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(a) Using \(v = u + at\) where \(u = 0\text{ m s}^{-1}\), \(a = 2.4\text{ m s}^{-2}\), and \(t = 3.0\text{ s}\): \(v = 0 + (2.4 \times 3.0) = 7.2\text{ m s}^{-1}\). (b) Distance in stage 1: \(s_1 = ut + \frac{1}{2}at^2 = 0 + 0.5 \times 2.4 \times 3.0^2 = 10.8\text{ m}\). Distance in stage 2: \(s_2 = v \times t = 7.2 \times 5.0 = 36.0\text{ m}\). Distance in stage 3: \(s_3 = 12.0\text{ m}\). Total distance \(s = 10.8 + 36.0 + 12.0 = 58.8\text{ m}\). (c) Using \(v^2 = u^2 + 2as\) for the final stage: \(0 = 7.2^2 + 2 \times a \times 12.0\). \(24.0 a = -51.84\), so \(a = -2.16\text{ m s}^{-2}\). The deceleration is \(2.16\text{ m s}^{-2}\).

(a) [1.25 marks] - Formula: \(v = u + at\) [0.5 marks] - Substitution and calculation leading to \(7.2\text{ m s}^{-1}\) [0.75 marks]. (b) [3.0 marks] - Stage 1 distance calculation: \(10.8\text{ m}\) [1.0 mark] - Stage 2 distance calculation: \(36.0\text{ m}\) [1.0 mark] - Total distance addition of all 3 stages: \(58.8\text{ m}\) [1.0 mark]. (c) [2.0 marks] - Formula: \(v^2 = u^2 + 2as\) and substitution [1.0 mark] - Correct calculation of deceleration magnitude as \(2.16\text{ m s}^{-2}\) [1.0 mark].

(a) Taking moments about pillar A: Weight of board \(W_b = 24 \times 9.81 = 235.44\text{ N}\) acting at \(1.8\text{ m}\). Weight of diver \(W_d = 70 \times 9.81 = 686.7\text{ N}\) acting at \(3.6\text{ m}\). Clockwise moments = Counter-clockwise moments: \((W_b \times 1.8) + (W_d \times 3.6) = F_B \times 1.2\). \((235.44 \times 1.8) + (686.7 \times 3.6) = 1.2 F_B\). \(423.79 + 2472.12 = 1.2 F_B\). \(2895.91 = 1.2 F_B \implies F_B = 2413.26\text{ N} \approx 2410\text{ N}\). (b) For vertical equilibrium: \(F_A + F_B = W_b + W_d\). \(F_A + 2413.26 = 235.44 + 686.7 = 922.14\text{ N}\). \(F_A = 922.14 - 2413.26 = -1491.12\text{ N}\). The magnitude is \(1490\text{ N}\) and the negative sign indicates the force acts vertically downwards.

(a) [3.25 marks] - Correct calculations of both weights: \(235.44\text{ N}\) and \(686.7\text{ N}\) [1.0 mark] - Taking moments about A: \((235.44 \times 1.8) + (686.7 \times 3.6) = 1.2 F_B\) [1.25 marks] - Correct value of \(F_B\): \(2410\text{ N}\) (accept \(2400\text{ N}\) to \(2415\text{ N}\)) [1.0 mark]. (b) [3.0 marks] - Use of vertical equilibrium condition: \(F_A + F_B = W_b + W_d\) [1.0 mark] - Correct magnitude: \(1490\text{ N}\) (accept \(1480\text{ N}\) to \(1500\text{ N}\)) [1.0 mark] - Correctly stating direction as downward [1.0 mark].

(a) Thrust force is given by \(F = \frac{\Delta p}{\Delta t} = \frac{\Delta m}{\Delta t} v_{rel}\). Substituting values: \(F = 0.25 \times 600 = 150\text{ N}\). (b) Change in momentum \( \Delta p = F \times \Delta t = 150 \times 8.0 = 1200\text{ N s}\) (or \(\text{kg m s}^{-1}\)). (c) Change in velocity \(\Delta v = \frac{\Delta p}{m} = \frac{1200}{150} = 8.0\text{ m s}^{-1}\). Final velocity \(v = u + \Delta v = 4.5 + 8.0 = 12.5\text{ m s}^{-1}\).

(a) [2.25 marks] - Formula: \(F = (\Delta m / \Delta t) \times v\) [1.0 mark] - Calculation: \(150\text{ N}\) [1.25 marks]. (b) [2.0 marks] - Formula: \(\Delta p = F \Delta t\) [1.0 mark] - Calculation: \(1200\text{ N s}\) (accept \(\text{kg m s}^{-1}\)) [1.0 mark]. (c) [2.0 marks] - Calculation of \(\Delta v = 8.0\text{ m s}^{-1}\) [1.0 mark] - Final velocity: \(12.5\text{ m s}^{-1}\) [1.0 mark].

(a) Force \(F = mg = 6.5 \times 9.81 = 63.765\text{ N}\). Tensile stress \(\sigma = \frac{F}{A} = \frac{63.765}{1.2 \times 10^{-6}} = 5.31375 \times 10^7\text{ Pa} \approx 5.3 \times 10^7\text{ Pa}\). (b) Since \(E = \frac{\sigma}{\epsilon} = \frac{\sigma}{\Delta L / L}\), we have \(\Delta L = \frac{\sigma L}{E} = \frac{5.31375 \times 10^7 \times 2.2}{1.1 \times 10^{11}} = 1.06275 \times 10^{-3}\text{ m} \approx 1.1 \times 10^{-3}\text{ m}\) (or \(1.1\text{ mm}\)). (c) Elastic strain energy \(E_{elastic} = \frac{1}{2} F \Delta L = 0.5 \times 63.765 \times 1.06275 \times 10^{-3} = 0.03388\text{ J} \approx 0.034\text{ J}\).

(a) [2.25 marks] - Force calculation: \(63.8\text{ N}\) [0.5 marks] - Stress formula and substitution: \(5.31 \times 10^7\text{ Pa}\) [1.75 marks]. (b) [2.0 marks] - Young modulus relation formula: \(E = \sigma L / \Delta L\) rearranged [1.0 mark] - Calculation of extension: \(1.1\text{ mm}\) [1.0 mark]. (c) [2.0 marks] - Energy formula: \(E = 0.5 F \Delta L\) [1.0 mark] - Calculation: \(0.034\text{ J}\) (accept \(0.033\text{ J}\) to \(0.035\text{ J}\)) [1.0 mark].

(a) The volume of a cylinder is \(V = \frac{\pi d^2 L}{4}\). The percentage uncertainty in volume is \(\frac{\Delta V}{V} = 2 \times \frac{\Delta d}{d} + \frac{\Delta L}{L}\). Percentage uncertainty in \(d\): \(\frac{0.2}{12.4} \times 100\% = 1.613\%\). Percentage uncertainty in \(L\): \(\frac{0.5}{52.0} \times 100\% = 0.962\%\). Therefore, \(\frac{\Delta V}{V} = 2 \times 1.613\% + 0.962\% = 3.226\% + 0.962\% = 4.188\% \approx 4.2\%\). (b) Density \(\rho = \frac{m}{V}\). The percentage uncertainty in density is \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta V}{V}\). Percentage uncertainty in mass \(m\): \(\frac{0.1}{48.2} \times 100\% = 0.207\%\). Total percentage uncertainty in density: \(0.207\% + 4.188\% = 4.395\% \approx 4.4\%\).

(a) [3.25 marks] - Identifying volume formula and fractional uncertainty expression: \(\Delta V / V = 2 (\Delta d/d) + \Delta L/L\) [1.25 marks] - Individual percentage uncertainties calculated correctly: \(1.61\%\) and \(0.96\%\) [1.0 mark] - Calculation showing sum is \(4.2\%\) [1.0 mark]. (b) [3.0 marks] - Identifying addition of mass and volume uncertainties: \(\Delta \rho/\rho = \Delta m/m + \Delta V/V\) [1.0 mark] - Calculation of percentage uncertainty in mass: \(0.21\%\) [1.0 mark] - Calculation of total percentage uncertainty: \(4.4\%\) (accept \(4.39\%\) to \(4.40\%\)) [1.0 mark].

(a) The nucleus of \({}^{14}_{6}\text{C}\) has 6 protons and 8 neutrons (since \(14 - 6 = 8\)). (b) At the nucleon level, beta-minus decay is: \(n \to p + e^{-} + \bar{\nu}_e\) (where \(\bar{\nu}_e\) is the electron antineutrino). (c) Specific charge is \(\frac{\text{charge}}{\text{mass}}\). Charge of the nucleus \(Q = 6 \times 1.60 \times 10^{-19}\text{ C} = 9.60 \times 10^{-19}\text{ C}\). Mass of the nucleus \(M \approx 14 \times 1.66 \times 10^{-27}\text{ kg} = 2.324 \times 10^{-26}\text{ kg}\). Specific charge \(= \frac{9.60 \times 10^{-19}}{2.324 \times 10^{-26}} = 4.13 \times 10^7\text{ C kg}^{-1} \approx 4.1 \times 10^7\text{ C kg}^{-1}\).

(a) [1.25 marks] - 6 protons and 8 neutrons correctly identified [1.25 marks]. (b) [2.0 marks] - Identification of neutron converting to proton and electron: \(n \to p + e^{-}\) [1.0 mark] - Correct presence of electron antineutrino: \(\bar{\nu}_e\) [1.0 mark]. (c) [3.0 marks] - Calculation of total charge: \(9.60 \times 10^{-19}\text{ C}\) [1.0 mark] - Calculation of total mass: \(2.32 \times 10^{-26}\text{ kg}\) (or \(2.34 \times 10^{-26}\text{ kg}\)) [1.0 mark] - Substitution and calculation showing \(4.1 \times 10^7\text{ C kg}^{-1}\) [1.0 mark].

(a) Mesons are quark-antiquark pairs. The positive kaon \(K^+\) has quark composition \(u \bar{s}\). The positive pion \(\pi^+\) has quark composition \(u \bar{d}\). (b) The strangeness of the positive kaon \(K^+\) is \(S = +1\) (because it contains an anti-strange quark). Pions are non-strange mesons, so both \(\pi^+\) and \(\pi^0\) have strangeness \(S = 0\). Total strangeness before decay is \(+1\) and after decay is \(0\). Thus, strangeness is not conserved; it changes by \(1\). This is permitted because the decay occurs via the weak interaction, which is the only force that can change flavor and violate strangeness conservation by \(\pm 1\).

