2024 AQA IAS-Level Physics (9630) 模擬試題連答案詳解

The volume of a cylindrical wire is given by \(V = \frac{\pi d^2 L}{4}\). The percentage uncertainty in the volume is: \(\%\Delta V = 2 \times \%\Delta d + \%\Delta L\). First, calculate the percentage uncertainty in the diameter: \(\%\Delta d = \frac{0.01\text{ mm}}{0.52\text{ mm}} \times 100\% \approx 1.923\%\). Double this contribution for the \(d^2\) term: \(2 \times 1.923\% = 3.846\%\). Next, calculate the percentage uncertainty in the length: \(\%\Delta L = \frac{0.1\text{ cm}}{84.5\text{ cm}} \times 100\% \approx 0.118\%\). Sum these to find the total percentage uncertainty: \(\%\Delta V = 3.846\% + 0.118\% = 3.964\% \approx 4.0\%\). To reduce the percentage uncertainty in the diameter, the student can take multiple readings of the diameter along different positions and perpendicular directions of the wire to calculate a mean diameter, and check for and subtract any zero error on the micrometer.

For part (a) [3.5 marks total]: 1.5 marks for calculating the percentage uncertainty in diameter (1.9%), 1.0 mark for doubling the percentage uncertainty for the d-squared term (3.8%), 1.0 mark for adding the percentage uncertainty of length to give a final answer of 4.0% (accept 3.96% to 4.0%). For part (b) [2.0 marks total]: 1.0 mark for suggesting taking multiple measurements along the wire and calculating a mean, 1.0 mark for checking for/correcting zero error on the micrometer.

(a) The Young modulus is defined as \(E = \frac{F L}{A \Delta L}\). Rearranging this formula for extension gives: \(\Delta L = \frac{F L}{A E}\). Substituting the given values: \(\Delta L = \frac{48 \times 2.4}{(3.2 \times 10^{-7}) \times (1.1 \times 10^{11})} = \frac{115.2}{35200} = 3.273 \times 10^{-3}\text{ m}\). In mm, this is \(3.27\text{ mm}\) (or \(3.3\text{ mm}\) to 2 significant figures). (b) Assuming Hooke's law is obeyed, the elastic strain energy stored is given by: \(E_s = \frac{1}{2} F \Delta L = 0.5 \times 48 \times 3.273 \times 10^{-3} = 0.0786\text{ J} \approx 0.079\text{ J}\) (or \(7.9 \times 10^{-2}\text{ J}\)).

For part (a) [3.5 marks total]: 1.5 marks for rearranging the Young modulus formula correctly to make extension the subject, 1.0 mark for correct substitution of values, 1.0 mark for calculating extension in mm (3.27 mm or 3.3 mm). For part (b) [2.0 marks total]: 1.0 mark for selecting and using the formula E_s = 0.5 * F * delta L, 1.0 mark for the correct calculation of strain energy as 0.079 J (or 0.0786 J).

(a) Since the heading is North and the wind is East, the two velocity vectors are perpendicular. The magnitude of the resultant velocity is found using Pythagoras' theorem: \(v = \sqrt{120^2 + 25^2} = \sqrt{14400 + 625} = \sqrt{15025} \approx 122.6\text{ m s}^{-1}\) (or \(123\text{ m s}^{-1}\)). (b) The angle \(\theta\) East of North is found using trigonometry: \(\tan \theta = \frac{25}{120} \implies \theta = \arctan(0.2083) \approx 11.8^\circ\). Expressed as a bearing, this is \(012^\circ\) (or \(12^\circ\)). (c) A scalar quantity has magnitude only, whereas a vector quantity has both magnitude and direction. An example of a scalar in mechanics is mass (or kinetic energy / work / time). An example of a vector is force (or displacement / acceleration / momentum).

For part (a) [2.0 marks]: 1.0 mark for applying Pythagoras' theorem, 1.0 mark for correct magnitude (123 m/s or 122.6 m/s). For part (b) [2.0 marks]: 1.0 mark for using tangent ratio, 1.0 mark for the correct bearing (011.8 or 012 degrees). For part (c) [1.5 marks]: 0.5 mark for defining scalar and vector, 0.5 mark for a valid scalar example, 0.5 mark for a valid vector example.

(a) Using the equation of motion \(v^2 = u^2 + 2as\) where \(u = 0\), \(v = 24\text{ m s}^{-1}\), and \(s = 120\text{ m}\): \(24^2 = 0 + 2 \times a \times 120 \implies 576 = 240 a \implies a = 2.4\text{ m s}^{-2}\). (b) Using \(v = u + at\): \(24 = 0 + 2.4 t \implies t = 10\text{ s}\). (Alternatively, \(s = \frac{u+v}{2} t \implies 120 = 12 t \implies t = 10\text{ s}\)). (c) The three parts of the velocity-time graph are described as follows: 1. A straight line with positive gradient starting from the origin \((0,0)\) up to \((10\text{ s}, 24\text{ m s}^{-1})\) representing uniform acceleration. 2. A horizontal line at \(24\text{ m s}^{-1}\) from \(10\text{ s}\) to \(15\text{ s}\) representing constant velocity. 3. A straight line with negative gradient from \((15\text{ s}, 24\text{ m s}^{-1})\) down to \((18\text{ s}, 0\text{ m s}^{-1})\) representing uniform deceleration.

For part (a) [2.0 marks]: 1.0 mark for choosing/using v^2 = u^2 + 2as, 1.0 mark for correct acceleration (2.4 m/s^2). For part (b) [2.0 marks]: 1.0 mark for using a correct suvat equation, 1.0 mark for the time of 10 s. For part (c) [1.5 marks]: 0.5 mark for describing each of the three phases with correct time intervals (0-10 s, 10-15 s, 15-18 s) and velocity values.

(a) Horizontal motion is at a constant velocity because air resistance is negligible. Using \(s_x = v_x \times t\): \(45 = 15 \times t \implies t = 3.0\text{ s}\). (b) Vertical motion starting from rest (\(u_y = 0\)): Using \(s_y = u_y t + \frac{1}{2} g t^2\): \(h = 0 + 0.5 \times 9.81 \times (3.0)^2 = 4.905 \times 9 = 44.15\text{ m} \approx 44\text{ m}\) (or \(44.1\text{ m}\)). (c) The vertical velocity component \(v_y\) is given by \(v_y = u_y + g t\). Substituting values: \(v_y = 0 + 9.81 \times 3.0 = 29.43\text{ m s}^{-1} \approx 29\text{ m s}^{-1}\) (or \(29.4\text{ m s}^{-1}\)).

For part (a) [1.5 marks]: 1.0 mark for using horizontal speed to find time, 0.5 mark for 3.0 s. For part (b) [2.0 marks]: 1.0 mark for using the vertical distance formula, 1.0 mark for calculating cliff height (44 m or 44.1 m). For part (c) [2.0 marks]: 1.0 mark for using v_y = gt, 1.0 mark for correct vertical velocity (29 m/s or 29.4 m/s).

(a) By conservation of momentum: \(m_A v_A + m_B v_B = (m_A + m_B) v\). Substituting values: \((1.2 \times 10^4 \times 4.0) + 0 = (1.2 \times 10^4 + 8.0 \times 10^3) v \implies 4.8 \times 10^4 = 2.0 \times 10^4 v \implies v = 2.4\text{ m s}^{-1}\). (b) Initial kinetic energy: \(E_{ki} = \frac{1}{2} m_A v_A^2 = 0.5 \times (1.2 \times 10^4) \times 4.0^2 = 9.60 \times 10^4\text{ J}\). Final kinetic energy: \(E_{kf} = \frac{1}{2} (m_A + m_B) v^2 = 0.5 \times (2.0 \times 10^4) \times 2.4^2 = 5.76 \times 10^4\text{ J}\). Loss in kinetic energy: \(\Delta E_k = E_{ki} - E_{kf} = 9.60 \times 10^4 - 5.76 \times 10^4 = 3.84 \times 10^4\text{ J}\) (or \(38.4\text{ kJ}\)). (c) The collision is inelastic, because kinetic energy is not conserved (it has decreased/transferred into thermal energy and sound).

For part (a) [2.0 marks]: 1.0 mark for applying conservation of momentum, 1.0 mark for correct final velocity (2.4 m/s). For part (b) [2.5 marks]: 1.0 mark for initial KE (96 kJ), 1.0 mark for final KE (57.6 kJ), 0.5 mark for the correct difference (38 kJ or 38.4 kJ). For part (c) [1.0 mark]: 1.0 mark for identifying the collision as inelastic because kinetic energy is not conserved.

(a) In a neutral Carbon-14 atom (\(^{14}_{6}\text{C}\)): the proton number \(Z = 6\), so there are 6 protons; the electron number is equal to the proton number, so there are 6 electrons; the neutron number \(N = A - Z = 14 - 6 = 8\) neutrons. (b) The specific charge is defined as the charge-to-mass ratio. For the nucleus: Charge \(Q = Z \times e = 6 \times 1.60 \times 10^{-19}\text{ C} = 9.60 \times 10^{-19}\text{ C}\). Mass \(M = A \times \text{nucleon mass} = 14 \times 1.67 \times 10^{-27}\text{ kg} = 2.338 \times 10^{-26}\text{ kg}\). Specific charge \(= \frac{Q}{M} = \frac{9.60 \times 10^{-19}}{2.338 \times 10^{-26}} = 4.106 \times 10^7\text{ C kg}^{-1}\) (or \(4.1 \times 10^7\text{ C kg}^{-1}\)). (c) Isotopes are nuclei/atoms of the same element with the same number of protons (atomic number) but a different number of neutrons (mass number).

