2025 AQA IAS-Level Physics (9630) 模擬試題連答案詳解

(a) Rearranging the equation h = 0.5 * g * t^2 gives g = 2h / t^2. Substituting the values: g = (2 * 1.250) / (0.50)^2 = 2.50 / 0.25 = 10 m s^-2. (b) The percentage uncertainty in h is (0.002 / 1.250) * 100% = 0.16%. The percentage uncertainty in t is (0.02 / 0.50) * 100% = 4.0%. Since g = 2h / t^2, the percentage uncertainty in g is % uncertainty in h + 2 * (% uncertainty in t) = 0.16% + 2 * 4.0% = 8.16% (or approximately 8.2%). (c) The absolute uncertainty in g is 8.16% of 10 m s^-2 = 0.816 m s^-2. Rounded to 1 significant figure, this is 0.8 m s^-2. The value of g must match the precision of the uncertainty, so the final value is written as (10.0 \pm 0.8) m s^-2.

(a) 1 mark for correct rearrangement of formula or substitution, 1 mark for calculating g = 10 m s^-2. (b) 1 mark for calculating percentage uncertainties of h (0.16%) and t (4.0%), 1 mark for adding them correctly with weight 2 for t to get 8.2% (or 8.16%). (c) 1 mark for calculating the absolute uncertainty as 0.8 m s^-2 (accept 0.82), 1 mark for expressing the final answer with correct precision and unit: (10.0 \pm 0.8) m s^-2.

(a) A neutral Carbon-14 atom has 6 protons (from atomic number 6), 6 electrons (to maintain neutrality), and 14 - 6 = 8 neutrons. (b) The beta-minus decay equation is: 14_6 C -> 14_7 N + e^- + \bar{
u}_e. Here, a neutron decays into a proton, an electron (beta particle), and an electron antineutrino to conserve lepton number. (c) Specific charge is the ratio of charge to mass. For the Carbon-14 nucleus: Charge Q = Z * e = 6 * 1.60 * 10^-19 C = 9.60 * 10^-19 C. Mass M = A * m_p = 14 * 1.67 * 10^-27 kg = 2.338 * 10^-26 kg. Specific charge = Q / M = 9.60 * 10^-19 / (2.338 * 10^-26) = 4.11 * 10^7 C kg^-1.

(a) 1 mark for identifying 6 protons and 6 electrons, 1 mark for 8 neutrons. (b) 1 mark for showing Nitrogen-14 and beta particle (e^-) with correct proton/nucleon numbers, 1 mark for including the electron antineutrino. (c) 1 mark for correct charge and mass of the Carbon-14 nucleus, 1 mark for calculating the specific charge as 4.1 * 10^7 C kg^-1 (accept 4.11 * 10^7 to 4.12 * 10^7 C kg^-1).

(a) The weight W of the uniform shelf is W = m * g = 4.5 * 9.81 = 44.15 N. Since the shelf is uniform, its centre of mass is at the midpoint, which is 0.40 m from the hinge. (b) Taking moments about the hinge: Clockwise moment = W * 0.40 = 44.15 * 0.40 = 17.66 N m. Anticlockwise moment = T * sin(35) * 0.80. In equilibrium, Clockwise moment = Anticlockwise moment: 17.66 = T * sin(35) * 0.80. T = 17.66 / (0.80 * sin(35)) = 38.48 N, which is approximately 38 N. (c) For horizontal equilibrium, the horizontal force H from the hinge must balance the horizontal component of the tension: H = T * cos(35) = 38.48 * cos(35) = 31.52 N (approximately 32 N acting away from the wall).

(a) 1 mark for calculating weight = 44 N (or 44.1 N), 1 mark for identifying the midpoint distance as 0.40 m. (b) 1 mark for set up of the moment equation about the hinge, 1 mark for calculating T = 38 N (accept range 38.0 N to 38.5 N). (c) 1 mark for identifying that horizontal force equals T * cos(35), 1 mark for calculating H = 31.5 N (accept 31 N to 32 N).

(a) The useful work done is equal to the gain in gravitational potential energy of the crate: W = m * g * h = 120 * 9.81 * 15 = 17658 J (approximately 1.8 * 10^4 J or 17.7 kJ). (b) The useful power output is the rate of doing useful work: P_out = W / t = 17658 / 8.0 = 2207 W (approximately 2.2 kW). (c) Efficiency = (P_out / P_in) * 100% = (2207 / 3100) * 100% = 71.2% (or 71%). The remaining 29% of the input energy is wasted (lost) mainly as thermal energy in the motor coils due to electrical resistance (Joule heating) and in the moving parts due to friction.

(a) 1 mark for using W = mgh, 1 mark for calculating 17.7 kJ (or 18 kJ). (b) 1 mark for using P = W/t, 1 mark for calculating 2.2 kW (accept 2.21 kW). (c) 1 mark for calculating efficiency = 71% (or 0.71), 1 mark for stating that the energy is lost as thermal energy due to friction/electrical resistance.

(a) The cross-sectional area is given by A = pi * d^2 / 4. Substituting d = 0.45 * 10^-3 m: A = pi * (0.45 * 10^-3)^2 / 4 = 1.59 * 10^-7 m^2, which is approximately 1.6 * 10^-7 m^2. (b) Using the resistivity formula R = rho * L / A, we rearrange for length: L = R * A / rho. Substituting the values: L = 15 * 1.59 * 10^-7 / (4.9 * 10^-7) = 4.87 m (approximately 4.9 m). (c) The area A is proportional to the square of the diameter (d^2). Since the diameter is doubled (from 0.45 mm to 0.90 mm), the cross-sectional area increases by a factor of 2^2 = 4. From R = rho * L / A, to keep resistance R and resistivity rho constant, the length L must be proportional to the cross-sectional area A. Therefore, the new wire must be 4 times longer than the original wire.

(a) 1 mark for using A = pi * d^2 / 4 and showing the calculation steps clearly. (b) 1 mark for rearranging R = rho * L / A to L = R * A / rho, 1 mark for calculating L = 4.9 m (accept 4.87 m to 4.90 m). (c) 1 mark for stating that doubling the diameter increases the area by a factor of 4, 1 mark for identifying that length is directly proportional to area for constant resistance, 1 mark for concluding the new wire must be 4 times longer.

(a) Using Snell's law: n_1 * sin(theta_1) = n_2 * sin(theta_2). Since the light travels from air (n_1 = 1.00) to glass (n_2 = 1.52): 1.00 * sin(42.0) = 1.52 * sin(theta_2). sin(theta_2) = sin(42.0) / 1.52 = 0.6691 / 1.52 = 0.4402. theta_2 = arcsin(0.4402) = 26.1 degrees. (b) The refractive index is defined as n = c / v. Rearranging for v: v = c / n = 3.00 * 10^8 / 1.52 = 1.97 * 10^8 m s^-1. (c) The relationship between critical angle theta_c and refractive index n is sin(theta_c) = 1 / n. Rearranging for n: n = 1 / sin(40.5) = 1 / 0.6494 = 1.54.

(a) 1 mark for using Snell's law, 1 mark for calculating 26.1 degrees. (b) 1 mark for using n = c / v, 1 mark for calculating 1.97 * 10^8 m s^-1. (c) 1 mark for using sin(theta_c) = 1 / n, 1 mark for calculating n = 1.54.

(a) Diffraction is a phenomenon unique to waves (where waves spread out and interfere constructively/destructively). The observation of concentric diffraction rings on the screen when electrons pass through the graphite grid can only be explained if the electrons behave as waves. (b) The kinetic energy gained by each electron is E_k = e * V = 1.60 * 10^-19 * 250 = 4.00 * 10^-17 J. Since E_k = p^2 / (2 * m), the momentum is p = sqrt(2 * m * E_k) = sqrt(2 * 9.11 * 10^-31 * 4.00 * 10^-17) = 8.537 * 10^-24 kg m s^-1. The de Broglie wavelength is lambda = h / p = 6.63 * 10^-34 / (8.537 * 10^-24) = 7.77 * 10^-11 m, which is approximately 7.8 * 10^-11 m. (c) Increasing the voltage increases the kinetic energy and momentum, which decreases the de Broglie wavelength. A shorter wavelength undergoes less diffraction, causing the diffraction rings on the screen to shrink (become smaller and closer to the central spot).

(a) 1 mark for stating diffraction is a wave property, 1 mark for linking the experimental observation of rings to constructive/destructive interference of electron waves. (b) 1 mark for calculating the kinetic energy (4.0 * 10^-17 J) or momentum (8.5 * 10^-24 kg m s^-1), 1 mark for using lambda = h / p, 1 mark for arriving at 7.8 * 10^-11 m. (c) 1 mark for explaining that higher voltage decreases wavelength, leading to smaller diffraction angles (rings contract).

(a) The Young modulus is defined as E = (F * L) / (A * delta L). Rearranging for extension delta L gives: delta L = (F * L) / (A * E). Substituting the values: delta L = (45 * 2.4) / (1.2 * 10^-7 * 2.0 * 10^11) = 108 / (2.4 * 10^4) = 4.5 * 10^-3 m (or 4.5 mm). (b) The elastic strain energy E_s stored in the wire is equal to the work done, which is the area under the force-extension graph: E_s = 0.5 * F * delta L = 0.5 * 45 * 4.5 * 10^-3 = 0.101 J (approximately 0.10 J). (c) Ultimate tensile strength is the maximum tensile stress that a material can withstand before breaking. Beyond the elastic limit, the wire undergoes plastic deformation, meaning it will no longer return to its original length when the load is removed; it will retain a permanent extension (permanent set).

