x 30 = 2.88 x 10^6 bit, i.e. 2.9 Mbit.
评分标准
B1 for C (2.9 Mbit). Distractors: A = answer in megabytes (0.36 MB) quoted as Mbit; B = 2 x correct value; D = extra factor 8 applied.
题目 15 · 選擇題
1 分
A thin converging lens has a focal length of \(10\ \text{cm}\). A small object is placed \(30\ \text{cm}\) from the lens. At what distance from the lens is the real image formed?
A.\(20\ \text{cm}\)
B.\(7.5\ \text{cm}\)
C.\(40\ \text{cm}\)
D.\(15\ \text{cm}\)
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解题
\(1/v = 1/u + 1/f\) with \(u = -30\ \text{cm}\) and \(f = +10\ \text{cm}\): \(1/v = -1/30 + 1/10 = 2/30\), so \(v = 15\ \text{cm}\).
评分标准
B1 for D (15 cm). Distractors: A = u - f; B = sign error giving 1/30 + 1/10; C = u + f.
题目 16 · 選擇題
1 分
An LED emits light of wavelength \(590\ \text{nm}\). What is the energy of one photon of this light?
B1 for B. Distractors: A = wavelength used as 590 m (power-of-ten slip, nm not converted to 10^-9 m); C = h/lambda (photon momentum value); D = power-of-ten slip in nm conversion the other way.
题目 17 · 選擇題
1 分
A potential divider is made from a 12 V supply of negligible internal resistance, a fixed resistor of resistance \(4.7\ \text{k}\Omega\) and a thermistor connected in series. The output is taken across the thermistor and is \(3.0\ \text{V}\). What is the resistance of the thermistor?
A.\(1.2\ \text{k}\Omega\)
B.\(1.6\ \text{k}\Omega\)
C.\(14\ \text{k}\Omega\)
D.\(19\ \text{k}\Omega\)
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解题
Using \(V_{out} = V_{in}\dfrac{R_2}{R_1+R_2}\): \(3.0 = 12\dfrac{R_2}{4700+R_2}\). So \(14100 + 3R_2 = 12R_2\), giving \(R_2 = 14100/9 = 1567\ \Omega \approx 1.6\ \text{k}\Omega\). Equivalently the fixed resistor has 9.0 V across it, so the ratio of resistances is 3.0 : 9.0 and \(R_2 = 4.7/3\). Answer B.
评分标准
1 mark for B. Method: potential divider equation or ratio of p.d.s equals ratio of resistances (V1/V2 = R1/R2). Distractor A: \(4.7/4\) (using 3 V across total ratio 1:4 wrongly). C: thermistor placed as the upper resistor (output across fixed resistor). D: \(4\times 4.7\). No partial credit for a multiple-choice item.
题目 18 · 選擇題
1 分
A trolley of mass \(0.80\ \text{kg}\) moves at \(1.5\ \text{m s}^{-1}\) along a level track and collides with a stationary trolley of mass \(0.40\ \text{kg}\). The trolleys stick together. How much kinetic energy is transferred to other forms in the collision?
A.\(0.15\ \text{J}\)
B.\(0.60\ \text{J}\)
C.\(0.30\ \text{J}\)
D.\(0.90\ \text{J}\)
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解题
Momentum conservation: \(0.80\times1.5 = 1.2\ v\), so \(v = 1.0\ \text{m s}^{-1}\). Initial KE \(=\tfrac12(0.80)(1.5)^2 = 0.90\ \text{J}\). Final KE \(=\tfrac12(1.2)(1.0)^2 = 0.60\ \text{J}\). Energy transferred \(=0.90-0.60 = 0.30\ \text{J}\). Answer C.
评分标准
1 mark for C. Method: conservation of momentum to find common speed 1.0 m/s; then difference in kinetic energies. Distractor A: half of the loss; B: final KE only; D: initial KE only (all energy lost).
题目 19 · 選擇題
1 分
A ball leaves the edge of a horizontal table-top platform moving horizontally at \(8.0\ \text{m s}^{-1}\). The platform is \(4.9\ \text{m}\) above level ground. Air resistance is negligible and \(g = 9.81\ \text{m s}^{-2}\). What is the horizontal distance travelled by the ball before it hits the ground?
A.\(8.0\ \text{m}\)
B.\(4.0\ \text{m}\)
C.\(16\ \text{m}\)
D.\(39\ \text{m}\)
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解题
Vertical motion: \(s = \tfrac12 g t^2\), so \(t = \sqrt{2\times4.9/9.81} = 1.0\ \text{s}\). Horizontal velocity is constant, so \(x = 8.0\times1.0 = 8.0\ \text{m}\). Answer A.
评分标准
1 mark for A. Method: independent vertical and horizontal motion; find time from vertical drop; multiply by horizontal speed. Distractor B: halves the time; C: doubles it; D: uses t = 4.9 s (height treated as time).
题目 20 · 選擇題
1 分
The resistivity of a wire is found using \(\rho = \dfrac{\pi d^2 R}{4L}\). The measurements are: diameter \(d = 0.50 \pm 0.01\ \text{mm}\), length \(L = 80.0 \pm 0.2\ \text{cm}\) and resistance \(R = 2.0 \pm 0.1\ \Omega\). What is the percentage uncertainty in the calculated resistivity?
A.7%
B.5%
C.14%
D.9%
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解题
Percentage uncertainties: \(d\): \(0.01/0.50 = 2\%\), counted twice because \(d\) is squared, giving 4%. \(L\): \(0.2/80.0 = 0.25\%\). \(R\): \(0.1/2.0 = 5\%\). Total \(=4+0.25+5 = 9.25\% \approx 9\%\). Answer D.
评分标准
1 mark for D. Method: add percentage uncertainties for products/quotients, doubling that of the squared quantity. Distractor A: forgets to double d (2+0.25+5); B: largest single uncertainty only; C: adds absolute-style errors incorrectly (e.g. doubling R as well).
Four short structured questions with calculations, a drawing task and a classification task.
9 题目 · 20 分
题目 1 · Short calculation
2 分
A greenhouse sensor circuit has a 9.0 V supply connected across two fixed resistors in series. The upper resistor has resistance 4.7 kΩ and the lower resistor has resistance 2.2 kΩ. The output is taken across the lower resistor.
Calculate the output potential difference. [2]
Answer = ..... V
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解题
\(V_{out} = V_{in} \times \frac{R_2}{R_1+R_2} = 9.0 \times \frac{2.2}{4.7+2.2} = 9.0 \times 0.3188 = 2.87\) V, i.e. 2.9 V.
评分标准
B1 (method): correct substitution into potential divider equation, \(9.0 \times 2.2/(4.7+2.2)\) (allow kΩ values used consistently). A1: 2.87 V or 2.9 V. Allow: use of \(I = 9.0/6900 = 1.30\) mA then \(V = IR\) for full marks. Do not allow: use of 4.7 kΩ as the lower resistor (gives 6.1 V) – 1 mark max (XP). POT error in resistances only if inconsistent: max 1 mark.
题目 2 · Short calculation
2 分
On a level air track, a glider of mass 0.80 kg moves at 1.5 m s\(^{-1}\) and collides with a stationary glider of mass 1.2 kg. The two gliders stick together after the collision.
Calculate the speed of the joined gliders immediately after the collision. [2]
Answer = ..... m s\(^{-1}\)
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解题
Momentum is conserved: \(0.80 \times 1.5 = (0.80+1.2)\,v\), so \(1.20 = 2.0\,v\) and \(v = 0.60\) m s\(^{-1}\).
评分标准
B1 (method): momentum before = momentum after, \(0.80 \times 1.5 = (0.80+1.2)v\). A1: 0.60 m s\(^{-1}\). Allow: 0.6 m s\(^{-1}\). Do not allow: 1.2 m s\(^{-1}\) (total mass omitted, XP) – 0 marks for A1; kinetic energy conserved – 0 marks. ECF for arithmetic slip in momentum before.
