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The power radiated by a black body is given by Stefan-Boltzmann law: \(P = \sigma A T^4\), where \(A = 4\pi R^2\) is the surface area. Thus, \(P = 4\pi \sigma R^2 T^4\). Comparing the two stars: \(P_X \propto R_X^2 T_X^4\) and \(P_Y \propto R_Y^2 T_Y^4 = \left(\frac{R_X}{2}\right)^2 (2T_X)^4 = \frac{R_X^2}{4} \times 16 T_X^4 = 4 R_X^2 T_X^4\). Therefore, the ratio of the power radiated by star X to star Y is \(\frac{P_X}{P_Y} = \frac{R_X^2 T_X^4}{4 R_X^2 T_X^4} = \frac{1}{4} = 0.25\).

Award [1] for the correct option. Award [0] otherwise.

The energy of an emitted photon is equal to the difference in energy levels. From \(E_3\) to \(E_1\): \(E_3 - E_1 = h f\). From \(E_2\) to \(E_1\): \(E_2 - E_1 = 0.75 h f\). For the transition from \(E_3\) to \(E_2\): \(E_3 - E_2 = (E_3 - E_1) - (E_2 - E_1) = h f - 0.75 h f = 0.25 h f\). Therefore, the frequency of the emitted photon is \(0.25 f\).

Award [1] for the correct option. Award [0] otherwise.

Unpolarized light of intensity \(I_0\) incident on the first polarizer becomes polarized with an intensity of \(I_1 = \frac{1}{2} I_0\). Using Malus's law for the second polarizer: \(I = I_1 \cos^2(\theta)\), where \(\theta = 30^\circ\). Thus, \(I = \frac{1}{2} I_0 \cos^2(30^\circ) = \frac{1}{2} I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} I_0 \times \frac{3}{4} = \frac{3}{8} I_0 = 0.375 I_0\).

Award [1] for the correct option. Award [0] otherwise.

For the net electric field to be zero, the individual electric fields due to the two charges must be equal in magnitude and opposite in direction. Since the charges have opposite signs, this point cannot lie between them. Since the magnitude of \(+q\) is smaller than that of \(-4q\), the point must be closer to \(+q\). Let this point be at a distance \(x\) to the left of \(+q\). The condition for zero electric field is: \(\frac{k q}{x^2} = \frac{k (4q)}{(x+d)^2}\). Simplifying this gives: \(\frac{1}{x^2} = \frac{4}{(x+d)^2}\). Taking the square root of both sides: \(\frac{1}{x} = \frac{2}{x+d}\), which yields \(x+d = 2x\), so \(x = d\). Thus, the point is at a distance \(d\) from \(+q\) on the side away from \(-4q\).

Award [1] for the correct option. Award [0] otherwise.

The terminal potential difference \(V\) is related to emf \(E\) and internal resistance \(r\) by \(V = \frac{E}{R+r} R\). For \(R = 2.0\ \Omega\): \(4.0 = \frac{E \times 2.0}{2.0+r} \implies 4.0(2.0+r) = 2.0E \implies E = 4.0 + 2.0r\). For \(R = 6.0\ \Omega\): \(6.0 = \frac{E \times 6.0}{6.0+r} \implies 6.0(6.0+r) = 6.0E \implies E = 6.0 + r\). Equating the two expressions for \(E\): \(4.0 + 2.0r = 6.0 + r \implies r = 2.0\ \Omega\).

Award [1] for the correct option. Award [0] otherwise.

After three half-lives, the fraction of remaining active nuclei is \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\). Therefore, the number of remaining nuclei is \(N = \frac{1}{8} N_0\). The number of decayed nuclei is \(N_{\text{decayed}} = N_0 - N = N_0 - \frac{1}{8} N_0 = \frac{7}{8} N_0\). The ratio of decayed to undecayed nuclei is \(\frac{N_{\text{decayed}}}{N} = \frac{7/8 N_0}{1/8 N_0} = 7\).

Award [1] for the correct option. Award [0] otherwise.

The observer is moving towards a stationary source. The formula for the observed frequency is \(f' = f_0 \left(1 + \frac{v}{v_s}\right)\). Since the speed of the observer is \(v = 0.10 v_s\), we get \(f' = f_0 (1 + 0.10) = 1.10 f_0\).

Award [1] for the correct option. Award [0] otherwise.

The escape speed from a planet's surface is given by \(v_{\text{esc}} = \sqrt{\frac{2GM}{R}}\). For the second planet: \(v'_{\text{esc}} = \sqrt{\frac{2G(2M)}{8R}} = \sqrt{\frac{4GM}{8R}} = \sqrt{\frac{1}{2} \frac{2GM}{R}} = \frac{1}{\sqrt{2}} v_{\text{esc}} \approx 0.71 v_{\text{esc}}\).

Award [1] for the correct option. Award [0] otherwise.

Using the Stefan-Boltzmann law, the power radiated by a black body is given by \(P = \sigma A T^4\). For body X, \(P_X = \sigma A T^4\). For body Y, \(P_Y = \sigma \left(\frac{A}{8}\right) (2T)^4 = \sigma \left(\frac{A}{8}\right) 16 T^4 = 2 \sigma A T^4\). Therefore, the ratio of the power radiated by Y to that of X is 2.

[1 mark] for correctly applying the Stefan-Boltzmann law to find the ratio of powers. Award [1] for the correct option C.

The wavelength is inversely proportional to the energy transition, \(\pi \propto \frac{1}{\Delta E}\). The energy transition from \(n=3\) to \(n=2\) is \(\Delta E_{3 \to 2} = -\frac{E_0}{9} - \left(-\frac{E_0}{4}\right) = \frac{5}{36}E_0\). The transition from \(n=2\) to \(n=1\) is \(\Delta E_{2 \to 1} = -\frac{E_0}{4} - (-E_0) = \frac{3}{4}E_0\). The ratio of the wavelengths is therefore \(\frac{\lambda_{3 \to 2}}{\lambda_{2 \to 1}} = \frac{\Delta E_{2 \to 1}}{\Delta E_{3 \to 2}} = \frac{3/4}{5/36} = \frac{27}{5}\).

[1 mark] for calculating the transition energies and taking the correct inverse ratio for the wavelengths. Award [1] for correct option C.

The first minimum in single slit diffraction occurs at an angle \(\theta \approx \frac{\lambda}{b}\). The angular width of the central maximum is \(2\theta = \frac{2\lambda}{b} = \theta_0\). Under the new conditions, the new angular width is \(\theta_{\text{new}} = \frac{2(2\lambda)}{b/2} = \frac{8\lambda}{b} = 4\theta_0\).

[1 mark] for relating the angular width to wavelength and slit width, and scaling correctly by a factor of 4. Award [1] for correct option C.

The potential due to a point charge is given by \(V = \frac{kQ}{r}\). Between the charges, at a distance \(x\) from \(+Q\), the potential is \(V = \frac{kQ}{x} - \frac{k(4Q)}{d-x} = 0\). Simplifying this yields \(\frac{1}{x} = \frac{4}{d-x}\), which leads to \(d - x = 4x \implies 5x = d \implies x = d/5\).

[1 mark] for writing the potential equation for both charges and solving for the location where the sum is zero. Award [1] for correct option A.

