2025 IB DP Physics 模擬試題連答案詳解

The total mechanical energy of the simple harmonic oscillator is given by \(E = \frac{1}{2} k X_0^2\). At a displacement of \(x = \frac{1}{3}X_0\), the elastic potential energy is \(E_p = \frac{1}{2} k x^2 = \frac{1}{2} k \left(\frac{1}{3}X_0\right)^2 = \frac{1}{9} \left(\frac{1}{2} k X_0^2\right) = \frac{1}{9}E\). The kinetic energy of the system at this point is \(E_k = E - E_p = E - \frac{1}{9}E = \frac{8}{9}E\). Therefore, the ratio of kinetic energy to potential energy is \(\frac{E_k}{E_p} = \frac{\frac{8}{9}E}{\frac{1}{9}E} = 8\), which corresponds to a ratio of \(8 : 1\).

Award 1 mark for the correct option B.

From Wien's displacement law, the temperature of a star is inversely proportional to its peak wavelength: \(T \propto \frac{1}{\lambda}\). Thus, \(\frac{T_X}{T_Y} = \frac{\lambda_Y}{\lambda_X} = \frac{800\text{ nm}}{400\text{ nm}} = 2\). The luminosity of a star is given by Stefan-Boltzmann's law: \(L = 4\pi R^2 \sigma T^4\). Therefore, the ratio of the luminosities is \(\frac{L_X}{L_Y} = \left(\frac{R_X}{R_Y}\right)^2 \left(\frac{T_X}{T_Y}\right)^4\). Substituting the given values: \(16 = \left(\frac{R_X}{R_Y}\right)^2 (2)^4 \implies 16 = \left(\frac{R_X}{R_Y}\right)^2 \times 16 \implies \left(\frac{R_X}{R_Y}\right)^2 = 1 \implies \frac{R_X}{R_Y} = 1\).

Award 1 mark for the correct option C.

The kinetic energy gained by a charge \(q\) accelerated through a potential \(V\) is \(qV = \frac{1}{2}mv^2\), which gives a velocity of \(v = \sqrt{\frac{2qV}{m}}\). When entering a perpendicular magnetic field, the magnetic force provides the centripetal force: \(qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB} = \frac{m}{qB}\sqrt{\frac{2qV}{m}} = \frac{1}{B}\sqrt{\frac{2mV}{q}}\). Since \(B\) and \(V\) are constant, the radius is proportional to \(\sqrt{\frac{m}{q}}\). Thus, the ratio is \(\frac{r_p}{r_{\alpha}} = \sqrt{\frac{m_p}{q_p} \times \frac{q_{\alpha}}{m_{\alpha}}} = \sqrt{\frac{m_p}{e} \times \frac{2e}{4m_p}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}\).

Award 1 mark for the correct option B.

The total energy of a satellite in a circular orbit of radius \(r\) is given by \(E_i = -\frac{GMm}{2r} = E\). When the orbit radius increases to \(3r\), the final total energy is \(E_f = -\frac{GMm}{2(3r)} = -\frac{GMm}{6r} = \frac{1}{3}E\). The change in total energy is \(\Delta E = E_f - E_i = \frac{1}{3}E - E = -\frac{2}{3}E\). Since \(E\) is a negative quantity, this change is positive, representing an increase in total energy.

Award 1 mark for the correct option A.

The time taken for the loop to be completely pulled out is \(t = \frac{s}{v}\). As it is pulled out, the rate of change of magnetic flux is \(\frac{\Delta \Phi}{\Delta t} = B \frac{\Delta A}{\Delta t} = Bsv\). By Faraday's Law, the magnitude of the induced electromotive force (EMF) is \(\varepsilon = Bsv\). The power dissipated in the resistance \(R\) is \(P = \frac{\varepsilon^2}{R} = \frac{B^2 s^2 v^2}{R}\). The total thermal energy dissipated is \(E = P \times t = \frac{B^2 s^2 v^2}{R} \times \frac{s}{v} = \frac{B^2 s^3 v}{R}\).

Award 1 mark for the correct option B.

Let \(S\) be the solar radiation power per unit area absorbed by the planet. Without an atmosphere, the surface energy balance is \(S = \sigma T_0^4\). With the single-layer atmosphere at temperature \(T_a\), the atmosphere absorbs all planetary radiation \(\sigma T_s^4\) and radiates both upwards to space and downwards to the surface. Thus, the atmosphere energy balance is \(\sigma T_s^4 = 2\sigma T_a^4 \implies \sigma T_a^4 = \frac{1}{2}\sigma T_s^4\). The surface energy balance is \(S + \sigma T_a^4 = \sigma T_s^4\). Substituting the atmospheric radiation: \(S + \frac{1}{2}\sigma T_s^4 = \sigma T_s^4 \implies S = \frac{1}{2}\sigma T_s^4 \implies \sigma T_s^4 = 2S\). Since \(S = \sigma T_0^4\), we have \(\sigma T_s^4 = 2\sigma T_0^4 \implies T_s^4 = 2 T_0^4 \implies T_s = 2^{1/4} T_0 = \sqrt[4]{2} T_0\).

Award 1 mark for the correct option C.

According to the observer on the space station, the length of the spacecraft is Lorentz-contracted: \(L = L_0 \sqrt{1 - \frac{v^2}{c^2}} = L_0 \sqrt{1 - 0.60^2} = 0.80 L_0\). The time interval for this length to pass a single point on the station at speed \(v = 0.60c\) is \(\Delta t = \frac{L}{v} = \frac{0.80 L_0}{0.60 c} = \frac{4 L_0}{3 c}\).

Award 1 mark for the correct option D.

We apply the first law of thermodynamics, \(\Delta U = Q - W\), to each step of the process. For the first step (constant volume), the work done is \(W_1 = 0\text{ J}\), so the change in internal energy is \(\Delta U_1 = Q_1 - W_1 = 400\text{ J} - 0\text{ J} = 400\text{ J}\). For the second step (constant pressure), the heat absorbed is \(Q_2 = 250\text{ J}\) and the work done by the gas is \(W_2 = 150\text{ J}\), so the change in internal energy is \(\Delta U_2 = Q_2 - W_2 = 250\text{ J} - 150\text{ J} = 100\text{ J}\). The total change in internal energy is \(\Delta U_{\text{total}} = \Delta U_1 + \Delta U_2 = 400\text{ J} + 100\text{ J} = 500\text{ J}\).

Award 1 mark for the correct option B.

The total mechanical energy of the simple harmonic oscillator is given by:
\( E_0 = \frac{1}{2} k A^2 \)

The potential energy \( E_P \) at a displacement \( x = \frac{A}{3} \) is:
\( E_P = \frac{1}{2} k x^2 = \frac{1}{2} k \left(\frac{A}{3}\right)^2 = \frac{1}{9} \left( \frac{1}{2} k A^2 \right) = \frac{1}{9} E_0 \)

Since the total energy is conserved:
\( E_0 = E_K + E_P \)
\( E_K = E_0 - E_P = E_0 - \frac{1}{9} E_0 = \frac{8}{9} E_0 \)

Award [1] for the correct answer C.

The lifetime \( T \) of a star is proportional to its available nuclear fuel (which is proportional to its mass \( M \)) divided by the rate at which it consumes fuel (its luminosity \( L \)):
\( T \propto \frac{M}{L} \)

Given that \( L \propto M^4 \), we substitute this into the relation:
\( T \propto \frac{M}{M^4} = M^{-3} \)

Therefore, the ratio of the lifetimes of star A and star B is:
\( \frac{T_A}{T_B} = \left(\frac{M_A}{M_B}\right)^{-3} = (2)^{-3} = \frac{1}{8} \)

Award [1] for the correct answer B.

