An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V3) Cambridge International A Level Physics (9702) paper. Not affiliated with or reproduced from Cambridge.
Paper 1 (Multiple Choice)
Answer all forty questions. For each question there are four possible answers A, B, C and D.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
An experiment is conducted to determine the resistivity \(\rho\) of a uniform metal wire using the formula: \(\rho = \frac{R \pi d^2}{4L}\). The measured values with their absolute uncertainties are: Resistance, \(R = (4.50 \pm 0.09)\ \Omega\); Diameter, \(d = (0.38 \pm 0.01)\ \text{mm}\); Length, \(L = (0.800 \pm 0.004)\ \text{m}\). What is the percentage uncertainty in the calculated value of the resistivity \(\rho\)?
A.5.1%
B.7.3%
C.7.8%
D.9.8%"},
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Worked solution
The percentage uncertainty in resistivity is given by the sum of the percentage uncertainties of the individual quantities, taking into account the power of each quantity: \(\frac{\Delta \rho}{\rho} \times 100\% = \frac{\Delta R}{R} \times 100\% + 2 \left(\frac{\Delta d}{d} \times 100\%\right) + \frac{\Delta L}{L} \times 100\%\). First, calculate each percentage uncertainty: For \(R\): \(\frac{0.09}{4.50} \times 100\% = 2.0\%\). For \(d\): \(\frac{0.01}{0.38} \times 100\% \approx 2.63\%\). For \(L\): \(\frac{0.004}{0.800} \times 100\% = 0.5\%\). Now combine them: Total percentage uncertainty = \(2.0\% + 2 \times 2.63\% + 0.5\% = 2.0\% + 5.26\% + 0.5\% = 7.76\%\), which rounds to \(7.8\%\).
Marking scheme
1 mark for the correct calculation of individual percentage uncertainties, doubling the diameter's percentage uncertainty, and summing them up to obtain 7.8% (Option C).
Question 2 · multiple-choice
1 marks
An experiment is performed to determine the density of a small metal sphere. The mass \(m\) of the sphere is measured as \((12.4 \pm 0.1)\ \text{g}\) and its diameter \(d\) is measured as \((1.42 \pm 0.02)\ \text{cm}\). What is the value of the density of the metal, with its associated absolute uncertainty, in \(\text{g\ cm}^{-3}\)?
A.\((8.3 \pm 0.1)\ \text{g\ cm}^{-3}\)
B.\((8.3 \pm 0.2)\ \text{g\ cm}^{-3}\)
C.\((8.3 \pm 0.4)\ \text{g\ cm}^{-3}\)
D.\((8.3 \pm 0.5)\ \text{g\ cm}^{-3}\)
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Worked solution
The volume of the sphere is \(V = \frac{1}{6}\pi d^3\). The density is \(\rho = \frac{m}{V} = \frac{6m}{\pi d^3}\). Calculating the density value: \(\rho = \frac{6 \times 12.4}{\pi \times 1.42^3} \approx 8.271\ \text{g\ cm}^{-3}\). The fractional uncertainty in density is: \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \frac{\Delta d}{d} = \frac{0.1}{12.4} + 3 \left(\frac{0.02}{1.42}\right) \approx 0.00806 + 3 \times 0.01408 = 0.00806 + 0.04225 = 0.05031\). The absolute uncertainty in density is: \(\Delta \rho = 8.271 \times 0.05031 \approx 0.416\ \text{g\ cm}^{-3}\). Since absolute uncertainties are conventionally quoted to one significant figure, \(\Delta \rho \approx 0.4\ \text{g\ cm}^{-3}\). The density value is then quoted to the same decimal place: \(\rho = (8.3 \pm 0.4)\ \text{g\ cm}^{-3}\).
Marking scheme
1 mark for calculating the density value to be 8.3 g/cm^3 and summing the mass and triple the diameter fractional uncertainties to obtain the absolute uncertainty of 0.4 g/cm^3.
Question 3 · multiple-choice
1 marks
A digital balance is used to measure the mass of a beaker. When empty, the balance reads \(0.0\ \text{g}\). When a standard \(100.0\ \text{g}\) calibration mass is placed on it, the balance reads \(101.5\ \text{g}\). A series of measurements of the mass of a beaker are made using this balance. The readings are: \(52.3\ \text{g}\), \(52.2\ \text{g}\), \(52.4\ \text{g}\), \(52.3\ \text{g}\). Which statement correctly describes the errors in these measurements?
A.The measurements have high precision and a zero error.
B.The measurements have high precision and a systematic calibration error.
C.The measurements have low precision and a zero error.
D.The measurements have low precision and a systematic calibration error.
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Worked solution
1. Precision: The measurements are close together (ranging from 52.2 g to 52.4 g), indicating a high degree of precision (small random error). 2. Systematic Error: The balance is calibrated incorrectly because it reads 101.5 g for a 100.0 g mass. Therefore, all measurements will have a systematic calibration error. 3. Zero Error: When empty, the balance reads 0.0 g, meaning there is no zero error (no offset at zero). Hence, the measurements have high precision and a systematic calibration error.
Marking scheme
1 mark for identifying that the small range of readings indicates high precision, and the incorrect reading for the standard mass without a zero offset indicates a systematic calibration error.
Question 4 · multiple-choice
1 marks
A beam of alpha particles (\(\text{He}^{2+}\)) and a beam of beta-minus particles (\(\text{e}^-\)) enter a region of uniform magnetic field. The magnetic field is directed perpendicular to and into the plane of the paper. Both types of particles enter the field with the same velocity \(v\), perpendicular to the boundary of the magnetic field. Which statement correctly describes their paths in the magnetic field?
A.Both the alpha particles and beta-minus particles deflect in the same direction, but the alpha particles follow a path of smaller radius of curvature.
B.Both the alpha particles and beta-minus particles deflect in opposite directions, and the alpha particles follow a path of smaller radius of curvature.
C.Both the alpha particles and beta-minus particles deflect in the same direction, but the beta-minus particles follow a path of smaller radius of curvature.
D.Both the alpha particles and beta-minus particles deflect in opposite directions, and the beta-minus particles follow a path of smaller radius of curvature.
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Worked solution
1. Direction of deflection: Since alpha particles are positively charged and beta-minus particles are negatively charged, the magnetic forces on them act in opposite directions. By Fleming's Left-Hand Rule, they will deflect in opposite directions. 2. Radius of curvature: The magnetic force provides the centripetal force: \(qvB = \frac{m v^2}{r} \implies r = \frac{m v}{q B}\). Since both enter with the same velocity \(v\) into the same field \(B\), the radius is proportional to the mass-to-charge ratio \(\frac{m}{q}\). For the alpha particle, \(\left(\frac{m}{q}\right)_{\alpha} \approx \frac{4\ \text{u}}{2e} = 2\ \frac{\text{u}}{e}\). For the beta-minus particle, \(\left(\frac{m}{q}\right)_{\beta} \approx \frac{1/1840\ \text{u}}{e} \approx 5.4 \times 10^{-4}\ \frac{\text{u}}{e}\). Since \(\left(\frac{m}{q}\right)_{\beta} \ll \left(\frac{m}{q}\right)_{\alpha}\), the beta-minus particles will follow a path with a much smaller radius of curvature.
Marking scheme
1 mark for identifying that opposite charges deflect in opposite directions and that the much smaller mass-to-charge ratio of beta-minus particles results in a smaller radius of curvature (Option D).
Question 5 · multiple-choice
1 marks
A rigid, straight metal wire of length \(0.40\ \text{m}\) carries a current of \(3.5\ \text{A}\) in a direction pointing due North. The wire is placed in a uniform magnetic field of flux density \(0.050\ \text{T}\) that is directed vertically upwards. What is the magnitude and direction of the magnetic force acting on the wire?
D.Zero, because the magnetic field is perpendicular to the horizontal plane.
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Worked solution
The magnitude of the magnetic force is given by \(F = B I L \sin\theta\). Here, the current points North (horizontal) and the magnetic field is directed vertically upwards, so the angle \(\theta = 90^\circ\). Thus, \(F = 0.050\ \text{T} \times 3.5\ \text{A} \times 0.40\ \text{m} \times \sin(90^\circ) = 0.070\ \text{N}\). To find the direction, apply Fleming's Left-Hand Rule: Point your first finger (Field) vertically UPWARDS. Point your second finger (Current) due NORTH (forward). Your thumb (Force) points due EAST (to the right). Thus, the force is \(0.070\ \text{N}\) pointing due East.
Marking scheme
1 mark for correctly calculating the force as 0.070 N and using Fleming's Left-Hand Rule to determine the direction is East (Option A).
Question 6 · multiple-choice
1 marks
Two long, straight, parallel wires, X and Y, are separated by a distance of \(8.0\ \text{cm}\) in a vacuum. Wire X carries a current of \(6.0\ \text{A}\) upwards and wire Y carries a current of \(4.0\ \text{A}\) downwards. What is the magnetic flux density \(B\) at a point P on the line joining the two wires, at a distance of \(2.0\ \text{cm}\) from wire X and \(6.0\ \text{cm}\) from wire Y?
A.\(1.3 \times 10^{-5}\ \text{T}\)
B.\(4.7 \times 10^{-5}\ \text{T}\)
C.\(6.0 \times 10^{-5}\ \text{T}\)
D.\(7.3 \times 10^{-5}\ \text{T}\)
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Worked solution
The magnetic flux density due to a long straight wire is given by \(B = \frac{\mu_0 I}{2\pi r}\). Point P lies between the two wires: 1. Field due to wire X: \(B_{\text{X}} = \frac{(4\pi \times 10^{-7}) \times 6.0}{2\pi \times 0.020} = 6.0 \times 10^{-5}\ \text{T}\). By the right-hand grip rule, since the current is upwards, the magnetic field at point P (to the right of wire X) points INTO the page. 2. Field due to wire Y: \(B_{\text{Y}} = \frac{(4\pi \times 10^{-7}) \times 4.0}{2\pi \times 0.060} \approx 1.33 \times 10^{-5}\ \text{T}\). By the right-hand grip rule, since the current is downwards, the magnetic field at point P (to the left of wire Y) also points INTO the page. Since both fields point in the same direction, we add them: \(B_{\text{total}} = B_{\text{X}} + B_{\text{Y}} = 6.0 \times 10^{-5} + 1.33 \times 10^{-5} \approx 7.3 \times 10^{-5}\ \text{T}\).
Marking scheme
1 mark for calculating the magnetic fields from both wires, recognizing they point in the same direction (into the page), and adding them to find 7.3 x 10^-5 T (Option D).
Question 7 · multiple-choice
1 marks
A particle undergoes simple harmonic motion with an amplitude \(A\) and period \(T\). The particle starts from the equilibrium position at time \(t = 0\). What is the minimum time taken for the displacement of the particle to become \(\frac{\sqrt{3}}{2} A\)?
A.\(\frac{T}{12}\)
B.\(\frac{T}{6}\)
C.\(\frac{T}{4}\)
D.\(\frac{T}{3}\)
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Worked solution
Since the particle starts from the equilibrium position at \(t = 0\), its displacement as a function of time is given by \(x = A \sin(\omega t)\), where \(\omega = \frac{2\pi}{T}\). We set \(x = \frac{\sqrt{3}}{2} A\): \(A \sin(\omega t) = \frac{\sqrt{3}}{2} A \implies \sin(\omega t) = \frac{\sqrt{3}}{2}\). The minimum positive angle satisfying this is \(\omega t = \frac{\pi}{3}\). Substituting \(\omega = \frac{2\pi}{T}\): \(\frac{2\pi}{T} t = \frac{\pi}{3} \implies t = \frac{T}{6}\).
Marking scheme
1 mark for setting up the displacement equation starting from equilibrium, solving for the phase angle to find pi/3, and converting this to the time t = T/6 (Option B).
Question 8 · multiple-choice
1 marks
A mass of \(0.20\ \text{kg}\) undergoes simple harmonic oscillations on a frictionless horizontal surface. The total energy of the oscillation is \(1.6\ \text{J}\). At a certain point in its motion, the displacement of the mass is equal to half of its amplitude. What are the kinetic energy and potential energy of the mass at this point?
A.Kinetic Energy = \(0.4\ \text{J}\), Potential Energy = \(1.2\ \text{J}\)
B.Kinetic Energy = \(0.8\ \text{J}\), Potential Energy = \(0.8\ \text{J}\)
C.Kinetic Energy = \(1.2\ \text{J}\), Potential Energy = \(0.4\ \text{J}\)
D.Kinetic Energy = \(1.4\ \text{J}\), Potential Energy = \(0.2\ \text{J}\)
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Worked solution
In simple harmonic motion, the potential energy \(E_p\) at displacement \(x\) is given by \(E_p = \frac{1}{2} k x^2\), where \(k\) is the force constant of the system. The total energy is the maximum potential energy, which occurs at maximum displacement (amplitude \(A\)): \(E_{\text{total}} = \frac{1}{2} k A^2 = 1.6\ \text{J}\). When the displacement is half the amplitude, \(x = \frac{1}{2}A\): \(E_p = \frac{1}{2} k \left(\frac{1}{2}A\right)^2 = \frac{1}{4} \left(\frac{1}{2} k A^2\right) = \frac{1}{4} E_{\text{total}} = \frac{1}{4} \times 1.6\ \text{J} = 0.4\ \text{J}\). By conservation of energy, the kinetic energy \(E_k\) at this point is: \(E_k = E_{\text{total}} - E_p = 1.6\ \text{J} - 0.4\ \text{J} = 1.2\ \text{J}\).
Marking scheme
1 mark for using the quadratic relationship between potential energy and displacement to determine that Ep is 1/4 of total energy (0.4 J), and subtracting from total energy to find Ek is 1.2 J (Option C).
Question 9 · Multiple Choice
1 marks
The density \(\rho\) of a uniform cylindrical wire is determined from measurements of its mass \(m\), length \(L\) and diameter \(d\). The measurements and their percentage uncertainties are shown:
Which of the following is the calculated value of \(R\) with its absolute uncertainty?
A.\((2.40 \pm 0.10)\ \Omega\)
B.\((2.40 \pm 0.14)\ \Omega\)
C.\((2.4 \pm 0.2)\ \Omega\)
D.\((2.4 \pm 0.3)\ \Omega\)
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Worked solution
First calculate \(R = \frac{P}{I^2} = \frac{15.0}{2.50^2} = 2.40\ \Omega\). Next, find the percentage uncertainties: \(\frac{\Delta P}{P} = \frac{0.6}{15.0} = 4.0\%\), and \(\frac{\Delta I}{I} = \frac{0.05}{2.50} = 2.0\%\). The percentage uncertainty in \(R\) is given by \(\frac{\Delta R}{R} = \frac{\Delta P}{P} + 2\frac{\Delta I}{I} = 4.0\% + 2(2.0\%) = 8.0\%\). The absolute uncertainty in \(R\) is \(\Delta R = 2.40 \times 0.080 = 0.192\ \Omega \approx 0.2\ \Omega\). Thus, \(R = (2.4 \pm 0.2)\ \Omega\).
Marking scheme
1 mark for the correct option C.
Question 11 · Multiple Choice
1 marks
A student performs an experiment to determine the acceleration of free fall \(g\) using a simple pendulum. The length \(L\) of the pendulum and the time \(t\) for 50 oscillations are measured as:
What is the percentage uncertainty in the calculated value of \(g\)?
A.1.1%
B.1.5%
C.1.7%
D.2.6%
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Worked solution
The formula for the acceleration of free fall is \(g = \frac{4\pi^2 L}{T^2}\), where \(T = t/50\) is the period of oscillation. The percentage uncertainty in the period \(T\) is equal to the percentage uncertainty in the total measured time \(t\), which is \(\frac{0.4}{90.0} \times 100\% \approx 0.444\%\). The percentage uncertainty in \(L\) is \(\frac{0.5}{80.0} \times 100\% = 0.625\%\). The percentage uncertainty in \(g\) is \(\frac{\Delta g}{g} \times 100\% = \frac{\Delta L}{L} \times 100\% + 2 \frac{\Delta T}{T} \times 100\% = 0.625\% + 2(0.444\%) = 1.513\% \approx 1.5\%\).
Marking scheme
1 mark for the correct option B.
Question 12 · Multiple Choice
1 marks
A beam of singly-charged positive ions of mass \(m\) enter a region of uniform magnetic field of flux density \(B\) at right angles to the field lines with speed \(v\). The radius of their circular path is \(r\). A second beam of doubly-charged positive ions of mass \(2m\) enter the same magnetic field at right angles with speed \(2v\). What is the radius of the circular path of these doubly-charged ions?