Using the equation \(E = I(R + r)\), we set up two expressions for the emf: \(E = 1.2(4.0 + r)\) and \(E = 0.60(9.0 + r)\). Equating them: \(1.2(4.0 + r) = 0.60(9.0 + r)\). Dividing by 0.60 gives \(2(4.0 + r) = 9.0 + r \implies 8.0 + 2r = 9.0 + r \implies r = 1.0\,\Omega\).

[1 mark] for constructing simultaneous circuit equations and solving for the internal resistance. Award [1] for correct option B.

In a one-dimensional elastic collision, the final velocity of the first mass is \(v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v = \frac{m - 2m}{m + 2m} v = -\frac{1}{3}v\). The final kinetic energy is proportional to the square of the final velocity: \(E_{kf} = \frac{1}{2}m\left(-\frac{1}{3}v\right)^2 = \frac{1}{9} E_{ki}\). Thus, the fraction of kinetic energy retained is \(1/9\).

[1 mark] for using the conservation laws for elastic collision to determine the final velocity and calculating the ratio of kinetic energy. Award [1] for correct option A.

After three half-lives, the number of active nuclei remaining is \(N = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8}\). The number of decayed nuclei is \(N_{\text{decayed}} = N_0 - \frac{N_0}{8} = \frac{7N_0}{8}\). The ratio of decayed nuclei to remaining active nuclei is \(\frac{7N_0/8}{N_0/8} = 7\).

[1 mark] for identifying the fraction of remaining and decayed nuclei and determining their ratio. Award [1] for correct option C.

The Doppler effect frequency for a source approaching a stationary observer is \(f' = f_0 \left(\frac{v}{v - v_s}\right)\). Substituting \(v_s = 0.10v\), we get \(f' = f_0 \left(\frac{v}{v - 0.10v}\right) = f_0 \left(\frac{v}{0.90v}\right) = \frac{1}{0.90} f_0 \approx 1.11 f_0\).

[1 mark] for selecting the correct Doppler formula for an approaching source and evaluating the final numerical factor. Award [1] for correct option C.

The mass-luminosity relation for main-sequence stars is \(L \propto M^{3.5}\). Therefore, the ratio of their luminosities is given by: \(\frac{L_B}{L_A} = \left(\frac{M_B}{M_A}\right)^{3.5}\). Substituting the given values: \(\frac{L_B}{L_A} = \left(\frac{4M}{M}\right)^{3.5} = 4^{3.5}\). We can calculate this as: \(4^{3.5} = 4^3 \times 4^{0.5} = 64 \times \sqrt{4} = 64 \times 2 = 128\).

[1 mark] Award for correct application of the mass-luminosity relation to find the ratio 128.

According to the Stefan-Boltzmann law, the power radiated by a black body is \(P = \sigma A T^4\). For a sphere of radius \(r\), the surface area is \(A = 4\pi r^2\), so \(P \propto r^2 T^4\). Let \(P'\) be the new power: \(P' \propto (2r)^2 \left(\frac{T}{2}\right)^4 = 4r^2 \times \frac{T^4}{16} = \frac{1}{4} (r^2 T^4)\). Therefore, \(P' = \frac{P}{4}\).

[1 mark] Award for determining that area increases by a factor of 4 and temperature factor decreases by a factor of 16, yielding a final ratio of 1/4.

The energy of the emitted photon is equal to the difference in energy between the two states: \(\Delta E = E_{\text{initial}} - E_{\text{final}} = -E_0 - (-3E_0) = 2E_0\). The relationship between photon energy and wavelength is \(\Delta E = \frac{hc}{\lambda}\). Substituting the energy difference: \(2E_0 = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{2E_0}\).

[1 mark] Award for identifying the energy difference as 2E_0 and solving for wavelength correctly.

When unpolarized light of intensity \(I_0\) passes through the first polarizing filter, its intensity is halved, so \(I_1 = \frac{I_0}{2}\). Using Malus's Law, the intensity after passing through the second filter is \(I_2 = I_1 \cos^2(\theta)\), where \(\theta = 60^\circ\). Since \(\cos(60^\circ) = 0.5\), we have \(I_2 = \frac{I_0}{2} \times (0.5)^2 = \frac{I_0}{2} \times 0.25 = \frac{I_0}{8}\).

[1 mark] Award for applying the 1/2 factor for the first unpolarized filter and Malus's law for the second filter.

Let \(x\) be the distance from charge \(Q\) where the electric field is zero. The distance from charge \(4Q\) is therefore \(L - x\). For the net electric field to be zero, the field strength due to both charges must be equal in magnitude: \(\frac{kQ}{x^2} = \frac{k(4Q)}{(L-x)^2}\). Simplifying this gives: \(\frac{1}{x^2} = \frac{4}{(L-x)^2}\). Taking the square root of both sides: \(\frac{1}{x} = \frac{2}{L-x} \implies L - x = 2x \implies L = 3x \implies x = \frac{L}{3}\).

[1 mark] Award for setting up the equation for equal field strengths and correctly solving for x.

The current in the circuit is given by \(I = \frac{\varepsilon}{R+r}\). As \(R\) is decreased towards zero, the total resistance of the circuit decreases, so the current \(I\) increases. The terminal potential difference across the cell is given by \(V = \varepsilon - Ir\). Since \(I\) increases, the lost volts term \(Ir\) increases, causing the terminal potential difference \(V\) to decrease.

[1 mark] Award for identifying that decreasing R increases I and decreases V.

By conservation of momentum: \(m v = (m + 3m) V \implies V = \frac{v}{4}\), where \(V\) is the common speed after the collision. The initial kinetic energy is \(E_{ki} = \frac{1}{2}mv^2\). The final kinetic energy is \(E_{kf} = \frac{1}{2}(4m)V^2 = 2m\left(\frac{v}{4}\right)^2 = \frac{2mv^2}{16} = \frac{1}{8}mv^2\). The fraction of kinetic energy remaining is \(\frac{E_{kf}}{E_{ki}} = \frac{1/8}{1/2} = \frac{1}{4}\). Therefore, the fraction of kinetic energy lost is \(1 - \frac{1}{4} = \frac{3}{4}\).

[1 mark] Award for finding the fraction of kinetic energy remaining (1/4) and subtracting from 1 to find the lost fraction (3/4).

Using the Doppler effect formula for a source moving away from a stationary observer: \(f' = f \left( \frac{v}{v + v_s} \right)\), where \(v\) is the speed of sound and \(v_s\) is the speed of the source. Given \(v_s = 0.2v\), we substitute this into the formula: \(f' = f \left( \frac{v}{v + 0.2v} \right) = f \left( \frac{v}{1.2v} \right) = \frac{1}{1.2}f = \frac{5}{6}f\).

[1 mark] Award for selecting the correct Doppler equation for a source moving away and correctly simplifying the fraction.

The power radiated by a blackbody is given by the Stefan-Boltzmann law: \(P = \sigma A T^4\). For a sphere, the surface area is \(A = 4\pi R^2\), so \(P \propto R^2 T^4\). For the second sphere, the power \(P'\) is proportional to \((2R)^2 (2T)^4 = 4 R^2 \times 16 T^4 = 64 R^2 T^4\). Therefore, \(P' = 64P\).

[1 mark] Award marks as follows: 1 mark for the correct choice D. No partial marks.