For a charged particle in a magnetic field, the centripetal force is provided by the magnetic force:
\( \frac{m v^2}{R} = q v B \implies R = \frac{m v}{q B} \)

We express momentum \( p = m v \) in terms of kinetic energy \( K \):
\( K = \frac{p^2}{2m} \implies p = \sqrt{2 m K} \)

Substituting this into the expression for radius:
\( R = \frac{\sqrt{2 m K}}{q B} \)

If the new kinetic energy is \( K' = 2K \) and the new magnetic field is \( B' = 2B \):
\( R' = \frac{\sqrt{2 m (2K)}}{q (2B)} = \frac{\sqrt{2} \sqrt{2 m K}}{2 q B} = \frac{\sqrt{2}}{2} R = \frac{R}{\sqrt{2}} \)

Award [1] for the correct answer C.

For a planet in a circular orbit of radius \( r \) around a star of mass \( M \), the gravitational force provides the centripetal force:
\( \frac{G M m}{r^2} =
\frac{m v^2}{r} \implies v = \sqrt{\frac{G M}{r}} \)

Thus, orbital speed is inversely proportional to the square root of the orbital radius:
\( v \propto \frac{1}{\sqrt{r}} \)

Therefore, the ratio of the orbital speeds is:
\( \frac{v_A}{v_B} = \sqrt{\frac{r_B}{r_A}} = \sqrt{4} = 2 \)

Award [1] for the correct answer A.

The induced electromotive force (EMF) \( \varepsilon \) in a loop moving at speed \( v \) perpendicular to a magnetic field \( B \) is given by:
\( \varepsilon = B L v \)

where \( L \) is the length of the side of the loop perpendicular to the motion. The power \( P \) dissipated as thermal energy in the loop of resistance \( R \) is:
\( P = \frac{\varepsilon^2}{R} = \frac{B^2 L^2 v^2}{R} \)

Since \( P \propto v^2 \), doubling the speed increases the power dissipation by a factor of \( 2^2 = 4 \):
\( P' = 4P \)

Award [1] for the correct answer C.

The solar constant \( S \) is the power per unit area incident on a flat disc perpendicular to the sun's rays. The total power absorbed by the planet is:
\( P_{\text{abs}} = S (1 - \alpha) \pi R^2 \)

where \( \alpha \) is the albedo and \( R \) is the radius of the planet. This absorbed power is distributed over the entire spherical surface area of the planet, which is \( 4 \pi R^2 \).

Thus, the average absorbed intensity \( I_{\text{avg}} \) is:
\( I_{\text{avg}} = \frac{P_{\text{abs}}}{4 \pi R^2} = \frac{S (1 - \alpha)}{4} \)

Substituting the values:
\( I_{\text{avg}} = \frac{1400 \times (1 - 0.30)}{4} = \frac{1400 \times 0.70}{4} = 245\text{ W m}^{-2} \)

Award [1] for the correct answer C.

From the perspective of the space station observer, the length of the spacecraft is contracted due to length contraction:
\( L = L_0 \sqrt{1 - \beta^2} = L_0 \sqrt{1 - (0.80)^2} = 0.60 L_0 \)

The time taken for this contracted length to pass the point-like beacon at speed \( v = 0.80c \) is:
\( \Delta t = \frac{L}{v} = \frac{0.60 L_0}{0.80c} = 0.75 \frac{L_0}{c} \)

Award [1] for the correct answer B.

The path difference \( \Delta r \) between the two waves arriving at \( P \) is:
\( \Delta r = 6.0\lambda - 4.5\lambda = 1.5\lambda \)

Since the path difference is a half-integer multiple of the wavelength (\( 1.5\lambda = (1 + \frac{1}{2})\lambda \)), the two waves arrive completely out of phase, leading to destructive interference.

The phase difference \( \Delta \phi \) in radians is:
\( \Delta \phi = \frac{2\pi}{\lambda} \Delta r = \frac{2\pi}{\lambda} (1.5\lambda) = 3\pi\text{ rad} \)

Award [1] for the correct answer B.

The total energy in simple harmonic motion is given by \(E = \frac{1}{2} k A^2\).

At a displacement of \(x = \frac{A}{2}\), the potential energy is:
\(E_p = \frac{1}{2} k x^2 = \frac{1}{2} k \left(\frac{A}{2}\right)^2 = \frac{1}{8} k A^2\)

By conservation of energy, the kinetic energy is:
\(E_k = E - E_p = \frac{1}{2} k A^2 - \frac{1}{8} k A^2 = \frac{3}{8} k A^2\)

We also know that \(E_k = \frac{1}{2} m v^2\). Setting these equal:
\[\frac{1}{2} m v^2 = \frac{3}{8} k A^2 \implies v^2 = \frac{3}{4} \frac{k}{m} A^2 \implies v = \frac{\sqrt{3}}{2}\sqrt{\frac{k}{m}} A\]
This corresponds to option B.

Award 1 mark for using conservation of mechanical energy to determine the kinetic energy and solving for the velocity correctly.

According to the Stefan-Boltzmann law, the luminosity of a star is given by:
\(L = 4\pi R^2 \sigma T^4\)

Therefore, we can write the ratio of the luminosities as:
\(\frac{L_2}{L_1} = \left(\frac{R_2}{R_1}\right)^2 \left(\frac{T_2}{T_1}\right)^4\)

Substituting the given ratios \(\frac{L_2}{L_1} = 16\) and \(\frac{T_2}{T_1} = 2\):
\(16 = \left(\frac{R_2}{R_1}\right)^2 \times 2^4\)
\(16 = \left(\frac{R_2}{R_1}\right)^2 \times 16\)
\(\left(\frac{R_2}{R_1}\right)^2 = 1 \implies \frac{R_2}{R_1} = 1\)

This corresponds to option A.

Award 1 mark for setting up the luminosity ratio equation using Stefan-Boltzmann's law and correctly solving for the radius ratio.

The magnetic force provides the centripetal acceleration:
\(q v B = \frac{m v^2}{r} \implies r = \frac{m v}{q B}\)

The period of the motion \(T\) is given by the distance of one full revolution divided by the speed:
\(T = \frac{2\pi r}{v} = \frac{2\pi \left(\frac{m v}{q B}\right)}{v} = \frac{2\pi m}{q B}\)

This formula shows that the period of the circular path is independent of the velocity \(v\). When the magnetic field strength is doubled from \(B\) to \(2B\), the period becomes:
\(T' = \frac{2\pi m}{q(2B)} = \frac{T}{2}\)

Since doubling the velocity does not affect the period, the new period is \(\frac{T}{2}\), which corresponds to option B.

Award 1 mark for deriving the formula for the period and recognizing that it is independent of velocity and inversely proportional to the magnetic field strength.

The total mechanical energy \(E\) of an orbit of radius \(r\) is the sum of its kinetic and potential energy:
\(E = -\frac{G M m}{2r}\)

For the initial orbit of radius \(r\):
\(E_1 = -\frac{G M m}{2r}\)

For the final orbit of radius \(2r\):
\(E_2 = -\frac{G M m}{2(2r)} = -\frac{G M m}{4r}\)

The energy required is the difference between the final and initial mechanical energies:
\(\Delta E = E_2 - E_1 = -\frac{G M m}{4r} - \left(-\frac{G M m}{2r}\right) = \frac{G M m}{2r} - \frac{G M m}{4r} = \frac{G M m}{4r}\)

This corresponds to option A.

Award 1 mark for writing correct expressions for total energy in both orbits and finding their difference.

