2024 AQA IAL Physics (9630) 模擬試題連答案詳解

(a) First, find the mean of the uncorrected readings: Mean uncorrected \(d = (0.42 + 0.45 + 0.41 + 0.44) / 4 = 0.43\text{ mm}\). Subtract the zero error to get the corrected mean: Corrected \(d = 0.43 - 0.02 = 0.41\text{ mm}\). (Alternatively, correct each reading first: \(0.40\text{ mm}\), \(0.43\text{ mm}\), \(0.39\text{ mm}\), \(0.42\text{ mm}\). Mean of these is \(0.41\text{ mm}\)). (b) The uncertainty in a repeated measurement is given by half the range of the readings. Range of readings \(= 0.45 - 0.41 = 0.04\text{ mm}\). Uncertainty \(= \text{Range} / 2 = 0.04 / 2 = 0.02\text{ mm}\). Note that the systematic zero error does not affect the random uncertainty (range of the readings). (c) Percentage uncertainty \(= (\text{Absolute uncertainty} / \text{Corrected mean}) \times 100\% = (0.02 / 0.41) \times 100\% \approx 4.88\% \approx 4.9\%\) (accept \(5\%\) if rounded to 1 sig fig).

(a) Calculates mean of uncorrected readings (\(0.43\text{ mm}\)) or corrects individual readings [1 mark]; subtracts zero error to obtain \(0.41\text{ mm}\) [0.5 marks]. (b) Identifies the range as \(0.04\text{ mm}\) [1 mark]; calculates absolute uncertainty as \(\text{range}/2 = 0.02\text{ mm}\) [1 mark]. (c) Recalls or uses \(\% \text{ uncertainty} = (\text{uncertainty} / \text{corrected mean}) \times 100\%\) [1 mark]; obtains \(4.9\%\) or \(4.88\%\) (accept \(5\%\)) [1 mark].

(a) Rearranging \(h = \frac{1}{2}gt^2\) gives \(g = \frac{2h}{t^2}\). Substituting the values: \(g = \frac{2 \times 1.20}{0.49^2} = \frac{2.40}{0.2401} \approx 10.0\text{ m s}^{-2}\). (b) \(\% \text{ uncertainty in } g = \% \text{ uncertainty in } h + 2 \times (\% \text{ uncertainty in } t)\). \(\% \text{ uncertainty in } h = (0.01 / 1.20) \times 100\% \approx 0.83\%\). \(\% \text{ uncertainty in } t = (0.02 / 0.49) \times 100\% \approx 4.08\%\). Total \(\% \text{ uncertainty in } g = 0.83\% + 2 \times 4.08\% = 9.0\%\) (or \(8.99\%\)). (c) Absolute uncertainty in \(g = 9.996 \times 0.08996 \approx 0.90\text{ m s}^{-2}\). Since the absolute uncertainty is \(0.9\text{ m s}^{-2}\) (1 decimal place), the final value of g should be quoted to 1 decimal place: \(g = 10.0 \pm 0.9\text{ m s}^{-2}\).

(a) Rearranges formula to \(g = \frac{2h}{t^2}\) [0.5 marks]; calculates \(g = 10.0\text{ m s}^{-2}\) [1 mark]. (b) Recalls or uses relationship \(\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t}\) [1 mark]; calculates individual percentage uncertainties correctly [0.5 marks]; obtains total percentage uncertainty of \(9.0\%\) (or \(9\%\)) [1 mark]. (c) Calculates absolute uncertainty as \(0.9\text{ m s}^{-2}\) [0.5 marks]; quotes final answer with correct units and consistent decimal places as \(10.0 \pm 0.9\text{ m s}^{-2}\) [1 mark].

(a) The volume of a cylinder is \(V = \pi R^2 L\). Therefore, density is \(\rho = \frac{M}{\pi R^2 L}\). (b) Substituting values: \(\rho = \frac{45.3}{\pi \times (1.20)^2 \times 5.0} = \frac{45.3}{22.619} \approx 2.003\text{ g cm}^{-3}\). Since length L is given to 2 significant figures, the density is rounded to 2 significant figures: \(\rho = 2.0\text{ g cm}^{-3}\). (c) The percentage uncertainty in density \(\rho\) is \(\% \text{ uncertainty in } \rho = \% \text{ uncertainty in } M + 2 \times (\% \text{ uncertainty in } R) + \% \text{ uncertainty in } L\). \(\% \text{ uncertainty in } M = (0.5 / 45.3) \times 100\% \approx 1.10\%\). \(\% \text{ uncertainty in } R = (0.02 / 1.20) \times 100\% \approx 1.67\%\). \(\% \text{ uncertainty in } L = (0.1 / 5.0) \times 100\% = 2.00\%\). Total \(\%\) uncertainty \(= 1.10\% + 2 \times (1.67\%) + 2.00\% = 6.44\% \approx 6.4\%\) (accept \(6.4\%\) or \(6\%\)).

(a) Expresses density correctly as \(\rho = \frac{M}{\pi R^2 L}\) [1 mark]. (b) Substitutes values into the formula correctly [0.5 marks]; obtains \(2.0\text{ g cm}^{-3}\) (or \(2.00\text{ g cm}^{-3}\)) [1 mark]. (c) Correctly identifies that the fractional uncertainty of R must be multiplied by 2 [1 mark]; calculates individual percentage uncertainties (M \(\approx 1.1\%\), R \(\approx 1.7\%\), L \(\approx 2.0\%\)) [1 mark]; sums components to obtain \(6.4\%\) (or \(6.44\%\)) [1 mark].

(a) The period T of a mass-spring system is \(T = 2\pi \sqrt{\frac{m}{k}}\). Squaring both sides and rearranging for k gives: \(k = \frac{4\pi^2 m}{T^2}\). Substituting the given values: \(k = \frac{4 \pi^2 \times 0.350}{0.780^2} \approx 22.71\text{ N m}^{-1}\), which is approximately \(23\text{ N m}^{-1}\). (b) The maximum speed of an object in simple harmonic motion is \(v_{\text{max}} = \omega A\), where \(\omega = \frac{2\pi}{T}\) and A is the amplitude. Here, \(A = 4.5\text{ cm} = 0.045\text{ m}\). The angular frequency is \(\omega = \frac{2\pi}{0.780} \approx 8.055\text{ rad s}^{-1}\). Thus, \(v_{\text{max}} = 8.055 \times 0.045 \approx 0.362\text{ m s}^{-1}\) (or \(0.36\text{ m s}^{-1}\)).

(a) Recalls \(T = 2\pi \sqrt{\frac{m}{k}}\) [1 mark]; rearranges to make k the subject [0.5 marks]; substitutes values and shows \(k = 22.7\text{ N m}^{-1}\) (must show at least 3 sig figs to justify 'approximately 23') [1 mark]. (b) Recalls \(v_{\text{max}} = \omega A\) [1 mark]; calculates \(\omega = \frac{2\pi}{0.780} = 8.06\text{ rad s}^{-1}\) [1 mark]; obtains \(v_{\text{max}} = 0.36\text{ m s}^{-1}\) (or \(0.362\text{ m s}^{-1}\)) [1 mark].

(a) The period of a simple pendulum is \(T = 2\pi \sqrt{\frac{L}{g}}\). Squaring both sides and rearranging for L gives: \(L = \frac{g T^2}{4\pi^2}\). Substituting \(T = 1.65\text{ s}\) and \(g = 9.81\text{ m s}^{-2}\): \(L = \frac{9.81 \times 1.65^2}{4\pi^2} \approx 0.677\text{ m}\) (or \(67.7\text{ cm}\)). (b) First, note that the period of a simple pendulum is independent of its mass. Changing the bob mass has no effect on the period. The new length is \(L_{\text{new}} = 0.85 L = 0.85 \times 0.677 = 0.575\text{ m}\). The new period is \(T_{\text{new}} = 2\pi \sqrt{\frac{L_{\text{new}}}{g}}\). Alternatively, using proportions: \(T_{\text{new}} = T \times \sqrt{0.85} = 1.65 \times \sqrt{0.85} \approx 1.52\text{ s}\).

(a) Recalls \(T = 2\pi \sqrt{\frac{L}{g}}\) and rearranges for L [1 mark]; obtains \(L = 0.68\text{ m}\) or \(0.677\text{ m}\) [1 mark]. (b) States that changing the mass of the bob has no effect on the period [1 mark]; identifies that the new length is \(0.85L\) [0.5 marks]; uses relationship \(T \propto \sqrt{L}\) or calculates new length and substitutes into the period formula [1 mark]; obtains \(1.52\text{ s}\) [1 mark].

(a) The energy supplied is given by \(E = P \times t = 800\text{ W} \times 110\text{ s} = 88000\text{ J}\) (or \(88\text{ kJ}\)). (b) Using \(Q = m c \Delta\theta\), where \(\Delta\theta = 85 - 20 = 65^\circ\text{C}\) (or \(65\text{ K}\)). Rearranging for c: \(c = \frac{Q}{m \Delta\theta} = \frac{88000}{1.2 \times 65} \approx 1128\text{ J kg}^{-1}\text{ K}^{-1}\) (or \(1.1 \times 10^3\text{ J kg}^{-1}\text{ K}^{-1}\)). (c) In reality, some thermal energy is lost to the surroundings (air, insulation, thermometer, etc.) rather than being absorbed entirely by the copper block. As a result, the energy actually absorbed by the copper is less than the calculated \(88000\text{ J}\). Since the formula assumes all supplied energy went into the copper, using an overestimate for Q leads to an overestimate for c.

(a) Uses \(E = P \times t\) [0.5 marks]; obtains \(88000\text{ J}\) [1 mark]. (b) Identifies \(\Delta\theta = 65\text{ K}\) and recalls \(Q = m c \Delta\theta\) [1 mark]; obtains \(1100\text{ J kg}^{-1}\text{ K}^{-1}\) (or \(1130\text{ J kg}^{-1}\text{ K}^{-1}\)) [1 mark]. (c) Explains that thermal energy is lost to the surroundings [1 mark]; states that this means the actual energy absorbed by the copper is less than the calculated energy supplied, so the calculated c is too high [1 mark].

(a) Work done by the gas during expansion at constant pressure is: \(W = p \Delta V\). Here, \(\Delta V = V_f - V_i = 3.8 \times 10^{-3} - 2.4 \times 10^{-3} = 1.4 \times 10^{-3}\text{ m}^3\). \(W = (1.5 \times 10^5\text{ Pa}) \times (1.4 \times 10^{-3}\text{ m}^3) = 210\text{ J}\). (b) According to the first law of thermodynamics, \(\Delta U = Q - W\) (where Q is thermal energy supplied to the gas, and W is work done BY the gas). Here, \(Q = +450\text{ J}\) and \(W = +210\text{ J}\). \(\Delta U = 450 - 210 = +240\text{ J}\). Since \(\Delta U\) is positive, the internal energy of the gas increased by \(240\text{ J}\).

(a) Calculates \(\Delta V = 1.4 \times 10^{-3}\text{ m}^3\) [0.5 marks]; recalls \(W = p \Delta V\) and substitutes values [1 mark]; obtains \(210\text{ J}\) [0.5 marks]. (b) Recalls the first law of thermodynamics (e.g., \(\Delta U = Q - W\)) with correct sign conventions [1 mark]; substitutes \(Q = +450\text{ J}\) and \(W = +210\text{ J}\) [1 mark]; calculates \(\Delta U = 240\text{ J}\) [1 mark]; states that the internal energy increased [0.5 marks].

(a) \(I = \frac{1}{2} M R^2 = 0.5 \times 2.50 \times (0.30)^2 = 0.1125\text{ kg m}^2\) (or \(0.11\text{ kg m}^2\)). (b) The torque applied is \(\tau = F \times R = 6.0\text{ N} \times 0.30\text{ m} = 1.8\text{ N m}\). Using \(\tau = I \alpha\): \(\alpha = \frac{\tau}{I} = \frac{1.8}{0.1125} = 16\text{ rad s}^{-2}\). (c) Using the angular kinematic equation starting from rest (\(\omega_0 = 0\)): \(\omega = \alpha t = 16 \times 4.0 = 64\text{ rad s}^{-1}\).

(a) Recalls \(I = \frac{1}{2} M R^2\) and substitutes values [0.5 marks]; obtains \(0.11\text{ kg m}^2\) (or \(0.113\text{ kg m}^2\)) [1 mark]. (b) Recalls \(\tau = F \times R\) and calculates \(\tau = 1.8\text{ N m}\) [1 mark]; recalls \(\tau = I \alpha\) and rearranges to find \(\alpha\) [0.5 marks]; obtains \(\alpha = 16\text{ rad s}^{-2}\) [1 mark]. (c) Recalls \(\omega = \alpha t\) [0.5 marks]; obtains \(\omega = 64\text{ rad s}^{-1}\) [1 mark].

(a) Since \(T = T_{20} / 20\), the division by a constant does not change the percentage uncertainty. Thus, the percentage uncertainty in \(T\) is equal to that of \(T_{20}\): \%\Delta T = \frac{0.4}{37.0} \times 100\% \approx 1.08\% (or 1.1\%). (b) Rearranging the formula for \(g\) yields: \(g = \frac{4\pi^2 L}{T^2}\). The percentage uncertainty in \(g\) is given by: \%\Delta g = \%\Delta L + 2 \times \%\Delta T. First, calculate the percentage uncertainty in \(L\): \%\Delta L = \frac{0.002}{0.850} \times 100\% \approx 0.235\%. Now, sum the uncertainties: \%\Delta g = 0.235\% + 2 \times 1.081\% = 2.397\%. Finally, find the absolute uncertainty in \(g\): \Delta g = 9.80 \times 0.02397 \approx 0.235\text{ m s}^{-2}\), which rounds to \(0.24\text{ m s}^{-2}\).

(a) [1.5 marks total]: 1 mark for expressing \%\Delta T = \frac{0.4}{37.0} \times 100\%. 0.5 marks for obtaining 1.1\% (or 1.08\%). (b) [4 marks total]: 1 mark for recognizing that percentage uncertainties add as \%\Delta g = \%\Delta L + 2\%\Delta T. 1 mark for calculating \%\Delta L = 0.235\% (or 0.24\%). 1 mark for obtaining total percentage uncertainty of 2.40\% (or 2.4\%). 1 mark for final absolute uncertainty of 0.24 m s^-2 (accept 0.23 to 0.24 m s^-2).

(a) A systematic error causes all readings to differ from the true value by a consistent, predictable amount in the same direction (e.g., the constant zero offset of \(+0.15\text{ g}\)). A random error causes readings to fluctuate unpredictably above and below the true value due to unpredictable variations (e.g., the limitation of digital resolution \(\pm 0.01\text{ g}\) in the final digit). (b) Corrected total mass of 10 spheres = \(48.25\text{ g} - 0.15\text{ g} = 48.10\text{ g}\). Corrected mass of 1 sphere = \(48.10\text{ g} / 10 = 4.810\text{ g}\). The uncertainty in the total mass measurement comes from combining the systematic accuracy and the resolution limit. Summing them linearly: \(\Delta m_{\text{total}} = 0.05\text{ g} + 0.01\text{ g} = 0.06\text{ g}\). (Alternatively, adding in quadrature: \(\sqrt{0.05^2 + 0.01^2} = 0.051\text{ g}\)). Dividing the total mass by 10 scales the absolute uncertainty down by a factor of 10. Therefore, the absolute uncertainty in a single sphere's mass is: \(\Delta m_{\text{single}} = 0.06\text{ g} / 10 = 0.006\text{ g}\) (or \(0.005\text{ g}\) in quadrature).

(a) [2 marks total]: 1 mark for defining systematic error with the zero offset example. 1 mark for defining random error with the resolution example. (b) [3.5 marks total]: 1 mark for calculating corrected single mass of 4.81 g. 1 mark for calculating combined total uncertainty as 0.06 g (linear) or 0.051 g (quadrature). 1.5 marks for showing that dividing by 10 reduces the absolute uncertainty, giving the final value of +/- 0.006 g (or +/- 0.005 g).

(a) The time period of a mass-spring system is given by \(T = 2\pi \sqrt{\frac{m}{k}}\). Squaring and rearranging for \(k\) gives: \(k = \frac{4\pi^2 m}{T^2}\). Substituting the values: \(k = \frac{4\pi^2 \times 0.45}{0.80^2} = \frac{17.765}{0.64} \approx 27.8\text{ N m}^{-1}\) (or \(28\text{ N m}^{-1}\)). (b) The maximum kinetic energy is equal to the total mechanical energy of the simple harmonic oscillator, which is given by \(E_{\text{total}} = \frac{1}{2} k A^2\). Rearranging for \(A\): \(A^2 = \frac{2 E_{k,\text{max}}}{k}\). Substituting the values: \(A^2 = \frac{2 \times 0.18}{27.76} \approx 0.01297\text{ m}^2\). Thus, \(A = \sqrt{0.01297} \approx 0.114\text{ m}\) (or \(0.11\text{ m}\)).

(a) [2 marks total]: 1 mark for rearranging the time period formula to make k the subject. 1 mark for correct calculation of k as 28 N m^-1 (or 27.8 N m^-1). (b) [3.5 marks total]: 1 mark for using the energy relationship E_k,max = 0.5 * k * A^2. 1.5 marks for correct substitution and rearranging to find A^2. 1 mark for calculating amplitude A = 0.11 m (accept 0.11 m to 0.114 m).

(a) The angular frequency of a simple pendulum is: \(\omega = \sqrt{\frac{g}{L}} = \sqrt{\frac{9.81}{1.2}} \approx 2.86\text{ rad s}^{-1}\). The maximum velocity occurs at equilibrium and is given by: \(v_{\text{max}} = \omega A = 2.859 \times 0.080 \approx 0.229\text{ m s}^{-1}\), which is approximately \(0.23\text{ m s}^{-1}\). (b) The potential energy at any displacement \(x\) is given by: \(E_p = \frac{1}{2} m \omega^2 x^2\). Substituting the values: \(E_p = \frac{1}{2} \times 0.15 \times (2.859)^2 \times (0.040)^2 = 0.075 \times 8.174 \times 0.00160 \approx 9.81 \times 10^{-4}\text{ J}\) (or \(0.98\text{ mJ}\)).