A.\(\frac{1}{2}r\)
B.\(r\)
C.\(2r\)
D.\(4r\)
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Worked solution
The magnetic force provides the centripetal force: \(q v B = \frac{m v^2}{r} \implies r = \frac{m v}{q B}\). For the second beam, \(m_2 = 2m\), \(v_2 = 2v\), and \(q_2 = 2q\). Thus, \(r_2 = \frac{(2m)(2v)}{(2q)B} = 2 \left(\frac{mv}{qB}\right) = 2r\).
Marking scheme
1 mark for the correct option C.
Question 13 · Multiple Choice
1 marks
A straight, horizontal copper wire of length \(0.15\text{ m}\) and mass \(4.5\text{ g}\) is suspended in a uniform, horizontal magnetic field of flux density \(0.20\text{ T}\). The direction of the magnetic field is perpendicular to the wire. What current must flow through the wire so that the magnetic force exactly balances the weight of the wire? (Take the acceleration of free fall to be \(9.81\text{ m s}^{-2}\))
A.0.15 A
B.1.5 A
C.6.8 A
D.15 A
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Worked solution
For the magnetic force to balance the weight: \(F_B = W \implies B I L = m g\). Solving for the current \(I\): \(I = \frac{m g}{B L}\). Substituting the values (converting mass to kilograms): \(I = \frac{4.5 \times 10^{-3} \text{ kg} \times 9.81 \text{ m s}^{-2}}{0.20 \text{ T} \times 0.15 \text{ m}} = \frac{0.044145}{0.030} \approx 1.47 \text{ A} \approx 1.5 \text{ A}\).
Marking scheme
1 mark for the correct option B.
Question 14 · Multiple Choice
1 marks
Two long, straight, parallel wires, \(X\) and \(Y\), are separated by a distance \(d\) in a vacuum. Wire \(X\) carries a current \(I\) and wire \(Y\) carries a current \(2I\) in the opposite direction. The magnitude of the magnetic field strength at the midpoint between the two wires is \(B_0\). The direction of the current in wire \(X\) is now reversed, whilst the magnitudes of the currents remain unchanged. What is the new magnetic field strength at the midpoint between the wires?
A.\(\frac{1}{3} B_0\)
B.\(\frac{1}{2} B_0\)
C.\(2 B_0\)
D.\(3 B_0\)
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Worked solution
Initially, the currents flow in opposite directions. Applying the right-hand grip rule, the magnetic fields produced by both wires at the midpoint point in the same direction. Therefore, they add up: \(B_0 = B_X + B_Y = \frac{\mu_0 I}{2\pi(d/2)} + \frac{\mu_0 (2I)}{2\pi(d/2)} = \frac{\mu_0 (3I)}{\pi d}\). When the current in wire \(X\) is reversed, the currents flow in the same direction, meaning their magnetic fields at the midpoint now oppose each other. The new magnetic field strength is the difference: \(B_{\text{new}} = B_Y - B_X = \frac{\mu_0 (2I)}{\pi d} - \frac{\mu_0 I}{\pi d} = \frac{\mu_0 I}{\pi d}\). Since \(B_0 = 3 \left(\frac{\mu_0 I}{\pi d}\right)\), the new field is \(B_{\text{new}} = \frac{1}{3} B_0\).
Marking scheme
1 mark for the correct option A.
Question 15 · Multiple Choice
1 marks
A body of mass \(m\) undergoes simple harmonic motion with amplitude \(A\) and frequency \(f\). What is the maximum acceleration of the body, and at what displacement magnitude \(x\) from the equilibrium position does it occur?
A.Maximum acceleration: \(2\pi f^2 A\), at displacement \(x = 0\)
B.Maximum acceleration: \(4\pi^2 f^2 A\), at displacement \(x = 0\)
C.Maximum acceleration: \(2\pi f^2 A\), at displacement \(x = A\)
D.Maximum acceleration: \(4\pi^2 f^2 A\), at displacement \(x = A\)
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Worked solution
In simple harmonic motion, the acceleration is given by \(a = -\omega^2 x\). The angular frequency is \(\omega = 2\pi f\), so \(\omega^2 = 4\pi^2 f^2\). The maximum acceleration occurs at the maximum displacement magnitude, i.e., at \(x = A\). Thus, \(a_{\text{max}} = \omega^2 A = 4\pi^2 f^2 A\) at displacement \(x = A\).
Marking scheme
1 mark for the correct option D.
Question 16 · Multiple Choice
1 marks
A particle of mass \(0.10\text{ kg}\) undergoes simple harmonic motion with an amplitude of \(5.0\text{ cm}\) and a period of \(0.40\text{ s}\). What is the total energy of the oscillator, and what is its kinetic energy when the displacement of the particle from the equilibrium position is \(3.0\text{ cm}\)?
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Worked solution
The angular frequency is \(\omega = \frac{2\pi}{T} = \frac{2\pi}{0.40} = 5\pi \text{ rad s}^{-1}\). The total energy is \(E_0 = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} \times 0.10 \times (5\pi)^2 \times (0.050)^2 \approx 0.0308\text{ J} \approx 31\text{ mJ}\). The kinetic energy at displacement \(x = 3.0\text{ cm} = 0.030\text{ m}\) is \(E_k = \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} \times 0.10 \times (5\pi)^2 \times (0.050^2 - 0.030^2) = 0.0197\text{ J} \approx 20\text{ mJ}\).
Marking scheme
1 mark for the correct option C.
Question 17 · Multiple Choice
1 marks
A student determines the resistivity \(\rho\) of a metal wire using the formula:
\(\rho = \frac{R \pi d^2}{4 L}\)
The measurements and their absolute uncertainties are:
Resistance \(R = 4.50 \pm 0.09\ \Omega\)
Diameter \(d = 0.40 \pm 0.01\ \text{mm}\)
Length \(L = 1.250 \pm 0.005\ \text{m}\)
What is the percentage uncertainty in the calculated resistivity?
A.\(2.9\%\)
B.\(4.9\%\)
C.\(7.4\%\)
D.\(10.4\%\)
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Worked solution
The percentage uncertainty in a combined quantity of the form \(Y = \frac{A B^2}{C}\) is given by adding the percentage uncertainties of each component, multiplying by power factors where appropriate:
Award [1 mark] for the correct answer of 7.4% (Option C).
Incorrect pathways: - 4.9% (Option B) represents neglecting the factor of 2 for the squared diameter \(d^2\). - 10.4% (Option D) represents using an incorrect power of 3 for diameter or other algebra error.
Question 18 · Multiple Choice
1 marks
A digital voltmeter is used to measure a steady potential difference. The manufacturer states that the meter always reads \(0.15\text{ V}\) higher than the true value, and has a random fluctuation of \(\pm 0.02\text{ V}\).
Which statement about the measurements is correct?
A.The systematic error can be corrected by subtracting \(0.15\text{ V}\) from each reading, while the random error cannot be corrected in this way.
B.The systematic error can be reduced to zero by taking a large number of readings and calculating the average.
C.The random error can be completely eliminated by subtracting \(0.02\text{ V}\) from each reading.
D.The random error is a result of poor calibration and the systematic error is due to parallax error.
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Worked solution
Systematic errors (such as a constant offset of \(0.15\text{ V}\)) can be eliminated or corrected by subtracting the offset value from each measurement. Random errors (such as the \(\pm 0.02\text{ V}\) fluctuation) represent unpredictable variations and cannot be corrected by subtracting a constant.
Marking scheme
Award [1 mark] for identifying that systematic error can be corrected by subtraction, but random error cannot (Option A).
Question 19 · Multiple Choice
1 marks
An experiment is performed to measure the acceleration of free fall \(g\) by timing a falling ball. The experimental result is expressed as \(g = 9.42 \pm 0.25\ \text{m s}^{-2}\).
The accepted value of \(g\) is \(9.81\ \text{m s}^{-2}\).
Which statement describes the accuracy and precision of this experimental result?
A.The result is inaccurate because the accepted value of \(9.81\ \text{m s}^{-2}\) lies outside the experimental range of values, and the precision is indicated by the uncertainty of \(\pm 0.25\ \text{m s}^{-2}\).
B.The result is accurate because the experimental range includes \(9.50\ \text{m s}^{-2}\).
C.The result is precise because the accepted value lies outside the range of the measured value.
D.The result is accurate and precise because the percentage uncertainty is approximately \(2.7\%\).
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Worked solution
The experimental range is \(9.42 \pm 0.25\ \text{m s}^{-2}\), which spans from \(9.17\ \text{m s}^{-2}\) to \(9.67\ \text{m s}^{-2}\).
Since the accepted value of \(9.81\ \text{m s}^{-2}\) lies outside this range, the result is inaccurate (it has a significant systematic error). Precision refers to the degree of agreement between repeated measurements, which is indicated by the size of the random uncertainty (\(\pm 0.25\ \text{m s}^{-2}\)).
Marking scheme
Award [1 mark] for the correct analysis of accuracy and precision (Option A).
Question 20 · Multiple Choice
1 marks
A beam of singly charged positive ions, each of mass \(m\) and charge \(q\), enters a region of uniform magnetic field of flux density \(B\). The velocity of the ions is perpendicular to the magnetic field, causing them to travel in a circular path of radius \(r\).
A second beam of ions consisting of doubly charged positive ions of mass \(2m\) and charge \(2q\) enters the same magnetic field with the same kinetic energy as the first beam.
What is the radius of the path of the second beam of ions?
A.\{\frac{r}{\sqrt{2}}\}
B.\(r\)
C.\(\sqrt{2} r\)
D.\(2 r\)
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Worked solution
For a charge \(q\) of mass \(m\) in a circular path of radius \(r\) in a magnetic field \(B\):
\(\frac{m v^2}{r} = B q v \implies r = \frac{m v}{B q}\)
Since the kinetic energy is \(E_k = \frac{1}{2} m v^2\), the momentum is \(p = m v = \sqrt{2 m E_k}\).
Substituting this into the radius formula:
\(r = \frac{\sqrt{2 m E_k}}{B q}\)
For the second beam, the mass is \(m' = 2m\), the charge is \(q' = 2q\), and the kinetic energy is \(E_k\):
\(r' = \frac{\sqrt{2 (2m) E_k}}{B (2q)} = \frac{\sqrt{2} \sqrt{2 m E_k}}{2 B q} = \frac{\sqrt{2}}{2} r = \frac{r}{\sqrt{2}}\)
Marking scheme
Award [1 mark] for the correct derivation showing the radius is scaled by a factor of \(1/\sqrt{2}\) (Option A).
Question 21 · Multiple Choice
1 marks
A straight horizontal wire of length \(0.50\text{ m}\) and mass \(15\text{ g}\) is suspended horizontally in a uniform horizontal magnetic field. The magnetic field of flux density \(0.40\text{ T}\) is perpendicular to the wire.
What is the current required in the wire to support its weight? (Take \(g = 9.81\text{ m s}^{-2}\).)
A.0.075 A
B.0.74 A
C.1.5 A
D.7.4 A
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Worked solution
For the wire to float, the upward magnetic force must equal the downward gravitational force (weight):
Award [1 mark] for the correct calculation of current \(I = 0.74\text{ A}\) (Option B).
Question 22 · Multiple Choice
1 marks
A flat circular coil of \(200\) turns and cross-sectional area \(1.5 \times 10^{-3}\ \text{m}^2\) is placed in a uniform magnetic field of flux density \(0.080\ \text{T\) so that the plane of the coil is perpendicular to the magnetic field.
The magnetic field is reduced to zero at a constant rate in a time of \(0.12\ \text{s}\).
What is the magnitude of the average electromotive force (e.m.f.) induced in the coil?
A.1.0 mV
B.0.20 V
C.1.2 V
D.130 V
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Worked solution
According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. is given by:
\(E = N \frac{\Delta \Phi}{\Delta t}\)
Since the field is perpendicular to the plane of the coil, the initial magnetic flux through a single turn is \(\Phi = B A\).
The change in flux is \(\Delta \Phi = B A - 0 = B A\).
Award [1 mark] for the correct calculation of the average induced e.m.f. (Option B).
Incorrect pathways: - 1.0 mV (Option A) represents neglecting the factor of \(N = 200\). - 1.2 V (Option C) represents omitting the area conversion or other math error.
Question 23 · Multiple Choice
1 marks
An object undergoing simple harmonic motion has displacement \(x\) at time \(t\) given by the equation:
\(x = 0.040 \cos(50\pi t)\)
where \(x\) is in metres and \(t\) is in seconds.
What is the maximum acceleration of the object, and its frequency of oscillation?
A.Maximum acceleration = \(990\ \text{m s}^{-2}\); Frequency = \(25\ \text{Hz}\)
B.Maximum acceleration = \(990\ \text{m s}^{-2}\); Frequency = \(50\ \text{Hz}\)
C.Maximum acceleration = \(6.3\ \text{m s}^{-2}\); Frequency = \(25\ \text{Hz}\)
D.Maximum acceleration = \(2.0\ \text{m s}^{-2}\); Frequency = \(50\pi\ \text{Hz}\)
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Worked solution
The general equation for simple harmonic motion is \(x = x_0 \cos(\omega t)\).
Comparing this to the given equation: Amplitude \(x_0 = 0.040\ \text{m}\) Angular frequency \(\omega = 50\pi\ \text{rad s}^{-1}\)
The frequency \(f\) of oscillation is: \(f = \frac{\omega}{2\pi} = \frac{50\pi}{2\pi} = 25\ \text{Hz}\)
Award [1 mark] for correctly calculating both maximum acceleration and frequency (Option A).
Question 24 · Multiple Choice
1 marks
A particle of mass \(0.15\ \text{kg}\) undergoes simple harmonic motion with an amplitude of \(8.0\ \text{cm}\) and a period of \(0.40\ \text{s}\).
What are the total energy of the oscillation and the kinetic energy of the particle when it is at a displacement of \(5.0\ \text{cm}\) from its equilibrium position?
A.Total energy = \(0.12\ \text{J}\); Kinetic energy = \(0.072\ \text{J}\)
B.Total energy = \(0.12\ \text{J}\); Kinetic energy = \(0.046\ \text{J}\)
C.Total energy = \(0.072\ \text{J}\); Kinetic energy = \(0.046\ \text{J}\)
D.Total energy = \(1.2\ \text{J}\); Kinetic energy = \(0.72\ \text{J}\)
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Award [1 mark] for correctly calculating the total energy and the kinetic energy (Option A).
Incorrect pathways: - Option B yields potential energy (0.046 J) instead of kinetic energy. - Option D represents a power of 10 failure.
Question 25 · Multiple Choice
1 marks
A student determines the resistivity \(\rho\) of a uniform metal wire using the formula \(\rho = \frac{R \pi d^2}{4 L}\). The measurements of resistance \(R\), diameter \(d\), and length \(L\) are given below with their absolute uncertainties: \(R = (25.0 \pm 0.5)\ \Omega\) \(d = (0.40 \pm 0.01)\text{ mm}\) \(L = (1.200 \pm 0.003)\text{ m}\) What is the percentage uncertainty in the calculated value of resistivity?
A.4.8%
B.7.3%
C.9.8%
D.12.3%
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Worked solution
To find the percentage uncertainty in \(\rho\), we sum the percentage uncertainties of each component, multiplying by their respective powers in the formula: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\) Calculate each fractional uncertainty: \(\frac{\Delta R}{R} = \frac{0.5}{25.0} = 0.020\) (or \(2.0\%\)) \(\frac{\Delta d}{d} = \frac{0.01}{0.40} = 0.025\) (or \(2.5\%\)) \(\frac{\Delta L}{L} = \frac{0.003}{1.200} = 0.0025\) (or \(0.25\%\)) Substitute these into the formula: \(\%\Delta \rho = 2.0\% + 2 \times 2.5\% + 0.25\% = 2.0\% + 5.0\% + 0.25\% = 7.25\%\) Rounding to two significant figures gives \(7.3\%\).
Marking scheme
1 mark for the correct answer B. Method: Identify that percentage uncertainties add, with the diameter uncertainty doubled due to the squared term. Accuracy: Correctly compute the final percentage uncertainty as 7.3%.
Question 26 · Multiple Choice
1 marks
The acceleration of free fall \(g\) is determined by measuring the length \(L\) of a simple pendulum and the time \(t\) for 50 complete oscillations. The formula used is \(g = \frac{4\pi^2 L}{T^2}\), where \(T\) is the period of one oscillation. The measurements are: \(L = (80.0 \pm 0.4)\text{ cm}\) \(t = (90.0 \pm 0.6)\text{ s}\) What is the percentage uncertainty in the calculated value of \(g\)?