Initially, the magnetic flux linkage through the coil is maximum because the field is perpendicular to the plane of the coil (angle \(\theta = 0^\circ\) between the normal and the magnetic field):
\(\Phi_1 = N B A\)

After rotating by \(90^\circ\), the plane of the coil is parallel to the magnetic field (angle \(\theta = 90^\circ\)), so the flux linkage is zero:
\(\Phi_2 = 0\)

The change in flux linkage is:
\(|\Delta \Phi| = |\Phi_2 - \Phi_1| = N B A\)

By Faraday's law of induction, the magnitude of the average induced emf is:
\(\varepsilon = \frac{|\Delta \Phi|}{\Delta t} = \frac{N B A}{\Delta t}\)

This corresponds to option B.

Award 1 mark for calculating the change in magnetic flux linkage and applying Faraday's law of induction.

In thermal equilibrium, the power absorbed from the star must equal the power radiated by the planet.

The power absorbed by the planet is:
\(P_{\text{abs}} = (1 - \alpha) S \pi R^2\)

where \(R\) is the radius of the planet and \(\pi R^2\) is the projected area intercepting sunlight.

The power radiated by the planet of radius \(R\) at temperature \(T\) is:
\(P_{\text{rad}} = \sigma (4\pi R^2) T^4\)

Equating the two:
\((1 - \alpha) S \pi R^2 = 4\pi R^2 \sigma T^4\)

Simplifying and solving for \(T\):
\((1 - \alpha) S = 4 \sigma T^4\)
\(T^4 = \frac{(1 - \alpha) S}{4 \sigma}\)
\(T^4 = \frac{(1 - 0.30) \times 1400}{4 \times 5.67 \times 10^{-8}} = \frac{0.70 \times 1400}{2.268 \times 10^{-7}} \approx 4.321 \times 10^9\text{ K}^4\)
\(T = (4.321 \times 10^9)^{1/4} \approx 256.5\text{ K}\)

This is closest to \(256\text{ K}\), which corresponds to option B.

Award 1 mark for equating the rate of solar energy absorption to the rate of black body emission, including the albedo term, and calculating the temperature correctly.

An astronaut in the spacecraft is at rest relative to it, and thus measures the proper length \(L_0\).

The length \(L\) measured from Earth is contracted according to:
\(L = \frac{L_0}{\gamma} = L_0 \sqrt{1 - \frac{v^2}{c^2}}\)

With \(v = 0.60c\):
\(\sqrt{1 - (0.60)^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.80\)

Therefore:
\(L = 0.80 L_0\)
\(80\text{ m} = 0.80 L_0 \implies L_0 = \frac{80}{0.80} = 100\text{ m}\)

This corresponds to option C.

Award 1 mark for applying the length contraction equation with the correct interpretation of proper length.

The work done *by* the gas is given by \(W = \int P \text{d}V\). The work done *on* the gas is the negative of the work done *by* the gas:
- In process I (isothermal expansion), the volume increases, so the work done *by* the gas is positive, meaning the work done *on* the gas is negative.
- In process II (isobaric compression), the volume decreases, so the work done *by* the gas is negative, meaning the work done *on* the gas is positive.
- In process III (isochoric heating), the volume is constant, so no work is done on or by the gas (\(W = 0\)).

Therefore, the work done on the gas is positive only during the isobaric compression. This corresponds to option B.

Award 1 mark for identifying that positive work on the gas requires a decrease in volume, which only occurs during the compression stage.

At \(t = 0\), the particle is at equilibrium, so the displacement is \(x(t) = A \sin(\omega t)\), with \(\omega = \frac{2\pi}{T}\). At \(t = \frac{T}{12}\), the phase is \(\omega t = \frac{2\pi}{T} \frac{T}{12} = \frac{\pi}{6}\). The displacement is \(x = A \sin(\frac{\pi}{6}) = \frac{A}{2}\). The potential energy is \(E_p = \frac{1}{2} k x^2 = \frac{1}{2} k \left(\frac{A}{2}\right)^2 = \frac{1}{4} \left(\frac{1}{2} k A^2\right) = \frac{1}{4} E_{\text{total}}\). Since total energy is conserved, the kinetic energy is \(E_k = E_{\text{total}} - E_p = \frac{3}{4} E_{\text{total}}\). Therefore, the ratio of kinetic energy to total energy is \(\frac{3}{4}\).

Award 1 mark for finding the correct potential energy fraction and calculating the kinetic energy fraction. Award 1 mark for final correct option C.

According to the Stefan-Boltzmann law, the luminosity of a star is given by \(L = 4\pi R^2 \sigma T^4\). Comparing the second star with the first star: \(\frac{L_2}{L_1} = \left(\frac{R_2}{R_1}\right)^2 \left(\frac{T_2}{T_1}\right)^4 = 2^2 \times 3^4 = 4 \times 81 = 324\). Therefore, \(L_2 = 324 L_0\).

Award 1 mark for using Stefan-Boltzmann law and calculating the correct factor of 324. Award 1 mark for final correct option D.

The kinetic energy gained by a charge \(q\) in potential difference \(V\) is \(qV = \frac{1}{2} m v^2\), so \(v = \sqrt{\frac{2qV}{m}}\). The radius in a magnetic field \(B\) is \(r = \frac{mv}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}\). Thus, \(r \propto \sqrt{\frac{m}{q}}\). For a proton, \(r_p \propto \sqrt{\frac{m_p}{q_p}}\). For an alpha particle, \(r_{\alpha} \propto \sqrt{\frac{4m_p}{2q_p}} = \sqrt{2} \sqrt{\frac{m_p}{q_p}}\). The ratio \(\frac{r_p}{r_{\alpha}} = \frac{1}{\sqrt{2}}\).

Award 1 mark for finding the dependency of radius on mass and charge and calculating the ratio of 1/sqrt(2). Award 1 mark for final correct option A.

For a circular orbit, the total energy is \(E = -\frac{GMm}{2r}\). To escape, the final total energy must be at least \(0\). The escape condition requires the total energy to be increased from \(E\) to \(0\). Therefore, the energy that must be supplied is \(E_{\text{escape}} = 0 - E = -E\). Since \(E\) is negative, \(-E\) is positive.

Award 1 mark for showing that escape requires total energy to reach zero, and therefore the energy needed is \(-E\). Award 1 mark for final correct option B.

The induced EMF as the loop leaves the field is \(\varepsilon = Bsv\). The electrical power dissipated as heat is \(P = \frac{\varepsilon^2}{R} = \frac{B^2 s^2 v^2}{R}\). The time taken for the loop of length \(s\) to exit is \(t = \frac{s}{v}\). The total thermal energy dissipated is \(E = P \cdot t = \frac{B^2 s^2 v^2}{R} \cdot \frac{s}{v} = \frac{B^2 s^3 v}{R}\).

Award 1 mark for expressing EMF and power, and multiplying by the time taken to find total energy. Award 1 mark for final correct option B.

The total power absorbed by Earth is \(P_{\text{abs}} = S(1-\alpha) \cdot \pi R^2\). In equilibrium, the total power radiated is \(P_{\text{rad}} = P_{\text{abs}} = S(1-\alpha)\pi R^2\). The average power per unit area radiated from Earth's entire surface area of \(4\pi R^2\) is \(I = \frac{P_{\text{rad}}}{4\pi R^2} = \frac{S(1-\alpha)}{4}\).

Award 1 mark for the derivation of average intensity by setting input and output energy equal and dividing by the total surface area of Earth. Award 1 mark for final correct option B.

The contracted length is \(L = \frac{L_0}{\gamma}\). With \(v = 0.80c\), the Lorentz factor is \(\gamma = \frac{1}{\sqrt{1 - 0.80^2}} = \frac{1}{0.60} = \frac{5}{3}\). Therefore, the proper length is \(L_0 = L \gamma = 30 \times \frac{5}{3} = 50\text{ m}\).