(a) 1 mark for rearranging the Young modulus equation for extension, 1 mark for calculating delta L = 4.5 * 10^-3 m. (b) 1 mark for using E_s = 0.5 * F * delta L, 1 mark for calculating energy = 0.10 J (accept 0.101 J). (c) 1 mark for defining ultimate tensile strength as maximum tensile stress before fracture, 1 mark for describing plastic deformation/permanent set beyond the elastic limit.

(a) The critical angle is the angle of incidence in the optically denser medium for which the angle of refraction in the less dense medium is \(90^\circ\).
Using the formula:
\(\sin \theta_c = \frac{1}{n}\)
\(\sin \theta_c = \frac{1}{1.52} = 0.6579\)
\(\theta_c = 41.1^\circ\)

(b) Since the angle of incidence of the ray is \(43.0^\circ\), which is greater than the critical angle of \(41.1^\circ\), and the light is travelling from a more optically dense medium (glass) to a less dense medium (air), the light undergoes total internal reflection (TIR). No light is transmitted into the air; all light is reflected back into the glass block at an angle of reflection equal to \(43.0^\circ\).

(a)
1 mark: Definition: critical angle is the angle of incidence (in the denser medium) that results in an angle of refraction of \(90^\circ\).
1 mark: Correct substitution into \(\sin \theta_c = \frac{1}{n}\).
1 mark: Correct calculation of critical angle \(\theta_c = 41.1^\circ\) (accept \(41^\circ\)).

(b)
1 mark: States that total internal reflection (TIR) occurs.
1 mark: Explains that this is because the angle of incidence (\(43.0^\circ\)) is greater than the critical angle (\(41.1^\circ\)).
1 mark: Mentions that the ray is travelling from a denser to a less dense medium (glass to air).

(a) The vertical component of the initial velocity \(u_y\) is given by:
\(u_y = u \sin \theta = 22.0 \times \sin(38.0^\circ) = 13.54\text{ m s}^{-1} \approx 13.5\text{ m s}^{-1}\).
At maximum height, the vertical component of velocity \(v_y = 0\).
Using the equation of motion:
\(v_y^2 = u_y^2 - 2g s\)
\(0 = (13.54)^2 - 2(9.81)s\)
\(s = \frac{183.33}{19.62} = 9.34\text{ m}\).

(b) First, calculate the total time of flight \(t\) by finding when vertical displacement \(s_y = 0\):
\(s_y = u_y t - \frac{1}{2}g t^2\)
\(0 = 13.54 t - 4.905 t^2\)
Since \(t \neq 0\):
\(t = \frac{13.54}{4.905} = 2.76\text{ s}\).
The horizontal component of velocity \(u_x\) remains constant:
\(u_x = u \cos \theta = 22.0 \times \cos(38.0^\circ) = 17.34\text{ m s}^{-1}\).
Horizontal range \(d = u_x \times t = 17.34 \times 2.76 = 47.9\text{ m}\) (or \(48.0\text{ m}\) using unrounded values).

(a)
1 mark: Calculates vertical velocity: \(u_y = 22.0 \sin(38.0^\circ) = 13.5\text{ m s}^{-1}\).
1 mark: Recalls or uses \(v^2 = u^2 + 2as\) with \(v = 0\) and \(a = -9.81\text{ m s}^{-2}\).
1 mark: Correctly calculates maximum height as \(9.34\text{ m}\) (accept \(9.3\text{ m}\)).

(b)
1 mark: Calculates total time of flight as \(t = 2.76\text{ s}\) (or \(2.8\text{ s}\)).
1 mark: Calculates horizontal velocity: \(u_x = 22.0 \cos(38.0^\circ) = 17.3\text{ m s}^{-1}\).
1 mark: Correctly calculates horizontal range as \(47.9\text{ m}\) or \(48.0\text{ m}\) (accept values in the range \(47.6\text{ m}\) to \(48.2\text{ m}\)).

(a) A neutral atom has equal numbers of protons and electrons.
For carbon-14 (\(^{14}_{6}\text{C}\)):
Protons = 6
Neutrons = 14 - 6 = 8
Electrons = 6

For nitrogen-14 (\(^{14}_{7}\text{N}\)):
Protons = 7
Neutrons = 14 - 7 = 7
Electrons = 7

(b) The nuclear equation is:
\(^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + ^{0}_{-1}\text{e} + \bar{\nu}_e\)
where \(\bar{\nu}_e\) is an electron antineutrino.
The decay is mediated by the weak interaction.

(a)
1 mark: Correct proton and neutron numbers for both Carbon-14 (6p, 8n) and Nitrogen-14 (7p, 7n).
1 mark: Correct electron numbers for both (6 for Carbon-14, 7 for Nitrogen-14).

(b)
1 mark: Shows reactant \(^{14}_{6}\text{C}\) and daughter nucleus \(^{14}_{7}\text{N}\) with correct proton and nucleon numbers.
1 mark: Shows beta particle with correct numbers: \(^{0}_{-1}\text{e}\) or \(\beta^-\).
1 mark: Shows electron antineutrino \(\bar{\nu}_e\) (must have the bar over the \(\nu\) to represent an anti-particle; reject neutrino \(\nu_e\)).
1 mark: Correctly identifies the fundamental interaction as the 'weak interaction' (or 'weak nuclear force').

(a) Diameter \(d = 0.280\text{ mm} = 0.280 \times 10^{-3}\text{ m}\).
Radius \(r = \frac{d}{2} = 0.140 \times 10^{-3}\text{ m}\).
Cross-sectional area \(A = \pi r^2 = \pi \times (0.140 \times 10^{-3}\text{ m})^2 = 6.1575 \times 10^{-8}\text{ m}^2 \approx 6.16 \times 10^{-8}\text{ m}^2\).

(b) Using the definition of resistivity \(R = \frac{\rho L}{A}\), we can rearrange for \(\rho\):
\(\rho = \frac{R A}{L}\)
\(\rho = \frac{31.5 \times 6.1575 \times 10^{-8}}{1.80} = 1.0776 \times 10^{-6}\ \Omega\text{ m}\).
Rounding to 3 significant figures gives \(1.08 \times 10^{-6}\ \Omega\text{ m}\).
The unit of resistivity is the ohm metre (\(\Omega\text{ m}\)).

(a)
1 mark: Converts diameter to radius correctly or uses \(A = \frac{\pi d^2}{4}\) with correct power of 10.
1 mark: Shows calculation to obtain at least 3 significant figures (e.g., \(6.158 \times 10^{-8}\text{ m}^2\)) before rounding to \(6.16 \times 10^{-8}\text{ m}^2\).

(b)
1 mark: Rearranges formula to make resistivity \(\rho\) the subject: \(\rho = \frac{R A}{L}\).
1 mark: Subsitutes values correctly: \(\rho = \frac{31.5 \times 6.16 \times 10^{-8}}{1.80}\) (accept unrounded area).
1 mark: Calculates correct numerical value: \(1.08 \times 10^{-6}\) (or \(1.07 \times 10^{-6}\) if using rounded area).
1 mark: Correctly states unit: \(\Omega\text{ m}\) (accept ohm-metre).

(a) A progressive wave travels along the wire and is reflected at the fixed boundary. This results in two progressive waves of the same frequency and amplitude travelling in opposite directions.
These two waves superpose (or overlap/interfere). At points where they meet in phase, constructive interference occurs, forming antinodes (points of maximum displacement). At points where they meet out of phase by \(180^\circ\), destructive interference occurs, forming nodes (points of zero displacement).

(b) For the third harmonic, the wire length \(L\) accommodates 3 loops (three half-wavelengths):
\(L = 3 \times \frac{\lambda}{2}\)
\(\lambda = \frac{2L}{3} = \frac{2 \times 1.20\text{ m}}{3} = 0.80\text{ m}\).
Using the wave equation \(v = f \lambda\):
\(f = \frac{v}{\lambda} = \frac{180\text{ m s}^{-1}}{0.80\text{ m}} = 225\text{ Hz}\).

(a)
1 mark: Mentions two progressive waves of identical frequency/wavelength travelling in opposite directions.
1 mark: States that these waves superpose or interfere.
1 mark: Explains that constructive interference forms antinodes and/or destructive interference forms nodes.

(b)
1 mark: States or uses the relationship \(L = 1.5\lambda\) or determines \(\lambda = 0.80\text{ m}\).
1 mark: Recalls or uses the wave speed formula \(v = f\lambda\).
1 mark: Obtains the correct frequency \(225\text{ Hz}\).

(a) The principle of moments states that for a system in rotational equilibrium, the sum of the clockwise moments about any pivot is equal to the sum of the anticlockwise moments about that same pivot.

