2023 IB DP Physics 模擬試題連答案詳解

The orbital speed \(v\) of a satellite in a circular orbit of radius \(r\) around a planet of mass \(M\) is given by \(v = \sqrt{\frac{GM}{r}}\). The initial orbital radius is \(r_1 = R + R = 2R\), so the initial speed is \(v = \sqrt{\frac{GM}{2R}}\). The new orbit is at a height of \(3R\), so the new orbital radius is \(r_2 = R + 3R = 4R\). The new orbital speed is \(v_2 = \sqrt{\frac{GM}{4R}} = \sqrt{\frac{GM}{2 \cdot 2R}} = \frac{1}{\sqrt{2}} \sqrt{\frac{GM}{2R}} = \frac{v}{\sqrt{2}}\).

Award [1] for the correct identification of the new speed. Correct answer is B.

After a time of \(4T\), the number of half-lives elapsed for nuclei \(X\) is \(n_X = \frac{4T}{T} = 4\). The number of remaining nuclei of \(X\) is \(N_X = N_0 \left(\frac{1}{2}\right)^4 = \frac{N_0}{16}\). For nuclei \(Y\), the number of half-lives elapsed is \(n_Y = \frac{4T}{2T} = 2\). The number of remaining nuclei of \(Y\) is \(N_Y = N_0 \left(\frac{1}{2}\right)^2 = \frac{N_0}{4}\). The ratio of the remaining nuclei is \(\frac{N_X}{N_Y} = \frac{N_0/16}{N_0/4} = \frac{4}{16} = \frac{1}{4}\).

Award [1] for the correct calculation of the ratio. Correct answer is C.

The fringe spacing \(s\) for a double-slit setup is given by the formula \(s = \frac{\lambda D}{d}\). When the wavelength becomes \(\lambda' = \frac{\lambda}{2}\) and the slit separation becomes \(d' = 2d\), the new fringe spacing is \(s' = \frac{\lambda' D}{d'} = \frac{(\lambda/2) D}{2d} = \frac{1}{4} \frac{\lambda D}{d} = \frac{s}{4}\).

Award [1] for the correct analysis using the double-slit formula. Correct answer is A.

The magnetic flux \(\Phi\) through the loop increases as it enters the field: \(\Phi = B L x\), where \(x\) is the distance the loop has entered. The magnitude of the induced electromotive force (emf) is given by Faraday's law: \(V = \frac{\Delta \Phi}{\Delta t} = B L v\). Thus, the magnitude of the induced current is \(I = \frac{V}{R} = \frac{BLv}{R}\). According to Lenz's law, the induced current must oppose the increase of magnetic flux directed into the page. To produce a magnetic field pointing out of the page, the current must flow in a counter-clockwise direction.

Award [1] for the correct current magnitude and direction. Correct answer is A.

The average force is given by Newton's second law as the rate of change of momentum: \(F_{\text{avg}} = \frac{\Delta p}{\Delta t}\). Taking the initial direction of motion as positive, the initial momentum is \(p_i = mu\) and the final momentum is \(p_f = -mv\). The change in momentum is \(\Delta p = p_f - p_i = -mv - mu = -m(u + v)\). The magnitude of this change in momentum is \(|\Delta p| = m(u + v)\). Thus, the average force is \(F_{\text{avg}} = \frac{m(u + v)}{\Delta t}\).

Award [1] for the correct force expression. Correct answer is C.

From the ideal gas equation \(PV = N k_B T\), the pressure is \(P = \frac{N k_B T}{V}\). If one-third of the molecules escape, the remaining number of molecules is \(N' = \frac{2}{3}N\). The new temperature is \(T' = 3T\). Since the volume \(V\) remains constant, the new pressure is \(P' = \frac{N' k_B T'}{V} = \frac{\left(\frac{2}{3}N\right) k_B (3T)}{V} = 2 \frac{N k_B T}{V} = 2P\).

Award [1] for the correct pressure change. Correct answer is C.

The energy of the emitted photon is equal to the difference in energy between the initial and final states: \(\Delta E = E_{\text{initial}} - E_{\text{final}} = -E_1 - (-E_2) = E_2 - E_1\). Using the relation \(\Delta E = h f\), where \(f\) is the frequency of the photon, we get \(f = \frac{E_2 - E_1}{h}\).

Award [1] for the correct frequency. Correct answer is B.

Using the relation \(\varepsilon = I(R + r)\), we can set up two equations for the two situations: \(1) \varepsilon = 2.0(3.0 + r)\) and \(2) \varepsilon = 1.0(8.0 + r)\). Equating the two expressions for \(\varepsilon\): \(2.0(3.0 + r) = 1.0(8.0 + r) \implies 6.0 + 2.0r = 8.0 + r \implies r = 2.0\ \Omega\).

Award [1] for the correct calculation of internal resistance. Correct answer is B.

The satellite orbits at a distance \(r = R + R = 2R\) from the center of the planet.

The gravitational potential energy of the satellite in this orbit is:
\(V_p = -\frac{GMm}{r} = -\frac{GMm}{2R}

For the satellite to escape to infinity, its total energy must be at least zero:
\)E_k + V_p = 0 \implies \frac{1}{2}mv_{\text{esc}}^2 - \frac{GMm}{2R} = 0\)

Solving for \(v_{\text{esc}}\):
\(v_{\text{esc}} = \sqrt{\frac{GM}{R}}\)

[1 mark] for correctly determining the orbital radius as 2R and setting the total mechanical energy to zero to find the escape speed.