For part (a) [1.5 marks total]: 0.5 mark for each correct value (6 protons, 8 neutrons, 6 electrons). For part (b) [3.0 marks total]: 1.0 mark for correct nuclear charge, 1.0 mark for correct nuclear mass, 1.0 mark for calculating correct specific charge with the correct unit (C kg^-1). For part (c) [1.0 mark]: 1.0 mark for defining isotopes clearly (same protons, different neutrons).

(a) The quark composition of a neutron is \(udd\) (one up quark, two down quarks) and the quark composition of a proton is \(uud\) (two up quarks, one down quark). (b) During beta-minus decay, one of the down quarks in the neutron changes into an up quark: \(d \to u\). (c) Conservation laws: 1. Baryon number (B): Neutron has \(B = +1\). Proton has \(B = +1\). Electron has \(B = 0\). Electron antineutrino has \(B = 0\). Total B before = \(+1\), Total B after = \(1 + 0 + 0 = 1\). Since \(1 = 1\), baryon number is conserved. 2. Lepton number (L): Neutron has \(L = 0\). Proton has \(L = 0\). Electron has \(L = +1\). Electron antineutrino has \(L = -1\). Total L before = \(0\), Total L after = \(0 + 1 + (-1) = 0\). Since \(0 = 0\), lepton number is conserved.

For part (a) [1.5 marks]: 0.75 mark for udd, 0.75 mark for uud. For part (b) [1.0 mark]: 1.0 mark for down to up quark change (d -> u). For part (c) [3.0 marks]: 1.5 marks for showing baryon number conservation with correct baryon numbers for all particles, 1.5 marks for showing lepton number conservation with correct lepton numbers (+1 for electron and -1 for antineutrino).

First, identify the forces acting on the shelf. The weight of the uniform shelf is \( W_s = 4.5 \text{ kg} \times 9.81 \text{ m s}^{-2} = 44.15 \text{ N} \), acting at its midpoint \( 0.40 \text{ m} \) from the hinge. The weight of the flowerpot is \( W_f = 12 \text{ kg} \times 9.81 \text{ m s}^{-2} = 117.72 \text{ N} \), acting at \( 0.60 \text{ m} \) from the hinge. Let \( T \) be the tension in the cable. The vertical component of tension is \( T \sin(35^\circ) \). Taking moments about the hinge: \( T \sin(35^\circ) \times 0.80 = (44.15 \times 0.40) + (117.72 \times 0.60) = 17.66 + 70.63 = 88.29 \text{ N m} \). Resolving this equation: \( T \times 0.4589 = 88.29 \), which yields \( T = 192.4 \text{ N} \). For part (b), resolve the forces vertically for equilibrium: \( V_h + T \sin(35^\circ) = W_s + W_f \). Since \( T \sin(35^\circ) = 110.36 \text{ N} \), we have \( V_h + 110.36 = 161.87 \text{ N} \), which gives \( V_h = 51.5 \text{ N} \).

(a) [3.5 marks total]: 1 mark for finding the sum of clockwise moments (\( 88.3 \text{ N m} \)). 1 mark for the correct moment equilibrium equation (\( T \sin(35^\circ) \times 0.80 = 88.3 \)). 1 mark for substitution and rearranging. 0.5 marks for correct tension (\( 192 \text{ N} \), accept range \( 190 \text{ N} \) to \( 193 \text{ N} \)). (b) [2 marks total]: 1 mark for vertical force equilibrium equation (\( V_h + T \sin(35^\circ) = W_s + W_f \)). 1 mark for correct final value of \( V_h \) (\( 51.5 \text{ N} \) or \( 52 \text{ N} \), accept range \( 51 \text{ N} \) to \( 52 \text{ N} \)).

For part (a): Carbon-14 has atomic number \( Z = 6 \), so it has 6 protons. The neutron number is \( N = 14 - 6 = 8 \). The specific charge is the ratio of charge to mass of the nucleus. Charge \( Q = 6 \times 1.60 \times 10^{-19} \text{ C} = 9.60 \times 10^{-19} \text{ C} \). Mass \( M = 14 \times 1.66 \times 10^{-27} \text{ kg} = 2.324 \times 10^{-26} \text{ kg} \). Specific charge = \( Q / M = (9.60 \times 10^{-19}) / (2.324 \times 10^{-26}) = 4.13 \times 10^7 \text{ C kg}^{-1} \). For part (b): Beta-minus decay converts a neutron into a proton. A neutron consists of one up quark and two down quarks (udd). A proton consists of two up quarks and one down quark (uud). Therefore, the overall change inside the nucleus is that one down quark converts into an up quark, with the emission of an electron and an electron antineutrino.

(a) [2.5 marks total]: 0.5 marks for proton number = 6. 0.5 marks for neutron number = 8. 1 mark for specific charge formula and substitution. 0.5 marks for correct specific charge value with unit (\( 4.1 \times 10^7 \text{ C kg}^{-1} \), accept \( 4.12 \times 10^7 \) to \( 4.14 \times 10^7 \)). (b) [3 marks total]: 1 mark for stating that a neutron changes to a proton. 1 mark for stating the correct quark configurations (udd for neutron, uud for proton). 1 mark for identifying that a down quark changes to an up quark.

For part (a): First calculate the cross-sectional area of the wire: \( A = \pi \times (d/2)^2 = \pi \times (0.14 \times 10^{-3} \text{ m})^2 = 6.1575 \times 10^{-8} \text{ m}^2 \). Using the Young modulus relation \( E = \frac{F L}{A \Delta L} \), rearrange to find the extension: \( \Delta L = \frac{F L}{A E} = \frac{55 \times 0.65}{6.1575 \times 10^{-8} \times 2.0 \times 10^{11}} = \frac{35.75}{12315} = 2.903 \times 10^{-3} \text{ m} \approx 2.9 \text{ mm} \). For part (b): Assuming Hooke's law is obeyed, the elastic strain energy is given by \( E_k = \frac{1}{2} F \Delta L = 0.5 \times 55 \times 2.903 \times 10^{-3} = 0.0798 \text{ J} \approx 0.080 \text{ J} \).

(a) [3.5 marks total]: 1 mark for correct cross-sectional area calculation. 1 mark for rearranging the Young modulus formula for extension. 1 mark for correct substitution. 0.5 marks for correct final extension (\( 2.9 \times 10^{-3} \text{ m} \) or \( 2.9 \text{ mm} \)). (b) [2 marks total]: 1 mark for the correct elastic strain energy formula (\( \frac{1}{2} F \Delta L \)). 1 mark for correct calculation of energy (\( 0.080 \text{ J} \), accept range \( 0.079 \text{ J} \) to \( 0.081 \text{ J} \)).

For part (a): The volume of the wire is given by \( V = \frac{\pi d^2 L}{4} \). The percentage uncertainty in volume is \( \frac{\Delta V}{V} \times 100 = 2 \times \left(\frac{\Delta d}{d} \times 100\right) + \left(\frac{\Delta L}{L} \times 100\right) \). Calculate percentage uncertainty in diameter: \( \frac{0.01}{0.84} \times 100 = 1.19\% \). Calculate percentage uncertainty in length: \( \frac{0.5}{82.5} \times 100 = 0.61\% \). Therefore, total percentage uncertainty in volume is \( 2 \times 1.19\% + 0.61\% = 2.99\% \approx 3.0\% \). For part (b): Density is given by \( \rho = \frac{m}{V} \). The percentage uncertainty in density is \( \frac{\Delta \rho}{\rho} \times 100 = \left(\frac{\Delta m}{m} \times 100\right) + \frac{\Delta V}{V}\% \). Calculate percentage uncertainty in mass: \( \frac{0.02}{4.18} \times 100 = 0.48\% \). Thus, total percentage uncertainty in density is \( 0.48\% + 2.99\% = 3.47\% \). Absolute uncertainty in density is \( 9.14 \times 10^3 \times 0.0347 = 317 \text{ kg m}^{-3} \approx 320 \text{ kg m}^{-3} \) (to 2 significant figures).

(a) [3.5 marks total]: 1 mark for stating that the percentage uncertainty in volume includes twice the percentage uncertainty in diameter plus the percentage uncertainty in length. 1 mark for calculating the percentage uncertainty in diameter (1.19%). 1 mark for calculating the percentage uncertainty in length (0.61%). 0.5 marks for correct sum (3.0% or 2.99%). (b) [2 marks total]: 1 mark for adding the percentage uncertainty of mass to get total percentage uncertainty in density (3.47% or 3.5%). 1 mark for calculating correct absolute uncertainty (\( 320 \text{ kg m}^{-3} \), accept range \( 310 \text{ kg m}^{-3} \) to \( 320 \text{ kg m}^{-3} \)).

To find the resultant force, resolve both forces into horizontal and vertical components. The horizontal component is \(F_x = 12 + 16 \cos(60^\circ) = 20\text{ N}\). The vertical component is \(F_y = 16 \sin(60^\circ) = 13.86\text{ N}\). The magnitude of the resultant is \(\sqrt{20^2 + 13.86^2} = 24.3\text{ N}\). The angle is \(\arctan(13.86 / 20) = 34.7^\circ\). Therefore, the correct option is A.

1 mark for the correct option A.

The component of the weight acting down the slope is \(W_{\parallel} = m g \sin(30^\circ) = 4.0 \times 9.81 \times \sin(30^\circ) = 19.62\text{ N}\). The net force acting down the slope is \(F = W_{\parallel} - F_{\text{friction}} = 19.62 - 6.0 = 13.62\text{ N}\). Using Newton's second law, \(a = F / m = 13.62 / 4.0 \approx 3.4\text{ m s}^{-2}\). Therefore, the correct option is B.

1 mark for the correct option B.