(a) [2.25 marks] - Composition of \(K^+\): \(u \bar{s}\) [1.0 mark] - Composition of \(\pi^+\): \(u \bar{d}\) [1.0 mark] - Mentioning both are quark-antiquark pairs [0.25 marks]. (b) [4.0 marks] - Stating strangeness of \(K^+\) is \(+1\) [1.0 mark] - Stating strangeness of pions is \(0\) [1.0 mark] - Stating strangeness changes by 1 / is not conserved [1.0 mark] - Identifying the weak interaction as responsible and noting it allows strangeness change [1.0 mark].

(a) Photon energy \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{380 \times 10^{-9}} = 5.234 \times 10^{-19}\text{ J}\). In electronvolts: \(E = \frac{5.234 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.271\text{ eV} \approx 3.27\text{ eV}\). (b) According to the photoelectric equation: \(E_k = hf - \phi = 3.271\text{ eV} - 2.36\text{ eV} = 0.911\text{ eV}\). In Joules: \(E_k = 0.911 \times 1.60 \times 10^{-19}\text{ J} = 1.458 \times 10^{-19}\text{ J} \approx 1.46 \times 10^{-19}\text{ J}\). (c) Maximum speed \(v = \sqrt{\frac{2 E_k}{m_e}} = \sqrt{\frac{2 \times 1.458 \times 10^{-19}}{9.11 \times 10^{-31}}} = \sqrt{3.20 \times 10^{11}} = 5.66 \times 10^5\text{ m s}^{-1}\).

(a) [2.25 marks] - Use of \(E = hc/\lambda\) to find energy in Joules [1.0 mark] - Conversion to eV by dividing by \(1.60 \times 10^{-19}\text{ J eV}^{-1}\) [1.25 marks]. (b) [2.0 marks] - Photoelectric equation formula and substitution: \(E_k = 3.27 - 2.36 = 0.91\text{ eV}\) [1.0 mark] - Conversion of \(E_k\) to Joules: \(1.46 \times 10^{-19}\text{ J}\) [1.0 mark]. (c) [2.0 marks] - Kinetic energy formula \(E_k = 0.5 m v^2\) rearranged for \(v\) [1.0 mark] - Correct substitution of electron mass \(9.11 \times 10^{-31}\text{ kg}\) and calculation: \(5.66 \times 10^5\text{ m s}^{-1}\) [1.0 mark].

Step-by-step solution: (a) Mean diameter: \(d_{\text{mean}} = \frac{0.38 + 0.39 + 0.37 + 0.38 + 0.38}{5} = 0.38\text{ mm}\). Range: \(0.39 - 0.37 = 0.02\text{ mm}\). Absolute uncertainty: \(\text{half-range} = \frac{0.02}{2} = 0.01\text{ mm}\). Thus, \(d = 0.38 \pm 0.01\text{ mm}\). (b) A positive zero error means the micrometer readings are systematically higher than the true value. Therefore, the measured diameter \(d\) is larger than the true diameter. Since the cross-sectional area \(A = \frac{\pi d^2}{4}\) is proportional to \(d^2\), the calculated area will be too large. Since resistivity \(\rho = R \frac{A}{L} = \text{gradient} \times A\), the calculated resistivity will be systematically overestimated (larger than the true value). (c) Area \(A = \frac{\pi (0.38 \times 10^{-3}\text{ m})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\). Resistivity \(\rho = \text{gradient} \times A = 4.25 \times 1.134 \times 10^{-7} = 4.82 \times 10^{-7}\text{ }\Omega\text{ m}\) (or \(4.8 \times 10^{-7}\text{ }\Omega\text{ m}\)). (d) Percentage uncertainty in diameter: \(\frac{0.01}{0.38} \times 100\% = 2.63\%\). Percentage uncertainty in area \(A\): \(2 \times 2.63\% = 5.26\%\). Percentage uncertainty in resistivity: \(\text{\% uncertainty in gradient} + \text{\% uncertainty in } A = 1.5\% + 5.26\% = 6.76\% \approx 6.8\%\).

Marks breakdown: (a) 1 mark for calculating correct mean diameter of 0.38 mm. 1 mark for calculating absolute uncertainty of 0.01 mm using the half-range method. (b) 1 mark for explaining that the measured diameter is larger than the actual diameter. 1 mark for concluding that the calculated resistivity is larger than the true value because resistivity is directly proportional to area. (c) 1 mark for calculating the correct cross-sectional area of \(1.13 \times 10^{-7}\text{ m}^2\). 1 mark for calculating the resistivity in the range \(4.8 \times 10^{-7}\text{ }\Omega\text{ m}\) to \(4.82 \times 10^{-7}\text{ }\Omega\text{ m}\) with correct unit. (d) 1 mark for calculating the percentage uncertainty in area as \(5.3\%\) (by doubling the diameter's percentage uncertainty of \(2.6\%\)). 1 mark for adding the percentage uncertainty of the gradient to find the total uncertainty as \(6.8\%\) (accept \(6.7\%\) to \(6.8\%\)).

Step-by-step solution: (a) Mean time for 20 oscillations: \(t = \frac{17.8 + 17.9 + 17.7}{3} = 17.8\text{ s}\). Absolute uncertainty in \(t\): \(\frac{\text{range}}{2} = \frac{17.9 - 17.7}{2} = 0.1\text{ s}\). Period \(T = \frac{t}{20} = \frac{17.8}{20} = 0.890\text{ s}\). Percentage uncertainty in \(T\) is equal to the percentage uncertainty in \(t\): \(\frac{0.1}{17.8} \times 100\% = 0.56\%\) (or \(0.6\%\)). (b) The equation for the period is \(T = 2\pi\sqrt{\frac{m}{k}}\). Squaring both sides gives \(T^2 = \left(\frac{4\pi^2}{k}\right)m\). This is in the linear form \(y = mx\), so plotting \(T^2\) against \(m\) gives a straight line through the origin. The gradient of this line is constant and equals \(\frac{4\pi^2}{k}\), from which \(k\) can be determined easily and accurately. In contrast, plotting \(T\) against \(m\) yields a curve, where finding the constant \(k\) is much more difficult. (c) Gradient \(= \frac{4\pi^2}{k} \implies k = \frac{4\pi^2}{\text{gradient}} = \frac{4\pi^2}{1.97} = 20.04\text{ N m}^{-1}\). Rounding to 3 significant figures gives \(k = 20.0\text{ N m}^{-1}\) (or \(\text{kg s}^{-2}\)).

Marks breakdown: (a) 1 mark for mean period \(T = 0.890\text{ s}\). 1 mark for finding the absolute uncertainty in time as \(0.1\text{ s}\) (half-range or stopwatch resolution). 1 mark for correct calculation of percentage uncertainty as \(0.56\%\) (or \(0.6\%\)). (b) 1 mark for explaining that \(T^2\) against \(m\) produces a straight line (through the origin) because \(T^2 \propto m\). 1 mark for stating that the gradient is constant and equal to \(\frac{4\pi^2}{k}\), which allows for straightforward determination of \(k\) (unlike a curve). (c) 1 mark for setting \(\text{gradient} = \frac{4\pi^2}{k}\). 1 mark for calculating \(k = 20.0\) (accept \(20\) or \(20.04\)). 1 mark for correct unit: \(\text{N m}^{-1}\) or \(\text{kg s}^{-2}\).

To find the percentage uncertainty in the calculated resistivity \( \rho \), we use the formula for compounding independent uncertainties: \( \frac{\Delta \rho}{\rho} = 2 \frac{\Delta d}{d} + \frac{\Delta R}{R} + \frac{\Delta L}{L} \). First, calculate individual percentage uncertainties: \( \%\Delta d = \frac{0.02}{0.82} \times 100\% \approx 2.44\% \), \( \%\Delta R = \frac{0.1}{2.4} \times 100\% \approx 4.17\% \), and \( \%\Delta L = \frac{0.01}{1.50} \times 100\% \approx 0.67\% \). Combining these gives: \( \%\Delta \rho = 2(2.44\%) + 4.17\% + 0.67\% = 4.88\% + 4.17\% + 0.67\% = 9.72\% \approx 9.7\% \).

1 mark for the correct calculation of percentage uncertainty leading to 9.7% (Option D).

The useful work done in lifting the load is the change in gravitational potential energy: \( W_{\text{out}} = mgh = 85 \times 9.81 \times 12 = 10006.2 \text{ J} \). The useful power output is \( P_{\text{out}} = \frac{W_{\text{out}}}{t} = \frac{10006.2}{8.0} = 1250.8 \text{ W} \). Since \( \text{efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} \), the electrical input power is \( P_{\text{in}} = \frac{P_{\text{out}}}{\text{efficiency}} = \frac{1250.8}{0.65} \approx 1924 \text{ W} \approx 1.9 \text{ kW} \).

1 mark for the correct calculation of electrical input power leading to 1.9 kW (Option C).

The work done during loading is the area under the loading force-extension graph: \( W_{\text{loading}} = \frac{1}{2} F_{\text{max}} \Delta x_{\text{loading}} = \frac{1}{2} \times 120 \times 4.0 \times 10^{-3} = 0.24 \text{ J} \). The work recovered during unloading is the area under the unloading graph: \( W_{\text{unloading}} = \frac{1}{2} F_{\text{max}} \Delta x_{\text{unloading}} = \frac{1}{2} \times 120 \times (4.0 - 1.0) \times 10^{-3} = 0.18 \text{ J} \). The net energy retained by the sample is \( W_{\text{loading}} - W_{\text{unloading}} = 0.24 - 0.18 = 0.060 \text{ J} \).

1 mark for the correct determination of retained energy of 0.060 J (Option A).

The resistance \( R \) of a cylindrical wire is given by \( R = \rho \frac{L}{A} = \rho \frac{4L}{\pi d^2} \), so \( R \propto \frac{L}{d^2} \). Therefore, the resistance of X is \( R_X \propto \frac{L}{d^2} \) and the resistance of Y is \( R_Y \propto \frac{2L}{(3d)^2} = \frac{2L}{9d^2} \). The ratio of their resistances is: \( \frac{R_X}{R_Y} = \frac{L / d^2}{2L / 9d^2} = \frac{9}{2} \).

1 mark for the correct ratio calculation leading to 9/2 (Option D).

Using the potential divider equation, \( V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}} \). In the dark: \( R_{\text{LDR}} = 12 \text{ k}\Omega \), so \( V_{\text{out, dark}} = 6.0 \times \frac{12}{4.0 + 12} = 4.5 \text{ V} \). In bright light: \( R_{\text{LDR}} = 0.8 \text{ k}\Omega \), so \( V_{\text{out, light}} = 6.0 \times \frac{0.8}{4.0 + 0.8} = 1.0 \text{ V} \). The change in the output voltage is \( |4.5 - 1.0| = 3.5 \text{ V} \).