(a) [2.5 marks total]: 1 mark for calculating angular frequency \(\omega = 2.86\text{ rad s}^{-1}\). 1 mark for using \(v_{\text{max}} = \omega A\). 0.5 marks for showing the result is approximately 0.23 m s^-1. (b) [3 marks total]: 1 mark for using \(E_p = \frac{1}{2} m \omega^2 x^2\) (or recognizing that potential energy is proportional to \(x^2\), so it is one-quarter of the total energy). 1 mark for correct substitution of values. 1 mark for final calculated answer of \(9.8 \times 10^{-4}\text{ J}\) (or \(9.81 \times 10^{-4}\text{ J}\)).

(a) Energy supplied during the heating stage is: \(Q = P \times t = 850\text{ W} \times 65\text{ s} = 55250\text{ J}\). Using \(Q = m c \Delta\theta\), where \(\Delta\theta = 78^\circ\text{C} - 20^\circ\text{C} = 58^\circ\text{C}\): \(55250 = 0.40 \times c \times 58\), which gives \(c = \frac{55250}{23.2} \approx 2381\text{ J kg}^{-1}\text{ K}^{-1}\). This value, when rounded to two significant figures, is \(2.4 \times 10^3\text{ J kg}^{-1}\text{ K}^{-1}\), matching the accepted value. (b) Energy supplied during the vaporisation stage is: \(Q_{\text{vap}} = P \times t_{\text{vap}} = 850\text{ W} \times 210\text{ s} = 178500\text{ J}\). Using \(Q_{\text{vap}} = m L_v\): \(178500 = 0.40 \times L_v\). Therefore, \(L_v = \frac{178500}{0.40} = 446250\text{ J kg}^{-1}\), which rounds to \(4.5 \times 10^5\text{ J kg}^{-1}\).

(a) [2 marks total]: 1 mark for calculating energy input \(Q = 55250\text{ J}\) and applying the heat formula. 1 mark for obtaining \(c \approx 2.4 \times 10^3\text{ J kg}^{-1}\text{ K}^{-1}\) and confirming it matches the accepted value. (b) [3.5 marks total]: 1 mark for calculating the energy input for the vaporisation phase: \(178500\text{ J}\). 1 mark for stating \(Q = m L_v\). 1.5 marks for correct computation to get \(L_v = 4.5 \times 10^5\text{ J kg}^{-1}\) (or \(4.46 \times 10^5\text{ J kg}^{-1}\)).

(a) The work done by an expanding gas at constant pressure is: \(W = p \Delta V\). Here, \(\Delta V = 4.5 \times 10^{-3} - 2.0 \times 10^{-3} = 2.5 \times 10^{-3}\text{ m}^3\). \(W = 1.5 \times 10^5 \times 2.5 \times 10^{-3} = 375\text{ J}\). (b) Using the First Law of Thermodynamics: \(\Delta U = Q - W\) (where \(Q\) is heat added to the gas and \(W\) is work done by the gas). Substituting the values: \(\Delta U = 800\text{ J} - 375\text{ J} = +425\text{ J}\). The positive sign indicates that the internal energy of the gas has increased by \(425\text{ J}\).

(a) [2 marks total]: 1 mark for calculating the change in volume \(\Delta V = 2.5 \times 10^{-3}\text{ m}^3\). 1 mark for calculating work done as \(375\text{ J}\). (b) [3.5 marks total]: 1.5 marks for stating the First Law of Thermodynamics and defining the sign convention (e.g. \(\Delta U = Q - W\) with \(W\) as work done by the gas). 1 mark for substituting the values correctly. 1 mark for calculating \(\Delta U = +425\text{ J}\) and clarifying that this represents an increase in internal energy.

(a) Using the formula: \(I = \frac{1}{2} M R^2\). Substituting the given values: \(I = \frac{1}{2} \times 12 \times (0.30)^2 = 6 \times 0.09 = 0.54\text{ kg m}^2\). (b) First, determine the angular acceleration \(\alpha\) from the kinematic equation \(\omega_f = \omega_i + \alpha t\): \(\alpha = \frac{45 - 0}{8.0} = 5.625\text{ rad s}^{-2}\). Now, apply Newton's second law for rotation: \(T = I \alpha\). Substituting the values: \(T = 0.54 \times 5.625 = 3.0375\text{ N m}\), which rounds to \(3.0\text{ N m}\) (or \(3.04\text{ N m}\)).

(a) [1.5 marks total]: 1 mark for substitution into the moment of inertia formula. 0.5 marks for finding \(I = 0.54\text{ kg m}^2\). (b) [4 marks total]: 1 mark for calculating the angular acceleration \(\alpha = 5.625\text{ rad s}^{-2}\). 1.5 marks for stating and using \(T = I \alpha\). 1.5 marks for correct final torque value of \(3.0\text{ N m}\) (or \(3.04\text{ N m}\)).

(a) The total kinetic energy is the sum of translational and rotational kinetic energies: \(E_k = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2\). Since the sphere rolls without slipping, its angular velocity is related to its linear velocity by \(\omega = \frac{v}{R}\). Substituting \(I = \frac{2}{3} M R^2\) and \(\omega = \frac{v}{R}\): \(E_k = \frac{1}{2} M v^2 + \frac{1}{2} \left(\frac{2}{3} M R^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{2} M v^2 + \frac{1}{3} M v^2 = \frac{5}{6} M v^2\). (b) Using the principle of conservation of energy: Loss in Gravitational Potential Energy = Gain in Total Kinetic Energy. \(M g h = \frac{5}{6} M v^2\). The mass \(M\) cancels out from both sides: \(g h = \frac{5}{6} v^2\). Rearranging for \(v\): \(v = \sqrt{\frac{6}{5} g h}\). Substituting the values: \(v = \sqrt{1.2 \times 9.81 \times 1.8} = \sqrt{21.1896} \approx 4.60\text{ m s}^{-1}\) (or \(4.6\text{ m s}^{-1}\)).

(a) [2.5 marks total]: 1 mark for stating the sum of translational and rotational kinetic energy. 1 mark for substituting \(\omega = v/R\) and \(I = \frac{2}{3} M R^2\). 0.5 marks for showing the final expression simplifies to \(\frac{5}{6} M v^2\). (b) [3 marks total]: 1 mark for equating the loss in gravitational potential energy to the total kinetic energy expression. 1 mark for isolating \(v\) as \(\sqrt{1.2 g h}\). 1 mark for calculating the final speed as \(4.6\text{ m s}^{-1}\) (accept \(4.60\text{ m s}^{-1}\)).

First, find the formula for volume of a cylinder: \(V = \frac{\pi d^2 L}{4}\).

The percentage uncertainty in \(V\) is:
\(\frac{\Delta V}{V} = 2 \times \frac{\Delta d}{d} + \frac{\Delta L}{L}\)
\(\frac{\Delta d}{d} = \frac{0.2}{12.4} \times 100\% \approx 1.61\%\)
\(\frac{\Delta L}{L} = \frac{0.5}{82.0} \times 100\% \approx 0.61\%\)

Percentage uncertainty in \(V = 2(1.61\%) + 0.61\% = 3.83\% \approx 3.8\%\).

For density \(\rho = \frac{m}{V}\):
Percentage uncertainty in \(\rho = \frac{\Delta m}{m} + \frac{\Delta V}{V}\)
\(\frac{\Delta m}{m} = \frac{0.1}{24.5} \times 100\% \approx 0.41\%\)

Total percentage uncertainty in \(\rho = 0.41\% + 3.83\% = 4.24\%\).

Nominal value of volume:
\(V = \frac{\pi (12.4 \times 10^{-3})^2 \times (82.0 \times 10^{-3})}{4} \approx 9.90 \times 10^{-6}\text{ m}^3\).

Nominal value of density:
\(\rho = \frac{24.5 \times 10^{-3}}{9.90 \times 10^{-6}} \approx 2475\text{ kg m}^{-3}\).

Absolute uncertainty in density:
\(\Delta \rho = 2475 \times 0.0424 \approx 105\text{ kg m}^{-3}\).

Rounding to appropriate significant figures, this is \(100\text{ kg m}^{-3}\) (or \(110\text{ kg m}^{-3}\)).

[1 mark] Recall or use of percentage uncertainty in volume formula: \(2 \times \% \Delta d + \% \Delta L\).
[1 mark] Correct calculation of percentage uncertainties in \(d\) (1.61%) and \(L\) (0.61%).
[1 mark] Correct calculation of volume percentage uncertainty: \(3.8\%\) (or \(3.83\%\)).
[1 mark] Calculates percentage uncertainty in mass (0.41%) and adds to volume uncertainty to find total density uncertainty: \(4.2\%\) (or \(4.24\%\)).
[1 mark] Correct calculation of nominal density: \(\rho \approx 2470\text{ kg m}^{-3}\) (or \(2.47\text{ g cm}^{-3}\)).
[0.5 mark] Correct absolute uncertainty: \(\pm 100\text{ kg m}^{-3}\) (or \(\pm 0.1\text{ g cm}^{-3}\)) quoted to 1 or 2 significant figures.

First, find the period \(T\):
\(T = \frac{36.2}{20} = 1.810\text{ s}\).

Absolute uncertainty in the period is:
\(\Delta T = \frac{0.2}{20} = 0.010\text{ s}\).

Next, calculate the nominal value of \(g\):
\(g = \frac{4\pi^2 (0.800)}{1.810^2} \approx 9.640\text{ m s}^{-2}\).

Now calculate the percentage uncertainties:
\(\% \Delta L = \frac{0.002}{0.800} \times 100\% = 0.25\%\).
\(\% \Delta T = \frac{0.010}{1.810} \times 100\% \approx 0.552\%\).

Combine these to find the percentage uncertainty in \(g\):
\(\% \Delta g = \% \Delta L + 2 \times (\% \Delta T) = 0.25\% + 2(0.552\%) = 1.35\% \approx 1.4\%\).

Absolute uncertainty in \(g\):
\(\Delta g = 9.640 \times 0.0135 = 0.13\text{ m s}^{-2}\).

[1 mark] Calculates period \(T = 1.81\text{ s}\).
[1 mark] Calculates absolute uncertainty in period \(\Delta T = 0.010\text{ s}\).
[1 mark] Calculates acceleration due to gravity \(g = 9.64\text{ m s}^{-2}\).
[1 mark] Calculates percentage uncertainties in length (0.25%) and period (0.55%).
[1 mark] Combines uncertainties: \(\% \Delta g = 0.25\% + 2 \times 0.55\% = 1.35\%\) (or \(1.4\%\)).
[0.5 mark] Correct absolute uncertainty in \(g\) shown as \(\pm 0.13\text{ m s}^{-2}\) (or \(\pm 0.1\text{ m s}^{-2}\)).

(a) For small angle oscillations of a simple pendulum, the displacement along the arc is \(x = L \theta\). The maximum linear displacement is the amplitude \(x_0 = L \theta_{\text{max}}\).
For simple harmonic motion, the maximum velocity is \(v_{\text{max}} = \omega x_0\).
Since \(\omega = \sqrt{\frac{g}{L}}\), we substitute to get:
\(v_{\text{max}} = \sqrt{\frac{g}{L}} \times (L \theta_{\text{max}}) = \theta_{\text{max}} \sqrt{g L}\).

(b) The maximum acceleration of a body in simple harmonic motion is given by:
\(a_{\text{max}} = \omega^2 x_0\).
Substituting \(\omega^2 = \frac{g}{L}\) and \(x_0 = L \theta_{\text{max}}\):
\(a_{\text{max}} = \frac{g}{L} (L \theta_{\text{max}}) = g \theta_{\text{max}}\).
Substituting the values:
\(a_{\text{max}} = 9.81 \times 0.120 = 1.1772\text{ m s}^{-2}\).
Rounding to three significant figures gives \(1.18\text{ m s}^{-2}\).

[1 mark] Identifies amplitude \(x_0 = L \theta_{\text{max}}\).
[1 mark] States or uses \(\omega = \sqrt{\frac{g}{L}}\) and \(v_{\text{max}} = \omega x_0\).
[1 mark] Shows complete algebraic steps to arrive at \(v_{\text{max}} = \theta_{\text{max}} \sqrt{g L}\).
[1 mark] Correctly deduces that \(a_{\text{max}} = g \theta_{\text{max}}\).
[1 mark] Substitutes values: \(9.81 \times 0.120 = 1.18\text{ m s}^{-2}\).
[0.5 mark] Correct unit of acceleration (\(\text{m s}^{-2}\)) and appropriate significant figures (3 s.f.).

(a) At equilibrium, tension equals the weight of the mass:
\(F = mg = k e\).
The extension \(e = 0.335 - 0.300 = 0.035\text{ m}\).
Rearranging for \(k\):
\(k = \frac{0.450 \times 9.81}{0.035} = 126.1\text{ N m}^{-1} \approx 126\text{ N m}^{-1}\).

(b) The amplitude of the simple harmonic motion is \(A = 0.040\text{ m}\).
The mass is released from the lowest position. It travels a total distance of \(0.120\text{ m}\).
One-half of a complete oscillation (from bottom to top) covers a distance of \(2A = 2 \times 0.040 = 0.080\text{ m}\) and takes time \(\frac{T}{2}\).
To cover the remaining \(0.120 - 0.080 = 0.040\text{ m}\) (which is exactly \(1A\)), it must travel from the top back to the equilibrium position, taking an additional time of \(\frac{T}{4}\).
Therefore, total time \(t = \frac{3}{4}T\).

Calculate period \(T\):
\(T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.450}{126.1}} \approx 0.3753\text{ s}\).

Calculate total time \(t\):
\(t = 0.75 \times 0.3753 \approx 0.2815\text{ s}\) (or \(0.282\text{ s}\) if using \(k = 126\text{ N m}^{-1}\)).

[1 mark] Identifies the extension as \(0.035\text{ m}\).
[1 mark] Sets up \(mg = k e\) and calculates \(k \approx 126\text{ N m}^{-1}\) clearly showing substitution.
[1 mark] Uses \(T = 2\pi \sqrt{\frac{m}{k}}\) to calculate \(T \approx 0.375\text{ s}\).
[1 mark] Analyzes distance traveled to show that a total distance of \(0.120\text{ m}\) corresponds to \(3A\) (or \(\frac{3}{4}\) of a cycle).
[1 mark] Calculates the time taken as \(\frac{3}{4}T = 0.28\text{ s}\) (or \(0.282\text{ s}\)).
[0.5 mark] Correct unit of time (s) and sensible precision (2 or 3 s.f.).

(a) The continuous-flow equation is:
\(P = \frac{\Delta m}{\Delta t} c \Delta \theta + H\)
where:
- \(P\) is the electrical power input to the heater (W)
- \(\frac{\Delta m}{\Delta t}\) is the mass of water flowing per second (kg s\(^{-1}\))
- \(c\) is the specific heat capacity of the fluid (J kg\(^{-1}\) K\(^{-1}\))
- \(\Delta \theta\) is the temperature rise of the water (K or \(^{\circ}\)C)
- \(H\) is the rate of heat loss to the surroundings (W).

(b) We are given:
\(P = 2400\text{ W}\)
\(\frac{\Delta m}{\Delta t} = 0.015\text{ kg s}^{-1}\)
\(\Delta \theta = 54.0 - 18.5 = 35.5\text{ }^{\circ}\text{C}\)
\(H = 120\text{ W}\)

Rearranging for \(c\):
\(c = \frac{P - H}{\frac{\Delta m}{\Delta t} \Delta \theta}\)
\(c = \frac{2400 - 120}{0.015 \times 35.5} = \frac{2280}{0.5325} \approx 4281.7\text{ J kg}^{-1}\text{ K}^{-1}\).
Rounding to 3 significant figures, we get \(c = 4280\text{ J kg}^{-1}\text{ K}^{-1}\).

[1 mark] Correct continuous-flow formula: \(P = \frac{\Delta m}{\Delta t} c \Delta \theta + H\).
[1 mark] Correctly defines all five terms in the equation.
[1 mark] Calculates temperature change \(\Delta \theta = 35.5\text{ }^{\circ}\text{C}\).
[1 mark] Rearranges equation to make \(c\) the subject: \(c = \frac{P - H}{\frac{\Delta m}{\Delta t} \Delta \theta}\).
[1 mark] Substitutes values and solves: \(c = 4280\text{ J kg}^{-1}\text{ K}^{-1}\).
[0.5 mark] Correct units (\(\text{J kg}^{-1}\text{ K}^{-1}\) or \(\text{kJ kg}^{-1}\text{ K}^{-1}\)) and correct number of significant figures (3 s.f.).

(a) Using the rotational kinematic equation:
\(\omega = \omega_0 + \alpha t\)
\(0 = 45 + \alpha (15) \implies \alpha = -3.0\text{ rad s}^{-2}\).
The magnitude of the angular deceleration is \(3.0\text{ rad s}^{-2}\).

The frictional torque is:
\(T = I \alpha\)
\(T = 0.64 \times 3.0 = 1.92\text{ N m}\).

(b) The angular displacement \(\theta\) is given by:
\(\theta = \frac{1}{2}(\omega_0 + \omega)t = \frac{1}{2}(45 + 0) \times 15 = 337.5\text{ rad}\).

The number of revolutions \(N\) is:
\(N = \frac{\theta}{2\pi} = \frac{337.5}{2\pi} \approx 53.7\text{ revolutions}\).

Since the question asks for complete revolutions, we round down to the nearest integer: 53 complete revolutions.

[1 mark] Recalls and uses \(\omega = \omega_0 + \alpha t\) to find \(\alpha = 3.0\text{ rad s}^{-2}\).
[1 mark] Recalls and uses \(T = I \alpha\) to calculate torque.
[1 mark] Obtains torque \(T = 1.92\text{ N m}\).
[1 mark] Calculates angular displacement \(\theta = 337.5\text{ rad}\) using appropriate kinematic equation.
[1 mark] Converts radians to revolutions: \(53.7\text{ revolutions}\).
[0.5 mark] Correctly states 53 complete revolutions.

(a) Systematic errors cause measurements to deviate from the true value by a consistent amount or in a consistent direction each time, whereas random errors cause measurements to vary unpredictably about the true value.
- Example of a systematic error: resistance in the connecting wires or an uncalibrated meter (like the zero error mentioned in part b).
- Example of a random error: temperature fluctuations in the wire (which vary the resistance during the experiment) or fluctuations in the power supply voltage.