The vertical component of the initial velocity is \(u_y = 25 \sin(40^\circ) \approx 16.07\text{ m s}^{-1}\). At maximum height, the vertical velocity is \(v_y = 0\). Using \(v_y^2 = u_y^2 - 2 g h\), we get \(0 = (16.07)^2 - 2 \times 9.81 \times h\), which gives \(h = 16.07^2 / (2 \times 9.81) \approx 13.2\text{ m}\). This rounds to \(13\text{ m}\), which corresponds to option A.

1 mark for the correct option A.

Using conservation of momentum: \(m_1 u_1 = (m_1 + m_2) v \implies 2.0 \times 6.0 = (2.0 + 4.0) v \implies v = 2.0\text{ m s}^{-1}\). The initial kinetic energy is \(E_i = 0.5 \times m_1 u_1^2 = 0.5 \times 2.0 \times 6.0^2 = 36.0\text{ J}\). The final kinetic energy is \(E_f = 0.5 \times (m_1 + m_2) v^2 = 0.5 \times 6.0 \times 2.0^2 = 12.0\text{ J}\). The loss in kinetic energy is \(36.0 - 12.0 = 24.0\text{ J}\). This corresponds to option C.

1 mark for the correct option C.

The total upward force must equal the total downward force: \(T_A + T_B = W \implies 45 + T_B = 120 \implies T_B = 75\text{ N}\). Taking moments about end \(A\): \(W \times d = T_B \times 4.0\), where \(d\) is the distance of the centre of gravity from \(A\). This gives \(120 \times d = 75 \times 4.0 = 300 \implies d = 2.5\text{ m}\). This corresponds to option D.

1 mark for the correct option D.

Extension is given by \(\Delta L = \frac{F L}{A E}\). Since both wires are made of the same material and support the same load, \(E\) and \(F\) are constant. The area \(A\) is proportional to \(d^2\), where \(d\) is the diameter. Thus, \(\Delta L \propto \frac{L}{d^2}\). For wire \(Y\), the ratio of extension to that of wire \(X\) is \(\frac{\Delta L_Y}{\Delta L_X} = \frac{L_Y}{L_X} \times \left(\frac{d_X}{d_Y}\right)^2 = 2 \times 2^2 = 8\). This corresponds to option C.

1 mark for the correct option C.

The specific charge of a nucleus is the charge-to-mass ratio. For a carbon-14 nucleus, the charge is \(Q = 6e = 6 \times 1.60 \times 10^{-19}\text{ C} = 9.60 \times 10^{-19}\text{ C}\). The mass is \(m \approx 14 \times 1.67 \times 10^{-27}\text{ kg} = 2.338 \times 10^{-26}\text{ kg}\). The specific charge is \(\frac{Q}{m} = \frac{9.60 \times 10^{-19}\text{ C}}{2.338 \times 10^{-26}\text{ kg}} \approx 4.11 \times 10^7\text{ C kg}^{-1}\). This matches option A.

1 mark for the correct option A.

During \(\beta^-\) decay, a neutron (\(udd\)) is converted into a proton (\(uud\)), meaning a down quark is converted into an up quark (\(d \rightarrow u\)). This process is mediated by the weak interaction with the emission of a \(W^-\) boson, which then decays into an electron and an electron antineutrino. This matches option D.

1 mark for the correct option D.

The resistivity \(\rho\) is calculated from the formula: \(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\). The fractional uncertainty in \(\rho\) is the sum of the fractional uncertainties of the independent variables: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\). Calculating the percentage uncertainties: Percentage uncertainty in \(R = \frac{0.6}{24.0} \times 100\% = 2.5\%\). Percentage uncertainty in \(d = \frac{0.02}{0.40} \times 100\% = 5.0\%\). Percentage uncertainty in \(L = \frac{0.03}{1.50} \times 100\% = 2.0\%\). Thus, percentage uncertainty in \(\rho = 2.5\% + 2(5.0\%) + 2.0\% = 14.5\%\).

1 mark: For the correct identification of the percentage uncertainty of each quantity and combining them correctly, including doubling the percentage uncertainty of the diameter.

The vector relation is: \(\vec{v}_{\text{resultant}} = \vec{v}_{\text{aircraft}} + \vec{v}_{\text{wind}}\). Since \(\vec{v}_{\text{resultant}}\) is due North (magnitude \(120\text{ m s}^{-1}\)) and \(\vec{v}_{\text{wind}}\) is due East (magnitude \(50\text{ m s}^{-1}\)), the aircraft's velocity vector relative to the air must form the hypotenuse of a right-angled triangle. \(v_{\text{aircraft}} = \sqrt{120^2 + 50^2} = 130\text{ m s}^{-1}\). To cancel the eastward wind, the aircraft must head at an angle \(\theta\) West of North: \(\tan \theta = \frac{50}{120} \implies \theta \approx 23^\circ\).

1 mark: For the correct application of Pythagoras' theorem to find the magnitude and trigonometric resolution to find the direction.

Stage 1 (Acceleration): \(v_{\text{max}} = a_1 t_1 = 2.0 \times 6.0 = 12\text{ m s}^{-1}\). Distance \(s_1 = \frac{1}{2} a_1 t_1^2 = 0.5 \times 2.0 \times 6.0^2 = 36\text{ m}\). Stage 3 (Deceleration): Time taken \(t_3 = \frac{v_{\text{max}}}{a_3} = \frac{12}{3.0} = 4.0\text{ s}\). Distance \(s_3 = \frac{1}{2} v_{\text{max}} t_3 = 0.5 \times 12 \times 4.0 = 24\text{ m}\). Stage 2 (Constant velocity): Distance \(s_2 = s_{\text{total}} - s_1 - s_3 = 240 - 36 - 24 = 180\text{ m}\). Time taken \(t_2 = \frac{s_2}{v_{\text{max}}} = \frac{180}{12} = 15\text{ s}\). Total time \(T = t_1 + t_2 + t_3 = 6.0 + 15 + 4.0 = 25\text{ s}\).

1 mark: For calculating the times and distances for all three stages and summing the times to find the correct total time of 25 seconds.

The tension \(F\) in a wire extended by \(e\) is given by \(F = \frac{E A e}{L}\). Since both wires are made of the same material, the Young modulus \(E\) is the same. The extension \(e\) is also identical. Since wire X has twice the diameter of wire Y, its cross-sectional area is four times larger: \(A_X = 4 A_Y\). Wire X has half the length of wire Y: \(L_X = 0.5 L_Y\). Therefore, the tension in wire X is: \(F_X = \frac{E (4 A_Y) e}{0.5 L_Y} = 8 \left(\frac{E A_Y e}{L_Y}\right) = 8 F_Y\). The elastic strain energy \(W\) stored in each wire is \(W = \frac{1}{2} F e\). Since \(e\) is identical for both wires, the ratio of the energy stored is \(\frac{W_X}{W_Y} = \frac{F_X}{F_Y} = 8\).

1 mark: For identifying the relationship between area, length and force, and correctly determining that the ratio of stored elastic energy is equal to the ratio of the forces.

The specific charge is defined as the ratio of charge to mass: \(\text{Specific charge} = \frac{Q}{m}\). A singly-ionized carbon-12 atom has lost one electron, giving it a net charge \(Q = +1e = 1.60 \times 10^{-19}\text{ C}\). The mass of the carbon-12 ion is approximately equal to the mass of its 12 nucleons: \(m \approx 12 \times 1.66 \times 10^{-27}\text{ kg} = 1.992 \times 10^{-26}\text{ kg}\). Therefore, \(\text{Specific charge} = \frac{1.60 \times 10^{-19}\text{ C}}{1.992 \times 10^{-26}\text{ kg}} \approx 8.03 \times 10^6\text{ C kg}^{-1}\).

1 mark: For identifying the charge of a singly ionized atom and using the total nucleon mass to calculate the specific charge.

The properties of the quarks are:
- Up quark (\(u\)): charge = \(+\frac{2}{3}e\), baryon number = \(+\frac{1}{3}\), strangeness = \(0\).
- Strange quark (\(s\)): charge = \(-\frac{1}{3}e\), baryon number = \(+\frac{1}{3}\), strangeness = \(-1\).

For the combination \(uss\):
- Total charge \(Q = \left(+\frac{2}{3} - \frac{1}{3} - \frac{1}{3}\right) e = 0\).
- Baryon number \(B = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1\).
- Strangeness \(S = 0 - 1 - 1 = -2\).

1 mark: For the correct determination of charge, baryon number, and strangeness of the three-quark system.

(a) The energy of an incident photon is given by:
\( E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{320 \times 10^{-9}} = 6.22 \times 10^{-19}\text{ J} \).

(b) First, convert the work function of lithium to joules:
\( \Phi = 2.30 \times 1.60 \times 10^{-19}\text{ J} = 3.68 \times 10^{-19}\text{ J} \).
Using the photoelectric equation:
\( E_{k,\text{max}} = E - \Phi = 6.22 \times 10^{-19} - 3.68 \times 10^{-19} = 2.54 \times 10^{-19}\text{ J} \).

(c) The stopping potential \( V_s \) is related to the maximum kinetic energy by:
\( E_{k,\text{max}} = e V_s \implies V_s = \frac{2.54 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ C}} = 1.59\text{ V} \).

(a) [2 marks]
- 1 mark for using \( E = \frac{hc}{\lambda} \) with correct substitution.
- 1 mark for correct value to 3 s.f. (\( 6.22 \times 10^{-19}\text{ J} \)).

(b) [2.6 marks]
- 1 mark for converting work function to joules: \( 3.68 \times 10^{-19}\text{ J} \).
- 1 mark for applying photoelectric equation: \( E_{k,\text{max}} = E - \Phi \).
- 0.6 marks for correct final value of kinetic energy (\( 2.54 \times 10^{-19}\text{ J} \), allow ecf from part a).