Let the initial number of nuclei be \(N_0\) for both. The decay constants are:
\(\lambda_X = \frac{\ln 2}{T}\) and \(\lambda_Y = \frac{\ln 2}{2T}\), so \(\lambda_X = 2\lambda_Y\).

After a time \(t = 2T\):
The number of remaining nuclei of \(X\) is:
\(N_X = N_0 \left(\frac{1}{2}\right)^{\frac{2T}{T}} = \frac{N_0}{4}

The number of remaining nuclei of \)Y\) is:
\(N_Y = N_0 \left(\frac{1}{2}\right)^{\frac{2T}{2T}} = \frac{N_0}{2}

The activity is given by \)A = \lambda N\):
\(A_X = \lambda_X N_X = (2\lambda_Y) \left(\frac{N_0}{4}\right) = \frac{\lambda_Y N_0}{2}
\)A_Y = \lambda_Y N_Y = \lambda_Y \left(\frac{N_0}{2}\right) = \frac{\lambda_Y N_0}{2}

Therefore, the ratio is:
\(\frac{A_X}{A_Y} = 1\)

[1 mark] for determining the active nuclei remaining after 2T and using the activity relationship with decay constants to find the ratio.

For single-slit diffraction, the first minimum occurs at:
\(\sin\theta \approx \theta = \frac{\lambda}{b}\)

When the width is halved to \(b' = \frac{b}{2}\) and the wavelength is doubled to \(\lambda' = 2\lambda\), the new angle \(\theta'\) is:
\(\theta' = \frac{\lambda'}{b'} = \frac{2\lambda}{\frac{b}{2}} = 4\left(\frac{\lambda}{b}\right) = 4\theta\)

[1 mark] for applying the single-slit diffraction formula and determining the change by a factor of 4.

As the loop is pulled out of the magnetic field, the changing magnetic flux induces an emf:
\(\varepsilon = BLv\)

The induced current is given by:
\(I = \frac{\varepsilon}{R} = \frac{BLv}{R}

The magnetic force opposing this motion acts on the section of the loop carrying current still within the magnetic field:
\)F = I L B = \left(\frac{BLv}{R}\right) L B = \frac{B^2 L^2 v}{R}

[1 mark] for using Faraday's law to find emf, Ohm's law to determine current, and the magnetic force formula to calculate the correct option.

Using conservation of momentum:
\(mv = (m + 2m)v_f \implies v_f = \frac{v}{3}

The initial kinetic energy is:
\)E_i = \frac{1}{2}mv^2

The final kinetic energy of the combined system is:
\(E_f = \frac{1}{2}(3m)v_f^2 = \frac{1}{2}(3m)\left(\frac{v}{3}\right)^2 = \frac{1}{6}mv^2 = \frac{1}{3}E_i

The energy dissipated as thermal energy is:
\)\Delta E = E_i - E_f = E_i - \frac{1}{3}E_i = \frac{2}{3}E_i

Therefore, the fraction of initial kinetic energy dissipated is \(\frac{2}{3}\).

[1 mark] for determining the final speed using conservation of momentum, calculating the final kinetic energy, and obtaining the ratio of energy loss to initial energy.

The root-mean-square speed of gas molecules is given by:
\(v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}

This indicates that the rms speed is directly proportional to the square root of the absolute temperature:
\)v_{\text{rms}} \propto \sqrt{T}

When the temperature is tripled to \(3T\), the new speed becomes:
\(v_{\text{rms}}' = \sqrt{3} \, v_{\text{rms}}\)

[1 mark] for recognizing that rms speed scales with \(\sqrt{T}\) and selecting the corresponding option.

The energy of the emitted photon is equal to the energy difference between the levels:
\(\Delta E = E_{\text{initial}} - E_{\text{final}} = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}

Using the relation between photon energy and wavelength:
\)E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E}

Substituting the values:
\(\lambda \approx \frac{1.24 \times 10^{-6}\text{ eV m}}{1.89\text{ eV}} \approx 6.56 \times 10^{-7}\text{ m}\)

[1 mark] for calculating the energy change of 1.89 eV and dividing the given value of hc by this energy to find the wavelength.

The power dissipated in the external resistor \(R\) is:
\(P = I^2 R = \left(\frac{\varepsilon}{R+r}\right)^2 R = \frac{\varepsilon^2 R}{(R+r)^2}

According to the Maximum Power Transfer Theorem, the power delivered to the load resistor is maximized when the resistance of the load resistor equals the internal resistance of the source, \)R = r\).

[1 mark] for correctly identifying that maximum power transfer occurs when load resistance is equal to the internal resistance.

For a circular orbit of radius \(r\), the kinetic energy is \(E_k = \frac{GMm}{2r}\) and the total energy is \(E_{\text{total}} = -\frac{GMm}{2r}\).

Given that the initial total energy at radius \(R\) is \(E_1 = -\frac{GMm}{2R}\), we have:
\(E_k(R) = \frac{GMm}{2R} = -E_1\).

When the orbit radius is changed to \(2R\), the new kinetic energy is:
\(E_k(2R) = \frac{GMm}{4R} = -\frac{1}{2} E_1\).