The energy of the emitted photon is equal to the transition energy difference: \(\Delta E = hc/\lambda\). For the first transition: \(\Delta E_{3\to2} = E_3 - E_2 = E_0\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5}{36}E_0\). Thus, \(\lambda = \frac{36hc}{5E_0}\). For the second transition: \(\Delta E_{4\to2} = E_4 - E_2 = E_0\left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3}{16}E_0\). Thus, \(\lambda' = \frac{16hc}{3E_0}\). The ratio of the wavelengths is \(\frac{\lambda'}{\lambda} = \frac{16/3}{36/5} = \frac{16}{3} \times \frac{5}{36} = \frac{20}{27}\), which yields \(\lambda' = \frac{20}{27}\lambda\).

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

For a single slit, the angle of the first minimum is given by \(\theta \approx \frac{\lambda}{b}\). Let the new angle be \(\theta'\). With the new wavelength \(\lambda' = \frac{\lambda}{2}\) and new slit width \(b' = 2b\), we have \(\theta' \approx \frac{\lambda'}{b'} = \frac{\lambda/2}{2b} = \frac{\lambda}{4b} = \frac{\theta}{4}\).

[1 mark] 1 mark for the correct choice D. No partial marks.

Initially, the force of attraction is \(F = \frac{k(6)(2)}{d^2} = \frac{12k}{d^2}\). When the identical spheres are brought into contact, the total charge is conserved and distributed equally. The total charge is \(+6\,\mu\text{C} + (-2\,\mu\text{C}) = +4\,\mu\text{C}\), so each sphere gets a charge of \(+2\,\mu\text{C}\). Since both are positive, they now repel. The new force is \(F' = \frac{k(2)(2)}{d^2} = \frac{4k}{d^2}\). Thus, \(F' = \frac{F}{3}\) (repulsive).

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

First, find the equivalent resistance of the three parallel resistors: \(\frac{1}{R_{parallel}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \implies R_{parallel} = \frac{R}{3}\). Next, add the series resistor: \(R_{total} = R + R_{parallel} = R + \frac{R}{3} = \frac{4}{3}R\). Using Ohm's law, the total current \(I\) is \(I = \frac{V}{R_{total}} = \frac{V}{4R/3} = \frac{3V}{4R}\).

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

By conservation of momentum, \(m v = (m + 3m) v'\), where \(v'\) is the final velocity of the combined mass. This gives \(v' = \frac{v}{4}\). The initial kinetic energy is \(E_k = \frac{1}{2}mv^2\). The final kinetic energy is \(E_k' = \frac{1}{2}(4m)(v')^2 = 2m \left(\frac{v}{4}\right)^2 = \frac{1}{8}mv^2\). The kinetic energy remaining is \(\frac{E_k'}{E_k} = \frac{1/8}{1/2} = 0.25 = 25\%\). Therefore, the percentage lost is \(100\% - 25\% = 75\%\).

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

For a moving source approaching a stationary observer, the observed frequency is \(f_1 = f \left(\frac{v}{v - v_s}\right) = f \left(\frac{v}{v - 0.20v}\right) = f \left(\frac{1}{0.80}\right) = 1.25f\). For a moving observer approaching a stationary source, the observed frequency is \(f_2 = f \left(\frac{v + v_o}{v}\right) = f \left(\frac{v + 0.20v}{v}\right) = 1.20f\). The ratio is \(\frac{f_1}{f_2} = \frac{1.25f}{1.20f} = \frac{125}{120} = \frac{25}{24}\).

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

The centripetal force is provided by the gravitational force: \(\frac{GMm}{r^2} = \frac{mv^2}{r}\), which simplifies to the orbital speed \(v = \sqrt{\frac{GM}{r}}\). Since the new radius \(r' = 4R\), the new orbital speed is \(v' = \sqrt{\frac{GM}{4R}} = \frac{1}{2}\sqrt{\frac{GM}{R}} = \frac{v}{2}\).

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

The luminosity of a star is given by Stefan-Boltzmann law: \(L = 4\pi R^2 \sigma T^4\). For the second star with radius \(2R\) and temperature \(T/2\), its luminosity \(L'\) is: \(L' = 4\pi (2R)^2 \sigma (T/2)^4 = 4\pi (4R^2) \sigma \frac{T^4}{16} = \frac{4}{16} (4\pi R^2 \sigma T^4) = 0.25 L\).

[1 mark] Award for calculating the ratio of the new luminosity to the old luminosity using the Stefan-Boltzmann law, leading to the correct factor of 0.25.

The energy of the emitted photon is given by \(\Delta E = \frac{hc}{\lambda}\), so wavelength is inversely proportional to the energy change: \(\lambda \propto \frac{1}{\Delta E}\). The energy transitions are \(\Delta E_{31} = E_3 - E_1 = -2.0 - (-10.0) = 8.0\text{ eV}\), and \(\Delta E_{21} = E_2 - E_1 = -5.0 - (-10.0) = 5.0\text{ eV}\). The ratio of the wavelengths is therefore \(\frac{\lambda_{31}}{\lambda_{21}} = \frac{\Delta E_{21}}{\Delta E_{31}} = \frac{5.0}{8.0} = 5/8\).

[1 mark] Award for finding the energy differences (8.0 eV and 5.0 eV) and using the inverse relationship between photon energy and wavelength to correctly identify the ratio as 5/8.

When unpolarized light of intensity \(I_0\) passes through the first polarizer, its intensity becomes \(I_1 = \frac{1}{2} I_0\). When this polarized light passes through the analyzer, Malus's Law states that the transmitted intensity is \(I_2 = I_1 \cos^2\theta = \frac{1}{2} I_0 \cos^2\theta\). We are given \(I_2 = \frac{3}{8} I_0\), so \(\frac{1}{2} \cos^2\theta = \frac{3}{8} \implies \cos^2\theta = \frac{3}{4} \implies \cos\theta = \frac{\sqrt{3}}{2}\). This corresponds to an angle of \(\theta = 30^\circ\).

[1 mark] Award for applying Malus's Law alongside the initial 50% reduction for unpolarized light to solve for \(\cos\theta = \frac{\sqrt{3}}{2}\), yielding 30 degrees.

Let the point of zero potential be at a distance \(x\) from the charge \(+q\) along the line segment connecting them. The distance from this point to the charge \(-2q\) is \(d - x\). The total electric potential \(V\) at this point is the sum of the potentials from both charges: \(V = \frac{kq}{x} + \frac{k(-2q)}{d-x} = 0\). Solving this equation gives \(\frac{kq}{x} = \frac{2kq}{d-x} \implies d - x = 2x \implies 3x = d \implies x = \frac{d}{3}\).

[1 mark] Award for setting up the electric potential equation for two point charges and solving for the distance \(x = d/3\).

Using the circuit equation \(\varepsilon = I(R + r)\): For the first case: \(\varepsilon = 1.0 \times (4.0 + r)\). For the second case: \(\varepsilon = 0.5 \times (9.0 + r)\). Since the emf \(\varepsilon\) remains constant, we can equate the two expressions: \(1.0(4.0 + r) = 0.5(9.0 + r) \implies 4.0 + r = 4.5 + 0.5r \implies 0.5r = 0.5 \implies r = 1.0\ \Omega\).

[1 mark] Award for setting up the two simultaneous equations using \(\varepsilon = I(R+r)\) and solving for \(r = 1.0\ \Omega\).