(b) Taking moments about the hinge at \(A\):
- The weight of the uniform shelf acts at its centre of gravity, which is at the midpoint: \(d_1 = 0.45\text{ m}\).
- The book of weight \(40\text{ N}\) acts at \(d_2 = 0.30\text{ m}\).
- The vertical cable tension \(T\) acts upwards at \(d_3 = 0.90\text{ m}\).

Clockwise moments = Anticlockwise moments:
\((15\text{ N} \times 0.45\text{ m}) + (40\text{ N} \times 0.30\text{ m}) = T \times 0.90\text{ m}\)
\(6.75 + 12.0 = 0.90 T\)
\(18.75 = 0.90 T\)
\(T = \frac{18.75}{0.90} = 20.83\text{ N} \approx 20.8\text{ N}\).

(c) For vertical equilibrium (translational equilibrium):
Total upward forces = Total downward forces
\(F_{\text{hinge}} + T = W_{\text{shelf}} + W_{\text{book}}\)
\(F_{\text{hinge}} + 20.83 = 15 + 40\)
\(F_{\text{hinge}} = 55 - 20.83 = 34.17\text{ N} \approx 34.2\text{ N}\) upwards.

(a)
1 mark: Mentions sum of clockwise moments equals sum of anticlockwise moments.
1 mark: Clearly specifies that the system must be in equilibrium / balanced.

(b)
1 mark: Correctly identifies shelf weight acts at \(0.45\text{ m}\) and constructs moments equation: \((15 \times 0.45) + (40 \times 0.30) = T \times 0.90\).
1 mark: Obtains \(T = 20.8\text{ N}\) (accept \(21\text{ N}\) or \(20.83\text{ N}\)).

(c)
1 mark: Uses translational equilibrium: \(F_{\text{hinge}} + T = 55\).
1 mark: Obtains \(34.2\text{ N}\) (accept \(34\text{ N}\) or \(34.17\text{ N}\)).

(a) Momentum \(p = m_e v\):
\(p = (9.11 \times 10^{-31}\text{ kg}) \times (3.20 \times 10^6\text{ m s}^{-1}) = 2.9152 \times 10^{-24}\text{ kg m s}^{-1} \approx 2.92 \times 10^{-24}\text{ kg m s}^{-1}\).

de Broglie wavelength \(\lambda = \frac{h}{p}\):
\(\lambda = \frac{6.63 \times 10^{-34}\text{ J s}}{2.9152 \times 10^{-24}\text{ kg m s}^{-1}} = 2.274 \times 10^{-10}\text{ m} \approx 2.27 \times 10^{-10}\text{ m}\).

(b) An electron beam is accelerated in an evacuated tube and directed towards a thin sheet of polycrystalline material, such as graphite. After passing through, the electrons strike a fluorescent screen at the end of the tube.
A pattern of concentric bright and dark rings is observed on the screen. This is a diffraction pattern. Since diffraction is a phenomenon unique to waves, this observation demonstrates that moving electrons have wave-like properties. The regular spacing of the carbon atoms in graphite acts as a diffraction grating.

(a)
1 mark: Calculates momentum \(p = 2.92 \times 10^{-24}\text{ kg m s}^{-1}\) showing substitution of numbers.
1 mark: Recalls or uses the de Broglie relationship \(\lambda = \frac{h}{p}\).
1 mark: Calculates wavelength \(\lambda = 2.27 \times 10^{-10}\text{ m}\).

(b)
1 mark: Describes the experimental setup: electron beam accelerated through a thin slice of graphite / crystalline target.
1 mark: Describes the observation: concentric rings or diffraction pattern on a screen.
1 mark: Explains that diffraction/interference is a wave property, indicating that the electrons have wave properties.

(a) Percentage uncertainty in \(L\):
\(\% \text{ uncertainty in } L = \frac{\Delta L}{L} \times 100\% = \frac{0.003}{0.950} \times 100\% = 0.316\% \approx 0.32\%\).

Time period \(T = \frac{t}{20} = \frac{39.2}{20} = 1.96\text{ s}\).
The percentage uncertainty in the period \(T\) is identical to the percentage uncertainty in the total measured time \(t\):
\(\% \text{ uncertainty in } T = \frac{\Delta t}{t} \times 100\% = \frac{0.4}{39.2} \times 100\% = 1.02\% \approx 1.0\%\).

(b) Calculate the value of \(g\):
\(g = \frac{4\pi^2 L}{T^2} = \frac{4\pi^2 \times 0.950}{(1.96)^2} = 9.762\text{ m s}^{-2}\).

Calculate the percentage uncertainty in \(g\):
Since \(g \propto \frac{L}{T^2}\):
\(\% \text{ uncertainty in } g = \% \text{ uncertainty in } L + 2 \times (\% \text{ uncertainty in } T)\)
\(\% \text{ uncertainty in } g = 0.316\% + 2 \times (1.02\%) = 0.316\% + 2.04\% = 2.356\%\).

Calculate the absolute uncertainty in \(g\):
\(\Delta g = 9.762 \times \frac{2.356}{100} = 0.230\text{ m s}^{-2} \approx 0.23\text{ m s}^{-2}\).

Therefore, \(g = 9.76 \pm 0.23\text{ m s}^{-2}\) (or \(g = 9.8 \pm 0.2\text{ m s}^{-2}\)).

(a)
1 mark: Correctly calculates percentage uncertainty in \(L\) as \(0.32\%\) (or \(0.316\%\)).
1 mark: Correctly identifies that the percentage uncertainty in \(T\) is the same as in \(t\), calculating it as \(1.02\%\) (or \(1.0\%\)).

(b)
1 mark: Calculates the correct value of \(g = 9.76\text{ m s}^{-2}\) (accept \(9.8\text{ m s}^{-2}\)).
1 mark: Combines percentage uncertainties correctly: \(\% \Delta g = \% \Delta L + 2 \times \% \Delta T = 2.36\%\) (or \(2.3\%\) to \(2.4\%\)).
1 mark: Calculates absolute uncertainty in \(g\) as \(0.23\text{ m s}^{-2}\) (or \(0.2\text{ m s}^{-2}\)).
1 mark: Formats the final answer correctly with appropriate significant figures and units: \(g = 9.76 \pm 0.23\text{ m s}^{-2}\) (or \(g = 9.8 \pm 0.2\text{ m s}^{-2}\)).

(a) Using de Broglie's equation: \(p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{1.8 \times 10^{-10}} = 3.68 \times 10^{-24}\text{ kg m s}^{-1} \approx 3.7 \times 10^{-24}\text{ kg m s}^{-1}\). (b) Kinetic energy \(E_k = \frac{p^2}{2m} = \frac{(3.68 \times 10^{-24})^2}{2 \times 9.11 \times 10^{-31}} = 7.44 \times 10^{-18}\text{ J}\). Since \(E_k = eV\), we have \(V = \frac{7.44 \times 10^{-18}}{1.60 \times 10^{-19}} = 46.5\text{ V}\).

(a) 1 mark for using \(p = \frac{h}{\lambda}\) with correct numbers. 1 mark for showing calculation steps leading to \(3.68 \times 10^{-24}\text{ kg m s}^{-1}\). (b) 1 mark for \(E_k = \frac{p^2}{2m}\) relation. 1 mark for calculating kinetic energy. 1 mark for \(V = \frac{E_k}{e}\). 1 mark for final potential difference in range 46 V to 47 V.

(a) Taking moments about hinge A: \(24\text{ N} \times 0.60\text{ m} = T \sin(35^\circ) \times 1.2\text{ m}\), which gives \(14.4 = 1.2 T \sin(35^\circ)\), so \(T = \frac{12}{\sin(35^\circ)} = 20.9\text{ N}\). (b) For vertical equilibrium: \(F_v + T \sin(35^\circ) - 24 = 0\). Substituting \(T \sin(35^\circ) = 12\text{ N}\) gives \(F_v = 12\text{ N}\).

(a) 1 mark for identifying weight acts at \(0.60\text{ m}\). 1 mark for writing correct moment equation. 1 mark for calculation of tension. (b) 1 mark for vertical equilibrium equation. 1 mark for substituting tension component. 1 mark for final vertical force of 12 N.

(a) Snell's law: \(1.00 \times \sin(48.0^\circ) = 1.55 \times \sin\theta_2\), so \(\sin\theta_2 = 0.4794\) which gives \(\theta_2 = 28.6^\circ\). (b) For the boundary between glass and liquid: \(\sin\theta_c = \frac{n_2}{n_1} = \frac{1.33}{1.55} = 0.8581\), which gives \(\theta_c = 59.1^\circ\).

(a) 1 mark for Snell's law equation. 1 mark for calculating \(28.6^\circ\). (b) 1 mark for recognizing denser-to-less-dense transition. 1 mark for critical angle formula. 1 mark for substituting refractive indices. 1 mark for calculating \(59.1^\circ\).

(a) Tension is the weight of the hanging mass: \(F = mg = 6.0 \times 9.81 = 58.86\text{ N} \approx 59\text{ N}\). (b) Using \(E = \frac{FL}{A\Delta L}\), we rearrange to find \(\Delta L = \frac{FL}{AE} = \frac{58.86 \times 2.50}{1.20 \times 10^{-7} \times 2.00 \times 10^{11}} = 6.13 \times 10^{-3}\text{ m}\). (c) The elastic strain energy is \(E_{\text{elastic}} = \frac{1}{2} F \Delta L = 0.5 \times 58.86 \times 6.13 \times 10^{-3} = 0.180\text{ J}\).