A.1.2%
B.1.8%
C.2.3%
D.3.2%
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Worked solution
First, we note that the period is \(T = t / 50\). The percentage uncertainty in \(T\) is identical to the percentage uncertainty in \(t\): \(\%\Delta T = \%\Delta t = \frac{0.6}{90.0} \times 100\% \approx 0.67\%\) The percentage uncertainty in length \(L\) is: \(\%\Delta L = \frac{0.4}{80.0} \times 100\% = 0.50\%\) Using the relation \(g = \frac{4\pi^2 L}{T^2}\), the percentage uncertainty in \(g\) is: \(\%\Delta g = \%\Delta L + 2 \%\Delta T = 0.50\% + 2 \times 0.67\% = 1.84\%\) This rounds to \(1.8\%\).
Marking scheme
1 mark for the correct answer B. Method: Realise that the divisor of 50 does not change the percentage uncertainty of the period, and double the percentage uncertainty of T. Accuracy: Correctly calculate the sum of the percentage uncertainties to get 1.8%.
Question 27 · Multiple Choice
1 marks
An analogue voltmeter is used to measure potential differences (p.d.). The voltmeter has a zero error and a scale-factor error. When the true p.d. across a component is zero, the meter reads \(+0.2\text{ V}\). For every increase of \(1.0\text{ V}\) in the true p.d., the meter's reading increases by only \(0.8\text{ V}\). Which statement correctly describes the graph of measured potential difference \(V_{\text{meas}}\) against true potential difference \(V_{\text{true}}\)
A.A straight line with a positive vertical intercept and a gradient less than 1.
B.A straight line with a positive vertical intercept and a gradient greater than 1.
C.A straight line passing through the origin with a gradient less than 1.
D.A straight line with a negative vertical intercept and a gradient greater than 1.
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Worked solution
The equation relating the measured voltage \(V_{\text{meas}}\) to the true voltage \(V_{\text{true}}\) is linear: \(V_{\text{meas}} = m V_{\text{true}} + C\). When \(V_{\text{true}} = 0\), \(V_{\text{meas}} = +0.2\text{ V}\), which corresponds to a positive vertical intercept (\(C = +0.2\text{ V}\)). The rate of change of measured voltage with respect to true voltage is the gradient \(m\): \(m = \frac{\Delta V_{\text{meas}}}{\Delta V_{\text{true}}} = \frac{0.8\text{ V}}{1.0\text{ V}} = 0.8\). Since \(0.8 < 1\), the gradient of the straight line is less than 1.
Marking scheme
1 mark for the correct answer A. Method: Formulate the linear relationship between the measured and true value to deduce the intercept and gradient. Accuracy: Identify that the intercept is positive (+0.2 V) and the gradient is less than 1 (0.8).
Question 28 · Multiple Choice
1 marks
A beam of singly-charged positive ions, each of mass \(m\) and charge \(q\), enters a region of uniform magnetic field with flux density \(B\). The magnetic field is directed perpendicular to the velocity of the ions, causing them to move in a circular path of radius \(R\). The kinetic energy of the incoming ions is now doubled, and the magnetic flux density is also doubled. What is the new radius of the circular path of the ions?
A.\(R / 2\)
B.\(R / \sqrt{2}\)
C.\(\sqrt{2} R\)
D.\(2 R\)
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Worked solution
For a charge \(q\) moving with velocity \(v\) perpendicular to a magnetic field \(B\), the magnetic force provides the centripetal force: \(q v B = \frac{m v^2}{R} \implies R = \frac{m v}{q B}\) The kinetic energy is given by \(E_k = \frac{1}{2} m v^2\), so the momentum is \(p = m v = \sqrt{2 m E_k}\). Thus, the radius can be written as: \(R = \frac{\sqrt{2 m E_k}}{q B}\) Therefore, the radius is proportional to \(\frac{\sqrt{E_k}}{B}\). When the kinetic energy is doubled (\(E_k \to 2 E_k\)) and the magnetic flux density is doubled (\(B \to 2 B\)), the new radius \(R'\) is: \(R' \propto \frac{\sqrt{2 E_k}}{2 B} = \frac{\sqrt{2}}{2} \frac{\sqrt{E_k}}{B} = \frac{1}{\sqrt{2}} R = \frac{R}{\sqrt{2}}\).
Marking scheme
1 mark for the correct answer B. Method: Express the radius in terms of kinetic energy and magnetic field. Accuracy: Find the scaling factor to be 1/sqrt(2) and apply it to the original radius R.
Question 29 · Multiple Choice
1 marks
A straight metal wire of length \(0.50\text{ m}\) carries a current of \(4.0\text{ A}\) through a uniform magnetic field of flux density \(0.15\text{ T}\). The direction of the current in the wire is at an angle of \(30^\circ\) to the magnetic field. What is the magnitude of the magnetic force acting on the wire?
A.0.15 N
B.0.26 N
C.0.30 N
D.0.60 N
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Worked solution
The magnetic force \(F\) on a straight current-carrying wire in a magnetic field is given by the formula: \(F = B I L \sin\theta\) Given values: \(B = 0.15\text{ T}\) \(I = 4.0\text{ A}\) \(L = 0.50\text{ m}\) \(\theta = 30^\circ\) Substitute these values into the formula: \(F = 0.15 \times 4.0 \times 0.50 \times \sin(30^\circ)\) \(F = 0.30 \times 0.5 = 0.15\text{ N}\).
Marking scheme
1 mark for the correct answer A. Method: Apply the formula F = B I L sin(theta) with the correct angle. Accuracy: Correctly compute the force to be 0.15 N.
Question 30 · Multiple Choice
1 marks
A flat, circular coil of 200 turns has a cross-sectional area of \(4.0 \times 10^{-4}\text{ m}^2\). The coil is positioned perpendicular to a uniform magnetic field. The magnetic flux density \(B\) of this field increases uniformly from \(0.10\text{ T}\) to \(0.50\text{ T}\) over a time interval of \(2.0\text{ s}\). What is the magnitude of the average electromotive force (e.m.f.) induced in the coil during this time?
A.0.080 mV
B.16 mV
C.20 mV
D.32 mV
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Worked solution
According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. \(E\) is: \(E = N \frac{\Delta \Phi}{\Delta t} = N A \frac{\Delta B}{\Delta t}\) where: \(N = 200\) \(A = 4.0 \times 10^{-4}\text{ m}^2\) \(\Delta B = 0.50\text{ T} - 0.10\text{ T} = 0.40\text{ T}\) \(\Delta t = 2.0\text{ s}\) Calculate \(E\): \(E = 200 \times (4.0 \times 10^{-4}) \times \frac{0.40}{2.0}\) \(E = 200 \times 4.0 \times 10^{-4} \times 0.20 = 0.016\text{ V} = 16\text{ mV}\).
Marking scheme
1 mark for the correct answer B. Method: Identify and apply Faraday's law for a multi-turn coil. Accuracy: Correctly determine the rate of change of flux density and compute the e.m.f. in millivolts (16 mV).
Question 31 · Multiple Choice
1 marks
A particle undergoes simple harmonic motion with an amplitude of \(5.0\text{ cm}\) and a time period of \(2.0\text{ s}\). What is the magnitude of the velocity of the particle when its displacement from the equilibrium position is \(3.0\text{ cm}\)?
A.\(6.3\text{ cm s}^{-1}\)
B.\(12.6\text{ cm s}^{-1}\)
C.\(15.7\text{ cm s}^{-1}\)
D.\(25.1\text{ cm s}^{-1}\)
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Worked solution
The velocity \(v\) of a particle in simple harmonic motion is related to its displacement \(x\) and amplitude \(x_0\) by: \(v = \pm \omega \sqrt{x_0^2 - x^2}\) First, calculate the angular frequency \(\omega\): \(\omega = \frac{2\pi}{T} = \frac{2\pi}{2.0} = \pi\text{ rad s}^{-1}\) Now, substitute the values into the velocity equation: \(v = \pi \sqrt{5.0^2 - 3.0^2} = \pi \sqrt{25.0 - 9.0} = \pi \sqrt{16.0} = 4\pi\text{ cm s}^{-1} \approx 12.57\text{ cm s}^{-1}\) This rounds to \(12.6\text{ cm s}^{-1}\).
Marking scheme
1 mark for the correct answer B. Method: Find the angular frequency and then apply the SHM velocity-displacement formula. Accuracy: Calculate the velocity magnitude as 12.6 cm/s.
Question 32 · Multiple Choice
1 marks
A system is undergoing forced oscillations. The frequency of the periodic driving force is varied, and the amplitude of the resulting steady-state oscillation is recorded to plot a resonance curve. The damping of the system is now increased. Which statement describes the changes to the resonance curve when the damping is increased?
A.The peak amplitude increases, and the resonance frequency increases slightly.
B.The peak amplitude increases, and the resonance frequency decreases slightly.
C.The peak amplitude decreases, and the resonance frequency increases slightly.
D.The peak amplitude decreases, and the resonance frequency decreases slightly.
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Worked solution
When damping is increased in a forced oscillation system: 1. More energy is dissipated per cycle, so the maximum (peak) amplitude of the oscillation decreases. 2. The frequency at which resonance (maximum amplitude) occurs, known as the resonance frequency, shifts to a slightly lower frequency (decreases slightly) due to the damping effect. Therefore, both the peak amplitude and the resonance frequency decrease.
Marking scheme
1 mark for the correct answer D. Method: Identify the two physical consequences of increasing damping on the resonance curve. Accuracy: Correctly conclude that both the peak amplitude decreases and the resonance frequency decreases slightly.
Question 33 · multiple_choice
1 marks
An experiment is carried out to determine the acceleration of free fall \( g \) by timing the fall of a steel ball through a vertical distance \( h \). The equation used is \( h = \frac{1}{2} g t^2 \). The measured values are: \( h = (1.20 \pm 0.01)\text{ m} \) and \( t = (0.49 \pm 0.02)\text{ s} \). What is the percentage uncertainty in the calculated value of \( g \)?
A.4.9%
B.8.2%
C.9.0%
D.9.8%
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Worked solution
To find the percentage uncertainty in \( g \), we first rearrange the equation: \( g = \frac{2h}{t^2} \). The percentage uncertainty in \( g \) is given by \( \frac{\Delta g}{g} \times 100\% = \left( \frac{\Delta h}{h} + 2 \frac{\Delta t}{t} \right) \times 100\% \). Calculating each term: \( \frac{\Delta h}{h} = \frac{0.01}{1.20} \approx 0.83\% \) and \( \frac{\Delta t}{t} = \frac{0.02}{0.49} \approx 4.08\% \). Therefore, the percentage uncertainty in \( g \) is \( 0.83\% + 2 \times 4.08\% = 0.83\% + 8.16\% = 8.99\% \approx 9.0\% \).
Marking scheme
1 mark for the correct answer C. Method: 1 mark for calculating the sum of fractional uncertainties with the time uncertainty doubled.
Question 34 · multiple_choice
1 marks
The power \( P \) dissipated in a resistor of resistance \( R \) when a current \( I \) passes through it is given by \( P = I^2 R \). The resistance is measured to be \( (47.0 \pm 0.5)\ \Omega \) and the current is measured to be \( (2.50 \pm 0.05)\text{ A} \). What is the absolute uncertainty in the calculated value of \( P \)?
A.0.6 W
B.9.0 W
C.12 W
D.15 W
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Worked solution
First, calculate the value of \( P \): \( P = I^2 R = (2.50)^2 \times 47.0 = 293.75\text{ W} \). The fractional uncertainty in \( P \) is \( \frac{\Delta P}{P} = 2\frac{\Delta I}{I} + \frac{\Delta R}{R} = 2 \left( \frac{0.05}{2.50} \right) + \frac{0.5}{47.0} = 0.040 + 0.01064 = 0.05064 \). The absolute uncertainty is \( \Delta P = 293.75 \times 0.05064 \approx 14.88\text{ W} \). To 2 significant figures, this is \( 15\text{ W} \).
Marking scheme
1 mark for the correct answer D. Method: 1 mark for adding fractional uncertainties correctly (doubling current uncertainty) and multiplying by the power value.
Question 35 · multiple_choice
1 marks
A student determines the density \( \rho \) of a solid sphere using the equation \( \rho = \frac{6M}{\pi d^3} \) where \( M \) is the mass and \( d \) is the diameter of the sphere. The mass is measured with a percentage uncertainty of \( 2.0\% \) and the diameter is measured with a percentage uncertainty of \( 1.5\% \). What is the percentage uncertainty in the calculated value of the density?
A.3.5%
B.5.0%
C.6.5%
D.15.5%
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Worked solution
The formula for density is \( \rho = \frac{6M}{\pi d^3} \). The constant factor \( \frac{6}{\pi} \) has no uncertainty. The percentage uncertainty in \( \rho \) is \( \frac{\Delta \rho}{\rho} \times 100\% = \frac{\Delta M}{M} \times 100\% + 3 \frac{\Delta d}{d} \times 100\% \). Substituting the values: \( 2.0\% + 3(1.5\%) = 2.0\% + 4.5\% = 6.5\% \).
Marking scheme
1 mark for the correct answer C. Method: 1 mark for multiplying the diameter percentage uncertainty by 3 and adding it to the mass percentage uncertainty.
Question 36 · multiple_choice
1 marks
A uniform magnetic field is directed perpendicularly into the plane of the page. An alpha particle (mass \( 4u \), charge \( +2e \)) enters the field with velocity \( v \) perpendicular to the field and describes a circular path of radius \( r_{\alpha} \). A proton (mass \( u \), charge \( +e \)) enters the same field with a velocity \( 2v \) perpendicular to the field, and describes a circular path of radius \( r_{p} \). What is the ratio \( \frac{r_{\alpha}}{r_{p}} \)?
A.0.5
B.1
C.2
D.4
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Worked solution
The magnetic force provides the centripetal force: \( Bqv = \frac{mv^2}{r} \implies r = \frac{mv}{Bq} \). For the alpha particle: \( r_{\alpha} = \frac{(4u)v}{B(2e)} = 2 \frac{uv}{Be} \). For the proton: \( r_{p} = \frac{u(2v)}{B(e)} = 2 \frac{uv}{Be} \). The ratio is \( \frac{r_{\alpha}}{r_{p}} = \frac{2}{2} = 1 \).
Marking scheme
1 mark for the correct answer B. Method: 1 mark for deriving \( r = \frac{mv}{Bq} \) and substituting the values of charge, mass and velocity for both particles.
Question 37 · multiple_choice
1 marks
A rigid straight wire of length \( 0.40\text{ m} \) carries a current of \( 3.0\text{ A} \) in a direction perpendicular to a uniform magnetic field. The magnetic force acting on this wire is \( 0.18\text{ N} \). The wire is now rotated so that it makes an angle of \( 30^\circ \) with the direction of the magnetic field, while the current remains unchanged. What is the new magnetic force acting on the wire?
A.0.090 N
B.0.16 N
C.0.18 N
D.0.36 N
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Worked solution
The magnetic force on a current-carrying wire is \( F = BIL\sin\theta \), where \( \theta \) is the angle between the wire and the magnetic field. Initially, \( \theta = 90^\circ \), so \( F = BIL = 0.18\text{ N} \). When the wire is rotated to \( \theta = 30^\circ \), the new force is \( F' = BIL\sin(30^\circ) = 0.18 \times 0.5 = 0.090\text{ N} \).
Marking scheme
1 mark for the correct answer A. Method: 1 mark for using the formula \( F = BIL\sin\theta \) and multiplying the maximum force by \( \sin(30^\circ) \).
Question 38 · multiple_choice
1 marks
An object is performing simple harmonic motion with an amplitude \( x_0 \) and a time period \( T \). The object starts from the position of maximum displacement at time \( t = 0 \). What is the displacement \( x \) of the object from its equilibrium position at time \( t = \frac{T}{6} \)?
A.0.50 x0
B.0.71 x0
C.0.87 x0
D.1.0 x0
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Worked solution
Since the object starts at maximum displacement at \( t = 0 \), its displacement is described by \( x = x_0 \cos(\omega t) \). The angular frequency is \( \omega = \frac{2\pi}{T} \). At \( t = \frac{T}{6} \), the phase angle is \( \omega t = \frac{2\pi}{T} \times \frac{T}{6} = \frac{\pi}{3} \text{ rad} \) (or \( 60^\circ \)). Therefore, the displacement is \( x = x_0 \cos\left(\frac{\pi}{3}\right) = 0.50 x_0 \).
Marking scheme
1 mark for the correct answer A. Method: 1 mark for using the cosine representation of SHM and calculating the phase angle at \( t = T/6 \).
Question 39 · multiple_choice
1 marks
A body of mass \( m \) is undergoing simple harmonic motion with frequency \( f \) and amplitude \( x_0 \). Which expression represents the total energy \( E \) of the oscillating body?