题目 3 · Short calculation
2 分
A thin converging lens of focal length 10 cm forms a real image of a small lamp placed 30 cm from the lens.
Calculate the distance from the lens to the image. [2]
Answer = ..... cm
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解题
Using \(\frac{1}{v} = \frac{1}{u} + \frac{1}{f}\) with \(u = -30\) cm and \(f = 10\) cm: \(\frac{1}{v} = -\frac{1}{30} + \frac{1}{10} = \frac{2}{30}\), so \(v = 15\) cm.
评分标准
B1 (method): correct substitution, \(1/v = 1/10 - 1/30\) (or \(1/u + 1/v = 1/f\) with \(u = 30\)). A1: 15 cm. Allow: 0.15 m. Do not allow: 7.5 cm (reciprocal not taken, AE/XP) or 40 cm (sign error: \(1/10 + 1/30\)) – max 0 for A1. Ignore: a negative sign on the answer if the image is clearly stated as real.
题目 4 · Short calculation
2 分
A voice signal contains frequencies up to 4.0 kHz. It is digitised by sampling 10 000 times per second, using 12 bits for each sample.
(a) State whether this sampling rate is suitable for the signal, giving a reason. [1]
(b) Calculate the number of bits transmitted in one second. [1]
Answer = ..... bit s\(^{-1}\)
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解题
(a) The minimum sampling rate must exceed \(2 \times 4.0\) kHz = 8.0 kHz; 10 kHz is greater, so it is suitable. (b) Transmission rate = samples per second \(\times\) bits per sample = \(10\,000 \times 12 = 1.2 \times 10^{5}\) bit s\(^{-1}\).
评分标准
(a) B1: suitable because sampling rate (10 kHz) is more than twice the maximum signal frequency (8 kHz). Do not allow 'suitable' with no comparison. (b) B1: \(1.2 \times 10^{5}\) bit s\(^{-1}\) (120 000). Allow 120 kbit s\(^{-1}\). Do not allow 1.0 x 10^4 (bits per sample omitted). ECF: none needed as parts are independent.
题目 5 · Short calculation
2 分
A cell of e.m.f. 1.5 V and internal resistance 0.40 Ω is connected to a resistor of resistance 2.6 Ω.
Calculate the power dissipated in the internal resistance of the cell. [2]
Answer = ..... W
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解题
Total resistance = 2.6 + 0.40 = 3.0 Ω, so \(I = \frac{1.5}{3.0} = 0.50\) A. Power in internal resistance \(P = I^2 r = 0.50^2 \times 0.40 = 0.10\) W.
评分标准
B1 (method): current found from \(\varepsilon/(R+r)\) = 0.50 A. A1: \(P = I^2 r = 0.10\) W. Allow: use of lost volts 0.20 V with \(P = IV\). Do not allow: use of 1.5 V with 0.40 Ω or power calculated for the 2.6 Ω resistor (0.65 W) – max 1 mark for the current. ECF from an incorrect current.
题目 6 · Explain / draw / classify
2.5 分
A weather buoy digitises the signal from a wave-height sensor. The highest frequency component of the signal is 4.5 kHz. Each sample is stored as a 10-bit number.
(a) State the minimum sampling rate needed to reproduce the signal without loss of information. [0.5]
(b) The buoy samples at 10 kHz. Calculate the transmission rate in bits per second. [1]
(c) Calculate the number of kilobytes of data produced in 30 s at this rate. (1 byte = 8 bits, 1 kB = 1000 bytes) [1]
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解题
(a) Sampling rate must exceed \(2 \times 4.5 = 9.0\) kHz. (b) Rate = samples per second \(\times\) bits per sample = \(10\,000 \times 10 = 1.0 \times 10^{5}\) bit s\(^{-1}\). (c) Bits in 30 s = \(1.0 \times 10^{5} \times 30 = 3.0 \times 10^{6}\) bits. Bytes = \(3.0\times10^{6}/8 = 3.75\times10^{5}\), so 375 kB.
评分标准
(a) Greater than 9 kHz (0.5). Allow 'more than twice the highest frequency'. Do not allow exactly 4.5 kHz. (b) Substitution \(10\,000 \times 10\) (method, 0.5); \(1.0\times10^5\) bit s\(^{-1}\) (accuracy, 0.5). Allow 100 kbit s\(^{-1}\). POT error loses the accuracy mark. (c) \(1.0\times10^5 \times 30\) (0.5); divide by 8 to give 375 kB (0.5). ECF from (b). Allow 3.75 x 10^2 kB.
题目 7 · Explain / draw / classify
2.5 分
A thin converging lens has focal length 6.0 cm. An object 2.0 cm tall stands upright on the principal axis, 9.0 cm from the lens.
(a) On a grid with the lens at the centre, scale 1 cm : 1 cm, draw a ray diagram with two construction rays to locate the image. Mark both principal foci (F) and label the image. [1.5]
(b) Calculate the image distance and the height of the image, and state whether the image is real or virtual. Use your answers to say what your diagram should show. [1]
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解题
(a) Ray 1: from the top of the object parallel to the axis, then refracted through the far focus F. Ray 2: from the top of the object straight through the centre of the lens, undeviated. The rays cross 18 cm beyond the lens, 4.0 cm below the axis; the image is drawn as an inverted arrow there. (b) \(1/v = 1/f - 1/u\) in the form \(1/v = 1/6.0 - 1/9.0 = 1/18\), so \(v = 18\) cm. \(m = v/u = 18/9.0 = 2.0\), so height = 4.0 cm. The image is real (rays actually meet) and inverted.
评分标准
(a) Ray parallel to axis then through F (0.5); ray through centre undeviated (0.5); foci marked at 6.0 cm each side and image labelled inverted at about 18 cm (0.5). Allow +/- 2 mm on the grid. (b) \(v = 18\) cm by correct use of the lens equation (0.5); height 4.0 cm and real (0.5). Allow ECF only if the diagram agrees with the calculated values. Do not allow a virtual image.
题目 8 · Explain / draw / classify
2.5 分
A technician tests three samples in a tensile rig: a copper wire, a glass rod and a polythene strip.
(a) Classify each sample as a metal, a ceramic or a polymer. [1.5]
(b) The glass rod fractures with almost no plastic deformation, whereas the copper wire stretches a long way before breaking. Explain this difference in terms of the internal structure of each material. [1]
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解题
(a) Copper is a metal, glass is a ceramic, polythene is a polymer (long-chain molecules). (b) In copper, dislocations move under stress so layers of atoms slip past each other, giving plastic flow (ductile). In glass the bonds are strong and rigid and there are no mobile dislocations, so slip cannot occur and cracks spread, causing brittle fracture.
评分标准
(a) One mark per correct pair of three, scaled: copper metal (0.5), glass ceramic (0.5), polythene polymer (0.5). (b) Copper has dislocations that can move / layers slip (0.5); glass has no mobile dislocations so it cannot deform plastically and fractures / is brittle (0.5). Allow 'atoms slide over one another' for slip. Ignore references to 'weak' or 'strong' on their own.
题目 9 · Explain / draw / classify
2.5 分
A light sensor uses a fixed resistor of 4.7 k\(\Omega\) in series with a light-dependent resistor (LDR) across a 6.0 V supply of negligible internal resistance. The output voltage is taken across the LDR.
(a) In room light the output voltage is 2.0 V. Calculate the resistance of the LDR. [1.5]
(b) The room is darkened. State and explain the change in the output voltage. [1]
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解题
(a) Voltage across the fixed resistor = \(6.0 - 2.0 = 4.0\) V. Current = \(4.0/4700 = 8.51\times10^{-4}\) A. LDR resistance = \(2.0/8.51\times10^{-4} = 2350\,\Omega \approx 2.4\) k\(\Omega\). (Equivalent: \(V_1/V_2 = R_1/R_2\), so \(R = 4700 \times 2.0/4.0\).) (b) In the dark the LDR resistance rises (fewer mobile charge carriers), so it takes a larger share of the supply voltage in the potential divider and the output voltage increases.