Award 1 mark for using the correct length contraction formula and solving for proper length. Award 1 mark for final correct option D.

The amplitude is proportional to the square root of intensity: \(A_1 \propto \sqrt{I}\) and \(A_2 \propto 2\sqrt{I}\). Constructive interference yields \(A_{\text{max}} = 3\sqrt{I}\), so \(I_{\text{max}} \propto (3\sqrt{I})^2 = 9I\). Destructive interference yields \(A_{\text{min}} = \sqrt{I}\), so \(I_{\text{min}} \propto (\sqrt{I})^2 = I\).

Award 1 mark for finding the relation between amplitude and intensity, combining amplitudes, and squaring them to find the correct intensity range. Award 1 mark for final correct option B.

The total energy of a simple harmonic oscillator is given by \(E_{\text{total}} = \frac{1}{2} m \omega^2 A^2\). The potential energy at displacement \(x\) is \(E_p = \frac{1}{2} m \omega^2 x^2\). The kinetic energy is given as \(E_k = 3 E_p\). Since \(E_{\text{total}} = E_k + E_p\), we have \(E_{\text{total}} = 3 E_p + E_p = 4 E_p\). Substituting the expressions gives \(\frac{1}{2} m \omega^2 A^2 = 4 \left(\frac{1}{2} m \omega^2 x^2\right)\). Simplifying this results in \(A^2 = 4 x^2\), which leads to \(x = \pm \frac{A}{2} = 0.50 A\).

Award 1 mark for identifying that the total energy is four times the potential energy and solving for displacement to obtain \(0.50 A\). This corresponds to option B.

Luminosity is given by the Stefan-Boltzmann law as \(L = 4\pi R^2 \sigma T^4\). Therefore, the ratio of the luminosities of Star Y and Star X is \(\frac{L_Y}{L_X} = \left(\frac{R_Y}{R_X}\right)^2 \left(\frac{T_Y}{T_X}\right)^4\). Substituting the given values gives \(\frac{L_Y}{L_X} = (3)^2 \times (2)^4 = 9 \times 16 = 144\).

Award 1 mark for calculating the ratio of luminosities to be 144. This corresponds to option C.

The magnetic force on a charged particle moving perpendicular to a magnetic field provides the centripetal force: \(q v B = \frac{m v^2}{r} \implies r = \frac{m v}{q B}\). Since momentum is related to kinetic energy by \(p = m v = \sqrt{2 m E_k}\), the radius is \(r = \frac{\sqrt{2 m E_k}}{q B}\). For the proton, \(R = \frac{\sqrt{2 m E_k}}{e B}\). For the helium-4 nucleus with mass \(4m\), charge \(2e\), and identical kinetic energy \(E_k\), the radius is \(r_{\text{He}} = \frac{\sqrt{2 (4m) E_k}}{2e B} = \frac{2 \sqrt{2 m E_k}}{2e B} = \frac{\sqrt{2 m E_k}}{e B} = R\).

Award 1 mark for showing that the radius remains \(R\) using the relationship between momentum, kinetic energy, and magnetic force. This corresponds to option B.

For a satellite in a circular orbit, the gravitational force provides the centripetal force: \(\frac{G M m}{r^2} = \frac{m v^2}{r}\), which simplifies to \(v = \sqrt{\frac{G M}{r}}\). Thus, the orbital speed is inversely proportional to the square root of the orbital radius, \(v \propto \frac{1}{\sqrt{r}}\). When the radius is increased to \(4r\), the new orbital speed becomes \(v' = \sqrt{\frac{G M}{4r}} = \frac{1}{2} \sqrt{\frac{G M}{r}} = 0.50 v\).

Award 1 mark for determining the correct orbital speed of \(0.50 v\) using orbital mechanics. This corresponds to option B.

According to Faraday's law of induction, the average induced electromotive force (emf) is given by the rate of change of magnetic flux linkage: \(\epsilon = \frac{\Delta \Phi}{\Delta t}\). The initial flux linkage when the coil is perpendicular to the field is \(\Phi_i = N B A\). The final flux linkage when the coil is parallel to the field is \(\Phi_f = 0\). The change in flux linkage is \(\Delta \Phi = \Phi_i - \Phi_f = N B A\). Thus, the average induced emf is \(\epsilon = \frac{N B A}{\Delta t}\).

Award 1 mark for obtaining \(\frac{N B A}{\Delta t}\) as the correct average induced emf. This corresponds to option C.

The total power incident on the planet's projected cross-sectional area \(\pi R^2\) is \(S \cdot \pi R^2\). Since the albedo is \(\alpha\), the fraction of power absorbed is \((1 - \alpha)\), making the total power absorbed equal to \((1 - \alpha) S \cdot \pi R^2\). In thermal equilibrium, the total power re-radiated from the planet's entire spherical surface area \(4 \pi R^2\) must equal the total power absorbed. Therefore, the average power re-radiated per unit surface area is \(\frac{(1 - \alpha) S \cdot \pi R^2}{4 \t\pi R^2} = \frac{(1 - \alpha) S}{4}\).

Award 1 mark for deriving \(\frac{(1 - \alpha) S}{4}\) as the correct average re-radiated power per unit area. This corresponds to option C.

The time interval measured on the spacecraft is the proper time interval \(\Delta t_0 = 15\text{ s}\) because the start and end of the light signal occur at the same position in the spacecraft's frame. According to special relativity, the time interval measured by the observer on Earth is dilated by the Lorentz factor \(\gamma\): \(\Delta t = \gamma \Delta t_0\). The Lorentz factor is \(\gamma = \frac{1}{\sqrt{1 - (v/c)^2}} = \frac{1}{\sqrt{1 - 0.8^2}} = \frac{1}{0.6} = \frac{5}{3}\). Therefore, the time interval measured on Earth is \(\Delta t = \frac{5}{3} \times 15\text{ s} = 25\text{ s}\).

Award 1 mark for calculating the Lorentz factor correctly as \(5/3\) and finding the dilated time to be \(25\text{ s}\). This corresponds to option C.

The useful work done in lifting the block is \(W_{\text{out}} = m g h = 5.0\text{ kg} \times 10\text{ m s}^{-2} \times 12\text{ m} = 600\text{ J}\). The useful power output of the motor is \(P_{\text{out}} = \frac{W_{\text{out}}}{t} = \frac{600\text{ J}}{4.0\text{ s}} = 150\text{ W}\). The efficiency of the motor is \(\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{150\text{ W}}{200\text{ W}} = 0.75\), which is \(75\%\).

Award 1 mark for calculating the useful power output as 150 W and the resulting efficiency as \(75\%\). This corresponds to option D.

(a) The period \(T\) is the time for one oscillation: \(T = \frac{t}{20} = \frac{36.0}{20} = 1.80\text{ s}\). The absolute uncertainty in \(T\) is \(\Delta T = \frac{\Delta t}{20} = \frac{0.4}{20} = 0.02\text{ s}\). Thus, \(T = (1.80 \pm 0.02)\text{ s}\).

(b) From the pendulum equation: \(T = 2\pi\sqrt{\frac{L}{g}} \Rightarrow g = \frac{4\pi^2 L}{T^2}\). Substituting the measured values: \(g = \frac{4\pi^2 \times 0.800}{1.80^2} = \frac{31.5827}{3.24} \approx 9.75\text{ m s}^{-2}\).