A.pi^2 m f^2 x0^2
B.2 pi^2 m f^2 x0^2
C.4 pi^2 m f^2 x0^2
D.0.5 pi^2 m f^2 x0^2
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Worked solution
The total energy in simple harmonic motion is equal to the maximum kinetic energy, which is \( E = \frac{1}{2} m v_{\text{max}}^2 \). Since \( v_{\text{max}} = \omega x_0 \), we have \( E = \frac{1}{2} m \omega^2 x_0^2 \). Substituting \( \omega = 2\pi f \) gives \( E = \frac{1}{2} m (2\pi f)^2 x_0^2 = 2 \pi^2 m f^2 x_0^2 \).
Marking scheme
1 mark for the correct answer B. Method: 1 mark for substituting \( \omega = 2\pi f \) into the standard SHM total energy equation.
Question 40 · multiple_choice
1 marks
A light mass on a spring is suspended in a beaker of liquid and driven by an external vibrator of variable frequency \( f \). A graph of the steady-state amplitude of oscillation against driving frequency is plotted. The viscosity of the liquid is now increased. Which statement correctly describes the changes to the maximum amplitude of the oscillation and the frequency at which this maximum amplitude occurs?
A.The maximum amplitude decreases, and the frequency of maximum amplitude decreases.
B.The maximum amplitude decreases, and the frequency of maximum amplitude increases.
C.The maximum amplitude increases, and the frequency of maximum amplitude decreases.
D.The maximum amplitude increases, and the frequency of maximum amplitude increases.
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Worked solution
Increasing the viscosity of the liquid increases the degree of damping on the system. For a more heavily damped system, the maximum steady-state amplitude decreases (the peak is flatter/lower), and the resonant frequency (the frequency at which the maximum amplitude occurs) shifts to a slightly lower frequency.
Marking scheme
1 mark for the correct answer A. Method: 1 mark for identifying that increased damping reduces maximum amplitude and decreases the resonant frequency.
Paper 2 (AS Level Structured)
Answer all questions. Write your answers in the spaces provided on the question paper.
8 Question · 60 marks
Question 1 · Structured
8 marks
A student determines the density of a solid sphere. The mass \(m\) of the sphere is measured as \(m = (124.5 \pm 0.5)\text{ g}\). The diameter \(d\) of the sphere is measured as \(d = (2.46 \pm 0.02)\text{ cm}\).
(a) Define *systematic error*. [1]
(b) Calculate the density \(\rho\) of the sphere in \(\text{g cm}^{-3}\). [3]
(c) Determine the absolute uncertainty in the calculated value of \(\rho\). Give your final density value with its uncertainty to the appropriate number of significant figures. [4]
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Worked solution
(a) A systematic error is an error that causes all readings to deviate from the true value in a consistent direction and magnitude.
(c) The fractional uncertainty is: \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \frac{\Delta d}{d}\) \(\frac{\Delta m}{m} = \frac{0.5}{124.5} \approx 0.00402\) \(\frac{\Delta d}{d} = \frac{0.02}{2.46} \approx 0.00813\) \(\frac{\Delta \rho}{\rho} = 0.00402 + 3(0.00813) = 0.02841\) \(\Delta \rho = 0.02841 \times 15.987 = 0.454\text{ g cm}^{-3}\) Rounding uncertainty to 1 significant figure gives \(\Delta \rho = 0.5\text{ g cm}^{-3}\). Rounding the density value to match the precision of the uncertainty gives: \(\rho = (16.0 \pm 0.5)\text{ g cm}^{-3}\).
Marking scheme
(a) [1] Constant error in all readings / error that deviates in one direction from the true value. (b) [1] Correct formula for volume \(V = \frac{1}{6}\pi d^3\) or correct radius calculation. [1] Correct substitution of numbers to get \(V = 7.79\text{ cm}^3\). [1] Correct calculation of density \(16.0\text{ g cm}^{-3}\) (or \(15.99\)). (c) [1] Correct fractional uncertainty expression with multiplication of \(3\) for \(d\). [1] Correct fractional uncertainty value (approx. \(0.028\)). [1] Correct absolute uncertainty calculated as \(0.45\) or \(0.5\text{ g cm}^{-3}\). [1] Final answer written to match decimal places: \((16.0 \pm 0.5)\text{ g cm}^{-3}\).
Question 2 · Structured
7 marks
An experiment is performed to determine the acceleration of free fall \(g\) by measuring the time \(t\) for a small metal ball to fall from rest through a vertical distance \(h\). The equation used is: \(h = \frac{1}{2} g t^2\) The percentage uncertainty in \(h\) is \(1.5\%\) and the percentage uncertainty in \(t\) is \(3.0\%\).
(a) State whether the uncertainty in \(h\) is a random or systematic uncertainty, and explain how a systematic uncertainty in \(h\) could occur. [2]
(b) Calculate the percentage uncertainty in the value of \(g\). [2]
(c) The calculated value of \(g\) is \(9.58\text{ m s}^{-2}\). Use your answer in (b) to find the absolute uncertainty in \(g\). Write down the value of \(g\) with its absolute uncertainty. [3]
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Worked solution
(a) A systematic uncertainty causes readings to be consistently too high or too low. This could occur if the ruler used to measure the vertical height is poorly calibrated, or if the ruler is not aligned vertically, or if measurement is consistently made from the wrong point of the ball.
(b) Rearranging the equation: \(g = \frac{2h}{t^2}\) Percentage uncertainty in \(g = \% \text{ uncertainty in } h + 2 \times (\% \text{ uncertainty in } t)\) Percentage uncertainty in \(g = 1.5\% + 2 \times (3.0\%) = 7.5\%\)
(c) Absolute uncertainty in \(g\): \(\Delta g = 9.58 \times \frac{7.5}{100} = 0.7185\text{ m s}^{-2} \approx 0.7\text{ m s}^{-2}\) Since \(\Delta g\) is given to 1 s.f. as \(0.7\text{ m s}^{-2}\), the value of \(g\) must be rounded to the same decimal place: \(g = (9.6 \pm 0.7)\text{ m s}^{-2}\).
Marking scheme
(a) [1] Recognising that systematic error causes consistent offset. [1] Correct example such as calibration error of the ruler or alignment error (ruler not vertical). (b) [1] Correct equation for combining percentage uncertainties: \(\%\Delta g = \%\Delta h + 2 \times \%\Delta t\). [1] Correct calculation of percentage uncertainty as \(7.5\%\). (c) [1] Absolute uncertainty calculation: \(9.58 \times 0.075 = 0.7185\text{ m s}^{-2}\). [1] Rounding absolute uncertainty to \(0.7\text{ m s}^{-2}\). [1] Expressing final answer correctly: \((9.6 \pm 0.7)\text{ m s}^{-2}\).
Question 3 · Structured
8 marks
The resistance \(R\) of a wire of length \(L\) and uniform circular cross-section of diameter \(D\) is given by the expression: \(R = \frac{4 \rho L}{\pi D^2}\) where \(\rho\) is the resistivity of the metal.
In an experiment, the following measurements are obtained: \(R = (4.50 \pm 0.05)\ \Omega\) \(L = (1.250 \pm 0.002)\text{ m}\) \(D = (0.38 \pm 0.01)\text{ mm}\)
(a) Show that the resistivity \(\rho\) is approximately \(4.1 \times 10^{-7}\ \Omega\text{ m}\). [3]
(b) Determine the percentage uncertainty in the resistivity \(\rho\). [3]
(c) Express the final value of \(\rho\), with its absolute uncertainty, in \(\Omega\text{ m}\). [2]
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(b) The formula for percentage uncertainty in \(\rho\) is: \(\%\Delta \rho = \%\Delta R + \%\Delta L + 2 \times \%\Delta D\) \(\%\Delta R = \frac{0.05}{4.50} \times 100\% \approx 1.11\%\) \(\%\Delta L = \frac{0.002}{1.250} \times 100\% \approx 0.16\%\) \(\%\Delta D = \frac{0.01}{0.38} \times 100\% \approx 2.63\%\) \(\%\Delta \rho = 1.11\% + 0.16\% + 2(2.63\%) = 1.27\% + 5.26\% = 6.53\% \approx 6.5\%\)
(c) Calculate the absolute uncertainty \(\Delta \rho\): \(\Delta \rho = 4.08 \times 10^{-7} \times 0.0653 \approx 2.66 \times 10^{-8}\ \Omega\text{ m} = 0.266 \times 10^{-7}\ \Omega\text{ m}\) To 1 significant figure, \(\Delta \rho = 0.3 \times 10^{-7}\ \Omega\text{ m}\). Thus, the final value of resistivity is: \(\rho = (4.1 \pm 0.3) \times 10^{-7}\ \Omega\text{ m}\).
Marking scheme
(a) [1] Rearranging formula correctly: \(\rho = \frac{R \pi D^2}{4 L}\). [1] Correct substitution of numbers including conversion of \(D\) to \(10^{-3}\text{ m}\). [1] Showing calculation leads to \(4.08 \times 10^{-7}\ \Omega\text{ m}\). (b) [1] Identifies that percentage uncertainty in \(D\) must be multiplied by 2. [1] Calculates individual percentage uncertainties correctly (\(1.11\%\), \(0.16\%\), \(2.63\%\)). [1] Obtains total percentage uncertainty of \(6.5\%\) (accept \(6.5\%\) to \(6.6\%\)). (c) [1] Calculates absolute uncertainty as \(0.27 \times 10^{-7}\) or \(0.3 \times 10^{-7}\ \Omega\text{ m}\). [1] Expresses final answer with correct format and matching precision: \((4.1 \pm 0.3) \times 10^{-7}\ \Omega\text{ m}\).
Question 4 · Structured
7 marks
An electron travels with speed \(v = 4.8 \times 10^6\text{ m s}^{-1}\) into a region of uniform magnetic field of flux density \(B = 3.5\text{ mT}\). The direction of the electron's velocity is perpendicular to the direction of the magnetic field.
(a) Explain why the path of the electron in the magnetic field is a circle. [2]
(b) Calculate the magnitude of the magnetic force acting on the electron. [2]
(c) Calculate the radius of the circular path. [3]
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Worked solution
(a) The magnetic force on a moving charge is always perpendicular to its velocity vector. This constant perpendicular force acts as a centripetal force, which changes only the direction of the velocity (not its magnitude), resulting in a circular path.
(b) The magnetic force is: \(F = B q v\) \(F = (3.5 \times 10^{-3}\text{ T}) \times (1.60 \times 10^{-19}\text{ C}) \times (4.8 \times 10^6\text{ m s}^{-1})\) \(F = 2.688 \times 10^{-15}\text{ N} \approx 2.7 \times 10^{-15}\text{ N}\)
(c) The centripetal force is provided by the magnetic force: \(\frac{m v^2}{r} = B q v \implies r = \frac{m v}{B q}\) \(r = \frac{9.11 \times 10^{-31}\text{ kg} \times 4.8 \times 10^6\text{ m s}^{-1}}{3.5 \times 10^{-3}\text{ T} \times 1.60 \times 10^{-19}\text{ C}}\) \(r = \frac{4.373 \times 10^{-24}}{5.60 \times 10^{-22}} = 7.81 \times 10^{-3}\text{ m} = 7.8\text{ mm}\).
Marking scheme
(a) [1] Magnetic force is perpendicular to velocity. [1] Perpendicular force acts as a centripetal force (constant speed, circular motion). (b) [1] Formula \(F = B q v\) stated or implied. [1] Correct calculation leading to \(2.7 \times 10^{-15}\text{ N}\). (c) [1] Equating centripetal force to magnetic force: \(\frac{m v^2}{r} = B q v\). [1] Correct substitution of electron mass \(9.11 \times 10^{-31}\text{ kg}\) and values. [1] Correct calculation to yield \(7.8 \times 10^{-3}\text{ m}\) (or \(7.8\text{ mm}\)).
Question 5 · Structured
8 marks
A straight copper wire of length \(12\text{ cm}\) and mass \(4.5\text{ g}\) is suspended horizontally in a uniform horizontal magnetic field of flux density \(0.15\text{ T}\). The magnetic field is perpendicular to the wire.
(a) State the direction of the magnetic force required to balance the weight of the wire. [1]
(b) State the rule used to determine the direction of the current required to produce this force. [1]
(c) Calculate the current \(I\) in the wire needed to keep the wire suspended horizontally. [3]
(d) The magnetic flux density is now increased by \(20\%\) while the current is kept constant. Describe and explain the subsequent motion of the wire. [3]
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Worked solution
(a) The magnetic force must act vertically upwards to oppose the gravitational weight of the wire.
(b) Fleming's left-hand rule.
(c) For horizontal suspension, the magnetic force must equal the weight: \(F = W \implies B I L = m g\) Convert mass to kg (\(4.5 \times 10^{-3}\text{ kg}\)) and length to m (\(0.12\text{ m}\)): \(0.15 \times I \times 0.12 = 4.5 \times 10^{-3} \times 9.81\) \(0.018 \times I = 0.044145\) \(I = \frac{0.044145}{0.018} \approx 2.45\text{ A} \approx 2.5\text{ A}\)
(d) When \(B\) increases, the upward magnetic force increases (as \(F = B I L\)) while the weight stays constant. There is now a net upward vertical force on the wire. By Newton's second law, the wire accelerates vertically upwards.
Marking scheme
(a) [1] Vertically upwards. (b) [1] Fleming's left-hand rule. (c) [1] Correct formula for weight \(W = m g\) used with correct conversion of units. [1] Correct setup of equation \(B I L = m g\). [1] Correct calculation of current as \(2.5\text{ A}\) (or \(2.45\text{ A}\)). (d) [1] States that magnetic force increases. [1] Explains there is now an unbalanced/net upward force. [1] Identifies that the wire accelerates upwards.
Question 6 · Structured
7 marks
A flat circular coil of \(250\) turns and radius \(3.2\text{ cm}\) is placed perpendicular to a uniform magnetic field. The magnetic flux density \(B\) of the field changes from \(0.45\text{ T}\) to \(0.15\text{ T}\) in a time interval of \(0.080\text{ s}\).
(a) Define *magnetic flux linkage*. [1]
(b) State Faraday's law of electromagnetic induction. [2]
(c) Calculate the magnitude of the average electromotive force (e.m.f.) induced in the coil. [4]
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Worked solution
(a) Magnetic flux linkage is the product of the magnetic flux through a coil and the number of turns in the coil, i.e., \(N \Phi = N B A\).
(b) Faraday's law states that the magnitude of the induced electromotive force (e.m.f.) is directly proportional to the rate of change of magnetic flux linkage.
(c) Area of the coil, \(A = \pi r^2 = \pi \times (0.032)^2 = 3.217 \times 10^{-3}\text{ m}^2\). Change in magnetic flux density, \(\Delta B = 0.45 - 0.15 = 0.30\text{ T}\). Change in magnetic flux linkage, \(\Delta(N \Phi) = N A \Delta B\): \(\Delta(N \Phi) = 250 \times (3.217 \times 10^{-3}) \times 0.30 = 0.2413\text{ Wb}\). Induced e.m.f. is: \(E = \frac{\Delta(N \Phi)}{\Delta t} = \frac{0.2413}{0.080} = 3.016\text{ V} \approx 3.0\text{ V}\).
Marking scheme
(a) [1] Product of number of turns and magnetic flux (or \(N B A\)). (b) [1] Induced e.m.f. is proportional to rate of change. [1] rate of change of magnetic flux linkage. (c) [1] Correct calculation of cross-sectional area \(A = 3.22 \times 10^{-3}\text{ m}^2\). [1] Correct change in magnetic flux density \(\Delta B = 0.30\text{ T}\). [1] Correct formula for induced e.m.f.: \(E = N A \frac{\Delta B}{\Delta t}\). [1] Correct calculation yielding \(3.0\text{ V}\) (or \(3.02\text{ V}\)).
Question 7 · Structured
8 marks
A small mass of \(0.12\text{ kg}\) undergoes simple harmonic motion with an amplitude of \(4.0\text{ cm}\). The frequency of the oscillations is \(2.5\text{ Hz}\).
(a) State the condition for an oscillation to be simple harmonic. [2]
(b) Calculate the maximum acceleration of the mass. [3]
(c) Calculate the maximum kinetic energy of the mass. [3]
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Worked solution
(a) The acceleration of the oscillating system must be directly proportional to its displacement from the equilibrium position and must always act in the opposite direction to the displacement (towards the equilibrium position).