评分标准
(a) Fixed resistor p.d. 4.0 V or current \(8.5\times10^{-4}\) A (0.5); correct method to find R from \(V/I\) or ratio of p.d.s (0.5); 2.4 k\(\Omega\) / 2350 \(\Omega\) (0.5). Allow 2.3 to 2.4 k\(\Omega\). POT error loses the final mark. (b) Output voltage increases (0.5); LDR resistance increases so larger fraction of the supply p.d. across it (0.5). Do not allow 'current increases'.
卷一 部分 C
Four longer structured questions combining recall, data analysis, graph work and practical evaluation.
4 题目 · 30 分
题目 1 · Structured multi-part questions
7.5 分
A greenhouse controller uses a potential divider to monitor temperature. A 5.0 V supply of negligible internal resistance is connected across a fixed resistor of resistance 4.7 k\(\Omega\) in series with an NTC thermistor. The output voltage \(V_{out}\) is taken across the fixed resistor.
(a) At 20 \(^\circ\)C the thermistor has resistance 6.8 k\(\Omega\). (i) Calculate \(V_{out}\) at 20 \(^\circ\)C. [2] Answer = ..... V (ii) At 35 \(^\circ\)C the thermistor resistance is 3.1 k\(\Omega\). Show that \(V_{out}\) is about 3.0 V. [1]
(b) Explain, in terms of charge carriers, why \(V_{out}\) increases as the temperature rises. [2]
(c) A heating element is made from a uniform nichrome wire of length 1.50 m and diameter 0.20 mm. The resistivity of nichrome is \(1.1\times10^{-6}\ \Omega\,\mathrm{m}\). Calculate the resistance of the wire. [3] Answer = ..... \(\Omega\)
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解题
(a)(i) \(V_{out}=5.0\times4.7/(4.7+6.8)=23.5/11.5=2.04\) V. (ii) \(V_{out}=5.0\times4.7/(4.7+3.1)=23.5/7.8=3.01\) V, about 3.0 V. (b) Temperature rise gives more mobile charge carriers (electrons freed) in the semiconductor, so its resistance falls. The thermistor's share of the supply pd falls (\(V_1/V_2=R_1/R_2\)), so the pd across the fixed resistor rises. (c) \(A=\pi(0.10\times10^{-3})^2=3.14\times10^{-8}\ \mathrm{m^2}\). \(R=\rho L/A=1.1\times10^{-6}\times1.50/3.14\times10^{-8}=52.5\ \Omega\).
评分标准
(a)(i) M1 correct substitution into potential divider equation 5.0 x 4.7/(4.7+6.8) [1]; A1 2.04 V (allow 2.0) [1]. (a)(ii) Full substitution shown with unrounded 3.01 (V) [1]. Do not allow a bare statement of 3.0. (b) B1 more (mobile) charge carriers / electrons released at higher temperature [1]; B1 thermistor resistance falls so (pd shares in ratio of resistances) larger fraction of pd across fixed resistor [1]. Ignore 'collisions' arguments. Do not allow 'more charge' without 'mobile carriers'. (c) C1 area = pi (d/2)^2 = 3.14e-8 m^2 [1] (using diameter rather than radius gives XP: lose this mark and ECF); C1 correct use of R = rho L / A [1]; A1 52.5 (ohm), allow 52 or 53 [1]. POT error in diameter: max 1 mark.
题目 2 · Structured multi-part questions
7.5 分
On a horizontal track a trolley of mass 0.80 kg moving at 0.60 m s\(^{-1}\) collides with a stationary trolley of mass 0.40 kg. The trolleys stick together and move off along the track.
(a) State the principle of conservation of momentum. [1]
(b) Calculate the speed of the joined trolleys just after the collision. [2] Answer = ..... m s\(^{-1}\)
(c) Calculate the kinetic energy lost in the collision and state what this shows about the type of collision. [2]
(d) The joined trolleys then slow uniformly to rest over a distance of 0.25 m because of friction. Calculate the frictional force. [2] Answer = ..... N
(e) The collision lasts 0.050 s. Calculate the mean force exerted on the 0.40 kg trolley during the collision. [1] Answer = ..... N
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解题
(a) In a closed system (no external force) the total momentum before an interaction equals the total momentum after. (b) \(0.80\times0.60=(0.80+0.40)v\), so \(v=0.48/1.2=0.40\ \mathrm{m\,s^{-1}}\). (c) KE before \(=\tfrac12\times0.80\times0.60^2=0.144\) J; after \(=\tfrac12\times1.20\times0.40^2=0.096\) J; loss \(=0.048\) J. Kinetic energy is not conserved so the collision is inelastic. (d) Work done by friction = KE lost: \(F\times0.25=0.096\), \(F=0.384\approx0.38\) N. (Check: \(a=v^2/2s=0.32\ \mathrm{m\,s^{-2}}\), \(F=1.2\times0.32=0.384\) N.) (e) \(F=\Delta p/\Delta t=0.40\times0.40/0.050=3.2\) N.
评分标准
(a) Total momentum (of a system) before = after [1] provided no external force / closed system [accept 'in the absence of resultant external force']. Allow B1 for the statement with condition; without condition 0. (b) C1 0.80 x 0.60 = 1.20 v [1]; A1 0.40 (m/s) [1]. Using 1.2 incorrectly as 0.8 on RHS: max 1. (c) C1 0.144 J and 0.096 J (or 0.048 J seen) [1]; A1 inelastic as KE is lost (0.048 J) [1]. ECF from (b). (d) C1 work done by friction = KE (0.096 J) = F x 0.25, or a = v^2/2s then F = ma [1]; A1 0.38 (0.384) N [1]. ECF from (b). Allow 0.4. (e) A1 F = change in momentum / time = 0.16/0.050 = 3.2 N [1]. Allow ECF from (b) (0.40 x v / 0.050). Do not allow use of total mass.
题目 3 · Structured multi-part questions
7.5 分
A student measures the acceleration of free fall by dropping a steel ball through different heights \(h\) and timing the fall \(t\) with a light gate and data-logger. Results:
(a) Calculate the missing value of \(t^2\) for \(h=1.000\) m. [1]
(b) The student plots a graph of \(h\) against \(t^2\). Explain why this should be a straight line through the origin and state what the gradient represents. [2]
(c) Use the first and last points in the table to determine the gradient, and hence a value for \(g\). [2] Answer = ..... m s\(^{-2}\)
(d) For the \(h = 0.400\) m drop, \(h\) has an uncertainty of \(\pm0.005\) m and \(t\) an uncertainty of \(\pm0.002\) s. Using \(g=2h/t^2\), calculate the percentage uncertainty in \(g\) for this drop and the absolute uncertainty. [2]
(e) Using your answer to (c) and the data given, state whether the results are consistent with \(g=9.81\ \mathrm{m\,s^{-2}}\) and justify your answer. [1]
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解题
(a) \(0.452^2=0.2043\ \mathrm{s^2}\). (b) \(h=\tfrac12 g t^2\) from \(s=ut+\tfrac12at^2\) with \(u=0\); this has the form \(y=mx\), so \(h\) against \(t^2\) is linear through the origin. Gradient \(=g/2\). (c) Gradient \(=(1.200-0.400)/(0.2450-0.0818)=0.800/0.1632=4.90\); \(g=2\times4.90=9.80\ \mathrm{m\,s^{-2}}\). (d) \(\%\Delta h=0.005/0.400=1.25\%\); \(\%\Delta t=0.002/0.286=0.70\%\), doubled for \(t^2\) gives 1.40%. Total \(=2.65\approx2.7\%\). \(g=2\times0.400/0.286^2=9.78\); absolute uncertainty \(=0.0265\times9.78=0.26\ \mathrm{m\,s^{-2}}\). (e) 9.81 lies within \(9.78\pm0.26\) (and the graph value 9.80 differs from 9.81 by about 0.1%), so the data are consistent with 9.81.