(c) [2 marks]
- 1 mark for using \( V_s = \frac{E_k}{e} \).
- 1 mark for correct stopping potential (\( 1.59\text{ V} \), allow range 1.58 V to 1.60 V depending on rounding, ecf from part b).

1. Calculate original cross-sectional area:
\( A = \pi r^2 = \pi (0.400 \times 10^{-3}\text{ m})^2 = 5.027 \times 10^{-7}\text{ m}^2 \).

2. Since volume \( V = A \times L \) is constant, if length increases by 15% (new length \( L' = 1.15 L = 1.725\text{ m} \)), the cross-sectional area must decrease by the same ratio:
\( A' = \frac{A}{1.15} = \frac{5.027 \times 10^{-7}}{1.15} = 4.371 \times 10^{-7}\text{ m}^2 \).

3. Calculate the new resistance \( R' \):
\( R' = \rho \frac{L'}{A'} = 1.70 \times 10^{-8} \times \frac{1.725}{4.371 \times 10^{-7}} = 0.0671\ \Omega \).

Alternatively, since \( R' = \rho \frac{L'}{A'} = \rho \frac{1.15 L}{A/1.15} = 1.15^2 R \):
\( R = 1.70 \times 10^{-8} \times \frac{1.50}{5.027 \times 10^{-7}} = 0.05073\ \Omega \).
\( R' = 1.15^2 \times 0.05073 = 1.3225 \times 0.05073 = 0.0671\ \Omega \).

[6.6 marks total]
- 1.5 marks for calculating original cross-sectional area or initial resistance.
- 1.5 marks for realizing volume conservation requires \( A' = A / 1.15 \) or that \( R' = 1.15^2 \times R \).
- 2.0 marks for setting up the equation for \( R' \) with correct values.
- 1.6 marks for final correct answer of \( 0.0671\ \Omega \) (or \( 0.067\ \Omega \)) with correct units.

(a) For the third harmonic of a wire fixed at both ends, the wire length contains 3 half-wavelengths:
\( L = \frac{3}{2}\lambda \implies \lambda = \frac{2}{3}L = \frac{2}{3} \times 0.850 = 0.567\text{ m} \).

(b) The speed \( v \) of the wave is:
\( v = f \lambda = 150 \times 0.567 = 85.0\text{ m s}^{-1} \).

(c) The mass per unit length \( \mu \) of the wire is:
\( \mu = \frac{m}{L} = \frac{3.40 \times 10^{-3}\text{ kg}}{0.850\text{ m}} = 4.00 \times 10^{-3}\text{ kg m}^{-1} \).
Using the formula for wave speed on a stretched string:
\( v = \sqrt{\frac{T}{\mu}} \implies T = v^2 \mu = 85.0^2 \times 4.00 \times 10^{-3} = 7225 \times 4.00 \times 10^{-3} = 28.9\text{ N} \).

(a) [2 marks]
- 1 mark for realizing \( \lambda = \frac{2}{3}L \).
- 1 mark for calculating \( 0.567\text{ m} \) (or \( 0.57\text{ m} \)).

(b) [2 marks]
- 1 mark for using \( v = f \lambda \).
- 1 mark for calculating \( 85.0\text{ m s}^{-1} \) (accept ecf from a).

(c) [2.6 marks]
- 1 mark for calculating mass per unit length \( \mu = 4.00 \times 10^{-3}\text{ kg m}^{-1} \).
- 1 mark for using \( T = v^2 \mu \).
- 0.6 marks for final answer of \( 28.9\text{ N} \) (accept range 28.8 N to 29.0 N, ecf allowed).

(a) Using the equation \( \varepsilon = I(R + r) \) for both scenarios:
For \( R_1 = 4.50\ \Omega \):
\( \varepsilon = 0.300(4.50 + r) = 1.35 + 0.300r \quad (1) \)
For \( R_2 = 12.00\ \Omega \):
\( \varepsilon = 0.120(12.00 + r) = 1.44 + 0.120r \quad (2) \)

Equating (1) and (2):
\( 1.35 + 0.300r = 1.44 + 0.120r \)
\( 0.180r = 0.090 \implies r = 0.50\ \Omega \).

(b) Substitute \( r = 0.50\ \Omega \) back into equation (1):
\( \varepsilon = 0.300(4.50 + 0.50) = 1.50\text{ V} \).

(c) Efficiency \( \eta \) of energy transfer is the ratio of useful output power to total power:
\( \eta = \frac{I^2 R_2}{I^2(R_2 + r)} = \frac{R_2}{R_2 + r} \)
\( \eta = \frac{12.00}{12.00 + 0.50} = \frac{12.00}{12.50} = 0.960 \text{ or } 96.0\% \).

(a) [2.5 marks]
- 1 mark for establishing two simultaneous equations using \( \varepsilon = I(R+r) \).
- 1 mark for algebraic manipulation to eliminate \( \varepsilon \).
- 0.5 marks for correct internal resistance of \( 0.50\ \Omega \).

(b) [2 marks]
- 1 mark for substituting \( r \) into either starting equation.
- 1 mark for correct electromotive force of \( 1.50\text{ V} \).

(c) [2.1 marks]
- 1 mark for expressing efficiency as \( \frac{V}{\varepsilon} \) or \( \frac{R}{R+r} \) or using \( P_{\text{out}}/P_{\text{total}} \).
- 1.1 marks for correct final percentage of \( 96.0\% \) (or 0.96, accept ecf).

(a) Using Snell's law at the first boundary:
\( n_1 \sin \theta_1 = n_2 \sin \theta_2 \)
\( 1.00 \sin(42.0^\circ) = 1.55 \sin \theta_2 \)
\( \sin \theta_2 = \frac{0.6691}{1.55} = 0.4317 \implies \theta_2 = 25.6^\circ \).

(b) For a triangular prism, the relation between the apex angle \( A \) and the internal angles is:
\( A = \theta_2 + \theta_3 \)
where \( \theta_3 \) is the angle of incidence on the second face inside the glass.
\( 60.0^\circ = 25.58^\circ + \theta_3 \implies \theta_3 = 34.42^\circ \approx 34.4^\circ \).

(c) The critical angle \( \theta_c \) at the glass-air boundary is given by:
\( \sin \theta_c = \frac{1}{n} = \frac{1}{1.55} = 0.6452 \implies \theta_c = 40.2^\circ \).
Comparing the angle of incidence at the second boundary (\( 34.4^\circ \)) with the critical angle (\( 40.2^\circ \)):
Since \( 34.4^\circ < 40.2^\circ \), the light ray does NOT undergo total internal reflection. It will refract out into the air.

(a) [2 marks]
- 1 mark for correct substitution into Snell's law: \( \sin \theta_2 = \frac{\sin(42.0)}{1.55} \).
- 1 mark for correct calculation of \( 25.6^\circ \) (allow 25.5° - 25.7°).

(b) [2 marks]
- 1 mark for utilizing prism geometry equation \( A = \theta_2 + \theta_3 \).
- 1 mark for showing \( \theta_3 = 60.0^\circ - 25.6^\circ = 34.4^\circ \).

(c) [2.6 marks]
- 1 mark for correct formula and calculation of critical angle \( \theta_c = 40.2^\circ \) (allow 40°).
- 1 mark for comparison statement: comparing internal incidence angle to critical angle.
- 0.6 marks for final correct conclusion that total internal reflection does not occur (ray escapes).

(a) Total resistance in the dark:
\( R_{\text{total}} = 2.20\text{ k}\Omega + 18.0\text{ k}\Omega = 20.20\text{ k}\Omega = 20200\ \Omega \).
The output voltage across the fixed resistor is:
\( V_{\text{out}} = V_{\text{in}} \left( \frac{R_1}{R_{\text{total}}} \right) = 9.00 \times \left( \frac{2200}{20200} \right) = 0.980\text{ V} \).

(b) Total resistance in bright light:
\( R_{\text{total}} = 2.20\text{ k}\Omega + 0.450\text{ k}\Omega = 2.65\text{ k}\Omega = 2650\ \Omega \).
The output voltage across the fixed resistor is:
\( V_{\text{out}} = 9.00 \times \left( \frac{2200}{2650} \right) = 7.47\text{ V} \).

(c) Voltage across the LDR in bright light:
\( V_{\text{LDR}} = V_{\text{in}} - V_{\text{out}} = 9.00 - 7.47 = 1.53\text{ V} \).
Alternatively, using current in bright light:
\( I = \frac{V_{\text{in}}}{R_{\text{total}}} = \frac{9.00}{2650} = 3.396 \times 10^{-3}\text{ A} \).
Power dissipated by the LDR:
\( P = I^2 R_{\text{LDR}} = (3.396 \times 10^{-3})^2 \times 450 = 5.19 \times 10^{-3}\text{ W} = 5.19\text{ mW} \).

(a) [2 marks]
- 1 mark for finding dark total resistance (\( 20.2\text{ k}\Omega \)) and setting up the ratio.
- 1 mark for correct calculation of \( 0.980\text{ V} \) (accept 0.98 V).

(b) [2 marks]
- 1 mark for finding bright total resistance (\( 2.65\text{ k}\Omega \)) and setting up the ratio.
- 1 mark for correct calculation of \( 7.47\text{ V} \) (accept 7.5 V).

(c) [2.6 marks]
- 1 mark for finding either the voltage across the LDR (\( 1.53\text{ V} \)) or the circuit current (\( 3.40\text{ mA} \)).
- 1 mark for applying a correct power formula: \( P = I^2 R \) or \( P = \frac{V^2}{R} \) or \( P = IV \).
- 0.6 marks for final answer of \( 5.19\text{ mW} \) (accept range 5.1 mW to 5.3 mW).

(a) Using the de Broglie relation:
\( p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s}}{0.120 \times 10^{-9}\text{ m}} = 5.525 \times 10^{-24}\text{ kg m s}^{-1} \approx 5.53 \times 10^{-24}\text{ kg m s}^{-1} \).