(b) The angular frequency \(\omega\) is: \(\omega = 2 \pi f = 2 \pi \times 2.5 = 15.71\text{ rad s}^{-1}\) Maximum acceleration \(a_0 = \omega^2 x_0\): \(a_0 = (15.71)^2 \times 0.040\) \(a_0 = 246.7 \times 0.040 = 9.87\text{ m s}^{-2} \approx 9.9\text{ m s}^{-2}\).
(c) Maximum velocity occurs at the equilibrium position: \(v_0 = \omega x_0 = 15.71 \times 0.040 = 0.628\text{ m s}^{-1}\) Maximum kinetic energy is: \(E_k = \frac{1}{2} m v_0^2\) \(E_k = \frac{1}{2} \times 0.12 \times (0.628)^2 = 0.060 \times 0.3944 = 0.0237\text{ J} \approx 0.024\text{ J}\).
Marking scheme
(a) [1] Acceleration is directly proportional to displacement. [1] Acceleration is in the opposite direction to displacement / directed towards equilibrium. (b) [1] Correct calculation of \(\omega = 15.71\text{ rad s}^{-1}\). [1] Uses \(a_0 = \omega^2 x_0\) with amplitude in metres. [1] Correct calculation of maximum acceleration as \(9.9\text{ m s}^{-2}\) (accept \(9.87\) or \(10\)). (c) [1] Calculates maximum velocity \(v_0 = 0.63\text{ m s}^{-1}\). [1] Uses kinetic energy formula \(\frac{1}{2} m v_0^2\). [1] Correct calculation of kinetic energy as \(0.024\text{ J}\) (or \(0.0237\text{ J}\)).
Question 8 · Structured
7 marks
A heavy mass suspended by a spring is driven by a periodic force of variable frequency \(f\).
(a) Explain what is meant by *resonance*. [2]
(b) Describe how the amplitude of the oscillations changes as the frequency \(f\) of the driving force is increased from a very low value, through the natural frequency \(f_0\) of the system, to a very high value. [3]
(c) State and explain the effect of increased damping on: (i) the maximum amplitude at resonance, [1] (ii) the frequency at which the maximum amplitude occurs. [1]
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Worked solution
(a) Resonance occurs when an oscillating system is driven by a periodic external force whose frequency is equal to (or very close to) the natural frequency of the system, resulting in a maximum amplitude of the oscillations.
(b) As \(f\) increases from a low value, the amplitude of the oscillations initially increases. As \(f\) approaches the natural frequency \(f_0\), the amplitude rises sharply to a maximum peak. As \(f\) continues to increase beyond \(f_0\), the amplitude decreases continuously toward zero.
(c) (i) Increased damping decreases the maximum amplitude at resonance because more energy is dissipated per cycle by resistive forces. (ii) Increased damping shifts the resonant peak slightly to a lower frequency.
Marking scheme
(a) [1] Driving frequency equals natural frequency of system. [1] Maximum amplitude of oscillation achieved. (b) [1] Amplitude increases as frequency increases towards the natural frequency. [1] Peak amplitude occurs at/near the natural frequency. [1] Amplitude decreases at frequencies above the natural frequency. (c) (i) [1] Maximum amplitude decreases (peak is flatter/lower). (ii) [1] The frequency of maximum amplitude shifts to a slightly lower frequency.
Paper 3 (Advanced Practical Skills)
Answer both questions. Record all your observations in the spaces provided as soon as they are made.
2 Question · 40 marks
Question 1 · Practical Investigation
20 marks
Question 1
Apparatus provided:
A wooden metre rule with a small hole drilled at the 5.0 cm mark.
A metal pin held securely in a boss and clamp on a retort stand to act as a horizontal pivot.
A 100 g mass and a small piece of adhesive putty (Blu-Tack) to attach it.
A stopwatch.
A second metre rule or half-metre rule.
An optical pin or pointer to act as a fiducial marker.
In this experiment, you will investigate how the period of oscillation of a suspended metre rule depends on the position of a sliding mass along its length.
Procedure:
(a) (i) Suspend the metre rule from the pivot through the hole at the 5.0 cm mark, ensuring it can swing freely in a vertical plane.
Using the adhesive putty, attach the 100 g mass to the rule such that its centre is at a distance \(d = 40.0\text{ cm}\) from the pivot hole. Record the value of \(d\).
(ii) Displace the bottom of the rule slightly to one side and release it so that it performs small oscillations. Measure and record the time \(t\) for 10 complete oscillations. Repeat this measurement to obtain a mean value of \(t\). Calculate the period \(T\) of the oscillations.
(b) Change the position of the 100 g mass to obtain six sets of readings of \(d\) and \(t\). The values of \(d\) should be in the range \(20.0\text{ cm} \le d \le 80.0\text{ cm}\). For each set, record \(d\), \(t\), \(T\), and calculate \(d^2\) and \(T^2 d\). Include your results in a table with appropriate column headings and units.
(c) (i) Plot a graph of \(T^2 d\) on the y-axis against \(d^2\) on the x-axis.
(ii) Draw the straight line of best fit.
(d) Determine the gradient and y-intercept of this line.
(e) The quantities \(T\) and \(d\) are related by the equation: \(T^2 d = p d^2 + q\) where \(p\) and \(q\) are constants. Use your answers from (d) to determine the values of \(p\) and \(q\). Include appropriate units for both constants.
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Worked solution
(a) (i) & (ii) Representative measurements: Let \(d = 40.0\text{ cm} = 0.400\text{ m}\). Time for 10 oscillations: \(t_1 = 13.4\text{ s}\), \(t_2 = 13.6\text{ s}\). Mean \(t = 13.5\text{ s}\). Period \(T = 1.35\text{ s}\).
(c) Graph: Plot \(T^2 d\) against \(d^2\). Both axes must have clear labels with units: \(T^2 d / \text{s}^2\text{ m}\) and \(d^2 / \text{m}^2\). Points are plotted precisely. A straight line of best fit is drawn.
(d) Gradient and y-intercept calculation: Using two points on the line of best fit: \((0.0800, 0.400)\) and \((0.4800, 2.300)\). \text{Gradient} = \frac{2.300 - 0.400}{0.4800 - 0.0800} = \frac{1.900}{0.4000} = 4.75\text{ s}^2\text{ m}^{-1}. Using \(y = mx + c\): \(0.400 = 4.75 \times 0.0800 + c \implies c = 0.400 - 0.380 = 0.020\text{ s}^2\text{ m}\). So, y-intercept = \(0.020\text{ s}^2\text{ m}\).
(e) Determining \(p\) and \(q\): From \(T^2 d = p d^2 + q\), we have: \(p = \text{gradient} = 4.75\text{ s}^2\text{ m}^{-1}\) (or \(0.0475\text{ s}^2\text{ cm}^{-1}\)). \(q = \text{y-intercept} = 0.020\text{ s}^2\text{ m}\) (or \(2.0\text{ s}^2\text{ cm}\)).
Marking scheme
(a) (i) [1 mark] Value of \(d\) to the nearest 1 mm with correct unit (e.g. \(40.0\text{ cm}\) or \(0.400\text{ m}\)). (a) (ii) [1 mark] Value of \(t\) measured to 0.1 s or 0.01 s with repeat readings. Correct period \(T\) calculated.
(b) Table [6 marks]: - Successful collection: Six sets of readings of \(d\) and \(t\) without helper assistance. [1] - Range: Range of \(d\) covers at least 45.0 cm (e.g. from 20.0 cm to 70.0 cm). [1] - Column headings: Headings must contain a quantity and a unit (e.g. \(d / \text{m}\), \(t / \text{s}\), \(T / \text{s}\), \(d^2 / \text{m}^2\), \(T^2 d / \text{s}^2\text{ m}\)). [1] - Consistency: All raw values of \(d\) recorded to the same precision (nearest mm) and all raw values of \(t\) recorded to the same precision (0.1 s or 0.01 s). [1] - Significant figures: Significant figures for \(T^2 d\) must be the same as, or one more than, the least number of significant figures in the raw values of \(T\) and \(d\). [1] - Quality: Points on the graph lie close to a straight line (minimum scatter). [1]
(c) Graph [5 marks]: - Axes: Linear scales where the plotted points occupy more than half the grid in both directions. Labels must have units. [1] - Plotting: All observations must be plotted to within half a small square. No thick blobs. [1] - Line of best fit: A single straight line representing the trend of all plotted points. Even distribution of points about the line. [1]
(d) Gradient and Intercept [2 marks]: - Gradient: Uses a triangle where the hypotenuse is at least 50% of the length of the drawn line. Correct calculation. [1] - Intercept: Read directly from the y-axis if the x-axis starts at 0, or calculated using coordinates of a point on the line in \(y = mx + c\). [1]
(e) Constants \(p\) and \(q\) [2 marks]: - Value of \(p\) equals the gradient with unit \(\text{s}^2\text{ m}^{-1}\) (or equivalent). [1] - Value of \(q\) equals the y-intercept with unit \(\text{s}^2\text{ m}\) (or equivalent). [1]
Question 2 · Practical Investigation
20 marks
Question 2
Apparatus provided:
A retort stand, boss, and clamp.
A length of thread (approx. 60 cm).
A pendulum bob (or 100 g mass hanger).
Two rectangular pieces of cardboard: - Card A of width \(w_1 \approx 4.0\text{ cm}\) and height \(h \approx 10.0\text{ cm}\) - Card B of width \(w_2 \approx 8.0\text{ cm}\) and height \(h \approx 10.0\text{ cm}\)
Adhesive tape.
A metre rule.
In this experiment, you will investigate how the damping of a simple pendulum depends on the width of a cardboard card attached to it.
Procedure:
(a) (i) Set up a simple pendulum with a length of approximately 50.0 cm from the clamp to the centre of the bob. Tape Card A of width \(w_1\) vertically to the thread just above the bob, ensuring that the plane of the card is perpendicular to the direction of swing. Measure and record the width \(w_1\) of Card A.
(ii) Position the metre rule horizontally on the bench directly behind the pendulum bob to act as a scale. Displace the bob horizontally from its equilibrium position by an initial amplitude \(x_0 = 15.0\text{ cm}\). Release the bob. Measure and record the amplitude \(x\) of the swing (maximum displacement from equilibrium) after exactly 5 complete oscillations.
(iii) Estimate the percentage uncertainty in your value of \(x\), showing your working.
(b) Remove Card A. Attach Card B to the thread just above the bob in the same manner.
(i) Measure and record the width \(w_2\) of Card B.
(ii) Displace the bob horizontally by the same initial amplitude \(x_0 = 15.0\text{ cm}\). Release the bob. Measure and record the amplitude \(x\) after exactly 5 complete oscillations.
(c) (i) It is suggested that the relationship between \(x\) and \(w\) is given by: \frac{x_0 - x}{x} = k w where \(k\) is a constant. Use your data to calculate two values of \(k\).
(ii) Explain whether your results support the suggested relationship. State a criterion for your judgment and show your reasoning clearly.
(d) Describe four sources of uncertainty or systematic/random errors in this experiment. For each, describe an improvement that could be made to improve the accuracy of the experiment. You may suggest the use of other apparatus or different procedures.
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Worked solution
(a) (i) & (ii) Representative measurements for Card A: Width \(w_1 = 4.0\text{ cm} = 0.040\text{ m}\). Initial amplitude \(x_0 = 15.0\text{ cm}\). Amplitude after 5 complete oscillations, \(x_1 = 11.2\text{ cm}\). (a) (iii) Percentage uncertainty calculation: Absolute uncertainty in measuring the amplitude \(x\) is typically \(\Delta x = 0.2\text{ cm}\) (due to reading a moving bob on a ruler). \text{Percentage uncertainty} = \frac{0.2}{11.2} \times 100\% \approx 1.8\%.
(c) (ii) Comparison and evaluation: Percentage difference between \(k_1\) and \(k_2\): \% \text{ difference} = \frac{|0.0982 - 0.0848|}{0.0848} \times 100\% = \frac{0.0134}{0.0848} \times 100\% \approx 15.8\%. Since the percentage difference of 15.8% is less than the typical experimental uncertainty limit of 20%, the results support the suggested relationship.
(d) Evaluation of limitations and improvements: 1. Limitation: Difficult to release the pendulum without lateral movement or a slight shove, which alters the initial amplitude \(x_0\). Improvement: Use a mechanical release mechanism, such as holding the bob with a thread and cutting the thread with scissors. 2. Limitation: Judging the exact maximum turning point (amplitude) after 5 oscillations is difficult as the bob is in motion for only a brief moment. Improvement: Record a video of the oscillations with a high-speed camera placed parallel to the scale, and analyze frame-by-frame to find the turning point. 3. Limitation: Two sets of readings are insufficient to conclusively prove a mathematical relationship. Improvement: Repeat the experiment for five or more cards of different widths and plot a graph of \(\frac{x_0 - x}{x}\) against \(w\). 4. Limitation: The cardboard card is flexible and bends or flutters during swing, altering the drag coefficient and the effective area. Improvement: Use a rigid plastic sheet of the same dimensions that does not deform under air resistance.
Marking scheme
(a) (i) [1 mark] Value of \(w_1\) recorded to the nearest 1 mm with unit. (a) (ii) [1 mark] Value of \(x_1\) recorded to the nearest 1 mm with unit, where \(x_1 < x_0\). (a) (iii) [1 mark] Correct calculation of percentage uncertainty in \(x_1\), showing an absolute uncertainty of \(0.1\text{ cm}\) to \(0.5\text{ cm}\) (usually \(0.2\text{ cm}\) is standard for a moving scale). (b) (i) [1 mark] Value of \(w_2\) recorded to the nearest 1 mm, with \(w_2 > w_1\). (b) (ii) [1 mark] Value of \(x_2\) recorded to the nearest 1 mm, where \(x_2 < x_1\).
(c) (i) [1 mark] Correct calculations of \(k_1\) and \(k_2\) with correct units (e.g., \(\text{cm}^{-1}\) or \(\text{m}^{-1}\)). (c) (ii) [1 mark] Valid conclusion based on the percentage difference between \(k_1\) and \(k_2\), compared to a stated threshold (e.g. 20%). If % difference \(\le\) threshold, relationship is supported; if not, it is not supported.
(d) Limitations and Improvements [8 marks total - 1 mark for each valid limitation, 1 mark for each corresponding improvement up to a maximum of 4 pairs]: 1. Release consistency: - Limitation: Human error in releasing the bob without giving it a side push / hand movement affecting \(x_0\). [1] - Improvement: Release method using electromagnet or cutting a holding string. [1] 2. Measuring amplitude: - Limitation: Difficult to read amplitude at the extreme of the swing because the bob is moving / parallax error. [1] - Improvement: Use video camera with frame-by-frame playback / smart sensor / scale positioned very close behind the bob. [1] 3. Range of data: - Limitation: Two readings of \(w\) are not enough to confirm the relationship. [1] - Improvement: Collect more data for different widths and plot a graph of \(\frac{x_0 - x}{x}\) against \(w\). [1] 4. Card stability: - Limitation: Card bends / flutters during oscillation, changing air resistance. [1] - Improvement: Use a rigid sheet of plastic / metal of same dimensions that does not bend. [1] 5. Background damping: - Limitation: Pendulum thread and bob experience air resistance even without the card. [1] - Improvement: Measure damping without any card attached and subtract this background damping value. [1]
Paper 4 (A Level Structured)
Answer all questions. You should show all your working and use appropriate units.
10 Question · 100 marks
Question 1 · A Level Structured
10 marks
An experiment is performed to determine the density \( d \) of a uniform metal sphere. The sphere is weighed using a digital balance, and its diameter is measured using a micrometer screw gauge. The experimental measurements are: - Mass of the sphere \( M = 34.2 \pm 0.1 \text{ g} \) - Diameter of the sphere \( D = 2.12 \pm 0.02 \text{ cm} \)
(a) Show that the density \( d \) is given by the expression \( d = \frac{6M}{\pi D^3} \). [2]
(b) Calculate the value of the density \( d \) in \( \text{g cm}^{-3} \). [2]
(c) Calculate: (i) the fractional uncertainty in the mass \( M \). [1] (ii) the fractional uncertainty in the diameter \( D \). [1]
(d) Calculate the absolute uncertainty in the density \( d \). Express the final value of the density with its uncertainty to an appropriate number of significant figures and with a suitable unit. [4]
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Worked solution
(a) The volume \( V \) of a sphere in terms of its diameter \( D \) is: \( V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left(\frac{D}{2}\right)^3 = \frac{4}{3}\pi \frac{D^3}{8} = \frac{\pi D^3}{6} \)
Density \( d \) is given by \( d = \frac{M}{V} \): \( d = \frac{M}{\frac{\pi D^3}{6}} = \frac{6M}{\pi D^3} \) [Proven]
(b) Substituting the measured values: \( d = \frac{6 \times 34.2}{\pi \times (2.12)^3} = \frac{205.2}{\pi \times 9.5281} = \frac{205.2}{29.9336} \approx 6.855 \text{ g cm}^{-3} \)
(c) (i) Fractional uncertainty in \( M \): \( \frac{\Delta M}{M} = \frac{0.1}{34.2} \approx 0.0029 \)
(ii) Fractional uncertainty in \( D \): \( \frac{\Delta D}{D} = \frac{0.02}{2.12} \approx 0.0094 \)
Absolute uncertainty in \( d \): \( \Delta d = 6.855 \times 0.0311 = 0.213 \text{ g cm}^{-3} \approx 0.2 \text{ g cm}^{-3} \)
We round the uncertainty to 1 significant figure (or 2 significant figures, e.g., \( 0.2 \) or \( 0.21 \)) and match the decimal places of the value. Thus, \( d = 6.9 \pm 0.2 \text{ g cm}^{-3} \) (or \( 6.86 \pm 0.21 \text{ g cm}^{-3} \)).