(a) 1 mark for correct weight force. (b) 1 mark for rearranging Young modulus equation. 1 mark for correct substitution. 1 mark for calculating extension of 6.1 mm (or \(6.13 \times 10^{-3}\text{ m}\). (c) 1 mark for elastic energy formula. 1 mark for calculating energy of \(0.18\text{ J}\).

(a) The cross-sectional area \(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.45 \times 10^{-3})^2}{4} = 1.59 \times 10^{-7}\text{ m}^2\). (b) The resistance \(R = \frac{\rho L}{A} = \frac{1.10 \times 10^{-6} \times 15.0}{1.59 \times 10^{-7}} = 103.8\ \Omega \approx 104\ \Omega\). (c) When the temperature increases, the amplitude of vibration of the positive metal ions increases. This increases the frequency of collisions between conduction electrons and lattice ions, resulting in an increased resistance.

(a) 1 mark for area formula. 1 mark for correct area value. (b) 1 mark for resistivity equation. 1 mark for calculating resistance of 104 ohms. (c) 1 mark for stating resistance increases. 1 mark for describing increased lattice vibrations and collision frequency.

(a) The first excited state energy level is \(E_2 = -3.40\text{ eV}\). The excitation energy is \(\Delta E = E_2 - E_1 = -3.40 - (-13.6) = 10.2\text{ eV}\). (b) Since the kinetic energy of the incoming electron (\(12.5\text{ eV}\)) is greater than the required excitation energy (\(10.2\text{ eV}\)), the electron can excite the atom by transferring exactly \(10.2\text{ eV}\) of its energy. The remaining kinetic energy of the electron after collision is \(12.5\text{ eV} - 10.2\text{ eV} = 2.3\text{ eV}\).

(a) 1 mark for identifying \(E_2 = -3.4\text{ eV}\) or similar. 1 mark for finding excitation energy of 10.2 eV. (b) 1 mark for comparing the incoming kinetic energy to the excitation energy. 1 mark for explaining that electrons can transfer a portion of their energy in collisions. 1 mark for subtracting energy. 1 mark for final kinetic energy of 2.3 eV.

(a) For the vertical motion: \(s_y = u_y t + \frac{1}{2} a_y t^2\). Since \(u_y = 0\), \(22.0 = 0.5 \times 9.81 \times t^2 \implies t^2 = 4.485 \implies t = 2.12\text{ s}\). (b) For the horizontal motion, velocity is constant: \(v_x = \frac{s_x}{t} = \frac{18.5\text{ m}}{2.12\text{ s}} = 8.73\text{ m s}^{-1}\).

(a) 1 mark for selecting \(s = \frac{1}{2}gt^2\). 1 mark for substituting values correctly. 1 mark for time of flight of 2.12 s. (b) 1 mark for using horizontal velocity relation. 1 mark for substitution of their time. 1 mark for horizontal speed of 8.7 m/s (or 8.73 m/s).

(a) The wave pattern consists of three loops along the string, with fixed nodes at both ends plus two nodes inside, and three antinodes. (b) For the third harmonic: \(L = 1.5\lambda\). Therefore, \(\lambda = \frac{2 L}{3} = \frac{2 \times 1.60}{3} = 1.07\text{ m}\). (c) Using the wave equation: \(f = \frac{v}{\lambda} = \frac{120\text{ m s}^{-1}}{1.07\text{ m}} = 112.5\text{ Hz} \approx 113\text{ Hz}\).

(a) 1 mark for drawing three loops with nodes at both ends. 1 mark for correctly labelling all nodes (N) and antinodes (A). (b) 1 mark for relationship \(L = 1.5 \lambda\). 1 mark for wavelength of 1.07 m. (c) 1 mark for using wave speed formula. 1 mark for frequency of 113 Hz.

(a) The Young modulus \(E\) is given by the formula:
\[E = \frac{\text{Stress}}{\text{Strain}} = \frac{F L}{A \Delta L}\]
Substitute the given values into the equation:
\(F = 45.0\text{ N}\)
\(L = 2.50\text{ m}\)
\(A = 1.20 \times 10^{-7}\text{ m}^2\)
\(\Delta L = 1.85 \times 10^{-3}\text{ m}\)

\[E = \frac{45.0 \times 2.50}{1.20 \times 10^{-7} \times 1.85 \times 10^{-3}}\]
\[E = \frac{112.5}{2.22 \times 10^{-10}} \approx 5.0676 \times 10^{11}\text{ Pa}\]
Rounded to 3 significant figures, the Young modulus is \(5.07 \times 10^{11}\text{ Pa}\).

(b) The elastic strain energy \(E_{\text{elastic}}\) stored in the wire is equal to the work done in stretching it:
\[E_{\text{elastic}} = \frac{1}{2} F \Delta L\]
\[E_{\text{elastic}} = 0.5 \times 45.0 \times 1.85 \times 10^{-3} = 0.041625\text{ J}\]
Rounded to 3 significant figures, the stored elastic strain energy is \(4.16 \times 10^{-2}\text{ J}\) (or \(0.0416\text{ J}\)).

(c) During plastic deformation, the stress applied is high enough to cause planes of atoms to slide over each other. This process involves the propagation of dislocations through the crystal lattice. Once the stress is removed, the atoms do not return to their initial positions because the metallic bonds have broken and reformed with new neighbouring atoms, resulting in a permanent change in shape.

(a) [2 Marks]
- M1: For substituting correct values into the Young modulus formula: \(E = \frac{F L}{A \Delta L}\) or calculating stress \(= 3.75 \times 10^8\text{ Pa}\) and strain \(= 7.40 \times 10^{-4}\).
- A1: Correct final answer: \(5.07 \times 10^{11}\text{ Pa}\) (accept range \(5.06 \times 10^{11}\) to \(5.1 \times 10^{11}\text{ Pa}\)). Must include correct unit (\(\text{Pa}\) or \(\text{N m}^{-2}\)).

(b) [2 Marks]
- M1: For using \(E_{\text{elastic}} = \frac{1}{2} F \Delta L\) with correct substitution of \(45.0\text{ N}\) and \(1.85 \times 10^{-3}\text{ m}\).
- A1: Correct final answer: \(4.16 \times 10^{-2}\text{ J}\) (or \(0.0416\text{ J}\) / \(0.042\text{ J}\)).

(c) [2 Marks]
- B1: For stating that planes of atoms slide past each other (due to movement of dislocations).
- B1: For explaining that atoms do not return to their original positions when the load is removed because bonds break and reform in new positions / permanent change of neighbouring atoms.

(a) (i)
- Use the ratchet to avoid over-tightening and squashing the wire.
- Measure the diameter in multiple positions along the length and at different orientations to account for non-circular cross-section (and average them).
- Check and adjust for any zero error on the micrometer before beginning.

(ii)
- Mean diameter \(d = \frac{0.38 + 0.39 + 0.37 + 0.38 + 0.38}{5} = 0.38\text{ mm}\)
- Absolute uncertainty is half the range: \(\frac{0.39 - 0.37}{2} = 0.01\text{ mm}\)
- So, \(d = 0.38 \pm 0.01\text{ mm}\).

(b) (i)
- From the resistivity equation, \(R = \frac{\rho L}{A}\). Since resistivity \(\rho\) and cross-sectional area \(A\) are constant, \(R \propto L\). This gives a straight line passing through the origin.
- The gradient is \(m = \frac{\rho}{A}\). Since \(A = \frac{\pi d^2}{4}\), the gradient \(m = \frac{4\rho}{\pi d^2}\).

(ii)
- Rearranging for resistivity: \(\rho = m \times \frac{\pi d^2}{4}\)
- Convert diameter to metres: \(d = 0.38 \times 10^{-3}\text{ m}\)
- \(A = \frac{\pi \times (0.38 \times 10^{-3})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\)
- \(\rho = 4.15 \times 1.134 \times 10^{-7} = 4.71 \times 10^{-7}\text{ }\Omega\text{ m}\) (accept \(4.7 \times 10^{-7}\text{ }\Omega\text{ m}\)).

(c)
- An increase in temperature causes the resistance/resistivity of the metal wire to increase.
- This means the resistance per unit length is no longer constant, making the graph non-linear or introducing systematic error.
- Minimise this by keeping the current low (using a variable resistor/reducing EMF) and turning off the power supply between taking readings.

(a) (i)
- Any two sensible precautions described clearly [1 mark each, max 2 marks]. E.g. Check for zero errors; Use ratchet; Measure at multiple positions/orientations.

(ii)
- Mean diameter = 0.38 mm [1 mark].
- Absolute uncertainty = 0.01 mm [1 mark].

(b) (i)
- Identifies \(R \propto L\) because \(\rho\) and \(A\) are constant [1 mark].
- Gradient expression: \(m = \frac{4\rho}{\pi d^2}\) or equivalent [1 mark].

(ii)
- Correct calculation of area \(A = 1.13 \times 10^{-7}\text{ m}^2\) [1 mark].
- Correct substitution to find \(\rho = 4.71 \times 10^{-7}\text{ }\Omega\text{ m}\) (allow 4.7) [1 mark].