1 mark for the correct calculation of the change in output voltage leading to 3.5 V (Option B).

First, calculate the energy of the incident photon: \( E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{380 \times 10^{-9}} = 5.234 \times 10^{-19} \text{ J} \). Convert this energy to electronvolts: \( E_{\text{eV}} = \frac{5.234 \times 10^{-19}}{1.60 \times 10^{-19}} \approx 3.27 \text{ eV} \). Using the photoelectric equation: \( E_{k,\text{max}} = hf - \Phi = 3.27 \text{ eV} - 2.30 \text{ eV} = 0.97 \text{ eV} \). Since the photon energy is greater than the work function, photoelectrons are emitted with this maximum kinetic energy.

1 mark for the correct calculation of maximum kinetic energy of 0.97 eV (Option A).

For a tube closed at one end (by the water surface), the first resonance occurs when the length of the air column is \( L = \frac{\lambda}{4} \). The wavelength \( \lambda \) of the sound wave is \( \lambda = \frac{v}{f} = \frac{340}{440} \approx 0.773 \text{ m} \). The length of the air column is therefore \( L = \frac{0.773}{4} \approx 0.193 \text{ m} \approx 0.19 \text{ m} \).

1 mark for the correct calculation of the length of the air column leading to 0.19 m (Option A).

The quark composition of the reactants and products are: \( \Sigma^+ = \text{uus} \), \( \text{p} = \text{uud} \), and \( \pi^0 \) contains quark-antiquark pairs (either \( \text{u}\bar{\text{u}} \) or \( \text{d}\bar{\text{d}} \)). In the weak decay, the strange quark \( \text{s} \) (charge \( -1/3e \)) decays into an up quark \( \text{u} \) (charge \( +2/3e \)) with the emission of a virtual \( \text{W}^- \) boson (charge \( -e \)). The \( \text{W}^- \) then decays into a \( \text{d} \) quark and an anti-up quark \( \bar{\text{u}} \). Thus, the net effect on quark flavor is that a strange quark decays into an up quark.

1 mark for identifying that a strange quark decays into an up quark (Option B).

First, calculate the electrical power input to the motor: \(P_{\text{in}} = V \times I = 24\text{ V} \times 5.0\text{ A} = 120\text{ W}\). Since the motor is 60% efficient, the useful mechanical power output is: \(P_{\text{out}} = 0.60 \times 120\text{ W} = 72\text{ W}\). The power output is also given by the formula \(P_{\text{out}} = F \times v = m g v\), where \(v\) is the constant speed of the load. Setting the two equal: \(72\text{ W} = 15\text{ kg} \times 9.81\text{ m s}^{-2} \times v\). Solving for \(v\) gives: \(v = \frac{72}{147.15} \approx 0.49\text{ m s}^{-1}\).

1 mark for the correct calculation of output power (72 W) and subsequent calculation of the constant speed resulting in 0.49 m/s (Option B).

Since the oil droplet is held stationary, the upward electric force balances the downward gravitational force: \(q E = m g\). The uniform electric field strength between the plates is given by: \(E = \frac{V}{d} = \frac{3.0 \times 10^3\text{ V}}{15 \times 10^{-3}\text{ m}} = 2.0 \times 10^5\text{ V m}^{-1}\). Now, solve for charge \(q\): \(q = \frac{m g}{E} = \frac{2.4 \times 10^{-14}\text{ kg} \times 9.81\text{ m s}^{-2}}{2.0 \times 10^5\text{ V m}^{-1}} = 1.18 \times 10^{-18}\text{ C}\), which rounds to \(1.2 \times 10^{-18}\text{ C}\).

1 mark for calculating the electric field strength and setting up the force balance to find the charge of 1.2 x 10^-18 C (Option A).

The total energy of a simple harmonic oscillator is the sum of its kinetic and potential energies: \(E_{\text{total}} = E_k + E_p\). Given that \(E_k = 3 E_p\), we can substitute this to get: \(E_{\text{total}} = 3 E_p + E_p = 4 E_p\). Since total energy is proportional to the square of the amplitude, \(E_{\text{total}} = \frac{1}{2} k A^2\), and potential energy is \(E_p = \frac{1}{2} k x^2\), we have: \dots \frac{1}{2} k A^2 = 4 \left( \frac{1}{2} k x^2 \right) \implies A^2 = 4 x^2 \implies x = \frac{A}{2}\). With \(A = 4.0\text{ cm}\), the displacement \(x\) is \(2.0\text{ cm}\).

1 mark for using energy conservation and SHM energy equations to find that displacement is half the amplitude, giving 2.0 cm (Option B).

Using the kinematic equation for displacement: \(s = u t + \frac{1}{2} a t^2\). Taking upwards as the positive direction, we have: \(u = +12.0\text{ m s}^{-1}\), \(a = -9.81\text{ m s}^{-2}\), and \(t = 4.50\text{ s}\). Substituting these values: \(s = (12.0 \times 4.50) + \frac{1}{2}(-9.81)(4.50)^2 = 54.0 - 99.33 = -45.33\text{ m}\). The negative sign indicates that the final position is 45.3 m below the starting point, so the height of the cliff is 45.3 m.

1 mark for substituting values into the constant acceleration formula with correct signs to obtain a height of 45.3 m (Option A).

Rearranging the equation for the acceleration due to gravity, we get: \(g = \frac{4\pi^2 L}{T^2}\). The percentage uncertainty in \(g\) is the sum of the percentage uncertainty in \(L\) and twice the percentage uncertainty in \(T\): \(\% \Delta g = \% \Delta L + 2(\% \Delta T)\). First, calculate the individual percentage uncertainties: \(\% \Delta L = \frac{0.8}{80.0} \times 100\% = 1.0\%\). \(\% \Delta T = \frac{0.03}{1.80} \times 100\% \approx 1.67\%\). Thus, \(\% \Delta g = 1.0\% + 2(1.67\%) = 1.0\% + 3.33\% = 4.33\%\), which rounds to 4.3%.

1 mark for applying the rules of combination of uncertainties correctly (including doubling the uncertainty contribution of T) to find 4.3% (Option C).

Using Einstein's photoelectric equation: \(E_k = \frac{h c}{\lambda} - \Phi\). From the first case: \(1.2 = \frac{h c}{\lambda} - \Phi \implies \frac{h c}{\lambda} = 1.2 + \Phi\). For the second case, the wavelength is halved, so the incident photon energy is doubled: \(4.8 = 2\left(\frac{h c}{\lambda}\right) - \Phi\). Substituting \(\frac{h c}{\lambda}\) from the first equation into the second: \(4.8 = 2(1.2 + \Phi) - \Phi = 2.4 + 2\Phi - ̑\Phi = 2.4 + \Phi\). Solving for \(\Phi\) gives: \(\Phi = 4.8 - 2.4 = 2.4\text{ eV}\).

1 mark for setting up two simultaneous photoelectric equations and solving for the work function to get 2.4 eV (Option C).

(a) Let the initial length be \(L_0\) and the initial area be \(A_0\). The initial resistance is given by:
\(R_0 = \rho \frac{L_0}{A_0}\)

Since the volume \(V = L \times A\) is constant:
\(L_1 = 1.120 L_0\)
\(A_1 = \frac{A_0}{1.120}\)

The new resistance is:
\(R_1 = \rho \frac{L_1}{A_1} = \rho \frac{1.120 L_0}{\frac{A_0}{1.120}} = 1.120^2 \left(\rho \frac{L_0}{A_0}\right) = 1.2544 R_0\)

Percentage increase in resistance:
\(\text{Percentage Increase} = \frac{R_1 - R_0}{R_0} \times 100 = (1.2544 - 1) \times 100 = 25.44\% \approx 25.4\%\)

(b) Since volume \(V\) is constant:
\(V = 1.88 \times 10^{-7}\text{ m}^3\)

The cross-sectional area of the stretched wire is:
\(A_1 = \frac{V}{L_1} = \frac{1.88 \times 10^{-7}\text{ m}^3}{1.68\text{ m}} \approx 1.119 \times 10^{-7}\text{ m}^2\)

Using the formula for resistivity:
\(\rho = \frac{R_1 A_1}{L_1} = \frac{3.84\ \Omega \times 1.119 \times 10^{-7}\text{ m}^2}{1.68\text{ m}} \approx 2.56 \times 10^{-7}\ \Omega\text{ m}\)

(a) [3 marks total]
- 1 mark for stating that area decreases by factor of 1.12 (or \(A_1 = A_0 / 1.12\))
- 1 mark for showing \(R \propto L^2\) or \(R_1 = 1.254 R_0\)
- 1 mark for calculating correct percentage increase of 25.4% (accept 25% or 25.44%)

(b) [2.55 marks total]
- 1 mark for calculating the new cross-sectional area \(A_1 = 1.12 \times 10^{-7}\text{ m}^2\)
- 1 mark for substituting values into \(\rho = \frac{R A}{L}\)
- 0.55 marks for final answer of \(2.56 \times 10^{-7}\ \Omega\text{ m}\) (accept range \(2.55 \times 10^{-7}\) to \(2.57 \times 10^{-7}\))

(a) The equation is:
\(V = \varepsilon - Ir\)
Comparing this to the equation of a straight line, \(y = mx + c\):
- A graph of \(V\) (on the y-axis) against \(I\) (on the x-axis) gives a straight line with a negative gradient.
- The y-intercept represents the electromotive force \(\varepsilon\).
- The magnitude of the gradient (slope) of the line represents the internal resistance \(r\).

(b) Using \(\varepsilon = I(R + r)\):
For the first setting:
\(\varepsilon = 0.30(4.5 + r) = 1.35 + 0.30r\) --- (Equation 1)

For the second setting:
\(\varepsilon = 0.15(9.5 + r) = 1.425 + 0.15r\) --- (Equation 2)

Equating Equation 1 and Equation 2 since \(\varepsilon\) is constant:
\(1.35 + 0.30r = 1.425 + 0.15r\)
\(0.15r = 0.075\)
\(r = 0.50\ \Omega\)

Substitute \(r\) back into Equation 1:
\(\varepsilon = 0.30(4.5 + 0.50) = 0.30(5.00) = 1.50\text{ V}\)

(a) [2.55 marks total]
- 1 mark for correct formula \(V = \varepsilon - Ir\)
- 1 mark for stating that \(\varepsilon\) is the y-intercept
- 0.55 marks for stating that \(r\) is the magnitude of the gradient / negative gradient

(b) [3 marks total]
- 1 mark for formulating two simultaneous equations
- 1 mark for calculating \(r = 0.50\ \Omega\)
- 1 mark for calculating \(\varepsilon = 1.50\text{ V}\)

(a) When light intensity decreases, the resistance of the LDR increases. In a series potential divider circuit, the output voltage across a component is directly proportional to its share of the total resistance:
\(V_{\text{out}} = V_{\text{supply}} \left(\frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\right)\)
Therefore, to obtain an output voltage that increases as resistance increases (in response to decreasing light), the voltmeter must be connected across the LDR.