评分标准
(a) 0.2043 or 0.204 [1]. (b) B1 h = (1/2) g t^2 (from s = ut + 1/2 at^2 with u = 0) so h proportional to t^2 [1]; B1 gradient = g/2 [1]. Do not allow gradient = g. (c) C1 (1.200 - 0.400)/(0.2450 - 0.0818) = 4.90 [1]; A1 g = 2 x gradient = 9.80 (m/s^2) [1]. Allow 9.8. Allow 4.9 as gradient. Using h/t^2 at one point only: max 1. (d) C1 percentages added: 1.25% + 2 x 0.70% [1]; A1 2.7% (allow 2.6 to 2.7) and 0.26 m s^-2 (allow 0.25 to 0.27) [1]. Doubling of time uncertainty missed: max 0 for A1 (ECF absolute). (e) B1 statement that 9.81 lies within the uncertainty range of the value (9.78 +/- 0.26) or that 9.80 is within 0.1% of 9.81, so consistent [1]. ECF from (c)/(d). Do not allow 'close' without comparison with uncertainty.
题目 4 · Structured multi-part questions
7.5 分
A digital camera has a thin converging lens of focal length 50 mm and an image sensor of 4000 \(\times\) 3000 pixels.
(a) The camera photographs a person standing 2.0 m from the lens. (i) Calculate the distance from the lens to the sensor needed for a sharp image. [2] Answer = ..... mm (ii) Calculate the magnification. [1] (iii) State the power of the lens. [1]
(b) Each pixel is stored using 12 bits. Calculate the number of bytes needed to store one uncompressed image. Give your answer in standard form. [2] Answer = ..... bytes
(c) The camera also records sound containing frequencies up to 15 kHz. (i) State the minimum sampling rate needed to digitise this sound without loss of information. [1] (ii) The sound is actually sampled at 44.1 kHz using 16 bits per sample. Calculate the transmission rate in kbit s\(^{-1}\) for one channel. [1]
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解题
(a)(i) \(1/v=1/u+1/f\) with \(u=-2.0\) m, \(f=0.050\) m: \(1/v=-0.50+20=19.5\), \(v=0.0513\) m \(=51.3\) mm. (ii) \(m=v/u=0.0513/(-2.0)=-0.026\) (image inverted, magnitude 0.026). (iii) \(P=1/f=1/0.050=20\) D (dioptres, \(\mathrm{m^{-1}}\)). (b) Pixels \(=4000\times3000=1.2\times10^7\); bits \(=1.2\times10^7\times12=1.44\times10^8\); bytes \(=1.44\times10^8/8=1.8\times10^7\). (c)(i) More than twice the highest frequency: greater than \(2\times15=30\) kHz. (ii) Rate \(=44100\times16=7.056\times10^5\) bit s\(^{-1}\) \(=706\) kbit s\(^{-1}\).
评分标准
(a)(i) C1 1/v = 1/u + 1/f with u = -2.0 (or 1/v = 1/0.050 - 1/2.0) [1]; A1 51.3 mm (allow 51) [1]. Wrong sign for u giving 48.8 mm: lose A1 (XP). Answer in metres 0.0513: allow if unit consistent. (ii) m = v/u = -0.026 [1]. Allow 0.026 or 0.0256; ECF from (i). (iii) 20 D (or m^-1) [1]. Unit required. Do not allow 20 mm^-1 or 0.02. (b) C1 4000 x 3000 x 12 = 1.44e8 bits [1]; A1 divide by 8 = 1.8 x 10^7 bytes [1]. Answer 1.44 x 10^8 (bits) scores 1. Standard form needed for A1. (c)(i) Greater than 30 kHz [1]. Allow 'at least 30 kHz'. Do not allow 15 kHz. (ii) 44100 x 16 = 7.06 x 10^5 bit/s = 706 kbit/s [1]. Allow 705.6 and 706; POT error scores 0.
卷二 甲部
Four practical-based questions on experiment methods, calculations, graphs and error analysis.
4 题目 · 26 分
题目 1 · Practical multi-part questions
7 分
A student determines the resistivity of a constantan wire. The wire is 1.000 m long and is connected in a circuit with a cell, an ammeter and a voltmeter connected across the wire. The potential difference across the wire is 1.20 V when the current is 0.150 A. A micrometer screw gauge is used to measure the diameter of the wire.
(a) Describe how the student should use the micrometer to obtain a reliable value for the mean diameter of the wire. [2]
(b) The mean diameter is 0.28 mm. (i) Calculate the resistivity of the constantan. [3]
Answer = ..... \(\Omega\) m
(ii) The uncertainty in the diameter is \(\pm 0.01\) mm. The percentage uncertainties in the resistance and in the length are 1% and 0.2% respectively. Calculate the percentage uncertainty in the resistivity and state the resistivity with its absolute uncertainty. [2]
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解题
(a) Measure the diameter at several (e.g. 5) positions along the wire, and at two perpendicular directions at each position, checking the zero error of the micrometer first; take the mean.
(ii) % uncertainty in d = 0.01/0.28 = 3.6%; in \(d^2\) = 7.1%. Total = 7.1 + 1 + 0.2 = 8.3% (about 8%). Absolute = 0.083 x 4.93e-7 = 0.41e-7. \(\rho = (4.9\pm0.4)\times10^{-7}\ \Omega\,\text{m}\).
评分标准
(a) Measurements at several places along the wire [1]; and in perpendicular directions / rotate the wire [1]; take the mean. Allow: check/correct for zero error (can replace one of the first two points if the other is given). Ignore: 'repeat readings' with no detail.
(b)(i) \(R = 1.20/0.150 = 8.0\ \Omega\) [1]; \(A=\pi d^2/4 = 6.2\times10^{-8}\ \text{m}^2\) (6.158e-8 unrounded) [1]; \(\rho = RA/L = 4.9\times10^{-7}\ \Omega\,\text{m}\) [1]. Allow 2-3 s.f. Using d as radius or d in mm without conversion: POT/XP, lose accuracy mark; ECF to (ii) not affected.
(b)(ii) Diameter % uncertainty doubled because d is squared: 7.1% [1]; sum with 1% and 0.2% = 8.3% (allow 8%) and \(\rho=(4.9\pm0.4)\times10^{-7}\) [1]. Allow ECF from (i) for the absolute value. Do not allow 3.6% without doubling.
题目 2 · Practical multi-part questions
6 分
A student tests conservation of momentum using two trolleys on a level runway. A 0.050 m long card on trolley A (mass 0.50 kg) passes through a light gate before the collision. Trolley A then hits stationary trolley B (mass 0.75 kg) and they couple together. The card on the coupled trolleys passes through a second light gate after the collision.
Time for the card to pass the first gate = 0.040 s Time for the card to pass the second gate = 0.103 s Each time has an uncertainty of \(\pm 0.001\) s.
(a) Calculate the velocity of trolley A before the collision. [1]
(b) Calculate the total momentum before and after the collision. [2]
(c) Determine whether the results are consistent with conservation of momentum, taking the uncertainties in the times into account. [2]
(d) Suggest one change to the apparatus which would reduce the difference between the two momenta. [1]
(c) Difference = 0.0182, i.e. 2.9% of 0.625. Uncertainty before = 0.001/0.040 = 2.5%; after = 0.001/0.103 = 1.0%; combined about 3.5%. Since 2.9% < 3.5%, momentum is conserved within the experimental uncertainty.
(d) Slightly tilt the runway to compensate for friction (or use a data-logger / longer card to reduce the percentage timing uncertainty).
评分标准
(a) 1.25 m/s [1]. Allow 1.3.
(b) Before = 0.625 [1]; after: v = 0.485 and p = 1.25 x 0.485 = 0.607 (allow 0.61) [1]. Allow ECF from (a). Do not allow using 0.50 kg only for the after momentum (XP).
