An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 (V3) Cambridge International A Level Physics (9702) paper. Not affiliated with or reproduced from Cambridge.
卷一
Answer all forty multiple choice questions.
40 題目 · 40 分
題目 1 · multiple_choice
1 分
A student determines the density of a metal cylinder by measuring its mass \(m\), diameter \(d\), and length \(l\).
A uniform sign of weight \(40\text{ N}\) and length \(1.2\text{ m}\) is hinged to a vertical wall at one end and held horizontally by a wire attached to the other end. The wire makes an angle of \(30^\circ\) with the horizontal sign.
What is the tension in the wire and the vertical force component exerted by the hinge on the sign?
A.Tension = 20 N, Hinge vertical force = 20 N
B.Tension = 40 N, Hinge vertical force = 20 N
C.Tension = 40 N, Hinge vertical force = 0 N
D.Tension = 80 N, Hinge vertical force = 40 N
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解題
Taking moments about the hinge in equilibrium: \(\sum M_{\text{hinge}} = 0 \implies (T \sin 30^\circ) \times L = W \times \frac{L}{2}\) \(T \times 0.5 = 40 \times 0.5 \implies T = 40\text{ N}\).
For vertical equilibrium of the forces on the sign: \(\sum F_y = 0 \implies F_v + T \sin 30^\circ = W\) \(F_v + 40 \times 0.5 = 40 \implies F_v = 20\text{ N}\).
評分準則
1 mark for the correct answer B.
題目 3 · multiple_choice
1 分
A potential divider circuit consists of a \(12.0\text{ V}\) power supply of negligible internal resistance, a fixed resistor of \(4.0\text{ k}\Omega\) and a light-dependent resistor (LDR) connected in series. The output voltage \(V_{\text{out}}\) is measured across the LDR.
In bright light, the resistance of the LDR is \(800\ \Omega\). In the dark, the resistance of the LDR is \(12.0\text{ k}\Omega\).
What is the change in the output voltage \(V_{\text{out}}\) when the conditions change from bright light to dark?
A steel wire and a brass wire are connected in series to support a load. The steel wire has length \(L\) and radius \(r\). The brass wire has length \(2L\) and radius \(2r\). The Young modulus of steel is twice that of brass.
What is the ratio \(\frac{\text{extension of steel wire}}{\text{extension of brass wire}}\) when they support the same load?
A.0.25
B.0.50
C.1.0
D.2.0
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解題
The extension is given by \(x = \frac{F L}{A E}\). Since they are in series, they experience the same tensile force \(F\). For steel: \(x_s = \frac{F L}{\pi r^2 E_s}\). For brass: \(x_b = \frac{F (2L)}{\pi (2r)^2 E_b} = \frac{2 F L}{4 \pi r^2 E_b} = \frac{F L}{2 \pi r^2 E_b}\). We are given \(E_s = 2E_b\), which means \(E_b = 0.5 E_s\). Substitute this into the expression for brass extension: \(x_b = \frac{F L}{2 \pi r^2 (0.5 E_s)} = \frac{F L}{\pi r^2 E_s}\). Therefore, \(x_s = x_b\), giving a ratio of \(1.0\).
評分準則
1 mark for the correct answer C.
題目 5 · multiple_choice
1 分
During \(\beta^-\) decay, a nucleon inside a nucleus decays to emit a lepton. Which row correctly describes the change in quark flavor and the classification of the emitted particle?
A.Quark change: \(u \to d\) | Emitted particle: electron (which is a hadron)
B.Quark change: \(d \to u\) | Emitted particle: electron (which is a lepton)
C.Quark change: \(u \to d\) | Emitted particle: electron antineutrino (which is a hadron)
D.Quark change: \(d \to u\) | Emitted particle: electron antineutrino (which is a baryon)
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解題
In \(\beta^-\) decay, a neutron (udd) decays to a proton (uud). This involves a down quark changing to an up quark (\(d \to u\)). The particles emitted are an electron (which is a fundamental lepton) and an electron antineutrino (also a lepton). Thus, option B is correct.
評分準則
1 mark for the correct answer B.
題目 6 · multiple_choice
1 分
A metal wire is stretched beyond its elastic limit but does not break. The load is then completely removed.
Which statement correctly describes the behavior of the wire and the energy changes involved?
A.The wire returns to its original length, and all the strain energy is recovered.
B.The wire retains a permanent extension, and some energy has been dissipated as thermal energy.
C.The wire returns to its original length, but some energy has been dissipated as thermal energy.
D.The wire retains a permanent extension, and all the work done in stretching the wire is stored as elastic potential energy.
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解題
Stretching a wire past its elastic limit causes plastic deformation. When the load is removed, the wire does not return to its original length, resulting in a permanent extension. The work done during plastic deformation is not fully stored as elastic potential energy; some of it is used to permanently rearrange atomic planes, which dissipates as heat (thermal energy).
評分準則
1 mark for the correct answer B.
題目 7 · multiple_choice
1 分
An experiment is performed to determine the acceleration of free fall \(g\) by measuring the distance \(h\) fallen by a ball from rest and the time \(t\) taken.
The relationship used is:
\(g = \frac{2h}{t^2}\)
The percentage uncertainty in the measurement of \(h\) is \(1.5\%\) and in the measurement of \(t\) is \(2.0\%\).
What is the percentage uncertainty in the calculated value of \(g\)?
A.3.5%
B.5.0%
C.5.5%
D.7.5%
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解題
Using the rules for combining uncertainties: \(\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t}\)
Substitute the percentage uncertainties: Percentage uncertainty in \(g = 1.5\% + 2 \times (2.0\%) = 5.5\%\).
評分準則
1 mark for the correct answer C.
題目 8 · multiple_choice
1 分
Two cylindrical wires, X and Y, are made of the same metal. Wire X has length \(L\) and diameter \(D\). Wire Y has length \(2L\) and diameter \(2D\).
What is the ratio \(\frac{\text{resistance of X}}{\text{resistance of Y}}\)?
A.0.5
B.1.0
C.2.0
D.4.0
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解題
Resistance of a uniform wire of resistivity \(\rho\), length \(l\) and cross-sectional area \(A\) is given by \(R = \rho \frac{l}{A} = \rho \frac{4l}{\pi d^2}\), where \(d\) is the diameter.
1 mark for identifying that percentage uncertainties are added and that the radius uncertainty must be multiplied by 3. 1 mark for obtaining 6.7%.
題目 10 · 選擇題
1 分
Two wires, \( X \) and \( Y \), are made of the same metal. Wire \( X \) has length \( L \) and diameter \( d \). Wire \( Y \) has length \( 2L \) and diameter \( 2d \). Both wires are suspended vertically and support the same load \( F \) within their elastic limits.
What is the ratio \( \frac{\text{strain in } X}{\text{strain in } Y} \)?
A.1
B.2
C.4
D.8
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解題
Since both wires are made of the same metal, their Young Modulus \( E \) is the same. Under elastic conditions:
\( E = \frac{\text{stress}}{\text{strain}} \implies \text{strain} = \frac{\text{stress}}{E} \)
Thus, strain is directly proportional to stress: \( \text{strain} \propto \text{stress} \).
Stress \( \sigma \) is given by \( \frac{F}{A} \), where \( A = \frac{\pi d^2}{4} \).
Therefore, stress is inversely proportional to the square of the diameter: \( \sigma \propto \frac{1}{d^2} \).
Since the force \( F \) is the same for both wires:
\( \frac{\text{strain in } X}{\text{strain in } Y} = \frac{\text{stress in } X}{\text{stress in } Y} = \frac{(2d)^2}{d^2} = 4 \)
評分準則
1 mark for showing that strain is independent of original length and depends only on stress for a given material, and correctly resolving the ratio to be 4.
題目 11 · 選擇題
1 分
A copper wire is stretched beyond its limit of proportionality to a maximum extension and then gradually unloaded. The force-extension graph shows a hysteresis loop.
Which statement about the work done and energy changes during this process is correct?
A.The area under the unloading curve represents the plastic work done on the wire.
B.The total area under the loading curve represents the elastic potential energy stored in the wire at maximum extension.
C.The difference in area between the loading and unloading curves represents the energy dissipated as heat in the wire.
D.No work is done during the unloading process since the force on the wire is decreasing.
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解題
When a wire is deformed plastically (beyond its limit of proportionality), some of the work done during loading is used to permanently rearrange the atoms. When unloaded, the wire contracts along a path parallel to the initial elastic region.
The area under the loading curve represents the total work done in stretching the wire. The area under the unloading curve represents the elastic potential energy released (recovered) during unloading. The difference between these two areas (the area enclosed by the loop) represents the work done that cannot be recovered, which is dissipated as thermal energy (heat) within the wire.