The change in kinetic energy is:
\(\Delta E_k = E_k(2R) - E_k(R) = -\frac{1}{2} E_1 - (-E_1) = \frac{1}{2} E_1\).

[1 mark] for identifying correct kinetic energy and total energy relations, performing the subtraction, and identifying option B as the correct answer.

The condition for the first minimum in single-slit diffraction is given by \(b \sin\theta = \lambda\). For a small angle, \(\sin\theta \approx \theta\), which gives \(\theta \approx \frac{\lambda}{b}\).

When the wavelength is halved (\(\lambda' = \frac{\lambda}{2}\)) and the slit width is doubled (\(b' = 2b\)), the new angle \(\theta'\) becomes:
\(\theta' \approx \frac{\lambda'}{b'} = \frac{\frac{\lambda}{2}}{2b} = \frac{\lambda}{4b} = \frac{\theta}{4}\).

[1 mark] for applying the single-slit diffraction formula, substituting the new values, and selecting option D.

According to Faraday's law of induction, the average electromotive force (emf) induced in the coil is:
\(\mathcal{E} = N \frac{\Delta \Phi}{\Delta t}\)

where \(\Delta \Phi = BA\) is the change in magnetic flux through a single turn of the coil since the initial flux is \(BA\) and the final flux is zero.

Using Ohm's law, the average current induced is:
\(I = \frac{\mathcal{E}}{R} = \frac{N BA}{R \Delta t}\)

The total charge \(q\) that flows through the coil is the product of the average current and the time interval:
\(q = I \Delta t = \left(\frac{N BA}{R \Delta t}\right) \Delta t = \frac{NBA}{R}\).

Notice that the charge is independent of the time interval \(\Delta t\).

[1 mark] for using Faraday's law and Ohm's law to relate current, time, and flux, and simplifying to find that charge is independent of time, matching option A.

The number of remaining active nuclei at any time \(t\) is given by \(N(t) = N_0 \left(\frac{1}{2}\right)^{t/T}\).

At \(t = T\) (one half-life):
\(N(T) = N_0 \left(\frac{1}{2}\right)^1 = \frac{N_0}{2}\).

At \(t = 3T\) (three half-lives):
\(N(3T) = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8}\).

The number of nuclei that decay during this interval is the difference between the number of active nuclei at the start and end of the interval:
\(\Delta N = N(T) - N(3T) = \frac{N_0}{2} - \frac{N_0}{8} = \frac{3}{8} N_0\).

[1 mark] for calculating the remaining active nuclei at 1 and 3 half-lives, finding the difference, and identifying option B as correct.

The energy levels of a hydrogen atom are proportional to \(-\frac{1}{n^2}\), so \(E_n = -\frac{E_0}{n^2}\) where \(E_0\) is the ionization energy.

The energy of the photon emitted in a transition from \(n = 4\) to \(n = 1\) is:
\(E_{4 \to 1} = -E_0 \left( \frac{1}{4^2} - \frac{1}{1^2} \right) = \frac{15}{16} E_0\).
Since \(E = hf\), we have \(hf = \frac{15}{16} E_0\).

The energy of the photon emitted in a transition from \(n = 2\) to \(n = 1\) is:
\(E_{2 \to 1} = -E_0 \left( \frac{1}{2^2} - \frac{1}{1^2} \right) = \frac{3}{4} E_0 = \frac{12}{16} E_0\).

Therefore, the new frequency \(f'\) is:
\(f' = \frac{E_{2 \to 1}}{E_{4 \to 1}} f = \frac{\frac{12}{16} E_0}{\frac{15}{16} E_0} f = \frac{12}{15} f = \frac{4}{5} f = 0.80f\).

[1 mark] for calculating the energy differences for both transitions, determining their ratio, and selecting option C.

Since the block moves with constant velocity, its acceleration is zero. By Newton's First Law, the net force acting on the block must be exactly zero.

Resolving the forces acting on the block:
- Parallel to the slope: The component of gravity pulling the block down is \(mg \sin\theta\), and the dynamic friction opposing this motion is \(f = \mu R\).
- Perpendicular to the slope: The normal reaction force is \(R = mg \cos\theta\).

Since there is no acceleration along the incline:
\(mg \sin\theta - f = 0 \implies mg \sin\theta = \mu mg \cos\theta\).

Solving for \(\mu\):
\(\mu = \frac{\sin\theta}{\cos\theta} = \tan\theta\).

[1 mark] for identifying that constant velocity implies zero net force, equating gravitational and frictional components to find the coefficient of friction, and choosing option B.

First, we must convert the temperatures from Celsius to Kelvin:
\(T_1 = 27 + 273 = 300\text{ K}\)
\(T_2 = 327 + 273 = 600\text{ K}\)

For an ideal gas at constant volume (isochoric process), pressure is directly proportional to absolute temperature (Gay-Lussac's Law):
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\)

Substituting the values:
\(P_2 = P_0 \times \frac{600}{300} = 2.0 P_0\).

[1 mark] for converting temperatures to Kelvin, using Gay-Lussac's law, and selecting option C.

Let's analyze the four possible configurations of three identical resistors of resistance \(R\):

1. **All three in series**:
\(R_{\text{eq}} = R + R + R = 3R\).