Marking scheme
(a) - States volume formula in terms of radius and substitutes \( r = D/2 \) [1] - Rearranges successfully to show \( d = \frac{6M}{\pi D^3} \) [1]
(b) - Substitutes values into the expression [1] - Obtains value \( 6.85 \) or \( 6.86 \) or \( 6.9 \text{ g cm}^{-3} \) [1]
(d) - Identifies correct formula for fractional uncertainty in \( d \): \( \frac{\Delta d}{d} = \frac{\Delta M}{M} + 3\frac{\Delta D}{D} \) [1] - Correctly calculates fractional uncertainty in \( d \) as \( 0.031 \) or \( 3.1\% \) [1] - Calculates absolute uncertainty \( \Delta d = 0.2 \text{ g cm}^{-3} \) (or \( 0.21 \text{ g cm}^{-3} \)) [1] - Expresses final answer as \( d = 6.9 \pm 0.2 \text{ g cm}^{-3} \) (or \( 6.86 \pm 0.21 \text{ g cm}^{-3} \)) with matching decimal places and correct units [1]
Question 2 · A Level Structured
10 marks
An alpha particle (\( \text{He}^{2+} \)) of mass \( 6.64 \times 10^{-27} \text{ kg} \) is accelerated from rest through a potential difference of \( 1.20 \times 10^5 \text{ V} \). It then enters a region of uniform magnetic field of flux density \( B = 0.450 \text{ T} \) acting perpendicular to its direction of motion.
(a) Calculate the speed of the alpha particle as it enters the magnetic field. [3]
(b) Explain why the path of the alpha particle in the magnetic field is a circular arc. [2]
(c) Calculate the radius of this circular path. [3]
(d) State the effect, if any, on the radius of the path if a beta-minus (\( \beta^- \)) particle with the same initial kinetic energy entered the same magnetic field. Give a qualitative comparison. [2]
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Worked solution
(a) Kinetic energy gained = electrical potential energy lost: \( \frac{1}{2} m v^2 = q V \) An alpha particle has a charge of \( q = 2e = 2 \times 1.60 \times 10^{-19} \text{ C} = 3.20 \times 10^{-19} \text{ C} \). \( v = \sqrt{\frac{2 q V}{m}} = \sqrt{\frac{2 \times 3.20 \times 10^{-19} \times 1.20 \times 10^5}{6.64 \times 10^{-27}}} \) \( v = \sqrt{\frac{7.68 \times 10^{-14}}{6.64 \times 10^{-27}}} = \sqrt{1.1566 \times 10^{13}} \approx 3.40 \times 10^6 \text{ m s}^{-1} \)
(b) The magnetic force on the charged particle is given by \( F = Bqv \) and is always perpendicular to its velocity (by Fleming's Left-Hand Rule). Because the force is perpendicular to the motion, it does no work, so the speed remains constant. A constant force acting perpendicular to the velocity provides the necessary centripetal force for circular motion.
(c) The magnetic force acts as the centripetal force: \( B q v = \frac{m v^2}{r} \implies r = \frac{m v}{B q} \) \( r = \frac{6.64 \times 10^{-27} \times 3.40 \times 10^6}{0.450 \times 3.20 \times 10^{-19}} \) \( r = \frac{2.258 \times 10^{-20}}{1.44 \times 10^{-19}} \approx 0.157 \text{ m} \) (or \( 15.7 \text{ cm} \))
(d) A beta-minus particle has a mass of \( 9.11 \times 10^{-31} \text{ kg} \) (about 7300 times smaller than the alpha particle) and a charge of magnitude \( 1e \) (half that of the alpha particle). Since \( E_k = \frac{p^2}{2m} \implies p = \sqrt{2m E_k} \), and \( r = \frac{p}{Bq} \): \( r = \frac{\sqrt{2m E_k}}{Bq} \). With the same kinetic energy \( E_k \), the much smaller mass \( m \) of the beta particle dominates, leading to a much smaller momentum \( p \) and therefore a much smaller radius of path. (It will also deflect in the opposite direction).
Marking scheme
(a) - Equates kinetic energy to electrical energy: \( \frac{1}{2}mv^2 = qV \) [1] - Recalls and uses charge of alpha particle as \( 2e \) (\( 3.20 \times 10^{-19} \text{ C} \)) [1] - Correct calculation of speed to 2 or 3 sig figs: \( 3.40 \times 10^6 \text{ m s}^{-1} \) (accept \( 3.4 \times 10^6 \)) [1]
(b) - States that magnetic force is perpendicular to velocity / direction of motion [1] - Explains that this perpendicular force provides the centripetal force (causing a constant-speed circular path) [1]
(c) - Equates magnetic force to centripetal force: \( Bqv = \frac{mv^2}{r} \) [1] - Correct rearrangement for \( r \) and substitution of values [1] - Obtains radius \( r = 0.157 \text{ m} \) (accept \( 0.16 \text{ m} \) or \( 15.7 \text{ cm} \)) [1]
(d) - States that the radius is much smaller for the beta particle [1] - Justifies using the fact that the beta particle has a much smaller mass (or momentum) for the same kinetic energy [1]
Question 3 · A Level Structured
10 marks
An experiment is conducted to determine the acceleration of free fall \( g \) by measuring the period of oscillation of a simple pendulum. The equation relating the time period \( T \) and length \( L \) of the pendulum is: \( T = 2\pi \sqrt{\frac{L}{g}} \) The experimental measurements are: - Length of pendulum \( L = 0.820 \pm 0.005 \text{ m} \) - Time for 20 oscillations \( t = 36.4 \pm 0.2 \text{ s} \)
(a) Determine the value of \( g \) from these measurements. [3]
(b) Calculate: (i) the percentage uncertainty in the length \( L \). [1] (ii) the percentage uncertainty in the period \( T \). [2]
(c) Calculate the absolute uncertainty in the calculated value of \( g \). Write down your final value of \( g \) with its absolute uncertainty to an appropriate number of significant figures. [4]
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Worked solution
(a) The time period \( T \) is the time for one oscillation: \( T = \frac{t}{20} = \frac{36.4}{20} = 1.82 \text{ s} \) Rearranging the formula for \( g \): \( T^2 = 4\pi^2 \frac{L}{g} \implies g = \frac{4\pi^2 L}{T^2} \) \( g = \frac{4\pi^2 \times 0.820}{1.82^2} = \frac{32.373}{3.3124} \approx 9.773 \text{ m s}^{-2} \)
(b) (i) Percentage uncertainty in \( L \): \( \frac{\Delta L}{L} \times 100\% = \frac{0.005}{0.820} \times 100\% \approx 0.610\% \)
(ii) Since \( T = t / 20 \), the percentage uncertainty in \( T \) is equal to the percentage uncertainty in the total measured time \( t \): \( \frac{\Delta T}{T} \times 100\% = \frac{\Delta t}{t} \times 100\% = \frac{0.2}{36.4} \times 100\% \approx 0.549\% \)
(c) The formula for \( g \) is \( g = \frac{4\pi^2 L}{T^2} \). Therefore: \( \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\left(\frac{\Delta T}{T}\right) \) \( \frac{\Delta g}{g} = 0.00610 + 2(0.00549) = 0.00610 + 0.01098 = 0.01708 \) Percentage uncertainty in \( g \) is \( 1.71\% \).
Absolute uncertainty in \( g \): \( \Delta g = g \times 0.01708 = 9.773 \times 0.01708 \approx 0.167 \text{ m s}^{-2} \) Rounding \( \Delta g \) to 1 significant figure gives \( 0.2 \text{ m s}^{-2} \) (or 2 significant figures as \( 0.17 \text{ m s}^{-2} \)).
Matching the decimal places of \( g \): \( g = 9.8 \pm 0.2 \text{ m s}^{-2} \) (or \( g = 9.77 \pm 0.17 \text{ m s}^{-2} \)).
Marking scheme
(a) - Calculates the time period \( T = 1.82 \text{ s} \) [1] - Rearranges equation to make \( g \) the subject: \( g = \frac{4\pi^2 L}{T^2} \) [1] - Calculates \( g = 9.77 \text{ m s}^{-2} \) (accept \( 9.8 \text{ m s}^{-2} \)) [1]
(b) - (i) Calculates percentage uncertainty in \( L \) as \( 0.61\% \) [1] - (ii) Identifies that fractional/percentage uncertainty of \( T \) is same as for \( t \) [1] - Calculates percentage uncertainty in \( T \) as \( 0.55\% \) [1]
(c) - Sums the fractional uncertainties: \( \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T} \) [1] - Calculates fractional/percentage uncertainty in \( g \) as \( 0.017 \) or \( 1.7\% \) [1] - Calculates absolute uncertainty as \( 0.17 \text{ m s}^{-2} \) (or \( 0.2 \text{ m s}^{-2} \)) [1] - Expresses final result with correct units and consistent significant figures: \( g = 9.8 \pm 0.2 \text{ m s}^{-2} \) or \( 9.77 \pm 0.17 \text{ m s}^{-2} \) [1]
Question 4 · A Level Structured
10 marks
A heavy pendulum is suspended from a support which is driven horizontally by a vibrator of variable frequency \( f \).
(a) Describe what is meant by: (i) natural frequency of a system. [1] (ii) resonance. [2]
(b) The amplitude of the pendulum's motion is recorded as the driving frequency \( f \) is varied. The experiment is repeated under three conditions: in air, in water, and in a highly viscous oil. (i) Sketch a graph on a single set of axes showing how the amplitude of oscillation varies with frequency \( f \) for the three conditions. Label each curve clearly. [3] (ii) State two effects of increasing the degree of damping on the resonance curve. [2]
(c) In one specific setup, the resonant frequency is \( 1.50 \text{ Hz} \). The maximum speed of the pendulum bob during resonance (under very light damping) is \( 0.480 \text{ m s}^{-1} \). Calculate the amplitude of this oscillation. [2]
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Worked solution
(a) (i) Natural frequency is the frequency at which a system oscillates when it is disturbed and then allowed to oscillate freely without any external driving force. (ii) Resonance occurs when a system is driven by an external periodic force whose frequency is equal to (or very close to) the natural frequency of the system. This results in maximum transfer of energy to the system and a maximum amplitude of oscillation.
(b) (i) The sketch should show three curves: - Light damping (air): sharp peak with a high maximum amplitude, with the peak occurring at the natural frequency \( f_0 \). - Medium damping (water): lower maximum amplitude, broader peak, with the peak shifted slightly to the left (lower frequency). - Heavy damping (oil): very flat curve, extremely low maximum amplitude, with no distinct sharp peak, or peak shifted significantly to the left.
(ii) As damping increases: 1. The peak amplitude of the oscillations decreases. 2. The peak becomes broader and less sharp. 3. The frequency at which the maximum amplitude occurs (resonant frequency) decreases slightly.
(c) For simple harmonic motion, the maximum speed is related to the amplitude \( A \) and angular frequency \( \omega_0 \) by: \( v_{\text{max}} = \omega_0 A = 2\pi f_0 A \) Given \( f_0 = 1.50 \text{ Hz} \) and \( v_{\text{max}} = 0.480 \text{ m s}^{-1} \): \( 0.480 = 2 \pi \times 1.50 \times A \) \( 0.480 = 3\pi A \implies A = \frac{0.480}{3\pi} \approx 0.05093 \text{ m} \approx 0.0510 \text{ m} \) (or \( 5.10 \text{ cm} \)).
Marking scheme
(a) - (i) Defines natural frequency as the free, undisturbed oscillation frequency of the system [1] - (ii) States driving frequency equals natural frequency [1] and results in maximum amplitude/energy transfer [1]
(b) - (i) Draws a graph with appropriate axes (Amplitude vs Frequency) showing: 1. Sharp peak for air (light damping) [1] 2. Broader, lower peak for water, shifted slightly left [1] 3. Very flat, low-amplitude curve for oil, labeled correctly [1] - (ii) States any two from: peak height decreases [1], peak width increases / peak broadens [1], resonant frequency decreases [1] (max 2 marks)
(c) - Recalls and uses \( v_{\text{max}} = 2\pi f A \) [1] - Calculates \( A = 0.0510 \text{ m} \) (accept \( 5.1 \text{ cm} \)) [1]
Question 5 · A Level Structured
10 marks
A rigid horizontal brass rod of length \( 15.0 \text{ cm} \) and mass \( 45.0 \text{ g} \) is suspended horizontally by two flexible conducting wires in a uniform horizontal magnetic field. The magnetic field is perpendicular to the rod.
(a) State the direction of the magnetic field required to balance the weight of the rod when the current in the rod is from left to right. Explain your reasoning. [3]
(b) The magnetic flux density \( B \) is \( 0.280 \text{ T} \). (i) Calculate the current \( I \) required in the rod to make the tension in the supporting wires zero. [3] (ii) If the current is increased to twice this calculated value, describe the resulting motion of the rod and calculate its initial acceleration. [4]
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Worked solution
(a) To make the tension in the wires zero, the magnetic force \( F_B \) must act vertically upwards to balance the weight of the rod acting downwards. According to Fleming's Left-Hand Rule: - The thumb points in the direction of the force (upwards). - The second finger points in the direction of the current (left to right). - Consequently, the first finger, which represents the magnetic field, must point horizontally at right angles to the rod, away from the observer (into the page).
(b) (i) For zero tension, magnetic force = weight: \( B I L = m g \) Where: \( B = 0.280 \text{ T} \) \( L = 15.0 \text{ cm} = 0.150 \text{ m} \) \( m = 45.0 \text{ g} = 0.0450 \text{ kg} \) \( g = 9.81 \text{ m s}^{-2} \)
\( 0.280 \times I \times 0.150 = 0.0450 \times 9.81 \) \( 0.0420 \times I = 0.44145 \implies I = \frac{0.44145}{0.0420} \times 10.51 \text{ A} \approx 10.5 \text{ A} \)
(ii) If the current is doubled to \( 2I \), the magnetic force becomes twice the weight of the rod: \( F_B = 2 m g \)
The net force \( F_{\text{net}} \) on the rod is: \( F_{\text{net}} = F_B - mg = 2mg - mg = mg \) acting vertically upwards.
Description of motion: The rod accelerates vertically upwards.
Initial acceleration \( a \): \( a = \frac{F_{\text{net}}}{m} = \frac{mg}{m} = g = 9.81 \text{ m s}^{-2} \).
Marking scheme
(a) - Identifies that the magnetic force must be directed vertically upwards to oppose weight [1] - Refers to Fleming's Left-Hand Rule to relate force, current, and field directions [1] - Correctly determines field direction is horizontally into the page / away from observer [1]
(b) - (i) Equates magnetic force to gravitational force: \( BIL = mg \) [1] - Substitutes SI units correctly: \( L = 0.150 \text{ m} \) and \( m = 0.0450 \text{ kg} \) [1] - Calculates current \( I = 10.5 \text{ A} \) [1] - (ii) States that the net upward force is equal to \( mg \) or calculates net force as \( 0.441 \text{ N} \) [1] - Identifies that the rod will accelerate vertically upwards [1] - Uses Newton's Second Law: \( a = F_{\text{net}}/m \) [1] - Obtains acceleration \( a = 9.81 \text{ m s}^{-2} \) (or \( 9.8 \text{ m s}^{-2} \)) [1]
Question 6 · A Level Structured
10 marks
An engineer measures the electrical resistivity \( \rho \) of a cylindrical wire. The formula used is: \( \rho = \frac{\pi d^2 R}{4 L} \) The following measurements are obtained: - Diameter of the wire \( d = 0.38 \pm 0.02 \text{ mm} \) - Resistance of the wire \( R = 12.4 \pm 0.3\ \Omega \) - Length of the wire \( L = 1.250 \pm 0.002 \text{ m} \)
(a) State whether each of the uncertainties in \( d \), \( R \), and \( L \) is likely to be a systematic error or a random error, and briefly justify your answer. [3]
(b) Calculate the value of the resistivity \( \rho \) of the wire. [2]
(c) Determine the percentage uncertainty in the calculated value of \( \rho \). [3]
(d) Explain which measurement contributes most to the uncertainty in \( \rho \), and suggest one improvement to the experimental method to reduce this uncertainty. [2]
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Worked solution
(a) - Uncertainty in diameter \( d \) is random, as it arises from variation in diameter along the wire or resolution limits of the micrometer. - Uncertainty in resistance \( R \) is random, arising from fluctuations in digital multimeter readings or contact resistance. - Uncertainty in length \( L \) can contain both, but is primarily random due to parallax errors in aligning the ruler, or systematic due to a poorly calibrated tape.