(b) Zero error is a type of systematic error.
To correct the data, the students should subtract \(0.15\text{ V}\) from every potential difference reading recorded before analysis.

[1.5 marks] Differentiates systematic error (consistent deviation) and random error (unpredictable variation about the mean).
[1 mark] Provides a valid systematic error example relevant to the resistance experiment.
[1 mark] Provides a valid random error example relevant to the resistance experiment.
[1 mark] Correctly identifies zero error as a systematic error.
[1 mark] Explicitly describes how to correct the data (subtracting \(0.15\text{ V}\) from each reading).

(a) The work done by a gas expanding at constant pressure is:
\(W = P \Delta V\)
\(\Delta V = V_{\text{final}} - V_{\text{initial}} = 4.10 \times 10^{-3} - 2.40 \times 10^{-3} = 1.70 \times 10^{-3}\text{ m}^3\).

Substituting values:
\(W = (1.50 \times 10^5\text{ Pa}) \times (1.70 \times 10^{-3}\text{ m}^3) = 255\text{ J}\).

(b) The first law of thermodynamics is written as:
\(\Delta U = Q - W\)
where:
- \(\Delta U\) is the change in internal energy
- \(Q\) is heat added to the gas (\(+420\text{ J}\))
- \(W\) is work done by the gas (\(+255\text{ J}\))

Substituting the values:
\(\Delta U = 420 - 255 = +165\text{ J}\).

Since the sign of \(\Delta U\) is positive, the internal energy of the gas increases.

[1 mark] Recalls and uses \(W = P \Delta V\).
[1 mark] Correctly calculates \(\Delta V = 1.70 \times 10^{-3}\text{ m}^3\).
[1 mark] Obtains work done \(W = 255\text{ J}\).
[1 mark] Recalls the first law of thermodynamics with correct signs: \(\Delta U = Q - W\).
[1 mark] Calculates \(\Delta U = +165\text{ J}\).
[0.5 mark] Correctly states that the internal energy increases.

The density of the wire is given by the formula \(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 L}\). The percentage uncertainty in density is calculated by adding the percentage uncertainties of each quantity: \(\frac{\Delta \rho}{\rho} \times 100\% = \left( \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L} \right) \times 100\%\). Calculating the individual percentage uncertainties: for mass, \(\frac{0.01}{0.88} \times 100\% \approx 1.14\%\); for diameter, \(\frac{0.02}{0.40} \times 100\% = 5.0\%\); for length, \(\frac{0.4}{80.0} \times 100\% = 0.5\%\). Thus, total percentage uncertainty = \(1.14\% + 2(5.0\%) + 0.5\% = 11.64\%\), which rounds to 11.6%.

1 mark: Correct calculation of the total percentage uncertainty showing the factor of 2 for diameter.

The elastic strain energy stored in a wire under tension is \(E = \frac{1}{2} F \Delta L\). Since the tension \(F\) is the same for both wires, the ratio of the energy stored is equal to the ratio of their extensions: \(\frac{E_X}{E_Y} = \frac{\Delta L_X}{\Delta L_Y}\). The extension is given by \(\Delta L = \frac{F L}{A E_y} = \frac{4 F L}{\pi d^2 E_y}\), where \(E_y\) is the Young modulus and \(d\) is the diameter. Therefore, \(\Delta L \propto \frac{L}{d^2 E_y}\). Comparing wire X to wire Y: \(\frac{\Delta L_X}{\Delta L_Y} = \frac{L_X}{L_Y} \times \left(\frac{d_Y}{d_X}\right)^2 \times \frac{E_{y,Y}}{E_{y,X}}\). Given that \(L_X = 2 L_Y\), \(d_X = 0.5 d_Y\), and \(E_{y,X} = 2 E_{y,Y}\), we substitute these values: \(\frac{E_X}{E_Y} = 2 \times (2)^2 \times \frac{1}{2} = 4\).

1 mark: Correct deduction of the ratio of elastic strain energy as 4.

The mass of a cylinder of density \(D\) is \(m = A L D\). For the second resistor, \(m_2 = 2m_1\) and \(L_2 = 0.5L_1\). Therefore, \(A_2 L_2 D = 2 A_1 L_1 D\), which simplifies to \(A_2 (0.5 L_1) = 2 A_1 L_1\), hence \(A_2 = 4 A_1\). The resistance of the second resistor is given by \(R_2 = \rho \frac{L_2}{A_2} = \rho \frac{0.5 L_1}{4 A_1} = \frac{1}{8} \left(\rho \frac{L_1}{A_1}\right) = \frac{R}{8}\).

1 mark: Correct determination of the cross-sectional area ratio and the resulting resistance ratio as 1/8.

According to Einstein's photoelectric equation, the maximum kinetic energy is \(E_{k,\text{max}} = hf - \Phi\). Since \(E_{k,\text{max}} = eV_s\), we can write \(eV_s = hf - \Phi\), or \(V_s = \left(\frac{h}{e}\right) f - \frac{\Phi}{e}\). Comparing this to the straight-line equation \(y = mx + c\), the gradient is \(m = \frac{h}{e}\). The x-intercept is the value of \(f\) when \(V_s = 0\), which corresponds to the threshold frequency \(f_0 = \frac{\Phi}{h}\).

1 mark: Correct identification of the gradient as h/e and the x-intercept as the threshold frequency.

Fringe spacing is given by \(w = \frac{\lambda D}{d}\). Since the screen distance \(D\) is constant, \(w \propto \frac{\lambda}{d}\). The ratio of the new fringe width to the original is \(\frac{w_2}{w_1} = \frac{\lambda_2}{\lambda_1} \times \frac{d_1}{d_2} = 1.2 \times \frac{1}{0.8} = 1.5\). Therefore, \(w_2 = 1.50 w_1\).

1 mark: Correct substitution into the fringe spacing ratio to obtain 1.50.

Initially, the potential difference across the fixed resistor is \(V_R = 8.0\text{ V}\). Therefore, the potential difference across the thermistor is \(V_{\text{th}} = 12.0 - 8.0 = 4.0\text{ V}\). Since \(V_{\text{th}}/V_R = R_{\text{th}}/R\), the initial resistance of the thermistor is \(R_{\text{th}} = R \times \frac{4.0}{8.0} = 2.0\text{ k}\Omega\). When the temperature increases, its resistance decreases to half, so the new resistance of the thermistor is \(R_{\text{th}}' = 1.0\text{ k}\Omega\). The new total resistance is \(4.0\text{ k}\Omega + 1.0\text{ k}\Omega = 5.0\text{ k}\Omega\). The new potential difference across the thermistor is \(V_{\text{th}}' = 12.0\text{ V} \times \frac{1.0\text{ k}\Omega}{5.0\text{ k}\Omega} = 2.4\text{ V}\).

1 mark: Correct calculation of the final potential difference across the thermistor.

Polarization can only be demonstrated by transverse waves, because it requires the wave oscillations to be restricted to a single plane perpendicular to the direction of energy transfer. Longitudinal waves cannot be polarized. However, both longitudinal and transverse waves can undergo diffraction and interference, which are general properties of all wave types.

1 mark: Correctly identifies that only diffraction and interference are common to both types of waves.

The pulling force \(F\) must balance both the component of gravity down the slope and the frictional force: \(F = mg\sin(30^\circ) + F_{\text{friction}} = 50 \times 9.81 \times 0.5 + 120 = 245.25 + 120 = 365.25\text{ N}\). The useful work output per second (power output) is \(P_{\text{out}} = F v = 365.25 \times 1.5 = 547.875\text{ W}\). Since the motor is only \(60\%\) efficient, the electrical power input is \(P_{\text{in}} = \frac{P_{\text{out}}}{0.60} = \frac{547.875}{0.60} \approx 913\text{ W}\), which is approximately \(910\text{ W}\) to two significant figures.

1 mark: Correct calculation of pulling force, power output, and the electrical power input.

The resistivity \(\rho\) is given by:
\$$\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\$$

Taking the fractional uncertainties:
\$$\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\$$

Calculate each percentage uncertainty:
\$$\frac{\Delta R}{R} \times 100\% = \frac{0.1}{4.8} \times 100\% \approx 2.08\%\$$
\$$\frac{\Delta d}{d} \times 100\% = \frac{0.01}{0.38} \times 100\% \approx 2.63\%\$$
\$$\frac{\Delta L}{L} \times 100\% = \frac{0.01}{1.25} \times 100\% = 0.80\%\$$

Total percentage uncertainty:
\$$\text{Percentage uncertainty} = 2.08\% + 2 \times 2.63\% + 0.80\% = 8.14\% \approx 8.1\%\$$

Award 1 mark for the correct option D.

- Method mark: Correct identification of the uncertainty equation: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\).
- Accuracy mark: Correct calculation of \(8.1\%\).

The negative kaon \(K^-\)) is a strange meson with strangeness \(S = -1\). Its quark composition is \(s\bar{u}\).

The final state consists of a muon and a muon antineutrino, both of which are leptons and have zero strangeness.

The change in strangeness is:
\$$\Delta S = S_{\text{final}} - S_{\text{initial}} = 0 - (-1) = +1\$$

Therefore, the change in strangeness is \(+1\) and the quark composition of \(K^-\)) is \(s\bar{u}\).

Award 1 mark for the correct option A.

- Reject options where strangeness is conserved (as this is a weak decay changing flavour).
- Reject options with incorrect quark structures of \(K^-\) (such as \(u\bar{s}\) which is \(K^+\)).

The strain energy stored in an elastically stretched wire is given by:
\$$E = \frac{1}{2} F \Delta L\$$

Since both wires support identical loads, the force \(F\) is the same for both. The extension \(\Delta L\) is:
\$$\Delta L = \frac{F L}{A E_Y}\$$
where \(E_Y\) is the Young modulus of the material and \(A\) is the cross-sectional area of the wire.

Since \(A \propto d^2\), the extension is proportional to \(\frac{L}{d^2}\):
\$$\Delta L \propto \frac{L}{d^2}\$$

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} = \frac{1}{2} \frac{L}{d^2}\$$

Thus, \(\Delta L_Q = \frac{1}{2} \Delta L_P\).

Comparing the strain energies stored:
\$$\frac{E_P}{E_Q} = \frac{\frac{1}{2} F \Delta L_P}{\frac{1}{2} F \Delta L_Q} = \frac{\Delta L_P}{\frac{1}{2} \Delta L_P} = 2\$$

Award 1 mark for the correct option C.

- Method mark: Deducing that \(\Delta L \propto \frac{L}{d^2}\) and therefore \(\Delta L_Q = 0.5 \Delta L_P\).
- Accuracy mark: Correctly calculating the ratio as \(2\).

The terminal potential difference \(V\) is given by:
\$$V = \varepsilon \left( \frac{R}{R + r} \right)\$$
As \(R\) increases, \(\frac{R}{R + r}\) increases continuously towards \(1\). Therefore, \(V\) increases continuously.

The power \(P\) dissipated in the external resistor is:
\$$P = I^2 R = \frac{\varepsilon^2 R}{(R + r)^2}\$$
According to the maximum power transfer theorem, the power delivered to the load is maximized when the load resistance equals the internal resistance of the source, i.e., when \(R = r\). For \(R < r\), \(P\) increases as \(R\) increases, and for \(R > r\), \(P\) decreases as \(R\) increases.

Award 1 mark for the correct option B.

- Identify that terminal potential difference \(V\) increases continuously with \(R\).
- Identify that power delivered to the external resistor peaks when \(R=r\).

First, calculate the critical angle \(\theta_c\) at the glass-air interface:
\$$\sin \theta_c = \frac{n_{\text{air}}}{n_{\text{glass}}} = \frac{1}{1.52} \approx 0.6579\$$
\$$\theta_c = \arcsin(0.6579) \approx 41.1^\circ\$$

For an angle of incidence \(\theta_i = 30^\circ\):
Since \(\theta_i < \theta_c\) (\(30^\circ < 41.1^\circ\)), the light ray undergoes refraction and exits into the air.

For an angle of incidence \(\theta_i = 50^\circ\):
Since \(\theta_i > \theta_c\) (\(50^\circ > 41.1^\circ\)), total internal reflection (TIR) occurs, and no light is refracted into the air.

Award 1 mark for the correct option C.

- Method mark: Calculate the critical angle as \(41.1^\circ\).
- Accuracy mark: Correctly compare \(30^\circ\) and \(50^\circ\) with \(41.1^\circ\) to determine refraction versus total internal reflection.

Taking the initial direction of motion as positive:

Initial velocity, \(u = +6.0\text{ m s}^{-1}\)

Final velocity, \(v = -4.0\text{ m s}^{-1}\)

Change in momentum, \(\Delta p = m(v - u)\):
\$$\Delta p = 0.15 \times (-4.0 - 6.0) = -1.5\text{ kg m s}^{-1}\$$

According to Newton's second law, the average force \(F\) is given by:
\$$F = \frac{\Delta p}{\Delta t} = \frac{-1.5\text{ kg m s}^{-1}}{0.080\text{ s}} = -18.75\text{ N}\$$

The magnitude of the average force is \(18.75\text{ N} \approx 19\text{ N}\).

Award 1 mark for the correct option C.

- Method mark: Calculate change in momentum by adding magnitudes due to reversal of direction (\(\Delta p = 1.5\text{ N s}\)).
- Accuracy mark: Correct division of momentum by contact time to obtain \(19\text{ N}\).

Einstein's photoelectric equation states:
\$$h f = \Phi + e V_s \implies e V_s = h f -
\Phi\$$

When the incident frequency is doubled to \(2f\), the new stopping potential \(V'_s\) satisfies:
\$$e V'_s = h(2f) - \Phi = 2(h f) - \Phi\$$

Substitute \(h f = e V_s + \Phi\) into this equation:
\$$e V'_s = 2(e V_s + \Phi) - \Phi = 2 e V_s + 2 \Phi - \Phi = 2 e V_s + \Phi\$$

Dividing by \(e\):
\$$V'_s = 2 V_s + \frac{\Phi}{e}\$$

Award 1 mark for the correct option B.

- Method mark: Set up the photoelectric equations for \(f\) and \(2f\).
- Accuracy mark: Correctly substitute and solve for \(V'_s\).

The resistance \(R\) of a wire of length \(L\), cross-sectional area \(A\), and resistivity \(\rho\) is:
\$$R = \rho \frac{L}{A}\$$

Since the volume \(V = A L\) remains constant, the area \(A\) can be written as \(A = \frac{V}{L}\). Substituting this into the resistance formula gives:
\$$R = \rho \frac{L^2}{V}\$$

Since \(\rho\) and \(V\) are constant, the resistance is directly proportional to the square of the length:
\$$R \propto L^2\$$

When the length increases by \(10\%\), the new length is \(L' = 1.10 L\).

Therefore, the new resistance \(R'\) is:
\$$R' \propto (1.10 L)^2 = 1.21 L^2 \implies R' = 1.21 R\$$

Award 1 mark for the correct option B.

- Method mark: Deducing that \(R \propto L^2\) because volume is constant.
- Accuracy mark: Correctly squaring \(1.10\) to get \(1.21\).

The formula for the Young modulus is:

\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L} = \frac{4 F L}{\pi d^2 \Delta L}\)

To find the percentage uncertainty in \(E\), we sum the percentage uncertainties of each independent variable, multiplying by their powers where applicable:

\(\%\Delta E = \%\Delta F + \%\Delta L + 2(\%\Delta d) + \%\Delta \Delta L\)

\(\%\Delta E = 2.0\% + 1.0\% + 2(1.5\%) + 3.0\% = 2.0\% + 1.0\% + 3.0\% + 3.0\% = 9.0\%\)

[1 mark] C - Correct identification and combination of percentage uncertainties using the formula for Young modulus.

The formula for the density \(\rho\) of a cylinder of mass \(m\), diameter \(d\), and height \(h\) is:

\(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 h}\)

The percentage uncertainty in density is given by:

\(\%\Delta \rho = \%\Delta m + 2(\%\Delta d) + \%\Delta h\)

First, calculate the individual percentage uncertainties:
- \(\%\Delta m = \frac{0.1}{24.0} \times 100\% \approx 0.417\%\)
- \(\%\Delta d = \frac{0.2}{12.0} \times 100\% \approx 1.667\%\)
- \(\%\Delta h = \frac{0.5}{30.0} \times 100\% \approx 1.667\%\)

Now combine them:

\(\%\Delta \rho = 0.417\% + 2(1.667\%) + 1.667\% = 0.417\% + 3.333\% + 1.667\% = 5.417\% \approx 5.4\%\)

[1 mark] C - Calculate individual percentage uncertainties correctly and combine them according to the density equation.

The time period of a simple pendulum is given by:

\(T = 2\pi \sqrt{\frac{L}{g}}\)

For the new pendulum on the other planet, the new period \(T'\) is:

\(T' = 2\pi \sqrt{\frac{2L}{\frac{1}{4}g}} = 2\pi \sqrt{\frac{8L}{g}} = \sqrt{8} \times 2\pi \sqrt{\frac{L}{g}} = 2\sqrt{2} T\)

[1 mark] C - Correct application of the pendulum period formula and simplifying the ratio.

The total energy of the simple harmonic oscillator is given by:

\(E = \frac{1}{2} k A^2\)

The elastic potential energy \(E_p\) at displacement \(x = \frac{2}{3}A\) is:

\(E_p = \frac{1}{2} k x^2 = \frac{1}{2} k \left(\frac{2}{3}A\right)^2 = \frac{4}{9} \left(\frac{1}{2} k A^2\right) = \frac{4}{9} E\)

The kinetic energy \(E_k\) is the difference between the total energy and the potential energy:

\(E_k = E - E_p = E - \frac{4}{9} E = \frac{5}{9} E\)

Therefore, the ratio is \(\frac{E_k}{E} = \frac{5}{9}\).

[1 mark] C - Correctly related potential energy and total energy with displacement amplitude, and subtracted to find kinetic energy ratio.

