2024 IB DP Physics Practice Paper with Answers

According to Faraday's Law of induction, the magnitude of the average induced emf is given by \(\varepsilon = N \frac{\Delta \Phi}{\Delta t}\). The magnetic flux through a single turn of the coil is \(\Phi = B \cdot A = B \cdot (\pi r^2)\). Since the field decreases at a constant rate to zero, the change in flux is \(\Delta \Phi = B \pi r^2\). Thus, the magnitude of the average induced emf is \(\varepsilon = \frac{\pi N B r^2}{\Delta t}\).

[1 mark] award for correct choice A. No partial marks.

By conservation of linear momentum: \(m v = (m + 2m) v_f\), which gives the final velocity \(v_f = \frac{1}{3}v\). The initial kinetic energy is \(E_k = \frac{1}{2}mv^2\). The final kinetic energy is \(E_f = \frac{1}{2}(3m)v_f^2 = \frac{3}{2}m\left(\frac{v}{3}\right)^2 = \frac{1}{6}mv^2\). The fraction of kinetic energy remaining is \(\frac{E_f}{E_k} = \frac{1}{3}\). Therefore, the fraction lost as thermal energy is \(1 - \frac{1}{3} = \frac{2}{3}\).

[1 mark] award for correct choice B. No partial marks.

For the series connection, the total equivalent resistance is \(R_S = 3R\), so the power is \(P_S = \frac{V^2}{3R}\). For the parallel connection, the equivalent resistance is \(R_P = \frac{R}{3}\), so the power is \(P_P = \frac{V^2}{R/3} = \frac{3V^2}{R}\). The ratio is \(\frac{P_P}{P_S} = \frac{3V^2/R}{V^2/(3R)} = 9\).

[1 mark] award for correct choice B. No partial marks.

The original fringe spacing is given by \(s = \frac{\lambda D}{d}\). When the screen distance becomes \(D' = 2D\) and the slit separation becomes \(d' = \frac{d}{2}\), the new fringe spacing is \(s' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{d/2} = 4 \frac{\lambda D}{d} = 4s\).

[1 mark] award for correct choice C. No partial marks.

According to Wien's displacement law, the peak wavelength is inversely proportional to the absolute temperature: \(\lambda_{\text{max}} T = \text{constant}\). Thus, \(\lambda_X T_X = \lambda_Y T_Y\), which gives \(\frac{\lambda_Y}{\lambda_X} = \frac{T_X}{T_Y} = \frac{6000\text{ K}}{3000\text{ K}} = 2\).

[1 mark] award for correct choice B. No partial marks.

As the square loop is pulled out of the magnetic field, a motional electromotive force (emf) of magnitude \(\varepsilon = B L v\) is induced across the side of the loop remaining inside the field. This results in an induced current \(I = \frac{\varepsilon}{R} = \frac{B L v}{R}\). The magnetic force acting on this wire segment carrying current \(I\) is \(F = I L B = \left(\frac{B L v}{R}\right) L B = \frac{B^2 L^2 v}{R}\). To maintain a constant speed, the external pulling force must be equal in magnitude to this magnetic force.

[1 mark] award for correct choice B. No partial marks.

The terminal potential difference \(V\) across the cell (which is measured by the ideal voltmeter connected across it) is related to the electromotive force \(E\), internal resistance \(r\), and current \(I\) by the equation \(V = E - I r\). Rearranging this equation to solve for the current yields \(I r = E - V\), which gives \(I = \frac{E - V}{r}\).

[1 mark] award for correct choice A. No partial marks.

When unpolarized light of intensity \(I_0\) passes through the first vertical polarizer, it becomes linearly polarized with half its initial intensity, so \(I_1 = \frac{I_0}{2}\). Then, using Malus's Law for the second polarizer: \(I_2 = I_1 \cos^2(60^\circ)\). Since \(\cos(60^\circ) = \frac{1}{2}\), we get \(I_2 = \frac{I_0}{2} \left(\frac{1}{2}\right)^2 = \frac{I_0}{2} \cdot \frac{1}{4} = \frac{I_0}{8}\).

[1 mark] award for correct choice C. No partial marks.

The angular position of the first minimum of a single slit diffraction pattern is given by \(\theta \approx \frac{\lambda}{b}\). The angular width of the central maximum is the angle between the first minima on either side, which is \(2\theta \approx \frac{2\lambda}{b}\). Let \(\theta_0 = \frac{2\lambda}{b}\). When the wavelength becomes \(2\lambda\) and the slit width becomes \(\frac{b}{2}\), the new angular width is \(\theta_{\text{new}} = \frac{2(2\lambda)}{b/2} = 4 \left(\frac{2\lambda}{b}\right) = 4\theta_0\).

Award 1 mark for identifying that the angular width is proportional to \(\lambda/b\), and another 1 mark for substituting the new values to obtain the factor of 4.

The induced electromotive force (emf) \(\varepsilon\) in the loop as it leaves the magnetic field is given by Faraday's law: \(\varepsilon = B L v\) (since only one side of length \(L\) cuts the magnetic field lines). The electrical power \(P\) dissipated in the loop of resistance \(R\) is given by \(P = \frac{\varepsilon^2}{R} = \frac{(B L v)^2}{R} = \frac{B^2 L^2 v^2}{R}\).

Award 1 mark for expressing the induced emf as \(BLv\) and 1 mark for squaring the emf and dividing by the resistance to find the electrical power.

The terminal potential difference \(V\) is related to the emf \(\varepsilon\) by \(V = \varepsilon - I r\). Since the current in the circuit is \(I = \frac{\varepsilon}{R + r}\), we can rewrite this as \(V = \varepsilon \frac{R}{R + r}\). When \(R\) is very small (approaching 0), the potential difference \(V\) approaches 0. As \(R\) becomes very large (approaching infinity), \(V\) approaches \(\varepsilon\). Thus, \(V\) increases continuously from approximately zero to \(\varepsilon\).

Award 1 mark for using the terminal potential difference formula and 1 mark for correctly analyzing the limits as \(R\) increases from zero to infinity.

Let the initial direction of motion be positive. The initial momentum is \(p_i = m u\) and the final momentum is \(p_f = -m v\) since it moves in the opposite direction. The change in momentum is \(\Delta p = p_f - p_i = -m v - m u = -m(u + v)\). The magnitude of the change in momentum is \(|\Delta p| = m(u + v)\). By Newton's second law, the magnitude of the average force is \(F_{\text{avg}} = \frac{|\Delta p|}{\Delta t} = \frac{m(u + v)}{\Delta t}\).

Award 1 mark for correct signs in the change of momentum calculation and 1 mark for dividing the magnitude of momentum change by \(\Delta t\).

The root-mean-square speed of ideal gas molecules is given by \(v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}\), where \(T\) is the absolute temperature. This implies \(v_{\text{rms}} \propto \sqrt{T}\). When the temperature is doubled, the root-mean-square speed increases by a factor of \(\sqrt{2}\).

Award 1 mark for relating the root-mean-square speed to the square root of the absolute temperature, and 1 mark for obtaining the ratio \(\sqrt{2}\).

The gravitational force provides the centripetal force for the circular orbit: \(\frac{G M m}{r^2} = \frac{m v^2}{r}\), which gives the kinetic energy \(E_k = \frac{1}{2} m v^2 = \frac{G M m}{2r}\). The gravitational potential energy is \(E_p = -\frac{G M m}{r\). The total mechanical energy is \(E = E_k + E_p = \frac{G M m}{2r} - \frac{G M m}{r} = -\frac{G M m}{2r}\).