(c)
- Identifies that resistivity/resistance increases with temperature [1 mark].
- Explains that this changes the slope/makes relationship non-linear [1 mark].
- Suggests using a switch to break circuit between readings / low voltage / high series resistance to limit current [1 mark].

(a) (i)
- The original length \(L\) should be measured from the clamping point at the top to the reference marker (e.g., a piece of taped paper or a marker) on the wire, using a tape measure or metre rule.
- The extension is measured by observing the movement of this marker relative to a fixed scale placed close to the wire.

(ii)
- A long wire increases the overall extension \(\Delta L\) for a given force.
- A thin wire has a small cross-sectional area \(A\), which also increases \(\Delta L\).
- Since the absolute uncertainty of the extension measuring tool is constant, a larger value of \(\Delta L\) results in a smaller percentage uncertainty.

(b) (i)
- \(\%\text{ uncertainty in } \Delta L = \frac{0.1\text{ mm}}{1.4\text{ mm}} \times 100\% \approx 7.14\%\) (accept 7.1% or 7.0%).

(ii)
- \(\%\text{ uncertainty in } L = \frac{0.01\text{ m}}{2.15\text{ m}} \times 100\% \approx 0.47\%\)
- \(\%\text{ uncertainty in } d = \frac{0.02\text{ mm}}{0.48\text{ mm}} \times 100\% \approx 4.17\%\)
- Since cross-sectional area is \(A = \frac{\pi d^2}{4}\), the percentage uncertainty in \(A\) is \(2 \times 4.17\% = 8.34\%\).
- Comparing the percentage uncertainties: \(L\) is 0.47%, \(\Delta L\) is 7.14%, and \(d\) contributes 8.34% (as \(d^2\)).
- Therefore, the diameter measurement contributes the most to the overall uncertainty.

(c)
- Sketch: A straight line starting from the origin up to a point labelled as the elastic limit.
- Young modulus is defined as \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F L}{A \Delta L}\).
- The gradient \(m\) of a force-extension graph is \(m = \frac{F}{\Delta L}\).
- Therefore, \(E = m \times \frac{L}{A} = m \times \frac{4L}{\pi d^2}\).

(a) (i)
- Measure length from top clamp/support to reference mark on wire using tape/rule [1 mark].
- Reference mark should move relative to a fixed scale [1 mark].

(ii)
- Long and thin wire yields a larger extension [1 mark].
- Larger extension reduces the percentage uncertainty (as absolute uncertainty is constant) [1 mark].

(b) (i)
- Percentage uncertainty = 7.1% (or 7.14%) [1 mark].

(ii)
- Calculated % uncertainty in \(L\) = 0.47% [1 mark].
- Calculated % uncertainty contribution from \(d\) (must be doubled for \(d^2\) or area) = 8.3% [1 mark].
- Compares all three and concludes that diameter contributes most [1 mark].

(c)
- Graph: Straight line from origin to elastic limit [1 mark].
- States that gradient \(m = \frac{F}{\Delta L}\) [1 mark].
- States final equation: \(E = m \times \frac{L}{A}\) or equivalent [1 mark].

(a) (i)
- The diagram must show a rectangular glass block.
- An incident ray from air entering the block, bending towards the normal inside the block.
- An emergent ray leaving the other side of the block, bending away from the normal.
- Labeled normal perpendicular to the glass-air boundary.
- Labeled angle of incidence \(i\) (between incident ray and normal in air) and angle of refraction \(r\) (between refracted ray and normal inside the block).

(ii)
- Trace the outline of the glass block on paper.
- Mark the paths of the incident and emergent rays using thin pencil marks/dots.
- Remove the block, draw the normal line and connect the entry and exit points to draw the refracted ray inside the block.
- Use a protractor to measure the angles of incidence and refraction relative to the normal line.

(b) (i)
- From Snell's Law: \(n_{\text{air}} \sin i = n_{\text{glass}} \sin r\). Since \(n_{\text{air}} \approx 1\), \(\sin i = n \sin r\).
- Plot \(\sin i\) on the y-axis and \(\sin r\) on the x-axis. The gradient of this straight line through the origin is the refractive index \(n\).

(ii)
- \(\sin(50^\circ) = 0.7660\)
- To find the uncertainty, calculate the values at the limits:
- \(\sin(51^\circ) = 0.7771\)
- \(\sin(49^\circ) = 0.7547\)
- Absolute uncertainty in \(\sin i \approx 0.7771 - 0.7660 = 0.0111\) (or using half-range \(\frac{0.7771 - 0.7547}{2} = 0.0112\)).
- \(\%\text{ uncertainty in } \sin i = \frac{0.0112}{0.7660} \times 100\% = 1.46\%\) (accept 1.5%).

(c)
- At small angles, the absolute reading error of the protractor (e.g. \(\pm 0.5^\circ\) or \(\pm 1^\circ\)) is a much larger fraction of the small measured angle \(r\), leading to very high percentage uncertainty.
- To improve reliability, the student should exclude data points where \(i < 15^\circ\) and focus on larger angles where the percentage uncertainty of measurements is significantly lower.

(a) (i)
- Neat diagram showing correct path of light bending towards normal inside glass [1 mark].
- Normal line drawn perpendicular to boundary [1 mark].
- Correctly labeled \(i\) and \(r\) [1 mark].

(ii)
- Trace block outline and mark ray paths with dots/pencil lines [1 mark].
- Remove block, construct normal, and use a protractor to measure angles relative to the normal [1 mark].

(b) (i)
- Plot \(\sin i\) on y-axis and \(\sin r\) on x-axis [1 mark].
- States gradient is equal to \(n\) because \(\sin i = n \sin r\) [1 mark].

(ii)
- Correct calculations of \(\sin i\) at limits (51 and/or 49 degrees) [1 mark].
- Correct final percentage uncertainty of 1.5% (or 1.46%) [1 mark].

(c)
- Explains that absolute protractor error is a larger fraction of a small angle [1 mark].
- Suggests excluding small angles (e.g., \(i < 15^\circ\)) to ensure data points have low percentage uncertainty [1 mark].

(a) (i)
- Progressive waves travel from the vibration generator along the string and reflect at the fixed end.
- These incident and reflected waves, which have equal frequency, wavelength, and speed, travel in opposite directions and superpose (interfere), forming nodes and antinodes.

(ii)
- For the first harmonic, \(\lambda = 2L\).

(b) (i)
- From \(v = f_1 \lambda\), we substitute \(\lambda = 2L\), giving \(v = 2 L f_1\).
- Equating this to the velocity equation: \(2 L f_1 = \sqrt{\frac{T}{\mu}}\).
- Since the tension is due to the hanging masses, \(T = Mg\). Substituting this gives:
\(f_1 = \frac{1}{2L} \sqrt{\frac{Mg}{\mu}}\).

(ii)
- The length \(L\) of the vibrating section of the string.
- The mass per unit length \(\mu\) of the string.

(c) (i)
- The gradient \(m\) of the graph of \(f_1\) vs \(\sqrt{M}\) is given by:
\(m = \frac{1}{2L} \sqrt{\frac{g}{\mu}}\)
- Rearranging for \(g\):
\(\sqrt{\frac{g}{\mu}} = 2 L m \implies g = 4 L^2 m^2 \mu\)
- Substitute the values:
\(g = 4 \times (1.20)^2 \times (72.5)^2 \times (3.20 \times 10^{-4})\)
\(g = 4 \times 1.44 \times 5256.25 \times (3.20 \times 10^{-4}) = 9.688 \approx 9.69\text{ m s}^{-2}\).

(ii)
- Formula for percentage uncertainty in \(g = 4 L^2 m^2 \mu\) is:
\(\% \Delta g = 2(\% \Delta L) + 2(\% \Delta m) + \% \Delta \mu\)
- \(\% \Delta L = \frac{0.01}{1.20} \times 100\% \approx 0.83\%\)
- \(\% \Delta m = 3.0\%\)
- \(\% \Delta \mu = 2.0\%\)
- \(\% \Delta g = 2(0.83\%) + 2(3.0\%) + 2.0\% = 1.66\% + 6.0\% + 2.0\% = 9.66\% \approx 9.7\%\) (accept 10%).

(a) (i)
- Waves travel along the string and reflect [1 mark].
- Superposition of coherent waves moving in opposite directions [1 mark].

(ii)
- \(\lambda = 2L\) [1 mark].

(b) (i)
- Links wave speed to frequency and wavelength, e.g. \(v = 2 L f_1\) [1 mark].
- Substitutes \(T = Mg\) and rearranges correctly to find the given formula [1 mark].

(ii)
- Length of string \(L\) [1 mark].
- Mass per unit length \(\mu\) [1 mark].

(c) (i)
- Equates gradient to \(\frac{1}{2L}\sqrt{\frac{g}{\mu}}\) [1 mark].
- Obtains \(g = 9.69\text{ m s}^{-2}\) (allow 9.7 or 9.68) [1 mark].

(ii)
- Calculates percentage uncertainty in length = 0.83% [1 mark].
- Combines uncertainties correctly (doubling for squared terms): \(2(0.83) + 2(3.0) + 2.0 = 9.7\%\) [1 mark].