The total thermal energy supplied by the heater is:

\(Q = P \times t = 50\text{ W} \times (4.0 \times 60\text{ s}) = 12,000\text{ J}\)

Since this energy is entirely used to change the state of the material without changing its temperature:

\(Q = m L_f\)

\(L_f = \frac{Q}{m} = \frac{12,000\text{ J}}{0.20\text{ kg}} = 60,000\text{ J kg}^{-1} = 6.0 \times 10^4\text{ J kg}^{-1}\)

[1 mark] C - Calculate thermal energy supplied in Joules and use the latent heat formula to find Specific Latent Heat.

Assuming no heat loss to the surroundings:

Heat lost by X = Heat gained by Y

\(m_X c_X (T_X - T_f) = m_Y c_Y (T_f - T_Y)\)

Substitute the given values:

\(0.10 \times 3000 \times (80 - T_f) = 0.20 \times 4500 \times (T_f - 20)\)

\(300(80 - T_f) = 900(T_f - 20)\)

Divide both sides by 300:

\(80 - T_f = 3(T_f - 20)\)

\(80 - T_f = 3T_f - 60\)

\(4T_f = 140 \implies T_f = 35^\circ\text{C}\)

[1 mark] A - Set up conservation of energy for thermal mixing and solve for the final temperature.

First, calculate the angular deceleration \(\alpha\) of the flywheel using Newton's second law for rotation:

\(\tau = I \alpha \implies \alpha = \frac{\tau}{I} = \frac{1.5}{0.50} = 3.0\text{ rad s}^{-2}\)

Using the rotational equation of motion \(\omega_f^2 = \omega_i^2 - 2\alpha\theta\) where \(\omega_f = 0\):

\(0 = 12^2 - 2(3.0)\theta\)

\(6.0\theta = 144 \implies \theta = 24\text{ rad}\)

[1 mark] C - Determined angular acceleration from torque and moment of inertia, then applied rotational equations of motion.

Using the conservation of energy, the loss in gravitational potential energy as the rod falls to the vertical position equals the gain in rotational kinetic energy.

The center of mass of the uniform rod is at its midpoint, \(\frac{L}{2}\) from the pivot. When the rod becomes vertical, the center of mass descends by \(\frac{L}{2}\).

\(\Delta E_p = Mg\frac{L}{2}\)

The gain in rotational kinetic energy is:

\(E_k = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{3}ML^2\right) \omega^2 = \frac{1}{6} ML^2 \omega^2\)

Equating loss in GPE to gain in rotational KE:

\(Mg\frac{L}{2} = \frac{1}{6} ML^2 \omega^2\)

Simplify the equation:

\(\frac{g}{2} = \frac{1}{6} L \omega^2 \implies \omega^2 = \frac{3g}{L} \implies \omega = \sqrt{\frac{3g}{L}}\)

[1 mark] C - Applied conservation of energy using the displacement of the center of mass of the rod, equated to rotational kinetic energy, and solved for angular velocity.

The percentage uncertainty in each measurement is calculated as follows: \(\text{Percentage uncertainty in } R = \frac{0.05}{4.50} \times 100\% \approx 1.11\%\), \(\text{Percentage uncertainty in } L = \frac{0.002}{1.200} \times 100\% \approx 0.17\%\), \(\text{Percentage uncertainty in } d = \frac{0.01}{0.38} \times 100\% \approx 2.63\%\). Since the formula for resistivity is \(\rho = \frac{\pi R d^2}{4 L}\), the percentage uncertainty in \(\rho\) is: \(\frac{\Delta \rho}{\rho} \times 100\% = \frac{\Delta R}{R} \times 100\% + 2 \left(\frac{\Delta d}{d} \times 100\%\right) + \frac{\Delta L}{L} \times 100\%\). Substituting the values gives: \(1.11\% + 2(2.63\%) + 0.17\% = 6.54\%\), which rounds to \(6.5\%\).

1 mark for the correct option C. Method: Award marks for calculating individual percentage uncertainties: R (1.11%), L (0.17%), and d (2.63%). Award marks for doubling the diameter uncertainty and adding all terms. Accuracy: Correctly computes 6.5%.

The extension of a wire is given by \(\Delta L = \frac{FL}{AE}\). The strain energy stored is \(W = \frac{1}{2} F \Delta L = \frac{F^2 L}{2AE}\). For the second wire, the length is \(2L\) and the area is \(0.5A\). Therefore, the new strain energy is \(W_2 = \frac{F^2 (2L)}{2(0.5A)E} = 4 \left(\frac{F^2 L}{2AE}\right) = 4W\).

1 mark for the correct option D. Method: Recalls the formula for strain energy in terms of force, length, area, and Young modulus. Substitute new values to show a fourfold increase. Accuracy: Correctly identifies 4W.

From Einstein's photoelectric equation, \(e V_s = \frac{hc}{\lambda} - \Phi\), which gives \(\frac{hc}{\lambda} = e V_s + \Phi\). For the second wavelength \(0.5\lambda\), the new equation is: \(e V_s' = \frac{hc}{0.5\lambda} - \Phi = 2\left(\frac{hc}{\lambda}\right) - \Phi\). Substituting \(\frac{hc}{\lambda}\): \(e V_s' = 2(e V_s + \Phi) - \Phi = 2e V_s + \Phi\). Dividing by \(e\) gives \(V_s' = 2 V_s + \frac{\Phi}{e}\).

1 mark for the correct option B. Method: Writes the photoelectric equation for both cases. Performs algebraic substitution to express V_s' in terms of V_s and work function. Accuracy: Finds the correct relation.

The resistance of a wire is \(R = \rho \frac{L}{A}\). Since the volume \(V = A L\) remains constant, the new area is \(A' = \frac{V}{1.10 L} = \frac{A}{1.10}\) when the length becomes \(L' = 1.10 L\). Therefore, the new resistance is \(R' = \rho \frac{1.10 L}{A / 1.10} = 1.10^2 \left(\rho \frac{L}{A}\right) = 1.21 R\).

1 mark for the correct option C. Method: Relates area change to length change due to constant volume. Expresses new resistance in terms of R. Accuracy: Finds the correct value 1.21 R.

1. The formula for the time period of a simple pendulum is:
\(T = 2\pi\sqrt{\frac{L}{g}}\)
Rearranging this for \(g\) gives:
\(g = \frac{4\pi^2 L}{T^2}\)

2. The percentage uncertainty in \(L\) is:
\(\% \Delta L = \frac{0.002}{0.800} \times 100\% = 0.25\%\)

3. The time period \(T\) is given by \(T = \frac{t}{20}\). The percentage uncertainty in the time period is equal to the percentage uncertainty in the total time \(t\):
\(\% \Delta T = \frac{0.2}{36.0} \times 100\% \approx 0.556\%\)

4. The percentage uncertainty in \(g\) is given by combining the uncertainties:
\(\% \Delta g = \% \Delta L + 2(\% \Delta T)\)
\(\% \Delta g = 0.25\% + 2(0.556\%) = 0.25\% + 1.112\% = 1.362\%\)

Rounding to 2 significant figures gives \(1.4\%\).

• [1 mark] Correct calculation of percentage uncertainty in \(L\) (\(0.25\%\)).
• [1.5 marks] Correct calculation of percentage uncertainty in \(T\) (1 mark for recognizing that uncertainty in \(T\) matches uncertainty in \(t\), and 0.5 marks for finding \(0.56\%\)).
• [1.5 marks] Stating or showing that \(\% \Delta g = \% \Delta L + 2(\% \Delta T)\) (1 mark for the factor of 2, 0.5 marks for the sum).
• [1.5 marks] Correct final calculation of the percentage uncertainty to 2 or 3 significant figures (\(1.4\%\) or \(1.36\%\)).

1. Express resistivity in terms of measured quantities:
\(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\)

2. Determine individual percentage uncertainties:
- Resistance: \(\% \Delta R = \frac{0.2}{12.4} \times 100\% \approx 1.613\%\)
- Length: \(\% \Delta L = \frac{0.01}{1.25} \times 100\% = 0.800\%\)
- Diameter: \(\% \Delta d = \frac{0.02}{0.38} \times 100\% \approx 5.263\%\)

3. Combine uncertainties:
\(\% \Delta \rho = \% \Delta R + 2(\% \Delta d) + \% \Delta L\)
\(\% \Delta \rho = 1.613\% + 2(5.263\%) + 0.800\%\)
\(\% \Delta \rho = 1.613\% + 10.526\% + 0.800\% = 12.939\%\)

Rounding to 2 significant figures gives \(13\%\).

• [1 mark] Expressing resistivity formula to show the dependence on the squared diameter: \(\rho \propto R d^2 / L\).
• [1.5 marks] Correct calculations of individual percentage uncertainties (0.5 marks each for \(\% \Delta R\), \(\% \Delta L\), and \(\% \Delta d\)).
• [1.5 marks] Showing correct combination formula, including the doubling of diameter's percentage uncertainty (1 mark for the factor of 2, 0.5 marks for summing).
• [1.5 marks] Correct final calculation of percentage uncertainty (\(13\%\) or \(12.9\%\)).

1. Maximum Kinetic Energy:
The maximum kinetic energy is equal to the total elastic potential energy at maximum displacement:
\(E_{\text{max}} = \frac{1}{2} k A^2\)
\(E_{\text{max}} = \frac{1}{2} (45.0\text{ N m}^{-1}) (0.080\text{ m})^2 = 0.144\text{ J}\)

2. Time Period of Oscillation:
\(T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.350}{45.0}} \approx 0.5543\text{ s}\)

3. Time to reach maximum kinetic energy:
The block is released from rest at maximum displacement (where KE is 0). It first reaches maximum kinetic energy at the equilibrium position. This corresponds to one-quarter of a full cycle:
\(t = \frac{T}{4} = \frac{0.5543}{4} \approx 0.1386\text{ s}\)

Rounding to 3 significant figures: \(t = 0.139\text{ s}\).

• [1.5 marks] Correct calculation of maximum kinetic energy (1 mark for formula/working, 0.5 mark for \(0.144\text{ J}\)).
• [1.5 marks] Correct calculation of time period \(T\) (1 mark for formula/substitution, 0.5 mark for \(0.55\text{ s}\)).
• [1 mark] Identifying that the block first reaches maximum kinetic energy at \(t = T/4\).
• [1.5 marks] Correct calculation of final time \(t = 0.139\text{ s}\) (or \(0.14\text{ s}\)) with appropriate unit.

1. Calculate angular frequency \(\omega\):
\(\omega = \sqrt{\frac{g}{L}} = \sqrt{\frac{9.81}{1.60}} \approx 2.476\text{ rad s}^{-1}\)

2. Calculate maximum velocity:
\(v_{\text{max}} = \omega A = 2.476 \times 0.12 = 0.297\text{ m s}^{-1} \approx 0.30\text{ m s}^{-1}\)

3. Calculate magnitude of acceleration at displacement \(x = 0.060\text{ m}\):
\(a = \omega^2 x\)
\(a = (2.476)^2 \times 0.060 \approx 6.13 \times 0.060 = 0.368\text{ m s}^{-2} \approx 0.37\text{ m s}^{-2}\)

• [1.5 marks] Correct calculation of \(\omega\) (1 mark for formula, 0.5 mark for value \(2.48\text{ rad s}^{-1}\)).
• [1.5 marks] Correct calculation of maximum velocity \(v_{\text{max}}\) (1 mark for formula, 0.5 mark for \(0.30\text{ m s}^{-1}\)).
• [1.5 marks] Correct calculation of acceleration magnitude (1 mark for formula/substitution, 0.5 mark for \(0.37\text{ m s}^{-2}\)).
• [1 mark] Correct SI units provided for both answers (\(\text{m s}^{-1}\) and \(\text{m s}^{-2}\)) (0.5 marks each).

1. Calculate energy required to heat the ice from \(-10^\circ\text{C}\) to \(0^\circ\text{C}\):
\(Q_1 = m c_{\text{ice}} \Delta T\)
\(Q_1 = 0.45\text{ kg} \times 2100\text{ J kg}^{-1}\text{ K}^{-1} \times 10\text{ K} = 9450\text{ J}\)

2. Calculate energy required to melt the ice at \(0^\circ\text{C}\):
\(Q_2 = m L_f\)
\(Q_2 = 0.45\text{ kg} \times 3.3 \times 10^5\text{ J kg}^{-1} = 148500\text{ J}\)

3. Calculate total energy required:
\(Q_{\text{total}} = Q_1 + Q_2 = 9450\text{ J} + 148500\text{ J} = 157950\text{ J}\)

4. Calculate time taken using the heater power:
\(t = \frac{Q_{\text{total}}}{P} = \frac{157950\text{ J}}{150\text{ W}} = 1053\text{ s}\)

Rounding to 3 significant figures gives \(1050\text{ s}\) (or \(1.1 \times 10^3\text{ s}\) to 2 significant figures).

• [1.5 marks] Correct calculation of energy to raise ice temperature, \(Q_1\) (1 mark for formula/working, 0.5 mark for \(9450\text{ J}\)).
• [1.5 marks] Correct calculation of energy to melt ice, \(Q_2\) (1 mark for formula/working, 0.5 mark for \(1.49 \times 10^5\text{ J}\)).
• [1 mark] Summing the two energy values correctly to obtain \(157950\text{ J}\).
• [1.5 marks] Using \(t = Q/P\) to calculate the final time (1 mark for method, 0.5 mark for correct answer of \(1053\text{ s}\) or \(1050\text{ s}\) with unit).

1. First Law of Thermodynamics statement:
\(Q = \Delta U + W\)
where \(Q\) is the heat added to the system, \(\Delta U\) is the change in internal energy, and \(W\) is the work done by the system.

2. Calculation:
Rearranging for \(\Delta U\):
\(\Delta U = Q - W = 850\text{ J} - 320\text{ J} = +530\text{ J}\)

3. Temperature explanation:
For an ideal gas, the internal energy is purely kinetic energy and is directly proportional to the absolute temperature (\(U \propto T\)). Since the internal energy increases (\(\Delta U > 0\)), the temperature of the gas must increase.

• [1.5 marks] Correct formulation/definition of the first law of thermodynamics with terms defined (1 mark for \(Q = \Delta U + W\) or equivalent, 0.5 mark for correct sign conventions).
• [1.5 marks] Correct calculation of \(\Delta U = +530\text{ J}\) (1 mark for method, 0.5 mark for final value and positive sign).
• [1 mark] Stating clearly that the temperature increases.
• [1.5 marks] Correct physical explanation linking internal energy to the temperature of an ideal gas (1 mark for stating \(U \propto T\) or kinetic energy only, 0.5 mark for conclusion).

1. Calculate the moment of inertia \(I\):
\(I = \frac{1}{2} M R^2 = \frac{1}{2} (12.0) (0.350)^2 = 0.735\text{ kg m}^2\)

2. Calculate the rotational kinetic energy:
\(E_k = \frac{1}{2} I \omega^2 = \frac{1}{2} (0.735) (45.0)^2 = 744.1875\text{ J} \approx 744\text{ J}\)

3. Calculate the required angular deceleration \(\alpha\):
\(\alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - 45.0}{8.00} = -5.625\text{ rad s}^{-2}\)

4. Calculate the torque magnitude \(T\):
\(T = I |\alpha| = 0.735 \times 5.625 = 4.134\text{ N m} \approx 4.13\text{ N m}\)

• [1.5 marks] Correct calculation of moment of inertia \(I = 0.735\text{ kg m}^2\) (1 mark for substitution, 0.5 mark for correct answer).
• [1.5 marks] Correct calculation of rotational kinetic energy (1 mark for formula, 0.5 mark for \(744\text{ J}\)).
• [1 mark] Correct calculation of angular acceleration magnitude \(\alpha = 5.63\text{ rad s}^{-2}\).
• [1.5 marks] Correct calculation of torque \(T = 4.13\text{ N m}\) (1 mark for method, 0.5 mark for value with unit).

1. Principle of Conservation of Angular Momentum:
Since no external torque acts on the system, the total angular momentum is conserved:
\(I_1 \omega_1 = I_f \omega_f\)

2. Calculate the moment of inertia of the dropped clay:
Treating the clay as a point mass:
\(I_{\text{clay}} = m r^2 = 0.120\text{ kg} \times (0.150\text{ m})^2 = 2.70 \times 10^{-3}\text{ kg m}^2\)

3. Calculate the total final moment of inertia of the system:
\(I_f = I_1 + I_{\text{clay}} = 3.50 \times 10^{-3} + 2.70 \times 10^{-3} = 6.20 \times 10^{-3}\text{ kg m}^2\)

4. Calculate the final angular speed \(\omega_f\):
\(\omega_f = \frac{I_1 \omega_1}{I_f} = \frac{(3.50 \times 10^{-3}\text{ kg m}^2) (8.20\text{ rad s}^{-1})}{6.20 \times 10^{-3}\text{ kg m}^2}\)
\(\omega_f = \frac{0.0287}{6.20 \times 10^{-3}} \approx 4.629\text{ rad s}^{-1}\)

Rounding to 3 significant figures gives \(4.63\text{ rad s}^{-1}\).

• [1 mark] Identifying conservation of angular momentum (\(I_1 \omega_1 = I_f \omega_f\)).
• [1.5 marks] Correct calculation of the moment of inertia of the clay (1 mark for formula \(m r^2\), 0.5 mark for \(2.70 \times 10^{-3}\text{ kg m}^2\)).
• [1 mark] Correctly summing moments of inertia to find \(I_f = 6.20 \times 10^{-3}\text{ kg m}^2\).
• [2 marks] Correct final calculation of the new angular speed (1.5 marks for substitution/calculation, 0.5 mark for final answer of \(4.63\text{ rad s}^{-1}\) with correct unit).