2. **All three in parallel**:
\(\frac{1}{R_{\text{eq}}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \implies R_{\text{eq}} = \frac{1}{3}R\) (Option A).

3. **Two in series, in parallel with the third**:
The series combination has resistance \(2R\). When in parallel with \(R\):
\(R_{\text{eq}} = \frac{2R \times R}{2R + R} = \frac{2}{3}R\) (Option B).

4. **Two in parallel, in series with the third**:
The parallel combination has resistance \(\frac{R}{2}\). When in series with \(R\):
\(R_{\text{eq}} = \frac{R}{2} + R = \frac{3}{2}R\) (Option D).

The value \(\frac{3}{4} R\) (Option C) cannot be produced by any network of three identical resistors.

[1 mark] for evaluating the possible equivalent resistances for three identical resistors and correctly identifying that C is the impossible value.

The total mechanical energy of a satellite in a circular orbit of radius \(r\) is given by \(E = -\frac{GMm}{2r}\). The initial energy is \(E_1 = -\frac{GMm}{2R}\). The final energy in the new orbit of radius \(3R\) is \(E_2 = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}\). The change in the total mechanical energy is \(\Delta E = E_2 - E_1 = -\frac{GMm}{6R} - \left(-\frac{GMm}{2R}\right) = \frac{GMm}{R} \left(-\frac{1}{6} + \frac{1}{2}\right) = \frac{GMm}{3R}\). Since the final energy is less negative, the mechanical energy of the satellite has increased by \(\frac{GMm}{3R}\).

Award 1 mark for the correct option B. Correctly identifies the formula for orbital mechanical energy (1 mark) and computes the positive energy difference (1 mark).

The angular half-width of the central maximum (the angle to the first minimum) in single-slit diffraction is given by \(\theta_{\text{min}} = \frac{\lambda}{b}\). The total angular width of the central maximum is \(\theta = 2\theta_{\text{min}} = \frac{2\lambda}{b}\). When the wavelength becomes \(2\lambda\) and the slit width becomes \(b/2\), the new angular width is \(\theta' = \frac{2(2\lambda)}{b/2} = 8\frac{\lambda}{b} = 4\left(\frac{2\lambda}{b}\right) = 4\theta\).

Award 1 mark for the correct option C. Recognizes that the angular width is proportional to the ratio of wavelength to slit width (1 mark) and determines the overall scaling factor of 4 (1 mark).

As the loop enters the magnetic field, a magnetic flux is swept through it. The induced electromotive force (emf) is given by Faraday's law: \(\varepsilon = BLv\). The resulting induced current in the loop is \(I = \frac{\varepsilon}{R} = \frac{BLv}{R}\). The magnetic force acting on the leading edge of the loop inside the field opposes its motion (by Lenz's law) and is given by \(F = BIL = B \left(\frac{BLv}{R}\right) L = \frac{B^2 L^2 v}{R}\).

Award 1 mark for the correct option C. Identifies the induced EMF as \(BLv\) (1 mark), the current as \(I = \frac{BLv}{R}\) (1 mark), and calculates the magnetic force as \(BIL = \frac{B^2 L^2 v}{R}\) (1 mark).

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

Award 1 mark for the correct option A. Identifies that 1/8 of the original sample remains (1 mark), calculates that 7/8 has decayed (1 mark), and computes the ratio as 7 (1 mark).

The energy of the emitted photon is equal to the difference in energy between the two levels: \(\Delta E = E_{\text{initial}} - E_{\text{final}} = -E_0 - (-4E_0) = 3E_0\). The energy of a photon is related to its wavelength by \(E = \frac{hc}{\lambda}\). Setting the two equal: \(3E_0 = \frac{hc}{\lambda}\), which gives \(\lambda = \frac{hc}{3E_0}\).

Award 1 mark for the correct option A. Determines the transition energy difference of \(3E_0\) (1 mark) and applies the formula \(E = \frac{hc}{\lambda}\) to solve for the wavelength (1 mark).

Using conservation of linear momentum: \(m v + 0 = (m + 3m) v_f\), so the final velocity of the combined system is \(v_f = \frac{v}{4}\). The initial kinetic energy is \(E_{k,i} = \frac{1}{2}mv^2\). The final kinetic energy is \(E_{k,f} = \frac{1}{2}(4m)v_f^2 = 2m \left(\frac{v}{4}\right)^2 = \frac{1}{8}mv^2\). The energy dissipated is \(\Delta E_k = E_{k,i} - E_{k,f} = \frac{1}{2}mv^2 - \frac{1}{8}mv^2 = \frac{3}{8}mv^2\). The fraction dissipated is \(\frac{\Delta E_k}{E_{k,i}} = \frac{\frac{3}{8}mv^2}{\frac{1}{2}mv^2} = \frac{3}{4}\).

Award 1 mark for the correct option D. Finds the final velocity using conservation of momentum (1 mark), calculates initial and final kinetic energies (1 mark), and computes the ratio of the change to initial kinetic energy (1 mark).

The temperature must be converted to kelvins. Initial temperature \(T_1 = 27 + 273 = 300\text{ K}\). Final temperature \(T_2 = 327 + 273 = 600\text{ K}\). According to Gay-Lussac's law for a constant volume, \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\). Therefore, \(P_2 = P_1 \frac{T_2}{T_1} = P \frac{600}{300} = 2P\).