Award 1 mark for finding the kinetic energy expression \(\frac{GMm}{2r}\) and 1 mark for combining it with potential energy to find the total mechanical energy.

The radius of curvature of a charged particle in a magnetic field is given by \(r = \frac{m v}{q B}\). For the proton, \(R = \frac{m v}{q B}\). For the alpha particle, the mass is \(4m\) and the charge is \(2q\), so its radius is \(r_{\alpha} = \frac{(4m) v}{(2q) B} = 2 \left(\frac{m v}{q B}\right) = 2R\).

Award 1 mark for using the circular motion formula in a magnetic field and 1 mark for comparing the ratio of mass-to-charge for both particles to find the radius.

For an open-open pipe of length \(L\), the fundamental wavelength is \(\lambda_{\text{open}} = 2L\), so the fundamental frequency is \(f_0 = \frac{v}{2L}\). For an open-closed pipe of the same length, the fundamental wavelength is \(\lambda_{\text{closed}} = 4L\), and the fundamental frequency is \(f_{\text{new}} = \frac{v}{4L} = \frac{1}{2} \left(\frac{v}{2L}\right) = \frac{f_0}{2\).

Award 1 mark for identifying the correct fundamental wavelength relationship for both types of boundary conditions, and 1 mark for computing the ratio of the frequencies.

For a single-slit diffraction pattern, the first minimum occurs at an angle \(\theta\) given by \(\sin\theta = \frac{\lambda}{b}\). Since the angle \(\theta\) is small, we can write \(\theta \approx \frac{\lambda}{b}\). When the wavelength is doubled to \(2\lambda\) and the slit width is halved to \(\frac{b}{2}\), the new angle \(\theta'\) is given by \(\theta' \approx \frac{2\lambda}{b/2} = 4\left(\frac{\lambda}{b}\right) = 4\theta\). Therefore, the new angle is \(4\theta\).

Award 1 mark for the correct option D. Reject all other options. The method requires applying the single-slit diffraction formula with the new parameters.

By Newton's second law, the average force \(F\) is equal to the rate of change of momentum of the ball: \(F = \frac{\Delta p}{\Delta t}\). Taking the direction to the right as positive, the initial momentum is \(p_i = mv\) and the final momentum is \(p_f = -\frac{2}{3}mv\). The change in momentum is \(\Delta p = p_f - p_i = -\frac{2}{3}mv - mv = -\frac{5}{3}mv\). The magnitude of this change in momentum is \(\frac{5}{3}mv\). Therefore, the magnitude of the average force is \(F = \frac{5mv}{3\Delta t}\).

Award 1 mark for the correct option C. Method: Apply the impulse-momentum theorem correctly taking into account the directions of initial and final velocities.

As the loop is pulled out of the magnetic field, the change in magnetic flux induces an electromotive force (emf) given by Faraday's law: \(\varepsilon = B L v\). The induced current in the loop is \(I = \frac{\varepsilon}{R} = \frac{B L v}{R}\). The magnetic force opposing the motion acts on the vertical side of the loop remaining inside the field and is given by \(F = I L B = \left(\frac{B L v}{R}\right) L B = \frac{B^2 L^2 v}{R}\).

Award 1 mark for the correct option B. Method: Determine induced emf, use Ohm's law to find induced current, and apply the magnetic force formula.

The terminal potential difference \(V\) of a cell with internal resistance \(r\) and emf \(E\) is related to the current \(I\) by the equation \(V = E - I r\). This equation represents a straight line with a negative slope of \(-r\) and a vertical intercept of \(E\).

Award 1 mark for the correct option B. Method: Identify the terminal voltage relationship for a real cell and relate it to a linear equation.

By Stefan-Boltzmann law, the luminosity of a star is \(L = 4\pi R^2 \sigma T^4\). Since the luminosities are equal, \(R_X^2 T_X^4 = R_Y^2 T_Y^4\). Substituting \(T_X = T\) and \(T_Y = 2T\) gives \(R_X^2 T^4 = R_Y^2 (2T)^4 = 16 R_Y^2 T^4\). Thus, \(R_X^2 = 16 R_Y^2\), which leads to \(\frac{R_X}{R_Y} = 4\).

Award 1 mark for the correct option D. Method: Apply Stefan-Boltzmann law to both stars, set their luminosities equal, and solve for the ratio of their radii.

The frequency of a wave does not change when it moves from one medium to another, so the frequency in medium 2 remains \(f\). The wavelength is given by \(\lambda = \frac{v}{f}\). Since the speed in medium 2 is \(v_2 = \frac{v_1}{1.5}\), the wavelength in medium 2 is \(\lambda_2 = \frac{v_2}{f} = \frac{v_1}{1.5f} = \frac{\lambda_1}{1.5}\).

Award 1 mark for the correct option C. Method: State that frequency is unchanged during boundary transition and apply the relationship between wave speed and wavelength.

The escape speed from the surface of a planet is given by the formula \(v_{\text{esc}} = \sqrt{\frac{2GM}{R}}\). For the second planet, the mass is \(4M\) and the radius is \(2R\). Substituting these values into the formula gives \(v' = \sqrt{\frac{2G(4M)}{2R}} = \sqrt{2 \cdot \frac{2GM}{R}} = \sqrt{2} v_{\text{esc}}\).

Award 1 mark for the correct option B. Method: Recall the escape speed formula and determine the factor by which the escape speed changes when mass is multiplied by 4 and radius is multiplied by 2.

For an isothermal process, the temperature \(T\) remains constant. Because the average kinetic energy of the gas molecules depends only on the temperature, it remains constant, which makes statement B incorrect. According to the ideal gas law, \(pV = nRT\). Since \(T\) is constant, pressure is inversely proportional to volume. Doubling the volume results in halving the pressure, making statement C correct and statement A incorrect. Work is done by the gas during expansion, and since the internal energy is constant, thermal energy must be absorbed by the gas, making statement D incorrect.

Award 1 mark for the correct option C. Method: Apply the ideal gas law for an isothermal process where the product of pressure and volume is constant.

According to the Rayleigh criterion for a circular aperture, the minimum angular separation \(\theta\) is given by \(\theta = 1.22 \frac{\lambda}{D}\), where \(\lambda\) is the wavelength of light and \(D\) is the diameter of the aperture. For small angles, the angular separation is also given by \(\theta \approx \frac{d}{L}\), where \(d\) is the separation of the headlights and \(L\) is the maximum distance. Equating the two expressions gives \(\frac{d}{L} = 1.22 \frac{\lambda}{D}\). Rearranging for \(L\) yields \(L = \frac{d D}{1.22 \lambda}\). Substituting the values: \(L = \frac{1.22 \times (5.0 \times 10^{-3}\text{ m})}{1.22 \times (500 \times 10^{-9}\text{ m})} = \frac{5.0 \times 10^{-3}}{5.0 \times 10^{-7}} = 10^4\text{ m} = 10\text{ km}\).

Award 1 mark for applying the Rayleigh criterion correctly and performing the unit conversions to arrive at the correct distance of 10 km.