(b) Let's calculate the minimum and maximum output voltages:

In bright light (minimum resistance, \(R_{\text{LDR}} = 1.5\text{ k}\Omega\)):
\(V_{\text{out, min}} = 9.0 \times \left(\frac{1.5}{12 + 1.5}\right) = 9.0 \times \frac{1.5}{13.5} = 1.0\text{ V}\)

In the dark (maximum resistance, \(R_{\text{LDR}} = 48\text{ k}\Omega\)):
\(V_{\text{out, max}} = 9.0 \times \left(\frac{48}{12 + 48}\right) = 9.0 \times \frac{48}{60} = 7.2\text{ V}\)

Range of the output voltage:
\(\text{Range} = V_{\text{out, max}} - V_{\text{out, min}} = 7.2\text{ V} - 1.0\text{ V} = 6.2\text{ V}\)

(a) [2.55 marks total]
- 1 mark for identifying that the voltmeter should be connected across the LDR
- 1 mark for explaining that LDR resistance increases when light intensity decreases
- 0.55 marks for explaining that a higher fraction of the total resistance results in a higher fraction of the supply voltage across the LDR

(b) [3 marks total]
- 1 mark for calculating \(V_{\text{out, min}} = 1.0\text{ V}\)
- 1 mark for calculating \(V_{\text{out, max}} = 7.2\text{ V}\)
- 1 mark for calculating the difference: \(7.2 - 1.0 = 6.2\text{ V}\)

(a) For the third harmonic on a string fixed at both ends, the length \(L\) contains exactly 1.5 wavelengths:
\(L = \frac{3}{2}\lambda\)
\(\lambda = \frac{2}{3} L = \frac{2}{3} \times 1.20\text{ m} = 0.800\text{ m}\)

(b) First, calculate the speed \(v\) of the wave:
\(v = f \lambda = 75.0\text{ Hz} \times 0.800\text{ m} = 60.0\text{ m s}^{-1}\)

The tension \(T\) in the string is equal to the weight of the suspended mass:
\(T = m g = 0.350\text{ kg} \times 9.81\text{ m s}^{-2} = 3.4335\text{ N}\)

The wave speed is related to tension and mass per unit length by:
\(v = \sqrt{\frac{T}{\mu}}\)
Squaring both sides:
\(v^2 = \frac{T}{\mu} \implies \mu = \frac{T}{v^2}\)
\(\mu = \frac{3.4335\text{ N}}{(60.0\text{ m s}^{-1})^2} = \frac{3.4335}{3600} \approx 9.5375 \times 10^{-4}\text{ kg m}^{-1}\)

Rounding to 3 significant figures gives \(9.54 \times 10^{-4}\text{ kg m}^{-1}\).

(a) [2.55 marks total]
- 1.55 marks for linking \(L = 1.5\lambda\) (or equivalent ratio)
- 1 mark for correct wavelength \(0.800\text{ m}\) (accept 0.8 m)

(b) [3 marks total]
- 1 mark for calculating the wave speed \(v = 60.0\text{ m s}^{-1}\)
- 1 mark for calculating tension \(T = 3.43\text{ N}\)
- 1 mark for calculating mass per unit length \(\mu = 9.54 \times 10^{-4}\text{ kg m}^{-1}\) (allow 9.53 to 9.55)

(a) A ray entering along the normal has an angle of incidence of \(\theta_1 = 0^\circ\). According to Snell's law (\(n_1 \sin \theta_1 = n_2 \sin \theta_2\)), since \(\sin 0^\circ = 0\), the angle of refraction \(\theta_2\) must also be \(0^\circ\). Thus, the ray does not refract or bend.

(b) At the critical angle \(\theta_c\) at the glass-to-water boundary:
\(\sin \theta_c = \frac{n_{\text{water}}}{n_{\text{glass}}}\)
\(\sin(61.5^\circ) = \frac{1.33}{n_{\text{glass}}}\)
\(n_{\text{glass}} = \frac{1.33}{\sin(61.5^\circ)} = \frac{1.33}{0.8788} \approx 1.513\)

Rounding to 3 significant figures, the refractive index of the glass is 1.51.

(c) The speed of light in glass is given by:
\(v = \frac{c}{n_{\text{glass}}} = \frac{3.00 \times 10^8\text{ m s}^{-1}}{1.513} \approx 1.983 \times 10^8\text{ m s}^{-1}\)
(If using \(n = 1.51\), \(v = 1.987 \times 10^8\text{ m s}^{-1} \approx 1.99 \times 10^8\text{ m s}^{-1}\)).

(a) [1.55 marks total]
- 1 mark for stating that the angle of incidence is zero degrees / ray is normal to boundary
- 0.55 marks for explaining that from Snell's law, angle of refraction is also zero / no deviation

(b) [2 marks total]
- 1 mark for using \(\sin \theta_c = n_2 / n_1\)
- 1 mark for correct refractive index 1.51 (accept 1.51 to 1.52)

(c) [2 marks total]
- 1 mark for using \(v = c / n\)
- 1 mark for correct speed \(1.98 \times 10^8\text{ m s}^{-1}\) or \(1.99 \times 10^8\text{ m s}^{-1}\)

(a) The energy \(E\) of a photon is given by:
\(E = \frac{hc}{\lambda}\)
\(E = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{240 \times 10^{-9}\text{ m}} = 8.2875 \times 10^{-19}\text{ J}\)
This is approximately \(8.3 \times 10^{-19}\text{ J}\), as shown.

(b) The work function of sodium in joules is:
\(\Phi = 2.28\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 3.648 \times 10^{-19}\text{ J}\)

Using Einstein's photoelectric equation:
\(E_{\text{k, max}} = E - \Phi\)
\(E_{\text{k, max}} = 8.2875 \times 10^{-19}\text{ J} - 3.648 \times 10^{-19}\text{ J} = 4.6395 \times 10^{-19}\text{ J} \approx 4.64 \times 10^{-19}\text{ J}\)
(If using the rounded value of \(8.3 \times 10^{-19}\text{ J}\), \(E_{\text{k, max}} = 4.65 \times 10^{-19}\text{ J}\))

(c) The threshold frequency \(f_0\) is the minimum frequency required to liberate electrons:
\(\Phi = h f_0 \implies f_0 = \frac{\Phi}{h}\)
\(f_0 = \frac{3.648 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} \approx 5.502 \times 10^{14}\text{ Hz} \approx 5.50 \times 10^{14}\text{ Hz}\)

(a) [2 marks total]
- 1 mark for substitution into \(E = hc/\lambda\)
- 1 mark for showing value of \(8.29 \times 10^{-19}\text{ J}\) before rounding

(b) [1.55 marks total]
- 1 mark for converting work function to joules (\(3.65 \times 10^{-19}\text{ J}\))
- 0.55 marks for final answer of \(4.64 \times 10^{-19}\text{ J}\) (accept \(4.65 \times 10^{-19}\text{ J}\))

(c) [2 marks total]
- 1 mark for using \(f_0 = \Phi / h\)
- 1 mark for correct threshold frequency of \(5.50 \times 10^{14}\text{ Hz}\) (accept range \(5.49 \times 10^{14}\) to \(5.52 \times 10^{14}\))

(a) Use the diffraction grating equation:
\(d \sin \theta = n \lambda\)
For \(n = 1\) and \(\theta = 18.1^\circ\):
\(d \sin(18.1^\circ) = 1 \times 589 \times 10^{-9}\text{ m}\)
\(d = \frac{589 \times 10^{-9}\text{ m}}{\sin(18.1^\circ)} = \frac{589 \times 10^{-9}}{0.31067} \approx 1.896 \times 10^{-6}\text{ m}\)

The number of lines per metre \(N\) is:
\(N = \frac{1}{d} = \frac{1}{1.896 \times 10^{-6}\text{ m}} \approx 5.274 \times 10^5\text{ lines m}^{-1}\)

Converting to lines per millimetre:
\(N_{\text{mm}} = \frac{5.274 \times 10^5}{1000} \approx 527.4 \approx 528\text{ lines mm}^{-1}\)

(b) For maximum observable order, the angle \(\theta\) cannot exceed \(90^\circ\):
\(\sin \theta \le 1\)
\(\frac{n \lambda}{d} \le 1 \implies n \le \frac{d}{\lambda}\)
\(n \le \frac{1.896 \times 10^{-6}\text{ m}}{589 \times 10^{-9}\text{ m}} \approx 3.22\)

Since \(n\) must be an integer, the highest observable order is \(n = 3\).

The total number of maxima is given by those observed on both sides of the central zero-order maximum:
\(2n + 1 = 2(3) + 1 = 7\) maxima (i.e. orders \(-3, -2, -1, 0, +1, +2, +3\)).

(a) [3 marks total]
- 1 mark for calculating slit spacing \(d = 1.90 \times 10^{-6}\text{ m}\)
- 1 mark for finding number of lines per metre \(N = 5.27 \times 10^5\text{ m}^{-1}\)
- 1 mark for converting to lines per mm to get \(528\text{ lines mm}^{-1}\) (allow 526 to 530)

(b) [2.55 marks total]
- 1 mark for setting \(\sin \theta = 1\) to find theoretical maximum order \(n = 3.22\)
- 1 mark for stating that highest integer order is \(n = 3\)
- 0.55 marks for final answer of 7 maxima

(a) As the voltage across the lamp increases, the current increases. This causes the temperature of the filament to rise. The metal ions in the lattice vibrate with greater amplitude. This increases the rate of collisions between the conduction electrons (charge carriers) and the vibrating lattice ions, which increases the resistance.

(b) First, calculate the theoretical current if the lamp's resistance remained constant at \(4.0\ \Omega\):
\(R_{\text{total}} = R_{\text{lamp}} + R_{\text{fixed}} = 4.0\ \Omega + 6.0\ \Omega = 10.0\ \Omega\)
\(I = \frac{V}{R_{\text{total}}} = \frac{12\text{ V}}{10.0\ \Omega} = 1.2\text{ A}\)

Under this assumption, the voltage across the lamp would be:
\(V_{\text{lamp}} = I \times R_{\text{lamp}} = 1.2\text{ A} \times 4.0\ \Omega = 4.8\text{ V}\)

Since \(4.8\text{ V}\) is much less than the rated \(12\text{ V}\), the filament lamp is operating at a lower voltage, and thus at a much lower temperature than its normal operating state.
Because its temperature is lower, its actual resistance is significantly less than \(4.0\ \Omega\).
Since the actual resistance of the lamp is lower than \(4.0\ \Omega\), the total resistance of the series circuit is less than \(10.0\ \Omega\). As a result, the actual current flowing through the circuit is larger than \(1.2\text{ A}\).