(c) Percentage (or absolute) uncertainties calculated, 2.5% and 1.0%, combined to about 3.5% [1]; compare with the 2.9% difference (or ranges overlap) and conclude consistent [1]. Allow ECF from (b). Allow comparison of ranges: before 0.610-0.640, after 0.601-0.613, which overlap. Do not allow a conclusion with no supporting numbers.
(d) One from: compensate for friction by tilting runway slightly; use a card of greater length / lower speed so that the percentage uncertainty in time is smaller; use data-logger with better time resolution; reduce friction between trolleys and track. [1] Ignore: 'repeat and average' alone, 'more careful'.
题目 3 · Practical multi-part questions
6 分
A student investigates a thin converging lens by placing an illuminated object at different distances from it and finding the position of a sharp real image on a screen. The Cartesian convention is used, so the object distance u is negative. The thin-lens equation is \(\frac{1}{v}=\frac{1}{u}+\frac{1}{f}\).
u / m: -0.250, -0.300, -0.400 v / m: 0.165, 0.152, 0.133
(a) Calculate 1/u and 1/v, in \(\text{m}^{-1}\), for the first pair of readings (u = -0.250 m, v = 0.165 m). [1]
(b) The student plots a graph of 1/v (y-axis) against 1/u (x-axis). Explain why the graph should be a straight line and state the gradient and the y-intercept it should have. [2]
(c) Use the second pair of readings (u = -0.300 m, v = 0.152 m) to calculate the focal length of the lens. [2]
Answer = ..... m
(d) The student finds it hard to judge the exact screen position at which the image is sharp. Suggest one way to reduce the effect of this on the value of v. [1]
(b) Rearranged, \(1/v = (1)(1/u) + 1/f\), which compares with \(y = mx + c\); so for fixed f the graph is a straight line with gradient +1 and y-intercept \(1/f\).
(d) Find the two screen positions where the image just becomes blurred on either side of the best focus and take the mean (or repeat and average; use a bright, sharp-edged object).
评分标准
(a) -4.00 and 6.06 both correct [1]. Allow 6.1. Do not allow a positive 1/u.
(b) Rearrangement to \(1/v = 1/u + 1/f\), compared with y = mx + c [1]; gradient = 1 (allow +1) and intercept = 1/f [1]. Allow the intercept stated as 'the power of the lens'.
(c) \(1/f = 1/0.152 - 1/(-0.300) = 9.91\) [1]; f = 0.101 m (0.1009 unrounded) [1]. Allow ECF for the second mark. Sign error giving 1/f = 3.25 (f = 0.31 m) scores 1 mark max (XP).
(d) Move the screen back and forth / approach the sharp position from both sides and average; or repeat and average; or use a darker room / brighter object with fine detail / use a smaller aperture. [1] Ignore: 'be more careful', 'use a better lens'.
题目 4 · Practical multi-part questions
7 分
A student measures the e.m.f. \(\varepsilon\) and internal resistance r of a cell. A variable resistor is connected across the cell, a voltmeter measures the terminal potential difference V and an ammeter measures the current I.
I / A: 0.10, 0.20, 0.30, 0.40, 0.50 V / V: 1.44, 1.38, 1.31, 1.26, 1.20
The terminal potential difference is related to the current by \(V = \varepsilon - Ir\).
(a) The student plots V against I. State what the gradient and the y-intercept of the graph represent. [2]
(b) Using the first and last data points as a best-fit line, calculate r and \(\varepsilon\). [3]
r = ..... \(\Omega\) \(\varepsilon\) = ..... V
(c) Calculate the power dissipated inside the cell when the current is 0.50 A. [1]
(d) The student switches off the circuit between readings. Explain why this improves the results. [1]
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解题
(a) \(V = \varepsilon - rI\) compares with \(y = c + mx\): gradient = \(-r\), y-intercept = \(\varepsilon\).
(c) \(P = I^2 r = 0.50^2\times0.60 = 0.15\ \text{W}\).
(d) A continuous current would warm and partly discharge the cell, changing its e.m.f. and internal resistance during the experiment; switching off keeps them constant.
评分标准
(a) Gradient = -r (allow magnitude of gradient = r) [1]; y-intercept = e.m.f. [1]. Do not allow 'gradient = resistance' without 'internal'.
(b) Gradient calculated as -0.24/0.40 [1]; r = 0.60 ohm [1]; e.m.f. = 1.50 V [1]. Allow tolerance for readings from a drawn best-fit line: r 0.55-0.65 ohm, e.m.f. 1.48-1.52 V. ECF for e.m.f. from the r value. A positive r from a sign slip: allow if magnitude correct.
(c) \(P = I^2 r = 0.15\) W [1]. Allow ECF from (b). Also allow \(P = (\varepsilon - V)I = 0.30\times0.50 = 0.15\) W. Do not allow \(P=VI\) = 0.60 W (power in the external resistor).
(d) Prevents the cell heating / discharging (so e.m.f. and r change/ remain constant) [1]. Ignore: 'to save the battery' with no reason.
卷二 乙部
Two applied questions: one on measurement uncertainty and resistor design, one on a sensor circuit with a 6-mark levels-of-response part.
3 题目 · 32 分
题目 1 · Applied multi-part question
13 分
A student investigates a resistance wire of length \(1.000\) m. The length is measured with a metre rule with an uncertainty of \(\pm 1\) mm. The resistance of the wire is measured with a digital meter as \(2.65\ \Omega\) with an uncertainty of \(\pm 0.05\ \Omega\).
The diameter of the wire is measured at five different positions using a micrometer screw gauge. The readings are:
\(0.42\) mm, \(0.43\) mm, \(0.41\) mm, \(0.42\) mm, \(0.44\) mm
(a) (i) Calculate the mean diameter of the wire and estimate its absolute uncertainty from the spread of the readings.
Mean diameter = ..... mm Uncertainty = ± ..... mm [2]
(ii) Calculate the percentage uncertainty in the mean diameter. [1]
(b) (i) Calculate the resistivity \(\rho\) of the wire material, using \(R = \rho L / A\).
\(\rho\) = ..... \(\Omega\) m [2]
(ii) Determine the percentage uncertainty in \(\rho\) and write the resistivity in the form \(\rho \pm \Delta\rho\). [2]
(iii) State which measurement is the largest source of uncertainty in \(\rho\) and give a reason. [1]
(c) The student now designs a heating resistor of resistance \(12.0\ \Omega\) from a different alloy wire of resistivity \(4.9 \times 10^{-7}\ \Omega\) m and diameter \(0.40\) mm.
(i) Calculate the length of wire required. [2]
(ii) The wire can safely dissipate no more than \(5.0\) W for each metre of its length. Calculate the maximum current the resistor can carry. [2]
(iii) The wire supplied is actually \(0.39\) mm in diameter. Explain whether the resistance of the finished resistor will be greater or smaller than \(12.0\ \Omega\). [1]
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解题
(a)(i) Mean = \((0.42+0.43+0.41+0.42+0.44)/5 = 2.12/5 = 0.424\) mm. Half range = \((0.44-0.41)/2 = 0.015\) mm. (ii) \(0.015/0.424 \times 100 = 3.5\%\). (b)(i) \(A = \pi d^2/4 = \pi (0.424\times10^{-3})^2/4 = 1.41\times10^{-7}\ \text{m}^2\). \(\rho = RA/L = 2.65 \times 1.412\times10^{-7}/1.000 = 3.74\times10^{-7}\ \Omega\) m. (ii) Percentage uncertainties: diameter counts twice, so \(2\times3.5 = 7.1\%\); resistance \(0.05/2.65 = 1.9\%\); length \(1/1000 = 0.1\%\). Total = \(9.1\%\). \(\Delta\rho = 0.091 \times 3.74\times10^{-7} = 0.34\times10^{-7}\), so \(\rho = (3.7 \pm 0.3)\times10^{-7}\ \Omega\) m. (iii) The diameter: its 3.5% is doubled to 7.1% because \(A \propto d^2\), which is far larger than the other contributions. (c)(i) \(A = \pi(0.40\times10^{-3})^2/4 = 1.257\times10^{-7}\ \text{m}^2\). \(L = RA/\rho = 12.0\times1.257\times10^{-7}/4.9\times10^{-7} = 3.08\) m. (ii) \(P_{max} = 5.0 \times 3.08 = 15.4\) W. \(I = \sqrt{P/R} = \sqrt{15.4/12.0} = 1.13\) A. (iii) Greater. A smaller diameter gives a smaller cross-sectional area, and \(R = \rho L/A\) so \(R\) increases (\(R \propto 1/d^2\)).