As the loop is pulled out, the magnetic flux through the loop decreases. The rate of change of the area of the loop within the magnetic field is \(\frac{\Delta A}{\Delta t} = x v\). The induced electromotive force (emf) \(\mathcal{E}\) in the loop is given by Faraday's law of induction: \(\mathcal{E} = \frac{\Delta \Phi}{\Delta t} = B \frac{\Delta A}{\Delta t} = B x v\). The power dissipated in the resistance of the loop is given by \(P = \frac{\mathcal{E}^2}{R} = \frac{(B x v)^2}{R} = \frac{B^2 x^2 v^2}{R}\).

Award 1 mark for obtaining the induced emf as \(B x v\) and applying the power equation \(P = \frac{\mathcal{E}^2}{R}\) to derive the correct expression.

In a time interval \(\Delta t\), the volume of water hitting the wall is \(\Delta V = A v \Delta t\). The mass of this water is \(\Delta m = \rho \Delta V = \rho A v \Delta t\). Since the water loses all of its momentum perpendicular to the wall, its change in momentum is \(\Delta p = \Delta m \cdot v = (\rho A v \Delta t) v = \rho A v^2 \Delta t\). According to Newton's second law, the average force exerted is \(F = \frac{\Delta p}{\Delta t} = \rho A v^2\). By Newton's third law, this is also the magnitude of the force exerted on the wall.

Award 1 mark for calculating the mass flow rate of water and applying the impulse-momentum theorem to determine the force.

The equivalent resistance of the parallel pair is \(R_{\text{parallel}} = \frac{R}{2}\). The total resistance of the circuit is \(R_{\text{total}} = R + R_{\text{parallel}} = \frac{3}{2}R\). The potential difference across the parallel combination can be found using the potential divider formula: \(V_{\text{parallel}} = V \frac{R_{\text{parallel}}}{R_{\text{total}}} = V \frac{R/2}{3R/2} = \frac{V}{3}\). The power dissipated in a single resistor of resistance \(R\) within the parallel combination is \(P = \frac{V_{\text{parallel}}^2}{R} = \frac{(V/3)^2}{R} = \frac{V^2}{9R}\).

Award 1 mark for finding the potential difference across the parallel combination as \(\frac{V}{3}\) and correctly computing the power dissipated.

According to the Stefan-Boltzmann law, the luminosity \(L\) of a star is given by \(L = 4 \pi R^2 \sigma T^4\), where \(R\) is the stellar radius and \(T\) is the surface temperature. Therefore, \(L \propto R^2 T^4\). For Star Y compared to Star X: \(\frac{L_{\text{Y}}}{L_{\text{X}}} = \left(\frac{R_{\text{Y}}}{R_{\text{X}}}\right)^2 \left(\frac{T_{\text{Y}}}{T_{\text{X}}}\right)^4 = (3)^2 \times (2)^4 = 9 \times 16 = 144\).

Award 1 mark for using the proportional relation \(L \propto R^2 T^4\) to correctly calculate the ratio as 144.

For a closed pipe of length \(L_{\text{c}}\), the fundamental wavelength is \(\lambda_{\text{c}} = 4 L_{\text{c}}\), so the fundamental frequency is \(f = \frac{v}{4 L_{\text{c}}}\). For an open pipe of length \(L_{\text{o}}\), the fundamental wavelength is \(\lambda_{\text{o}} = 2 L_{\text{o}}\). Given that \(L_{\text{o}} = 2 L_{\text{c}}\), the fundamental wavelength of the open pipe is \(\lambda_{\text{o}} = 2(2 L_{\text{c}}) = 4 L_{\text{c}}\). Therefore, its fundamental frequency is \(f_{\text{o}} = \frac{v}{\lambda_{\text{o}}} = \frac{v}{4 L_{\text{c}}} = f\).

Award 1 mark for setting up the fundamental wavelength equations for both pipe types and demonstrating that they have the same frequency.

The ideal gas equation can be written as \(P V = N k_{\text{B}} T\), where \(N\) is the number of molecules, \(V\) is the volume, and \(T\) is the absolute temperature. Rearranging for pressure gives \(P = \frac{N k_{\text{B}} T}{V}\). Since the volume \(V\) is constant, the final pressure is given by \(P_{\text{final}} = \frac{N_{\text{final}} k_{\text{B}} T_{\text{final}}}{V} = \frac{(0.5 N) k_{\text{B}} (2 T)}{V} = \frac{N k_{\text{B}} T}{V} = P\). The pressure remains unchanged.

Award 1 mark for applying the ideal gas law to show that the effects of doubling temperature and halving the number of molecules cancel out, leaving the pressure unchanged.

The total kinetic energy of a rolling object is the sum of its translational and rotational kinetic energies: \(E_{\text{total}} = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2\). Since the objects roll without slipping, their angular velocity is \(\omega = \frac{v}{R}\). For the sphere: \(E_{\text{sphere}} = \frac{1}{2} M v^2 + \frac{1}{2} \left(\frac{2}{5} M R^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{2} M v^2 + \frac{1}{5} M v^2 = \frac{7}{10} M v^2\). For the hoop: \(E_{\text{hoop}} = \frac{1}{2} M v^2 + \frac{1}{2} (M R^2) \left(\frac{v}{R}\right)^2 = \frac{1}{2} M v^2 + \frac{1}{2} M v^2 = M v^2\). The ratio is \(\frac{E_{\text{sphere}}}{E_{\text{hoop}}} = \frac{7/10 M v^2}{M v^2} = 0.7\).

Award 1 mark for expressing total kinetic energy as the sum of translational and rotational kinetic energies, correctly evaluating both for the sphere and the hoop, and calculating the ratio to be 0.7.

The induced electromotive force (emf) \(\varepsilon\) in the loop as it leaves the magnetic field is given by Faraday's law: \(\varepsilon = BLv\).

The power dissipated as thermal energy in the resistor is:
\(P = \frac{\varepsilon^2}{R} = \frac{B^2 L^2 v^2}{R}\).

The time \(t\) taken for the loop of length \(L\) to completely exit the field at a constant speed \(v\) is:
\(t = \frac{L}{v}\).

The total thermal energy \(E\) dissipated is:
\(E = P \cdot t = \frac{B^2 L^2 v^2}{R} \cdot \frac{L}{v} = \frac{B^2 L^3 v}{R}\).

[1 mark] for identifying the correct expression for energy dissipated. Award [1/1] for the correct answer B, and [0/1] for other options.

The condition for a maximum in a diffraction grating is given by \(d \sin\theta = n\lambda\).
For the first-order maximum (\(n = 1\)), the initial condition is:
\(\sin\theta = \frac{\lambda}{d}\).

For the modified setup, the new wavelength is \(\lambda' = 1.5\lambda\) and the new slit spacing is \(d' = 0.75d\). Therefore, the sine of the new first-order angle \(\theta'\) is:
\(\sin\theta' = \frac{\lambda'}{d'} = \frac{1.5\lambda}{0.75d} = 2 \left(\frac{\lambda}{d}\right) = 2\sin\theta\).

[1 mark] Award [1/1] for the correct calculation leading to D. Other options are incorrect.

Using the conservation of linear momentum in 1D:
\(p_{\text{initial}} = p_{\text{final}}\)
\(m v = m (-0.2v) + 3m v_2\)
\(1.2 m v = 3m v_2 \implies v_2 = 0.4v\)

Now, calculate the initial and final kinetic energies:
\(E_{ki} = \frac{1}{2} m v^2\)
\(E_{kf} = \frac{1}{2} m (-0.2v)^2 + \frac{1}{2} (3m) (0.4v)^2 = \frac{1}{2} m v^2 [0.04 + 3 \times 0.16] = \frac{1}{2} m v^2 [0.52]\)

The kinetic energy remaining is \(52\%\) of the initial value, meaning the fraction of kinetic energy lost is:
\(1 - 0.52 = 0.48\).