1. Calculate the volume of the cylinder:
\(V = \pi \left(\frac{d}{2}\right)^2 L = \frac{\pi d^2 L}{4} = \frac{\pi \times (1.42)^2 \times 8.54}{4} = 13.525\text{ cm}^3\)

2. Calculate the density:
\(\rho = \frac{m}{V} = \frac{124.3}{13.525} = 9.191\text{ g cm}^{-3}\)

3. Calculate the percentage uncertainties for each quantity:
- Percentage uncertainty in mass: \(\%m = \frac{0.1}{124.3} \times 100\% = 0.080\%\)
- Percentage uncertainty in length: \(\%L = \frac{0.05}{8.54} \times 100\% = 0.585\%\)
- Percentage uncertainty in diameter: \(\%d = \frac{0.02}{1.42} \times 100\% = 1.408\%\)

4. Calculate the percentage uncertainty in the density:
\(\%\rho = \%m + \%L + 2(\%d) = 0.080\% + 0.585\% + 2(1.408\%) = 3.481\%\)

5. Calculate the absolute uncertainty in the density:
\(\Delta\rho = 9.191 \times 0.03481 = 0.320\text{ g cm}^{-3}\)

6. Rounding to appropriate significant figures:
Since the absolute uncertainty is \(0.3\text{ g cm}^{-3}\) (1 s.f.), the density should be rounded to the same decimal place:
\(\rho = 9.2 \pm 0.3\text{ g cm}^{-3}\).

Award marks as follows:
- [1 mark] Correct calculation of the volume \(V = 13.5\text{ cm}^3\) and density \(\rho = 9.19\text{ g cm}^{-3}\).
- [1 mark] Calculation of percentage uncertainties for mass (\(0.08\%\)) and length (\(0.59\%\)).
- [1 mark] Calculation of percentage uncertainty for diameter (\(1.41\%\)) and doubling it for the \(d^2\) term (\(2.82\%\)).
- [1 mark] Summing the percentage uncertainties to find the overall percentage uncertainty in density (\(3.48\%\) or \(3.5\%\)).
- [1 mark] Calculating the absolute uncertainty as \(0.32\text{ g cm}^{-3}\).
- [0.5 marks] Expressing the final answer to 1 decimal place matching the uncertainty: \(9.2 \pm 0.3\text{ g cm}^{-3}\).

(a) The angular frequency \(\omega\) is given by:
\(\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{14.0}{0.350}} = \sqrt{40} = 6.325\text{ rad s}^{-1}\)
The maximum acceleration is:
\(a_{\max} = \omega^2 A = 40.0 \times 0.085 = 3.40\text{ m s}^{-2}\)
This is approximately \(3.4\text{ m s}^{-2}\).

(b) The maximum kinetic energy is equal to the total mechanical energy stored in the spring at maximum displacement:
\(E_k = \frac{1}{2} k A^2 = 0.5 \times 14.0 \times (0.085)^2 = 0.050575\text{ J} \approx 0.051\text{ J}\).

(c) The resistive force does work against the motion of the glider. Since the force is resistive, it always acts in the direction opposite to the velocity. Energy is dissipated as thermal energy. During one full oscillation of path length \(4A\), the work done against friction is \(W = F \times 4A = 0.050 \times 4(0.085) = 0.017\text{ J}\). Therefore, the total mechanical energy decreases by this amount over one full cycle.

(a) [1.5 marks total]:
- [1 mark] Correctly calculating \(\omega^2 = 40\text{ rad}^2\text{ s}^{-2}\) or \(\omega = 6.33\text{ rad s}^{-1}\).
- [0.5 marks] Showing \(a_{\max} = 3.40\text{ m s}^{-2}\).

(b) [2 marks total]:
- [1 mark] Using \(E = \frac{1}{2} k A^2\) or finding \(v_{\max} = \omega A = 0.538\text{ m s}^{-1}\) and using \(\frac{1}{2} m v^2\).
- [1 mark] Arriving at \(0.051\text{ J}\) (or \(0.0506\text{ J}\)).

(c) [2 marks total]:
- [1 mark] Explaining that work is done against the resistive force, converting mechanical energy into thermal energy / heat.
- [1 mark] Quantifying that energy decreases continuously, or identifying that the reduction in energy depends on the total distance travelled.

(a) The change in internal energy \(\Delta U = 0\text{ J}\). For an ideal gas, the internal energy is purely kinetic and is directly proportional to its absolute temperature \(T\). Since the expansion is isothermal (constant temperature), the internal energy does not change.

(b) Using the First Law of Thermodynamics, \(\Delta U = Q - W\) (where \(W\) is the work done by the gas). Since \(\Delta U = 0\), we have:
\(Q = W = 800\text{ J}\).
Therefore, \(800\text{ J}\) of heat energy is transferred to the gas.

(c) As the gas expands, the gas molecules collide with the moving boundaries (the piston) and do work. During these collisions, molecules rebound with lower speeds, which would cause a decrease in their average kinetic energy and thus a drop in temperature. To maintain a constant temperature, the average kinetic energy of the molecules must remain constant. Heat energy must be supplied from the surroundings to replace the energy lost as work, keeping the average molecular speed constant.

(a) [2 marks total]:
- [1 mark] Correctly stating that \(\Delta U = 0\text{ J}\).
- [1 mark] Explaining that the internal energy of an ideal gas depends only on its temperature, which is constant.

(b) [1.5 marks total]:
- [0.5 marks] Stating or showing the first law of thermodynamics formula \(Q = \Delta U + W\).
- [1 mark] Deducing that \(Q = 800\text{ J}\).

(c) [2 marks total]:
- [1 mark] Explaining that expanding molecules lose kinetic energy as they do work against the surroundings / colliding with a moving piston.
- [1 mark] Explaining that incoming heat maintains/restores the average kinetic energy (and thus temperature) of the molecules.

(a) First, calculate the moment of inertia \(I\) of the flywheel:
\(I = \frac{1}{2} M R^2 = \frac{1}{2} \times 8.0 \times (0.25)^2 = 4.0 \times 0.0625 = 0.25\text{ kg m}^2\)
Now, calculate the rotational kinetic energy:
\(E_k = \frac{1}{2} I \omega^2 = \frac{1}{2} \times 0.25 \times (42)^2 = 0.125 \times 1764 = 220.5\text{ J} \approx 220\text{ J}\).

(b) The angular deceleration \(\alpha\) is determined by the torque:
\(\tau = I \alpha \implies 0.65 = 0.25 \times \alpha \implies \alpha = \frac{0.65}{0.25} = 2.6\text{ rad s}^{-2}\)
Using the rotational kinematic equation \(\omega = \omega_0 - \alpha t\) where \(\omega = 0\):
\(0 = 42 - 2.6 t \implies t = \frac{42}{2.6} = 16.15\text{ s} \approx 16\text{ s}\).

(c) The angular displacement \(\theta\) during deceleration is:
\(\theta = \omega_{\text{avg}} \times t = \left(\frac{42 + 0}{2}\right) \times 16.15 = 21 \times 16.15 = 339.2\text{ rad}\)
The number of revolutions \(N\) is:
\(N = \frac{\theta}{2\pi} = \frac{339.2}{2\pi} = 53.98\text{ revolutions}\)
Therefore, the flywheel makes 54 complete revolutions.

(a) [2 marks total]:
- [1 mark] Correctly calculating the moment of inertia \(I = 0.25\text{ kg m}^2\).
- [1 mark] Calculating the rotational kinetic energy \(220\text{ J}\) (allow 221 J).

(b) [2.5 marks total]:
- [1 mark] Using \(\tau = I \alpha\) to find the angular acceleration \(\alpha = 2.6\text{ rad s}^{-2}\).
- [1 mark] Recalling and using \(\omega = \omega_0 + \alpha t\) (or equivalent deceleration relation).
- [0.5 marks] Finding the time \(t = 16\text{ s}\) (allow 16.2 s).

(c) [1 mark total]:
- [1 mark] Calculating the total angle \(\theta = 339\text{ rad}\) and dividing by \(2\pi\) to obtain 54 complete revolutions.

(a) The formula for resistivity is:
\(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\)
Substitute the given values (converting \(d\) to meters):
\(d = 0.38 \times 10^{-3}\text{ m}\)
\(\rho = \frac{4.25 \times \pi \times (0.38 \times 10^{-3})^2}{4 \times 1.200}\)
\(\rho = \frac{4.25 \times 4.5365 \times 10^{-7}}{4.800} = 4.016 \times 10^{-7}\ \Omega\text{ m} \approx 4.0 \times 10^{-7}\ \Omega\text{ m}\).

(b) Calculate individual percentage uncertainties:
- \(\%R = \frac{0.05}{4.25} \times 100\% = 1.176\%\)
- \(\%L = \frac{0.002}{1.200} \times 100\% = 0.167\%\)
- \(\%d = \frac{0.01}{0.38} \times 100\% = 2.632\%\)

The formula for percentage uncertainty in \(\rho\) is:
\(\%\rho = \%R + \%L + 2(\%d)\)
\(\%\rho = 1.176\% + 0.167\% + 2(2.632\%) = 1.176\% + 0.167\% + 5.263\% = 6.606\% \approx 6.6\%\).

(c) To reduce the uncertainty in diameter, the student can:
1. Measure the diameter at multiple different positions and orientations along the wire and calculate an average value.
2. Use a digital micrometer with a higher precision / smaller resolution.

(a) [2 marks total]:
- [1 mark] Correct rearrangement of formula to \(\rho = \frac{R \pi d^2}{4 L}\).
- [1 mark] Calculation of \(4.0 \times 10^{-7}\ \Omega\text{ m}\) (allow \(4.02 \times 10^{-7}\)).

(b) [2.5 marks total]:
- [1 mark] Correct calculation of percentage uncertainties of \(R\) (\(1.18\%\)) and \(L\) (\(0.17\%\)).
- [1 mark] Calculation of percentage uncertainty of \(d\) (\(2.63\%\)) and doubling it (\(5.26\%\)).
- [0.5 marks] Summing these to get \(6.6\%\) (accept range \(6.5\% - 6.7\%\)).

(c) [1 mark total]:
- [1 mark] Suggesting measuring at several different points/orientations along the wire to take an average OR using an instrument with higher resolution (e.g., digital micrometer).

(a) The time period \(T\) of a simple pendulum is given by:
\(T = 2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{1.15}{9.81}} = 2.152\text{ s} \approx 2.15\text{ s}\).

(b) The total energy \(E\) of a simple harmonic oscillator is proportional to the square of its amplitude \(A^2\):
\(E \propto A^2\)
The fraction of energy remaining is:
\(\text{Fraction} = \frac{E_{\text{final}}}{E_{\text{initial}}} = \frac{A_{\text{final}}^2}{A_{\text{initial}}^2} = \frac{(4.5)^2}{(6.0)^2} = \frac{20.25}{36.0} = 0.5625 \approx 0.56\) (or \(56\%\)).

(c) This type of damping is light damping (or underdamping). In simple harmonic motion, the time period (and thus the frequency) is independent of the amplitude. Therefore, the frequency of the pendulum remains constant as the amplitude decreases.

(a) [1.5 marks total]:
- [1 mark] Using \(T = 2\pi \sqrt{\frac{l}{g}}\).
- [0.5 marks] Calculating \(T = 2.15\text{ s}\).

(b) [2 marks total]:
- [1 mark] Stating that energy is proportional to amplitude squared (\(E \propto A^2\)).
- [1 mark] Calculating the remaining fraction as \(0.56\) (or \(56.3\%\)).

(c) [2 marks total]:
- [1 mark] Identifying the damping as 'light damping' (or 'underdamping').
- [1 mark] Stating that the frequency remains constant because the period of a pendulum is independent of amplitude (for small angles).

(a) The total thermal energy supplied is:
\(E = P \times t = 45\text{ W} \times (5.0 \times 60\text{ s}) = 45 \times 300 = 13500\text{ J}\).

(b) Using the heat transfer formula:
\(Q = m c \Delta T\)
where:
\(\Delta T = 58.5^\circ\text{C} - 20.0^\circ\text{C} = 38.5\text{ K}\)
\(13500 = 0.75 \times c \times 38.5\)
\(13500 = 28.875 \times c\)
\(c = \frac{13500}{28.875} = 467.5\text{ J kg}^{-1}\text{ K}^{-1} \approx 470\text{ J kg}^{-1}\text{ K}^{-1}\).

(c) The experimental value (\(470\text{ J kg}^{-1}\text{ K}^{-1}\)) is higher than the accepted value because some thermal energy is lost to the surroundings. Consequently, more energy is required to raise the temperature of the block by the measured amount. To improve accuracy, the copper block should be wrapped in an insulating material (lagging) to reduce thermal energy losses.

(a) [1.5 marks total]:
- [1 mark] Using \(E = P \times t\) with correct unit conversion for time.
- [0.5 marks] Calculating \(13500\text{ J}\).

(b) [2 marks total]:
- [1 mark] Recalling and rearranging \(Q = m c \Delta T\) to solve for \(c\).
- [1 mark] Calculating \(470\text{ J kg}^{-1}\text{ K}^{-1}\) (accept \(468\text{ J kg}^{-1}\text{ K}^{-1}\)).

(c) [2 marks total]:
- [1 mark] Explaining that heat lost to the surroundings means the calculated \(c\) is too high (since actual energy entering the block is less than \(13500\text{ J}\)).
- [1 mark] Suggesting wrapping the block in lagging/insulating material or using a lid.

(a) The initial angular speed \(\omega_0\) is:
\(\omega_0 = 2 \pi f = 2 \pi \times 1.5 = 3.0 \pi \approx 9.425\text{ rad s}^{-1} \approx 9.4\text{ rad s}^{-1}\).

(b) By conservation of angular momentum:
\(I_i \omega_i = I_f \omega_f\)
The initial moment of inertia is \(I_i = 0.12\text{ kg m}^2\).
The moment of inertia of the clay (treating it as a point mass) is:
\(I_{\text{clay}} = m r^2 = 0.18 \times (0.22)^2 = 0.18 \times 0.0484 = 0.008712\text{ kg m}^2\)
The final moment of inertia of the system is:
\(I_f = I_i + I_{\text{clay}} = 0.12 + 0.008712 = 0.128712\text{ kg m}^2\)

Now, calculate the final angular speed:
\(\omega_f = \frac{I_i \omega_i}{I_f} = \frac{0.12 \times 9.425}{0.128712} = 8.787\text{ rad s}^{-1} \approx 8.8\text{ rad s}^{-1}\).

(c) Kinetic energy is not conserved. This is a completely inelastic collision where the clay sticks to the turntable. Some of the rotational kinetic energy is converted into thermal energy (and sound) during the deformation and sticking of the clay.

(a) [1.5 marks total]:
- [1 mark] Using \(\omega = 2\pi f\).
- [0.5 marks] Calculating \(9.4\text{ rad s}^{-1}\) (or \(3.0\pi\)).

(b) [3 marks total]:
- [1 mark] Calculating the moment of inertia of the clay point mass \(I_{\text{clay}} = 0.0087\text{ kg m}^2\).
- [1 mark] Stating the principle of conservation of angular momentum and writing the expression \(I_i \omega_i = I_f \omega_f\).
- [1 mark] Correct calculation of the final angular speed \(8.8\text{ rad s}^{-1}\) (allow \(8.7\text{ rad s}^{-1}\) to \(8.8\text{ rad s}^{-1}\)).

(c) [1 mark total]:
- [1 mark] Stating that kinetic energy is not conserved because the process is an inelastic collision where energy is lost as heat/sound.

(a) Calculate cross-sectional area:
\( A = \frac{\pi d^2}{4} = \frac{\pi \times (0.46 \times 10^{-3}\text{ m})^2}{4} = 1.6619 \times 10^{-7}\text{ m}^2 \)

Calculate resistivity:
\( \rho = \frac{R A}{L} = \frac{4.3 \times 1.6619 \times 10^{-7}}{1.250} = 5.717 \times 10^{-7}\ \Omega\text{ m} \)

(b) Calculate percentage uncertainties:
- \( \%\Delta d = \frac{0.01}{0.46} \times 100\% = 2.174\% \)
- \( \%\Delta A = 2 \times 2.174\% = 4.348\% \)
- \( \%\Delta L = \frac{0.002}{1.250} \times 100\% = 0.160\% \)
- \( \%\Delta R = \frac{0.2}{4.3} \times 100\% = 4.651\% \)

Total percentage uncertainty in \( \rho \):
\( \%\Delta\rho = 4.348\% + 0.160\% + 4.651\% = 9.159\% \)

Absolute uncertainty in \( \rho \):
\( \Delta\rho = 5.717 \times 10^{-7} \times 0.09159 = 5.236 \times 10^{-8}\ \Omega\text{ m} \approx 0.5 \times 10^{-7}\ \Omega\text{ m} \)

Therefore, \( \rho = (5.7 \pm 0.5) \times 10^{-7}\ \Omega\text{ m} \).

(a) [3 marks total]:
- Correct calculation of area \( A = 1.66 \times 10^{-7}\text{ m}^2 \) (1 mark)
- Correct substitution of values into resistivity formula yielding \( 5.7 \times 10^{-7} \) (1 mark)
- Correct unit \( \Omega\text{ m} \) (1 mark)

(b) [2.5 marks total]:
- Doubling percentage uncertainty of \( d \) to get percentage uncertainty in \( A \) (4.35%) (1 mark)
- Correct sum of all three percentage uncertainties (9.16% or 9.2%) (1 mark)
- Correct absolute uncertainty to 1 significant figure: \( 0.5 \times 10^{-7}\ \Omega\text{ m} \) (or \( 5 \times 10^{-8}\ \Omega\text{ m} \)) (0.5 marks)

(a) The measured temperature change is:
\( \Delta\theta_{\text{measured}} = 82.3 - 45.1 = 37.2\ ^\circ\text{C} \)

The true temperatures are:
- Initial: \( 82.3 - 1.5 = 80.8\ ^\circ\text{C} \)
- Final: \( 45.1 - 1.5 = 43.6\ ^\circ\text{C} \)

The true temperature change is:
\( \Delta\theta_{\text{true}} = 80.8 - 43.6 = 37.2\ ^\circ\text{C} \)

Therefore, the systematic error has no effect on the calculated temperature change because it cancels out.

(b) Advantages of the digital sensor:
- Resolution: The digital sensor has a higher resolution (\( 0.1\ ^\circ\text{C} \)) compared to the mercury-in-glass thermometer (\( 1\ ^\circ\text{C} \)), resulting in a smaller uncertainty due to resolution.
- Parallax error: The digital sensor provides a direct numerical readout on a screen, which completely eliminates parallax error.
- Ease of data collection: The digital sensor can be connected to a data logger to record temperatures automatically at precise intervals.