(b) The kinetic energy \( E_k \) of the electron of mass \( m_e = 9.11 \times 10^{-31}\text{ kg} \) is:
\( E_k = \frac{p^2}{2 m_e} = \frac{(5.525 \times 10^{-24})^2}{2 \times 9.11 \times 10^{-31}} = \frac{3.053 \times 10^{-47}}{1.822 \times 10^{-30}} = 1.676 \times 10^{-17}\text{ J} \approx 1.68 \times 10^{-17}\text{ J} \).

(c) The kinetic energy gained is equal to the work done by the accelerating potential difference:
\( E_k = e V \implies V = \frac{E_k}{e} = \frac{1.676 \times 10^{-17}\text{ J}}{1.60 \times 10^{-19}\text{ C}} = 104.7\text{ V} \approx 105\text{ V} \).

(a) [2 marks]
- 1 mark for using \( p = \frac{h}{\lambda} \) with correct units conversion (nm to m).
- 1 mark for correct answer of \( 5.53 \times 10^{-24}\text{ kg m s}^{-1} \).

(b) [2 marks]
- 1 mark for using \( E_k = \frac{p^2}{2m} \) or \( E_k = \frac{1}{2}mv^2 \) by first finding velocity.
- 1 mark for correct calculation of \( 1.68 \times 10^{-17}\text{ J} \) (accept ecf from a).

(c) [2.6 marks]
- 1 mark for utilizing \( E_k = eV \).
- 1 mark for rearranging to \( V = E_k / e \).
- 0.6 marks for final correct potential difference of \( 105\text{ V} \) (accept 104 V to 105 V, ecf from b).

(a) The fringe width \( w \) is given by:
\( w = \frac{\lambda D}{s} \)
\( w = \frac{632.8 \times 10^{-9}\text{ m} \times 2.40\text{ m}}{0.250 \times 10^{-3}\text{ m}} = 6.075 \times 10^{-3}\text{ m} \approx 6.08\text{ mm} \).

(b) The distance to the dark fringes from the central maximum is given by:
\( y_m = \left(m + \frac{1}{2}\right) w \)
For the first dark fringe, \( m = 0 \) (distance \( 0.5 w \)).
For the second dark fringe, \( m = 1 \) (distance \( 1.5 w \)).
For the third dark fringe, \( m = 2 \) (distance \( 2.5 w \)).
So the distance is:
\( d = 2.5 \times 6.075 \times 10^{-3}\text{ m} = 1.519 \times 10^{-2}\text{ m} \approx 15.2\text{ mm} \).

(c) The formula for fringe width is \( w = \frac{\lambda D}{s} \). Since \( s \) is in the denominator, if \( s \) is halved, the fringe width \( w \) will double. The bright and dark bands will become wider and more widely spaced.

(a) [2 marks]
- 1 mark for using \( w = \frac{\lambda D}{s} \) with correct unit conversions.
- 1 mark for correct calculation of \( 6.08\text{ mm} \).

(b) [2.6 marks]
- 1 mark for stating that the third dark fringe is at a distance of \( 2.5 w \) (or 2.5 fringe widths).
- 1 mark for a clear calculation steps multiplying by 2.5.
- 0.6 marks for final correct distance of \( 15.2\text{ mm} \) (accept ecf from a).

(c) [2 marks]
- 1 mark for stating that the fringe separation doubles (or increases by a factor of 2).
- 1 mark for explaining that the pattern becomes wider / more spread out.

**Part (a)**
The cross-sectional area \(A\) of the wire is:
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.38 \times 10^{-3}\text{ m})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\)

The resistivity \(\rho\) is given by:
\(\rho = \frac{R A}{L} = \frac{15.2\ \Omega \times 1.134 \times 10^{-7}\text{ m}^2}{2.4\text{ m}} = 7.18 \times 10^{-7}\ \Omega\text{ m}\)
This is approximately \(7.2 \times 10^{-7}\ \Omega\text{ m}\).

**Part (b)**
The new length of the wire is:
\(L' = 1.15 \times L\)

Since the volume \(V = A \times L\) is constant:
\(A' \times L' = A \times L \implies A' = \frac{A}{1.15}\)

The new resistance \(R'\) is:
\(R' = \rho \frac{L'}{A'} = \rho \frac{1.15 L}{A / 1.15} = 1.15^2 \left(\rho \frac{L}{A}\right) = 1.15^2 R\)
\(R' = 1.3225 \times 15.2\ \Omega = 20.102\ \Omega \approx 20.1\ \Omega\)

**Part (a) [3.0 Marks]**
* **[1 mark]** For calculating or using the correct cross-sectional area of the wire: \(A = 1.13 \times 10^{-7}\text{ m}^2\).
* **[1 mark]** For rearranging the resistivity equation and substituting values: \(\rho = \frac{15.2 \times 1.13 \times 10^{-7}}{2.4}\).
* **[1 mark]** For obtaining the final value of \(\approx 7.2 \times 10^{-7}\ \Omega\text{ m}\) (must show calculation with at least 3 significant figures: \(7.18 \times 10^{-7}\ \Omega\text{ m}\)).

**Part (b) [3.6 Marks]**
* **[1.6 marks]** For applying the conservation of volume to find the new area: \(A' = \frac{A}{1.15}\) (or finding \(L' = 2.76\text{ m}\) and \(A' = 9.86 \times 10^{-8}\text{ m}^2\)).
* **[1.0 mark]** For substituting the new length and area into the resistance formula, or showing that \(R' = 1.15^2 \times R\).
* **[1.0 mark]** For calculating the correct new resistance: \(20.1\ \Omega\) (accept answers in range \(20.0\ \Omega\) to \(20.1\ \Omega\)).

**Part (a)**
First, calculate the energy \(E\) of the incident photons:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{320 \times 10^{-9}\text{ m}} = 6.216 \times 10^{-19}\text{ J}\)

Next, convert the work function \(\Phi\) of caesium from \(\text{eV}\) to Joules (\(\text{J}\)):
\(\Phi = 2.14\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 3.424 \times 10^{-19}\text{ J}\)

Using Einstein’s photoelectric equation:
\(E_{k,\text{max}} = E - \Phi\)
\(E_{k,\text{max}} = 6.216 \times 10^{-19}\text{ J} - 3.424 \times 10^{-19}\text{ J} = 2.792 \times 10^{-19}\text{ J} \approx 2.79 \times 10^{-19}\text{ J}\)

**Part (b)**
The stopping potential \(V_s\) is related to the maximum kinetic energy by:
\(E_{k,\text{max}} = e V_s\)

Therefore:
\(V_s = \frac{E_{k,\text{max}}}{e} = \frac{2.792 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ C}} = 1.745\text{ V} \approx 1.74\text{ V}\) (or \(1.75\text{ V}\))

**Part (a) [3.6 Marks]**
* **[1.6 marks]** For calculating the energy of the photon using \(E = \frac{hc}{\lambda}\) to get \(6.22 \times 10^{-19}\text{ J}\).
* **[1.0 mark]** For converting the work function to Joules: \(\Phi = 3.42 \times 10^{-19}\text{ J}\).
* **[1.0 mark]** For calculating the correct maximum kinetic energy: \(2.79 \times 10^{-19}\text{ J}\) (allow \(2.8 \times 10^{-19}\text{ J}\)).

**Part (b) [3.0 Marks]**
* **[1.0 mark]** For stating or using the relation \(E_{k,\text{max}} = e V_s\).
* **[1.0 mark]** For substituting the values: \(V_s = \frac{2.79 \times 10^{-19}}{1.60 \times 10^{-19}}\) (or finding \(3.88\text{ eV} - 2.14\text{ eV} = 1.74\text{ eV}\)).
* **[1.0 mark]** For the correct stopping potential: \(1.74\text{ V}\) or \(1.75\text{ V}\) (accept answers in range \(1.7\text{ V}\) to \(1.8\text{ V}\)).

Let the original length be L0, original cross-sectional area be A0, and the original resistance be R = \rho \frac{L_0}{A_0}. When the wire is stretched, its length increases to L1 = 1.10 L0. Since the volume V = A \times L remains constant, we have A0 L0 = A1 L1, which gives A1 = A0 \frac{L_0}{L_1} = \frac{A_0}{1.10}. The new resistance R1 is given by R1 = \rho \frac{L_1}{A_1} = \rho \frac{1.10 L_0}{A_0 / 1.10} = 1.10^2 \times \rho \frac{L_0}{A_0} = 1.21 R.

1 mark for the correct answer C. Volume conservation implies the cross-sectional area is inversely proportional to the length, leading to the resistance being proportional to the square of the length.

From Einstein's photoelectric equation, we have Ek = \frac{hc}{\lambda} - \Phi, where \Phi is the work function of the metal. When the wavelength is halved, the new photon energy is \frac{hc}{0.5\lambda} = 2\frac{hc}{\lambda}. The new maximum kinetic energy Ek' is given by Ek' = 2\frac{hc}{\lambda} - \Phi = 2(Ek + \Phi) - \Phi = 2Ek + \Phi. Since the work function \Phi > 0, it follows that Ek' > 2Ek.

1 mark for the correct answer C. Applying Einstein's photoelectric equation shows that halving the wavelength doubles the incident photon energy, resulting in a maximum kinetic energy of 2Ek + \Phi, which is greater than 2Ek.

Using the potential divider formula, the output voltage Vout across the LDR is Vout = Vin \times \frac{RLDR}{Rfixed + RLDR}. In bright light, Vout, bright = 12.0 \times \frac{2.0}{4.0 + 2.0} = 4.0 V. In darkness, Vout, dark = 12.0 \times \frac{8.0}{4.0 + 8.0} = 8.0 V. The change in the voltmeter reading is \Delta V = 8.0 V - 4.0 V = 4.0 V.