After three half-lives, the remaining number of active nuclei is \(N = N_0 \left(\frac{1}{2}\right)^3 = \frac{1}{8} N_0\). The number of nuclei that have decayed is \(N_{\text{decayed}} = N_0 - N = N_0 - \frac{1}{8} N_0 = \frac{7}{8} N_0\). The ratio of decayed nuclei to remaining active nuclei is \(\frac{N_{\text{decayed}}}{N} = \frac{7/8 N_0}{1/8 N_0} = 7\).

[1 mark] Award for identifying the remaining active nuclei fraction (1/8) and decayed nuclei fraction (7/8), and finding their ratio to be 7.

According to the Doppler effect for a moving source: For approach: \(f_{\text{approach}} = f_0 \frac{c}{c - v} = f_0 \frac{c}{c - 0.2c} = f_0 \frac{1}{0.8} = 1.25 f_0\). For recession: \(f_{\text{recede}} = f_0 \frac{c}{c + v} = f_0 \frac{c}{c + 0.2c} = f_0 \frac{1}{1.2} = 0.833 f_0\). The ratio is \(\frac{f_{\text{approach}}}{f_{\text{recede}}} = \frac{1/0.8}{1/1.2} = \frac{1.2}{0.8} = 1.5\).

[1 mark] Award for applying the correct Doppler effect equations for both moving source scenarios and calculating the correct ratio of 1.5.

The total mechanical energy \(E\) of a satellite of mass \(m\) in a circular orbit of radius \(r\) around a planet of mass \(M\) is given by \(E = -\frac{GMm}{2r}\). Thus, the total energy is inversely proportional to the radius of the orbit: \(E \propto \frac{1}{r}\). When the radius increases from \(R\) to \(2R\), the total energy becomes \(E' = \frac{E}{2} = 0.50 E\).

[1 mark] Award for recognizing that total mechanical energy of an orbiting satellite is inversely proportional to orbit radius \(r\) and correctly deducing that doubling the radius halves the energy value to \(0.50 E\).

(a) From Stefan-Boltzmann law, \(L = 4\pi R^2 \sigma T^4\). Rearranging for radius: \(R = \sqrt{\frac{L}{4\pi \sigma T^4}}\). Substituting values: \(R = \sqrt{\frac{3.85 \times 10^{26}}{4\pi \times (5.67 \times 10^{-8}) \times 5780^4}} = \sqrt{\frac{3.85 \times 10^{26}}{7.953 \times 10^8}} = 6.96 \times 10^8 \text{ m} \approx 7.0 \times 10^8 \text{ m}\).

(b) Apparent brightness \(b = \frac{L}{4\pi d^2}\). Substituting the values: \(b = \frac{3.85 \times 10^{26}}{4\pi \times (1.50 \times 10^{11})^2} = \frac{3.85 \times 10^{26}}{2.827 \times 10^{23}} = 1.36 \times 10^3 \text{ W m}^{-2}\).

(c) The nuclear fusion equation is \(4\text{p} \rightarrow \text{He-4} + 2\text{e}^+ + 2\nu\). Reactant mass: \(4 \times 1.007276 \text{ u} = 4.029104 \text{ u}\). Product mass: \(4.001506 + 2 \times 0.000549 = 4.002604 \text{ u}\). Mass defect \(\Delta m = 4.029104 - 4.002604 = 0.026500 \text{ u}\). Energy released \(E = 0.026500 \text{ u} \times 931.5 \text{ MeV u}^{-1} = 24.68 \text{ MeV}\).

(d) Hydrostatic equilibrium is maintained by the balance between the inward gravitational force of the star's mass and the outward radiation and thermal pressure produced by the nuclear fusion reactions in its core.

(a) [3 marks]
- Award [1] for writing Stefan-Boltzmann equation rearranged for R.
- Award [1] for correct substitution of values.
- Award [1] for obtaining \(6.96 \times 10^8 \text{ m}\) and showing it is \(\approx 7.0 \times 10^8 \text{ m}\).

(b) [3 marks]
- Award [1] for using the inverse square law equation for brightness.
- Award [1] for substituting \(L\) and \(d\) correctly.
- Award [1] for the final answer with correct units (\(1.36 \times 10^3 \text{ W m}^{-2}\)).

(c) [3.25 marks]
- Award [1] for calculating correct mass of reactants (\(4.029104 \text{ u}\)).
- Award [1] for calculating correct mass of products (\(4.002604 \text{ u}\)).
- Award [1] for calculating mass defect \(\Delta m = 0.026500 \text{ u}\).
- Award [0.25] for final energy value of \(24.68 \text{ MeV}\).

(d) [2 marks]
- Award [1] for identifying inward force as gravity.
- Award [1] for identifying outward force as radiation/gas pressure from fusion.

(a) Incident power \(P_{\text{in}} = I \times A = 850 \times 2.40 = 2040 \text{ W}\). Useful thermal power absorbed \(P_{\text{out}} = \eta \times P_{\text{in}} = 0.650 \times 2040 = 1326 \text{ W} \approx 1330 \text{ W}\).

(b) Rate of heat transfer is given by \(\frac{Q}{t} = \frac{m}{t} c \Delta T\). Substituting values: \(1326 = 0.0350 \times 4180 \times \Delta T\). Therefore, \(\Delta T = \frac{1326}{146.3} = 9.06 \text{ K}\) (or \(9.06^\circ\text{C}\)).

(c) Total energy required to heat the tank: \(Q = m c \Delta T = 120 \times 4180 \times (50.0 - 20.0) = 1.5048 \times 10^7 \text{ J}\). Time taken \(t = \frac{Q}{P_{\text{out}}} = \frac{1.5048 \times 10^7}{1326} = 11348 \text{ s} \approx 1.13 \times 10^4 \text{ s}\).

(d) Thermal radiation: The black surface of the panel is at a higher temperature than the surroundings, so it emits infrared radiation to the atmosphere according to Stefan-Boltzmann's law. Alternatively, convection: warm air in contact with the panel rises and is replaced by cooler air, carrying heat away.

(a) [3 marks]
- Award [1] for calculating total incident power (\(2040 \text{ W}\)).
- Award [1] for applying efficiency.
- Award [1] for final answer of \(1326 \text{ W}\) (or \(1330 \text{ W}\)).

(b) [3 marks]
- Award [1] for using the rate form of the heating equation.
- Award [1] for correct substitution of values.
- Award [1] for final temperature rise of \(9.06 \text{ K}\) (accept \(9.06^\circ\text{C}\) or \(9.1 \text{ K}\)).

(c) [3.25 marks]
- Award [1] for calculating correct energy required (\(1.5 \times 10^7 \text{ J}\)).
- Award [1] for setting up the equation \(t = Q / P\).
- Award [1] for calculation of \(11300 \text{ s}\).
- Award [0.25] for expressing the final answer to 3 significant figures.

(d) [2 marks]
- Award [1] for stating a valid heat loss mechanism (e.g., radiation, convection, conduction).
- Award [1] for explaining how it occurs in this context (e.g., emitting infrared because of temperature difference, or air movement carrying heat).