(c) The fractional uncertainty in \(g\) is given by: \(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}\). Percentage uncertainty in \(L\): \(\frac{0.005}{0.800} \times 100\\% = 0.625\\%\). Percentage uncertainty in \(T\): \(\frac{0.02}{1.80} \times 100\\% \approx 1.111\\%\). Total percentage uncertainty in \(g\): \(\\%\Delta g = 0.625\\% + 2 \times 1.111\\% = 2.847\\% \approx 2.8\\%\). Absolute uncertainty in \(g\): \(\Delta g = 9.75 \times 0.02847 \approx 0.28\text{ m s}^{-2}\). Standard practice is to quote the uncertainty to 1 significant figure, i.e., \\pm 0.3\\text{ m s}^{-2}\, which matches the decimal precision of \(9.7\text{ m s}^{-2}\). Hence, \(g = (9.7 \pm 0.3)\text{ m s}^{-2}\).

- Award [1] mark for calculating \(T = 1.80\text{ s}\) and [1] mark for \(\Delta T = 0.02\text{ s}\).
- Award [2] marks for correctly rearranging for \(g\) and calculating \(g \approx 9.75\text{ m s}^{-2}\) (allow 9.7 or 9.8 if rounded early, but penalize lack of working).
- Award [1] mark for summing fractional/percentage uncertainties correctly (\(0.625\\% + 2 \times 1.11\\% = 2.8\\%\) or \(2.9\\%\)).
- Award [1.67] marks for calculating \(\Delta g \approx 0.3\text{ m s}^{-2}\) and expressing the final answer as \(9.7 \pm 0.3\text{ m s}^{-2}\) (allow \(9.75 \pm 0.28\text{ m s}^{-2}\)).

(a) The ideal gas law is \(pV = nRT\), which can be rewritten as \(p = (nRT) \frac{1}{V}\). Comparing this to \(y = mx\), the gradient is \(m = nRT\).

(b) Rearranging for \(n\): \(n = \frac{m}{RT}\). Substituting the values: \(n = \frac{4.20}{8.31 \times 293} = \frac{4.20}{2434.83} \approx 1.725 \times 10^{-3}\text{ mol}\).

(c) Since uncertainty in \(T\) is negligible, the fractional uncertainty in \(n\) is equal to the fractional uncertainty in \(m\): \(\frac{\Delta n}{n} = \frac{\Delta m}{m} = \frac{0.15}{4.20} \approx 0.0357\) (or \(3.57\\%\)). The absolute uncertainty in \(n\) is \(\Delta n = 1.725 \times 10^{-3} \times 0.0357 \approx 0.0616 \times 10^{-3}\text{ mol}\). Rounded to 1 significant figure, this is \(0.06 \times 10^{-3}\text{ mol}\). Thus, the final value is \(n = (1.73 \pm 0.06) \times 10^{-3}\text{ mol}\).

- Award [1] mark for identifying \(m = nRT\).
- Award [2] marks for calculating \(n \approx 1.73 \times 10^{-3}\text{ mol}\) (allow equivalent representations, e.g., \(0.0017\text{ mol}\)).
- Award [2] marks for calculating the fractional or percentage uncertainty in \(m\) (\(3.6\\%\) or \(0.036\)).
- Award [1.67] marks for calculating the absolute uncertainty \(\Delta n \approx 0.06 \times 10^{-3}\text{ mol}\) and expressing the final answer with consistent decimal places.

(a) The cross-sectional area is \(A = \frac{\pi d^2}{4}\). Using \(d = 0.38 \times 10^{-3}\text{ m}\): \(A = \frac{\pi (0.38 \times 10^{-3})^2}{4} \approx 1.134 \times 10^{-7}\text{ m}^2\). The percentage uncertainty in \(A\) is twice the percentage uncertainty in \(d\): \(\\:%\Delta A = 2 \times \frac{\Delta d}{d} \times 100\\% = 2 \times \frac{0.02}{0.38} \times 100\\% \approx 10.53\\%\).

(b) The formula for resistivity is \(\rho = \frac{R A}{L}\). Substituting the calculated and measured values: \(\rho = \frac{4.50 \times 1.134 \times 10^{-7}}{1.250} \approx 4.082 \times 10^{-7}\\ \Omega\text{ m}\).

(c) The fractional uncertainty in \(\rho\) is given by: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + \frac{\Delta A}{A} + \frac{\Delta L}{L}\). Percentage uncertainty in \(R\): \(\frac{0.10}{4.50} \times 100\\% \approx 2.22\\%\). Percentage uncertainty in \(L\): \(\frac{0.002}{1.250} \times 100\\% = 0.16\\%\). Total percentage uncertainty in \(\rho\): \(\\%\Delta \rho = 2.22\\% + 10.53\\% + 0.16\\% = 12.91\\%\). Absolute uncertainty in \(\rho\): \(\Delta \rho = 4.082 \times 10^{-7} \times 0.1291 \approx 0.527 \times 10^{-7}\\ \Omega\text{ m}\). Rounded to 1 significant figure, \(\Delta \rho \approx 0.5 \times 10^{-7}\\ \Omega\text{ m}\). The final value is \(\rho = (4.1 \pm 0.5) \times 10^{-7}\\ \Omega\text{ m}\).

- Award [1] mark for calculating \(A \approx 1.13 \times 10^{-7}\text{ m}^2\) and [1] mark for \(\\%\Delta A \approx 10.5\\%\) (or 11%).
- Award [2] marks for calculating \(\rho \approx 4.1 \times 10^{-7}\\ \Omega\text{ m}\).
- Award [1] mark for summing percentage uncertainties correctly (\(2.2\\% + 10.5\\% + 0.2\\% \approx 13\\%\)).
- Award [1.67] marks for calculating the absolute uncertainty \(\Delta \rho \approx 0.5 \times 10^{-7}\\ \Omega\text{ m}\) and stating the final answer with appropriate significant figures.

(a) The period of oscillation of a mass-spring system is given by:
\(T = 2\pi \sqrt{\frac{m}{k}}\)
Substituting the given values:
\(T = 2\pi \sqrt{\frac{0.40}{160}} = 2\pi \sqrt{\frac{1}{400}} = \frac{2\pi}{20} = \frac{\pi}{10} \approx 0.314\text{ s} \approx 0.31\text{ s}\)

(b) Kinetic energy is given by:
\(E_k = \frac{1}{2}k(A^2 - x^2)\)
At \(x = 0\), the maximum kinetic energy is:
\(E_{k,\max} = \frac{1}{2} \times 160 \times 0.050^2 = 80 \times 0.0025 = 0.20\text{ J}\)
At \(x = \pm A = \pm 0.050\text{ m}\), the kinetic energy is 0.
The graph is a downward-opening parabola with its vertex at \((0, 0.20\text{ J})\) and horizontal intercepts at \(x = -0.050\text{ m}\) and \(x = 0.050\text{ m}\).

(c)
- Underdamped: The system oscillates about the equilibrium position with an exponentially decaying amplitude over time.
- Critically damped: The system returns to the equilibrium position in the minimum possible time without any oscillations.
- Overdamped: The system returns to equilibrium without oscillating, but takes a longer time than the critically damped system because of the high resistive force.