1 mark for the correct answer B. Correctly calculating the potential difference across the LDR in both states and finding their difference.

The de Broglie wavelength is \lambda = \frac{h}{p}. Since their wavelengths are equal, their momenta must be equal, so pe = pp. The non-relativistic kinetic energy is given by E = \frac{p^2}{2m}. Since they have equal momenta and the mass of the electron is much smaller than the mass of the proton (me < mp), the kinetic energy of the electron must be greater than that of the proton: Ee > Ep.

1 mark for the correct answer B. Recognizing that equal de Broglie wavelength implies equal momentum, and using E = p^2/2m to compare kinetic energies based on mass.

By Snell's law at the air-glass interface, n_air \sin(\theta_air) = n_glass \sin(\theta_glass). At the glass-liquid interface, n_glass \sin(\theta_glass) = n_liquid \sin(\theta_liquid). Combining these gives n_air \sin(\theta_air) = n_liquid \sin(\theta_liquid). Substituting the given values, we get 1.00 \times \sin(45.0 degrees) = 1.33 \times \sin(\theta_liquid), which yields \sin(\theta_liquid) = \frac{\sin(45.0 degrees)}{1.33} \approx 0.5317, leading to \theta_liquid \approx 32.1 degrees.

1 mark for the correct answer B. Showing that the intermediate glass layer does not affect the final angle of refraction and solving the relation between the air and liquid boundaries.

The position of the m-th order bright fringe in a double-slit system is given by ym = \frac{m \lambda D}{d}. For both fringes to be at the same position, we have \frac{5 \lambda_1 D}{d} = \frac{4 \lambda_2 D}{d}, which simplifies to 5 \lambda_1 = 4 \lambda_2. Solving for the new wavelength \lambda_2, we get \lambda_2 = \frac{5}{4} \times 600 nm = 750 nm.

1 mark for the correct answer D. Formulating the relation between order number and wavelength and solving for the second wavelength.

The equivalent resistance of the two parallel resistors is Rp = \frac{R}{2}. Since this is in series with the third resistor of resistance R, the total equivalent resistance of the circuit is Rtotal = Rp + R = 1.5 R = \frac{3}{2} R. The total power P dissipated in the circuit is P = \frac{E^2}{Rtotal} = \frac{E^2}{\frac{3}{2} R} = \frac{2 E^2}{3 R}.

1 mark for the correct answer A. Finding the correct equivalent resistance of the network and using it to calculate total power.

A baryon contains three quarks. Since its strangeness is -1, it must contain exactly one strange (s) quark, which has a charge of -\frac{1}{3}e. Let the charges of the remaining two quarks be q1 and q2. The total charge of the baryon is +1e, so q1 + q2 - \frac{1}{3}e = +1e, which simplifies to q1 + q2 = +\frac{4}{3}e. Since up (u) quarks have charge +\frac{2}{3}e and down (d) quarks have charge -\frac{1}{3}e, the only combination that can yield a sum of +\frac{4}{3}e is two up quarks (uu). Thus, the quark composition is u u s.

1 mark for the correct answer C. Correctly identifying the quark content using strangeness and electric charge conservation.

The resistance of a wire is given by \(R = \rho \frac{L}{A}\). Since the volume \(V = A L\) is constant, we can write the cross-sectional area as \(A = \frac{V}{L}\). Substituting this into the resistance formula gives \(R = \rho \frac{L^2}{V}\). Since resistivity \(\rho\) and volume \(V\) remain constant, the resistance is directly proportional to the square of the length: \(R \propto L^2\). If the length increases by \(2.0\%\), the new length is \(1.02 L\). Therefore, the new resistance is \(R' = 1.02^2 R = 1.0404 R\). The percentage increase in resistance is \((1.0404 - 1) \times 100\% = 4.04\%\), which is approximately \(4.0\%\).

1 mark for the correct option C. Deduct 0 marks for incorrect choices.

According to Einstein's photoelectric equation, \(E_{\text{max}} = hf - \Phi\), where \(\Phi\) is the work function of the metal. When the frequency of the incident light is doubled to \(2f\), the new maximum kinetic energy \(E_{\text{max}}'\) is \(E_{\text{max}}' = h(2f) - \Phi = 2hf - \Phi\). Substituting \(hf = E_{\text{max}} + \Phi\) into this equation yields \(E_{\text{max}}' = 2(E_{\text{max}} + \Phi) - \Phi = 2E_{\text{max}} + \Phi\). Since the work function \(\Phi\) is a positive physical quantity, \(E_{\text{max}}'\) is greater than \(2E_{\text{max}}\).

1 mark for selecting the correct option C.

The fringe spacing in a double-slit setup is given by the formula \(w = \frac{\lambda D}{d}\). If the new slit separation is \(d' = \frac{d}{2}\) and the new screen distance is \(D' = 2D\), the new fringe width \(w'\) becomes \(w' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{d/2} = 4 \frac{\lambda D}{d} = 4w\).

1 mark for the correct answer option D.

The negative pion (\(\pi^-\)) is a meson, which means it consists of a quark and an antiquark. It has a charge of \(-1e\) and a strangeness of \(0\). The charge of an up quark \(u\) is \(+2/3 e\), so an anti-up quark \(\bar{u}\) has a charge of \(-2/3 e\). The charge of a down quark \(d\) is \(-1/3 e\). Combining these two gives \(-2/3 e - 1/3 e = -1 e\). Therefore, the quark structure of \(\pi^-\) is \(\bar{u}d\).

1 mark for identifying the correct quark structure in option B.

First, calculate the voltmeter reading (the potential difference across the LDR) in bright light: \(V_{\text{bright}} = V_{\text{supply}} \times \frac{R_{\text{LDR}}}{R_{\text{fixed}} + R_{\text{LDR}}} = 12.0\text{ V} \times \frac{1.0\text{ k}\Omega}{4.0\text{ k}\Omega + 1.0\text{ k}\Omega} = 2.4\text{ V}\). Next, calculate the voltmeter reading in the dark: \(V_{\text{dark}} = V_{\text{supply}} \times \frac{R_{\text{LDR}}}{R_{\text{fixed}} + R_{\text{LDR}}} = 12.0\text{ V} \times \frac{20.0\text{ k}\Omega}{4.0\text{ k}\Omega + 20.0\text{ k}\Omega} = 10.0\text{ V}\). The change in the voltmeter reading is \(V_{\text{dark}} - V_{\text{bright}} = 10.0\text{ V} - 2.4\text{ V} = 7.6\text{ V}\).

1 mark for the correct answer option C.

For a tube closed at one end, stationary waves form with a node at the closed end and an antinode at the open end. The shortest resonance length is \(L_1 = \frac{\lambda}{4}\), and the next shortest is \(L_2 = \frac{3\lambda}{4}\). The difference between these two consecutive resonance lengths is \(\Delta L = L_2 - L_1 = \frac{\lambda}{2}\). Using the wave equation \(v = f \lambda\), we find the wavelength: \(\lambda = \frac{v}{f} = \frac{340\text{ m s}^{-1}}{340\text{ Hz}} = 1.0\text{ m}\). The difference between the two shortest lengths is \(\Delta L = \frac{1.0\text{ m}}{2} = 0.50\text{ m}\).

1 mark for calculating the correct length difference of 0.50 m (option B).

(a) Simple harmonic motion is defined by the relation \(a \propto -x\), where acceleration is directly proportional to displacement from the equilibrium position and is always directed towards that equilibrium position.

(b)(i) Using the equation for maximum acceleration: \(a_{\max} = \omega^2 A = (2\pi f)^2 A\).
Rearranging for \(f\):
\(\omega^2 = \frac{a_{\max}}{A} = \frac{5.4}{0.085} = 63.53\text{ s}^{-2}\)
\(\omega = \sqrt{63.53} = 7.97\text{ rad s}^{-1}\)
\(f = \frac{7.97}{2\pi} = 1.27\text{ Hz}\)

(b)(ii) The spring constant \(k\) can be found using \(\omega^2 = \frac{k}{m}\):
\(k = m \omega^2 = 0.35 \times 63.53 = 22.2\text{ N m}^{-1}\)

(c) The graph of acceleration against displacement is a straight line passing through the origin with a negative gradient. The extreme points (ends of the line) are \((-0.085\text{ m}, 5.4\text{ m s}^{-2})\) and \((0.085\text{ m}, -5.4\text{ m s}^{-2})\).

(a) Acceleration is directly proportional to displacement [1 mark]; acceleration is directed towards the equilibrium position (or opposite in direction to displacement) [1 mark].
(b)(i) Recall of \(a_{\max} = \omega^2 A\) or equivalent [1 mark]; calculation of \(\omega = 7.97\text{ rad s}^{-1}\) [1 mark]; correct frequency \(1.27\text{ Hz}\) or \(1.3\text{ Hz}\) [1 mark].
(b)(ii) Recall of \(\omega^2 = \frac{k}{m}\) or \(T = 2\pi \sqrt{\frac{m}{k}}\) [1 mark]; correct substitution of values [1 mark]; correct spring constant \(22\text{ N m}^{-1}\) or \(22.2\text{ N m}^{-1}\) with correct unit [1 mark].
(c) Straight line through the origin with a negative gradient [1 mark]; both axes correctly labelled with quantities and units (\(a/\text{m s}^{-2}\) and \(x/\text{m}\)) [1 mark]; limits of line correctly marked at \(\pm 0.085\text{ m}\) and \(\mp 5.4\text{ m s}^{-2}\) [1 mark].