The ray enters face AB normally, meaning it passes into the glass without deviation.

Inside the prism, the ray hits the face AC. Since the angle of the prism at A is \(55.0^\circ\) and the angle at B is \(90.0^\circ\), the angle of incidence \(\theta\) at face AC is found by geometry to be:
\(\theta = 55.0^\circ\).

For total internal reflection at the face AC to occur, the angle of incidence must be greater than or equal to the critical angle \(\theta_c\):
\(\sin\theta \ge \sin\theta_c\)

Since \(\sin\theta_c = \frac{n}{1.55}\), we have:
\(\sin(55.0^\circ) \ge \frac{n}{1.55}\)

\(n \le 1.55 \sin(55.0^\circ) = 1.55 \times 0.8192 = 1.27\).

Thus, the maximum value of \(n\) is \(1.27\).

1 mark for the correct answer B.

Each alpha decay reduces the nucleon number \(A\) by 4 and the proton number \(Z\) by 2. Each beta-minus decay keeps \(A\) unchanged and increases \(Z\) by 1.

For 3 alpha decays and 2 beta-minus decays:
Change in \(A\): \(\Delta A = 3 \times (-4) = -12\)
Change in \(Z\): \(\Delta Z = 3 \times (-2) + 2 \times (+1) = -6 + 2 = -4\)

New nucleon number: \(A = 226 - 12 = 214\)
New proton number: \(Z = 88 - 4 = 84\)

1 mark for the correct answer A.

Taking moments about the pivot:

1. The weight of the uniform beam (\(100\text{ N}\)) acts at its center of gravity, which is at the midpoint \(1.50\text{ m}\) from either end. The distance from the pivot (at \(0.60\text{ m}\) from the left) to the center of gravity is \(1.50 - 0.60 = 0.90\text{ m}\) to the right. This creates a clockwise moment:
\(\text{Moment}_{\text{beam}} = 100\text{ N} \times 0.90\text{ m} = 90\text{ N m}\).

2. The downward force of \(25\text{ N}\) acts at the right end of the beam, which is at a distance of \(3.0 - 0.60 = 2.40\text{ m}\) from the pivot. This also creates a clockwise moment:
\(\text{Moment}_{\text{force}} = 25\text{ N} \times 2.40\text{ m} = 60\text{ N m}\).

3. The load of weight \(W\) acts at the left end of the beam, at a distance of \(0.60\text{ m}\) to the left of the pivot. This creates a counter-clockwise moment:
\(\text{Moment}_{\text{load}} = W \times 0.60\text{ m}\).

For horizontal equilibrium, the sum of clockwise moments must equal the sum of counter-clockwise moments:
\(0.60 W = 90 + 60\)
\(0.60 W = 150\)
\(W = 250\text{ N}\).

1 mark for the correct answer C.

The resistance of a wire is given by \(R = \rho \frac{L}{A}\), where \(\rho\) is the resistivity of the material.

Since the volume \(V = A L\) of the wire remains constant during stretching, when the length is increased to \(1.20 L\), the cross-sectional area \(A'\) must decrease such that:
\(V = A' (1.20 L) = A L \implies A' = \frac{A}{1.20}\).

The new resistance \(R'\) is:
\(R' = \rho \frac{1.20 L}{A'} = \rho \frac{1.20 L}{A / 1.20} = (1.20)^2 \rho \frac{L}{A} = 1.44 R\).

1 mark for the correct answer C.

The extension of a wire is given by \(\Delta L = \frac{F L}{A E}\), where \(F\) is the applied load, \(L\) is the original length, \(A\) is the cross-sectional area, and \(E\) is the Young modulus.

Since both wires are made of the same material (same \(E\)) and support equal loads (same \(F\)):
\(\Delta L \propto \frac{L}{A} \propto \frac{L}{d^2}\) (since \(A = \pi d^2 / 4\)).

For wire P: \(\Delta L_P \propto \frac{L}{d^2}\)

For wire Q: \(\Delta L_Q \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = 0.5 \frac{L}{d^2}\)

Therefore, the ratio is:
\(\frac{\Delta L_P}{\Delta L_Q} = \frac{1}{0.5} = 2.0\).

1 mark for the correct answer C.

The kinetic energy \(E_k\) gained by an electron is \(E_k = eV\).

The momentum \(p\) is related to the kinetic energy by:
\(p = \sqrt{2 m_e E_k} = \sqrt{2 m_e e V}\).

The de Broglie wavelength \(\lambda\) is given by:
\(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m_e e V}}\).

Thus, \(\lambda \propto \frac{1}{\sqrt{V}}\).

If the potential difference is increased to \(4V\), the new wavelength \(\lambda'\) is:
\(\lambda' \propto \frac{1}{\sqrt{4V}} = \frac{1}{2 \sqrt{V}} = 0.50 \lambda\).

1 mark for the correct answer B.

First, find the acceleration \(a\) of the body using Newton's second law:
\(a = \frac{F}{m} = \frac{60\text{ N}}{5.0\text{ kg}} = 12\text{ m s}^{-2}\).

Next, find the final velocity \(v\) using the equation of motion:
\(v = u + at = 4.0\text{ m s}^{-1} + (12\text{ m s}^{-2} \times 3.0\text{ s}) = 4.0 + 36 = 40\text{ m s}^{-1}\).

Finally, calculate the final kinetic energy \(E_k\):
\(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 5.0\text{ kg} \times (40\text{ m s}^{-1})^2 = 2.5 \times 1600 = 4000\text{ J} = 4.0\text{ kJ}\).

1 mark for the correct answer D.

For a string fixed at both ends, the boundaries must be nodes.

- The first harmonic (fundamental) has 2 nodes at the ends and 1 antinode in the middle.
- The second harmonic has 3 nodes (including ends) and 2 antinodes.
- The third harmonic has 3 vibrating loops, which means it has 3 antinodes and 4 nodes (one at each end, and two internal nodes between the loops).

Therefore, there are 4 nodes and 3 antinodes.

1 mark for the correct answer C.

The light ray is incident normally on one of the equal faces, so it enters without deviation. It meets the hypotenuse face at an angle of incidence of \(45^\circ\). For total internal reflection (TIR) to occur at the hypotenuse boundary, the angle of incidence must be greater than or equal to the critical angle \(\theta_c\). Thus, \(\sin(45^\circ) \ge \sin \theta_c = \frac{n_l}{n}\). Rearranging gives \(n_l \le n \sin(45^\circ) = 1.52 \times 0.7071 = 1.075\). Therefore, the maximum value of \(n_l\) is approximately 1.07.

1 mark for the correct option A. Reject all other options.

At equilibrium, \(mg = ke\), so the spring constant is \(k = \frac{mg}{e} = \frac{0.25 \times 9.81}{0.040} = 61.3\text{ N m}^{-1}\). The mass is pulled down an additional 2.0 cm, which is the amplitude of the subsequent oscillations, so \(A = 0.020\text{ m}\). The maximum kinetic energy equals the total mechanical energy of the system: \(E_{k,\max} = \frac{1}{2} k A^2 = \frac{1}{2} \times 61.3 \times (0.020)^2 = 1.23 \times 10^{-2}\text{ J}\), which rounds to \(1.2 \times 10^{-2}\text{ J}\).

1 mark for the correct option A. Reject all other options.

By conservation of linear momentum, the total momentum before decay is zero, so the alpha particle and the Radium nucleus have equal and opposite momenta: \(p_{\alpha} = -p_{Ra}\). The kinetic energy of each particle can be expressed as \(E = \frac{p^2}{2m}\). Since their momenta are equal in magnitude, the kinetic energies are inversely proportional to their masses: \(\frac{E_{\alpha}}{E_{Ra}} = \frac{m_{Ra}}{m_{\alpha}} \approx \frac{224}{4} = 56\). Therefore, \(E_{\alpha} = 56 E_{Ra}\). Since the total energy released is \(Q = E_{\alpha} + E_{Ra} = E_{\alpha} + \frac{E_{\alpha}}{56} = E_{\alpha} \left(\frac{57}{56}\right)\), we find \(E_{\alpha} = \frac{56}{57} Q = \frac{224}{228} Q\).

1 mark for the correct option B. Reject all other options.

The energy stored in a capacitor is given by \(E = \frac{1}{2} C V^2\). Since energy is proportional to the square of the potential difference, if the energy decreases to 25% of its initial value, the potential difference must decrease to \(\sqrt{0.25} = 0.5\) (or 50%) of its initial value. Thus, \(V = 0.5 V_0\). During discharge, \(V = V_0 e^{-t/RC}\). Substituting \(V\) gives \(0.5 V_0 = V_0 e^{-t/RC} \implies e^{-t/RC} = 0.5 \implies e^{t/RC} = 2\). Taking the natural logarithm of both sides yields \(\frac{t}{RC} = \ln 2 \implies t = RC \ln 2\).

1 mark for the correct option A. Reject all other options.