(a) The energy of level 3 is \(E_3 = -\frac{13.6}{9} = -1.511 \text{ eV}\). The energy of level 2 is \(E_2 = -\frac{13.6}{4} = -3.400 \text{ eV}\). The energy difference \(\Delta E = E_3 - E_2 = -1.511 - (-3.400) = 1.889 \text{ eV}\). Converting to Joules: \(\Delta E = 1.889 \times 1.60 \times 10^{-19} \text{ J} = 3.022 \times 10^{-19} \text{ J}\). Since \(\lambda = \frac{hc}{\Delta E}\), we have \(\lambda = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{3.022 \times 10^{-19}} = 6.58 \times 10^{-7} \text{ m} \approx 6.6 \times 10^{-7} \text{ m}\).

(b) The energy of the incident photon is \(E_{\text{photon}} = 1.89 \text{ eV}\). The work function is \(\Phi = 1.80 \text{ eV}\). Photoelectric emission occurs if \(E_{\text{photon}} > \Phi\). Since \(1.89 \text{ eV} > 1.80 \text{ eV}\), photoelectric emission will occur.

(c) (i) Most alpha particles passed straight through the foil undeflected, indicating that the atom is mostly empty space. A tiny fraction of alpha particles were deflected by angles greater than \(90^\circ\). This large-angle deflection requires a very large repulsive force, which can only be explained if the positive charge and nearly all the mass of the atom are concentrated in a tiny central region (the nucleus).

(ii) The electrostatic force (or Coulomb repulsion force) of repulsion between the positively charged alpha particle and the positively charged gold nucleus.

(a) [3.25 marks]
- Award [1] for calculating both energy levels \(E_3 = -1.51 \text{ eV}\) and \(E_2 = -3.40 \text{ eV}\).
- Award [1] for finding the energy difference \(1.89 \text{ eV}\) and converting to Joules (\(3.02 \times 10^{-19} \text{ J}\)).
- Award [1] for using \(\lambda = hc / \Delta E\).
- Award [0.25] for showing the final value \(6.58 \times 10^{-7} \text{ m}\).

(b) [3 marks]
- Award [1] for stating the condition for photoelectric emission (\(E > \Phi\)).
- Award [1] for comparing the calculated photon energy \(1.89 \text{ eV}\) with the work function \(1.80 \text{ eV}\).
- Award [1] for concluding that emission will occur.

(c)(i) [3 marks]
- Award [1] for noting that most particles pass undeflected showing empty space.
- Award [1] for noting that some particles deflect at large angles.
- Award [1] for linking this to a concentrated positive charge/mass (the nucleus).

(c)(ii) [2 marks]
- Award [2] for stating "electrostatic force" or "Coulomb repulsion". (Award [1] for just "electrical force").

(a) For single slit diffraction, the first minimum occurs at \(\theta = \frac{\lambda}{b}\). The angular width of the central maximum is \(2\theta = \frac{2\lambda}{b}\). Substituting values: \(2\theta = \frac{2 \times 633 \times 10^{-9} \text{ m}}{0.120 \times 10^{-3} \text{ m}} = 1.055 \times 10^{-2} \text{ rad} \approx 1.06 \times 10^{-2} \text{ rad}\).

(b) The physical half-width is \(x = D \tan\theta \approx D \theta\). The total physical width of the central maximum is \(W = 2x = 2D\theta = D \times (2\theta)\). Substituting values: \(W = 2.00 \times 1.055 \times 10^{-2} = 2.11 \times 10^{-2} \text{ m}\) (or \(2.11 \text{ cm}\)).

(c) (i) Fringe spacing in double-slit interference is given by \(s = \frac{\lambda D}{d}\). Substituting values: \(s = \frac{633 \times 10^{-9} \times 2.00}{0.500 \times 10^{-3}} = 2.532 \times 10^{-3} \text{ m} \approx 2.53 \text{ mm}\).

(ii) Increasing the width of the individual slits \(b\) narrows the single-slit diffraction envelope. This causes the intensity of the double-slit fringes to drop off more rapidly away from the center, so fewer interference fringes are visible within the central maximum.

(a) [3 marks]
- Award [1] for recalling \(\theta = \lambda / b\).
- Award [1] for doubling this to find the total angular width \(2\theta = 2\lambda/b\).
- Award [1] for final answer of \(1.06 \times 10^{-2} \text{ rad}\) (or \(1.05 \times 10^{-2} \text{ rad}\)).

(b) [3 marks]
- Award [1] for writing the relationship between physical width and angular width, \(W = 2 D \theta\) (or \(W = 2 D \tan \theta\)).
- Award [1] for substituting values correctly.
- Award [1] for final answer of \(2.11 \times 10^{-2} \text{ m}\) (or \(2.1 \text{ cm}\)).

(c)(i) [3.25 marks]
- Award [1] for stating the formula \(s = \lambda D / d\).
- Award [1] for substituting values with correct powers of ten.
- Award [1] for calculating \(2.53 \times 10^{-3} \text{ m}\).
- Award [0.25] for clear units.

(c)(ii) [2 marks]
- Award [1] for stating that the single-slit diffraction envelope narrows.
- Award [1] for concluding that fewer interference fringes are visible (or they fade out faster).

(a) The electric field magnitude is \(E = \frac{V}{d} = \frac{120}{15.0 \times 10^{-3}} = 8000 \text{ V m}^{-1}\) (or \(\text{N C}^{-1}\)). Since the upper plate is positive and the lower plate is at \(0 \text{ V}\), the electric field points vertically downwards.

(b) (i) For the oil drop to remain stationary, the upward electric force must balance the downward gravitational force: \(F_e = F_g \Rightarrow q E = mg\). Solving for magnitude: \(q = \frac{mg}{E} = \frac{2.61 \times 10^{-15} \times 9.81}{8000} = 3.20 \times 10^{-18} \text{ C}\). Since the electric field points downwards, an upward electric force requires a negative charge. Hence, \(q = -3.20 \times 10^{-18} \text{ C}\).

(ii) Since the charge is negative, there is an excess of electrons. Number of excess electrons \(N = \frac{|q|}{e} = \frac{3.20 \times 10^{-18}}{1.60 \times 10^{-19}} = 20.0\) electrons.

(c) The magnetic force acts as a centripetal force: \(q v B = \frac{m v^2}{r} \Rightarrow r = \frac{m v}{q B}\). Substituting the mass of an electron (\(9.11 \times 10^{-31} \text{ kg}\)) and charge (\(1.60 \times 10^{-19} \text{ C}\)): \(r = \frac{9.11 \times 10^{-31} \times 2.50 \times 10^6}{1.60 \times 10^{-19} \times 0.450} = \frac{2.2775 \times 10^{-24}}{7.20 \times 10^{-20}} = 3.16 \times 10^{-5} \text{ m}\).

(a) [3 marks]
- Award [1] for using \(E = V/d\).
- Award [1] for correct value \(8000 \text{ V m}^{-1}\) (or \(\text{N C}^{-1}\)).
- Award [1] for stating the direction as vertically downwards.

(b)(i) [3.25 marks]
- Award [1] for equating electric force to gravitational force (\(q E = mg\)).
- Award [1] for substituting values correctly to find magnitude \(3.20 \times 10^{-18} \text{ C}\).
- Award [1] for stating that the charge is negative.
- Award [0.25] for final answer formatted clearly.

(b)(ii) [2 marks]
- Award [1] for dividing charge by elementary charge to get 20.
- Award [1] for identifying this as an "excess" of electrons.