評分準則
1 mark for identifying that the area enclosed by the hysteresis loop represents energy dissipated as heat.
題目 12 · 選擇題
1 分
A uniform picture frame of weight \( 24\text{ N} \) is suspended in equilibrium from a single nail by a symmetric string. The tension in each half of the string is \( 15\text{ N} \).
What is the angle \( \theta \) that each half of the string makes with the horizontal?
A.37°
B.53°
C.74°
D.106°
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解題
For the picture frame to be in equilibrium, the vertical components of the tension in the two halves of the string must balance the weight of the picture frame:
\( 2 T \sin\theta = W \)
Substitute the given values:
\( 2(15) \sin\theta = 24 \)
\( 30 \sin\theta = 24 \)
\( \sin\theta = \frac{24}{30} = 0.80 \)
\( \theta = \sin^{-1}(0.80) \approx 53^\circ \)
評分準則
1 mark for setting up the vertical equilibrium equation \( 2T \sin\theta = W \) and solving for \( \theta = 53^\circ \).
題目 13 · 選擇題
1 分
During a beta-plus (\( \beta^+ \)) decay, a proton in a nucleus decays into a neutron, a positron, and an electron neutrino.
Which row correctly describes the changes in the total number of up quarks and down quarks in the nucleus?
A.Up quarks decrease by 1, down quarks increase by 1
B.Up quarks increase by 1, down quarks decrease by 1
C.Up quarks decrease by 2, down quarks increase by 2
D.Up quarks increase by 2, down quarks decrease by 2
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解題
A proton has the quark composition \( uud \). A neutron has the quark composition \( udd \).
In a \( \beta^+ \) decay, a proton turns into a neutron:
\( uud \rightarrow udd + e^+ + \nu_e \)
This means that one up (\( u \)) quark is converted into one down (\( d \)) quark. Therefore, the total number of up quarks in the nucleus decreases by 1, and the total number of down quarks in the nucleus increases by 1.
評分準則
1 mark for correctly identifying that a \( \beta^+ \) decay converts an up quark to a down quark, resulting in a decrease of 1 up quark and an increase of 1 down quark.
題目 14 · 選擇題
1 分
A potential divider circuit consists of a fixed resistor of resistance \( R \) connected in series with a light-dependent resistor (LDR). A constant potential difference \( V_{\text{in}} \) is applied across the series combination. The output voltage \( V_{\text{out}} \) is measured across the LDR.
How do the resistance of the LDR and the output voltage \( V_{\text{out}} \) change when the intensity of the light incident on the LDR increases?
A.Resistance of LDR increases, output voltage increases
B.Resistance of LDR increases, output voltage decreases
C.Resistance of LDR decreases, output voltage increases
D.Resistance of LDR decreases, output voltage decreases
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解題
When the light intensity on an LDR increases, the number of free charge carriers increases, which causes the resistance of the LDR to decrease.
As \( R_{\text{LDR}} \) decreases, it represents a smaller fraction of the total resistance in the circuit, so the potential difference across it (\( V_{\text{out}} \)) decreases.
評分準則
1 mark for correctly stating that the resistance of the LDR decreases with light intensity and that this causes a corresponding decrease in the output voltage.
題目 15 · 選擇題
1 分
A cylindrical copper wire has a resistance \( R \). The wire is stretched uniformly until its length is doubled while its volume remains constant.
What is the new resistance of the wire?
A.R
B.2R
C.4R
D.8R
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解題
The volume \( V \) of a cylinder of length \( L \) and cross-sectional area \( A \) is \( V = A L \).
If the length is doubled to \( 2L \) and volume remains constant, the cross-sectional area must be halved to \( A/2 \).
Resistance is given by \( R = \rho \frac{L}{A} \), where \( \rho \) is the resistivity.
1 mark for determining that the area is halved when the length is doubled to conserve volume, leading to a new resistance of \( 4R \).
題目 16 · 選擇題
1 分
A uniform beam \( AB \) of length \( 3.0\text{ m} \) and weight \( 120\text{ N} \) is pivoted at end \( A \). The beam is held horizontal by a vertical wire attached at a distance of \( 2.0\text{ m} \) from \( A \).
What is the tension in the wire?
A.60 N
B.80 N
C.90 N
D.180 N
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解題
Since the beam is uniform, its center of gravity is at its midpoint, \( 1.5\text{ m} \) from end \( A \). The weight acts vertically downwards at this point.
Taking moments about the pivot \( A \) for rotational equilibrium:
1 mark for correctly taking moments about pivot A (placing the weight at 1.5 m) and calculating the tension to be 90 N.
題目 17 · 選擇題
1 分
A solid metal cylinder has a measured mass of \(45.0 \pm 0.2\text{ g}\), a length of \(8.0 \pm 0.1\text{ cm}\), and a diameter of \(1.20 \pm 0.02\text{ cm}\). What is the percentage uncertainty in the calculated density of the metal?
A.\(3.4\%\)
B.\(4.6\%\)
C.\(5.0\%\)
D.\(6.2\%\)
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解題
The formula for the density \(\rho\) of a cylinder is given by: \(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 L}\)
The fractional uncertainty in density is the sum of the fractional uncertainties of the independent measurements, with the diameter term multiplied by 2 because it is squared: \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\)
Substitute the given values into the equation: - \(\frac{\Delta m}{m} = \frac{0.2}{45.0} \approx 0.00444\) (or \(0.44\%\)) - \(\frac{\Delta d}{d} = \frac{0.02}{1.20} \approx 0.01667\) (or \(1.67\%\)) - \(\frac{\Delta L}{L} = \frac{0.1}{8.0} = 0.0125\) (or \(1.25\%\))
Now, calculate the total percentage uncertainty: \(\frac{\Delta \rho}{\rho} \times 100\% = 0.44\% + 2(1.67\%) + 1.25\% = 5.03\% \approx 5.0\%\)
評分準則
- Recall and apply the uncertainty combining rule for products and powers: 0.5 marks - Accurate calculation to obtain \(5.0\%\): 0.5 marks
題目 18 · 選擇題
1 分
A uniform picture frame of weight \(24\text{ N}\) is suspended symmetrically from a hook on a wall by a light wire of length \(1.0\text{ m}\). The two ends of the wire are attached to the top corners of the frame, which are separated by a horizontal distance of \(0.60\text{ m}\). What is the tension in the wire?
A.\(12\text{ N}\)
B.\(15\text{ N}\)
C.\(20\text{ N}\)
D.\(24\text{ N}\)
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解題
Since the frame is uniform and suspended symmetrically: - The wire is divided into two halves, each of length \(0.50\text{ m}\). - The horizontal distance from each corner to the vertical line of symmetry is \(0.30\text{ m}\).
This forms a right-angled triangle where: - The hypotenuse (wire half) is \(0.50\text{ m}\). - The adjacent horizontal side is \(0.30\text{ m}\). - The vertical height is \(\sqrt{0.50^2 - 0.30^2} = 0.40\text{ m}\).
Let \(\theta\) be the angle between each half of the wire and the horizontal. Therefore: \(\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{0.40}{0.50} = 0.80\)
For vertical equilibrium of the picture frame: \(2 T \sin\theta = W\) \(2 T (0.80) = 24\text{ N}\) \(1.6 T = 24\text{ N}\) \(T = 15\text{ N}\)
評分準則
- Use geometry to find the sine of the angle of inclination of the wire: 0.5 marks - Set up the vertical force equilibrium equation and solve for tension: 0.5 marks
題目 19 · 選擇題
1 分
A wire of length \(L\) and cross-sectional area \(A\) is suspended vertically. A load \(F\) is applied to its lower end, causing a tensile stress \(\sigma\) and a tensile strain \(\varepsilon\). The material of the wire has Young modulus \(E\). Which expression gives the strain energy per unit volume stored in the wire?
A.\(\frac{\sigma^2}{E}\)
B.\(\frac{\sigma^2}{2E}\)
C.\(\frac{E\sigma^2}{2}\)
D.\(\frac{\sigma \varepsilon}{2L}\)
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解題
The strain energy \(U\) stored in a stretched wire is given by: \(U = \frac{1}{2} F e\) where \(e\) is the extension.
To find the strain energy per unit volume \(u\), we divide by the volume \(V = A L\): \(u = \frac{U}{V} = \frac{\frac{1}{2} F e}{A L} = \frac{1}{2} \left(\frac{F}{A}\right) \left(\frac{e}{L}\right) = \frac{1}{2} \sigma \varepsilon\)
By definition, the Young modulus is \(E = \frac{\sigma}{\varepsilon}\), which means \(\varepsilon = \frac{\sigma}{E}\).