Limitations of the digital sensor:
- It requires a power source/battery to operate, unlike the mercury thermometer.
- It has a systematic error (calibration error) of \( +1.5\ ^\circ\text{C} \), which must be corrected for absolute temperature measurements, whereas the mercury thermometer might not have such an offset.

(a) [2 marks total]:
- Calculation showing measured change is \( 37.2\ ^\circ\text{C} \) and true change is \( 37.2\ ^\circ\text{C} \) (1 mark)
- Explanation that systematic error cancels out when taking a difference (1 mark)

(b) [3.5 marks total]:
- Highlighting better resolution of digital sensor (\( 0.1\ ^\circ\text{C} \) vs \( 1\ ^\circ\text{C} \)) (1 mark)
- Highlighting elimination of parallax error due to digital display (1 mark)
- Mentioning automatic logging / remote sensing capability (1 mark)
- Mentioning a limitation (e.g., dependence on power, or needing calibration to remove the systematic error) (0.5 marks)

(a) In equilibrium, the upward spring force equals the downward weight:
\( k \Delta x = mg \)
\( k = \frac{mg}{\Delta x} = \frac{0.350\text{ kg} \times 9.81\text{ m s}^{-2}}{0.085\text{ m}} = 40.39\text{ N m}^{-1} \approx 40\text{ N m}^{-1} \)

(b) The angular frequency \( \omega \) is:
\( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{40.39}{0.350}} = 10.74\text{ rad s}^{-1} \)

The maximum speed \( v_{\text{max}} \) is given by:
\( v_{\text{max}} = \omega A \)

Since the mass is pulled down by a further \( 0.040\text{ m} \) from equilibrium and released, the amplitude is \( A = 0.040\text{ m} \).

\( v_{\text{max}} = 10.74\text{ rad s}^{-1} \times 0.040\text{ m} = 0.430\text{ m s}^{-1} \)

(c) The total distance travelled is \( 0.12\text{ m} \).
Since \( A = 0.040\text{ m} \), the total distance is exactly \( 3A \).

Starting from maximum displacement:
- It travels \( 1A \) (0.04 m) to the equilibrium position.
- It travels \( 2A \) (0.04 m) to the opposite displacement extreme.
- It travels \( 3A \) (0.04 m) back to the equilibrium position.

This corresponds to exactly three-quarters of a time period, \( t = \frac{3}{4}T \).

Calculate the time period \( T \):
\( T = \frac{2\pi}{\omega} = \frac{2\pi}{10.74} = 0.585\text{ s} \)

Now calculate the time taken:
\( t = \frac{3}{4} \times 0.585\text{ s} = 0.439\text{ s} \approx 0.44\text{ s} \)

(a) [1.5 marks total]:
- Equating spring force to weight: \( k \Delta x = mg \) (1 mark)
- Substituting correct values to get \( 40.4\text{ N m}^{-1} \) (0.5 marks)

(b) [2 marks total]:
- Correct calculation of \( \omega \) (or equivalent method using formula) (1 mark)
- Correct value of \( v_{\text{max}} = 0.43\text{ m s}^{-1} \) (1 mark)

(c) [2 marks total]:
- Deducing that the travel distance of \( 0.12\text{ m} \) is equivalent to \( \frac{3}{4}T \) (1 mark)
- Correct calculation of \( T = 0.59\text{ s} \) and final time \( t = 0.44\text{ s} \) (1 mark)

(a) Method 1: Using gravitational potential energy at maximum height.
\( h = L(1 - \cos\theta) = 1.60\text{ m} \times (1 - \cos 7.5^\circ) \)
\( \cos 7.5^\circ = 0.99144 \)
\( h = 1.60\text{ m} \times 0.008555 = 0.01369\text{ m} \)

Total energy \( E = mgh = 0.120\text{ kg} \times 9.81\text{ m s}^{-2} \times 0.01369\text{ m} = 0.01611\text{ J} \approx 1.6 \times 10^{-2}\text{ J} \)

Method 2: Using SHM energy formula.
\( \omega = \sqrt{\frac{g}{L}} = \sqrt{\frac{9.81}{1.60}} = 2.476\text{ rad s}^{-1} \)
\( A = L \theta_{\text{rad}} = 1.60 \times \left(7.5 \times \frac{\pi}{180}\right) = 0.2094\text{ m} \)
\( E = \frac{1}{2} m \omega^2 A^2 = 0.5 \times 0.120 \times (2.476)^2 \times (0.2094)^2 = 0.01613\text{ J} \approx 1.6 \times 10^{-2}\text{ J} \)

(b) Under light damping:
- The total mechanical energy of the pendulum decreases over time.
- This is because work is done by the bob against resistive (frictional/drag) forces of the medium.
- Energy is transferred from mechanical energy of the pendulum to thermal energy of the surroundings.
- The rate of energy loss decreases as amplitude decreases, leading to an exponential decay of energy over time.

(a) [3 marks total]:
- Determination of maximum height \( h \) or amplitude \( A \) in appropriate units (1 mark)
- Correct formula for energy (either \( mgh \) or \( \frac{1}{2}m\omega^2 A^2 \)) (1 mark)
- Correct evaluation yielding \( 1.6 \times 10^{-2}\text{ J} \) (allow \( 1.61 \times 10^{-2}\text{ J} \)) (1 mark)

(b) [2.5 marks total]:
- Mentioning that mechanical energy decreases (0.5 marks)
- Explanation that work is done against resistive forces / drag (1 mark)
- Explanation that energy is converted to thermal energy / dissipated to surroundings (1 mark)

(a) The thermal power transferred to the water is:
\( P = \frac{\Delta Q}{\Delta t} = \frac{m c \Delta\theta}{\Delta t} \)

Where:
- \( P = 8.5 \times 10^3\text{ W} \)
- \( c = 4180\text{ J kg}^{-1}\text{ K}^{-1} \)
- \( \Delta\theta = 38.0 - 14.0 = 24.0\ ^\circ\text{C} \)

Let \( R_m = \frac{m}{\Delta t} \) be the mass flow rate in \( \text{kg s}^{-1} \):
\( 8.5 \times 10^3 = R_m \times 4180 \times 24.0 \)
\( R_m = \frac{8500}{100320} = 0.08473\text{ kg s}^{-1} \)

To find the mass flowing per minute, multiply by 60:
\( \text{Mass per minute} = 0.08473\text{ kg s}^{-1} \times 60\text{ s} = 5.084\text{ kg} \approx 5.1\text{ kg} \)

(b) If energy is lost to the surroundings:
- The useful power transferred to the water is less than the total electrical power input of \( 8.5\text{ kW} \).
- Since \( P_{\text{useful}} = R_m c \Delta\theta \), and \( \Delta\theta \) must remain at \( 24.0\ ^\circ\text{C} \), a smaller useful power \( P_{\text{useful}} \) means the mass flow rate \( R_m \) must be smaller.
- Thus, the actual mass flow rate must decrease.

(a) [3.5 marks total]:
- Correct calculation of temperature rise \( \Delta\theta = 24.0\ ^\circ\text{C} \) (0.5 marks)
- Correct substitution into power equation to find mass flow rate per second \( 0.085\text{ kg s}^{-1} \) (2 marks)
- Correct conversion to mass flow rate per minute: \( 5.1\text{ kg min}^{-1} \) (1 mark)

(b) [2 marks total]:
- Explanation that energy loss reduces the useful power transferred to water (1 mark)
- Concluding that mass flow rate must be reduced to achieve the same temperature rise (1 mark)

(a) Since the process is isothermal, the temperature \( T \) of the ideal gas remains constant.
The internal energy of an ideal gas depends solely on its absolute temperature (as there are no intermolecular forces, hence potential energy is zero, and mean kinetic energy is proportional to \( T \)).
Therefore, \( \Delta U = 0\text{ J} \).

(b) According to the first law of thermodynamics:
\( \Delta U = Q + W \)
where \( W \) is the work done on the gas.

Substituting \( \Delta U = 0 \) and \( W = +420\text{ J} \) (work done on the gas):
\( 0 = Q + 420 \)
\( Q = -420\text{ J} \)

The negative sign indicates that \( 420\text{ J} \) of thermal energy is transferred out of the gas to the surroundings (i.e., it is released).

(a) [2 marks total]:
- Stating \( \Delta U = 0\text{ J} \) (1 mark)
- Reasoning: internal energy of an ideal gas depends only on temperature, and temperature is constant in an isothermal process (1 mark)

(b) [3.5 marks total]:
- Correctly stating the First Law of Thermodynamics formula (1 mark)
- Correct substitution of values with proper signs (1.5 marks)
- Clearly stating that \( 420\text{ J} \) of heat is released/transferred to the surroundings (1 mark)

(a) First, calculate the moment of inertia \( I \) of the cylinder:
\( I = \frac{1}{2} M R^2 = 0.5 \times 4.5\text{ kg} \times (0.12\text{ m})^2 = 0.0324\text{ kg m}^2 \)

Next, calculate the torque \( T \) exerted by the tangential force:
\( T = F R = 6.0\text{ N} \times 0.12\text{ m} = 0.72\text{ N m} \)

Now, use Newton's second law for rotation to find the angular acceleration \( \alpha \):
\( T = I \alpha \)
\( \alpha = \frac{T}{I} = \frac{0.72\text{ N m}}{0.0324\text{ kg m}^2} = 22.22\text{ rad s}^{-2} \approx 22\text{ rad s}^{-2} \)

(b) Calculate the angular displacement \( \theta \) in radians:
\( \theta = 5.0\text{ revolutions} \times 2\pi\text{ rad/rev} = 10\pi\text{ rad} \approx 31.42\text{ rad} \)

Using the rotational work-energy theorem, the rotational kinetic energy \( E_k \) gained is equal to the work done by the torque:
\( W = T \theta = 0.72\text{ N m} \times 31.42\text{ rad} = 22.62\text{ J} \approx 23\text{ J} \)

Alternatively, calculate the final angular velocity \( \omega \):
\( \omega^2 = \omega_0^2 + 2 \alpha \theta = 0 + 2 \times 22.22 \times 31.42 = 1396\text{ rad}^2\text{ s}^{-2} \)
\( E_k = \frac{1}{2} I \omega^2 = 0.5 \times 0.0324 \times 1396 = 22.6\text{ J} \approx 23\text{ J} \)

(a) [3 marks total]:
- Correct calculation of moment of inertia \( I = 0.0324\text{ kg m}^2 \) (1 mark)
- Correct calculation of torque \( T = 0.72\text{ N m} \) (1 mark)
- Correct calculation of angular acceleration \( \alpha = 22\text{ rad s}^{-2} \) (1 mark)

(b) [2.5 marks total]:
- Correct conversion of 5.0 revolutions to radians (\( 10\pi \) or \( 31.4\text{ rad} \)) (1 mark)
- Use of work-energy relation or kinematics to find rotational energy (1 mark)
- Correct final kinetic energy of \( 23\text{ J} \) (or \( 22.6\text{ J} \)) (0.5 marks)

(a) Moment of inertia of the disk:
\( I_{\text{disk}} = \frac{1}{2} M R^2 = 0.5 \times 120\text{ kg} \times (2.0\text{ m})^2 = 240\text{ kg m}^2 \)

Initial moment of inertia of the child at the edge (\( r_i = 2.0\text{ m} \)):
\( I_{\text{child, i}} = m r_i^2 = 40\text{ kg} \times (2.0\text{ m})^2 = 160\text{ kg m}^2 \)

Total initial moment of inertia:
\( I_i = I_{\text{disk}} + I_{\text{child, i}} = 240 + 160 = 400\text{ kg m}^2 \)

(b) Final moment of inertia of the child at \( r_f = 0.80\text{ m} \):
\( I_{\text{child, f}} = m r_f^2 = 40\text{ kg} \times (0.80\text{ m})^2 = 25.6\text{ kg m}^2 \)

Total final moment of inertia of the system:
\( I_f = I_{\text{disk}} + I_{\text{child, f}} = 240 + 25.6 = 265.6\text{ kg m}^2 \)

Since there are no external torques, the angular momentum of the system is conserved:
\( I_i \omega_i = I_f \omega_f \)

Substitute the known values:
\( 400 \times 1.5 = 265.6 \times \omega_f \)

\( 600 = 265.6 \times \omega_f \)

\( \omega_f = \frac{600}{265.6} = 2.259\text{ rad s}^{-1} \approx 2.3\text{ rad s}^{-1} \)

(a) [2 marks total]:
- Calculation of disk moment of inertia (240 kg m^2) and child initial moment of inertia (160 kg m^2) (1 mark)
- Correct sum to give total initial moment of inertia of \( 400\text{ kg m}^2 \) (1 mark)

(b) [3.5 marks total]:
- Calculation of final total moment of inertia (265.6 kg m^2) (1 mark)
- Stating the principle of conservation of angular momentum (1 mark)
- Correct formulation of the equation \( 400 \times 1.5 = 265.6 \times \omega_f \) (1 mark)
- Correct evaluation to give final angular velocity of \( 2.3\text{ rad s}^{-1} \) (0.5 marks)

The formula for resistivity is \(\rho = \frac{R A}{L} = \frac{\pi R d^2}{4 L}\). The fractional uncertainty in \(\rho\) is given by: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\). Calculating the percentage uncertainties of each variable: \(\frac{\Delta R}{R} = \frac{0.2}{5.0} \times 100\% = 4.0\%\), \(\frac{\Delta d}{d} = \frac{0.01}{0.50} \times 100\% = 2.0\%\), \(\frac{\Delta L}{L} = \frac{0.005}{1.000} \times 100\% = 0.5\%\). Summing these terms with the correct powers: \(\frac{\Delta \rho}{\rho} = 4.0\% + 2(2.0\%) + 0.5\% = 8.5\%\). Thus, the percentage uncertainty is 8.5%.

1 mark for calculating the correct percentage uncertainty of 8.5% (C).

A typical family car has a mass \(m \approx 1500\text{ kg}\). A speed of \(110\text{ km h}^{-1}\) is equivalent to approximately \(30.6\text{ m s}^{-1}\). The kinetic energy is calculated as: \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 1500\text{ kg} \times (30.6\text{ m s}^{-1})^2 \approx 7.0 \times 10^5\text{ J}\). This value is closest to \(10^6\text{ J}\) in terms of order of magnitude.

1 mark for identifying 10^6 J (C) as the correct order of magnitude based on standard estimations.

The time period of a simple pendulum is given by \(T = 2\pi\sqrt{\frac{L}{g}}\). For the new setup, the length is \(L_2 = \frac{L}{2}\) and the gravitational acceleration is \(g_2 = \frac{g}{8}\). This gives: \(T_2 = 2\pi\sqrt{\frac{L/2}{g/8}} = 2\pi\sqrt{\frac{8L}{2g}} = 2\pi\sqrt{\frac{4L}{g}} = 2 \times 2\pi\sqrt{\frac{L}{g}} = 2 T_1\). Thus, the period is doubled.

1 mark for deriving the relation and identifying the correct answer as 2T_1 (D).

Increasing the damping of a system undergoing forced oscillations reduces the maximum amplitude at resonance because more energy is dissipated per cycle. Furthermore, the resonant frequency (the peak frequency of the amplitude response curve) shifts slightly to a lower value as damping increases. Therefore, both the maximum amplitude and the resonant frequency decrease.

1 mark for identifying that both the maximum amplitude and the resonant frequency decrease (D).

By the first law of thermodynamics, \(\Delta U = Q - W\), where \(Q = +140\text{ J}\) is the heat supplied. The work done \(W\) by the gas during constant pressure expansion is: \(W = p\Delta V = (1.5 \times 10^5\text{ Pa}) \times (1.0 \times 10^{-3}\text{ m}^3 - 6.0 \times 10^{-4}\text{ m}^3) = 1.5 \times 10^5 \times 4.0 \times 10^{-4} = 60\text{ J}\). Substituting this into the first law equation: \(\Delta U = 140\text{ J} - 60\text{ J} = 80\text{ J}\).

1 mark for calculating work done as 60 J and applying the first law to get 80 J (B).

Convert the temperatures from Celsius to Kelvin: \(T_{\text{initial}} = 27 + 273.15 = 300\text{ K}\) and \(T_{\text{final}} = 327 + 273.15 = 600\text{ K}\). The root-mean-square speed of gas molecules is given by \(c_{\text{rms}} = \sqrt{\frac{3RT}{M}}\), which means \(c_{\text{rms}} \propto \sqrt{T}\). Therefore: \(\frac{c_{\text{rms, final}}}{c_{\text{rms, initial}}} = \sqrt{\frac{T_{\text{final}}}{T_{\text{initial}}}} = \sqrt{\frac{600}{300}} = \sqrt{2} \approx 1.41\).

1 mark for converting Celsius to Kelvin and calculating the correct ratio of 1.41 (B).

The relationship between torque, moment of inertia, and angular acceleration is \(\tau = I \alpha\). The time \(t\) to stop the cylinder is \(t = \frac{\omega_0}{\alpha} = \frac{\omega_0 I}{\tau}\). Since \(\omega_0\) and \(\tau\) are constant, \(t\) is directly proportional to \(I\). The initial moment of inertia is \(I = \frac{1}{2} M R^2\). The new moment of inertia with mass \(2M\) and radius \(\frac{R}{2}\) is: \(I' = \frac{1}{2} (2M) \left(\frac{R}{2}\right)^2 = \frac{1}{2} \times 2M \times \frac{R^2}{4} = \frac{1}{2} \left( \frac{1}{2} M R^2 \right) = \frac{1}{2} I\). Therefore, the new time taken is \(t' = \frac{1}{2} t\).

1 mark for demonstrating that the moment of inertia is halved and thus the time taken is halved (B).

At the highest point of the bridge, the vertical forces acting on the car are gravity \(mg\) (downwards) and the normal reaction \(N\) (upwards). The resultant force towards the center of rotation (downwards) provides the necessary centripetal acceleration: \(F_{\text{net}} = mg - N = \frac{m v^2}{r}\). Solving for \(N\) yields: \(N = mg - \frac{m v^2}{r}\).

1 mark for setting up the dynamic equation for centripetal force and resolving to find N (C).