(c) [3 marks]
- Award [1] for equating magnetic force to centripetal force to obtain \(r = mv / qB\).
- Award [1] for using correct mass and charge values for an electron.
- Award [1] for final answer of \(3.16 \times 10^{-5} \text{ m}\).

(a) Electromotive force (emf) is the work done per unit charge by a power source in moving charge around a complete circuit (or the total energy per unit charge converted from other forms to electrical energy).

(b) (i) Since they are in series, total resistance is \(R_{\text{total}} = R_{\text{thermistor}} + R_{\text{fixed}} + r = 14.3 + 8.20 + 1.50 = 24.0\ \Omega\).

(ii) The current in the circuit is \(I = \frac{\varepsilon}{R_{\text{total}}} = \frac{12.0}{24.0} = 0.500 \text{ A}\). The potential difference across the fixed resistor is \(V_R = I \times R_{\text{fixed}} = 0.500 \times 8.20 = 4.10 \text{ V}\).

(iii) The power dissipated in the internal resistance is \(P_r = I^2 r = (0.500)^2 \times 1.50 = 0.250 \times 1.50 = 0.375 \text{ W}\).

(c) When the temperature of an NTC thermistor increases, its resistance decreases. This reduces the total resistance of the circuit, which causes the current \(I\) to increase. The terminal potential difference of the battery is given by \(V = \varepsilon - Ir\). Since \(I\) increases and \(r\) is constant, the potential drop across the internal resistance (\(Ir\)) increases, which causes the terminal potential difference \(V\) to decrease.

(a) [2 marks]
- Award [1] for "work done per unit charge" or "energy converted per unit charge".
- Award [1] for specifying "around a complete circuit" or "by the source".

(b)(i) [2 marks]
- Award [1] for summing the series resistances.
- Award [1] for final answer of \(24.0\ \Omega\).

(b)(ii) [3.25 marks]
- Award [1] for calculating circuit current \(I = 0.500 \text{ A}\).
- Award [1] for using \(V = I R\).
- Award [1] for final answer \(4.10 \text{ V}\).
- Award [0.25] for writing correct units.

(b)(iii) [2 marks]
- Award [1] for using \(P = I^2 r\).
- Award [1] for final answer \(0.375 \text{ W}\).

(c) [2 marks]
- Award [1] for noting that higher temperature decreases thermistor resistance, thereby increasing circuit current.
- Award [1] for stating that this increases the internal voltage drop \(Ir\), causing the terminal potential difference (\(V = \varepsilon - Ir\)) to decrease.

(a) The total linear momentum of an isolated system remains constant, provided no external forces act on the system.

(b) (i) From conservation of momentum: \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\). Substituting values: \(1.20 \times 4.50 + 0 = (1.20 + 2.30) \times v \Rightarrow 5.40 = 3.50 v\). Thus, \(v = \frac{5.40}{3.50} = 1.543 \text{ m s}^{-1} \approx 1.54 \text{ m s}^{-1}\).

(ii) Initial kinetic energy: \(E_{ki} = \frac{1}{2} m_1 u_1^2 = 0.5 \times 1.20 \times (4.50)^2 = 12.15 \text{ J}\). Final kinetic energy: \(E_{kf} = \frac{1}{2} (m_1 + m_2) v^2 = 0.5 \times 3.50 \times (1.543)^2 = 4.167 \text{ J}\). Loss of kinetic energy: \(\Delta E_k = E_{ki} - E_{kf} = 12.15 - 4.167 = 7.983 \text{ J} \approx 7.98 \text{ J}\).

(c) The average force is given by Newton's second law: \(F = \frac{\Delta p}{\Delta t}\). For block 2, the change in momentum is \(\Delta p_2 = m_2 v - 0 = 2.30 \times 1.543 = 3.549 \text{ kg m s}^{-1}\). Average force \(F = \frac{3.549}{0.0850} = 41.75 \text{ N} \approx 41.8 \text{ N}\). (Alternatively, analyzing block 1 yields the same magnitude but in the opposite direction).

(a) [2 marks]
- Award [1] for "total momentum remains constant".
- Award [1] for clarifying "in a closed/isolated system" or "no external force acts".

(b)(i) [3 marks]
- Award [1] for stating conservation of momentum equation.
- Award [1] for correct substitution.
- Award [1] for final answer \(1.54 \text{ m s}^{-1}\).

(b)(ii) [3.25 marks]
- Award [1] for calculating initial kinetic energy (\(12.15 \text{ J}\)).
- Award [1] for calculating final kinetic energy (\(4.17 \text{ J}\)).
- Award [1] for subtracting to find the difference.
- Award [0.25] for final answer \(7.98 \text{ J}\) with units.

(c) [3 marks]
- Award [1] for using \(F = \Delta p / \Delta t\).
- Award [1] for calculating change in momentum of either block (\(3.55 \text{ kg m s}^{-1}\)).
- Award [1] for final force value \(41.8 \text{ N}\) (accept \(41.7 \text{ N}\) due to rounding).

(a) Half-life is the time required for half of the radioactive nuclei in a sample to decay (or the time for the activity of the sample to decrease to half of its initial value).

(b) The relationship between decay constant \(\lambda\) and half-life \(T_{1/2}\) is \(\lambda = \frac{\ln 2}{T_{1/2}}\). Substituting values: \(\lambda = \frac{0.69315}{1.25 \times 10^9 \text{ years}} = 5.545 \times 10^{-10} \text{ year}^{-1} \approx 5.5 \times 10^{-10} \text{ year}^{-1}\).

(c) (i) Let the initial number of nuclei of X be \(N_0\). At time \(t\), the number of remaining nuclei of X is \(N_X = N_0 e^{-\lambda t}\). Since each decayed nucleus of X becomes a nucleus of Y, the number of Y nuclei is \(N_Y = N_0 - N_X = N_0(1 - e^{-\lambda t})\). We are given \(\frac{N_Y}{N_X} = 3.00 \Rightarrow \frac{N_0(1 - e^{-\lambda t})}{N_0 e^{-\lambda t}} = e^{\lambda t} - 1 = 3.00\). This simplifies to \(e^{\lambda t} = 4.00 \Rightarrow \lambda t = \ln(4.00) = 1.3863\). Therefore, \(t = \frac{1.3863}{5.545 \times 10^{-10} \text{ year}^{-1}} = 2.50 \times 10^9 \text{ years}\). (Alternatively, since \(N_Y/N_X = 3\), X is 1/4 of the total, which corresponds to exactly 2 half-lives: \(t = 2 \times 1.25 \times 10^9 = 2.50 \times 10^9 \text{ years}\)).

(ii) If some of the daughter isotope Y had escaped, the measured number of Y atoms, and therefore the measured ratio \(\frac{N_Y}{N_X}\), would be lower than it should be. Since a lower ratio corresponds to less decay, the calculation would yield a smaller age, resulting in an underestimate of the rock's true age.

(a) [2 marks]
- Award [1] for "time taken for number of active nuclei to halve".
- Award [1] for "or time taken for activity to halve".

(b) [3 marks]
- Award [1] for stating \(\lambda = \ln 2 / T_{1/2}\).
- Award [1] for correct substitution of \(T_{1/2}\).
- Award [1] for obtaining \(5.54 \times 10^{-10} \text{ year}^{-1}\) and showing it is \(\approx 5.5 \times 10^{-10} \text{ year}^{-1}\).