(a) [2.55 marks total]
- 1 mark for stating that resistance increases as voltage/current increases
- 1 mark for stating that temperature increases, causing greater lattice vibrations
- 0.55 marks for linking to more frequent collisions of charge carriers / electrons

(b) [3 marks total]
- 1 mark for calculating the assumed current of \(1.2\text{ A}\) and voltage across the lamp of \(4.8\text{ V}\)
- 1 mark for explaining that the lamp is at a lower temperature because \(V < 12\text{ V}\)
- 1 mark for stating that a lower temperature leads to a lower actual resistance, which decreases total circuit resistance and increases actual current above \(1.2\text{ A}\)

**(a)**
When light intensity increases, the number of charge carriers in the semi-conducting material of the LDR increases, which causes its resistance \(R_{\text{LDR}}\) to decrease.

Since \(V_{\text{out}}\) is measured across the LDR, we use the potential divider relationship:
\[V_{\text{out}} = V_{\text{s}} \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\]
As \(R_{\text{LDR}}\) decreases, the ratio of the LDR's resistance to the total resistance of the circuit decreases. Therefore, the output voltage \(V_{\text{out}}\) decreases.

**(b)**
Using the potential divider formula:
\[V_{\text{out}} = V_{\text{s}} \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\]
Substitute the given values:
\[4.0 = 12.0 \times \frac{8.0}{R + 8.0}\]
\[\frac{4.0}{12.0} = \frac{8.0}{R + 8.0}\]
\[\frac{1}{3} = \frac{8.0}{R + 8.0}\]
\[R + 8.0 = 24.0\]
\[R = 16.0\text{ k}\Omega\]
*(This matches the required value.)*

**(c)**
With \(R = 16.0\text{ k}\Omega\) and the new \(R_{\text{LDR}} = 4.0\text{ k}\Omega\):
\[V_{\text{out}} = 12.0 \times \frac{4.0}{16.0 + 4.0}\]
\[V_{\text{out}} = 12.0 \times \frac{4.0}{20.0}\]
\[V_{\text{out}} = 12.0 \times 0.20 = 2.4\text{ V}\]

**(a)**
* **M1:** State that the resistance of the LDR decreases as light intensity increases. [1]
* **A1:** State that the output voltage \(V_{\text{out}}\) decreases (with valid explanation, e.g., the LDR takes a smaller share of the total resistance or total supply voltage). [1]

**(b)**
* **M1:** State or use the potential divider equation \(V_{\text{out}} = V_{\text{s}} \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\) OR use the ratio method \(\frac{R}{R_{\text{LDR}}} = \frac{V_{R}}{V_{\text{out}}}\). [1]
* **A1:** Correct substitution and rearrangement leading to \(R = 16.0\text{ k}\Omega\). Must show clear working to gain this mark. [1]

**(c)**
* **M1:** Correct substitution of \(R = 16.0\text{ k}\Omega\) and \(R_{\text{LDR}} = 4.0\text{ k}\Omega\) into the potential divider equation. [1]
* **A1:** \(2.4\text{ V}\) (or \(2.40\text{ V}\)). [1]

(a) The cross-sectional area of the wire is: \(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.46 \times 10^{-3}\text{ m})^2}{4} = 1.6619 \times 10^{-7}\text{ m}^2\). The Young modulus is: \(E = \frac{F L}{A e} = \frac{28.5 \times 1.850}{1.6619 \times 10^{-7} \times 1.4 \times 10^{-3}} = 2.266 \times 10^{11}\text{ Pa} \approx 2.3 \times 10^{11}\text{ Pa}\). (b)(i) Because \(A \propto d^2\), the percentage uncertainty in \(A\) is: \(2 \times \frac{\Delta d}{d} \times 100\% = 2 \times \frac{0.01}{0.46} \times 100\% = 4.35\% \approx 4.3\%\). (b)(ii) The percentage uncertainty in \(E\) is the sum of the individual percentage uncertainties: \(\%\Delta F = \frac{0.2}{28.5} \times 100\% = 0.70\%\), \(\%\Delta L = \frac{0.002}{1.850} \times 100\% = 0.11\%\), \(\%\Delta e = \frac{0.1}{1.4} \times 100\% = 7.14\%\). Therefore, \(\%\Delta E = 0.70\% + 0.11\% + 4.35\% + 7.14\% = 12.3\% \approx 12\%\). (c) The measurement of extension, \(e\), contributes most to the uncertainty because it has the largest individual percentage uncertainty (\(7.1\%\)). This uncertainty could be reduced by using a longer or thinner wire to produce a larger extension for the same force, or by using a higher precision instrument (such as a travelling microscope) to measure the extension with a smaller absolute uncertainty.

Part (a) [2 Marks]: 1 Mark for calculating area \(A = 1.66 \times 10^{-7}\text{ m}^2\) or substituting all values correctly into \(E = \frac{FL}{Ae}\). 1 Mark for correct final value \(E = 2.3 \times 10^{11}\text{ Pa}\) (or \(2.27 \times 10^{11}\text{ Pa}\)) with correct unit (\(\text{Pa}\) or \(\text{N m}^{-2}\)). Part (b) [4 Marks]: 1 Mark for establishing that percentage uncertainty in \(A\) is twice that of \(d\). 1 Mark for correct percentage uncertainty in \(A = 4.3\%\) (or \(4.35\%\)). 1 Mark for adding all individual percentage uncertainties. 1 Mark for correct percentage uncertainty in \(E = 12\%\) (or \(12.3\%\)). Part (c) [2 Marks]: 1 Mark for identifying extension \(e\) as the main contributor with correct reasoning (highest percentage uncertainty). 1 Mark for a valid suggestion to reduce this uncertainty (e.g. use a longer/thinner wire or a travelling microscope).

(a) The electromotive force \(\varepsilon\) is the total energy supplied per unit charge. This is equal to the terminal potential difference \(V\) plus the 'lost volts' across the internal resistance \(Ir\): \(\varepsilon = V + Ir\). Rearranging this formula gives: \(V = -rI + \varepsilon\). (b)(i) Substituting both data points into the equation: \(1.24 = \varepsilon - 0.420r\) (Equation 1) and \(0.98 = \varepsilon - 0.840r\) (Equation 2). Subtracting Equation 2 from Equation 1: \(0.26 = 0.420r \Rightarrow r = \frac{0.26}{0.420} = 0.619\ \Omega \approx 0.62\ \Omega\). (b)(ii) Substituting \(r\) back into Equation 1: \(\varepsilon = 1.24 + (0.420 \times 0.619) = 1.50\text{ V}\). (c) The expression for internal resistance is \(r = \frac{\Delta V}{\Delta I} = \frac{V_1 - V_2}{I_2 - I_1}\). Let \(y = V_1 - V_2 = 0.26\text{ V}\). The absolute uncertainty is \(\Delta y = 0.02 + 0.02 = 0.04\text{ V}\), so the percentage uncertainty in \(y\) is \(\frac{0.04}{0.26} \times 100\% = 15.38\%\). Let \(x = I_2 - I_1 = 0.420\text{ A}\). The absolute uncertainty is \(\Delta x = 0.005 + 0.005 = 0.010\text{ A}\), so the percentage uncertainty in \(x\) is \(\frac{0.010}{0.420} \times 100\% = 2.38\%\). The total percentage uncertainty in \(r\) is \(15.38\% + 2.38\% = 17.76\%\). The absolute uncertainty in \(r\) is \(0.619 \times 17.76\% = 0.11\ \Omega\).

Part (a) [1 Mark]: 1 Mark for showing the starting relation \(\varepsilon = V + Ir\) and rearranging it correctly to \(V = -rI + \varepsilon\). Part (b) [4 Marks]: 1 Mark for setting up two simultaneous equations or writing the gradient formula \(r = -\frac{\Delta V}{\Delta I}\). 1 Mark for calculating \(r = 0.62\ \Omega\) (accept \(0.619\ \Omega\)). 1 Mark for a correct method of substituting \(r\) back to find \(\varepsilon\). 1 Mark for calculating \(\varepsilon = 1.50\text{ V}\) (accept \(1.5\text{ V}\)). Part (c) [3 Marks]: 1 Mark for calculating the absolute uncertainties in the difference values: \(\Delta y = 0.04\text{ V}\) and \(\Delta x = 0.010\text{ A}\). 1 Mark for summing the fractional/percentage uncertainties of \(y\) and \(x\): \(\frac{0.04}{0.26} + \frac{0.010}{0.420} \approx 17.8\%\) (or \(0.178\)). 1 Mark for calculating the correct absolute uncertainty in \(r = 0.11\ \Omega\) (accept answers in the range \(0.10\ \Omega\) to \(0.11\ \Omega\)).

The useful output power of the motor is given by: \(P_{\text{out}} = F v = m g v = 45 \times 9.81 \times 1.2 = 529.74\text{ W}\). Since the efficiency of the motor is \(65\%\) (or \(0.65\)), the electrical input power \(P_{\text{in}}\) is: \(P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{529.74}{0.65} \approx 815\text{ W}\). To two significant figures, this is \(810\text{ W}\).

1 mark for the correct input power calculation (C).

From Einstein's photoelectric equation: \(E_k = \frac{hc}{\lambda} - \Phi\) where \(\Phi\) is the work function of the metal. For light of wavelength \(\frac{\lambda}{2}\), the new maximum kinetic energy \(E_k'\) is: \(E_k' = \frac{hc}{\lambda/2} - \Phi = 2\frac{hc}{\lambda} - \Phi\). Substituting \(\frac{hc}{\lambda} = E_k + \Phi\) into the equation: \(E_k' = 2(E_k + \Phi) - \Phi = 2 E_k + \Phi\). Since the work function \(\Phi\) must be positive, \(E_k' > 2 E_k\).

1 mark for identifying that the new maximum kinetic energy is greater than twice the original value (B).