(d) The maximum acceleration of the block occurs at maximum displacement and is given by:
\(a_{\max} = \omega^2 A = \frac{k}{m} A\)
Substituting the values:
\(a_{\max} = \frac{160}{0.40} \times 0.050 = 400 \times 0.050 = 20\text{ m s}^{-2}\)

(a)
- Correct formula or substitution [1]
- Correct final answer with units: 0.31 s or 0.314 s [1]

(b)
- Parabolic shape opening downwards symmetric about the y-axis [1]
- Correct y-intercept (maximum KE) labeled as 0.20 J [1]
- Correct x-intercepts labeled as +/- 0.050 m [1]

(c)
- Underdamped description: oscillates with decreasing amplitude [1]
- Critically damped description: returns to equilibrium in shortest time without oscillation [1]
- Overdamped description: returns to equilibrium slowly without oscillation [1]

(d)
- Correct formula used (e.g., kA/m or omega^2 A) [1]
- Correct calculation: 20 m s^-2 [1]

(a) Wien's displacement law is given by:
\(\lambda_{\max} T = 2.90 \times 10^{-3}\text{ m K}\)
Substituting \(T = 5800\text{ K}\):
\(\lambda_{\max} = \frac{2.90 \times 10^{-3}}{5800} = 5.0 \times 10^{-7}\text{ m}\)

(b) The luminosity of a black body is given by Stefan-Boltzmann law:
\(L = 4\pi R^2 \sigma T^4\)
Rearranging for radius \(R\):
\(R = \sqrt{\frac{L}{4\pi \sigma T^4}}\)
Using \(\sigma = 5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}\):
\(R = \sqrt{\frac{4.0 \times 10^{26}}{4\pi \times 5.67 \times 10^{-8} \times (5800)^4}}\)
\((5800)^4 \approx 1.1316 \times 10^{15}\text{ K}^4\)
\(4\pi \sigma T^4 \approx 4\pi \times 5.67 \times 10^{-8} \times 1.1316 \times 10^{15} \approx 8.06 \times 10^8\text{ W m}^{-2}\)
\(R = \sqrt{\frac{4.0 \times 10^{26}}{8.06 \times 10^8}} = \sqrt{4.96 \times 10^{17}} \approx 7.04 \times 10^8\text{ m} \approx 7.0 \times 10^8\text{ m}\)

(c) The fusion reaction is:
\(4p \rightarrow ^4_2\text{He} + 2e^+ + 2\nu_e\)
Mass of reactants:
\(m_{\text{reactants}} = 4 \times 1.007276\text{ u} = 4.029104\text{ u}\)
Mass of products (neglecting neutrino mass):
\(m_{\text{products}} = 4.001506\text{ u} + 2 \times 0.000549\text{ u} = 4.002604\text{ u}\)
Mass defect:
\(\Delta m = 4.029104\text{ u} - 4.002604\text{ u} = 0.026500\text{ u}\)
Using the conversion factor \(1\text{ u} = 931.5\text{ MeV}\):
\(E = 0.026500 \times 931.5\text{ MeV} \approx 24.68\text{ MeV} \approx 24.7\text{ MeV}\)

(d) The main sequence lifetime of a star \(\tau\) depends inversely on its mass, typically as \(\tau \propto M^{-2.5}\) (or accept any relation showing inverse dependence, e.g., \(\tau \propto M^{-2}\) to \(\tau \propto M^{-3}\)). This is because high-mass stars have much higher core temperatures and pressures, leading to much higher luminosities (energy consumption rates) that scale as \(L \propto M^{3.5}\), which far outweighs the additional fuel mass.

(a)
- Correct formula used [1]
- Correct peak wavelength: 5.0 * 10^-7 m [1]

(b)
- Correct formula for luminosity of a star [1]
- Correct substitution of values [1]
- Correct final radius: 7.0 * 10^8 m (accept 7.04 * 10^8 m) [1]

(c)
- Correct calculation of reactant mass and product mass [1]
- Correct mass defect: 0.0265 u [1]
- Correct energy conversion to MeV: 24.7 MeV (accept 24.68 MeV) [1]

(d)
- Lifetime decreases as mass increases / inverse relationship [1]
- Explanation that higher mass stars consume nuclear fuel much faster due to significantly higher core temperatures / luminosities [1]

(a) The work done by the electric field equals the gain in kinetic energy:
\(q V = \frac{1}{2} m v^2\)
Rearranging for speed \(v\):
\(v = \sqrt{\frac{2 q V}{m}}\)
Substituting the values:
\(v = \sqrt{\frac{2 \times (1.60 \times 10^{-19}\text{ C}) \times (2400\text{ V})}{5.81 \times 10^{-26}\text{ kg}}} = \sqrt{\frac{7.68 \times 10^{-16}}{5.81 \times 10^{-26}}} \approx 1.15 \times 10^5\text{ m s}^{-1}\)

(b) The magnetic force provides the centripetal acceleration:
\(F_B = q v B = \frac{m v^2}{r}\)
Rearranging for radius \(r\):
\(r = \frac{m v}{q B}\)
Substituting the values:
\(r = \frac{(5.81 \times 10^{-26}\text{ kg}) \times (1.15 \times 10^5\text{ m s}^{-1})}{(1.60 \times 10^{-19}\text{ C}) \times (0.150\text{ T})}
\)r = \frac{6.68 \times 10^{-21}}{2.40 \times 10^{-20}} \approx 0.278\text{ m}\)

(c)
- In a velocity selector, an electric field and a magnetic field are set up perpendicular to each other and to the direction of motion of the ions.
- The electric force on the ion is \(F_E = qE\) and the magnetic force is \(F_B = qvB\).
- These two forces act in opposite directions.
- For ions of a specific speed, these forces are equal in magnitude and cancel out, allowing the ions to pass through undeflected: \(qE = qvB \Rightarrow v = \frac{E}{B}\).
- Ions with other speeds will have unbalanced forces and will be deflected, colliding with the walls.

(a)
- Equating work done to kinetic energy change: qV = 0.5 * m * v^2 [1]
- Correct substitution of V = 2400 V and other constants [1]
- Correct speed: 1.15 * 10^5 m s^-1 [1]

(b)
- Equating magnetic force to centripetal force: qvB = m v^2 / r [1]
- Correct substitution of calculated speed and field values [1]
- Correct radius: 0.278 m (or 0.28 m) [1]

(c)
- Statement that electric and magnetic fields are perpendicular [1]
- Formulas for both forces: F_E = qE and F_B = qvB acting in opposite directions [1]
- Equating forces for undeflected path: qE = qvB [1]
- Deriving final equation: v = E/B [1]

(a) The orbital radius \(r\) is the sum of the planet's radius and the altitude: \(r = R + R = 2R = 2 \times 6.4 \times 10^6\text{ m} = 1.28 \times 10^7\text{ m}\).
The gravitational force provides the centripetal force:
\(\frac{G M m}{r^2} = \frac{m v^2}{r} \Rightarrow v = \sqrt{\frac{G M}{r}}\)
Substituting the values:
\(v = \sqrt{\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{1.28 \times 10^7}} = \sqrt{3.1265 \times 10^7} \approx 5590\text{ m s}^{-1} \approx 5.6\text{ km s}^{-1}\). (Shown)

(b) The gravitational potential energy is given by:
\(E_p = -\frac{G M m}{r}\)
Substituting the values:
\(E_p = -\frac{(6.67 \times 10^{-11}) \times (6.0 \times 10^{24}) \times 1200}{1.28 \times 10^7} \approx -3.75 \times 10^{10}\text{ J}\)

(c) The total energy \(E_{\text{total}}\) of the satellite in orbit is the sum of its kinetic and potential energies:
\(E_k = \frac{1}{2} m v^2 = \frac{G M m}{2r} = 1.875 \times 10^{10}\text{ J}\)
\(E_{\text{total}} = E_k + E_p = 1.875 \times 10^{10}\text{ J} - 3.75 \times 10^{10}\text{ J} = -1.875 \times 10^{10}\text{ J}\)
To escape to infinity (where total energy is 0), the additional energy required is:
\(E_{\text{required}} = E_f - E_i = 0 - (-1.875 \times 10^{10}\text{ J}) = 1.875 \times 10^{10}\text{ J} \approx 1.88 \times 10^{10}\text{ J}\)

(d) The gravitational potential at a point is defined as the work done per unit mass in bringing a small test mass from infinity to that point in the gravitational field.