[1 mark] Award [1/1] for the correct identification of the lost kinetic energy fraction. Correct answer is B.

The current in the circuit is given by \(I = \frac{E}{R + r}\).
The power dissipated in the external resistor \(R\) is \(P = I^2 R = \frac{E^2 R}{(R+r)^2}\).

For \(R = r\):
\(P = \frac{E^2 r}{(2r)^2} = \frac{E^2}{4r}\).

For \(R' = 3r\):
\(P' = \frac{E^2 (3r)}{(3r+r)^2} = \frac{3E^2 r}{16r^2} = \frac{3 E^2}{16r}\).

Expressing \(P'\) in terms of \(P\):
\(P' = \frac{3}{16} \left(4P\right) = \frac{12}{16} P = 0.75P\).

[1 mark] Award [1/1] for finding the correct ratio of power resulting in C. All other options are incorrect.

According to Wien's displacement law, the temperature \(T\) of a star is inversely proportional to its peak wavelength \(\lambda\):
\(T \propto \frac{1}{\lambda}\)
Since \(\lambda_X = 2\lambda_Y\), the temperature relation is:
\(T_X = \frac{1}{2} T_Y\).

The Stefan-Boltzmann law states that luminosity \(L = 4\pi R^2 \sigma T^4\). Since both stars have the same luminosity (\(L_X = L_Y\)):
\(R_X^2 T_X^4 = R_Y^2 T_Y^4 \implies \left(\frac{R_X}{R_Y}\right)^2 = \left(\frac{T_Y}{T_X}\right)^4\)

Substituting \(\frac{T_Y}{T_X} = 2\):
\(\left(\frac{R_X}{R_Y}\right)^2 = 2^4 = 16 \implies \frac{R_X}{R_Y} = 4\).

[1 mark] Award [1/1] for combining Wien's law and Stefan-Boltzmann law correctly to find \(R_X/R_Y = 4\) (option D).

Without the diode, the power is dissipated during both halves of the AC cycle. The average power is \(P = \frac{V_{\text{rms}}^2}{R} = \frac{V_0^2}{2R}\).

When a diode is connected in series, current only flows during one half of the cycle (when the diode is forward-biased), and is zero during the other half. Because the voltage shape across the resistor during the active half-cycle is identical to the original unrectified half-cycle, the energy delivered per full cycle is exactly halved.

Since power is energy delivered per unit time and the period remains the same, the average power dissipated is halved: \(P' = 0.50P\).

[1 mark] Award [1/1] for realizing that half-wave rectification halves the total energy per period, hence halving the average power.

Let's examine the phase changes upon reflection:
1. At the air-to-oil interface (\(n_{\text{air}} = 1.0 \to n_{\text{oil}} = 1.4\)): Since light goes from a lower to a higher refractive index, a phase shift of \(\pi\) radians (equivalent to a path difference of \(\frac{\lambda}{2}\)) occurs.
2. At the oil-to-glass interface (\(n_{\text{oil}} = 1.4 \to n_{\text{glass}} = 1.5\)): Since light again goes from a lower to a higher refractive index, a phase shift of \(\pi\) radians occurs.

Since both reflected rays undergo the same phase change of \(\pi\), their relative phase shift due to reflection is zero. Therefore, to achieve destructive interference, the path difference in the film must introduce an odd half-wavelength phase difference:
\(2 n_{\text{oil}} t = \left(m + \frac{1}{2}\right) \lambda\) for \(m = 0, 1, 2, \dots\)

For the minimum non-zero thickness (\(m = 0\)):
\(2 n_{\text{oil}} t = \frac{\lambda}{2}\)
\(t = \frac{\lambda}{4 n_{\text{oil}}} = \frac{\lambda}{4 \times 1.4} = \frac{\lambda}{5.6}\).

[1 mark] Award [1/1] for correctly analyzing the phase changes and calculating the thickness to be \(\frac{\lambda}{5.6}\).

The total impulse \(J\) delivered to the body is equal to the area under the force-time graph.

The force-time graph forms a triangle with a base of \(t = 5.0\text{ s}\) and a peak height of \(F = 12\text{ N}\).

\(J = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5.0\text{ s} \times 12\text{ N} = 30\text{ N s}\).

According to the impulse-momentum theorem:
\(J = \Delta p = m \Delta v\)

Since the body starts from rest (\(u = 0\)):
\(30\text{ N s} = (2.0\text{ kg}) \cdot v\)
\(v = 15\text{ m s}^{-1}\).

[1 mark] Award [1/1] for obtaining the correct impulse (30 N s) and calculating the final speed as 15 m s\(^{-1}\).

(a) An emf is induced when there is a change in magnetic flux linkage. As the loop enters the field, the area inside the field increases, increasing the flux linkage and inducing an emf. Once fully inside, the flux linkage is constant, so no emf is induced. (b) The induced emf is \(\varepsilon = B l v = 0.45 \times 0.12 \times 3.2 = 0.1728\text{ V}\). The current is \(I = \varepsilon / R = 0.1728 / 2.5 = 0.06912\text{ A}\) (or \(6.9 \times 10^{-2}\text{ A}\)). (c) The magnetic force on the loop is \(F = B I l = 0.45 \times 0.06912 \times 0.12 = 0.003732\text{ N}\). For constant speed, the external force must equal this magnetic force: \(F_{\text{ext}} = 3.7 \times 10^{-3}\text{ N}\).

(a) Award [1] for stating that emf depends on the rate of change of flux linkage, and [1] for explaining that flux linkage only changes while the area in the field is changing. (b) Award [1] for calculating emf, [1] for using Ohm's law, and [1] for correct final current with units. (c) Award [1] for the formula for magnetic force, [1] for linking the external force to the magnetic force (constant velocity), and [1] for correct numerical answer.

(a) A phase change of \(\pi\) occurs when reflection is off a medium of higher refractive index. The air-oil interface reflects light going from air (n=1.00) to oil (n=1.45), so a \(\pi\) phase change occurs. The oil-water interface reflects light going from oil (n=1.45) to water (n=1.33), so no phase change occurs. (b) With one phase change, the condition for constructive interference is \(2 n_{\text{oil}} d = (m + 0.5)\lambda\). For minimum thickness, \(m = 0\), so \(d = \lambda / (4 n_{\text{oil}}) = 580\text{ nm} / (4 \times 1.45) = 100\text{ nm}\). (c) When white light is used, different wavelengths satisfy the constructive interference condition at different film thicknesses. As a result, a pattern of colored bands is observed, where each band corresponds to a region where a specific color is strongly reflected.

(a) Award [1] for identifying air-oil boundary and [1] for explaining that no phase change occurs at oil-water boundary because the refractive index decreases. (b) Award [1] for using the correct constructive interference condition with one phase change, [1] for substituting values, and [1] for final correct thickness (100 nm). (c) Award [1] for stating that colored bands are observed, [1] for explaining that white light consists of multiple wavelengths, and [1] for stating that different wavelengths interfere constructively at different thicknesses.