The total stopping distance consists of the thinking distance and the braking distance. Thinking distance: \(s_{\text{thinking}} = v \times t = 24 \times 0.50 = 12\text{ m}\). Braking distance: using \(v^2 = u^2 + 2 a s\), we have \(0 = 24^2 - 2 \times 6.0 \times s_{\text{braking}}\), which simplifies to \(12 s_{\text{braking}} = 576\) and gives \(s_{\text{braking}} = 48\text{ m}\). Total stopping distance = \(12 + 48 = 60\text{ m}\).

1 mark for the correct total stopping distance of 60 m (B).

The relationship between current and drift velocity is \(I = n A e v\), which gives \(v = \frac{I}{n A e}\). For the second wire, the diameter is halved, so its cross-sectional area is \(A' = \frac{A}{4}\). The current is \(2I\). Therefore, the new drift velocity \(v'\) is: \(v' = \frac{2I}{n \left(\frac{A}{4}\right) e} = 8 \left(\frac{I}{n A e}\right) = 8v\).

1 mark for the correct drift velocity calculation of 8v (C).

The terminal potential difference \(V\) is related to the e.m.f. \(E\), internal resistance \(r\), and current \(I\) by the equation: \(V = E - I r\). Rearranging into the linear form \(y = m x + c\), we get \(V = (-r) I + E\). Thus, a graph of \(V\) against \(I\) is a straight line with gradient \(-r\) and y-intercept \(E\).

1 mark for identifying the correct linear relationship and parameters of the V-I graph (B).

The grating spacing \(d\) is calculated as: \(d = \frac{1}{N} = \frac{1}{5.00 \times 10^5\text{ m}^{-1}} = 2.00 \times 10^{-6}\text{ m}\). The grating equation is \(d \sin \theta = n \lambda\). For the maximum order that can be observed, the angle of diffraction \(\theta\) must satisfy \(\sin \theta \le 1\). This gives: \(n \le \frac{d}{\lambda} = \frac{2.00 \times 10^{-6}}{633 \times 10^{-9}} \approx 3.16\). Since \(n\) must be an integer, the highest observable order is \(3\).

1 mark for the correct calculation of maximum order (B).

The strain energy \(W\) stored in a wire under tension is given by: \(W = \frac{1}{2} F \Delta L\). The Young modulus \(E\) is defined as: \(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L}\). Rearranging this for \(F\) gives: \(F = \frac{E A \Delta L}{L}\). Substituting this back into the strain energy formula: \(W = \frac{1}{2} \left(\frac{E A \Delta L}{L}\right) \Delta L = \frac{E A (\Delta L)^2}{2 L}\).

1 mark for the correct algebraic derivation of the strain energy (B).

In simple harmonic motion, the maximum speed \(v_{\text{max}}\) is related to the angular frequency \(\omega\) and amplitude \(A\) by: \(v_{\text{max}} = \omega A\). The angular frequency \(\omega\) is related to the time period \(T\) by: \(\omega = \frac{2\pi}{T}\). Substituting \(\omega\) into the speed equation gives: \(v_{\text{max}} = \frac{2\pi A}{T}\).

1 mark for the correct expression of maximum speed (B).

For a car climbing a slope at constant speed \( v \), there is no acceleration, so the net force along the direction of motion is zero. The component of the weight acting down the slope is \( mg \sin\theta \). The resistive force acting down the slope is \( R \). Therefore, the forward driving force \( F \) provided by the engine must balance both of these forces: \( F = R + mg \sin\theta \). Power \( P \) is defined as the product of the driving force and the velocity: \( P = F v = v(R + mg \sin\theta) \).

1 mark for the correct option A.

In one complete cycle of simple harmonic motion, the object starts at a position (for example, equilibrium), moves to one maximum displacement \( +A \), travels back through equilibrium to the opposite maximum displacement \( -A \), and returns to its starting position. The total distance travelled in one cycle is \( s = A + 2A + A = 4A \). The time taken for one complete cycle is the period \( T \). Therefore, the average speed is given by dividing the total distance by the total time: \( \text{Average speed} = \frac{4A}{T} \).

1 mark for the correct option B.

From Einstein's photoelectric equation, \( hf = \Phi + E_k \), which gives \( hf = E_k + \Phi \). When the frequency of the light is doubled to \( 2f \), the new maximum kinetic energy \( E'_k \) of the emitted photoelectrons is given by \( E'_k = h(2f) - \Phi = 2hf - \Phi \). Substituting the expression for \( hf \) into this equation gives \( E'_k = 2(E_k + \Phi) - \Phi = 2E_k + 2\Phi - \Phi = 2E_k + \Phi \).

1 mark for the correct option B.

When the light intensity on an LDR increases, its resistance decreases. The potential divider equation for the output voltage across the LDR is \( V_{\text{out}} = V \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}} \). As \( R_{\text{LDR}} \) decreases relative to the fixed resistance \( R \), the fraction of the total potential difference dropped across the LDR also decreases. Therefore, both the resistance of the LDR and the output voltage \( V_{\text{out}} \) decrease.

1 mark for the correct option A.

The density \( \rho \) of a cylinder is calculated as \( \rho = \frac{m}{V} = \frac{4m}{\pi d^2 L} \). The percentage uncertainty in the density is the sum of the percentage uncertainty in mass, twice the percentage uncertainty in diameter, and the percentage uncertainty in length: \( \frac{\Delta \rho}{\rho} \times 100\% = \left( \frac{\Delta m}{m} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L} \right) \times 100\% \). Substituting the values, we find: \( \frac{\Delta m}{m} \times 100\% = \frac{0.01}{0.40} \times 100\% = 2.5\% \); \( 2 \frac{\Delta d}{d} \times 100\% = 2 \times \frac{0.02}{0.80} \times 100\% = 5.0\% \); and \( \frac{\Delta L}{L} \times 100\% = \frac{0.5}{100.0} \times 100\% = 0.5\% \). Summing these gives \( 2.5\% + 5.0\% + 0.5\% = 8.0\% \).

1 mark for the correct option C.

At the bottom of the loop, the net centripetal force acts upwards (towards the center of the loop). The normal contact force \( N \) acts upwards while the weight \( mg \) acts downwards. Therefore, \( N - mg = \frac{m v^2}{r} \), which gives \( N = m\left(\frac{v^2}{r} + g\right) \). At the top of the loop, both forces act downwards (towards the center): \( N + mg = \frac{m v^2}{r} \), which gives \( N = m\left(\frac{v^2}{r} - g\right) \). Thus, the normal contact force has its maximum value at the bottom of the loop.

1 mark for the correct option B.

(a) The time period of a mass-spring system is given by \( T = 2\pi \sqrt{\frac{m}{k}} \). Substituting the given values: \( T = 2\pi \sqrt{\frac{0.350}{15.0}} \approx 0.9598\text{ s} \). The frequency is the reciprocal of the period: \( f = \frac{1}{T} = \frac{1}{0.9598} \approx 1.042\text{ Hz} \), which is approximately \( 1.04\text{ Hz} \). (b) The maximum speed in simple harmonic motion is given by \( v_{\max} = \omega A = 2\pi f A \). Using the unrounded value of \( f \): \( v_{\max} = 2\pi \times 1.0419 \times 0.0800 = 0.524\text{ m s}^{-1} \). (c) The magnitude of the maximum acceleration is given by \( a_{\max} = \omega^2 A = \frac{k}{m} A \). Substituting the values: \( a_{\max} = \frac{15.0}{0.350} \times 0.0800 \approx 3.43\text{ m s}^{-2} \).

(a) [3 Marks] - Formula \( T = 2\pi \sqrt{\frac{m}{k}} \) used or substituted correctly (1 mark) - \( T = 0.96\text{ s} \) or \( f = 1.042\text{ Hz} \) calculated (1 mark) - Final value shown to at least 3 significant figures: \( 1.04\text{ Hz} \) (1 mark) (b) [3 Marks] - \( v_{\max} = \omega A \) or \( v_{\max} = 2\pi f A \) stated (1 mark) - Correct substitution of values (1 mark) - Correct final speed with units: \( 0.524\text{ m s}^{-1} \) [accept range 0.522 to 0.524] (1 mark) (c) [3.28 Marks] - \( a_{\max} = \omega^2 A \) or \( a_{\max} = \frac{k}{m} A \) stated (1 mark) - Correct substitution of values (1 mark) - Correct final acceleration with units: \( 3.43\text{ m s}^{-2} \) (1.28 marks)

(a) The electric field strength is given by \( E = \frac{V}{d} \). Converting values to SI units: \( V = 1.20 \times 10^3\text{ V} \) and \( d = 50.0 \times 10^{-3}\text{ m} \). Thus, \( E = \frac{1.20 \times 10^3}{50.0 \times 10^{-3}} = 2.40 \times 10^4\text{ V m}^{-1} \). (b) Since the sphere is in equilibrium, the forces are balanced. Resolving vertically: \( T \cos\theta = m g \). Resolving horizontally: \( T \sin\theta = F_e \). Combining these gives \( F_e = m g \tan\theta \). Substituting the values: \( F_e = (1.50 \times 10^{-5}\text{ kg}) \times (9.81\text{ m s}^{-2}) \times \tan(7.50^\circ) \approx 1.4715 \times 10^{-4} \times 0.13165 \approx 1.937 \times 10^{-5}\text{ N} \approx 1.94 \times 10^{-5}\text{ N} \). (c) The electrostatic force is given by \( F_e = E Q \). Rearranging for charge: \( Q = \frac{F_e}{E} = \frac{1.937 \times 10^{-5}}{2.40 \times 10^4} \approx 8.07 \times 10^{-10}\text{ C} \).

(a) [3 Marks] - Stating formula \( E = V/d \) (1 mark). - Correct conversion of units (1.20 kV to 1200 V and 50 mm to 0.050 m) (1 mark). - Calculation showing \( 2.40 \times 10^4\text{ V m}^{-1} \) clearly (1 mark). (b) [3 Marks] - Stating equations for vertical and horizontal equilibrium (1 mark). - Deducing \( F_e = mg \tan\theta \) (1 mark). - Correct calculation of force as \( 1.94 \times 10^{-5}\text{ N} \) [accept range 1.93 to 1.95] (1 mark). (c) [3.28 Marks] - Stating formula \( F_e = EQ \) or rearrangement (1 mark). - Correct substitution of previously calculated force and field strength (1 mark). - Correct calculation of charge with units: \( 8.07 \times 10^{-10}\text{ C} \) (1.28 marks).