评分标准
(a)(i) Mean = 0.424 mm [1]. Uncertainty = half the range = 0.015 mm [1]. Allow 0.02 mm. Allow 0.43 mm for mean only if working is shown. (a)(ii) \(0.015/0.424 \times 100 = 3.5\%\) [1]. ECF from (a)(i). Allow 4% if the uncertainty was 0.02 mm. (b)(i) \(A = \pi d^2/4 = 1.41\times10^{-7}\ \text{m}^2\) [1]. \(\rho = RA/L = 3.7\times10^{-7}\ \Omega\) m (unrounded 3.74) [1]. Do not allow POT error in mm to m conversion (one mark lost). ECF from (a). (b)(ii) Adds percentages with diameter doubled: \(7.1 + 1.9 + 0.1 = 9.1\%\) [1]. Absolute uncertainty \(\pm 0.3\times10^{-7}\ \Omega\) m, quoted with \(\rho\) to the same decimal place [1]. Allow 9% and ±0.3. Allow recalculation using extreme values if correct. Do not allow if the diameter is not squared (XP). (b)(iii) Diameter [1] with reason: its percentage uncertainty is doubled / largest percentage contribution (7.1%). Reason needed for the mark. Ignore references to the micrometer being less accurate. (c)(i) \(A = 1.26\times10^{-7}\ \text{m}^2\) [1]. \(L = RA/\rho = 3.08\) m [1]. Allow 3.1 m. (c)(ii) \(P = 5.0 \times 3.08 = 15.4\) W [1]. \(I = \sqrt{P/R} = 1.13\) A [1]. ECF from (c)(i). Allow 1.1 A. (c)(iii) Greater [1] because smaller area / \(R \propto 1/d^2\) (the explanation must be given). Do not allow 'thinner wire is longer'. Significant-figure errors penalised once only across the question.
题目 2 · Applied multi-part question with levels-of-response part
13 分
A greenhouse monitor uses a negative temperature coefficient (NTC) thermistor in a potential divider. The thermistor is connected in series with a fixed resistor \(R\) across a \(5.0\) V supply of negligible internal resistance. The output voltage \(V_{out}\) is taken across the fixed resistor. The resistance of the thermistor is \(4.7\ \text{k}\Omega\) at \(20\ ^\circ\)C and \(0.90\ \text{k}\Omega\) at \(60\ ^\circ\)C.
(a) Explain, in terms of charge carriers, why the resistance of the thermistor decreases as its temperature rises. [2]
(b) The fixed resistor has \(R = 2.2\ \text{k}\Omega\).
(i) Show that \(V_{out}\) is about \(1.6\) V at \(20\ ^\circ\)C. [2]
(ii) Calculate \(V_{out}\) at \(60\ ^\circ\)C.
\(V_{out}\) = ..... V [1]
(c) A voltmeter of resistance \(10\ \text{k}\Omega\) is connected across the fixed resistor \(R = 2.2\ \text{k}\Omega\) while the thermistor is at \(20\ ^\circ\)C. Calculate the reading on the voltmeter. [2]
*(d) The monitor must give the largest possible change in \(V_{out}\) between \(20\ ^\circ\)C and \(60\ ^\circ\)C. A technician has fixed resistors of \(220\ \Omega\), \(2.2\ \text{k}\Omega\) and \(22\ \text{k}\Omega\). The output will be read by the \(10\ \text{k}\Omega\) voltmeter.
Discuss which resistor is most suitable. You may include calculations in your answer. [6]
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解题
(a) The thermistor is a semiconductor. As temperature rises, more charge carriers are released (the number density of mobile carriers increases), so for the same dimensions the conductance rises and resistance falls. (b)(i) \(V_{out} = V_{in} R/(R_T + R) = 5.0 \times 2.2/(4.7+2.2) = 11/6.9 = 1.59\) V. (ii) \(V_{out} = 5.0 \times 2.2/(0.90+2.2) = 11/3.1 = 3.55\) V. (c) Voltmeter in parallel with R: \(R_p = (2.2\times10)/(2.2+10) = 1.80\ \text{k}\Omega\). \(V = 5.0 \times 1.80/(4.7+1.80) = 1.39\) V. (d) Unloaded values: 220 Ω gives 0.22 V at 20 °C and 0.98 V at 60 °C, a change of 0.76 V. 2.2 kΩ gives 1.59 V and 3.55 V, a change of 1.95 V. 22 kΩ gives 4.12 V and 4.80 V, a change of 0.68 V. So 2.2 kΩ gives the largest change; \(R\) close to the thermistor's mid-range resistance (\(\sqrt{4.7\times0.9} \approx 2.1\ \text{k}\Omega\)) is best. For 22 kΩ, a 10 kΩ voltmeter is lower in resistance than R, so loading reduces the reading and the range substantially (parallel resistance is 6.9 kΩ, so the voltage falls from 4.12 V to about 2.97 V at 20 °C and from 4.80 V to about 4.42 V at 60 °C, a change of only about 1.45 V). For 2.2 kΩ, loading lowers the 20 °C reading from 1.59 V to 1.39 V, which is acceptable. For 220 Ω, loading is negligible but the voltage change is small. Conclusion: 2.2 kΩ.
评分标准
(a) Thermistor is a semiconductor / has few free charge carriers [1]. Higher temperature releases more charge carriers (greater number density), so resistance falls [1]. Allow 'more electrons freed'. Do not allow 'electrons move faster so fewer collisions' or 'atoms vibrate less'. (b)(i) Substitution \(5.0\times2.2/(4.7+2.2)\) [1]. Unrounded 1.594 V, quoted 1.59 V [1]. (b)(ii) 3.55 V [1] (unrounded 3.548). Allow 3.5. (c) Parallel resistance 1.80 k\(\Omega\) [1] (allow 1.8). Voltage 1.39 V [1] (allow 1.4). ECF from the parallel resistance. Do not allow 1.59 V. (d) Levels of response (6 marks). Level 3 (5-6 marks): Clear and logically structured discussion. Calculates the change in \(V_{out}\) for at least two resistors, correctly identifies 2.2 kΩ as best, AND considers the loading effect of the voltmeter on at least one option with a supporting calculation or a clear comparison of resistances. Sustained reasoning with a justified conclusion. Level 2 (3-4 marks): Calculations or reasoning for at least two resistors with some correct values, and a conclusion, but the loading effect is missing or only qualitative, or there are some errors in the reasoning. Level 1 (1-2 marks): Some relevant points (e.g. that \(V_{out}\) depends on the resistor ratio, or one correct calculation) but no valid comparison or conclusion. Level 0 (0 marks): No creditworthy response. Indicative content (may include): - Divider equation used for each resistor at both temperatures. - 220 Ω: 0.22 V to 0.98 V, change 0.76 V (small). - 2.2 kΩ: 1.59 V to 3.55 V, change 1.95 V (largest). - 22 kΩ: 4.12 V to 4.80 V, change 0.68 V (small, because R is much larger than the thermistor resistance). - Best when R is similar to the thermistor's resistance (around 2 kΩ). - Voltmeter in parallel with R lowers the effective resistance; the effect is largest for 22 kΩ (parallel resistance 6.9 kΩ) and smaller for 2.2 kΩ (1.80 kΩ). - 2.2 kΩ is the most suitable. Allow ECF throughout. Allow alternative valid arguments, e.g. sensitivity (V per °C), if supported.