(c)(i) [3.25 marks]
- Award [1] for identifying that \(\frac{N_Y}{N_X} = 3\) means \(1/4\) of original X remains (or using the exponential formulation).
- Award [1] for identifying that this represents exactly two half-lives (or solving \(\lambda t = \ln 4\)).
- Award [1] for calculating \(2.50 \times 10^9 \text{ years}\).
- Award [0.25] for correct unit.

(c)(ii) [3 marks]
- Award [1] for noting that escape of Y decreases the measured ratio \(\frac{N_Y}{N_X}\).
- Award [1] for explaining that a lower ratio implies less decay of X has apparently occurred.
- Award [1] for concluding that the calculated age of the rock will be an underestimate.

(a) The cross-sectional area \(A\) is:
\[A = \frac{\pi d^2}{4} = \frac{\pi (0.38 \times 10^{-3} \text{ m})^2}{4} = 1.134 \times 10^{-7} \text{ m}^2 \approx 1.1 \times 10^{-7} \text{ m}^2\]

The percentage uncertainty in \(d\) is:
\[\frac{\Delta d}{d} \times 100\% = \frac{0.02}{0.38} \times 100\% = 5.26\%\]

Since \(A \propto d^2\), the percentage uncertainty in \(A\) is:
\[2 \times 5.26\% = 10.52\% \approx 11\%\]

(b) The resistance is given by \(R = \frac{\rho L}{A}\), so \(\frac{R}{L} = \frac{\rho}{A}\). The gradient \(m = \frac{R}{L}\), which gives:
\[\rho = m \times A = 9.2 \times 1.134 \times 10^{-7} = 1.043 \times 10^{-6} \text{ } \Omega\text{ m}\]

The percentage uncertainty in \(\rho\) is:
\[\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta A}{A} = \frac{0.3}{9.2} \times 100\% + 10.52\% = 3.26\% + 10.52\% = 13.78\% \approx 14\%\]

The absolute uncertainty is:
\[\Delta \rho = 13.78\% \times 1.043 \times 10^{-6} = 0.144 \times 10^{-6} \text{ } \Omega\text{ m} \approx 0.1 \times 10^{-6} \text{ } \Omega\text{ m}\]

Thus, the final value is:
\[\rho = (1.0 \pm 0.1) \times 10^{-6} \text{ } \Omega\text{ m} \text{ (or } (1.04 \pm 0.14) \times 10^{-6} \text{ } \Omega\text{ m)}\]

(c) (i) A constant contact resistance acts as a systematic error that adds to all resistance measurements. This shifts the entire line vertically upwards, creating a positive, non-zero y-intercept.

(ii) Since the contact resistance is constant, it does not change as the length of the wire is varied. Therefore, the gradient of the graph remains unchanged, and the calculated resistivity is unaffected.

**Part (a): [3 marks]**
* \(A = 1.13 \times 10^{-7} \text{ m}^2\) (or \(1.1 \times 10^{-7} \text{ m}^2\)) [1]
* Fractional uncertainty in \(d = 0.053\) (or \(5.3\%\)) [1]
* Percentage uncertainty in \(A = 11\%\) (or \(10.5\%\)) [1] (Award [3] for correct final answer).

**Part (b): [3 marks]**
* \(\rho = 1.0 \times 10^{-6} \text{ } \Omega\text{ m}\) (or \(1.04 \times 10^{-6}\)) [1]
* Sum of percentage uncertainties used to find total percentage uncertainty (\(\approx 14\%\)) [1]
* Absolute uncertainty calculated and final value given with correct unit: \(\pm 0.1 \times 10^{-6} \text{ } \Omega\text{ m}\) (or \(\pm 0.14 \times 10^{-6}\)) [1]

**Part (c): [2 marks]**
* (i) Non-zero/positive y-intercept [1]
* (ii) No change to gradient, so resistivity is unaffected [1]

(a) It takes time for the thermal energy to conduct from the heater through the aluminum block and establish thermal equilibrium with the temperature probe.

(b) The energy balance equation is given by:
\[P = m c \frac{\Delta\theta}{\Delta t}\]

where \(\frac{\Delta\theta}{\Delta t}\) is the gradient of the temperature-time graph.
\[c = \frac{P}{m \times \text{gradient}} = \frac{45.0}{1.10 \times 0.046} = 889.3 \text{ J kg}^{-1}\text{ K}^{-1} \approx 890 \text{ J kg}^{-1}\text{ K}^{-1}\]

(c) The relation is \(c = \frac{P}{m \cdot \text{gradient}}\). The overall percentage uncertainty is the sum of the individual percentage uncertainties:
\[\frac{\Delta P}{P} = \frac{0.5}{45.0} \times 100\% = 1.11\%\]
\[\frac{\Delta m}{m} = \frac{0.01}{1.10} \times 100\% = 0.91\%\]
\[\frac{\Delta \text{gradient}}{\text{gradient}} = \frac{0.002}{0.046} \times 100\% = 4.35\%\]

Total percentage uncertainty:
\[\% \Delta c = 1.11\% + 0.91\% + 4.35\% = 6.37\% \approx 6.4\% \text{ (or } 6\% \text{ to } 1 \text{ sig fig)}\]

(d) Surround the block with fiberglass insulation (or other suitable thermal insulators) to minimize heat transfer to the ambient air.

**Part (a): [1 mark]**
* Time delay for thermal conduction / thermal equilibrium between the heater, block, and probe. [1]

**Part (b): [2 marks]**
* Correct rearrangement: \(c = \frac{P}{m \times \text{gradient}}\) [1]
* Specific heat capacity value: \(890 \text{ J kg}^{-1}\text{ K}^{-1}\) (or \(889 \text{ J kg}^{-1}\text{ K}^{-1}\)) [1]

**Part (c): [3 marks]**
* Calculation of individual percentage uncertainties (\(1.11\%\), \(0.91\%\), \(4.35\%\)) [1]
* Adding percentage uncertainties together [1]
* Final calculated value: \(6.4\%\) (accept \(6.37\%\) or \(6\%\)) [1]

**Part (d): [1 mark]**
* Wrap the block in insulation / apply grease to improve thermal contact between the heater and the block [1]

(a) Distance in parsecs:
\(d = \frac{1}{p} = \frac{1}{0.040} = 25\text{ pc}\)

Distance in meters:
\(d = 25 \times 3.09 \times 10^{16}\text{ m} = 7.73 \times 10^{17}\text{ m}\) (accept \(7.7 \times 10^{17}\text{ m}\))

(b) Using the brightness formula \(b = \frac{L}{4\pi d^2}\):
\(L = 4\pi d^2 b\)
\(L = 4\pi \times (7.725 \times 10^{17})^2 \times 2.5 \times 10^{-11}\)
\(L = 1.88 \times 10^{26}\text{ W}\) (accept \(1.9 \times 10^{26}\text{ W}\))

(c) Two limitations:
1. Earth's atmosphere causes distortion (blurring/seeing limit), making it difficult to measure extremely small angles from ground-based telescopes.
2. For stars very far away (typically greater than \(100\text{ pc}\) for ground telescopes, or a few kiloparsecs for space-based telescopes), the parallax angle becomes too small to measure with sufficient accuracy.