Substituting this back into the expression for strain energy per unit volume: \(u = \frac{1}{2} \sigma \left(\frac{\sigma}{E}\right) = \frac{\sigma^2}{2E}\)
評分準則
- Relate strain energy per unit volume to stress and strain: 0.5 marks - Substitute Young's modulus to obtain the correct expression in terms of stress and Young's modulus: 0.5 marks
題目 20 · 選擇題
1 分
A metal wire is stretched elastically up to its elastic limit, where the tensile force is \(80\text{ N}\) and the extension is \(2.0\text{ mm}\). Beyond this point, the wire deforms plastically, and the force increases linearly with extension up to a maximum value of \(120\text{ N}\) at an extension of \(5.0\text{ mm}\). The wire is then completely unloaded. The unloading line is straight and parallel to the initial elastic loading line. What is the net work done in plastically deforming the wire?
A.\(0.18\text{ J}\)
B.\(0.20\text{ J}\)
C.\(0.30\text{ J}\)
D.\(0.38\text{ J}\)
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解題
1. **Work done during loading (total work input):** - For the elastic region (extension \(0\) to \(2.0\text{ mm}\)): \(W_{\text{elastic}} = \frac{1}{2} \times F \times x = \frac{1}{2} \times 80\text{ N} \times 2.0 \times 10^{-3}\text{ m} = 0.080\text{ J}\) - For the plastic region (extension \(2.0\text{ mm}\) to \(5.0\text{ mm}\)): \(W_{\text{plastic}} = \frac{80\text{ N} + 120\text{ N}}{2} \times (5.0 - 2.0) \times 10^{-3}\text{ m} = 100\text{ N} \times 3.0 \times 10^{-3}\text{ m} = 0.300\text{ J}\) - Total work input \(W_{\text{total}} = 0.080\text{ J} + 0.300\text{ J} = 0.380\text{ J}\).
2. **Elastic energy recovered during unloading:** - The initial elastic line has a gradient (force constant) of: \(k = \frac{80\text{ N}}{2.0\text{ mm}} = 40\text{ N\,mm}^{-1}\) - Since the unloading line is parallel, it has the same gradient. For a force drop from \(120\text{ N}\) to \(0\text{ N}\), the decrease in extension is: \(\Delta x = \frac{120\text{ N}}{40\text{ N\,mm}^{-1}} = 3.0\text{ mm}\) - The energy recovered (area under the unloading line) is: \(W_{\text{recovered}} = \frac{1}{2} \times 120\text{ N} \times 3.0 \times 10^{-3}\text{ m} = 0.180\text{ J}\)
- Calculate total work input during loading: 0.5 marks - Determine elastic energy recovered and subtract it to find net work: 0.5 marks
題目 21 · 選擇題
1 分
A Sigma-minus baryon (\(\Sigma^-\)), which has a quark composition of \(dds\), decays via the weak interaction into a neutron (\(udd\)) and a negative pion (\(\pi^-\)), which has a quark composition of \(d\bar{u}\). Which change in quark flavor occurs during this decay, and what is the mediating boson?
A.\(d \rightarrow u\) and \(W^-\
B.\(s \rightarrow u\) and \(W^+\
C.\(s \rightarrow u\) and \(W^-\
D.\(s \rightarrow d\) and \(W^-\
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解題
To analyze the decay: - The initial state is the baryon \(\Sigma^- = dds\). - The final state contains a neutron \(n = udd\) and a pion \(\pi^- = d\bar{u}\).
Comparing the quark content of the initial and final states, we see that one strange quark (\(s\)) is replaced by an up quark (\(u\)). Therefore, the quark flavor transition is: \(s \rightarrow u\)
To determine the mediating virtual boson, we apply conservation of charge at the interaction vertex: - Charge of the initial strange quark is \(-\frac{1}{3}e\). - Charge of the resulting up quark is \(+\frac{2}{3}e\).
Let \(Q\) be the charge of the emitted virtual W boson: \(-\frac{1}{3} = +\frac{2}{3} + Q \implies Q = -1e\)
Thus, the mediating boson is the \(W^-\).
評分準則
- Identify the correct quark transition (s to u): 0.5 marks - Determine the charge of the mediating W boson using conservation laws: 0.5 marks
題目 22 · 選擇題
1 分
A student uses a digital micrometer to measure the diameter of a uniform metal wire. The micrometer has a zero error of \(-0.02\text{ mm}\) that the student fails to notice. Five independent measurements of the diameter along the wire are taken: \(1.42\text{ mm}\), \(1.44\text{ mm}\), \(1.41\text{ mm}\), \(1.43\text{ mm}\), and \(1.45\text{ mm}\). Which statement correctly describes the precision and accuracy of these raw measurements, and the correct diameter of the wire after adjusting for the zero error?
A.The measurements are precise but inaccurate; correcting the zero error gives a diameter of \(1.41\text{ mm}\).
B.The measurements are precise but inaccurate; correcting the zero error gives a diameter of \(1.45\text{ mm}\).
C.The measurements are accurate but imprecise; correcting the zero error gives a diameter of \(1.41\text{ mm}\).
D.The measurements are accurate but imprecise; correcting the zero error gives a diameter of \(1.45\text{ mm}\).
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解題
1. **Precision:** The measurements have a small spread (from \(1.41\text{ mm}\) to \(1.45\text{ mm}\)) around their mean, indicating that the measurements are precise. 2. **Accuracy:** Due to the uncorrected zero error, the measurements are consistently shifted from the true value. Thus, they are inaccurate. 3. **Correction:** The average of the raw measurements is: \(\text{Mean Reading} = \frac{1.42 + 1.44 + 1.41 + 1.43 + 1.45}{5} = 1.43\text{ mm}\)
Since the zero error is \(-0.02\text{ mm}\), this value must be subtracted from the raw reading to find the actual value: \(\text{True Value} = \text{Reading} - \text{Zero Error} = 1.43\text{ mm} - (-0.02\text{ mm}) = 1.45\text{ mm}\)
評分準則
- Classify the quality of the measurements as precise but inaccurate: 0.5 marks - Subtract the systematic zero error correctly to obtain \(1.45\text{ mm}\): 0.5 marks
題目 23 · 選擇題
1 分
A battery of e.m.f. \(10.0\text{ V}\) and internal resistance \(1.0\ \Omega\) is connected in parallel with a second battery of e.m.f. \(6.0\text{ V}\) and internal resistance \(2.0\ \Omega\). This parallel combination is then connected across an external resistor of resistance \(2.0\ \Omega\) such that both batteries drive current in the same direction through the external resistor. What is the current in the external resistor?
A.\(1.5\text{ A}\)
B.\(3.0\text{ A}\)
C.\(3.25\text{ A}\)
D.\(3.33\text{ A}\)
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解題
Let \(I_1\) be the current leaving the positive terminal of the \(10.0\text{ V}\) battery, and \(I_2\) be the current leaving the positive terminal of the \(6.0\text{ V}\) battery. By Kirchhoff’s first law, the current in the external resistor \(R\) is: \(I = I_1 + I_2\)
Using Kirchhoff's second law on the loop containing the first battery and the external resistor: \(10.0 - I_1(1.0) - I(2.0) = 0 \implies I_1 = 10.0 - 2I\)
Using Kirchhoff's second law on the loop containing the second battery and the external resistor: \(6.0 - I_2(2.0) - I(2.0) = 0 \implies 2I_2 = 6.0 - 2I \implies I_2 = 3.0 - I\)
Substitute the expressions for \(I_1\) and \(I_2\) into the current junction equation: \(I = I_1 + I_2\) \(I = (10.0 - 2I) + (3.0 - I)\) \(I = 13.0 - 3I\) \(4I = 13.0\) \(I = 3.25\text{ A}\)
評分準則
- Set up simultaneous equations using Kirchhoff's first and second laws: 0.5 marks - Solve the equations accurately for the load current: 0.5 marks
題目 24 · 選擇題
1 分
A potential divider circuit consists of a battery of electromotive force (e.m.f.) \(9.0\text{ V}\) and negligible internal resistance, connected in series with a thermistor and a fixed resistor of resistance \(1200\ \Omega\). The output voltage \(V_{\text{out}}\) is taken across the fixed resistor. Initially, at a temperature of \(15^\circ\text{C}\), the resistance of the thermistor is \(2400\ \Omega\). The temperature is then increased to \(35^\circ\text{C}\), causing the resistance of the thermistor to decrease by \(60\%\). What is the change in the output voltage \(V_{\text{out}}\?