题目 3 · Level-of-response extended answer
6 分
A greenhouse over-temperature warning unit uses the circuit described below (not to scale). A 6.0 V supply of negligible internal resistance is connected across a negative temperature coefficient (NTC) thermistor in series with a fixed resistor of resistance 6.8 k\(\Omega\). The output voltage \(V_{out}\) is taken across the fixed resistor and is connected to a switching circuit that sounds a buzzer when \(V_{out}\) rises above 3.0 V.
The resistance of the thermistor was measured at four temperatures:
The grower wants the buzzer to sound at a temperature of about 20 \(^\circ\)C, and asks whether the unit will do this.
*Explain how the circuit responds to a rise in temperature and use the data to judge whether the buzzer will sound at about 20 \(^\circ\)C. Include calculations in your answer. Then explain how the unit could be modified so that the buzzer sounds at a higher temperature, without changing the supply or the thermistor. [6]*
Qualitative: NTC thermistor resistance falls as temperature rises (more charge carriers released). Total resistance falls, current rises, and the p.d. across the fixed resistor rises while that across the thermistor falls. \(V_{out}\) increases with temperature.
At 18 \(^\circ\)C: \(V_{out} = 6.0\times 6.8/(6.8+7.5) = 40.8/14.3 = 2.853\) V = 2.85 V < 3.0 V, so no buzzer. At 22 \(^\circ\)C: \(V_{out} = 6.0\times 6.8/(6.8+6.2) = 40.8/13.0 = 3.138\) V = 3.14 V > 3.0 V, so buzzer sounds. The switching point is between 18 and 22 \(^\circ\)C, so consistent with about 20 \(^\circ\)C.
Threshold: \(3.0 = 6.0\times 6.8/(6.8+R_{th})\) gives \(R_{th}\) = 6.8 k\(\Omega\) (when \(V_{out}\) is half of \(V_{in}\) the two resistances are equal). This lies between 7.5 and 6.2 k\(\Omega\), i.e. between 18 and 22 \(^\circ\)C.
Modification: for the threshold \(V_{out}\) = 3.0 V the thermistor resistance must equal the fixed resistance. A higher switching temperature needs a lower thermistor resistance, so reduce the fixed resistor. E.g. 4.0 k\(\Omega\) gives switching when \(R_{th}\) = 4.0 k\(\Omega\), about 30 \(^\circ\)C. (Alternatively a variable resistor can be used to adjust the set point.)
评分标准
Levels of response (6 marks). Mark on the quality and sustained logic of the reasoning as well as the number of indicative points.
Level 3 (5-6 marks): Clear, logically structured explanation covering the potential divider action, correct calculations of \(V_{out}\) at (at least) both 18 and 22 \(^\circ\)C with a valid conclusion about 20 \(^\circ\)C, AND a correct, justified modification. Few or no errors.
Level 2 (3-4 marks): Explains the effect of temperature on the thermistor and on \(V_{out}\) and attempts calculations (may have one slip, ECF allowed) leading to a conclusion; the modification may be missing or weakly justified. Reasoning is mostly coherent.
Level 1 (1-2 marks): Some relevant statements, e.g. thermistor resistance falls with temperature or \(V_{out}\) rises, or one correct calculation, with little linking or conclusion.
Level 0 (0 marks): No creditable response.
Indicative scientific points that may be made: - NTC thermistor: resistance decreases as temperature increases (more charge carriers released). - Series circuit: same current; p.d. shares in ratio of resistances. - As thermistor resistance falls, p.d. across it falls and p.d. across the fixed resistor (\(V_{out}\)) rises. - Use of \(V_{out} = V_{in}R_2/(R_1+R_2)\) (method mark). - 18 \(^\circ\)C: 2.85 V, below 3.0 V so no buzzer. - 22 \(^\circ\)C: 3.14 V, above 3.0 V so buzzer sounds. - Therefore switching occurs between 18 and 22 \(^\circ\)C, so about 20 \(^\circ\)C is plausible. - Alternatively: threshold requires \(R_{th}\) = 6.8 k\(\Omega\), lying between 7.5 and 6.2 k\(\Omega\). - Modification: reduce the fixed resistance (or use a variable resistor), so a smaller thermistor resistance, hence a higher temperature, is needed to give 3.0 V. A quantified example (e.g. 4.0 k\(\Omega\) giving about 30 \(^\circ\)C) is creditable.
Allow ECF from an incorrect \(V_{out}\) value into the conclusion. Do not allow increasing the fixed resistor to raise the switching temperature. Significant figure errors penalised once only.
卷二 部分 C
One extended context question with calculations, an iterative table and a final 6-mark levels-of-response comparison.
2 题目 · 24 分
题目 1 · Extended context question
18 分
A student investigates the motion of a dynamics trolley of mass \(0.50\,\text{kg}\) on a horizontal, friction-compensated track. A constant horizontal force from a pulley-and-string system gives a constant resultant force of \(0.35\,\text{N}\) on the trolley, which starts from rest. The student models the motion in discrete time steps of \(\Delta t = 0.50\,\text{s}\). In the model, the velocity at the start of each step is used to find the displacement in that step: \(\Delta s = v\,\Delta t\), and the velocity at the end of the step is \(v + a\,\Delta t\).
(a) Calculate the acceleration of the trolley. [2]
Answer = ........................ m s\(^{-2}\)
(b) Complete the iterative table. The rows for \(t = 0\), \(0.50\,\text{s}\) and \(1.00\,\text{s}\) are already complete. [4]
| t / s | v / m s\(^{-1}\) | displacement s / m | |---|---|---| | 0.00 | 0.00 | 0.000 | | 0.50 | 0.35 | 0.000 | | 1.00 | 0.70 | 0.175 | | 1.50 | ........ | ........ | | 2.00 | ........ | ........ |
(c) (i) Use an equation of uniformly accelerated motion to calculate the exact displacement at \(t = 2.00\,\text{s}\), and hence the percentage difference between the model and the exact value. [2]
Answer = ........................ %
(ii) Explain why the model gives a displacement that is too small, and state one change that would make it more accurate. [2]
(d) To check the speed at \(t = 2.00\,\text{s}\), the student fits a card of length \(100\,\text{mm} \pm 1\,\text{mm}\) to the trolley. A light gate records the card taking \(0.071\,\text{s} \pm 0.001\,\text{s}\) to pass. Calculate the average speed of the trolley while the card passes the gate, and the percentage uncertainty in this speed. [2]
Answer = ........................ m s\(^{-1}\) \(\pm\) ........................ %
(e)* A second student proposes a different method: using a hand-held stopwatch to time the trolley over the \(1.40\,\text{m}\) it travels from rest, and then using \(v = 2s/t\) to find the final speed. The student's reaction-time uncertainty is about \(0.2\,\text{s}\).
Compare the suitability of the two methods (light gate with card, and stopwatch) for testing the predicted speed of the trolley. You may include calculations in your answer. Reach a reasoned conclusion. [6]
(c)(i) \(s = \tfrac12 a t^2 = 0.5\times0.70\times2.00^2 = 1.40\,\text{m}\). Difference \(=(1.40-1.050)/1.40 = 25\%\). (ii) The velocity is increasing, so using the velocity at the start of each step underestimates the average velocity in that step, so each \(\Delta s\) is too small. Use a smaller \(\Delta t\) (or use the average of the start and end velocities).