(a) Using conservation of momentum: \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\). Substituting the values: \(0.60 \times 3.5 + 0 = 0.60 \times 1.1 + 0.40 \times v_2\). This gives \(2.1 = 0.66 + 0.40 v_2\), leading to \(0.40 v_2 = 1.44\), so \(v_2 = 3.6\text{ m s}^{-1}\). (b) Initial kinetic energy: \(E_{k,i} = 0.5 \times 0.60 \times 3.5^2 = 3.675\text{ J}\). Final kinetic energy: \(E_{k,f} = 0.5 \times 0.60 \times 1.1^2 + 0.5 \times 0.40 \times 3.6^2 = 0.363 + 2.592 = 2.955\text{ J}\). Since \(E_{k,f} < E_{k,i}\), kinetic energy is lost, hence the collision is inelastic. (c) Impulse is defined as the change in momentum: \(J = \Delta p = m_2 v_2 - m_2 u_2 = 0.40 \times 3.6 = 1.44\text{ N s}\) (or \(\text{kg m s}^{-1}\)).

(a) Award [1] for writing the momentum equation, [1] for correct substitution, and [1] for obtaining 3.6 m s^-1. (b) Award [1] for calculating initial kinetic energy, [1] for calculating final kinetic energy, and [1] for concluding it is inelastic because kinetic energy decreases. (c) Award [1] for the definition of impulse (change in momentum), and [1] for calculating 1.44 N s.

(a) Electromotive force (emf) is the total energy transferred (or work done) per unit charge by the cell in moving charge around a complete circuit. (b) Maximum power is delivered when the external resistance equals the internal resistance: \(R = r = 1.2\ \Omega\). The current is then \(I = \varepsilon / (R+r) = 6.0 / (1.2 + 1.2) = 2.5\text{ A}\). The maximum power is \(P = I^2 R = 2.5^2 \times 1.2 = 7.5\text{ W}\). (c) The terminal potential difference is \(V = \varepsilon - I r\). This is a straight line with a negative gradient of \(-1.2\ \Omega\). The vertical (V) intercept is at \((0, 6.0\text{ V})\) and the horizontal (I) intercept is at \((5.0\text{ A}, 0)\).

(a) Award [1] for stating work done per unit charge, and [1] for specifying complete circuit (or total energy supplied). (b) Award [1] for identifying R = r = 1.2 ohms, [1] for calculating current (2.5 A), and [1] for power (7.5 W). (c) Award [1] for showing a straight line with negative gradient, [1] for identifying the V-intercept as 6.0 V, and [1] for identifying the I-intercept as 5.0 A.

(a) Using Wien's displacement law: \(\lambda_{\text{max}} = 2.90 \times 10^{-3} / T = 2.90 \times 10^{-3} / 3780 = 7.67 \times 10^{-7}\text{ m}\). (b) Using the Stefan-Boltzmann law: \(L = \sigma (4 \pi R^2) T^4 = 5.67 \times 10^{-8} \times 4 \pi \times (3.6 \times 10^8)^2 \times 3780^4 \approx 1.88 \times 10^{25}\text{ W}\). (c) Main sequence stars are stable due to hydrostatic equilibrium, where the inward force of gravity is balanced by the outward radiation pressure. The core energy source maintaining this pressure is the nuclear fusion of hydrogen into helium via the proton-proton chain.

(a) Award [1] for Wien's displacement law formula, and [1] for correct peak wavelength with units. (b) Award [1] for Stefan-Boltzmann formula, [1] for calculating surface area, and [1] for correct luminosity value with units. (c) Award [1] for identifying hydrostatic equilibrium, [1] for describing gravitational force balanced by outward pressure, and [1] for naming hydrogen fusion (or proton-proton chain).

(a) (i) The displacement standing wave pattern has a node at the closed end and an antinode at the open end. (ii) For a closed-open pipe, \(L = \lambda / 4 \implies \lambda = 4 L = 4 \times 0.68 = 2.72\text{ m}\). The frequency is \(f_1 = v / \lambda = 340 / 2.72 = 125\text{ Hz}\). (b) (i) For an open-open pipe, \(L = \lambda / 2 \implies \lambda = 2 L = 2 \times 0.68 = 1.36\text{ m}\). The frequency is \(f_1' = v / \lambda = 340 / 1.36 = 250\text{ Hz}\). (ii) At the open ends, the tube is open to the atmosphere, meaning the pressure variation is zero (a pressure node). Since pressure variation and air displacement are 90 degrees out of phase, a pressure node corresponds to a displacement antinode, allowing maximum amplitude of physical vibration of the air molecules.

(a)(i) Award [1] for stating node at closed end and antinode at open end. (a)(ii) Award [1] for calculating wavelength (2.72 m) and [1] for frequency (125 Hz). (b)(i) Award [1] for calculating wavelength (1.36 m) and [1] for frequency (250 Hz). (b)(ii) Award [1] for mentioning open ends are at atmospheric pressure (pressure node), [1] for relating pressure node to maximum displacement, and [1] for concluding a displacement antinode forms.

(a) From \(PV = nRT\), \(P = nRT / V = 0.18 \times 8.31 \times 300 / (1.5 \times 10^{-3}) = 2.9916 \times 10^5\text{ Pa} \approx 3.0 \times 10^5\text{ Pa}\). (b) (i) Since volume doubles at constant pressure, Charles's Law gives \(T_2 = 2 T_1 = 600\text{ K}\). (ii) Work done is \(W = P \Delta V = P (V_2 - V_1) = 2.9916 \times 10^5 \times 1.5 \times 10^{-3} = 448.74\text{ J} \approx 450\text{ J}\). (c) In an ideal gas, particles are assumed to have zero potential energy (no intermolecular forces) and negligible size. For real gases at low temperatures, particles slow down, meaning attractive intermolecular forces are no longer negligible, and the volume occupied by the particles themselves becomes significant relative to the gas volume.

(a) Award [1] for rearranging ideal gas law, and [1] for correct calculation of pressure with units. (b)(i) Award [1] for identifying volume-temperature proportionality, and [1] for final temperature (600 K). (b)(ii) Award [1] for work formula \(W = P\Delta V\) or \(W=nR\Delta T\), and [1] for correct work (449 J or 450 J). (c) Award [1] for mentioning intermolecular forces in real gases, and [1] for mentioning finite particle volume.

(a) The activity ratio is \(105 / 840 = 0.125 = (1/2)^3\), which represents 3 half-lives. Thus, the half-life is \(18.0\text{ hours} / 3 = 6.0\text{ hours}\). (b) Converting the half-life to seconds: \(T_{1/2} = 6.0 \times 3600 = 21600\text{ s}\). The decay constant is \(\lambda = \ln 2 / T_{1/2} = 0.693 / 21600 = 3.209 \times 10^{-5}\text{ s}^{-1}\). (c) The initial activity is \(A_0 = \lambda N_0\). Rearranging gives \(N_0 = A_0 / \lambda = 840 / (3.209 \times 10^{-5}) = 2.618 \times 10^7\) nuclei.

(a) Award [1] for identifying that 3 half-lives have elapsed, and [1] for calculating half-life of 6.0 hours. (b) Award [1] for converting hours to seconds, [1] for the formula relating decay constant and half-life, and [1] for correct decay constant with units. (c) Award [1] for the formula \(A = \lambda N\), [1] for substituting values, and [1] for correct final number of nuclei.