Award 1 mark for the correct option B. Converts temperatures to absolute kelvin scale (1 mark) and applies Gay-Lussac's law to find the new pressure (1 mark).

For the series connection, the total equivalent resistance is \(R_s = R + R = 2R\). The total power dissipated is \(P_s = \frac{V^2}{R_s} = \frac{V^2}{2R}\). For the parallel connection, the total equivalent resistance is \(R_p = \left(\frac{1}{R} + \frac{1}{R}\right)^{-1} = \frac{R}{2}\). The total power dissipated is \(P_p = \frac{V^2}{R_p} = \frac{2V^2}{R}\). The ratio of parallel to series power is \(\frac{P_p}{P_s} = \frac{2V^2/R}{V^2/(2R)} = 4\).

Award 1 mark for the correct option D. Finds equivalent resistances for series and parallel circuits (1 mark), expresses power for both configurations (1 mark), and computes the ratio (1 mark).

The lifetime \(\tau\) of a main-sequence star is proportional to the ratio of its mass (fuel supply) to its luminosity (rate of fuel consumption): \(\tau \propto \frac{M}{L}\).

Given the mass-luminosity relationship is \(L \propto M^{3.5}\), we can substitute this into the lifetime relation:
\(\tau \propto \frac{M}{M^{3.5}} = M^{-2.5}\)

Therefore, the ratio of the lifetimes of the two stars is:
\(\frac{\tau_X}{\tau_Y} = \left(\frac{M_X}{M_Y}\right)^{-2.5} = (4)^{-2.5}\)

We can calculate this value without a calculator:
\(4^{-2.5} = (2^2)^{-2.5} = 2^{-5} = \frac{1}{32}\)

Award [1] for the correct answer: A.
Award [0] for incorrect answers.

Let the initial activity of Y be \(A_0\). Therefore, the initial activity of X is \(2A_0\).

After a time interval \(t = 4T\):
- For nuclide X, the elapsed time represents 4 half-lives (since its half-life is \(T\)). Its activity becomes:
\(A_X = 2A_0 \times \left(\frac{1}{2}\right)^4 = \frac{2A_0}{16} = \frac{A_0}{8}\)

- For nuclide Y, the elapsed time represents 2 half-lives (since its half-life is \(2T\)). Its activity becomes:
\(A_Y = A_0 \times \left(\frac{1}{2}\right)^2 = \frac{A_0}{4}\)

Taking the ratio of their activities after \(4T\):
\(\frac{A_X}{A_Y} = \frac{A_0 / 8}{A_0 / 4} = \frac{4}{8} = \frac{1}{2}\)

Award [1] for the correct answer: B.
Award [0] for incorrect answers.

The angle for the first minimum in single-slit diffraction is given by the formula:
\(\theta = \frac{\lambda}{b}\)

For the modified setup, the new wavelength is \(\lambda' = 2\lambda\) and the new slit width is \(b' = \frac{b}{2}\).

The new angle for the first minimum is:
\(\theta' = \frac{\lambda'}{b'} = \frac{2\lambda}{\frac{b}{2}} = 4 \left(\frac{\lambda}{b}\right) = 4\theta\)

Award [1] for the correct answer: C.
Award [0] for incorrect answers.

The gravitational potential energy \(E_p\) of a mass \(m\) at a distance \(r\) from a mass \(M\) is:
\(E_p = -\frac{GMm}{r}\)

The initial potential energy is:
\(E_{p,i} = -\frac{GMm}{r}\)

The final potential energy at radius \(2r\) is:
\(E_{p,f} = -\frac{GMm}{2r}\)

The change in gravitational potential energy \(\Delta E_p\) is:
\(\Delta E_p = E_{p,f} - E_{p,i} = -\frac{GMm}{2r} - \left(-\frac{GMm}{r}\right) = -\frac{GMm}{2r} + \frac{GMm}{r} = +\frac{GMm}{2r}\)

Award [1] for the correct answer: B.
Award [0] for incorrect answers.

According to Faraday's law of induction, the magnitude of the average induced emf \(\varepsilon\) is given by:
\(\varepsilon = \left| \frac{\Delta \Phi}{\Delta t} \right|\)
where \(\Phi\) is the magnetic flux linkage.

Initially, the magnetic field is perpendicular to the plane of the coil, so the initial flux linkage is:
\(\Phi_i = N B A\)

After a \(180^\circ\) rotation, the coil has been flipped, so the magnetic field is now in the opposite direction relative to the normal vector of the coil area. The final flux linkage is:
\(\Phi_f = -N B A\)

The magnitude of the change in magnetic flux linkage is:
\(|\Delta \Phi| = |\Phi_f - \Phi_i| = |-N B A - N B A| = 2 N B A\)

Thus, the average induced emf is:
\(\varepsilon = \frac{2 N B A}{\Delta t}\)

Award [1] for the correct answer: C.
Award [0] for incorrect answers.

First, define a coordinate system where the initial direction of motion is positive.
- Initial momentum: \(p_i = mv\)
- Final momentum: \(p_f = -m\left(\frac{v}{2}\right)\)

The change in momentum (impulse) is:
\(\Delta p = p_f - p_i = -\frac{mv}{2} - mv = -\frac{3}{2}mv\)

The magnitude of the change in momentum is \(\frac{3}{2}mv\).