The percentage uncertainty in a quantity calculated using multiplication and division is found by adding the percentage uncertainties of each term. When a term is raised to a power, its percentage uncertainty is multiplied by that power. Here, \(\rho \propto \frac{R d^2}{L}\). Therefore, the percentage uncertainty in \(\rho\) is given by: \(\%\Delta \rho = \%\Delta R + 2 \times (\%\Delta d) + \%\Delta L\). Substituting the given values: \(\%\Delta \rho = 1.5\% + 2 \times 2.0\% + 1.0\% = 1.5\% + 4.0\% + 1.0\% = 6.5\%\).

1 mark for the correct calculation of the combined percentage uncertainty, correctly multiplying the uncertainty of diameter by 2.

First, calculate the temperature rise: \(\Delta \theta = \theta_2 - \theta_1 = 43.8 - 21.5 = 22.3 ^\circ\text{C}\). For addition or subtraction, the absolute uncertainties of the individual measurements are added together to find the absolute uncertainty of the result: \(\text{Absolute uncertainty} = 0.2 + 0.3 = 0.5 ^\circ\text{C}\). Next, calculate the percentage uncertainty: \(\frac{0.5}{22.3} \times 100\% \approx 2.24\%\), which rounds to \(2.2\%\).

1 mark for correctly adding the absolute uncertainties to get \(0.5 ^\circ\text{C}\) and dividing by the temperature rise to get \(2.2\%\).

The stiffness (spring constant) of a spring is inversely proportional to its length. When a spring of constant \(k\) is cut in half, each half has a spring constant of \(k' = 2k\). The period of a mass-spring system is given by \(T = 2\pi \sqrt{\frac{m}{k}}\). With the new spring constant, the new period is: \(T' = 2\pi \sqrt{\frac{m}{2k}} = \frac{1}{\sqrt{2}} \left(2\pi \sqrt{\frac{m}{k}}\right) = \frac{T}{\sqrt{2}}\).

1 mark for identifying that the spring constant doubles and correctly applying this to the time period formula to get \(T/\sqrt{2}\).

The potential energy is given by \(E_p = \frac{1}{2} k x^2\). The total mechanical energy is \(E_T = \frac{1}{2} k A^2\). We are given that \(E_p = 3 E_k\). Since total energy is the sum of kinetic and potential energy, we have: \(E_T = E_k + E_p = \frac{1}{3} E_p + E_p = \frac{4}{3} E_p\). Substituting the expressions: \(\frac{1}{2} k A^2 = \frac{4}{3} \left( \frac{1}{2} k x^2 \right)\). This simplifies to: \(A^2 = \frac{4}{3} x^2 \implies x^2 = \frac{3}{4} A^2 \implies x = \pm \frac{\sqrt{3}A}{2}\).

1 mark for establishing the relationship between total energy and potential energy at this point, and solving for the displacement \(x\).

According to the first law of thermodynamics: \(\Delta U = Q + W\), where \(\Delta U\) is the change in internal energy, \(Q\) is the heat energy added to the system, and \(W\) is the work done on the system. Here, the gas absorbs heat, so \(Q = +1200\text{ J}\). Work is done on the gas, so \(W = +2100\text{ J}\). Therefore, \(\Delta U = 1200 + 2100 = +3300\text{ J}\).

1 mark for correctly applying the sign convention of the first law of thermodynamics to determine that both quantities add to increase internal energy.

The electrical energy supplied by the heater is \(E = P \times t = 40\text{ W} \times (5.0 \times 60\text{ s}) = 12000\text{ J}\). Assuming no heat loss to the surroundings, this energy increases the thermal energy of the liquid: \(E = m c \Delta \theta\), where \(m = 0.25\text{ kg}\) and \(\Delta \theta = 60 - 20 = 40\text{ K}\). Rearranging for \(c\): \(c = \frac{E}{m \Delta \theta} = \frac{12000}{0.25 \times 40} = \frac{12000}{10} = 1200\text{ J kg}^{-1} \text{ K}^{-1}\).

1 mark for calculating energy supplied in joules, calculating the temperature difference, and solving for the specific heat capacity.

First, calculate the moment of inertia \(I\) of the cylinder: \(I = \frac{1}{2} M R^2 = 0.5 \times 4.0\text{ kg} \times (0.15\text{ m})^2 = 2.0 \times 0.0225 = 0.045\text{ kg m}^2\). Next, determine the torque \(\tau\) produced by the tangential force: \(\tau = F R = 6.0\text{ N} \times 0.15\text{ m} = 0.90\text{ N m}\). Using Newton's second law for rotation, \(\tau = I \alpha\), solve for angular acceleration \(\alpha\): \(\alpha = \frac{\tau}{I} = \frac{0.90}{0.045} = 20\text{ rad s}^{-2}\).

1 mark for calculating the correct moment of inertia and torque, leading to the correct angular acceleration.

Since there are no external torques acting on the system, the total angular momentum is conserved: \(L_{\text{initial}} = L_{\text{final}}\). The initial angular momentum is \(L_{\text{initial}} = I_1 \omega_1 = 0.80\text{ kg m}^2 \times 15\text{ rad s}^{-1} = 12\text{ kg m}^2 \text{s}^{-1}\). After dropping the second disc, the total moment of inertia is \(I_{\text{total}} = I_1 + I_2 = 0.80 + 0.40 = 1.20\text{ kg m}^2\). Using conservation of angular momentum: \(I_{\text{total}} \omega_2 = L_{\text{initial}} \implies 1.20 \times \omega_2 = 12 \implies \omega_2 = 10\text{ rad s}^{-1}\).

1 mark for applying the conservation of angular momentum and correctly determining the new angular speed.

The resistivity \(\rho\) is given by:

\(\rho = \frac{R A}{L} = \frac{\pi R d^2}{4 L}\)

The fractional uncertainty in the resistivity is:

\(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\)

Calculate each percentage uncertainty:

- Percentage uncertainty in \(R\):
\(\frac{0.05}{4.50} \times 100\% \approx 1.11\%\)

- Percentage uncertainty in \(d\):
\(\frac{0.01}{0.42} \times 100\% \approx 2.38\%\)

- Percentage uncertainty in \(L\):
\(\frac{0.002}{1.500} \times 100\% \approx 0.13\%\)

Summing these up with the weightings:

\(\text{Total percentage uncertainty} = 1.11\% + 2(2.38\%) + 0.13\% = 6.00\%\)

1 mark for calculating the percentage uncertainty in each quantity, doubling the diameter's percentage uncertainty, and summing them up to arrive at 6.0%.

The time period of a simple pendulum is given by:

\(T = 2\pi \sqrt{\frac{L}{g}}\)

When the length is increased by \(44\%\), the new length is \(1.44 L\).
The new gravitational field strength is \(2.25 g\).

The new period \(T'\) is:

\(T' = 2\pi \sqrt{\frac{1.44 L}{2.25 g}} = T \sqrt{\frac{1.44}{2.25}} = T \left( \frac{1.2}{1.5} \right) = 0.80 T\)

1 mark for using the pendulum period formula, substituting the new values for length and gravitational field strength, and simplifying to find 0.80T.

According to the first law of thermodynamics:

\(\Delta Q = \Delta U + \Delta W\)

where:
- \(\Delta Q\) is the thermal energy transferred to the gas (\(+950\text{ J}\)).
- \(\Delta W\) is the work done by the gas.
- \(\Delta U\) is the change in internal energy.

Calculate the work done during the isobaric expansion:

\(\Delta W = P \Delta V = 1.5 \times 10^5\text{ Pa} \times (6.0 \times 10^{-3} - 2.0 \times 10^{-3})\text{ m}^3\)
\(\Delta W = 1.5 \times 10^5 \times 4.0 \times 10^{-3} = 600\text{ J}\)

Rearranging to find the change in internal energy:

\(\Delta U = \Delta Q - \Delta W = 950\text{ J} - 600\text{ J} = 350\text{ J}\)

1 mark for using the work formula for isobaric expansion, calculating work as 600 J, and using the first law of thermodynamics to find the internal energy change of 350 J.

Rotational kinetic energy is given by:

\(E_k = \frac{1}{2} I \omega^2\)

Under constant torque, the angular acceleration \(\alpha\) is constant:

\(\tau = I \alpha\)

Since it starts from rest, the angular speed \(\omega\) after time \(t\) is:

\(\omega = \alpha t \implies \alpha = \frac{\omega}{t}\)

Substitute \(\alpha\) into the torque expression:

\(\tau = I \left( \frac{\omega}{t} \right) \implies I = \frac{\tau t}{\omega}\)

Substitute this expression for \(I\) back into the kinetic energy equation:

\(E_k = \frac{1}{2} \left( \frac{\tau t}{\omega} \right) \omega^2 = \frac{1}{2} \tau \omega t\)

1 mark for relating torque to moment of inertia and angular acceleration, linking angular speed to time, and substituting to find the correct expression.

For a simple harmonic oscillator, the maximum acceleration magnitude is:

\(a_{\text{max}} = \omega^2 A\)

where \(\omega\) is the angular frequency and \(A\) is the amplitude.

We know:

\(\omega^2 = \frac{k}{m} = \frac{40\text{ N m}^{-1}}{0.25\text{ kg}} = 160\text{ rad}^2\text{ s}^{-2}\)

Given \(A = 0.080\text{ m}\):

\(a_{\text{max}} = 160 \times 0.080 = 12.8\text{ m s}^{-2}\)

1 mark for calculating the angular frequency squared and multiplying by the amplitude to obtain 12.8 m s^-2.

First, find the absolute uncertainty in the time of 20 oscillations, \(t_{20}\), using half the range:

\(\text{Range} = \text{Maximum value} - \text{Minimum value} = 34.6\text{ s} - 34.1\text{ s} = 0.5\text{ s}\)

\(\text{Uncertainty in } t_{20} = \frac{\text{Range}}{2} = \frac{0.5\text{ s}}{2} = 0.25\text{ s}\)

Since this uncertainty (0.25 s) is greater than the resolution of the stopwatch (0.1 s), we use 0.25 s.

Now find the uncertainty in a single oscillation period, \(T\):

\(\Delta T = \frac{\Delta t_{20}}{20} = \frac{0.25\text{ s}}{20} = 0.0125\text{ s}\)

Rounding to two significant figures gives \(0.013\text{ s}\).

1 mark for calculating the range of the times, dividing by 2 to find the uncertainty in 20 oscillations, and dividing by 20 to find the single oscillation uncertainty.

By conservation of energy, the heat lost by the metal equals the heat gained by the water:

\(Q_{\text{lost}} = Q_{\text{gained}}\)

\(m_{\text{metal}} c_{\text{metal}} \Delta T_{\text{metal}} = m_{\text{water}} c_{\text{water}} \Delta T_{\text{water}}\)

Substitute the known values:
- \(m_{\text{metal}} = 0.50\text{ kg}\)
- \(\Delta T_{\text{metal}} = 120^\circ\text{C} - 25^\circ\text{C} = 95\text{ K}\)
- \(m_{\text{water}} = 0.80\text{ kg}\)
- \(c_{\text{water}} = 4200\text{ J kg}^{-1}\text{ K}^{-1}\)
- \(\Delta T_{\text{water}} = 25^\circ\text{C} - 20^\circ\text{C} = 5\text{ K}\)

\(0.50 \times c_{\text{metal}} \times 95 = 0.80 \times 4200 \times 5\)

\(47.5 \times c_{\text{metal}} = 16800\)

\(c_{\text{metal}} = \frac{16800}{47.5} \approx 354\text{ J kg}^{-1}\text{ K}^{-1}\)

Which rounds to \(350\text{ J kg}^{-1}\text{ K}^{-1}\).

1 mark for establishing the thermal equilibrium energy balance and solving for the specific heat capacity of the metal to get approximately 350 J kg^-1 K^-1.

The rotational dynamics equation is:

\(\tau = I \alpha\)

When the rod is horizontal, the gravitational force acts at its center of mass (distance of \(\frac{L}{2}\) from the pivot). The torque \(\tau\) is:

\(\tau = M g \left( \frac{L}{2} \right)\)

The moment of inertia \(I\) is given as \(\frac{1}{3} M L^2\).

Equating these:

\(M g \left( \frac{L}{2} \right) = \left( \frac{1}{3} M L^2 \right) \alpha\)

Divide both sides by \(M L\):

\(\frac{g}{2} = \frac{L}{3} \alpha\)

Rearranging for \(\alpha\):

\(\alpha = \frac{3 g}{2 L}\)

1 mark for identifying the torque about the pivot, equating it to I times angular acceleration, and correctly solving for alpha.

The percentage uncertainty in \(\rho\) is determined by adding the percentage uncertainties of the individual terms, multiplying the diameter uncertainty by 2 because it is raised to the second power. This yields: percentage uncertainty = \(1.5\% + 2 \times 2.0\% + 1.0\% = 6.5\%\). Therefore, the correct option is C.

1 mark for adding the percentage uncertainties correctly, including doubling the percentage uncertainty of the diameter, to obtain 6.5%.

The time period of a simple pendulum is \(T = 2\pi \sqrt{\frac{L}{g}}\). When accelerating upwards at \(0.25 g\), the effective gravitational acceleration becomes \(g_{eff} = g + a = 1.25 g\). The new time period is \(T' = 2\pi \sqrt{\frac{L}{1.25 g}} = \frac{T}{\sqrt{1.25}} \approx 0.89 T\). Thus, the correct option is B.

1 mark for identifying the new effective gravitational field strength and calculating the resulting ratio to find 0.89 T.

The useful power supplied to the water is \(1200\text{ W} \times 0.85 = 1020\text{ W}\). In one minute (60 seconds), the useful heat energy delivered is \(Q = 1020\text{ W} \times 60\text{ s} = 61200\text{ J}\). The temperature change is \(\Delta \theta = 40\text{ }^\circ\text{C} - 15\text{ }^\circ\text{C} = 25\text{ }^\circ\text{C}\). Using \(Q = mc\Delta\theta\), we find \(61200 = m \times 4200 \times 25\), which gives \(m = \frac{61200}{105000} \approx 0.58\text{ kg}\). Therefore, the correct option is B.

1 mark for calculating the useful energy supplied in one minute and applying the thermal energy transfer equation to obtain 0.58 kg.

By conservation of energy, the gravitational potential energy lost is equal to the sum of translational and rotational kinetic energies: \(Mgh = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2\). Since there is no slipping, we substitute \(\omega = \frac{v}{R}\) and \(I = \frac{1}{2} M R^2\) to get \(Mgh = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2\). Rearranging gives \(v^2 = \frac{4}{3} gh\), so \(v = \sqrt{\frac{4}{3} gh}\). Therefore, the correct option is B.

1 mark for setting up the conservation of energy equation with both translational and rotational kinetic energy terms and solving for the linear speed.

The temperature rise is \(\Delta \theta = 43.0\text{ }^\circ\text{C} - 21.5\text{ }^\circ\text{C} = 21.5\text{ }^\circ\text{C}\). When subtracting two independent measurements, their absolute uncertainties add together: \(0.5\text{ }^\circ\text{C} + 0.5\text{ }^\circ\text{C} = 1.0\text{ }^\circ\text{C}\). The percentage uncertainty is \(\frac{1.0}{21.5} \times 100\% \approx 4.7\%\). Therefore, the correct option is C.

1 mark for adding the absolute uncertainties of both temperature readings to find the total absolute uncertainty and expressing it as a percentage of the temperature change.

The angular acceleration is given by \(\alpha = \frac{T}{I} = \frac{-2.0\text{ N m}}{0.40\text{ kg m}^2} = -5.0\text{ rad s}^{-2}\). Using the rotational equation of motion \(\omega_f^2 = \omega_i^2 + 2\alpha\theta\) with \(\omega_f = 0\), we get \(0 = 30^2 + 2(-5.0)\theta\), which simplifies to \(10\theta = 900\). Thus, \(\theta = 90\text{ rad}\). This corresponds to option B.

1 mark for calculating the angular acceleration from torque and moment of inertia, and using it in the angular motion equation to determine the displacement.

(a) Measure the length from the bottom of the split-cork to the centre of the steel bob using a metre ruler with a set square to ensure the ruler is vertical. Use a fiducial marker placed at the equilibrium position to improve timing accuracy.

(b) Mean time \(t_{\text{mean}} = \frac{37.12 + 37.35 + 37.21}{3} = 37.227\text{ s}\).
Mean period \(T = \frac{37.227}{20} = 1.861\text{ s} \approx 1.86\text{ s}\).
Absolute uncertainty in \(t\): \(\Delta t = \frac{\text{range}}{2} = \frac{37.35 - 37.12}{2} = 0.115\text{ s}\).
Absolute uncertainty in \(T\): \(\Delta T = \frac{\Delta t}{20} = \frac{0.115}{20} = 0.00575\text{ s} \approx 0.006\text{ s}\).
Percentage uncertainty in \(T\): \(\% \Delta T = \frac{0.00575}{1.861} \times 100\% = 0.31\%\).

(c) Rearranging the period equation gives \(g = \frac{4\pi^2 l}{T^2}\).
Percentage uncertainty in \(g\) is \(\% \Delta g = \% \Delta l + 2 \times \% \Delta T\).
\(\% \Delta l = \frac{0.002}{0.850} \times 100\% = 0.235\%\).
\(\% \Delta g = 0.235\% + 2 \times 0.309\% = 0.235\% + 0.618\% = 0.853\% \approx 0.85\%\).

(a)
1 mark: Mention of measuring from the underside of the split cork to the centre of gravity/centre of the bob.
1 mark: Mention of using a fiducial marker at the equilibrium position or using a set square to keep the ruler vertical.

(b)
1 mark: Mean period calculated correctly as \(1.86\text{ s}\) (or \(1.861\text{ s}\)).
1 mark: Absolute uncertainty in time calculated as half the range (\(0.115\text{ s}\) or \(0.12\text{ s}\)).
1 mark: Absolute uncertainty in period calculated as \(\Delta T = 0.006\text{ s}\) (accept \(0.0057\text{ s}\) to \(0.0060\text{ s}\)).
1 mark: Percentage uncertainty in \(T\) calculated as \(0.31\%\) (accept \(0.30\%\) to \(0.32\%\)).

(c)
1 mark: Recall that \(\% \Delta g = \% \Delta l + 2 \% \Delta T\).
1 mark: Correct calculation of \(\% \Delta l = 0.235\%\) (or \(0.24\%\)).
1 mark: Correct addition of uncertainties to find \(\% \Delta g = 0.85\%\) (accept \(0.83\%\) to \(0.87\%\) depending on rounding in part b).