(a)(i) Comparing the given equation to the standard form \(y = A \cos(\omega t)\), the amplitude is \(A = 0.035\text{ m}\).
(a)(ii) The angular frequency is \(\omega = 6.2\text{ rad s}^{-1}\).
The period \(T\) is:
\(T = \frac{2\pi}{\omega} = \frac{2\pi}{6.2} = 1.01\text{ s}\).
(a)(iii) The maximum speed is:
\(v_{\max} = \omega A = 6.2 \times 0.035 = 0.217\text{ m s}^{-1}\)
The maximum kinetic energy is:
\(E_{k,\max} = \frac{1}{2} m v_{\max}^2 = 0.5 \times 0.045 \times (0.217)^2 = 1.06 \times 10^{-3}\text{ J} \approx 1.1 \times 10^{-3}\text{ J}\).

(b) Over one complete cycle starting from \(t = 0\):
- At \(t = 0\): Displacement is at its positive maximum (\(+0.035\text{ m}\)), velocity is zero, and kinetic energy is zero.
- At \(t = T/4 \approx 0.25\text{ s}\): The test tube passes through its equilibrium position, so displacement is zero, velocity is at its negative maximum (\(-0.217\text{ m s}^{-1}\)), and kinetic energy is at its maximum (\(1.1 \times 10^{-3}\text{ J}\)).
- At \(t = T/2 \approx 0.50\text{ s}\): Displacement is at its negative maximum (\(-0.035\text{ m}\)), velocity is zero, and kinetic energy is zero.
- At \(t = 3T/4 \approx 0.75\text{ s}\): The test tube passes through its equilibrium position moving upwards, so displacement is zero, velocity is at its positive maximum (\(+0.217\text{ m s}^{-1}\)), and kinetic energy is at its maximum.
- At \(t = T \approx 1.01\text{ s}\): The system returns to its starting state with positive maximum displacement, zero velocity, and zero kinetic energy.

(a)(i) Identifies amplitude as \(0.035\text{ m}\) [1 mark].
(a)(ii) Identifies \(\omega = 6.2\text{ rad s}^{-1}\) [1 mark]; calculates \(T = 1.0\text{ s}\) or \(1.01\text{ s}\) [1 mark].
(a)(iii) Calculates maximum speed \(v_{\max} = 0.217\text{ m s}^{-1}\) [1 mark]; substitutes values into kinetic energy formula [1 mark]; calculates max kinetic energy as \(1.1 \times 10^{-3}\text{ J}\) or \(1.06 \times 10^{-3}\text{ J}\) [1 mark].
(b) Correctly relates states at \(t=0\) (\(y=+A\), \(v=0\), \(E_k=0\)) [1 mark]; correctly relates states at \(t=T/4\) (\(y=0\), \(v=-v_{\max}\), \(E_k=\max\)) [1 mark]; correctly relates states at \(t=T/2\) (\(y=-A\), \(v=0\), \(E_k=0\)) [1 mark]; correctly relates states at \(t=3T/4\) (\(y=0\), \(v=+v_{\max}\), \(E_k=\max\)) [1 mark]; correctly relates states at \(t=T\) (return to start) [1 mark].

(a) The block remains in contact as long as there is a positive normal contact force \(N > 0\). The boundary condition for losing contact is when \(N = 0\), which occurs when the required downward acceleration of the platform equals or exceeds the acceleration due to gravity, \(g\).

(b) At the highest point of the vertical SHM, the platform's acceleration is directed downwards and is at its maximum value \(a_{\max} = \omega^2 A\). The only downward force acting on the block to accelerate it downwards is its weight (\(mg\)), giving a maximum possible downward acceleration of \(g\). If the platform accelerates downwards at a rate greater than \(g\), the block cannot keep up and loses contact.

(c) For the block to just stay in contact, the maximum acceleration must not exceed \(g\):
\(a_{\max} = \omega^2 A = g \implies (2\pi f)^2 A = g\)
Rearranging for \(A\):
\(A = \frac{g}{4\pi^2 f^2} = \frac{9.81}{4\pi^2 \times (4.5)^2} = \frac{9.81}{799.4} = 0.01227\text{ m} \approx 0.012\text{ m}\).

(d) The maximum normal contact force \(N_{\max}\) occurs at the lowest point of the oscillation, where the acceleration is directed upwards at its maximum value.
Using \(F_{\text{net}} = m a \implies N - mg = m a_{\max} \implies N_{\max} = m(g + a_{\max})\).
With \(A = 0.0080\text{ m}\) and \(f = 4.5\text{ Hz}\):
\(a_{\max} = (2\pi \times 4.5)^2 \times 0.0080 = 799.4 \times 0.0080 = 6.40\text{ m s}^{-2}\).
\(N_{\max} = 0.050 \times (9.81 + 6.40) = 0.050 \times 16.21 = 0.81\text{ N}\).

(a) Contact is lost when normal force \(N = 0\) [1 mark]; this happens when required downward acceleration \(a \ge g\) [1 mark].
(b) States that at the highest point, acceleration is downwards and maximum [1 mark]; weight is the only force accelerating the block downwards, so its max acceleration is \(g\) [1 mark]; explains that if platform acceleration exceeds \(g\), the platform falls faster than the block, causing separation [1 mark].
(c) Identifies \(a_{\max} = g\) [1 mark]; substitutes \(\omega = 2\pi \times 4.5 = 28.3\text{ rad s}^{-1}\) [1 mark]; calculates \(A = 0.0123\text{ m}\) which is \(\approx 0.012\text{ m}\) [1 mark].
(d) Recognises maximum normal force occurs at the bottom of the cycle [1 mark]; uses \(N = m(g + \omega^2 A)\) [1 mark]; calculates \(0.81\text{ N}\) [1 mark].

(a) Measuring 20 oscillations reduces the percentage uncertainty in the time measurement. It also minimizes the impact of human reaction time errors (which occur at the start and stop of timing) by spreading the error over 20 cycles instead of just one.

(b)(i) Period \(T = \frac{t}{20} = \frac{37.0}{20} = 1.85\text{ s}\).
The absolute uncertainty in \(T\) is \(\Delta T = \frac{\Delta t}{20} = \frac{0.2}{20} = 0.01\text{ s}\).
So, \(T = 1.85 \pm 0.01\text{ s}\).

(b)(ii) Percentage uncertainty in \(L = \frac{\Delta L}{L} \times 100\% = \frac{0.002}{0.850} \times 100\% = 0.235\% \approx 0.24\%\$.\n\n(b)(iii) Percentage uncertainty in \)T = \frac{\Delta T}{T} \times 100\% = \frac{0.01}{1.85} \times 100\% = 0.54\%\$.\n\n(c)(i) Rearranging the formula: \(g = \frac{4\pi^2 L}{T^2}\).
\(g = \frac{4 \pi^2 \times 0.850}{1.85^2} = 9.805\text{ m s}^{-2} \approx 9.81\text{ m s}^{-2}\).

(c)(ii) The percentage uncertainty in \(g\) is:
\(\% \Delta g = \% \Delta L + 2 \times (\% \Delta T) = 0.24\% + 2 \times 0.54\% = 1.32\% \approx 1.3\%\).
The absolute uncertainty in \(g\) is:
\(\Delta g = 9.81 \times 0.0132 = 0.13\text{ m s}^{-2}\).

(a) Reduces the percentage uncertainty in the timing [1 mark]; reduces the relative effect of human reaction time errors [1 mark].
(b)(i) Correct period \(1.85\text{ s}\) [1 mark]; correct absolute uncertainty \(0.01\text{ s}\) [1 mark]; expresses both to appropriate precision [1 mark].
(b)(ii) Calculates percentage uncertainty in \(L\) as \(0.24\%\) (or \(0.2\%\)) [1 mark].
(b)(iii) Identifies percentage uncertainty formula [1 mark]; calculates \(0.54\%\) [1 mark].
(c)(i) Rearranges formula and calculates \(g = 9.81\text{ m s}^{-2}\) (or \(9.8\text{ m s}^{-2}\)) [1 mark].
(c)(ii) Sums percentage uncertainties correctly: \(\% \Delta L + 2 \times \% \Delta T = 1.3\%\) [1 mark]; calculates absolute uncertainty as \(0.13\text{ m s}^{-2}\) (or \(0.1\text{ m s}^{-2}\)) [1 mark].

(a)(i) Vernier callipers are suitable for measuring lengths around \(60\text{ mm}\) with a resolution of \(0.1\text{ mm}\).
(a)(ii) A micrometer screw gauge is suitable for measuring diameters around \(20\text{ mm}\) with a resolution of \(0.01\text{ mm}\).

(b) Converting measurements to cm: \(d = 2.004\text{ cm}\), \(h = 6.24\text{ cm}\).
\(V = \pi r^2 h = \pi \left(\frac{d}{2}\right)^2 h = \pi \times (1.002)^2 \times 6.24 = 19.68\text{ cm}^3 \approx 19.7\text{ cm}^3\).

(c) \(\rho = \frac{M}{V} = \frac{154.6}{19.68} = 7.856\text{ g cm}^{-3} \approx 7.86\text{ g cm}^{-3}\).

(d)(i) The volume is \(V = \frac{\pi}{4} d^2 h\).
The percentage uncertainty in \(V\) is:
\(\% \Delta V = 2 \times (\% \Delta d) + \% \Delta h\)
\(\% \Delta d = \frac{0.02}{20.04} \times 100\% = 0.0998\%\)
\(\% \Delta h = \frac{0.1}{62.4} \times 100\% = 0.1603\%\)
\(\% \Delta V = 2(0.0998\%) + 0.1603\% = 0.3599\% \approx 0.36\%\).

(d)(ii) The percentage uncertainty in density is:
\(\% \Delta \rho = \% \Delta M + \% \Delta V\)
\(\% \Delta M = \frac{0.1}{154.6} \times 100\% = 0.0647\%\)
\(\% \Delta \rho = 0.0647\% + 0.3599\% = 0.4246\% \approx 0.43\%\).