- Calculate the initial output voltage correctly: 0.5 marks - Determine the new thermistor resistance and calculate the final voltage change: 0.5 marks
題目 25 · 選擇題
1 分
A student measures the resistivity \(\rho\) of a wire using the formula \(\rho = \frac{R \pi d^2}{4 L}\). The measured values with their absolute uncertainties are:
Resistance \(R = (25.0 \pm 0.5) \ \Omega\)
Diameter \(d = (0.40 \pm 0.02) \ \text{mm}\)
Length \(L = (1.50 \pm 0.03) \ \text{m}\)
What is the percentage uncertainty in the calculated value of the resistivity?
A.5.0%
B.9.0%
C.12.0%
D.14.0%
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解題
To find the percentage uncertainty in the resistivity \(\rho\), we use the relation for propagating fractional uncertainties:
1 mark for the correct calculation of the individual percentage uncertainties and sum with the factor of 2 for diameter, leading to 14.0%.
題目 26 · 選擇題
1 分
A small sphere of weight \(8.0 \ \text{N}\) is suspended from a ceiling by a light vertical string. A horizontal wind exerts a constant force \(F\) on the sphere, causing the string to hang at an angle of \(35^\circ\) to the vertical. What is the tension in the string?
A.4.6 N
B.6.6 N
C.8.0 N
D.9.8 N
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解題
The forces acting on the sphere are in equilibrium. Resolving the forces vertically:
\(T \cos(35^\circ) = W\)
where \(T\) is the tension in the string and \(W = 8.0 \ \text{N}\) is the weight of the sphere.
1 mark for resolving forces vertically and solving for tension, yielding 9.8 N.
題目 27 · 選擇題
1 分
Two wires, X and Y, are made of the same metal. Wire X has twice the length and half the diameter of wire Y. Both wires are suspended vertically from a rigid support and each carries the same load at its lower end. What is the ratio \(\frac{\text{strain in wire X}}{\text{strain in wire Y}}\)?
A.1
B.2
C.4
D.8
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解題
Young Modulus \(E\) is defined as \(E = \frac{\text{stress}}{\text{strain}}\). Since both wires are made of the same metal, they have the same Young Modulus \(E\). Thus:
\(\frac{\text{strain in X}}{\text{strain in Y}} = \left(\frac{d_{\text{Y}}}{d_{\text{X}}}\right)^2 = \left(\frac{1}{0.5}\right)^2 = 4\)
評分準則
1 mark for identifying that strain is inversely proportional to cross-sectional area (and diameter squared) since F and E are identical, leading to a ratio of 4.
題目 28 · 選擇題
1 分
A free neutron decays into a proton, an electron, and an electron antineutrino. Which change in quark flavour describes this decay, and what is the mediating exchange boson?
A.up to down (u -> d), mediated by a W+ boson
B.down to up (d -> u), mediated by a W- boson
C.up to down (u -> d), mediated by a W- boson
D.down to up (d -> u), mediated by a W+ boson
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解題
A neutron has a quark composition of \(udd\) and a proton has a quark composition of \(uud\). During beta-minus decay (where a neutron becomes a proton), a down quark (\(d\)) is transformed into an up quark (\(u\)). To conserve charge at the quark decay vertex:
\(d \rightarrow u + W^-\)
(where the charge of \(d\) is \(-\frac{1}{3}e\), the charge of \(u\) is \(+\frac{2}{3}e\), and the charge of the \(W^-\)\ boson is \(-1e\)). Therefore, the mediating boson is the \(W^-\)\ boson.
評分準則
1 mark for identifying the correct quark transformation (down to up) and the correct mediating exchange particle (W- minus boson).
題目 29 · 選擇題
1 分
A potential divider circuit consists of a fixed resistor of resistance \(2.0 \ \text{k}\Omega\) connected in series with a Light Dependent Resistor (LDR) across a \(9.0 \ \text{V}\) power supply of negligible internal resistance. The output voltage \(V_{\text{out}}\) is measured across the LDR. In bright light, the resistance of the LDR is \(500 \ \Omega\). In darkness, its resistance is \(8.0 \ \text{k}\Omega\). What is the change in the output voltage \(V_{\text{out}}\) when the conditions change from bright light to darkness?
A.1.8 V
B.5.4 V
C.7.2 V
D.9.0 V
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解題
In bright light, \(R_{\text{LDR}} = 0.5 \ \text{k}\Omega\). The output voltage is:
1 mark for calculating both output voltages correctly and finding the difference to get 5.4 V.
題目 30 · 選擇題
1 分
A metal wire of original length \(2.0 \ \text{m}\) and cross-sectional area \(1.5 \times 10^{-6} \ \text{m}^2\) is stretched within its elastic limit. The metal has a Young Modulus of \(2.0 \times 10^{11} \ \text{Pa}\). When a tensile force is applied, the elastic strain energy stored in the wire is \(0.30 \ \text{J}\). What is the extension of the wire?
A.0.10 mm
B.2.0 mm
C.4.0 mm
D.14 mm
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解題
The elastic strain energy \(E_{\text{str}}\) stored in a stretched wire is given by:
\(E_{\text{str}} = \frac{1}{2} F x\)
Using the Young Modulus equation, \(F = \frac{E A x}{L}\), we substitute \(F\) to get:
1 mark for relating elastic strain energy to Young Modulus parameters, solving for extension, and obtaining 2.0 mm.
題目 31 · 選擇題
1 分
A student plots a graph of current \(I\) against potential difference \(V\) for a metal wire resistor. Due to a zero error in the ammeter, it always records a current that is \(0.05 \ \text{A}\) higher than the true value. The voltmeter is calibrated correctly. What is the effect of this zero error on the gradient of the graph of \(I\) against \(V\), and on the calculated value of the resistance \(R = \frac{V}{I}\) at any given point?
A.The gradient is increased and the resistance is underestimated.
B.The gradient is unchanged and the resistance is underestimated.
C.The gradient is unchanged and the resistance is overestimated.
D.The gradient is decreased and the resistance is overestimated.
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解題
The relationship between measured current \(I_{\text{meas}}\) and potential difference \(V\) is:
Since the zero error is a constant value of \(+0.05 \ \text{A}\), the line of best fit on the graph of \(I\) against \(V\) is shifted vertically upwards but its gradient (which is \(\frac{1}{R}\)) remains **unchanged**.
However, at any single point, the calculated resistance is \(R_{\text{calc}} = \frac{V}{I_{\text{meas}}}\). Since \(I_{\text{meas}}\) is greater than the true current, the denominator is larger, and therefore \(R_{\text{calc}}\) is **underestimated**.
評分準則
1 mark for identifying that a constant zero offset does not change the gradient of the straight-line graph but causes the calculated resistance (V/I) at any point to be underestimated.
題目 32 · 選擇題
1 分
A uniform copper wire of resistance \(R\) is stretched such that its length increases by \(10\%\). The volume of the copper remains constant during the stretching process. What is the new resistance of the wire in terms of \(R\)?
A.1.10 R
B.1.20 R
C.1.21 R
D.1.44 R
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解題
The resistance is given by \(R = \rho \frac{L}{A}\), where \(\rho\) is the resistivity, \(L\) is the length, and \(A\) is the cross-sectional area. Since volume \(V = A \cdot L\) is constant:
\(A = \frac{V}{L}\)
Substituting this into the resistance formula gives:
\(R = \rho \frac{L^2}{V}\)
Since \(\rho\) and \(V\) are constant, the resistance is proportional to the square of the length: \(R \propto L^2\).
If the length increases by \(10\%\), the new length is \(1.10 L\). The new resistance \(R'\) is:
\(R' \propto (1.10 L)^2 = 1.21 L^2\)
Thus, \(R' = 1.21 R\).
評分準則
1 mark for deducing that resistance is proportional to length squared under constant volume conditions, leading to a new resistance of 1.21R.
題目 33 · 選擇題
1 分
A student determines the density \(\rho\) of a uniform metal wire using the formula \(\rho = \frac{4m}{\pi d^2 L}\). The measurements and their absolute uncertainties are: mass \(m = (1.24 \pm 0.02)\text{ g}\), diameter \(d = (0.80 \pm 0.02)\text{ mm}\), and length \(L = (15.0 \pm 0.1)\text{ cm}\). What is the percentage uncertainty in the calculated value of the density \(\rho\)?