(a) Faraday's law states that the magnitude of the induced electromotive force (emf) in a circuit is directly proportional to (or equal to) the rate of change of magnetic flux linkage through the circuit: \(\varepsilon = -\frac{\Delta (N\Phi)}{\Delta t}\).

(b) First, calculate the area of the square coil:
\(A = s^2 = (0.050\text{ m})^2 = 2.5 \times 10^{-3}\text{ m}^2\).

Next, determine the initial magnetic flux through a single turn:
\(\Phi_i = B \cdot A = 0.15\text{ T} \times 2.5 \times 10^{-3}\text{ m}^2 = 3.75 \times 10^{-4}\text{ Wb}\).

The initial flux linkage for \(N\) turns is:
\(N\Phi_i = 250 \times 3.75 \times 10^{-4}\text{ Wb} = 0.09375\text{ Wb}\).

Since the final magnetic field is zero, the final flux linkage \(N\Phi_f = 0\).

The average induced emf is:
\(\varepsilon = \frac{\Delta(N\Phi)}{\Delta t} = \frac{0.09375\text{ Wb} - 0}{0.080\text{ s}} = 1.172\text{ V} \approx 1.2\text{ V}\).

(c) Assuming the magnetic field decreases at a constant rate, the induced emf remains constant at \(\varepsilon = 1.172\text{ V}\).
The rate of thermal energy dissipation (power) is:
\(P = \frac{\varepsilon^2}{R} = \frac{(1.171875\text{ V})^2}{4.5\ \Omega} = 0.305\text{ W}\).

The total thermal energy dissipated during \(\Delta t = 0.080\text{ s}\) is:
\(E = P \cdot \Delta t = 0.305\text{ W} \times 0.080\text{ s} = 0.0244\text{ J} \approx 0.024\text{ J}\) (or \(2.4 \times 10^{-2}\text{ J}\)).

(a)
- Induced emf is proportional to / equal to the rate of change of magnetic flux linkage. [1]
- Clearly defines variables or mentions "magnetic flux linkage" (product of flux and number of turns). [1]

(b)
- Correct calculation of area \(A = 2.5 \times 10^{-3}\text{ m}^2\). [1]
- Correct formulation of change in flux linkage \(\Delta (N\Phi) = 0.094\text{ Wb}\). [1]
- Final answer \(1.2\text{ V}\) (must be 2 significant figures to match data). [1]

(c)
- Correct use of \(P = \frac{V^2}{R}\) or \(E = \frac{V^2}{R}t\). [1]
- Final answer \(0.024\text{ J}\) (or \(2.4 \times 10^{-2}\text{ J}\)) consistent with their (b). [1]
- States the assumption that the rate of change of magnetic field is constant / uniform (allowing constant emf). [1]

(a)
(i) Front surface: Light travels from air (lower refractive index) to the film (higher refractive index). Therefore, reflection occurs at a optically denser boundary, resulting in a phase change of \(\pi\) radians (or \(180^\circ\)).
(ii) Back surface: Light travels from the film to air (lower refractive index). Therefore, reflection occurs at an optically less dense boundary, resulting in no phase change (\(0\) radians).

(b)
Let \(d\) be the thickness of the film.
The path difference between the wave reflected from the front surface and the wave reflected from the back surface is \(2dn\).
Because there is a \(\pi\) phase change at the first reflection and no phase change at the second, the relative phase change from reflection is \(\pi\) (equivalent to a path difference of \(\frac{\lambda}{2}\)).
For destructive interference to occur, the total phase difference must be an odd multiple of \(\pi\). This means the path difference must satisfy:
\(2dn = m\lambda\) where \(m = 1, 2, 3, \dots\)

For the minimum non-zero thickness, we choose \(m = 1\):
\(2dn = \lambda\)
\(d = \frac{\lambda}{2n} = \frac{632.8 \times 10^{-9}\text{ m}}{2 \times 1.33} = 2.3789 \times 10^{-7}\text{ m} \approx 2.38 \times 10^{-7}\text{ m}\) (or \(238\text{ nm}\)).

(c)
- The film will show horizontal alternate bright and dark bands (fringes).
- This is because as we move down the wedge, the thickness \(d\) increases continuously, causing the path difference to periodically satisfy the conditions for constructive and destructive interference.
- At the very top where the film is extremely thin (thickness \(d \approx 0\)), the path difference is negligible. However, due to the \(\pi\) phase change at the front surface, the two reflected waves are out of phase by \(\pi\), resulting in destructive interference. Thus, the top edge of the film appears completely dark.

(a)
(i) \(\pi\) rad (or \(180^\circ\) or half-cycle). [1]
(ii) Zero (or no phase change). [1]

(b)
- Correct condition for destructive interference: \(2dn = m\lambda\) (or equivalent reasoning showing awareness of the phase change). [1]
- Substitution of values: \(2 \times d \times 1.33 = 632.8 \times 10^{-9}\). [1]
- Final answer \(2.4 \times 10^{-7}\text{ m}\) or \(238\text{ nm}\) (accept \(2.38 \times 10^{-7}\text{ m}\)). [1]

(c)
- Mentions horizontal alternate bright and dark bands/fringes. [1]
- Explains that the bands correspond to progressive changes in film thickness yielding alternate constructive and destructive interference. [1]
- Explains that the top of the wedge is dark because the thickness is nearly zero, meaning the phase difference is solely due to the \(\pi\) phase change on reflection. [1]

(a) By the law of conservation of linear momentum (since no external horizontal forces act during the collision):
\(p_{\text{initial}} = p_{\text{final}}\)
\(m_A v_0 = (m_A + m_B) v\)

Substitute the given values:
\(1.5\text{ kg} \times 4.0\text{ m s}^{-1} = (1.5\text{ kg} + 2.5\text{ kg}) \times v\)
\(6.0\text{ kg m s}^{-1} = 4.0\text{ kg} \times v\)
\(v = \frac{6.0}{4.0} = 1.5\text{ m s}^{-1}\). (This is exactly as shown.)

(b) Calculate initial kinetic energy:
\(E_{k,i} = \frac{1}{2} m_A v_0^2 = \frac{1}{2} (1.5\text{ kg}) (4.0\text{ m s}^{-1})^2 = 0.5 \times 1.5 \times 16 = 12.0\text{ J}\).

Calculate final kinetic energy:
\(E_{k,f} = \frac{1}{2} (m_A + m_B) v^2 = \frac{1}{2} (4.0\text{ kg}) (1.5\text{ m s}^{-1})^2 = 0.5 \times 4.0 \times 2.25 = 4.5\text{ J}\).

Loss of kinetic energy:
\(\Delta E_k = E_{k,i} - E_{k,f} = 12.0\text{ J} - 4.5\text{ J} = 7.5\text{ J}\).

(c)
(i) The normal reaction force \(R\) on the combined mass is:
\(R = (m_A + m_B) g = 4.0\text{ kg} \times 9.81\text{ m s}^{-2} = 39.24\text{ N}\).

The frictional force \(F_f\) is:
\(F_f = \mu_d R = 0.30 \times 39.24\text{ N} = 11.77\text{ N} \approx 12\text{ N}\).

(ii) Using the work-energy theorem, the work done by friction equals the change in kinetic energy of the combined blocks:
\(W = -F_f \cdot d = -E_{k,f}\)
\(11.77\text{ N} \times d = 4.5\text{ J}\)
\(d = \frac{4.5\text{ J}}{11.77\text{ N}} = 0.382\text{ m} \approx 0.38\text{ m}\).