(a) For a circular orbit, the gravitational force provides the centripetal force: \( \frac{G M m}{r^2} = \frac{m v^2}{r} \). Rearranging for orbital speed gives \( v = \sqrt{\frac{G M}{r}} \). Substituting the values: \( v = \sqrt{\frac{6.67 \times 10^{-11} \times 4.80 \times 10^{24}}{9.50 \times 10^6}} \approx 5.805 \times 10^3\text{ m s}^{-1} \), which is approximately \( 5.8 \times 10^3\text{ m s}^{-1} \). (b) The orbital period \( T \) is the distance traveled in one orbit divided by speed: \( T = \frac{2\pi r}{v} = \frac{2\pi \times 9.50 \times 10^6}{5.805 \times 10^3} \approx 1.028 \times 10^4\text{ s} \). In hours, \( T = \frac{1.028 \times 10^4}{3600} \approx 2.86\text{ hours} \). (c) The kinetic energy is given by \( E_k = \frac{1}{2} m v^2 \). Substituting the values: \( E_k = \frac{1}{2} \times 1.20 \times 10^3 \times (5.805 \times 10^3)^2 \approx 2.02 \times 10^{10}\text{ J} \).

(a) [3 Marks] - Equating gravitational and centripetal force (1 mark). - Formula \( v = \sqrt{\frac{G M}{r}} \) derived or stated (1 mark). - Calculation showing \( v = 5.81 \times 10^3\text{ m s}^{-1} \) (1 mark). (b) [3 Marks] - Formula \( T = \frac{2\pi r}{v} \) stated (1 mark). - Calculating time period in seconds (1.03 x 10^4 s) (1 mark). - Conversion of seconds to hours to give \( 2.86\text{ hours} \) (1 mark). (c) [3.28 Marks] - Formula \( E_k = \frac{1}{2} m v^2 \) or \( E_k = \frac{G M m}{2r} \) stated (1 mark). - Substitution of values (1 mark). - Calculation of kinetic energy as \( 2.02 \times 10^{10}\text{ J} \) (1.28 marks).

(a) Coulomb's law states \( F = \frac{1}{4\pi\varepsilon_0} \frac{|Q_1 Q_2|}{r^2} \). Here, \( r = 0.120\text{ m} \), \( Q_1 = 4.50 \times 10^{-9}\text{ C} \), and \( Q_2 = 3.00 \times 10^{-9}\text{ C} \). Substituting values: \( F = \frac{8.99 \times 10^9 \times 4.50 \times 10^{-9} \times 3.00 \times 10^{-9}}{0.120^2} = 8.428 \times 10^{-6}\text{ N} \approx 8.43 \times 10^{-6}\text{ N} \). (b) Since \( Q_3 \) is at the midpoint, the distance to each charge is \( r' = 0.060\text{ m} \). The force on \( Q_3 \) due to \( Q_1 \) is repulsive (pointing away from \( Q_1 \) towards \( Q_2 \)): \( F_{13} = \frac{8.99 \times 10^9 \times 4.50 \times 10^{-9} \times 2.00 \times 10^{-9}}{0.060^2} = 2.2475 \times 10^{-5}\text{ N} \). The force on \( Q_3 \) due to \( Q_2 \) is attractive (pointing towards \( Q_2 \)): \( F_{23} = \frac{8.99 \times 10^9 \times 3.00 \times 10^{-9} \times 2.00 \times 10^{-9}}{0.060^2} = 1.4983 \times 10^{-5}\text{ N} \). Both forces act in the same direction, so the resultant force is \( F_{\text{res}} = F_{13} + F_{23} = 2.2475 \times 10^{-5} + 1.4983 \times 10^{-5} = 3.746 \times 10^{-5}\text{ N} \approx 3.75 \times 10^{-5}\text{ N} \). (c) A field of force is a region of space in which a particle experiences a non-contact force. Electric field strength is defined as force per unit positive charge (\( E = \frac{F}{Q} \)). Because force is a vector (has direction), electric field strength also has a defined direction, making it a vector quantity.

(a) [3 Marks] - Coulomb's Law formula stated (1 mark). - Correct substitutions including unit conversion of distances to meters (1 mark). - Force calculated correctly as \( 8.43 \times 10^{-6}\text{ N} \) (1 mark). (b) [3.28 Marks] - Calculating force from \( Q_1 \) on \( Q_3 \) (1 mark). - Calculating force from \( Q_2 \) on \( Q_3 \) (1 mark). - Realizing both forces are in the same direction and adding them to obtain \( 3.75 \times 10^{-5}\text{ N} \) (1.28 marks). (c) [3 Marks] - Defining field of force as a region where a body experiences a force (1 mark). - Defining electric field strength as force per unit charge (1 mark). - Explaining that it is a vector because force has direction (1 mark).

(a) The period of a simple pendulum is given by \( T = 2\pi \sqrt{\frac{L}{g}} \). Substituting the given values: \( T = 2\pi \sqrt{\frac{1.25}{9.81}} \approx 2.243\text{ s} \), which is approximately \( 2.24\text{ s} \). (b) (i) The amplitude is the maximum horizontal displacement: \( A = L \sin\theta \) or \( A = L \theta \) (where \( \theta \) is in radians). Using \( A = L \sin(5.00^\circ) = 1.25 \times \sin(5.00^\circ) \approx 0.1089\text{ m} \approx 0.109\text{ m} \). (Or using \( A = L \theta = 1.25 \times (5.00 \times \frac{\pi}{180}) \approx 0.1091\text{ m} \approx 0.109\text{ m} \)). (ii) The height \( h \) gained by the bob is \( h = L(1 - \cos\theta) = 1.25 \times (1 - \cos(5.00^\circ)) = 1.25 \times (1 - 0.99619) \approx 0.004756\text{ m} \). The potential energy is \( E_p = m g h = 0.180 \times 9.81 \times 0.004756 \approx 8.40 \times 10^{-3}\text{ J} \) (or \( 8.40\text{ mJ} \)). Alternatively, using SHM approximation: \( E_p = \frac{1}{2} m \omega^2 A^2 \) where \( \omega = \frac{2\pi}{T} = \sqrt{\frac{g}{L}} \approx 2.801\text{ rad s}^{-1} \). \( E_p = \frac{1}{2} \times 0.180 \times (2.801)^2 \times (0.109)^2 \approx 8.41 \times 10^{-3}\text{ J} \).

(a) [3 Marks] - Formula \( T = 2\pi \sqrt{\frac{L}{g}} \) stated (1 mark). - Correct substitution with \( g = 9.81\text{ m s}^{-2} \) (1 mark). - Showing final value of \( 2.24\text{ s} \) (at least 3 s.f.) (1 mark). (b) (i) [3 Marks] - Formula \( A = L \sin\theta \) or \( A = L \theta \) stated (1 mark). - Conversion of angle to radians or direct substitution (1 mark). - Amplitude calculated as \( 0.109\text{ m} \) (1 mark). (b) (ii) [3.28 Marks] - Stating \( E_p = m g h \) or \( E_p = \frac{1}{2} m \omega^2 A^2 \) (1 mark). - Correct calculation of height \( h \) or angular frequency \( \omega \) (1 mark). - Final potential energy calculated as \( 8.40 \times 10^{-3}\text{ J} \) or \( 8.41 \times 10^{-3}\text{ J} \) (1.28 marks).

(a) Gravitational potential is given by \( V_g = -\frac{G M}{R_E} \). Substituting the given values: \( V_g = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^6} \approx -6.251 \times 10^7\text{ J kg}^{-1} \), which is approximately \( -6.25 \times 10^7\text{ J kg}^{-1} \). (b) To escape the gravitational field, the kinetic energy must equal the magnitude of the gravitational potential energy at the surface: \( \frac{1}{2} m v_{\text{esc}}^2 = \frac{G M m}{R_E} \), which simplifies to \( v_{\text{esc}} = \sqrt{\frac{2GM}{R_E}} = \sqrt{2 |V_g|} \). Substituting values: \( v_{\text{esc}} = \sqrt{2 \times 6.251 \times 10^7} \approx 1.118 \times 10^4\text{ m s}^{-1} \approx 1.12 \times 10^4\text{ m s}^{-1} \) (or \( 11.2\text{ km s}^{-1} \)). (c) The gravitational potential at the orbit is \( V_{\text{orbit}} = -\frac{G M}{r_{\text{orbit}}} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{4.22 \times 10^7} \approx -9.436 \times 10^6\text{ J kg}^{-1} \). The change in gravitational potential energy is \( \Delta E_p = m (V_{\text{orbit}} - V_{\text{surface}}) = 1.50 \times 10^3 \times (-9.436 \times 10^6 - (-6.251 \times 10^7)) = 1.50 \times 10^3 \times 5.3074 \times 10^7 \approx 7.961 \times 10^{10}\text{ J} \approx 7.96 \times 10^{10}\text{ J} \).

(a) [3 Marks] - Stating gravitational potential formula \( V_g = -\frac{G M}{r} \) (1 mark). - Correct substitutions of \( G \), \( M \), and \( R_E \) (1 mark). - Showing final value of \( -6.25 \times 10^7\text{ J kg}^{-1} \) (1 mark). (b) [3.28 Marks] - Equating kinetic energy and potential energy (1 mark). - Deriving escape velocity formula \( v_{\text{esc}} = \sqrt{\frac{2 G M}{R_E}} \) (1 mark). - Calculation of escape velocity as \( 1.12 \times 10^4\text{ m s}^{-1} \) (1.28 marks). (c) [3 Marks] - Calculating gravitational potential at the orbit (1 mark). - Using \( \Delta E_p = m \Delta V_g \) (1 mark). - Correct final potential energy change \( 7.96 \times 10^{10}\text{ J} \) (1 mark).

(a) The work done on the charge is the electrical potential energy lost, which becomes kinetic energy: \( E_k = W = q V \). For an alpha particle, \( q = 2 e = 2 \times 1.60 \times 10^{-19}\text{ C} = 3.20 \times 10^{-19}\text{ C} \). Substituting values: \( E_k = 3.20 \times 10^{-19}\text{ C} \times 25.0 \times 10^3\text{ V} = 8.00 \times 10^{-15}\text{ J} \). (b) Kinetic energy is given by \( E_k = \frac{1}{2} m v^2 \). Solving for speed: \( v = \sqrt{\frac{2 E_k}{m}} = \sqrt{\frac{2 \times 8.00 \times 10^{-15}}{6.64 \times 10^{-27}}} \approx \sqrt{2.41 \times 10^{12}} \approx 1.55 \times 10^6\text{ m s}^{-1} \). (c) At the distance of closest approach \( d \), all initial kinetic energy of the alpha particle is converted into electrostatic potential energy: \( E_k = \frac{1}{4\pi\varepsilon_0} \frac{q_{\alpha} q_{\text{gold}}}{d} \). Here, \( q_{\alpha} = 2 e \) and \( q_{\text{gold}} = 79 e \). Rearranging for \( d \): \( d = \frac{1}{4\pi\varepsilon_0} \frac{158 e^2}{E_k} = \frac{8.99 \times 10^9 \times 158 \times (1.60 \times 10^{-19})^2}{8.00 \times 10^{-15}} \approx 4.545 \times 10^{-12}\text{ m} \approx 4.55 \times 10^{-12}\text{ m} \).