A.3.1%
B.4.8%
C.7.3%
D.9.8%
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解題
The formula for the fractional uncertainty in \(\rho\) is given by: \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\). Substituting the measured values and absolute uncertainties: \(\frac{\Delta m}{m} = \frac{0.02}{1.24} \approx 0.0161\), \(2\frac{\Delta d}{d} = 2 \times \frac{0.02}{0.80} = 0.0500\), and \(\frac{\Delta L}{L} = \frac{0.1}{15.0} \approx 0.0067\). Adding these fractional uncertainties: \(\frac{\Delta \rho}{\rho} = 0.0161 + 0.0500 + 0.0067 = 0.0728\). Expressed as a percentage, this is \(7.28\%\), which rounds to \(7.3\%\).
評分準則
1 mark for the correct option C. 0 marks for incorrect options.
題目 34 · 選擇題
1 分
A uniform beam of length \(2.0\text{ m}\) and weight \(120\text{ N}\) is hinged to a vertical wall at one end and held horizontally by a cable attached to the other end. The cable makes an angle of \(30^\circ\) with the horizontal beam. A load of \(80\text{ N}\) is hung from the outer end of the beam. What is the tension in the cable?
A.140 N
B.200 N
C.280 N
D.560 N
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解題
Taking moments about the hinge at the wall for rotational equilibrium: Clockwise moments = \((120\text{ N} \times 1.0\text{ m}) + (80\text{ N} \times 2.0\text{ m}) = 120\text{ N m} + 160\text{ N m} = 280\text{ N m}\). Counter-clockwise moments = \(T \sin(30^\circ) \times 2.0\text{ m} = T \times 0.5 \times 2.0\text{ m} = T \times 1.0\text{ m}\). Equating clockwise and counter-clockwise moments: \(T = 280\text{ N}\).
評分準則
1 mark for the correct option C. 0 marks for incorrect options.
題目 35 · 選擇題
1 分
Two wires \(X\) and \(Y\) are made of the same metal. Wire \(X\) has twice the length and half the diameter of wire \(Y\). Both wires are subjected to the same tensile force. What is the ratio \(\frac{\text{strain in } X}{\text{strain in } Y}\)?
A.1
B.2
C.4
D.8
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解題
Young modulus is given by \(E = \frac{\text{stress}}{\text{strain}}\), so \(\text{strain} = \frac{\text{stress}}{E} = \frac{F}{A E}\). Since both wires are made of the same metal, they have the same Young modulus \(E\). Both are subjected to the same tensile force \(F\). Thus, strain is inversely proportional to the cross-sectional area: \(\text{strain} \propto \frac{1}{A} \propto \frac{1}{d^2}\). The length of the wire does not affect the strain under these conditions. Wire \(X\) has half the diameter of wire \(Y\) (\(d_X = 0.5 d_Y\)), which means its area \(A_X = 0.25 A_Y\). The ratio of the strains is \(\frac{\text{strain}_X}{\text{strain}_Y} = \frac{A_Y}{A_X} = \frac{1}{0.25} = 4\).
評分準則
1 mark for the correct option C. 0 marks for incorrect options.
題目 36 · 選擇題
1 分
A polymeric specimen is stretched and then unstretched. The area under the loading force-extension curve is \(12.0\text{ J}\), and the area under the unloading force-extension curve is \(8.5\text{ J}\). Which statement about this process is correct?
A.The maximum elastic potential energy stored in the fully stretched material is 3.5 J.
B.The thermal energy dissipated during the loading-unloading cycle is 3.5 J.
C.The work done by the material during unloading is 12.0 J.
D.The total work done on the material during the entire cycle is 20.5 J.
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解題
The area under the loading curve represents the work done on the material to stretch it (\(12.0\text{ J}\)). The area under the unloading curve represents the work done by the material during contraction (the recovered elastic energy, \(8.5\text{ J}\)). The difference between these two areas, \(12.0\text{ J} - 8.5\text{ J} = 3.5\text{ J}\), represents the work done that is not recovered, which is dissipated as thermal energy in the material due to elastic hysteresis.
評分準則
1 mark for the correct option B. 0 marks for incorrect options.
題目 37 · 選擇題
1 分
A uniform cylindrical copper wire of resistance \(R\) is stretched such that its length increases by \(10\%\). The volume of the copper wire remains constant during the stretching process. What is the new resistance of the wire in terms of \(R\)?
A.1.00 R
B.1.10 R
C.1.21 R
D.1.33 R
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解題
Resistance of a wire is given by \(R = \rho \frac{L}{A}\). Since volume \(V = A L\) is constant, we can write the area as \(A = \frac{V}{L}\), giving \(R = \rho \frac{L^2}{V}\). When the length increases by \(10\%\), the new length becomes \(L' = 1.10 L\). The new resistance is \(R' = \rho \frac{(1.10 L)^2}{V} = 1.21 \left(\rho \frac{L^2}{V}\right) = 1.21 R\).
評分準則
1 mark for the correct option C. 0 marks for incorrect options.
題目 38 · 選擇題
1 分
A potential divider circuit consists of a \(6.0\text{ V}\) d.c. supply of negligible internal resistance, a fixed resistor of resistance \(3.0\text{ k}\Omega\), and a light-dependent resistor (LDR) connected in series. The output voltage \(V_{\text{out}}\) is measured across the LDR. In bright light, the LDR has a resistance of \(1.5\text{ k}\Omega\). In darkness, the LDR has a resistance of \(12\text{ k}\Omega\). What is the change in the output voltage \(V_{\text{out}}\) when the conditions change from bright light to darkness?
A.1.2 V
B.2.8 V
C.3.0 V
D.4.8 V
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解題
The output voltage is given by the potential divider equation: \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R_{\text{fixed}} + R_{\text{LDR}}}\). In bright light, \(V_{\text{out}} = 6.0 \times \frac{1.5}{3.0 + 1.5} = 2.0\text{ V}\). In darkness, \(V_{\text{out}} = 6.0 \times \frac{12}{3.0 + 12} = 4.8\text{ V}\). The change in the output voltage is \(4.8\text{ V} - 2.0\text{ V} = 2.8\text{ V}\).
評分準則
1 mark for the correct option B. 0 marks for incorrect options.
題目 39 · 選擇題
1 分
In a beta-plus (\(\beta^+\)) decay, a proton inside a nucleus decays into a neutron, a positron, and an electron neutrino: \(\text{p} \rightarrow \text{n} + \text{e}^+ + \nu_{\text{e}}\). Which quark change occurs, and what is the change in the total strangeness of the particles?
A.an up quark changes to a down quark, and the change in strangeness is 0
B.an up quark changes to a down quark, and the change in strangeness is +1
C.a down quark changes to an up quark, and the change in strangeness is 0
D.a down quark changes to an up quark, and the change in strangeness is -1
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解題
A proton has a quark structure of \(\text{uud}\) and a neutron has a quark structure of \(\text{udd}\). During beta-plus decay, one of the up quarks changes into a down quark (\(\text{u} \rightarrow \text{d}\)). Since none of the involved particles contain strange quarks, the strangeness of all particles is 0, so the change in the total strangeness of the particles is 0.
評分準則
1 mark for the correct option A. 0 marks for incorrect options.
題目 40 · 選擇題
1 分
In the alpha-particle scattering experiment, a beam of \(\alpha\)-particles is directed at a thin gold foil. Which experimental observation leads to the correct conclusion about the structure of the atom?
A.The vast majority of \(\alpha\)-particles passing straight through concludes that the nucleus is positively charged.
B.The vast majority of \(\alpha\)-particles passing straight through concludes that most of the mass of the atom is concentrated in the nucleus.
C.A very small fraction of \(\alpha\)-particles being deflected through large angles concludes that most of the atom is empty space.
D.A very small fraction of \(\alpha\)-particles being deflected through large angles concludes that the nucleus is extremely small and contains most of the mass and positive charge of the atom.
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解題
The observation that a very small fraction of \(\alpha\)-particles are deflected through large angles (greater than \(90^\circ\)) indicates that they experienced a strong electrostatic repulsive force from a highly concentrated positive charge that contains almost all the mass of the atom, which is the nucleus. The observation that most particles pass straight through with no deflection shows that most of the atom is empty space.
評分準則
1 mark for the correct option D. 0 marks for incorrect options.
卷二
Answer all six structured theoretical questions on the question paper.
6 題目 · 60 分
題目 1 · Structured
10 分
(a) Distinguish between systematic errors and random errors in experimental measurements. [2]
(b) A student measures the diameter \(d\) of a uniform metal wire using a micrometer screw gauge.