(e) Light gate: speed 1.41 m/s measured directly, uncertainty about 2.4% (about 0.03 m/s), so the range 1.38 to 1.44 m/s is consistent with the predicted speed of 1.40 m/s. Card gives an average speed over its length, slightly less than the instantaneous speed at the front, a small systematic effect; a shorter card reduces this but raises the percentage uncertainty in length and time. Stopwatch: \(t = \sqrt{2s/a} = \sqrt{2\times1.40/0.70} = 2.0\,\text{s}\); reaction-time uncertainty 0.2 s gives 10% uncertainty in t, so \(v = 2s/t = 1.4\,\text{m s}^{-1}\) has about 10% (about 0.14 m/s) uncertainty. Also \(v = 2s/t\) assumes constant acceleration, which is the very thing being tested, so it is not an independent check. The stopwatch also has a random and systematic start/stop error. Conclusion: the light gate method is more precise, more objective and independent, so more suitable.
评分标准
(a) C1 \(a = F/m\) [1]; A1 0.70 [1]. Allow 0.7.
(b) 1.05 [1]; 0.525 [1]; 1.40 [1]; 1.050 or 1.05 [1]. ECF within the table for consistent use of the model. Allow 1.4 for 1.40.
(c)(i) 1.40 m from \(\tfrac12 a t^2\) or \(\tfrac12 (0+1.4)\times 2\) [1]; 25% [1] (allow 25 to 26; ECF from (b)). Do not allow 0.25 without a % sign being implied. (ii) Velocity increases during each step / start-of-step velocity is less than the average velocity [1]. Smaller time step [1]. Allow using mean velocity. Ignore 'human error'.
(d) \(v = 0.100/0.071 = 1.41\) m/s [1]; % uncertainty 2.4% (allow 2%, 2.5%) [1]. Allow absolute uncertainty 0.03 m/s. Do not allow 0.4% (arithmetic error from adding absolute values).
(e) Levels of response (AO3). Level 3 (5-6 marks): a clear, well-structured comparison of both methods, supported by at least two quantitative points (for example about 2.4% uncertainty for the light gate and about 10% for the stopwatch via \(t \approx 2.0\) s and 0.2 s), with a reasoned conclusion that the light gate is better. Level 2 (3-4 marks): comparison of both methods with some quantitative support or some reasoning, but incomplete or one-sided; conclusion may be weakly supported. Level 1 (1-2 marks): isolated statements about one method or generic statements (e.g. 'light gate more accurate') with little reasoning or no values. 0 marks: no creditworthy response. Indicative points that may be included: light gate gives time resolution 0.001 s; percentage uncertainty about 2.4%; speed 1.41 m/s consistent with the predicted 1.40 m/s within uncertainty; card gives average rather than instantaneous speed, a small systematic effect, and a shorter card would reduce it; stopwatch \(t = 2.0\) s so 0.2 s is 10%; resulting speed uncertainty about 0.14 m/s, so small discrepancies in the prediction cannot be detected;\(v = 2s/t\) assumes constant acceleration so not independent; reaction time has both start and finish errors; repeats would reduce random error of the stopwatch but not an assumed-model error; the light gate is automatic and objective. Allow ECF from (d). Ignore comments on cost unless linked to suitability.
题目 2 · Level-of-response extended answer
6 分
*A rescue sled of mass \(1.5\,\text{kg}\) (a small model used in a lab test) is released from rest on a level, frictionless ice surface. A constant horizontal resultant force of \(6.0\,\text{N}\) acts on it.
A student models the motion in discrete time steps of \(\Delta t = 0.50\,\text{s}\). In each step the model uses the velocity at the START of the step to find the change in displacement: \(\Delta s = v_{\text{start}}\,\Delta t\) and \(\Delta v = a\,\Delta t\).
Part of the iterative table is shown.
| t / s | a / m s\(^{-2}\) | v / m s\(^{-1}\) | s / m | |---|---|---|---| | 0 | 4.0 | 0 | 0 | | 0.50 | 4.0 | 2.0 | 0 | | 1.00 | 4.0 | 4.0 | 1.0 | | 1.50 | | | | | 2.00 | | | |
The student also calculates the displacement at each time using the equation \(s = ut + \tfrac{1}{2}at^2\).
Compare the results of the iterative model with those of the equation of motion. In your answer you should: - complete the table for \(t = 1.50\,\text{s}\) and \(t = 2.00\,\text{s}\), - calculate the displacement at \(t = 2.00\,\text{s}\) from the equation and find the percentage difference from the model, - explain the reason for the difference, and suggest and justify one change that would make the model agree more closely with the equation.
[6]
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解题
Acceleration: \(a = F/m = 6.0/1.5 = 4.0\,\text{m s}^{-2}\) (constant, so a column entry of 4.0 each row).
Iteration: \(v\) at 1.50 s = \(4.0 + 4.0\times0.50 = 6.0\,\text{m s}^{-1}\); \(s\) at 1.50 s = \(1.0 + 4.0\times0.50 = 3.0\,\text{m}\) (uses v at 1.00 s). \(v\) at 2.00 s = \(8.0\,\text{m s}^{-1}\); \(s\) at 2.00 s = \(3.0 + 6.0\times0.50 = 6.0\,\text{m}\).
Equation: \(s = 0 + \tfrac{1}{2}\times4.0\times(2.00)^2 = 8.0\,\text{m}\). (Other values: 0.50 m, 2.0 m, 4.5 m at 0.50, 1.00, 1.50 s.)
Difference: \(8.0 - 6.0 = 2.0\,\text{m}\), i.e. \(2.0/8.0\times100 = 25\%\) lower than the equation.
Explanation: the velocity increases throughout each step, but the model assumes it equals the (smaller) start value for the whole step, so every step's displacement is underestimated and the errors accumulate.
Improvement: use the average velocity in each step, \(\Delta s = \tfrac{1}{2}(v_{\text{start}}+v_{\text{end}})\Delta t\); for constant acceleration this reproduces the equation exactly (e.g. \(s = 0.50, 2.00, 4.50, 8.00\,\text{m}\)). Alternatively use a smaller \(\Delta t\) (e.g. 0.10 s), which reduces the velocity change within a step, so the error falls, but with more rows.
评分标准
Levels of response (6 marks, assessed on a sustained and coherent line of reasoning).
Level 3 (5-6 marks): Table correctly completed AND equation value 8.0 m AND percentage difference (25%) correctly calculated AND a clear explanation of why the model underestimates (start-of-step velocity used while v increases) AND a justified improvement. Calculations and reasoning are logically linked and correct. Allow ECF for the percentage if the table is wrong.
Level 2 (3-4 marks): Most calculations correct (e.g. table and equation value) but percentage difference missing/incorrect, OR the explanation or the improvement is missing, incomplete or not justified. Reasoning mostly clear.
Level 1 (1-2 marks): Some correct relevant statement or calculation (e.g. a = 4.0 m s^-2, v values, or s = 8.0 m) with little or no comparison or explanation.
Level 0 (0 marks): No response or no creditable physics.
Indicative scientific points that may be included: - a = F/m = 6.0/1.5 = 4.0 m s^-2 (constant) - t = 1.50 s: v = 6.0 m s^-1, s = 3.0 m; t = 2.00 s: v = 8.0 m s^-1, s = 6.0 m - equation: s = 1/2 x 4.0 x 2.00^2 = 8.0 m (accept 0.50, 2.0, 4.5 m at earlier times) - difference 2.0 m; 25% (of the equation value). Allow 33% if difference is taken relative to the model value, if clearly stated - velocity increases during each step, but the model uses the lower start value so each displacement step is too small; error accumulates (the velocity values themselves agree with v = at) - improvement: use average velocity of the step (gives exact agreement for constant a), OR reduce time step (error decreases as step shrinks, more rows/computation); with a reason
Allow: use of v = u + at to confirm the velocities. Allow end-of-step velocity only if it is described as overestimating and an average is then suggested. Do not allow: attributing the difference to friction, air resistance or rounding. Do not allow 'make the time step bigger'. Ignore: minor arithmetic slips that are carried forward consistently (ECF); significant figure errors penalised once only.
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