From Newton's second law, the average force is the rate of change of momentum:
\(F_{\text{avg}} = \frac{|\Delta p|}{\Delta t} = \frac{3mv}{2\Delta t}\)

Award [1] for the correct answer: C.
Award [0] for incorrect answers.

From the ideal gas law: \(P V = N k_B T\).
Since the volume \(V\) of the container is fixed and \(k_B\) is a constant, the pressure is directly proportional to the product of the number of molecules and the absolute temperature:
\(P \propto N T\)

Initially:
\(P_1 = P\), \(N_1 = N\), and \(T_1 = T\).

Finally:
\(N_2 = 0.5 N\) and \(T_2 = 1.5 T\).

The new pressure \(P_2\) is:
\(P_2 = P_1 \times \left(\frac{N_2}{N_1}\right) \times \left(\frac{T_2}{T_1}\right) = P \times 0.5 \times 1.5 = 0.75 P = \frac{3}{4} P\)

Award [1] for the correct answer: A.
Award [0] for incorrect answers.

The energy of the emitted photon is equal to the difference in energy between the two states:
\(\Delta E = E_{\text{initial}} - E_{\text{final}} = -E_0 - (-4E_0) = 3E_0\)

The relationship between the energy of a photon and its wavelength \(\lambda\) is given by:
\(\Delta E = \frac{hc}{\lambda}\)

Rearranging for \(\lambda\):
\(\lambda = \frac{hc}{\Delta E} = \frac{hc}{3E_0}\)

Award [1] for the correct answer: B.
Award [0] for incorrect answers.

(a) When light reflects from an interface with a medium of higher optical density (higher refractive index), it undergoes a phase shift of \(\pi\) radians. This is a boundary condition effect, analogous to a mechanical wave on a string reflecting off a fixed end, where the boundary forces the wave amplitude to zero, resulting in a phase inversion.

(b) (i) Since \(n_{\text{air}} < n_f < n_g\), phase changes of \(\pi\) occur at both the air-coating and the coating-glass boundaries. Therefore, the relative phase change from reflection alone is zero. For destructive interference (minimum reflection), the path difference must be half a wavelength in the film: \(2d = \frac{\lambda_f}{2} = \frac{\lambda}{2n_f}\). Solving for thickness: \(d = \frac{\lambda}{4n_f} = \frac{550 \times 10^{-9}}{4 \times 1.38} = 9.96 \times 10^{-8}\text{ m} \approx 100\text{ nm}\).
(ii) The coating is optimized for the center of the visible spectrum (green-yellow). Other wavelengths (like red and violet/blue) undergo less destructive interference and are reflected more, making the lens appear purple.

(c) The condition for the first diffraction minimum is \(\theta = \frac{\lambda}{b}\). The half-width of the central maximum on the screen is \(y = D \tan\theta \approx D \theta = \frac{D\lambda}{b}\). The total width of the central maximum is \(2y = \frac{2 D \lambda}{b} = \frac{2 \times 2.0 \times 6.33 \times 10^{-7}}{1.2 \times 10^{-4}} = 0.0211\text{ m} = 21\text{ mm}\).

(a)
- Reflection from a denser medium results in a phase inversion / \(\pi\) shift. [1]
- Boundary condition analogy (e.g., reflection at a fixed boundary). [1]

(b)(i)
- Recognition that reflections at both boundaries have a \(\pi\) phase change. [1]
- Correct equation: \(2d = \frac{\lambda}{2n_f}\) or \(d = \frac{\lambda}{4n_f}\). [1]
- Correct substitution resulting in \(9.96 \times 10^{-8}\text{ m}\) or \(99.6\text{ nm}\). [1]

(b)(ii)
- Red and blue/violet light are not fully cancelled by destructive interference and reflect back. [1]

(c)
- Recalls single slit minimum formula: \(\theta = \frac{\lambda}{b}\). [1]
- States relation for central half-width: \(y = D\theta\) or similar. [1]
- Identifies that total width is \(2y\). [1]
- Calculates \(21\text{ mm}\) (accept range \(21 - 21.1\text{ mm}\)). [1]

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

(b) (i) The gravitational force provides the centripetal force: \(\frac{G M m}{r^2} = \frac{m v^2}{r}\), where \(r = R + h = 3.4 \times 10^6 + 1.2 \times 10^6 = 4.6 \times 10^6\text{ m}\).
Solving for speed: \(v = \sqrt{\frac{G M}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{4.6 \times 10^6}} = \sqrt{9.28 \times 10^6} = 3.046 \times 10^3\text{ m s}^{-1} \approx 3.0 \times 10^3\text{ m s}^{-1}\).

(ii) The total mechanical energy is \(E_k + E_p = \frac{1}{2}mv^2 - \frac{GMm}{r} = -\frac{GMm}{2r}\).
\(E = -\frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23} \times 450}{2 \times 4.6 \times 10^6} = -2.09 \times 10^9\text{ J} \approx -2.1 \times 10^9\text{ J}\).

(c) The resistive force does negative work on the satellite, so its total mechanical energy decreases (becomes more negative). As the total energy decreases, the satellite spirals to a lower orbit (smaller \(r\)), and because \(v = \sqrt{\frac{GM}{r}}\), its orbital speed actually increases.