(i) Calculate the average diameter \(d\) and estimate its absolute uncertainty. Explain how you determined the uncertainty. [3]
(ii) The student also measures the following quantities for the same wire: - Current \(I = (1.20 \pm 0.05)\text{ A}\) - Potential difference \(V = (3.4 \pm 0.1)\text{ V}\) - Length \(L = (75.0 \pm 0.2)\text{ cm}\)
Calculate the resistivity \(\rho\) of the material of the wire. [3]
(iii) Calculate the percentage uncertainty in the value of \(\rho\). [2]
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解題
(a) - Systematic errors cause measurements to be consistently shifted from the true value in one direction. They cannot be eliminated by repeating and averaging. - Random errors cause measurements to fluctuate unpredictably above and below the true value. They can be reduced by taking multiple readings and calculating a mean.
(b)(i) - Average diameter: \(d = \frac{0.38 + 0.36 + 0.37 + 0.39 + 0.35}{5} = 0.37\text{ mm}\) - Absolute uncertainty (using half the range): \(\Delta d = \frac{0.39 - 0.35}{2} = 0.02\text{ mm}\) (Alternatively, using the resolution of the micrometer, \(\Delta d = 0.01\text{ mm}\) if justified, but half-range is standard for fluctuating data). Thus, \(d = (0.37 \pm 0.02)\text{ mm}\).
(iii) - Since \(\rho = \frac{V \pi d^2}{4 I L}\), the fractional uncertainty formula is: \(\frac{\Delta \rho}{\rho} = \frac{\Delta V}{V} + \frac{\Delta I}{I} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\) - Substituting the values: \(\frac{\Delta \rho}{\rho} = \frac{0.1}{3.4} + \frac{0.05}{1.20} + 2\left(\frac{0.02}{0.37}\right) + \frac{0.2}{75.0}\) \(\frac{\Delta \rho}{\rho} \approx 0.0294 + 0.0417 + 0.1081 + 0.0027 = 0.1819\) - Percentage uncertainty \(= 18.2\%\) (or \(18\%\) to 2 s.f.). (If \(\Delta d = 0.01\text{ mm}\) was used, \(\frac{\Delta \rho}{\rho} \approx 0.128\), giving \(13\%\)).
評分準則
(a) - M1: Systematic errors consistently depart from the true value in one direction / cannot be reduced by averaging. - A1: Random errors vary randomly above and below the true value / can be reduced by averaging.
(b)(i) - C1: Mean diameter calculated correctly as 0.37 mm. - A1: Uncertainty correctly identified as 0.02 mm (half range) [accept 0.01 mm if justified by instrument limit]. - B1: Clear justification given (e.g., "calculated as half the range of the readings").
(ii) - C1: Correct calculation of R (2.83 ohms) or area A (1.08 * 10^-7 m^2). - C1: Correct formula for resistivity used with consistent SI units. - A1: Correct final value 4.1 * 10^-7 ohm m (or 4.06 * 10^-7 ohm m).
(iii) - C1: Correct formulation of the percentage uncertainty expression (with factor of 2 for d). - A1: Correct final percentage uncertainty of 18% (or 13% if absolute uncertainty of 0.01 mm was used).
題目 2 · Structured
10 分
(a) State the two conditions required for a rigid body to be in static equilibrium. [2]
(b) A uniform metal shelf of length \(1.20\text{ m}\) and weight \(45\text{ N}\) is hinged to a vertical wall at point \(P\). The shelf is maintained in a horizontal position by a light cable attached to the outer edge of the shelf at point \(Q\) and anchored to the wall at a point \(R\) directly above \(P\). The cable makes an angle of \(35^\circ\) with the horizontal shelf.
(i) By taking moments about \(P\), calculate the tension \(T\) in the cable. [3]
(ii) Determine the magnitude and direction of the force exerted on the shelf by the hinge at \(P\). [5]
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解題
(a) 1. The resultant (or net) force acting on the body must be zero in any direction. 2. The resultant (or net) moment about any pivot point must be zero.
(b)(i) - Let \(L = 1.20\text{ m}\) be the length of the shelf. Since it is uniform, its weight \(W = 45\text{ N}\) acts at its midpoint, \(0.60\text{ m}\) from \(P\). - Taking moments about \(P\): \(\text{Clockwise moment} = W \times 0.60 = 45 \times 0.60 = 27.0\text{ N m}\) \(\text{Counter-clockwise moment} = (T \sin(35^\circ)) \times 1.20 = 1.20 T \sin(35^\circ)\) - For equilibrium: \(1.20 T \sin(35^\circ) = 27.0\) \(T = \frac{27.0}{1.20 \sin(35^\circ)} = 39.22\text{ N}\) (or \(39\text{ N}\) to 2 s.f.).
(b)(ii) - Let \(H\) and \(V\) be the horizontal and vertical components of the hinge force at \(P\). - For horizontal equilibrium: \(H = T \cos(35^\circ) = 39.22 \cos(35^\circ) \approx 32.13\text{ N}\) (acting away from the wall, to the right). - For vertical equilibrium: \(V + T \sin(35^\circ) = W\) \(V + 39.22 \sin(35^\circ) = 45\) \(V = 45 - 22.5 = 22.5\text{ N}\) (acting upwards). - The magnitude of the hinge force is: \(F_P = \sqrt{H^2 + V^2} = \sqrt{32.13^2 + 22.5^2} = \sqrt{1032.3 + 506.25} = 39.22\text{ N}\) (or \(39\text{ N}\)). - The angle \(\theta\) of the force with the horizontal is: \(\theta = \tan^{-1}\left(\frac{V}{H}\right) = \tan^{-1}\left(\frac{22.5}{32.13}\right) = 35^\circ\) above the horizontal.
評分準則
(a) - B1: No resultant / net force in any direction. - B1: No resultant / net moment about any point.
(b)(i) - C1: Correct moment equation about P: e.g., 45 * 0.6 = T sin(35) * 1.2 - C1: Correct calculation of weight moment (27 N m) or perpendicular tension distance. - A1: Tension calculated as 39 N (or 39.2 N).
(b)(ii) - C1: Resolve forces horizontally: H = T cos(35) = 32.1 N. - C1: Resolve forces vertically: V = 45 - T sin(35) = 22.5 N. - C1: Calculate magnitude of resultant force using Pythagoras theorem. - A1: Resultant force = 39 N (or 39.2 N). - A1: Direction = 35 degrees above the horizontal (pointing upwards and rightwards away from wall).
題目 3 · Structured
10 分
(a) Define: (i) tensile stress [1] (ii) tensile strain [1]
(b) A lift (elevator) of mass \(850\text{ kg}\) is supported by a single vertical steel cable of length \(40\text{ m}\). The Young modulus of steel is \(2.0 \times 10^{11}\text{ Pa}\).
The lift accelerates upwards from rest with an acceleration of \(1.6\text{ m s}^{-2}\).
Calculate: (i) the tension \(T\) in the cable. [2] (ii) the minimum diameter of the cable if the tensile stress in the cable is not to exceed \(1.2 \times 10^{8}\text{ Pa}\). [3] (iii) the extension of the cable under this tension when the lift is accelerating, using the minimum diameter calculated in (ii). [3]
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解題
(a)(i) Tensile stress is the force per unit cross-sectional area, perpendicular to the force: \(\sigma = \frac{F}{A}\). (a)(ii) Tensile strain is the extension per unit original length: \(\epsilon = \frac{x}{L}\).
(b)(i) - Using Newton's second law for the upward accelerating lift: \(T - mg = ma\) \(T = m(g + a) = 850(9.81 + 1.6) = 850(11.41) = 9698.5\text{ N}\) \(T \approx 9.7 \times 10^3\text{ N}\) (or \(9700\text{ N}\)).
(a)(i) - B1: Force / cross-sectional area (with mention of force being normal to area). (a)(ii) - B1: Extension / original length.
(b)(i) - C1: Correct formula incorporating Newton's 2nd Law: T - mg = ma (or T = m(g+a)) - A1: T = 9.7 * 10^3 N (or 9700 N, accept 9698.5 N)
(b)(ii) - C1: Correctly relates Stress = T/A to find area A. - C1: Area A = 8.08 * 10^-5 m^2. - A1: Diameter d = 1.0 * 10^-2 m (or 1.01 * 10^-2 m; accept 1.0 cm).
(b)(iii) - C1: Use of E = Stress / Strain to obtain Strain. - C1: Strain = 6.0 * 10^-4. - A1: Extension = 2.4 * 10^-2 m (or 2.4 cm; accept 0.024 m).
題目 4 · Structured
10 分
(a) State Hooke's Law and identify the feature of a force-extension graph that shows a material obeys this law. [2]
(b) A spring is stretched elastically by a force \(F\). The work done in stretching the spring to an extension \(x_1\) is \(W_1\).