Since the charges have opposite signs, the point where the net electric field is zero must lie outside the region between the charges, and closer to the charge of smaller magnitude (the \(-1.0\text{ }\mu\text{C}\) charge). Let this point be at a distance \(d\) to the outer side of the \(-1.0\text{ }\mu\text{C}\) charge. Equating the magnitudes of the electric field strengths at this point: \(\frac{k Q_1}{(1.2 + d)^2} = \frac{k Q_2}{d^2} \implies \frac{4.0 \times 10^{-6}}{(1.2 + d)^2} = \frac{1.0 \times 10^{-6}}{d^2} \implies \frac{4}{(1.2 + d)^2} = \frac{1}{d^2}\). Taking the square root of both sides gives \(\frac{2}{1.2 + d} = \frac{1}{d} \implies 2d = 1.2 + d \implies d = 1.2\text{ m}\).

1 mark for the correct option C. Reject all other options.

The de Broglie wavelength is given by \(\lambda = \frac{h}{p}\), where \(p\) is the momentum. Since the electron and proton have the same wavelength, they must have the same momentum. The kinetic energy is related to momentum by \(E = \frac{p^2}{2m}\). Since momentum is constant, kinetic energy is inversely proportional to mass. Because the mass of the electron is much smaller than the mass of the proton (\(m_e \ll m_p\)), it follows that \(E_e > E_p\).

1 mark for the correct option B. Reject all other options.

For a uniform beam, the weight of 150 N acts at its centre of gravity, which is at its midpoint (2.0 m from the left end). Since the pivot is 1.0 m from the left end, the distance from the pivot to the weight is \(2.0 - 1.0 = 1.0\text{ m}\). This produces a clockwise moment of \(150\text{ N} \times 1.0\text{ m} = 150\text{ N m}\). The vertical upward force \(F\) is applied at the right-hand end (4.0 m from the left end), so its distance from the pivot is \(4.0 - 1.0 = 3.0\text{ m}\). This produces an anticlockwise moment of \(F \times 3.0\text{ m}\). Applying the principle of moments: \(3.0 F = 150 \implies F = 50\text{ N}\).

1 mark for the correct option B. Reject all other options.

At the maximum height of its trajectory, the vertical component of the projectile's velocity is zero (\(v_y = 0\)). The horizontal component of the velocity remains constant throughout the flight because there is no horizontal acceleration. This component is \(v_x = u \cos \theta = 25 \times \cos(30^\circ) = 21.65\text{ m s}^{-1}\). The speed of the projectile at maximum height is the magnitude of the velocity vector, which is simply \(v_x \approx 21.7\text{ m s}^{-1}\).

1 mark for the correct option C. Reject all other options.

The critical angle \(\theta_c\) is related to the refractive indices by \(\sin \theta_c = \frac{n_{\text{liquid}}}{n_{\text{glass}}}\). Rearranging this formula to find the refractive index of the liquid gives: \(n_{\text{liquid}} = n_{\text{glass}} \sin \theta_c = 1.52 \times \sin(58.5^\circ) = 1.52 \times 0.8526 = 1.296 \approx 1.30\).

1 mark for the correct answer C.

First convert energy to Joules: \(E_k = 120 \times 1.60 \times 10^{-19}\text{ J} = 1.92 \times 10^{-17}\text{ J}\). The momentum of the electron is given by \(p = \sqrt{2 m_e E_k} = \sqrt{2 \times 9.11 \times 10^{-31}\text{ kg} \times 1.92 \times 10^{-17}\text{ J}} \approx 5.91 \times 10^{-24}\text{ kg m s}^{-1}\). The de Broglie wavelength is \(\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}\text{ J s}}{5.91 \times 10^{-24}\text{ kg m s}^{-1}} \approx 1.12 \times 10^{-10}\text{ m} = 0.11\text{ nm}\).

1 mark for the correct answer B.

Since the beam is uniform, its weight of \(120\text{ N}\) acts at its centre of gravity, which is \(1.5\text{ m}\) from either end. The distance from the pivot (at \(1.0\text{ m}\) from the left) to the centre of gravity is \(1.5\text{ m} - 1.0\text{ m} = 0.5\text{ m}\). Taking moments about the pivot: Clockwise moment = \(120\text{ N} \times 0.5\text{ m} = 60\text{ N m}\). Anticlockwise moment = \(W \times 1.0\text{ m}\). For equilibrium, \(W \times 1.0 = 60\text{ N m}\), which gives \(W = 60\text{ N}\).

1 mark for the correct answer B.

The Young modulus is given by \(E = \frac{F L}{A \Delta L}\), so \(\Delta L = \frac{F L}{A E}\). For the first wire, \(\Delta L = \frac{F L}{A E}\). For the second wire, \(\Delta L_2 = \frac{F_2 (2L)}{(3A) E} = \frac{2 F_2 L}{3 A E}\). Since both extensions are equal, \(\frac{2 F_2 L}{3 A E} = \frac{F L}{A E}\). Simplifying this gives \(\frac{2}{3} F_2 = F\), or \(F_2 = \frac{3}{2}F\).

1 mark for the correct answer C.

Let the initial length be \(L\) and initial area be \(A\). The new length is \(L' = 1.20 L\). Since the volume \(V = A L\) remains constant, the new area \(A'\) is \(A' = \frac{A}{1.20}\). The resistance is \(R = \rho \frac{L}{A}\). The new resistance is \(R' = \rho \frac{1.20 L}{A / 1.20} = 1.44 \rho \frac{L}{A} = 1.44 R\).

1 mark for the correct answer C.

At the maximum height, the vertical component of the velocity is zero. Using the equation of motion \(v_y = u_y - g t\) where \(u_y = v \sin \theta\) and \(v_y = 0\), we get \(0 = v \sin \theta - g t\). Solving for \(t\) gives \(t = \frac{v \sin \theta}{g}\).

1 mark for the correct answer B.

The relationship can be written as \(T = 2\pi L^{1/2} g^{-1/2}\). The percentage uncertainty in \(T\) is given by \(\frac{\Delta T}{T} \times 100 = \frac{1}{2} \left( \frac{\Delta L}{L} \times 100 \right) + \frac{1}{2} \left( \frac{\Delta g}{g} \times 100 \right) = \frac{1}{2}(1.5\%) + \frac{1}{2}(2.5\%) = 0.75\% + 1.25\% = 2.0\%\).

1 mark for the correct answer B.

The atomic number \(Z = 12\) gives the number of protons, which is 12. The mass number \(A = 25\) is the sum of protons and neutrons, so the number of neutrons is \(25 - 12 = 13\). The ion has a charge of \(2+\), which means it has lost 2 electrons. A neutral magnesium atom has 12 electrons, so this ion has \(12 - 2 = 10\) electrons.

1 mark for the correct answer A.

For total internal reflection to occur at the boundary, the angle of incidence \(\theta_i\) must be greater than the critical angle \(\theta_c\). Therefore, \(\sin\theta_i > \sin\theta_c\). Since \(\sin\theta_c = \frac{n_l}{n_g}\), we have \(\sin(54.0^\circ) > \frac{n_l}{1.62}\). This gives \(n_l < 1.62 \times \sin(54.0^\circ) = 1.3106\). Thus, the maximum possible value for the refractive index of the liquid is 1.31.

Award 1 mark for the correct answer A. No partial marks.

In the process \(n \rightarrow p + e^- + \bar{\nu}_e\): Lepton number before is 0, after is \(0 + 1 - 1 = 0\) (conserved). Baryon number before is 1, after is \(1 + 0 + 0 = 1\) (conserved). Charge before is 0, after is \(+1 - 1 + 0 = 0\) (conserved). Therefore, all three quantities are conserved.

Award 1 mark for the correct answer A. No partial marks.

Let \(T_R\) be the tension in the right-hand string. Taking moments about the left end: \(\sum M_{\text{clockwise}} = \sum M_{\text{anticlockwise}}\). The weight of the uniform beam acts at its center of gravity (1.0 m from either end). Thus, \((80\text{ N} \times 0.50\text{ m}) + (120\text{ N} \times 1.0\text{ m}) = T_R \times 2.0\text{ m}\), which gives \(40 + 120 = 2.0 T_R\), hence \(T_R = 80\text{ N}\).

Award 1 mark for the correct answer B. No partial marks.

The resistance is given by \(R = \frac{\rho L}{A}\), where the cross-sectional area \(A = \frac{\pi d^2}{4}\). The percentage uncertainty in \(A\) is \(2 \times \\%\Delta d = 2 \times 1\\% = 2\\%\). The total percentage uncertainty in \(R\) is the sum of the percentage uncertainties of \(\rho\), \(L\), and \(A\), which equals \(2\\% + 1.5\\% + 2\\% = 5.5\\%\).

Award 1 mark for the correct answer B. No partial marks.

The kinetic energy gained by the electron is \(E_k = eV\). The momentum of the electron is related to its kinetic energy by \(p = \sqrt{2m E_k} = \sqrt{2meV}\). The de Broglie wavelength is \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}\).

Award 1 mark for the correct answer A. No partial marks.

Resistance \(R\) is proportional to \(\frac{L}{A}\), which is proportional to \(\frac{L}{d^2}\). For wire X, \(R_X \propto \frac{L}{d^2}\). For wire Y, \(R_Y \propto \frac{3L}{(2d)^2} = \frac{3L}{4d^2}\). Thus, \(R_Y = \frac{3}{4} R_X\).

Award 1 mark for the correct answer A. No partial marks.