(a) [3 Marks] - Stating \( W = q V \) (1 mark). - Substituting charge of alpha particle \( q = 2e \) (1 mark). - Correct calculation showing \( 8.00 \times 10^{-15}\text{ J} \) (1 mark). (b) [3.28 Marks] - Stating \( E_k = \frac{1}{2} m v^2 \) (1 mark). - Substituting values correctly (1 mark). - Speed calculated as \( 1.55 \times 10^6\text{ m s}^{-1} \) (1.28 marks). (c) [3 Marks] - Equating kinetic energy to electrostatic potential energy (1 mark). - Stating charges of alpha particle and gold nucleus in terms of \( e \) (1 mark). - Calculation of distance of closest approach as \( 4.55 \times 10^{-12}\text{ m} \) [accept range 4.50 to 4.56] (1 mark).

The rate of doing work against gravity to raise the sand is:
ewline P_{\text{gravity}} = \frac{dE_p}{dt} = g h \frac{dM}{dt} = \mu g h.
ewline The rate of kinetic energy delivered to the sand to accelerate it to speed \(v\) is:
ewline P_{\text{kinetic}} = \frac{dE_k}{dt} = \frac{1}{2} v^2 \frac{dM}{dt} = \frac{1}{2} \mu v^2.
ewline Therefore, the total minimum power is the sum of these two terms:
ewline P_{\text{total}} = \mu g h + \frac{1}{2} \mu v^2 = \mu\left(g h + \frac{1}{2} v^2\right).

Correct option selected: C (1 mark).

The potential energy of the particle is:
ewline E_p = \frac{1}{2} m \omega^2 x^2
ewline At displacement \(x = \frac{1}{2}A\):
ewline E_p = \frac{1}{2} m \omega^2 \left(\frac{1}{2}A\right)^2 = \frac{1}{8} m \omega^2 A^2
ewline The total energy is:
ewline E_{\text{total}} = \frac{1}{2} m \omega^2 A^2
ewline The kinetic energy is:
ewline E_k = E_{\text{total}} - E_p = \frac{1}{2} m \omega^2 A^2 - \frac{1}{8} m \omega^2 A^2 = \frac{3}{8} m \omega^2 A^2
ewline The ratio \(\frac{E_k}{E_p}\) is:
ewline \frac{E_k}{E_p} = \frac{\frac{3}{8} m \omega^2 A^2}{\frac{1}{8} m \omega^2 A^2} = 3.

Correct option selected: D (1 mark).

The equation of motion for a sphere of mass \(m\) falling in a viscous fluid is:
ewline m g - k v = m a
ewline where \(k\) is a constant.
ewline Solving for acceleration \(a\):
ewline a = g - \frac{k}{m} v
ewline This represents a linear relationship between \(a\) and \(v\) with a negative gradient. Thus, the graph is a straight line with a negative gradient.

Correct option selected: A (1 mark).

The fractional uncertainty in resistivity is given by:
ewline \frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}
ewline Substituting the percentage uncertainties:
ewline \% \Delta \rho = 2.0\% + 2 \times 1.5\% + 1.0\% = 2.0\% + 3.0\% + 1.0\% = 6.0\%.

Correct option selected: C (1 mark).

From the photoelectric equation:
ewline e V_s = h f - \Phi \implies h f = e V_s + \Phi
ewline For frequency \(2f\), the stopping potential \(V'_s\) is given by:
ewline e V'_s = h(2f) - \Phi = 2(h f) - \Phi
ewline Substitute \(h f = e V_s + \Phi\) into the equation:
ewline e V'_s = 2(e V_s + \Phi) - \Phi = 2 e V_s + \Phi
ewline Dividing by \(e\):
ewline V'_s = 2 V_s + \frac{\Phi}{e}.

Correct option selected: B (1 mark).

First, find the internal resistance \(r\) of the cell.
ewline The current in the first circuit is:
ewline I_1 = \frac{V_1}{R_1} = \frac{5.0\text{ V}}{5.0\ \Omega} = 1.0\text{ A}
ewline Using the formula for emf:
ewline \varepsilon = V_1 + I_1 r \implies 6.0\text{ V} = 5.0\text{ V} + (1.0\text{ A}) r \implies r = 1.0\ \Omega
ewline Next, calculate the current in the second circuit with \(R_2 = 2.0\ \Omega\):
ewline I_2 = \frac{\varepsilon}{R_2 + r} = \frac{6.0\text{ V}}{2.0\ \Omega + 1.0\ \Omega} = 2.0\text{ A}
ewline The terminal potential difference \(V_2\) is:
ewline V_2 = I_2 R_2 = 2.0\text{ A} \times 2.0\ \Omega = 4.0\text{ V}.

Correct option selected: C (1 mark).

An increase in light intensity on an LDR causes more charge carriers to be released, which decreases its resistance.
ewline In a potential divider circuit, the output voltage across the LDR is:
ewline V_{\text{out}} = V_{\text{supply}} \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}
ewline As the resistance of the LDR (\(R_{\text{LDR}}\)) decreases, the fraction of the total resistance occupied by the LDR decreases. Therefore, the output voltage \(V_{\text{out}}\) decreases.

Correct option selected: D (1 mark).

The resistance of a wire is given by:
ewline R = \rho \frac{L}{A} = \rho \frac{4 L}{\pi d^2}
ewline For wire P:
ewline R_P = \rho_P \frac{4 L}{\pi d^2}
ewline For wire Q:
ewline R_Q = \rho_Q \frac{4 (2L)}{\pi (2d)^2} = \rho_Q \frac{8 L}{4 \pi d^2} = \rho_Q \frac{2 L}{\pi d^2}
ewline Since the resistances are equal (\(R_P = R_Q\)):
ewline \rho_P \frac{4 L}{\pi d^2} = \rho_Q \frac{2 L}{\pi d^2}
ewline 4 \rho_P = 2 \rho_Q \implies \frac{\rho_P}{\rho_Q} = \frac{2}{4} = \frac{1}{2}.

Correct option selected: B (1 mark).

The energy \(E\) stored in a capacitor at any time \(t\) during discharge is given by \(E = E_0 e^{-2t/RC}\), where \(E_0\) is the initial energy. We want to find the time \(t\) when \(E = 0.10 E_0\). Thus, \(0.10 E_0 = E_0 e^{-2t/RC}\), which simplifies to \(0.10 = e^{-2t/RC}\). Taking the natural logarithm of both sides gives \(\ln(0.10) = -2t/RC\). This leads to \(-2.30 = -2t/RC\), so \(t = 1.15 RC\).

1 mark for the correct option C.

The resistance of a wire is given by \(R = \rho \frac{L}{A} = \rho \frac{4L}{\pi d^2}\). For wire X, \(L_X = 2 L_Y\) and \(d_X = 0.5 d_Y\). Therefore, \(R_X = \rho \frac{4 (2 L_Y)}{\pi (0.5 d_Y)^2} = \rho \frac{8 L_Y}{0.25 \pi d_Y^2} = 32 \rho \frac{L_Y}{\pi d_Y^2}\). Since \(R_Y = \rho \frac{4 L_Y}{\pi d_Y^2}\), we have \(\frac{R_X}{R_Y} = \frac{32}{4} = 8\).

1 mark for the correct option C.

The total resistance of the circuit is \(R_{\text{total}} = R + r = 2.60\ \Omega + 0.40\ \Omega = 3.00\ \Omega\). The current \(I\) in the circuit is \(I = \frac{\varepsilon}{R_{\text{total}}} = \frac{1.50\text{ V}}{3.00\ \Omega} = 0.50\text{ A}\). The power dissipated in the internal resistance is \(P = I^2 r = (0.50\text{ A})^2 \times 0.40\ \Omega = 0.25 \times 0.40 = 0.10\text{ W}\).

1 mark for the correct option A.

Rearranging the equation for \(g\) gives \(g = \frac{2h}{t^2}\). The percentage uncertainty in \(g\) is given by the sum of the percentage uncertainty in \(h\) and twice the percentage uncertainty in \(t\): \(\frac{\Delta g}{g} \times 100\% = (\frac{\Delta h}{h} + 2\frac{\Delta t}{t}) \times 100\%\). Calculating the values: \(\frac{\Delta h}{h} \times 100\% = \frac{0.02}{2.00} \times 100\% = 1.0\%\) and \(\frac{\Delta t}{t} \times 100\% = \frac{0.02}{0.64} \times 100\% = 3.125\%\). Therefore, the percentage uncertainty in \(g\) is \(1.0\% + 2 \times 3.125\% = 1.0\% + 6.25\% = 7.25\%\), which rounds to \(7.3\%\).

1 mark for the correct option C.

The useful power output \(P_{\text{out}}\) required to lift the crate is \(P_{\text{out}} = F v = m g v = 80\text{ kg} \times 9.81\text{ m s}^{-2} \times 1.5\text{ m s}^{-1} = 1177.2\text{ W}\). The efficiency \(\eta\) of the motor is \(60\%\) or \(0.60\), so \(P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{1177.2\text{ W}}{0.60} = 1962\text{ W} \approx 2.0\text{ kW}\).

1 mark for the correct option C.

The maximum kinetic energy is given by \(E_{k,\text{max}} = \frac{1}{2} m v_{\text{max}}^2\), where \(v_{\text{max}} = 2\pi f A\). Given \(m = 0.20\text{ kg}\), \(A = 0.050\text{ m}\), and \(f = 4.0\text{ Hz}\), we find \(v_{\text{max}} = 2\pi \times 4.0 \times 0.050 = 0.40\pi \approx 1.257\text{ m s}^{-1}\). Thus, \(E_{k,\text{max}} = \frac{1}{2} \times 0.20 \times 1.257^2 \approx 0.158\text{ J}\), which rounds to \(0.16\text{ J}\).

1 mark for the correct option B.

According to Einstein's photoelectric equation, \(h f = \Phi + E_k\). For the frequency \(2f\), the equation becomes \(h(2f) = \Phi + 3E_k\), which is \(2(h f) = \Phi + 3E_k\). Substituting the first equation into the second gives \(2(\Phi + E_k) = \Phi + 3E_k\). This simplifies to \(2\Phi + 2E_k = \Phi + 3E_k\), yielding \(\Phi = E_k\) (or \(1.0\,E_k\)).

1 mark for the correct option B.