Alternatively, using kinematics:
\(a = -\frac{F_f}{m} = -\frac{11.77}{4.0} = -2.94\text{ m s}^{-2}\).
Using \(v_f^2 = u^2 + 2ad\) with \(v_f = 0\) and \(u = 1.5\text{ m s}^{-1}\):
\(0 = 1.5^2 + 2(-2.94)d\)
\(d = \frac{2.25}{5.88} \approx 0.38\text{ m}\).

(a)
- Applies conservation of momentum equation: \(m_A v_0 = (m_A + m_B) v\). [1]
- Substitutes values correctly to show \(v = 1.5\text{ m s}^{-1}\). [1]

(b)
- Calculates initial kinetic energy (\(12\text{ J}\)) and final kinetic energy (\(4.5\text{ J}\)). [1]
- Subtracts values to get \(7.5\text{ J}\). [1]

(c)
(i)
- Determines normal force \(R = 39.2\text{ N}\) (or \(40\text{ N}\) if \(g=10\) is used). [1]
- Calculates frictional force \(F_f = 11.8\text{ N}\) (or \(12\text{ N}\)). [1]
(ii)
- Sets up work-energy balance (\(F_f d = \Delta E_k\)) or uses correct kinematic equations. [1]
- Calculates distance \(d = 0.38\text{ m}\) (accept answers in range \(0.38\text{ m}\) to \(0.39\text{ m}\) depending on rounding of friction force/g). [1]

(a) Rearranging the ideal gas law: \(P(V + V_{\text{tubing}}) = nRT \implies V + V_{\text{tubing}} = \frac{nRT}{P} \implies V = nRT\left(\frac{1}{P}\right) - V_{\text{tubing}}\).

(b) The equation is of the form \(y = mx + c\), where the y-intercept \(c = -V_{\text{tubing}}\). First, find the gradient: \(m = \frac{48.0 - 8.0}{5.0 \times 10^{-5} - 1.0 \times 10^{-5}} = \frac{40.0}{4.0 \times 10^{-5}} = 1.0 \times 10^6\text{ cm}^3\text{ Pa}\). Using \(V = m\left(\frac{1}{P}\right) - V_{\text{tubing}}\) at the first point: \(8.0 = (1.0 \times 10^6)(1.0 \times 10^{-5}) - V_{\text{tubing}} \implies 8.0 = 10.0 - V_{\text{tubing}} \implies V_{\text{tubing}} = 2.0\text{ cm}^3\).

(c) The gradient \(m = 1.0 \times 10^6\text{ cm}^3\text{ Pa}\). Converting \(\text{cm}^3\) to \(\text{m}^3\): \(1\text{ cm}^3 = 10^{-6}\text{ m}^3\). Therefore, \(m = 1.0 \times 10^6 \times 10^{-6}\text{ m}^3\text{ Pa} = 1.0\text{ Pa m}^3\) or \(1.0\text{ J}\).

(d) The gradient is equal to \(nRT\). Thus: \(nRT = 1.0\text{ J} \implies n = \frac{1.0}{8.31 \times 290} = 4.15 \times 10^{-4}\text{ mol}\). To two significant figures, \(n = 4.2 \times 10^{-4}\text{ mol}\).

(a) Award [1] for the correct rearranged expression: \(V = nRT\left(\frac{1}{P}\right) - V_{\text{tubing}}\).

(b) Award [1] for calculating the gradient or setting up the linear equation, and [1] for the correct value of the volume of the tubing: \(2.0\text{ cm}^3\) (allow \(-2.0\text{ cm}^3\) if vertical intercept is quoted as the final answer).

(c) Award [1] for calculating the numerical gradient value (e.g., \(1.0 \times 10^6\) or \(1.0\)), [1] for performing the correct unit conversion (from \(\text{cm}^3\) to \(\text{m}^3\)), and [1] for the correct SI unit (\(\text{J}\) or \(\text{Pa m}^3\) or \(\text{N m}\)).

(d) Award [1] for setting \(nRT = \text{gradient}\) in SI units, and [1] for the correct final answer: \(4.2 \times 10^{-4}\text{ mol}\) (accept range \(4.1 \times 10^{-4}\) to \(4.2 \times 10^{-4}\)).

(a) When the frequency is zero, the magnetic field is static, meaning there is no change in magnetic flux over time. By Faraday's law, the induced emf must be zero. Thus, the curve must pass through \((0,0)\).

(b) The gradient is: \(m = \frac{\Delta \varepsilon_0}{\Delta f} = \frac{60.0 \times 10^{-3}\text{ V} - 0}{500\text{ Hz} - 0} = 1.20 \times 10^{-4}\text{ V s}\) (or \(\text{V Hz}^{-1}\)).

(c) The theoretical formula gives \(\varepsilon_0 = (2\pi N A B_0) f\), so the gradient \(m = 2\pi N A B_0\). Rearranging for \(B_0\): \(B_0 = \frac{m}{2\pi N A} = \frac{1.20 \times 10^{-4}}{2\pi \times 500 \times 2.0 \times 10^{-4}} = \frac{1.20 \times 10^{-4}}{0.6283} = 1.91 \times 10^{-4}\text{ T}\). To two significant figures, \(B_0 = 1.9 \times 10^{-4}\text{ T}\).

(d) The percentage uncertainty is \(\frac{3.0\text{ mV}}{36.0\text{ mV}} \times 100\% \approx 8.3\%\). To find the uncertainty in the gradient, the student should draw a worst-fit line (either the steepest or shallowest line of fit) that still passes through the error bars of all the data points. The absolute uncertainty in the gradient is then calculated as the absolute difference between the gradient of the best-fit line and the gradient of this worst-fit line: \(\Delta m = |m_{\text{best}} - m_{\text{worst}}|\).

(a) Award [1] for explaining that zero frequency means a constant magnetic field/flux, so the rate of change of flux is zero, leading to zero induced emf.

(b) Award [1] for a correct calculation of the gradient magnitude (\(1.2 \times 10^{-4}\)), and [1] for the correct SI unit (\(\text{V s}\) or \(\text{V/Hz}\) or \(\text{kg m}^2 \text{A}^{-1} \text{s}^{-2}\)).

(c) Award [1] for equating the gradient to \(2\pi N A B_0\) and rearranging correctly, and [1] for the correct final value with unit: \(1.9 \times 10^{-4}\text{ T}\) (accept \(1.91 \times 10^{-4}\text{ T}\); allow ECF from (b)).

(d) Award [1] for \(8.3\%\) (accept range \(8.3\%\) to \(8.4\%\)). Award [1] for describing the worst-fit line as the steepest or shallowest line through the error bars, and calculating the uncertainty as the difference between the best-fit and worst-fit gradients.

(a) Distance in parsecs:
\(d = \frac{1}{p} = \frac{1}{0.125} = 8.0\text{ pc}\).

Conversion to meters:
\(1\text{ pc} = 3.09 \times 10^{16}\text{ m}\), so
\(d = 8.0 \times 3.09 \times 10^{16}\text{ m} = 2.47 \times 10^{17}\text{ m}\) (accept \(2.5 \times 10^{17}\text{ m}\)).