For the third harmonic on a string fixed at both ends, the length \(L\) contains 1.5 wavelengths: \(L = \frac{3\lambda}{2} \implies \lambda = \frac{2L}{3}\). The distance between adjacent nodes is half a wavelength: \(\frac{\lambda}{2} = \frac{L}{3}\).

Award 1 mark for the correct answer A. No partial marks.

The horizontal displacement is \(s_x = u_x t = 15 \times 2.0 = 30\text{ m}\). The vertical displacement is \(s_y = \frac{1}{2} g t^2 = 0.5 \times 9.81 \times (2.0)^2 = 19.62\text{ m}\). The magnitude of the resultant displacement is \(s = \sqrt{s_x^2 + s_y^2} = \sqrt{30^2 + 19.62^2} \approx 35.8\text{ m}\) which rounds to \(36\text{ m}\) to two significant figures.

Award 1 mark for the correct answer B. No partial marks.

First, we use Snell's Law to find the angle of refraction \(\theta_2\) in the glass:

\(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\)

Since the ray is incident from air, \(n_1 = 1.00\):

\(1.00 \times \sin(35.0^\circ) = 1.52 \times \sin(\theta_2)\)

\(\sin(\theta_2) = \frac{\sin(35.0^\circ)}{1.52} \approx 0.3774\)

\(\theta_2 = \arcsin(0.3774) \approx 22.17^\circ\)

The angle of deviation \(\theta_d\) is the angle between the original direction of the ray and its new refracted path:

\(\theta_d = \theta_1 - \theta_2 = 35.0^\circ - 22.17^\circ \approx 12.8^\circ\)

1 mark for selecting the correct option (A).
- Reject other options: B represents the angle of refraction; D represents the sum of the angles.

Since alpha particles are positively charged and beta-minus particles are negatively charged, they experience forces in opposite directions, causing them to deflect in opposite directions. The magnetic force acts as a centripetal force: \(Bqv = \frac{mv^2}{r}\), which gives the radius of curvature as \(r = \frac{mv}{Bq}\). Since beta-minus particles have a much smaller mass-to-charge ratio than alpha particles, they have a much smaller radius of curvature and are therefore deflected more sharply for the same speed. Gamma radiation is uncharged and thus passes through undeflected.

1 mark for selecting the correct option (C).
- Reject options stating alpha deflects more (mass-to-charge ratio makes beta deflect more for equivalent speeds), or that gamma is deflected.

The energy stored in a capacitor is given by \(E = \frac{1}{2}CV^2\).
During discharge, the potential difference across the capacitor decreases exponentially with time according to:
\(V = V_0 e^{-t/RC\)}

At \(t = RC\), the voltage is:
\(V = V_0 e^{-1}\)

The energy remaining in the capacitor is:
\(E = \frac{1}{2} C (V_0 e^{-1})^2 = \left(\frac{1}{2} C V_0^2\right) e^{-2} = E_0 e^{-2}\)

Therefore, the fraction of the initial energy remaining is \(e^{-2}\).

1 mark for selecting the correct option (B).
- Reject options based on linear decrease or incorrect power of the exponential factor.

For the particle to remain in equilibrium, the upward electric force must balance the downward gravitational force:

\(F_E = F_g\)

\(q E = m g\)

The electric field strength \(E\) between two parallel plates is related to the potential difference \(V\) and separation \(d\) by:

\(E = \frac{V}{d}\)

Substituting this into the equilibrium equation:

\(q \left(\frac{V}{d}\right) = m g\)

Solving for \(V\):

\(V = \frac{m g d}{q}\)

1 mark for selecting the correct option (C).
- Reject options where variables are in the wrong algebraic relationships.

The volume of a cylinder is given by \(V = \pi r^2 h\).

The percentage uncertainty in the volume is:

\(\frac{\Delta V}{V} \times 100\% = \left(2 \frac{\Delta r}{r} + \frac{\Delta h}{h}\right) \times 100\%

First, calculate the percentage uncertainty in \)r\):

\(\frac{\Delta r}{r} \times 100\% = \frac{0.05}{2.00} \times 100\% = 2.5\%\)

Next, calculate the percentage uncertainty in \(h\):

\(\frac{\Delta h}{h} \times 100\% = \frac{0.2}{10.0} \times 100\% = 2.0\%\)

Now combine them:

\(\frac{\Delta V}{V} \times 100\% = 2(2.5\%) + 2.0\% = 5.0\% + 2.0\% = 7.0\%\)

1 mark for selecting the correct option (C).
- Reject A (just adding percentage uncertainties without doubling for \(r^2\)).
- Reject B (adding absolute uncertainties directly).

The kinetic energy \(E_k\) gained by a particle of charge \(q\) accelerated through a potential difference \(V\) is \(E_k = q V\). Since both electron and proton have charge of magnitude \(e\), they gain the same kinetic energy:

\(E_k = e V\)

The momentum \(p\) of a particle of mass \(m\) is given by:

\(p = \sqrt{2 m E_k}\)

The de Broglie wavelength is:

\(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m e V}}\)

Therefore, the ratio of the wavelengths is:

\(\frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2 m_e e V}}}{\frac{h}{\sqrt{2 m_p e V}}} = \sqrt{\frac{m_p}{m_e}}\)

Given \(m_p = 1800 m_e\):

\(\frac{\lambda_e}{\lambda_p} = \sqrt{1800} \approx 42.4\)

1 mark for selecting the correct option (C).
- Reject A (using the mass ratio directly without square root).
- Reject D (the inverse ratio).

Take moments about the pivot.

The weight \(W\) acts at the midpoint of the uniform beam (distance \(\frac{1}{2}L\) from the pivot), producing a clockwise moment:
\(\text{Clockwise Moment} = W \times \frac{1}{2}L\)

The vertical tension \(T\) acts at a distance of \(\frac{3}{4}L\) from the pivot, producing an anticlockwise moment:
\(\text{Anticlockwise Moment} = T \times \frac{3}{4}L\)

For rotational equilibrium, the sum of clockwise moments must equal the sum of anticlockwise moments:
\(T \times \frac{3}{4}L = W \times \frac{1}{2}L\)

Solving for \(T\):
\(T = W \times \frac{1}{2} \times \frac{4}{3} = \frac{2}{3}W\)

1 mark for selecting the correct option (B).
- Reject A (assumes weight acts at the end of the beam).
- Reject C and D (incorrect balance of moments).

Let the initial velocity be \(u\). The initial kinetic energy is:

\(E_{k,\text{initial}} = \frac{1}{2} m u^2\)

At the highest point, the vertical component of the velocity is zero. The horizontal component of the velocity remains constant throughout the flight because there is no horizontal acceleration:

\(v_x = u \cos(60^\circ) = 0.5 u\)

Thus, at the highest point, the total velocity is equal to the horizontal component, \(v = 0.5 u\). The kinetic energy at the highest point is:

\(E_{k,\text{highest}} = \frac{1}{2} m (0.5 u)^2 = 0.25 \left(\frac{1}{2} m u^2\right) = 0.25 E_{k,\text{initial}}\)

The ratio is \(0.25\).

1 mark for selecting the correct option (A).
- Reject B (ratio of velocities rather than kinetic energy).

The resistance of a wire is given by \(R = \rho \frac{L}{A}\), where the cross-sectional area \(A = \frac{\pi d^2}{4}\).

Thus, \(R \propto \frac{L}{d^2}\).

For wire \(X\): \(R_X = R\).

For wire \(Y\): \(R_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = 0.5 \frac{L}{d^2}\). Thus, \(R_Y = 0.5 R\).

Since the wires are connected in parallel across a constant voltage \(V\), the power dissipated in each wire is given by \(P = \frac{V^2}{R}\).

Therefore, the power dissipated in \(X\) is \(P_X = \frac{V^2}{R_X} = \frac{V^2}{R}\).

The power dissipated in \(Y\) is \(P_Y = \frac{V^2}{R_Y} = \frac{V^2}{0.5R} = \frac{2V^2}{R}\).

The ratio is \(\frac{P_X}{P_Y} = \frac{V^2/R}{2V^2/R} = 0.50\).

1 mark for the correct answer B.

The kinetic energy gained by the electron is \(E_k = eV = \frac{p^2}{2m}\), where \(p\) is its momentum.

Rearranging for momentum gives \(p = \sqrt{2meV}\).

The de Broglie wavelength is \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}\).

When the potential difference is increased to \(4V\), the new wavelength \(\lambda'\) is:

\(\lambda' = \frac{h}{\sqrt{2me(4V)}} = \frac{1}{\sqrt{4}} \frac{h}{\sqrt{2meV}} = \frac{1}{2}\lambda = 0.50\lambda\).

1 mark for the correct answer B.

For a wire of length \(L\) fixed at both ends, stationary waves are formed with nodes at both ends.

The condition for stationary waves is \(L = n \frac{\lambda}{2}\), where \(n\) is the harmonic number.

For the third harmonic, \(n = 3\), so:

\(L = 3 \frac{\lambda}{2} \implies \lambda = \frac{2L}{3}\).

Using the wave equation \(v = f\lambda\), the frequency is:

\(f = \frac{v}{\lambda} = \frac{v}{\frac{2L}{3}} = \frac{3v}{2L}\).

1 mark for the correct answer C.