(d)(iii) The absolute uncertainty in \(\rho\) is:
\(\Delta \rho = 7.86 \times 0.00425 = 0.033\text{ g cm}^{-3} \approx 0.03\text{ g cm}^{-3}\).

(a)(i) Vernier callipers [1 mark].
(a)(ii) Micrometer screw gauge [1 mark].
(b) Uses correct formula \(V = \pi r^2 h\) with consistent units [1 mark]; calculates \(V = 19.7\text{ cm}^3\) [1 mark].
(c) Uses \(\rho = M/V\) [1 mark]; calculates \(\rho = 7.86\text{ g cm}^{-3}\) (accept \(7.85\text{ to } 7.86\)) [1 mark].
(d)(i) Recalls percentage uncertainty formula for volume: \(2 \times \%\Delta d + \%\Delta h\) [1 mark]; calculates individual percentage uncertainties correctly [1 mark]; calculates total \(\% \Delta V = 0.36\%\) [1 mark].
(d)(ii) Sums percentage uncertainties: \(\% \Delta M + \% \Delta V = 0.43\%\) (or \(0.4\%\)) [1 mark].
(d)(iii) Calculates absolute uncertainty as \(0.03\text{ g cm}^{-3}\) [1 mark].

(a) The kinetic energy gained by the electron is equal to the work done by the accelerating potential difference:
\(eV = \frac{1}{2} m v^2 \implies v = \sqrt{\frac{2eV}{m}}\)
\(v = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 1200}{9.11 \times 10^{-31}}} = \sqrt{4.215 \times 10^{14}} = 2.053 \times 10^7\text{ m s}^{-1} \approx 2.1 \times 10^7\text{ m s}^{-1}\).

(b) According to Fleming's left-hand rule, the magnetic force acting on a moving charge is always perpendicular to its velocity. Since the force is perpendicular to the velocity, no work is done on the electron, so its speed remains constant. A constant force always perpendicular to the velocity acts as a centripetal force, forcing the electron into a circular path.

(c)(i) The centripetal force is provided by the magnetic force:
\(\frac{m v^2}{r} = B e v \implies r = \frac{m v}{B e}\)
Using the calculated value of \(v = 2.053 \times 10^7\text{ m s}^{-1}\):
\(r = \frac{9.11 \times 10^{-31} \times 2.053 \times 10^7}{4.5 \times 10^{-3} \times 1.60 \times 10^{-19}} = \frac{1.870 \times 10^{-23}}{7.20 \times 10^{-22}} = 0.02597\text{ m} \approx 0.026\text{ m}\) (or \(2.6\text{ cm}\)).

(c)(ii) The frequency \(f\) is given by:
\(f = \frac{v}{2\pi r} = \frac{2.053 \times 10^7}{2\pi \times 0.02597} = 1.258 \times 10^8\text{ Hz} \approx 1.26 \times 10^8\text{ Hz}\) (or \(1.3 \times 10^8\text{ Hz}\)).
Alternatively, using \(f = \frac{B e}{2\pi m} = \frac{4.5 \times 10^{-3} \times 1.60 \times 10^{-19}}{2\pi \times 9.11 \times 10^{-31}} = 1.257 \times 10^8\text{ Hz}\).

(a) Equates work done to kinetic energy: \(eV = \frac{1}{2}mv^2\) [1 mark]; substitutes numbers correctly [1 mark]; shows calculated value is \(2.05 \times 10^7\text{ m s}^{-1}\) (must show at least 3 s.f. before rounding to \(2.1 \times 10^7\)) [1 mark].
(b) States force is perpendicular to velocity [1 mark]; explains that this perpendicular force changes direction but not speed [1 mark]; identifies this as centripetal force leading to circular motion [1 mark].
(c)(i) Equates magnetic force to centripetal force: \(B e v = \frac{m v^2}{r}\) [1 mark]; rearranges to make \(r\) the subject [1 mark]; calculates \(r = 0.026\text{ m}\) (or \(2.6\text{ cm}\)) [1 mark].
(c)(ii) Uses \(f = \frac{v}{2\pi r}\) or \(f = \frac{Be}{2\pi m}\) [1 mark]; calculates \(f = 1.26 \times 10^8\text{ Hz}\) (or \(1.3 \times 10^8\text{ Hz}\)) [1 mark].

Using the relationship between velocity and displacement in simple harmonic motion:

\(v = \pm \omega \sqrt{A^2 - x^2}\)

Substituting \(x = \frac{3}{5}A\):

\(v = \omega \sqrt{A^2 - \left(\frac{3}{5}A\right)^2}\)
\(v = \omega \sqrt{A^2 - \frac{9}{25}A^2}\)
\(v = \omega \sqrt{\frac{16}{25}A^2} = \frac{4}{5}\omega A\)

1 mark for the correct answer C.

The maximum speed of an object in SHM is given by \(v_{\text{max}} = \omega A\) and the maximum acceleration is given by \(a_{\text{max}} = \omega^2 A\).

Dividing the two equations gives the angular frequency:
\omega = \frac{a_{\text{max}}}{v_{\text{max}}} = \frac{8.0\text{ m s}^{-2}}{2.0\text{ m s}^{-1}} = 4.0\text{ rad s}^{-1}

The time period \(T\) is:
T = \frac{2\pi}{\omega} = \frac{2\pi}{4.0} \approx 1.57\text{ s}

Rounding to 2 significant figures gives \(1.6\text{ s}\).

1 mark for selecting the correct option B.
- Award 1 mark for correct calculation of \omega = 4.0\text{ rad s}^{-1} and time period \(T = 1.6\text{ s}\).

The magnetic force provides the centripetal force:
Bev = \frac{mv^2}{R} \implies R = \frac{mv}{Be}

Since the kinetic energy is \(E = \frac{1}{2}mv^2\), the velocity is \(v = \sqrt{\frac{2E}{m}}\).

Substituting \(v\) into the radius equation gives:
R = \frac{m}{Be} \sqrt{\frac{2E}{m}} = \frac{1}{B e} \sqrt{2 m E}

Therefore, \(R \propto \sqrt{E}\).

If the kinetic energy is increased to \(3E\), the new radius \(R'\) is:
R' = R \times \sqrt{3} = 12\text{ cm} \times \sqrt{3} \approx 20.8\text{ cm} \approx 21\text{ cm}.

1 mark for selecting the correct option B.
- Award 1 mark for identifying that \(R \propto \sqrt{E}\) and calculating the new radius.

First, calculate the percentage uncertainty in the length \(L\):
\%\Delta L = \frac{0.004}{0.800} \times 100\% = 0.5\%

The period \(T\) has the same percentage uncertainty as the total measured time because dividing the total time and its uncertainty by 20 leaves the ratio unchanged:
\%\Delta T = \frac{0.2}{36.0} \times 100\% \approx 0.556\%

From the equation \(g = \frac{4\pi^2 L}{T^2}\), the percentage uncertainty in \(g\) is:
\%\Delta g = \%\Delta L + 2 \times \%\Delta T = 0.5\% + 2 \times 0.556\% = 1.61\%

Rounding to two significant figures gives \(1.6\%\).

1 mark for selecting the correct option B.
- Award 1 mark for calculating \%\Delta L = 0.5\% and \%\Delta T = 0.56\% and combining them to get 1.6%.

For the block to rotate in circular motion without slipping, the friction force must provide the necessary centripetal force.

The maximum frictional force is:
F_{\text{friction}} = \mu m g

The centripetal force required is:
F_{\text{centripetal}} = m \omega^2 r

At the limit of slipping, these two forces are equal:
m \omega^2 r = \mu m g

Solving for \(\omega\):
\omega^2 = \frac{\mu g}{r} \implies \omega = \sqrt{\frac{\mu g}{r}}

1 mark for selecting the correct option A.
- Award 1 mark for equating the maximum frictional force to centripetal force and solving for \omega.

The total energy \(E_T\) in simple harmonic motion is given by:
E_T = \frac{1}{2} k A^2

The potential energy at displacement \(x\) is:
E_p = \frac{1}{2} k x^2

The kinetic energy is:
E_k = E_T - E_p = \frac{1}{2} k (A^2 - x^2)

We are given that \(E_k = 3 E_p\):
\frac{1}{2} k (A^2 - x^2) = 3 \left(\frac{1}{2} k x^2\right)

A^2 - x^2 = 3 x^2 \implies A^2 = 4 x^2

x = \pm \frac{A}{2} = 0.50 A

1 mark for selecting the correct option C.
- Award 1 mark for setting up the energy equation and finding \(x = 0.50 A\).

For the ions to pass through undeflected, the electric force and the magnetic force acting on them must be equal in magnitude and opposite in direction:
F_E = F_B

q E = q v B

Notice that the charge \(q\) cancels out of the equation. This means the required speed is independent of the charge of the ions:
v = \frac{E}{B} = \frac{4.0 \times 10^4\text{ V m}^{-1}}{0.20\text{ T}} = 2.0 \times 10^5\text{ m s}^{-1}

1 mark for selecting the correct option B.
- Award 1 mark for using the condition \(v = E/B\) and calculating the correct speed.

A zero error of \(-0.02\text{ mm}\) means that the instrument reads \(-0.02\text{ mm}\) when it should read zero. As a result, all measured values will be smaller than the true values by \(0.02\text{ mm}\). This is a systematic error because it affects all measurements in the same way. Since the average diameter is underestimated, the cross-sectional area (which is calculated from the diameter using \(A = \frac{\pi d^2}{4}\)) will also be systematically underestimated (smaller than the true value).

1 mark for selecting the correct option A.
- Award 1 mark for correctly identifying that a negative zero error systematically underestimates both diameter and area.