(b) Using the luminosity-brightness equation:
\(b = \frac{L}{4\pi d^2}\)

\(L = b \times 4\pi d^2 = (2.5 \times 10^{-10}\text{ W m}^{-2}) \times 4\pi \times (2.47 \times 10^{17}\text{ m})^2 = 1.91 \times 10^{26}\text{ W}\) (accept range \(1.9 \times 10^{26}\text{ W}\) to \(2.0 \times 10^{26}\text{ W}\)).

(c) The Earth's atmosphere causes 'seeing' or turbulence, which blurs the images of stars. This limits the resolution of the parallax angle measurement from the ground to around \(0.01\text{ arcseconds}\), restricting the maximum distance that can be measured from the ground to about \(100\text{ pc}\).

(a)
[1] For correctly calculating \(d = 8.0\text{ pc}\).
[1] For correctly converting to meters to get \(2.47 \times 10^{17}\text{ m}\) (or \(2.5 \times 10^{17}\text{ m}\)).

(b)
[1] For showing correct substitution into the formula \(L = 4\pi d^2 b\).
[1] For final value of \(1.9 \times 10^{26}\text{ W}\) (allow ECF from part a).

(c)
[1] For identifying atmospheric turbulence / refractive index fluctuations / atmospheric 'seeing' effect.
[1] For connecting this to a limit in angular resolution or stating that this limits ground-based measurements to distances less than \(100\text{ pc}\) (or angles greater than \(0.01\text{ arcsec}\)).

(a) The variation is caused by the periodic expansion and contraction of the star's outer layers. When the star is compressed, helium in its atmosphere becomes doubly ionized and more opaque, trapping radiation and increasing internal pressure. This pressure causes the outer layers to expand. As they expand, they cool, helium recombines to become singly ionized and more transparent, allowing radiation to escape. The pressure drops, and gravity pulls the layers back inward, repeating the cycle.

(b)
1. Observe the Cepheid variable over time and measure its period of pulsation.
2. Use the known Period-Luminosity relationship (established from nearby Cepheids) to find the average luminosity \(L\) of the star.
3. Measure the average apparent brightness \(b\) of the star as received on Earth.
4. Use the inverse-square law \(b = \frac{L}{4\pi d^2}\) to calculate the distance \(d\) to the Cepheid (and therefore the host galaxy).

(a)
[1] For identifying the periodic expansion and contraction / pulsation of the outer layers.
[1] For explaining the role of changing opacity / ionization of helium trapping and releasing energy.

(b)
[1] For stating that the period of pulsation of the Cepheid is measured.
[1] For stating that the luminosity \(L\) is found using the Period-Luminosity relationship.
[1] For stating that the apparent brightness \(b\) is measured.
[1] For stating that the distance \(d\) is calculated using \(b = \frac{L}{4\pi d^2}\) (equation must be explicitly stated or described).

(a) Redshift \(z\) is related to the scale factors by \(1 + z = \frac{R_{\text{now}}}{R_{\text{then}}}\).
Therefore, \(\frac{R_{\text{then}}}{R_{\text{now}}} = \frac{1}{1 + z} = \frac{1}{1 + 0.40} = \frac{1}{1.40} \approx 0.71\).

(b) Using the definition of redshift: \(z = \frac{\lambda_{\text{obs}} - \lambda_0}{\lambda_0} \implies \lambda_{\text{obs}} = \lambda_0 (1 + z)\).
\(\lambda_{\text{obs}} = 656\text{ nm} \times (1 + 0.40) = 918.4\text{ nm} \approx 918\text{ nm}\).

(c) The Big Bang model predicts that the early universe was extremely hot, dense, and filled with high-energy radiation. As the universe expanded, this radiation cooled and its wavelength stretched (cosmological redshift). The CMB is this remnant radiation, which is highly isotropic and homogeneous, with a perfect blackbody spectrum at a temperature of about \(2.7\text{ K}\), matching the predictions of an expanding universe.

(a)
[1] For using the relationship \(1 + z = \frac{R_{\text{now}}}{R_{\text{then}}}\).
[1] For calculating the ratio as \(0.71\) (or \(5/7\)).

(b)
[1] For using \(\lambda_{\text{obs}} = \lambda_0 (1 + z)\).
[1] For calculating \(918\text{ nm}\) (or \(9.18 \times 10^{-7}\text{ m}\)).

(c)
[1] For stating that CMB radiation is highly isotropic / homogeneous / has a blackbody spectrum at \(2.7\text{ K}\).
[1] For explaining that it represents the highly redshifted (cooled) remnant of radiation from the early, hot, dense universe.

(a) Star A (low mass, \(< 8 M_{\odot}\)) will end its life cycle as a white dwarf.
Star B (high mass, \(> 8 M_{\odot}\)) will end its life cycle as a neutron star (or potentially a black hole after a supernova explosion, though a \(15 M_{\odot}\) main sequence star typically leaves a neutron star remnant).

(b) In Star A's remnant (white dwarf), gravity acts to contract the core. Because the electrons are squeezed extremely close together, the Pauli exclusion principle prevents them from occupying the same quantum states. This creates an outward electron degeneracy pressure that exactly balances the inward gravitational force, preventing further collapse.

(c) The Oppenheimer-Volkoff limit (approximately \(2\text{ to }3 M_{\odot}\)) is the maximum mass that a neutron star can support against gravitational collapse. If the mass of Star B's remnant core exceeds this limit, neutron degeneracy pressure is insufficient to halt gravity, and the core will collapse past the neutron star stage to form a black hole.

(a)
[1] Star A: White dwarf.
[1] Star B: Neutron star (accept Black Hole).

(b)
[1] For stating that the Pauli exclusion principle prevents electrons from occupying the same quantum state when compressed.
[1] For explaining that this creates an outward pressure (electron degeneracy pressure) that balances the inward gravitational pull.

(c)
[1] For defining the Oppenheimer-Volkoff limit as the maximum mass of a neutron star.
[1] For stating that if the core mass exceeds this limit, the remnant collapses into a black hole.

(a) The orbital speed \(v\) is expected to decrease with distance \(r\) according to a Keplerian decline: \(v \propto r^{-1/2}\) (or \(v \propto \frac{1}{\sqrt{r}}\)).

(b)
1. The observed rotation curves are flat / orbital speeds remain constant (or slightly increase) at large distances from the galactic center.
2. This flat curve implies that the mass of the galaxy \(M(r)\) enclosed within radius \(r\) continues to increase with distance (specifically \(M(r) \propto r\)), even far beyond the visible edge of the galaxy.
3. Therefore, there must be a massive halo of non-luminous (dark) matter surrounding the visible galaxy.

(c) 'WIMPs' stands for Weakly Interacting Massive Particles. These are hypothetical subatomic particles that possess mass (and thus exert gravitational force) but do not interact via the electromagnetic force (they are dark / emit no light) and only interact through gravity and the weak nuclear force. This allows them to cluster in galactic halos without being visible, perfectly explaining the missing mass.

(a)
[1] For stating that speed is expected to decrease with distance (or \(v \propto r^{-1/2}\)).

(b)
[1] For stating that observed speed remains constant / flat with distance.
[1] For stating that this implies the mass \(M\) increases with radius \(r\) (or \(M \propto r\)).
[1] For concluding that there is a large amount of invisible / dark matter in a halo surrounding the galaxy.

(c)
[1] For identifying the acronym: Weakly Interacting Massive Particles.
[1] For explaining that they have mass (provide gravitational pull) but do not interact electromagnetically (thus emit/absorb no light).