(i) Show that the elastic potential energy \(E_p\) stored in the spring is given by \(E_p = \frac{1}{2} k x_1^2\, where \)k\) is the spring constant. [2]
(ii) The spring is now stretched further to an extension \(x_2 = 2 x_1\). Determine, in terms of \(W_1\), the work done in stretching the spring from extension \(x_1\) to \(x_2\). [3]
(c) Distinguish between the elastic limit and plastic deformation, describing what happens to the microscopic structure of a metal when it undergoes plastic deformation. [3]
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解題
(a) - Hooke's Law: The extension is directly proportional to the applied force, provided the limit of proportionality is not exceeded. - Feature: The graph is a straight line passing through the origin.
(b)(i) - The work done in stretching a spring is given by the area under the force-extension graph. - For a spring obeying Hooke's Law, the graph is linear, so the area of the triangular region under the graph is: \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} x_1 F\) - Since \(F = k x_1\), substituting this in gives: \(E_p = \frac{1}{2} k x_1^2\) (as required).
(b)(ii) - Work done to stretch the spring to extension \(x_1\) is: \(W_1 = \frac{1}{2} k x_1^2\) - Total work done to stretch the spring to extension \(x_2 = 2 x_1\) is: \(W_2 = \frac{1}{2} k x_2^2 = \frac{1}{2} k (2 x_1)^2 = 4 \left(\frac{1}{2} k x_1^2\right) = 4 W_1\) - The work done to stretch the spring *from* \(x_1\) to \(x_2\) is: \(\Delta W = W_2 - W_1 = 4 W_1 - W_1 = 3 W_1\).
(c) - Elastic limit is the maximum force/extension up to which the material will return to its original length when the load is removed. - Plastic deformation occurs when the material is permanently deformed and does not return to its original shape/length after the load is removed. - Microscopic description: Atomic planes slide past one another (movement of dislocations) and do not return to their original lattice positions.
評分準則
(a) - B1: Force is directly proportional to extension (up to proportionality limit). - B1: Graph is a straight line through the origin.
(b)(i) - B1: Explains that work done is the area under the force-extension graph. - B1: Uses F = kx to show Area = 1/2 * x * (kx) = 1/2 k x^2.
(b)(ii) - C1: Relates total work at x_2 to W_1, i.e., W_2 = 4 W_1. - C1: Identifies the required work is the difference between W_2 and W_1. - A1: Correctly derives 3 W_1.
(c) - B1: Defines elastic limit (returns to original shape if load removed). - B1: Defines plastic deformation (permanent stretch/deformation). - B1: Mentions planes of atoms sliding past one another / movement of dislocations.
題目 5 · Structured
10 分
(a) State Kirchhoff's first and second laws, and name the conservation law associated with each. [4]
(b) A circuit consists of two batteries connected in parallel with each other across an external resistor \(R = 4.0\ \Omega\).
Battery 1 has an electromotive force (e.m.f.) \(E_1 = 12.0\text{ V}\) and internal resistance \(r_1 = 2.0\ \Omega\). Battery 2 has an e.m.f. \(E_2 = 6.0\text{ V}\) and internal resistance \(r_2 = 4.0\ \Omega\).
(i) Write down three equations, based on Kirchhoff's laws, that relate the currents \(I_1\) in Battery 1, \(I_2\) in Battery 2, and the current \(I\) in resistor \(R\). Assume all currents flow towards the positive potential junction. [3]
(ii) Solve these equations to find the current \(I\) through the \(4.0\ \Omega\) resistor and the potential difference \(V\) across it. [3]
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解題
(a) - Kirchhoff's First Law: The sum of currents entering any junction is equal to the sum of currents leaving that junction. It is a consequence of the conservation of charge. - Kirchhoff's Second Law: The algebraic sum of e.m.f.s in any closed loop is equal to the algebraic sum of the potential differences (or \(I R\) drops) in that loop. It is a consequence of the conservation of energy.
(b)(i) - Let the currents flowing from the batteries be \(I_1\) (from Battery 1) and \(I_2\) (from Battery 2), joining at the top junction to form current \(I\) flowing through resistor \(R\): 1. Junction equation: \(I_1 + I_2 = I\) 2. Loop 1 (outer loop containing Battery 1 and resistor \(R\)): \(12.0 - 2.0 I_1 - 4.0 I = 0 \implies 12.0 = 2.0 I_1 + 4.0 I\) 3. Loop 2 (containing Battery 2 and resistor \(R\)): \(6.0 - 4.0 I_2 - 4.0 I = 0 \implies 6.0 = 4.0 I_2 + 4.0 I\)
(b)(ii) - From (2): \(2.0 I_1 = 12.0 - 4.0 I \implies I_1 = 6.0 - 2.0 I\) - From (3): \(4.0 I_2 = 6.0 - 4.0 I \implies I_2 = 1.5 - I\) - Substitute into the junction equation (1): \((6.0 - 2.0 I) + (1.5 - I) = I\) \(7.5 - 3.0 I = I\) \(4.0 I = 7.5 \implies I = 1.875\text{ A}\) (or \(1.9\text{ A}\) to 2 s.f.). - The potential difference \(V\) across the resistor \(R\) is: \(V = I R = 1.875 \times 4.0 = 7.5\text{ V}\).
評分準則
(a) - B1: First law defined as sum of currents in = sum of currents out. - B1: First law linked to conservation of charge. - B1: Second law defined as sum of e.m.f.s = sum of p.d.s in a closed loop. - B1: Second law linked to conservation of energy.
(b)(ii) - C1: Substitution/elimination method shown to solve for I. - A1: Current I = 1.88 A or 1.9 A. - A1: Potential difference V = 7.5 V.
題目 6 · Structured
10 分
(a) State the names of the two main families of hadrons. [2]
(b) A free neutron decays into a proton, an electron, and an electron antineutrino.
(i) Write a complete nuclear equation representing this decay. [2]
(ii) Describe this decay in terms of the quark composition of the nucleons involved. [2]
(iii) Name the fundamental force responsible for this decay, and identify the class of exchange particles (gauge bosons) involved. [2]
(c) In a high-energy particle collision, an electron and a proton can interact to produce a neutron and an electron neutrino: \(e^- + p \rightarrow n + \nu_e\)
Show that this reaction is possible by verifying the conservation of charge, baryon number, and lepton number. [2]
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解題
(a) Baryons and Mesons.
(b)(i) - The equation for beta-minus decay is: \({^1_0\text{n}} \rightarrow {^1_1\text{p}} + {^0_{-1}\text{e}} + \overline{\nu}_e\) (Accept \(\text{n} \rightarrow \text{p} + \text{e}^- + \overline{\nu}_e\)).
(b)(ii) - A neutron consists of quarks: \(udd\). - A proton consists of quarks: \(uud\). - During this decay, a down (\(d\)) quark transitions into an up (\(u\)) quark.
(b)(iii) - The fundamental force is the weak nuclear force (or weak interaction). - The gauge boson is the \(W^-\) boson.
(c) We test the conservation laws for: \(e^- + p \rightarrow n + \nu_e\) 1. Charge \(Q\): - Left side: \((-1) + (+1) = 0\) - Right side: \(0 + 0 = 0\) - Conserved.
2. Baryon number \(B\): - Left side: \(0 + 1 = 1\) (proton is a baryon) - Right side: \(1 + 0 = 1\) (neutron is a baryon) - Conserved.
3. Lepton number \(L\): - Left side: \(1 + 0 = 1\) (electron is a lepton) - Right side: \(0 + 1 = 1\) (electron neutrino is a lepton) - Conserved. Since all three quantities are conserved, the reaction is possible.
評分準則
(a) - B1: Baryons - B1: Mesons
(b)(i) - B1: Correct symbols for neutron, proton, and electron. - B1: Correct symbol for the electron antineutrino (with the bar above the nu).
(b)(ii) - B1: State neutron is udd and proton is uud. - B1: State that a down quark changes to an up quark.
(b)(iii) - B1: Weak interaction / Weak nuclear force. - B1: W- boson (or W / Z bosons).
(c) - B1: Show charge is conserved (0 = 0) and baryon number is conserved (1 = 1). - B1: Show lepton number is conserved (1 = 1) and state that since all are conserved, the reaction is possible.
Paper 3
Perform both practical tasks, collecting data, plotting a graph, and outlining limitations.
2 題目 · 40 分
題目 1 · Practical
20 分
**Practical Task 1: Investigation of an Electrical Circuit containing a Resistance Wire**
In this experiment, you will investigate how the current in a circuit depends on the length of a resistance wire connected in series with a fixed resistor.