(e) The orbital period would remain unchanged because the period depends only on the mass of the central body \(M\) and the orbital radius \(r\), not on the mass of the satellite \(m\).

(a)
- Correct formula for orbital radius: r = 2R = 1.28 * 10^7 m [1]
- Correct calculation leading to 5.6 * 10^3 m s^-1 [1]

(b)
- Correct formula for gravitational potential energy with negative sign [1]
- Correct calculation: -3.75 * 10^10 J [1]

(c)
- Correct kinetic energy or total energy equation: E_total = -G M m / 2r [1]
- Correct total energy: -1.88 * 10^10 J [1]
- Correct minimum work required: +1.88 * 10^10 J [1]

(d)
- Work done per unit mass [1]
- in bringing a small test mass from infinity to that point [1]

(e)
- Correct statement: no change [1]

(a) The area of the coil is:
\(A = \pi r^2 = \pi \times (0.040)^2 \approx 5.03 \times 10^{-3}\text{ m}^2\)
The initial magnetic flux linkage is:
\(\Phi_i = N B_i A = 250 \times 0.80 \times 5.03 \times 10^{-3} \approx 1.01\text{ Wb-turns} \approx 1.01\text{ Wb}\)

(b) According to Faraday's law of induction, the average induced emf is:
\(\text{emf} = -\frac{\Delta \Phi}{\Delta t} = -N A \frac{\Delta B}{\Delta t}\)
Here, \(\Delta B = B_f - B_i = 0.20 - 0.80 = -0.60\text{ T}\).
\(\text{emf} = -250 \times (5.03 \times 10^{-3}\text{ m}^2) \times \frac{-0.60\text{ T}}{0.15\text{ s}} = 5.03\text{ V} \approx 5.0\text{ V}\)

(c) The average induced current is:
\(I = \frac{\text{emf}}{R} = \frac{5.03\text{ V}}{5.0\ \Omega} \approx 1.01\text{ A} \approx 1.0\text{ A}\)
The thermal energy dissipated is:
\(E = I^2 R \Delta t = (1.006\text{ A})^2 \times 5.0\ \Omega \times 0.15\text{ s} \approx 0.76\text{ J}\)

(d) Lenz's law states that the direction of the induced current is such that it opposes the change in magnetic flux that produces it.
Since the external magnetic field is decreasing, the magnetic flux through the coil is decreasing. The induced current must produce its own magnetic field in the same direction as the external field to try to maintain the original flux level.

(a)
- Correct calculation of area: A = 5.03 * 10^-3 m^2 [1]
- Correct initial flux linkage: 1.0 Wb or 1.01 Wb [1]

(b)
- Correct statement/use of Faraday's Law: emf = -delta(Phi)/delta(t) [1]
- Correct calculation of change in flux linkage: 0.75 Wb-turns [1]
- Correct average emf: 5.0 V [1]

(c)
- Correct current: 1.0 A (or 1.01 A) [1]
- Correct formula for thermal energy (I^2 R t or V^2 t / R) [1]
- Correct thermal energy: 0.76 J [1]

(d)
- Statement that the induced field opposes the change in flux [1]
- Explanation that the induced field is in the same direction as the external field to oppose the decrease [1]

(a) The average incoming solar radiation over the surface of a spherical planet is \(\frac{S}{4}\).
Taking albedo \(\alpha\) into account, the average absorbed intensity is:
\(I_{\text{absorbed}} = \frac{S}{4}(1 - \alpha) = \frac{1400}{4}(1 - 0.30) = 350 \times 0.70 = 245\text{ W m}^{-2}\)

(b) Let \(I_s\) be the radiation intensity emitted by the surface and \(I_a\) be the radiation intensity emitted by the single-layer atmosphere (both upwards and downwards).
At the top of the atmosphere, the outgoing radiation equals the incoming absorbed solar radiation:
\(I_a = I_{\text{absorbed}} = 245\text{ W m}^{-2}\)
At the surface, the total incoming radiation (absorbed solar + downward atmospheric) must equal the emitted radiation:
\(I_s = I_{\text{absorbed}} + I_a = 245 + 245 = 490\text{ W m}^{-2}\)
According to Stefan-Boltzmann law:
\(I_s = \sigma T_s^4 \Rightarrow T_s = \left(\frac{I_s}{\sigma}\right)^{1/4}\)
Using \(\sigma = 5.67 \times 10^{-8}\text{ W m}^{-2}\text{ K}^{-4}\):
\(T_s = \left(\frac{490}{5.67 \times 10^{-8}}\right)^{1/4} = (8.642 \times 10^9)^{1/4} \approx 305.1\text{ K} \approx 305\text{ K}\). (Shown)

(c)
- Greenhouse gas molecules (such as \(\text{CO}_2\) and \(\text{H}_2\text{O}\)) have multiple atoms that are bound by flexible chemical bonds, giving rise to quantized vibrational energy levels.
- The differences in energy between these vibrational levels correspond to the energy of photons in the infrared region of the electromagnetic spectrum.
- When an infrared photon with a frequency matching the molecular vibrational natural frequency is incident, the molecule absorbs the photon's energy and transitions to a higher vibrational energy state.
- This phenomenon is called resonance, which is highly selective to specific wavelengths/frequencies of infrared radiation.

(a)
- Correct formula for average solar energy: S/4 * (1 - alpha) [1]
- Correct calculation: 245 W m^-2 [1]

(b)
- Balance at top of atmosphere: I_a = 245 W m^-2 [1]
- Balance at surface: I_s = 490 W m^-2 [1]
- Correct Stefan-Boltzmann relation: T_s = (I_s / sigma)^0.25 [1]
- Correct calculation showing T_s = 305 K [1]

(c)
- Greenhouse gases have vibrational energy states [1]
- Energy difference between states corresponds to infrared photon energy / frequency [1]
- Absorption happens when photon frequency matches natural frequency / resonance [1]
- Leads to excitation / higher vibrational state [1]

(a) The Lorentz factor \(\gamma\) is given by:
\(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\)
Substituting \(v = 0.80c\):
\(\gamma = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.60} = \frac{5}{3} \approx 1.67\)

(b) Due to length contraction, the length \(L\) measured in the station's frame is:
\(L = \frac{L_0}{\gamma} = 120 \times 0.60 = 72\text{ m}\)

(c)
(i) In the spacecraft's frame, the light travels a distance of \(L_0 = 120\text{ m}\) at the speed of light \(c = 3.0 \times 10^8\text{ m s}^{-1}\):
\(\Delta t' = \frac{L_0}{c} = \frac{120}{3.0 \times 10^8} = 4.0 \times 10^{-7}\text{ s}\)

(ii) In the station's frame, the spacecraft has length \(L = 72\text{ m}\). The light travels to the left at speed \(c\), while the back of the spacecraft moves to the right at speed \(v = 0.80c\).
Since they travel towards each other, the distance closed is the sum of their displacements in time \(\Delta t\):
\(c \Delta t + v \Delta t = L\)
\(\Delta t (c + v) = L \Rightarrow \Delta t = \frac{L}{c + v}\)
\(\Delta t = \frac{72\text{ m}}{3.0 \times 10^8 \times (1 + 0.80)} = \frac{72}{5.4 \times 10^8} \approx 1.33 \times 10^{-7}\text{ s}\)

Alternatively, using Lorentz transformation:
Let emission be at \(x'_1 = 0, t'_1 = 0\), and absorption at the back be at \(x'_2 = -L_0 = -120\text{ m}\), \(t'_2 = L_0/c = 4.0 \times 10^{-7}\text{ s}\).
\(t_1 = 0\)
\(t_2 = \gamma \left( t'_2 + \frac{v x'_2}{c^2} \right) = \gamma \left( \frac{L_0}{c} - \frac{v L_0}{c^2} \right) = \gamma \frac{L_0}{c} \left(1 - \frac{v}{c}\right)\)
\(\Delta t = \frac{5}{3} \times (4.0 \times 10^{-7}\text{ s}) \times (1 - 0.80) = \frac{5}{3} \times 4.0 \times 10^{-7} \times 0.20 = 1.33 \times 10^{-7}\text{ s}\).