(c) The fractional uncertainty relation is: \( \frac{\Delta \rho}{\rho} = 2\left(\frac{\Delta d}{d}\right) + \frac{\Delta R}{R} + \frac{\Delta L}{L} \) Let's calculate the individual percentage uncertainties: - For \( d \): \( \frac{0.02}{0.38} \times 100\% \approx 5.26\% \) - For \( R \): \( \frac{0.3}{12.4} \times 100\% \approx 2.42\% \) - For \( L \): \( \frac{0.002}{1.250} \times 100\% \approx 0.16\% \)
Now combine them: \( \frac{\Delta \rho}{\rho} = 2(5.26\%) + 2.42\% + 0.16\% = 10.52\% + 2.42\% + 0.16\% = 13.1\% \) So, the percentage uncertainty is approximately \( 13\% \).
(d) The measurement of diameter \( d \) contributes the most to the uncertainty because it has the largest individual percentage uncertainty (\( 5.3\% \)) and its effect is doubled (\( 10.5\% \) contribution) because \( d \) is squared in the formula. Improvement: Use a digital micrometer to measure the diameter at multiple different positions and orientations along the wire, then find the average, which reduces the random uncertainty.
Marking scheme
(a) - Correctly identifies and briefly justifies random/systematic nature for any of the measurements (e.g., diameter is random due to variations in shape or scale reading) [1] - Do. for second measurement [1] - Do. for third measurement [1]
(b) - Substitutes values correctly in the resistivity formula in SI units [1] - Obtains \( 1.1 \times 10^{-6}\ \Omega \text{ m} \) or \( 1.13 \times 10^{-6}\ \Omega \text{ m} \) [1]
(c) - Identifies the formula for combining fractional uncertainties, multiplying the diameter term by 2 [1] - Calculates percentage uncertainty for each term: \( d \) is \( 5.26\% \), \( R \) is \( 2.42\% \), \( L \) is \( 0.16\% \) [1] - Sums correctly to obtain \( 13\% \) (accept \( 13.1\% \)) [1]
(d) - States that diameter \( d \) contributes most because it has the largest uncertainty and its fractional uncertainty is doubled [1] - Suggests measuring the diameter at multiple positions and orientations, calculating a mean [1]
Question 7 · A Level Structured
10 marks
A block of mass \( 0.250 \text{ kg} \) on a frictionless horizontal surface is attached to a spring of spring constant \( k = 40.0 \text{ N m}^{-1} \). The block is pulled a distance of \( 8.00 \text{ cm} \) from its equilibrium position and released from rest.
(a) Show that the block performs simple harmonic motion, and calculate its frequency of oscillation. [3]
(b) Calculate: (i) the maximum kinetic energy of the block. [2] (ii) the potential energy of the spring when the displacement of the block is \( 4.00 \text{ cm} \). [2]
(c) On a single set of axes, sketch graphs to show how the kinetic energy \( E_k \), potential energy \( E_p \), and total energy \( E_{\text{total}} \) of the system vary with the displacement \( x \) from the equilibrium position. Label the curves and critical values on both axes clearly. [3]
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Worked solution
(a) The restoring force of the spring is given by Hooke's Law: \( F = -kx \). By Newton's second law, \( F = ma \), so: \( ma = -kx \implies a = -\left(\frac{k}{m}\right) x \) Since the acceleration \( a \) is directly proportional to the displacement \( x \) and is in the opposite direction (directed towards the equilibrium position), the motion is simple harmonic (with \( \omega^2 = \frac{k}{m} \)).
(b) (i) The maximum kinetic energy occurs at the equilibrium position where all energy is kinetic. This is equal to the total energy stored in the spring when fully extended (at amplitude \( x_0 = 8.00 \text{ cm} = 0.0800 \text{ m} \)): \( E_{k,\text{max}} = \frac{1}{2} k x_0^2 = \frac{1}{2} \times 40.0 \times (0.0800)^2 = 20.0 \times 0.00640 = 0.128 \text{ J} \)
(ii) The potential energy \( E_p \) of the spring at displacement \( x = 4.00 \text{ cm} = 0.0400 \text{ m} \) is: \( E_p = \frac{1}{2} k x^2 = \frac{1}{2} \times 40.0 \times (0.0400)^2 = 20.0 \times 0.00160 = 0.0320 \text{ J} \)
(c) The sketch should contain: - Horizontal axis \( x \) going from \( -8.0 \text{ cm} \) to \( +8.0 \text{ cm} \). - Vertical axis representing Energy up to \( 0.128 \text{ J} \). - \( E_{\text{total}} \): A horizontal straight line at \( 0.128 \text{ J} \). - \( E_p \): A parabola curved upwards, passing through \( (0,0) \) and peaking at \( 0.128 \text{ J} \) at \( x = \pm 8.0 \text{ cm} \). - \( E_k \): A parabola curved downwards, peaking at \( 0.128 \text{ J} \) at \( x = 0 \) and falling to zero at \( x = \pm 8.0 \text{ cm} \).
Marking scheme
(a) - States \( F = -kx \) and uses \( F = ma \) to show \( a = -\frac{k}{m}x \) and explains SHM definition [1] - Recalls and uses \( \omega = \sqrt{\frac{k}{m}} \) [1] - Calculates frequency \( f = 2.01 \text{ Hz} \) (accept \( 2.0 \text{ Hz} \)) [1]
(b) - (i) Recalls and uses \( E_{k,\text{max}} = \frac{1}{2}kx_0^2 \) [1] - Calculates maximum kinetic energy = \( 0.128 \text{ J} \) (or \( 0.13 \text{ J} \)) [1] - (ii) Recalls and uses \( E_p = \frac{1}{2}kx^2 \) [1] - Calculates potential energy = \( 0.0320 \text{ J} \) (or \( 0.032 \text{ J} \)) [1]
(c) - Draws horizontal line for \( E_{\text{total}} \) and labels peak value \( 0.128 \text{ J} \) on y-axis and limits \( \pm 8.0 \text{ cm} \) on x-axis [1] - Draws \( E_p \) as a symmetrical upward-opening parabola with minimum at 0 and maxima at \( \pm 8.0 \text{ cm} \) [1] - Draws \( E_k \) as a symmetrical downward-opening parabola with maximum at 0 and minima at \( \pm 8.0 \text{ cm} \) [1]
Question 8 · A Level Structured
10 marks
A flat circular coil consists of 120 turns of wire and has a mean radius of \( 2.50 \text{ cm} \). The coil is placed in a uniform magnetic field of flux density \( B = 0.150 \text{ T} \) such that the plane of the coil is perpendicular to the magnetic field.
(a) State Faraday's law of electromagnetic induction. [2]
(b) Calculate the magnetic flux linkage of the coil in this position. [2]
(c) The magnetic field is reduced to zero in a uniform manner over a time interval of \( 0.080 \text{ s} \). (i) Calculate the average electromotive force (e.m.f.) induced in the coil. [2] (ii) State Lenz's law and explain how it determines the direction of the induced current in the coil as the magnetic field is decreasing. [4]
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Worked solution
(a) Faraday's law of electromagnetic induction states that the magnitude of the induced electromotive force (e.m.f.) is directly proportional to the rate of change of magnetic flux linkage.
(b) The magnetic flux linkage \( \Phi \) is given by: \( \Phi = N B A \) Where: - \( N = 120 \) - \( B = 0.150 \text{ T} \) - \( A = \pi r^2 = \pi \times (0.0250)^2 = 1.9635 \times 10^{-3} \text{ m}^2 \)
(c) (i) The average induced e.m.f. \( E \) is given by: \( E = \frac{\Delta \Phi}{\Delta t} \) Since the field drops to zero, the change in flux linkage is equal to the initial flux linkage: \( E = \frac{0.03534}{0.080} \approx 0.4418 \text{ V} \approx 0.442 \text{ V} \) (or \( 0.44 \text{ V} \))
(ii) Lenz's law states that the direction of the induced e.m.f. (and current) is such that it opposes the change in magnetic flux that produces it.
As the external magnetic field is reduced to zero, the magnetic flux passing through the coil decreases. To oppose this decrease, the induced current in the coil must create its own magnetic field in the same direction as the original external magnetic field. By using the right-hand grip rule, this means the induced current must flow in a direction that supports the dwindling external field (e.g., if the original field pointed into the face of the coil, the induced current flows clockwise to create a field pointing into the face).
Marking scheme
(a) - States that magnitude of induced e.m.f. is proportional to rate of change [1] - States of magnetic flux linkage / magnetic flux [1]
(c) - (i) Uses \( E = \frac{\Delta \Phi}{\Delta t} \) [1] - Calculates induced e.m.f. \( E = 0.44 \text{ V} \) (or \( 0.442 \text{ V} \)) [1] - (ii) States Lenz's law correctly (induced e.m.f. opposes the change in flux) [1] - States that the external flux through the coil is decreasing [1] - Explains that the induced current must create a magnetic field in the same direction as the original field to try to maintain the flux [1] - Explains how this determines the direction of current (e.g. clockwise/counter-clockwise relative to the field direction) [1]
Question 9 · A Level Structured
10 marks
A long solenoid has \(1200\) turns per meter. An alternating current \(I\) in the solenoid is given by the equation: \(I = I_0 \sin(\omega t)\) where \(I_0 = 4.5\text{ A}\) and \(\omega = 100\pi\text{ rad s}^{-1}\).
A small flat search coil of area \(3.5 \times 10^{-4}\text{ m}^2\) and containing \(250\) turns is positioned at the center of the solenoid. The plane of the search coil is perpendicular to the longitudinal axis of the solenoid.
(a) State Faraday's law of electromagnetic induction. [2]
(b) (i) Show that the magnetic flux linkage \(\Phi\) of the search coil is given by \(\Phi = \mu_0 n N A I\), where the symbols have their standard meanings. [2] (ii) Calculate the maximum magnetic flux linkage of the search coil. [2]
(c) (i) Explain why the electromotive force (e.m.f.) induced in the search coil is out of phase by \(90^\circ\) (\(\pi/2\) radians) with the current in the solenoid. [2] (ii) Calculate the maximum e.m.f. induced in the search coil. [2]
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Worked solution
(a) Faraday's law states that the magnitude of the induced electromotive force (e.m.f.) is directly proportional to the rate of change of magnetic flux linkage.
(b) (i) The magnetic field \(B\) at the center of a long solenoid is given by \(B = \mu_0 n I\), where \(n\) is the number of turns per unit length. Since the search coil has \(N\) turns, each of area \(A\), and its plane is perpendicular to the magnetic field, the total magnetic flux linkage \(\Phi\) is: \(\Phi = N \phi = N B A = N (\mu_0 n I) A = \mu_0 n N A I\).
(ii) The maximum flux linkage \(\Phi_0\) occurs when the current has its maximum value \(I_0 = 4.5\text{ A}\): \(\Phi_0 = \mu_0 n N A I_0\) \(\Phi_0 = (4\pi \times 10^{-7}\text{ H m}^{-1}) \times 1200\text{ m}^{-1} \times 250 \times (3.5 \times 10^{-4}\text{ m}^2) \times 4.5\text{ A}\) \(\Phi_0 \approx 5.94 \times 10^{-4}\text{ Wb}\) (or \(\text{V s}\)). To 2 significant figures, this is \(5.9 \times 10^{-4}\text{ Wb}\).
(c) (i) The induced e.m.f. \(E\) is given by \(E = -\frac{d\Phi}{dt}\). Since \(I = I_0 \sin(\omega t)\), the flux linkage is \(\Phi = \Phi_0 \sin(\omega t)\). Differentiating this with respect to time gives \(E = -\omega \Phi_0 \cos(\omega t)\). The cosine function is out of phase with the sine function by \(90^\circ\) (or \(\pi/2\) radians). Alternatively, the rate of change of flux linkage (and thus induced e.m.f.) is maximum when the flux linkage itself (and current) is zero, and is zero when the flux linkage is maximum.
(ii) The maximum induced e.m.f. \(E_0\) is: \(E_0 = \omega \Phi_0 = (100\pi\text{ rad s}^{-1}) \times (5.938 \times 10^{-4}\text{ Wb}) \approx 0.187\text{ V}\). To 2 significant figures, this is \(0.19\text{ V}\).
Marking scheme
(a) - Induced e.m.f. is directly proportional to [1] - rate of change of magnetic flux linkage [1]
(b) (i) - Expression for magnetic field of a solenoid: \(B = \mu_0 n I\) [1] - Linkage \(\Phi = N B A\), leading to \(\Phi = \mu_0 n N A I\) [1]
(c) (i) - \(E = -d\Phi/dt\) and since \(\Phi = \Phi_0 \sin(\omega t)\), then \(E \propto -\cos(\omega t)\) [1] - State that \(\sin\) and \(\cos\) curves are \(90^\circ\) out of phase / rate of change is maximum when current is zero [1]
(ii) - Use of \(E_0 = \omega \Phi_0\) [1] - Value: \(0.19\text{ V}\) (accept range \(0.18\text{ V}\) to \(0.19\text{ V}\)) [1]
Question 10 · A Level Structured
10 marks
A block of mass \(0.35\text{ kg}\) is attached to a horizontal spring of spring constant \(45\text{ N m}^{-1}\). The block is pulled a distance of \(8.0\text{ cm}\) from its equilibrium position on a frictionless horizontal surface and released from rest at time \(t = 0\).
(a) Show that the block undergoes simple harmonic motion with an angular frequency \(\omega \approx 11.3\text{ rad s}^{-1}\). [2]
(b) (i) Calculate the total energy of the oscillating system. [2] (ii) Determine the displacement of the block when its kinetic energy is equal to its potential energy. [2]
(c) (i) Calculate the speed of the block when its displacement from equilibrium is \(4.0\text{ cm}\). [2] (ii) Calculate the time taken for the block to move from its initial position of \(8.0\text{ cm}\) to a displacement of \(4.0\text{ cm}\) for the first time. [2]
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Worked solution
(a) For a mass on a spring on a horizontal frictionless surface, the restoring force is \(F = -kx\). Using Newton's second law, \(F = ma \implies ma = -kx \implies a = -\left(\frac{k}{m}\right)x\). Since acceleration \(a\) is directly proportional to displacement \(x\) and is directed towards the equilibrium position (represented by the negative sign), the motion is simple harmonic. The acceleration equation of simple harmonic motion is \(a = -\omega^2 x\). Therefore, \(\omega^2 = \frac{k}{m} \implies \omega = \sqrt{\frac{k}{m}}\). Substituting the given values: \(\omega = \sqrt{\frac{45}{0.35}} = \sqrt{128.57} \approx 11.34\text{ rad s}^{-1} \approx 11.3\text{ rad s}^{-1}\).
(b) (i) The total energy \(E_T\) of the system is the maximum potential energy of the spring: \(E_T = \frac{1}{2} k x_0^2\), where \(x_0 = 8.0\text{ cm} = 0.080\text{ m}\). \(E_T = \frac{1}{2} \times 45 \times (0.080)^2 = 22.5 \times 0.0064 = 0.144\text{ J}\). To 2 significant figures, this is \(0.14\text{ J}\).
(ii) Let \(E_k\) be the kinetic energy and \(E_p\) be the potential energy. \(E_T = E_k + E_p\) When \(E_k = E_p\), we have \(E_T = 2 E_p\). \(\frac{1}{2} k x_0^2 = 2 \left(\frac{1}{2} k x^2\right)\) \(x^2 = \frac{1}{2} x_0^2 \implies x = \frac{x_0}{\sqrt{2}}\) \(x = \frac{8.0}{\sqrt{2}} \approx 5.66\text{ cm}\). To 2 significant figures, this is \(5.7\text{ cm}\) (or \(0.057\text{ m}\)).