**Apparatus provided**: - 1.5 V cell - Switch - Ammeter (or digital multimeter configured to measure current up to 2 A d.c.) - 1.00 m length of resistance wire taped to a metre rule, with its cross-sectional area labeled as \(A = 1.13 \times 10^{-7} \text{ m}^2\) - Fixed resistor labeled \(5.6 \ \Omega\) - Connecting leads and crocodile clips
**Procedure**: 1. Connect the cell, the switch, the ammeter, the fixed resistor, and a length \(x\) of the resistance wire in a single loop series circuit. 2. Position the crocodile clip on the resistance wire so that the length \(x\) of wire connected in the circuit is \(0.200 \text{ m}\). 3. Close the switch, measure the current \(I\), and record this value. Open the switch immediately. 4. Change the position of the crocodile clip to vary the length \(x\) of the wire in the circuit. Measure and record the current \(I\) for at least six different values of \(x\) in the range \(0.200 \text{ m} \le x \le 0.800 \text{ m}\). 5. Record all your measurements and calculated values of \(1/I\) in a single table. Include headings and units for all columns. 6. Plot a graph of \(1/I\) on the y-axis against \(x\) on the x-axis. Draw the straight line of best fit. 7. Determine the gradient and the y-intercept of your line of best fit. Show your working clearly. 8. The variables \(I\) and \(x\) are related by the equation: \[\frac{1}{I} = P x + Q\] where \(P\) and \(Q\) are constants. Use your results from (7) to determine the values of \(P\) and \(Q\). Include appropriate units for both constants.
**Graph Calculations**: Using coordinates from the line of best fit: \((0.200, 4.65)\) and \((0.800, 6.37)\): Gradient \(P = \frac{6.37 - 4.65}{0.800 - 0.200} = \frac{1.72}{0.600} \approx 2.87 \text{ A}^{-1} \text{ m}^{-1}\).
**Collection of Data (5 marks)**: - Six sets of readings of \(x\) and \(I\) recorded with correct units (5 marks). Deduct 1 mark for each set fewer than 6, down to 0. - Range of \(x\) values must cover at least \(0.500 \text{ m}\) (1 mark).
**Table Quality and Presentation (4 marks)**: - Column headings must contain a quantity and a unit (e.g., \(x / \text{ m}\), \(I / \text{ A}\), \(1/I / \text{ A}^{-1}\)) (1 mark). - Consistency of raw readings: \(x\) recorded to the nearest millimeter (\(0.001 \text{ m}\)) and \(I\) recorded to the nearest \(0.001 \text{ A}\) or \(0.01 \text{ A}\) consistently (1 mark). - Significant figures: \(1/I\) calculated to the same number of significant figures as, or one more than, the least number of significant figures in the raw current \(I\) (1 mark). - Correct calculation of \(1/I\) values (1 mark).
**Graph (4 marks)**: - Linear axes with appropriate scale (no factor of 3, etc.). Plotted points occupy more than half the grid in both directions (1 mark). - Plotting: All observations plotted to within half a small square (1 mark). - Best-fit straight line drawn, with balanced scatter of points about the line (1 mark). - Quality: Scatter is small, with no point further than \(0.1 \text{ A}^{-1}\) from the best-fit line (1 mark).
**Gradient and Intercept (2 marks)**: - Gradient: Determined from a triangle where the hypotenuse is at least half the length of the drawn line, with coordinates read to within half a small square (1 mark). - y-intercept: Read directly from the y-axis (if \(x=0\) scale is on the page) or calculated using \(y = mx + c\) with a point on the line (1 mark).
**Analysis and Constants (5 marks)**: - Value of \(P\) matches gradient, with correct unit (\(\text{ A}^{-1} \text{ m}^{-1}\) or \(\text{ V}^{-1} \Omega \text{ m}^{-1}\)) (2 marks). - Value of \(Q\) matches y-intercept, with correct unit (\(\text{ A}^{-1}\) or \(\text{ V}^{-1} \Omega\)) (2 marks). - Values must be written to 2 or 3 significant figures (1 mark).
題目 2 · Practical
20 分
**Practical Task 2: Investigation of the Bending of a Clamped Rule**
In this experiment, you will investigate how the vertical depression of a clamped wooden rule depends on the load suspended from its free end.
**Apparatus provided**: - Wooden metre rule - G-clamp and two wooden blocks to clamp the rule - Stand, boss, and clamp - Mass hanger and slotted masses (to make up \(100 \text{ g}\) and \(300 \text{ g}\)) - Micrometer screw gauge - Second metre rule to measure heights
**Procedure**: 1. Use the G-clamp and wooden blocks to clamp one end of the first metre rule to the edge of the bench. The rule should project horizontally beyond the edge of the bench, with a length \(L = 0.800 \text{ m}\) extending from the clamp. 2. Use the micrometer screw gauge to measure the thickness \(d\) of the rule. Record your measurements and determine the mean value of \(d\). 3. Estimate the absolute uncertainty in your value of \(d\) and calculate its percentage uncertainty. 4. Attach the mass hanger to the free end of the clamped rule at the \(0.790 \text{ m}\) mark. 5. Measure the initial vertical height \(h_0\) of the bottom edge of the free end of the rule from the floor. 6. Place a mass \(m_1 = 100 \text{ g}\) on the hanger. Measure the new height \(h_1\) of the free end. Calculate the vertical depression \(y_1 = h_0 - h_1\). 7. Remove the mass and place a mass \(m_2 = 300 \text{ g}\) on the hanger. Measure the new height \(h_2\) of the free end. Calculate the vertical depression \(y_2 = h_0 - h_2\). 8. It is suggested that the relationship between the depression \(y\) and mass \(m\) is: \[y = k \cdot m\] where \(k\) is a constant. Calculate two values of \(k\) (\(k_1\) and \(k_2\)) for your measurements. 9. State whether your results support the suggested relationship. Justify your answer by comparing the percentage difference between your two values of \(k\) with a specified limit of experimental accuracy. 10. Describe four sources of difficulty or limitations in this experiment, and suggest four improvements to reduce these difficulties. Present your findings in a table.
**Comparison**: Percentage difference between \(k_1\) and \(k_2\): \(\frac{|0.227 - 0.22|}{0.22} \times 100\% = 3.2\%\). Since \(3.2\%\) is less than the standard experimental limit of \(20\%\), the relationship is strongly supported.
**Evaluation Table**: 1. **Limitation**: Error due to parallax when reading the scale of the vertical rule to measure heights. *Improvement*: Use a set square or align a horizontal reference pointer to eliminate parallax. 2. **Limitation**: Oscillation of the rule after placing weights makes it difficult to measure height. *Improvement*: Introduce a damping mechanism or wait a specified period of time for the oscillations to die down. 3. **Limitation**: Only two values of mass were investigated, which is not enough to confirm a proportional relationship. *Improvement*: Take measurements with multiple other masses and plot a graph of \(y\) against \(m\). 4. **Limitation**: Small slipping of the rule at the G-clamp support under larger loads. *Improvement*: Use protective rubber pads or a second clamp to ensure a completely rigid support.
評分準則
**Measurements and Uncertainties (6 marks)**: - Measurement of \(d\) to nearest \(0.01 \text{ mm}\) with repeat readings (1 mark). - Correct unit for \(d\) shown in raw table/results (1 mark). - Absolute uncertainty of \(d\) estimated to be within \(\pm 0.01 \text{ mm}\) to \(\pm 0.05 \text{ mm}\) (1 mark). - Percentage uncertainty of \(d\) calculated correctly with working (1 mark). - Heights \(h_0\), \(h_1\), \(h_2\) recorded to nearest \(1 \text{ mm}\) (1 mark). - Quality: \(y_2 > y_1\) (1 mark).
**Calculations and Analysis (4 marks)**: - Correct calculation of \(k_1\) and \(k_2\) with appropriate units (e.g., \(\text{m kg}^{-1}\) or \(\text{cm g}^{-1}\)) (2 marks). - Correct percentage difference calculation (1 mark). - Clear conclusion comparing percentage difference to 20% limit (or candidate's own suggested limit) to determine if the relation is supported (1 mark).
**Evaluation (10 marks)**: - Identify 4 distinct sources of difficulties/uncertainties (maximum 5 marks, 1 per valid point): 1. Parallax error in measuring vertical height. 2. Metre rule vibrates/oscillates after loading. 3. Two sets of mass values are insufficient to draw a firm conclusion. 4. Slipping of the ruler at the pivot/clamp. - Suggest 4 corresponding improvements (maximum 5 marks, 1 per valid point): 1. Use a set square on the table/floor against the metre rule / use a marker on the ruler. 2. Wait for oscillations to die down / use a liquid damping system. 3. Take more readings for other masses and plot a graph. 4. Use rubber jaws on the clamp to prevent slipping.
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