(a)
- Work done per unit mass. [1]
- In bringing a test mass from infinity to the point. [1]

(b)(i)
- Equating gravitational force to centripetal force: \(\frac{GMm}{r^2} = \frac{mv^2}{r}\). [1]
- Correct orbital radius \(r = 4.6 \times 10^6\text{ m}\). [1]
- Correct substitution showing \(3.05 \times 10^3\text{ m s}^{-1}\). [1]

(b)(ii)
- Using \(E = -\frac{GMm}{2r}\) or summing kinetic and potential energy. [1]
- Correct substitution of values. [1]
- Final answer: \(-2.1 \times 10^9\text{ J}\) (negative sign is essential). [1]

(c)
- Total mechanical energy decreases. [1]
- Orbital speed increases (as radius decreases). [1]

(a) The decay constant is the probability of decay of a nucleus per unit time.

(b) (i) \(T_{1/2} = 5730\text{ years} = 5730 \times 365.25 \times 24 \times 3600\text{ s} = 1.808 \times 10^{11}\text{ s}\).
\(\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.69315}{1.808 \times 10^{11}} = 3.83 \times 10^{-12}\text{ s}^{-1} \approx 3.8 \times 10^{-12}\text{ s}^{-1}\).

(ii) The fraction of carbon-14 remaining is \(\frac{N}{N_0} = \frac{2.5 \times 10^{10}}{2.0 \times 10^{11}} = 0.125 = \frac{1}{8}\).
Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly 3 half-lives have elapsed.
Age = \(3 \times 5730\text{ years} = 17190\text{ years} \approx 1.7 \times 10^4\text{ years}\).

(c) Activity is the rate at which unstable nuclei decay in the entire source (measured in becquerels). The count rate is the number of decay particles detected by a measurement device (like a Geiger-Muller tube) per unit time, which is much lower than the activity due to detector geometry, absorption, and efficiency, but includes background radiation.

(a)
- Probability of decay of a nucleus [1]
- Per unit time. [1]

(b)(i)
- Correct conversion of years to seconds (approx. \(1.81 \times 10^{11}\text{ s}\)). [1]
- Recalls \(\lambda = \frac{\ln 2}{T_{1/2}}\). [1]
- Correct calculation to get \(3.8 \times 10^{-12}\text{ s}^{-1}\). [1]

(b)(ii)
- Finds fraction remaining is \(0.125\) or \(1/8\). [1]
- Identifies that this corresponds to 3 half-lives. [1]
- Calculates age: \(1.7 \times 10^4\text{ years}\) (or \(17190\text{ years}\)). [1]

(c)
- Activity is the total number of decays per second in the source. [1]
- Count rate is the number of decays detected/registered by a detector, which is smaller than activity. [1]

(a) Faraday’s law states that the induced electromotive force (emf) in a circuit is directly proportional to the rate of change of magnetic flux linkage through the circuit.

(b) (i) \(\Phi_{\text{initial}} = B A = 0.25 \times 4.5 \times 10^{-3} = 1.125 \times 10^{-3}\text{ Wb}\).
\(\Delta \Phi = -1.125 \times 10^{-3}\text{ Wb}\).
\(\text{emf} = N \frac{\Delta \Phi}{\Delta t} = 150 \times \frac{1.125 \times 10^{-3}}{0.12} = 1.406\text{ V} \approx 1.4\text{ V}\).

(ii) Power \(P = \frac{V^2}{R} = \frac{1.406^2}{8.0} = 0.247\text{ W} \approx 0.25\text{ W}\).

(c) Lenz's law states that the induced current will flow in a direction that opposes the change in magnetic flux that created it. Here, the upward magnetic flux is decreasing. To oppose this decrease, the induced current must create its own upward magnetic field. By the right-hand grip rule, looking from above, a counter-clockwise current is required to produce an upward magnetic field.

(a)
- Induced emf is proportional to / equal to the rate of change of flux linkage. [1]
- Defines symbols or explicitly mentions "magnetic flux linkage". [1]

(b)(i)
- Calculates initial magnetic flux: \(1.125 \times 10^{-3}\text{ Wb}\). [1]
- Applies \(V = N \frac{\Delta \Phi}{\Delta t}\). [1]
- Yields \(1.4\text{ V}\). [1]

(b)(ii)
- Uses \(P = \frac{V^2}{R}\) or calculates current first (\(I = 0.176\text{ A}\)) then uses \(P = I^2R\). [1]
- Yields \(0.25\text{ W}\). [1]

(c)
- Opposes the decrease in upward flux by producing upward flux. [1]
- Right-hand rule link established. [1]
- Deduces direction is counter-clockwise (when viewed from above). [1]

(a) Assumptions include: (1) The intermolecular forces are negligible except during collisions; (2) The volume of the molecules is negligible compared to the volume of the gas container; (3) Collisions are perfectly elastic; (4) The molecules are in continuous, rapid, random motion.

(b) (i) Since pressure is constant, Charles’s Law applies: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\).
As \(V_2 = 2V_1\), \(T_2 = 2T_1 = 2 \times 300 = 600\text{ K}\).