(a) [3 marks]
- Correct calculation of parsecs: \(d = 25\text{ pc}\) [1]
- Correct conversion factor used (\(1\text{ pc} = 3.09 \times 10^{16}\text{ m}\)) [1]
- Correct distance in meters: \(7.7 \times 10^{17}\text{ m}\) (accept range \(7.70 \times 10^{17}\) to \(7.73 \times 10^{17}\)) [1]

(b) [2 marks]
- Correct formula or substitution: \(L = 4\pi (7.7 \times 10^{17})^2 \times 2.5 \times 10^{-11}\) [1]
- Final answer: \(1.9 \times 10^{26}\text{ W}\) (accept \(1.88 \times 10^{26}\text{ W}\)) [1]

(c) [2 marks]
- Award [1] for each valid limitation, up to [2].
- Acceptable points: atmospheric absorption/turbulence, parallax angle too small for distant stars (beyond a certain limit), equipment resolution limits, etc.

(a) Cepheid variables have a known relationship between their period of pulsation and their absolute luminosity (the period-luminosity relationship). By measuring the period of pulsation of a Cepheid, its luminosity can be determined. By measuring its apparent brightness from Earth and using the relation \(b = \frac{L}{4\pi d^2}\), the distance to the galaxy containing the Cepheid can be calculated.

(b)(i)
\(L = 1.2 \times 10^{4} \times 3.8 \times 10^{26}\text{ W} = 4.56 \times 10^{30}\text{ W}\) (accept \(4.6 \times 10^{30}\text{ W}\))

(b)(ii)
Using \(b = \frac{L}{4\pi d^2}\):
\(d = \sqrt{\frac{L}{4\pi b}} = \sqrt{\frac{4.56 \times 10^{30}}{4\pi \times 4.8 \times 10^{-17}}}\)
\(d = \sqrt{7.56 \times 10^{45}} = 8.69 \times 10^{22}\text{ m}\)
Converting to Mpc:
\(d = \frac{8.69 \times 10^{22}}{3.09 \times 10^{22}} = 2.81\text{ Mpc}\)

(a) [3 marks]
- Identify that Cepheids have a period of pulsation that is directly related to their luminosity / period-luminosity relationship [1]
- Explain that measuring the period allows the determination of the absolute luminosity [1]
- Explain that by measuring the apparent brightness, the distance can be calculated using \(b = \frac{L}{4\pi d^2}\) [1]

(b)(i) [1 mark]
- Correct calculation: \(4.6 \times 10^{30}\text{ W}\) or \(4.56 \times 10^{30}\text{ W}\) [1]

(b)(ii) [4 marks]
- Recall or use of \(d = \sqrt{\frac{L}{4\pi b}}\) [1]
- Correct substitution of values [1]
- Correct calculation of distance in meters: \(8.7 \times 10^{22}\text{ m}\) [1]
- Correct conversion to Mpc: \(2.8\text{ Mpc}\) (accept \(2.81\text{ Mpc}\)) [1]

(a)(i) Vertical axis: Luminosity (or absolute magnitude) increasing upwards. Horizontal axis: Temperature (or spectral class) increasing to the left (from high to low temperature).
(a)(ii) Main sequence is a band running from top-left (hot, bright) to bottom-right (cool, dim). Red Giants are in the upper-right (cool, bright). White Dwarfs are in the lower-left (hot, dim).

(b)(i) After leaving the main sequence:
- The star expands and cools to become a Red Supergiant as it fuses heavier elements in shells around the core.
- When iron forms in the core and fusion ceases, the core collapses rapidly, resulting in a supernova explosion.
- The remnant left behind is either a neutron star or a black hole (depending on the remaining core mass).

(b)(ii) The Chandrasekhar limit (\(1.4 M_{\odot}\)) is the maximum mass at which electron degeneracy pressure can support a star's core against gravitational collapse. Since this star has an initial mass of \(15 M_{\odot}\), its remaining core mass after outer layers are blown away will exceed this limit. Thus, gravity overcomes the electron degeneracy pressure, collapsing the core beyond the white dwarf stage.

(a)(i) [1 mark]
- Luminosity on y-axis (increasing upwards) AND Temperature on x-axis (increasing to the left) [1]
(a)(ii) [2 marks]
- Main sequence correctly identified as diagonal band [1]
- Red Giants in upper right AND White Dwarfs in lower left [1]

(b)(i) [3 marks]
- Identifies expansion to Red Supergiant / shell fusion [1]
- Identifies core collapse leading to a supernova explosion [1]
- Identifies the remnant as a neutron star or black hole [1]

(b)(ii) [2 marks]
- Defines or references Chandrasekhar limit as \(1.4 M_{\odot}\) (the maximum mass of a core supported by electron degeneracy pressure) [1]
- States that the core mass of a \(15 M_{\odot}\) star exceeds this limit, so it must collapse further (or cannot be supported as a white dwarf) [1]

(a) Redshift:
\(z = \frac{\Delta \lambda}{\lambda_0} = \frac{685.8 - 656.3}{656.3} = \frac{29.5}{656.3} \approx 0.04495\) (accept \(0.045\))

(b) Recession velocity (non-relativistic approximation since \(z \ll 1\)):
\(v = z c = 0.04495 \times 3.00 \times 10^8\text{ m s}^{-1} = 1.35 \times 10^7\text{ m s}^{-1}\) (or \(1.35 \times 10^4\text{ km s}^{-1}\))

(c)(i) Distance \(d\):
\(d = \frac{v}{H_0} = \frac{1.35 \times 10^4\text{ km s}^{-1}}{70\text{ km s}^{-1}\text{ Mpc}^{-1}} \approx 192.8\text{ Mpc}\) (accept \(193\text{ Mpc}\) or \(190\text{ Mpc}\))

(c)(ii) Age of the universe \(T \approx \frac{1}{H_0}\):
\(H_0 = \frac{70 \times 10^3\text{ m s}^{-1}}{3.09 \times 10^{22}\text{ m}} = 2.265 \times 10^{-18}\text{ s}^{-1}\)
\(T = \frac{1}{2.265 \times 10^{-18}} \approx 4.41 \times 10^{17}\text{ s}\)
In years:
\(T = \frac{4.41 \times 10^{17}}{365.25 \times 24 \times 3600} \approx 1.4 \times 10^{10}\text{ years}\) (or \(14\text{ billion years}\))

Assumption: The expansion rate of the universe (Hubble constant) has been constant over time since the Big Bang.

(a) [1 mark]
- Correct calculation of redshift: \(z = 0.045\) (or \(0.0449\)) [1]

(b) [2 marks]
- Recall of \(v = z c\) [1]
- Correct calculation: \(1.35 \times 10^7\text{ m s}^{-1}\) (or \(1.35 \times 10^4\text{ km s}^{-1}\)) [1]

(c)(i) [2 marks]
- Recall of Hubble's Law \(v = H_0 d\) [1]
- Correct calculation: \(190\text{ Mpc}\) (or \(193\text{ Mpc}\)) [1]

(c)(ii) [2 marks]
- Calculation of age as \(T \approx \frac{1}{H_0} \approx 1.4 \times 10^{10}\text{ years}\) [1]
- Stating assumption: Constant expansion rate of the universe / constant Hubble constant [1]