(c) (i) The speed \(v\) of a particle in simple harmonic motion is given by: \(v = \omega \sqrt{x_0^2 - x^2}\), where \(x = 4.0\text{ cm} = 0.040\text{ m}\). Using \(\omega = 11.34\text{ rad s}^{-1}\): \(v = 11.34 \times \sqrt{0.080^2 - 0.040^2} = 11.34 \times \sqrt{0.0048} \approx 11.34 \times 0.06928 \approx 0.786\text{ m s}^{-1}\). To 2 significant figures, this is \(0.79\text{ m s}^{-1}\) (accept \(0.78\text{ m s}^{-1}\) if using \(\omega = 11.3\text{ rad s}^{-1}\)).
(ii) Since the block is released from rest at \(t = 0\) when \(x = x_0\), the displacement is modeled by: \(x = x_0 \cos(\omega t)\) At \(x = 4.0\text{ cm}\) and \(x_0 = 8.0\text{ cm}\): \(4.0 = 8.0 \cos(\omega t) \implies \cos(\omega t) = 0.5\) The smallest positive angle for which \(\cos(\theta) = 0.5\) is \(\theta = \frac{\pi}{3}\text{ rad}\). \(\omega t = \frac{\pi}{3}\) \(t = \frac{\pi}{3\omega} = \frac{\pi}{3 \times 11.34} \approx 0.0924\text{ s}\). To 2 significant figures, this is \(0.092\text{ s}\) (or \(0.093\text{ s}\) using \(\omega = 11.3\text{ rad s}^{-1}\)).
Marking scheme
(a) - Deduces \(a = -(k/m)x\) from \(F = ma\) and \(F = -kx\), explaining why this represents SHM [1] - Calculates \(\omega = \sqrt{45/0.35} = 11.3\text{ rad s}^{-1}\) [1]
(b) (i) - Use of \(E_T = \frac{1}{2} k x_0^2\) (or \(E_T = \frac{1}{2} m \omega^2 x_0^2\)) [1] - Calculates \(0.14\text{ J}\) (or \(0.144\text{ J}\)) [1]
(c) (i) - Use of \(v = \omega \sqrt{x_0^2 - x^2}\) [1] - Calculates \(0.79\text{ m s}^{-1}\) (or \(0.78\text{ m s}^{-1}\)) [1]
(ii) - Use of \(x = x_0 \cos(\omega t)\) to obtain \(\omega t = \pi/3\) [1] - Calculates \(0.092\text{ s}\) (or \(0.093\text{ s}\)) [1]
Paper 5 (Planning, Analysis and Evaluation)
Answer both questions. Write your answers in the spaces provided.
2 Question · 30 marks
Question 1 · planning
15 marks
A student is investigating the electromagnetic induction between a long solenoid and a small coaxial search coil placed at the centre of the solenoid. An alternating current of peak value \( I_0 \) and frequency \( f \) is passed through the solenoid. This induces an alternating electromotive force (e.m.f.) of peak value \( V \) in the search coil. It is suggested that \( V \) is related to \( f \) by the equation: \( V = \frac{\mu_0 \pi^2 d^2 N_1 N_2 I_0 f}{L} \) where: \( d \) is the diameter of the search coil, \( N_1 \) is the number of turns of the solenoid, \( L \) is the length of the solenoid, \( N_2 \) is the number of turns of the search coil, and \( \mu_0 \) is the permeability of free space. Design a laboratory experiment to investigate how \( V \) depends on \( f \). You should draw a diagram showing the arrangement of your apparatus. In your description, pay particular attention to: the independent and dependent variables, the control of variables, the apparatus to be used, how the peak current \( I_0 \) is kept constant, how the frequency \( f \) and peak e.m.f. \( V \) are measured, how \( \mu_0 \) is determined from the results, including any graph to be plotted, and safety precautions.
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Worked solution
To investigate the relationship \( V = \frac{\mu_0 \pi^2 d^2 N_1 N_2 I_0 f}{L} \): 1. Arrange a long solenoid in series with a signal generator, an AC ammeter, and a rheostat. 2. Place a small circular search coil coaxially at the centre of the solenoid. 3. Connect the terminals of the search coil to the input of a cathode-ray oscilloscope (CRO). 4. Measure the physical parameters of the coils: - Measure the diameter \( d \) of the search coil using a micrometer screw gauge. Measure at different angles and average the readings. - Measure the length \( L \) of the solenoid winding using a metre rule. - Record the number of turns \( N_1 \) of the solenoid and \( N_2 \) of the search coil. 5. Set the frequency \( f \) of the signal generator to a low value. Adjust the rheostat until the peak current reaches a designated value \( I_0 \) (as measured on the AC ammeter or calculated from the peak-to-peak voltage across a known series resistor on the CRO). 6. Observe the induced voltage waveform on the CRO. Measure the peak-to-peak voltage \( V_{\text{p-p}} \) and calculate the peak e.m.f. \( V = V_{\text{p-p}} / 2 \). 7. Measure the period \( T \) of the alternating current waveform on the CRO and calculate the frequency \( f = 1 / T \). 8. Vary the frequency \( f \) of the signal generator to obtain several different values. At each frequency, adjust the rheostat to ensure the peak current \( I_0 \) remains constant, and measure the corresponding peak e.m.f. \( V \). 9. Plot a graph of \( V \) on the y-axis against \( f \) on the x-axis. A straight line passing through the origin confirms the suggested relationship. 10. The gradient \( m \) of the line of best fit is: \( m = \frac{\mu_0 \pi^2 d^2 N_1 N_2 I_0}{L} \). The permeability of free space is determined as: \( \mu_0 = \frac{m L}{\pi^2 d^2 N_1 N_2 I_0} \). 11. Safety: Solenoids carry high currents and can become very hot. Turn off the power supply between taking measurements to prevent overheating and potential burns.
Marking scheme
Defining the problem (2 marks): - M1: Identify \( f \) as the independent variable and \( V \) as the dependent variable. - M2: State that peak current \( I_0 \) is kept constant (accept also that \( d \), \( N_1 \), \( N_2 \), and \( L \) are kept constant).
Methods of data collection (4 marks): - M3: Draw a clear, labelled circuit diagram showing a signal generator connected in series with a solenoid, an AC ammeter, and a variable resistor; and the coaxial search coil connected directly to a cathode-ray oscilloscope (CRO). - M4: Describe how the peak e.m.f. \( V \) is measured using a CRO (e.g., measuring peak-to-peak height and dividing by 2). - M5: Describe how frequency \( f \) is measured using the CRO time-base (e.g., measuring period \( T \) and calculating \( f = 1/T \)). - M6: Describe measuring the diameter \( d \) of the search coil using a micrometer screw gauge or Vernier caliper, taking measurements along multiple diameters and taking the average.
Method of analysis (3 marks): - M7: Plot a graph of \( V \) against \( f \). - M8: State that the relationship is valid if the graph is a straight line through the origin. - M9: Express \( \mu_0 \) in terms of the gradient \( m \) as \( \mu_0 = \frac{m L}{\pi^2 d^2 N_1 N_2 I_0} \).
Safety considerations (1 mark): - M10: Precaution to avoid skin burns by switching off the current/signal generator between readings to prevent the solenoid from overheating.
Additional details (5 marks max): - A1: Use a variable resistor (rheostat) to adjust and keep the current constant as frequency is varied (since inductive reactance varies with frequency). - A2: Align the central axis of the search coil coaxially with the axis of the solenoid to ensure maximum magnetic flux linkage. - A3: Use shielded / coaxial cables for connections to the CRO to reduce high-frequency external noise / interference. - A4: Use a CRO with high input impedance to prevent loading of the search coil. - A5: Measure the length \( L \) of the solenoid winding using a metre rule.
Question 2 · practical-analysis
15 marks
A student investigates how the amplitude of swing of a magnetic pendulum is damped due to electromagnetic induction in a nearby conductor. A small aluminum plate is suspended from a rigid support and swings as a pendulum between the poles of an electromagnet. The electromagnet is connected to a variable direct current (d.c.) power supply. The amplitude \( A \) of the oscillations of the pendulum after a fixed time \( t = 10.0\text{ s} \) is measured for different currents \( I \) in the electromagnet. The relationship between \( A \) and \( I \) is suggested to be: \( A = A_0 e^{-k I^2} \) where \( A_0 \) is the initial amplitude at \( I = 0 \), and \( k \) is a constant.\ \ (a) State the equation of the straight line graph that should be plotted to test this relationship. Explain how the values of \( A_0 \) and \( k \) can be determined from the y-intercept and gradient of this graph. (2 marks)\ \ (b) Values of \( I \) and \( A \) are given in the table below. The absolute uncertainty in \( A \) is \( \pm 0.2\text{ cm} \). Calculate and record values of \( I^2 / \text{A}^2 \) and \( \ln(A / \text{cm}) \) in the table. Include the absolute uncertainties in \( \ln(A / \text{cm}) \). (3 marks)\ \ | \( I / \text{A} \) | \( I^2 / \text{A}^2 \) | \( A / \text{cm} \) | \( \ln(A / \text{cm}) \) |\ | :---: | :---: | :---: | :---: |\ | 0.50 | | 11.6 | |\ | 1.00 | | 10.3 | |\ | 1.50 | | 8.6 | |\ | 2.00 | | 6.6 | |\ | 2.50 | | 4.7 | |\ | 3.00 | | 3.1 | |\ \ (c) (i) Plot a graph of \( \ln(A / \text{cm}) \) against \( I^2 / \text{A}^2 \). Include error bars for \( \ln(A / \text{cm}) \). (2 marks)\ (ii) Draw the straight line of best fit and a worst acceptable straight line on your graph. Both lines should be clearly labelled. (1 mark)\ (iii) Determine the gradient of the line of best fit and the uncertainty in this gradient. (2 marks)\ (iv) Determine the y-intercept of the line of best fit and the uncertainty in this y-intercept. (2 marks)\ \ (d) Using your answers from (c)(iii) and (c)(iv), determine the values of \( k \) and \( A_0 \). Include appropriate units and absolute uncertainties for both constants. (3 marks)
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Worked solution
(a) Taking natural logarithms: \( \ln(A) = -k I^2 + \ln(A_0) \). By plotting \( \ln(A / \text{cm}) \) on the y-axis against \( I^2 / \text{A}^2 \) on the x-axis, we obtain a straight line where: - Gradient \( m = -k \) (or \( k = -m \)) - y-intercept \( c = \ln(A_0 / \text{cm}) \) (or \( A_0 = e^c \)).\ \ (b) The completed table values: - For \( I = 0.50 \): \( I^2 = 0.25 \), \( A = 11.6 \), \( \ln(A / \text{cm}) = 2.45 \pm 0.02 \) (uncertainty: \( \ln(11.8) - \ln(11.6) = 0.017 \approx 0.02 \)) - For \( I = 1.00 \): \( I^2 = 1.00 \), \( A = 10.3 \), \( \ln(A / \text{cm}) = 2.33 \pm 0.02 \) (uncertainty: \( \ln(10.5) - \ln(10.3) = 0.019 \approx 0.02 \)) - For \( I = 1.50 \): \( I^2 = 2.25 \), \( A = 8.6 \), \( \ln(A / \text{cm}) = 2.15 \pm 0.02 \) (uncertainty: \( \ln(8.8) - \ln(8.6) = 0.023 \approx 0.02 \)) - For \( I = 2.00 \): \( I^2 = 4.00 \), \( A = 6.6 \), \( \ln(A / \text{cm}) = 1.89 \pm 0.03 \) (uncertainty: \( \ln(6.8) - \ln(6.6) = 0.030 \approx 0.03 \)) - For \( I = 2.50 \): \( I^2 = 6.25 \), \( A = 4.7 \), \( \ln(A / \text{cm}) = 1.55 \pm 0.04 \) (uncertainty: \( \ln(4.9) - \ln(4.7) = 0.042 \approx 0.04 \)) - For \( I = 3.00 \): \( I^2 = 9.00 \), \( A = 3.1 \), \( \ln(A / \text{cm}) = 1.13 \pm 0.06 \) (uncertainty: \( \ln(3.3) - \ln(3.1) = 0.063 \approx 0.06 \)).\ \ (c) (i) Plotting: Points plotted accurately at (0.25, 2.45), (1.00, 2.33), (2.25, 2.15), (4.00, 1.89), (6.25, 1.55), (9.00, 1.13). Vertical error bars drawn of length \( \pm 0.02 \) to \( \pm 0.06 \).\ (ii) Lines: The best-fit line is drawn smoothly. The worst acceptable line is drawn passing through the bottom of the first error bar and the top of the last error bar (or vice versa).\ (iii) Gradient: - Gradient of best-fit line: \( m = \frac{1.13 - 2.45}{9.00 - 0.25} = \frac{-1.32}{8.75} = -0.151\text{ A}^{-2} \). - Worst acceptable gradient: \( m_{\text{worst}} = \frac{1.19 - 2.43}{9.00 - 0.25} = \frac{-1.24}{8.75} = -0.142\text{ A}^{-2} \). - Uncertainty in gradient: \( \Delta m = |m_{\text{best}} - m_{\text{worst}}| = 0.151 - 0.142 = 0.009\text{ A}^{-2} \) (or \( \pm 0.01\text{ A}^{-2} \).\ (iv) y-intercept: - y-intercept of best-fit line: \( c = 2.45 - (-0.151 \times 0.25) = 2.49 \). - y-intercept of worst acceptable line: \( c_{\text{worst}} = 2.43 - (-0.142 \times 0.25) = 2.47 \). - Uncertainty in y-intercept: \( \Delta c = |c_{\text{best}} - c_{\text{worst}}| = 2.49 - 2.47 = 0.02 \).\ \ (d) Determination of constants: - \( k = -m = 0.151 \pm 0.009\text{ A}^{-2} \) (or \( 0.15 \pm 0.01\text{ A}^{-2} \)). - \( A_0 = e^c = e^{2.49} = 12.1\text{ cm} \). - Uncertainty in \( A_0 \): \( \Delta A_0 = e^{c + \Delta c} - e^c = e^{2.51} - e^{2.49} = 12.3 - 12.1 = 0.2\text{ cm} \) (or \( \pm 0.3\text{ cm} \) using worst intercept difference). Thus, \( A_0 = 12.1 \pm 0.3\text{ cm} \).
Marking scheme
Part (a) (2 marks): - M1: State the equation \( \ln(A) = -k I^2 + \ln(A_0) \). - M2: State that \( k = -\text{gradient} \) and \( A_0 = e^{\text{y-intercept}} \).
Part (b) (3 marks): - M3: All values of \( I^2 \) calculated correctly: 0.25, 1.00, 2.25, 4.00, 6.25, 9.00 (allow 2 or 3 decimal places). - M4: All values of \( \ln(A / \text{cm}) \) calculated correctly: 2.45, 2.33, 2.15, 1.89, 1.55, 1.13. - M5: Absolute uncertainties in \( \ln(A / \text{cm}) \) calculated correctly: \( \pm 0.02, \pm 0.02, \pm 0.02, \pm 0.03, \pm 0.04, \pm 0.06 \).
Part (c) (7 marks): - M6: Plotting: All 6 points plotted correctly within half a small square. Error bars plotted correctly on the y-axis with correct lengths. - M7: Line of Best Fit: Drawn as a single straight line passing through all error bars. - M8: Worst Acceptable Line: Drawn as a straight line passing through the top of the first error bar and the bottom of the last error bar, or vice versa. - M9: Gradient of best-fit line calculated correctly using points on the line separated by at least half the length of the line. Expected value: \( -0.151 \pm 0.005 \). - M10: Uncertainty in gradient calculated correctly as \( |\text{best gradient} - \text{worst gradient}| \). Expected value: \( \pm 0.009 \) (or \( \pm 0.01 \)). - M11: y-intercept of best-fit line determined correctly. Expected value: \( 2.49 \pm 0.03 \). - M12: Uncertainty in y-intercept determined correctly as \( |\text{best intercept} - \text{worst intercept}| \). Expected value: \( \pm 0.02 \) to \( \pm 0.03 \).
Part (d) (3 marks): - M13: Value of \( k \) given with unit: \( 0.151 \pm 0.009\text{ A}^{-2} \) (or \( 0.15 \pm 0.01\text{ A}^{-2} \)). Unit must be \( \text{A}^{-2} \). - M14: Value of \( A_0 \) calculated correctly: \( 12.1\text{ cm} \) (allow range \( 11.9 \) to \( 12.2 \)). - M15: Uncertainty in \( A_0 \) calculated correctly using \( e^{c_{\text{best}}} - e^{c_{\text{worst}}} \) or equivalent. Expected value: \( \pm 0.3\text{ cm} \) (allow range \( \pm 0.2 \) to \( \pm 0.4\text{ cm} \)).
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