(a)
- Correct formula for Lorentz factor [1]
- Correct calculation showing 1.67 (or 5/3) [1]

(b)
- Correct length contraction formula: L = L_0 / gamma [1]
- Correct calculation: 72 m [1]

(c)(i)
- Statement that light travels at speed c over proper distance L_0 [1]
- Correct calculation: 4.0 * 10^-7 s [1]

(c)(ii)
- Recognizing the relative motion or set up for Lorentz transformation [1]
- Substitution of correct variables (e.g., L = 72 m and v = 0.80c, or x' = -120 m) [1]
- Correct algebraic formulation [1]
- Correct final calculation: 1.33 * 10^-7 s [1]

(a) The work done during an isobaric expansion is:
\(W_{1\rightarrow 2} = P_1 \Delta V = P_1 (V_2 - V_1)\)
\(W_{1\rightarrow 2} = 2.0 \times 10^5\text{ Pa} \times (8.0 \times 10^{-3} - 4.0 \times 10^{-3}\text{ m}^3) = 2.0 \times 10^5 \times 4.0 \times 10^{-3} = 800\text{ J}\)

(b) Using the ideal gas law at state 2:
\(P_2 V_2 = n R T_2 \Rightarrow T_2 = \frac{P_2 V_2}{n R}\)
\(T_2 = \frac{2.0 \times 10^5\text{ Pa} \times 8.0 \times 10^{-3}\text{ m}^3}{0.50\text{ mol} \times 8.31\text{ J K}^{-1}\text{ mol}^{-1}} = \frac{1600}{4.155} \approx 385.1\text{ K} \approx 385\text{ K}\). (Shown)

(c) The change in internal energy \(\Delta U\) is zero. This is because the process is isothermal (constant temperature) and, for an ideal gas, internal energy depends solely on the temperature.

(d) The net work done during the cycle is the sum of the work done in each of the three steps:
- Step 1 (isobaric expansion): \(W_{1\rightarrow 2} = +800\text{ J}\)
- Step 2 (isochoric cooling): \(W_{2\rightarrow 3} = 0\text{ J}\) (since volume is constant)
- Step 3 (isothermal compression):
At state 3, \(V_3 = 8.0 \times 10^{-3}\text{ m}^3\) and \(P_3 = 1.0 \times 10^5\text{ Pa}\).
The temperature at state 3 is:
\(T_3 = T_1 = \frac{P_1 V_1}{n R} = \frac{2.0 \times 10^5 \times 4.0 \times 10^{-3}}{0.50 \times 8.31} = \frac{800}{4.155} \approx 192.5\text{ K}\)
The work done BY the gas during this isothermal compression is:
\(W_{3\rightarrow 1} = n R T_1 \ln\left(\frac{V_1}{V_3}\right) = 0.50 \times 8.31 \times 192.5 \times \ln\left(\frac{4.0 \times 10^{-3}}{8.0 \times 10^{-3}}\right) = 800 \ln(0.5) \approx 800 \times (-0.6931) = -554.5\text{ J}\)
Therefore, the net work done by the gas is:
\(W_{\text{net}} = W_{1\rightarrow 2} + W_{2\rightarrow 3} + W_{3\rightarrow 1} = 800 + 0 - 554.5 = 245.5\text{ J} \approx 246\text{ J}\)

(a)
- Correct formula: W = P * delta(V) [1]
- Correct calculation: 800 J [1]

(b)
- Correct formula: T = PV / (nR) [1]
- Correct substitution and calculation leading to 385 K [1]

(c)
- Statement that Delta U = 0 [1]
- Justification: Internal energy of ideal gas depends only on temperature, which is constant [1]

(d)
- Statement that work done in step 2 is zero [1]
- Correct temperature for isothermal compression: 192.5 K (or calculation of nRT = 800 J) [1]
- Correct calculation of isothermal work: -555 J (accept -554 J) [1]
- Correct net work: 246 J (accept 245 J to 247 J) [1]

### Part (a)
- As the loop is pulled out of the magnetic field, the magnetic flux linkage through the loop decreases.
- According to Faraday's law, this rate of change of magnetic flux induces an electromotive force (emf) in the loop.
- According to Lenz's law, the induced current flows in such a direction as to oppose the change in flux that produced it. Thus, it creates an induced magnetic field pointing into the page to oppose the decreasing inward flux. By the right-hand rule, this corresponds to a **clockwise** current.

### Part (b)
- The induced emf \(\varepsilon\) is given by:
$$\varepsilon = B L v$$
- Substitute the given values:
$$\varepsilon = 0.40\text{ T} \times 0.050\text{ m} \times 2.0\text{ m s}^{-1} = 0.040\text{ V}$$

### Part (c)
- The induced current \(I\) is:
$$I = \frac{\varepsilon}{R} = \frac{0.040\text{ V}}{0.20\ \Omega} = 0.20\text{ A}$$
- The magnetic force \(F_B\) acting on the vertical side of the loop remaining inside the field is:
$$F_B = B I L = 0.40\text{ T} \times 0.20\text{ A} \times 0.050\text{ m} = 4.0 \times 10^{-3}\text{ N}$$
- Since the loop moves at a constant speed, the net force on it is zero. Therefore, the external force \(F_{\text{ext}}\) must be equal in magnitude and opposite in direction to the magnetic force:
$$F_{\text{ext}} = 4.0 \times 10^{-3}\text{ N}$$

### Part (d)
- The rate of work done by the external force (mechanical power input) is:
$$P_{\text{work}} = F_{\text{ext}} v = (4.0 \times 10^{-3}\text{ N}) \times (2.0\text{ m s}^{-1}) = 8.0 \times 10^{-3}\text{ W}$$
- The rate of thermal energy dissipation in the loop (electrical power output) is:
$$P_{\text{thermal}} = I^2 R = (0.20\text{ A})^2 \times 0.20\ \Omega = 8.0 \times 10^{-3}\text{ W}$$
- Comparison: The mechanical power supplied by the external force is exactly equal to the electrical power dissipated as heat in the resistor. This is consistent with the principle of conservation of energy.

**Part (a) [3 marks]**
- Award [1] for stating that the magnetic flux linkage decreases.
- Award [1] for linking the change in flux to the induced emf / current using Faraday's law.
- Award [1] for using Lenz's law to conclude that the induced current must be clockwise to oppose the change (maintain the inward magnetic flux).

**Part (b) [2 marks]**
- Award [1] for writing the formula \(\varepsilon = B L v\) or showing a correct substitution.
- Award [1] for obtaining \(0.040\text{ V}\) (or \(4.0 \times 10^{-2}\text{ V}\)).

**Part (c) [2 marks]**
- Award [1] for calculating the induced current \(I = 0.20\text{ A}\).
- Award [1] for substituting into \(F_B = B I L\) to yield \(4.0 \times 10^{-3}\text{ N}\), stating that this equals \(F_{\text{ext}}\) for constant velocity.

**Part (d) [3 marks]**
- Award [1] for calculating the power of the external force as \(8.0 \times 10^{-3}\text{ W}\) (or \(8.0\text{ mW}\)).
- Award [1] for calculating the electrical power dissipation as \(8.0 \times 10^{-3}\text{ W}\) (or \(8.0\text{ mW}\)).
- Award [1] for stating that the two rates are equal, which is a direct consequence of the conservation of energy.