(a) Close the jaws of the micrometer using the friction ratchet until they touch. If the zero line on the thimble does not align with the datum line on the sleeve, record this reading as the zero error. Subtract this zero error from all subsequent measurements.

(b) Measuring at multiple positions along the length checks for non-uniformity of the cross-sectional area. Measuring at different orientations (at 90 degrees) checks if the cross-section is elliptical rather than perfectly circular, allowing an average diameter to be computed.

(c) \(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 L}\).
Substituting \(m = 1.38 \times 10^{-3}\text{ kg}\), \(d = 0.46 \times 10^{-3}\text{ m}\), and \(L = 1.042\text{ m}\):
\(\rho = \frac{4 \times (1.38 \times 10^{-3})}{\pi \times (0.46 \times 10^{-3})^2 \times 1.042} = 7969\text{ kg m}^{-3} \approx 7970\text{ kg m}^{-3}\).

(d) \(\% \Delta \rho = \% \Delta m + 2 \times \% \Delta d + \% \Delta L\).
\(\% \Delta m = \frac{0.01}{1.38} \times 100\% = 0.725\%\).
\(\% \Delta d = \frac{0.02}{0.46} \times 100\% = 4.348\%\).
\(\% \Delta L = \frac{0.002}{1.042} \times 100\% = 0.192\%\).
\(\% \Delta \rho = 0.725\% + 2 \times (4.348\%) + 0.192\% = 9.613\% \approx 9.6\%\).

(a)
1 mark: Mention of closing the jaws using the ratchet to read the zero value on the thimble/sleeve.
1 mark: Correctly explaining how to subtract/add the zero error from/to the final measured value.

(b)
1 mark: Measuring at multiple positions to check for non-uniformity along the wire.
1 mark: Measuring at different orientations to account for a non-circular/elliptical cross-section.

(c)
1 mark: Recall and correct use of density formula \(\rho = \frac{4m}{\pi d^2 L}\).
1 mark: Correct conversion of units (mass to kg, diameter to m).
1 mark: Correct calculation of density as \(7970\text{ kg m}^{-3}\) (accept \(7900\) to \(8000\text{ kg m}^{-3}\)).

(d)
1 mark: Correct formula for percentage uncertainty: \(\% \Delta \rho = \% \Delta m + 2\% \Delta d + \% \Delta L\).
1 mark: Calculation of individual percentage uncertainties (\(\% \Delta m = 0.725\%\), \(\% \Delta d = 4.35\%\), \(\% \Delta L = 0.19\%\)).
1 mark: Correct summation to get \(9.6\%\) (accept \(9.5\%\) to \(9.7\%\)).

(a) Taking natural logarithms of both sides of \(v = k r^n\) gives \(\ln v = n \ln r + \ln k\). Comparing this to \(y = mx + c\), a graph of \(\ln v\) on the y-axis against \(\ln r\) on the x-axis yields a straight line where the gradient is \(n\) and the vertical intercept is \(\ln k\).

(b) The gradient directly gives \(n\), so \(n = 1.95 \pm 0.05\).
The vertical intercept is \(\ln k = 3.40\), which gives \(k = e^{3.40} = 29.96 \approx 30.0\).
The unit of \(k\) is \(\text{mm}^{1-n}\text{ s}^{-1}\), which is approximately \(\text{mm}^{-0.95}\text{ s}^{-1}\) (or assuming \(n=2\), \(\text{mm}^{-1}\text{ s}^{-1}\)).

(c) Yes, the experimental value is consistent with the theory because the theoretical value of \(2\) lies within the experimental range of uncertainty of \(n\) which is \([1.90, 2.00]\).
A potential systematic error is the cylinder wall effect, where the proximity of the container walls slows down the falling sphere. This effect systematically reduces the measured velocity \(v\) especially for larger spheres, resulting in a systematically lower gradient \(n\).

(a)
1 mark: Correctly taking logarithms to get \(\ln v = n \ln r + \ln k\).
1 mark: Correctly identifying that \(\ln v\) is plotted on the vertical axis and \(\ln r\) on the horizontal axis.
1 mark: Correctly identifying that the gradient is \(n\) and the vertical intercept is \(\ln k\).

(b)
1 mark: Stating \(n = 1.95\).
1 mark: Correct calculation of \(k = e^{3.40} = 30.0\) (accept \(29.9\) to \(30.1\)).
1 mark: Correct unit of \(k\) as \(\text{mm}^{-1}\text{ s}^{-1}\) or \(\text{mm}^{-0.95}\text{ s}^{-1}\).

(c)
1 mark: Stating that it is consistent because \(2.0\) lies within the experimental uncertainty range \([1.90, 2.00]\).
1 mark: Identification of a valid systematic error (e.g., wall effect, or temperature fluctuations affecting viscosity).
1 mark: Explanation of the effect on the results (e.g., wall effects reduce \(v\) more for larger spheres, decreasing the gradient).

(a) Squaring both sides of the equation gives \(T^2 = \frac{4\pi^2}{k}(M + m_s) = \frac{4\pi^2}{k}M + \frac{4\pi^2 m_s}{k}\). Therefore, plotting \(T^2\) on the vertical axis against \(M\) on the horizontal axis yields a straight line with a gradient of \(\frac{4\pi^2}{k}\).

(b) Let's use two points from the table to find the gradient. First convert \(M\) to kg:
Point 1: \((M_1, T^2_1) = (0.100\text{ kg}, 0.635\text{ s}^2)\)
Point 2: \((M_2, T^2_2) = (0.500\text{ kg}, 2.952\text{ s}^2)\)
\(\text{Gradient} = \frac{2.952 - 0.635}{0.500 - 0.100} = \frac{2.317}{0.400} = 5.7925\text{ s}^2\text{ kg}^{-1} \approx 5.79\text{ s}^2\text{ kg}^{-1}\) (accept \(5.70\) to \(5.90\)).

(c) Since \(\text{Gradient} = \frac{4\pi^2}{k}\), we have:
\(k = \frac{4\pi^2}{\text{Gradient}} = \frac{4\pi^2}{5.7925} = 6.82\text{ N m}^{-1}\) (accept \(6.69\) to \(6.93\text{ N m}^{-1}\)).

(d) The horizontal intercept occurs where \(T^2 = 0\):
\(0 = 5.79 M + C \implies M = -m_s\).
Using \(T^2 = 5.7925 M + C\):
At \(M = 0.100\text{ kg}\), \(0.635 = 5.7925 \times 0.100 + C \implies C = 0.635 - 0.579 = 0.056\text{ s}^2\).
When \(T^2 = 0 \implies M = -0.056 / 5.7925 = -0.0097\text{ kg} = -9.7\text{ g}\).
Thus, \(m_s = 9.7\text{ g}\) (accept \(8.0\text{ g}\) to \(12.0\text{ g}\)).
Its physical significance is that it represents the effective mass contribution of the spring itself as it oscillates, which is approximately \(\frac{1}{3}\) of the total mass of the spring.

(a)
1 mark: Correctly squaring the equation to get \(T^2 = \frac{4\pi^2}{k}M + \frac{4\pi^2 m_s}{k}\).
1 mark: Stating that \(T^2\) is on the y-axis and \(M\) is on the x-axis.

(b)
1 mark: Correct conversion of mass \(M\) from grams to kilograms.
1 mark: Use of a large triangle or selecting distant coordinates to calculate the gradient.
1 mark: Correct gradient value between \(5.70\) and \(5.90\text{ s}^2\text{ kg}^{-1}\) with appropriate unit.

(c)
1 mark: Correctly relating the gradient to \(\frac{4\pi^2}{k}\).
1 mark: Correct calculation of \(k\) as \(6.8\text{ N m}^{-1}\) or \(6.82\text{ N m}^{-1}\).
1 mark: Correct unit of \(\text{N m}^{-1}\) or \(\text{kg s}^{-2}\).

(d)
1 mark: Correct value for \(m_s\) in the range \(8.0\text{ g}\) to \(12.0\text{ g}\) (or \(0.008\text{ kg}\) to \(0.012\text{ kg}\)).
1 mark: Stating that \(m_s\) represents the effective mass contribution of the oscillating spring.

(a) Independent variable: number of oscillations \(n\).
Dependent variable: amplitude \(\theta_n\) (or maximum displacement).
Controlled variables: length of the pendulum, mass of the bob, size/shape/mass of the cardboard card, initial release angle.

(b) Place an angular scale (or protractor) behind the pendulum bob at the equilibrium position. To minimise parallax error, place a flat mirror directly behind the scale, and ensure that the pendulum cord is aligned with its reflection before reading the maximum displacement. Alternatively, use a high-frame-rate video camera placed directly in front of the scale and play back the video in slow motion.

(c) Taking natural logarithms of \(\theta_n = \theta_0 e^{-\gamma n}\) gives \(\ln\theta_n = \ln\theta_0 - \gamma n\).
Thus, the gradient of a plot of \(\ln(\theta_n)\) against \(n\) is equal to \(-\gamma\).
Using two points from the table, e.g., \((0, 3.81)\) and \((10, 2.30)\):
\(\text{Gradient} = \frac{2.30 - 3.81}{10 - 0} = \frac{-1.51}{10} = -0.151\).
Thus, \(\gamma = 0.151\) (accept \(0.145\) to \(0.155\)).

(d) A larger cardboard card increases the cross-sectional area and thus increases the air resistance (drag force) acting on the pendulum bob. This leads to a faster rate of energy dissipation, causing the amplitude to decay more rapidly, which increases the damping constant \(\gamma\).

(a)
1 mark: Correctly identifying the independent variable as \(n\) and dependent variable as amplitude \(\theta_n\).
1 mark: Identifying at least one relevant controlled variable (e.g., length of string, mass of bob).

(b)
1 mark: Describing placing an angular scale/protractor behind the bob.
1 mark: Describing a clear method to minimise parallax (mirror method or high-speed camera).

(c)
1 mark: Stating that the gradient of \(\ln(\theta_n)\) against \(n\) is equal to \(-\gamma\).
1 mark: Showing a correct calculation of the gradient using coordinates from the table.
1 mark: Finding \(\gamma = 0.151\) (accept \(0.145\) to \(0.155\)).

(d)
1 mark: Explaining that a larger card increases the air resistance/drag.
1 mark: Concluding that this increases the rate of energy dissipation, leading to a larger damping constant \(\gamma\).

(a) The diagram should show:
1. An aluminium block with two pre-drilled holes, one containing the electrical immersion heater and the other containing the thermometer or temperature sensor.
2. The heater connected to a low-voltage DC power supply, with an ammeter connected in series and a voltmeter connected in parallel across the heater terminals.
3. Insulation (lagging) wrapped completely around the aluminium block to minimise thermal energy losses to the surroundings.

(b)
1. Heat loss to the surroundings: Even with insulation, heat is lost to the air. This can be minimised by starting the experiment with the block at a temperature below room temperature (e.g., \(5^\circ\text{C}\) below) and continuing until the temperature is the same amount above room temperature (e.g., \(5^\circ\text{C}\) above). This ensures net heat transfer to the surroundings is approximately zero.
2. Thermal contact resistance: Poor thermal contact between the heater/thermometer and the block results in a lag in temperature reading. This can be minimised by adding a small amount of thermal grease or glycerol into the pre-drilled holes before inserting the heater and thermometer.

(c) The electrical power supplied by the heater is:
\(P = V \times I = 12.0\text{ V} \times 4.15\text{ A} = 49.8\text{ W}\).
The heat energy equation is \(P \Delta t = M c \Delta \theta \implies P = M c \frac{\Delta \theta}{\Delta t}\).
Since the gradient of the \(\theta\) against \(t\) graph is \(\frac{\Delta \theta}{\Delta t} = 0.048\text{ K s}^{-1}\), we have:
\(c = \frac{P}{M \times \text{gradient}} = \frac{49.8}{1.05 \times 0.048} = \frac{49.8}{0.0504} = 988.1\text{ J kg}^{-1}\text{ K}^{-1} \approx 988\text{ J kg}^{-1}\text{ K}^{-1}\) (or \(990\text{ J kg}^{-1}\text{ K}^{-1}\) to 2 s.f.).

(a)
1 mark: Drawing the aluminium block with two holes, one for the heater and one for the thermometer.
1 mark: Correctly connecting the ammeter in series and the voltmeter in parallel across the heater.
1 mark: Showing insulation/lagging surrounding the block.

(b)
1 mark: Identifying heat loss to the surroundings as a source of error.
1 mark: Describing how to minimise this (insulation, or starting below room temperature and ending above).
1 mark: Identifying poor thermal contact/thermal resistance as a source of error.
1 mark: Describing how to minimise this (use of thermal paste/conductive grease or glycerol).

(c)
1 mark: Correct calculation of power \(P = 49.8\text{ W}\).
1 mark: Correctly relating power and temperature rate of change: \(P = M c \times \text{gradient}\).
1 mark: Correct calculation of \(c = 988\text{ J kg}^{-1}\text{ K}^{-1}\) (accept \(980\) to \(1000\text{ J kg}^{-1}\text{ K}^{-1}\)).

(a) Room temperature can vary over the course of the experiment due to ambient drafts or the heat given off by the hot liquid itself. If they differ, the average room temperature during the experiment should be calculated and used in the analysis, or the experiment should be conducted in a larger, temperature-controlled environment.

(b) Taking the natural logarithm of both sides of the cooling equation yields:
\(\ln(\theta - \theta_{\text{room}}) = -kt + \ln(\theta_{\text{initial}} - \theta_{\text{room}})\).
This is a linear relationship where a plot of \(\ln(\theta - \theta_{\text{room}})\) against \(t\) has a gradient of \(-k\).
Using the first and last data points from the table:
\(\text{Gradient} = \frac{2.833 - 4.151}{600 - 0} = \frac{-1.318}{600} = -2.197 \times 10^{-3}\text{ s}^{-1}\).
Thus, \(k = 2.20 \times 10^{-3}\text{ s}^{-1}\) (accept \(2.15 \times 10^{-3}\) to \(2.25 \times 10^{-3}\text{ s}^{-1}\)).
The unit of \(k\) is \(\text{s}^{-1}\).

(c)
1. Set up two identical beakers, one sealed with a lid and the other left open (uncovered).
2. Fill both beakers with the exact same volume (or mass) of hot water initially at the same temperature (e.g., \(90^\circ\text{C}\)).
3. Place both beakers in the same room away from drafts, ensuring they experience the same room temperature \(\theta_{\text{room}}\).
4. Record the temperature of the water in both beakers at regular time intervals (e.g., every 60 seconds) for the same total duration.
5. Plot cooling curves for both beakers. A slower cooling rate or a smaller cooling constant \(k\) for the beaker with a lid confirms the student's claim.

(a)
1 mark: Explaining that room temperature can change due to heat from the beaker or environmental changes.
1 mark: Stating that the mean room temperature should be used, or the room temperature kept constant.

(b)
1 mark: Showing the linearised equation \(\ln(\theta - \theta_{\text{room}}) = -kt + \text{constant}\).
1 mark: Correctly calculating the gradient using points from the table.
1 mark: Finding \(k = 2.20 \times 10^{-3}\text{ s}^{-1}\) (accept \(2.15 \times 10^{-3}\) to \(2.25 \times 10^{-3}\)).
1 mark: Correct unit of \(\text{s}^{-1}\).

(c)
1 mark: Describing using two identical beakers (one with a lid, one without).
1 mark: Controlling the initial water volume/mass and initial water temperature.
1 mark: Keeping environmental conditions identical (same room temperature, no drafts).
1 mark: Describing how to compare results (e.g., comparing the cooling curves or calculated \(k\) values).

(a) The net torque \(\tau_{\text{net}}\) on the rotating cylinder is the torque due to the tension in the string minus the opposing frictional torque:
\(\tau_{\text{net}} = T R - \tau_f\).
Using Newton's second law for rotational motion, \(\tau_{\text{net}} = I \alpha\):
\(T R - \tau_f = I \alpha \implies \alpha = \frac{T R - \tau_f}{I}\).

(b) Assuming no slipping, the linear acceleration \(a\) of the string (and the falling mass) is related to the angular acceleration \(\alpha\) of the cylinder by:
\(a = \alpha R\) or \(\alpha = \frac{a}{R}\).

(c)(i) Rearranging the equation from part (a):
\(T R = I \alpha + \tau_f\).
Comparing this to \(y = mx + c\), where \(y = T R\) and \(x = \alpha\):
- The gradient represents the moment of inertia \(I\) of the cylinder.
- The y-intercept represents the frictional torque \(\tau_f\) at the axle.

(c)(ii) The gradient is \(I = 0.045\text{ kg m}^2\).
We are given \(I = \frac{1}{2} M R^2\).
Substituting \(I = 0.045\text{ kg m}^2\) and \(R = 0.15\text{ m}\):
\(0.045 = \frac{1}{2} M (0.15)^2\)
\(0.045 = 0.5 \times M \times 0.0225 \implies 0.045 = 0.01125 M\)
\(M = \frac{0.045}{0.01125} = 4.0\text{ kg}\).

(c)(iii) Frictional torque can be reduced by applying a low-viscosity lubricant (such as machine oil) to the axle bearings, or by replacing sleeve bearings with high-quality ball bearings.

(a)
1 mark: Writing the torque balance equation \(\tau_{\text{net}} = T R - \tau_f\).
1 mark: Correctly substituting \(\tau_{\text{net}} = I \alpha\) and solving for \(\alpha\).

(b)
1 mark: Correct relationship \(a = \alpha R\) (or \(\alpha = a/R\)).

(c)(i)
1 mark: Correctly identifying the gradient as the moment of inertia \(I\).
1 mark: Correctly identifying the y-intercept as the frictional torque \(\tau_f\).

(c)(ii)
1 mark: Stating the formula \(I = \frac{1}{2} M R^2\).
1 mark: Substituting the values \(I = 0.045\) and \(R = 0.15\).
1 mark: Correct calculation of mass \(M = 4.0\text{ kg}\).

(c)(iii)
1 mark: Identifying the source of friction as the axle/bearings.
1 mark: Suggesting a valid practical method to reduce friction (lubrication or using ball bearings).