(ii) Using the ideal gas equation: \(p V = n R T\).
\(n = \frac{p_1 V_1}{R T_1} = \frac{1.2 \times 10^5 \times 3.5 \times 10^{-3}}{8.31 \times 300} = \frac{420}{2493} = 0.168\text{ mol} \approx 0.17\text{ mol}\).

(iii) The work done by the gas during isobaric expansion is \(W = p \Delta V\).
\(\Delta V = V_2 - V_1 = 3.5 \times 10^{-3}\text{ m}^3\).
\(W = 1.2 \times 10^5 \times 3.5 \times 10^{-3} = 420\text{ J}\).

(a)
- Award [1] per valid assumption (max 2), e.g., point masses, elastic collisions, no intermolecular forces. [2]

(b)(i)
- Recognizes relationship \(V \propto T\) at constant pressure. [1]
- Yields \(600\text{ K}\). [1]

(b)(ii)
- Uses \(pV = nRT\). [1]
- Correct substitution of values. [1]
- Calculates \(0.17\text{ mol}\) (accept \(0.168\text{ mol}\)). [1]

(b)(iii)
- Recalls \(W = p\Delta V\). [1]
- Computes \(\Delta V = 3.5 \times 10^{-3}\text{ m}^3\). [1]
- Yields \(420\text{ J}\). [1]

(a) Electrons inside an atom can only occupy specific, quantized energy states. When an electron transitions from a higher energy level to a lower energy level, it loses energy and emits a single photon. The energy of this photon is exactly equal to the difference in energy between the two states: \(E_{\text{photon}} = \Delta E = hf\). Since the energy levels are discrete, only specific photon energies (and thus specific frequencies and wavelengths) can be emitted, resulting in separate lines on the emission spectrum.

(b) (i) Energy of state \(n=3\): \(E_3 = -\frac{13.6}{3^2} = -1.511\text{ eV}\).
Energy of state \(n=2\): \(E_2 = -\frac{13.6}{2^2} = -3.400\text{ eV}\).
Energy difference: \(\Delta E = E_3 - E_2 = -1.511 - (-3.400) = 1.889\text{ eV}\).
Convert energy to joules: \(1.889\text{ eV} \times 1.60 \times 10^{-19}\text{ J/eV} = 3.022 \times 10^{-19}\text{ J} \approx 3.0 \times 10^{-19}\text{ J}\).

(ii) Using \(E = \frac{hc}{\lambda}\), we get:
\(\lambda = \frac{hc}{E} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{3.022 \times 10^{-19}} = 6.582 \times 10^{-7}\text{ m} \approx 6.6 \times 10^{-7}\text{ m}\) (or \(660\text{ nm}\)).

(a)
- Electrons transition from high to low energy levels and emit a photon. [1]
- Photon energy equals difference in energy levels (\(E = hf\)). [1]
- Since levels are discrete, only specific frequencies/lines are produced. [1]

(b)(i)
- Computes energies of \(n=3\) and \(n=2\) levels in eV. [1]
- Calculates transition energy as \(1.89\text{ eV}\). [1]
- Multiplies by charge of electron to convert to J. [1]
- Obtains \(3.0 \times 10^{-19}\text{ J}\). [1]

(b)(ii)
- Recalls \(E = \frac{hc}{\lambda}\). [1]
- Correct substitution of values. [1]
- Obtains \(6.6 \times 10^{-7}\text{ m}\) (accept \(6.58 \times 10^{-7}\text{ m}\)). [1]

(a) Electromotive force (emf) is the total energy converted from other forms (e.g., chemical) to electrical energy per unit charge by the cell. Potential difference (pd) is the electrical energy converted to other forms of energy (e.g., thermal) per unit charge in the external component. Emf measures total energy supplied to the circuit per coulomb, whereas terminal pd is the work available per coulomb outside the cell (lower than emf due to internal resistance).

(b) (i) Using \(\varepsilon = I(R + r)\):
Case 1: \(\varepsilon = 1.5(4.0 + r) = 6.0 + 1.5r\)
Case 2: \(\varepsilon = 0.80(9.0 + r) = 7.2 + 0.8r\)
Set them equal to solve for \(r\):
\(6.0 + 1.5r = 7.2 + 0.8r \implies 0.7r = 1.2 \implies r = 1.71\text{ }\Omega \approx 1.7\text{ }\Omega\).
Now solve for \(\varepsilon\):
\(\varepsilon = 6.0 + 1.5(1.714) = 8.57\text{ V} \approx 8.6\text{ V}\).

(ii) For \(R = 4.0\text{ }\Omega\), the current is \(I = 1.5\text{ A}\).
Power dissipated in internal resistance: \(P = I^2 r = 1.5^2 \times 1.714 = 2.25 \times 1.714 = 3.86\text{ W} \approx 3.9\text{ W}\).

(a)
- Emf is work done per unit charge (energy conversion) in the whole circuit / inside source. [1]
- Terminal pd is work done per unit charge in the external circuit only. [1]
- Terminal pd is less than emf due to "lost volts" / internal resistance when current flows. [1]

(b)(i)
- Writes \(\varepsilon = I(R + r)\) or equivalent circuit loop equation. [1]
- Forms two correct simultaneous equations: \(\varepsilon = 1.5(4.0 + r)\) and \(\varepsilon = 0.8(9.0 + r)\). [1]
- Equates and solves for \(r\): \(1.7\text{ }\Omega\) (accept range \(1.7 - 1.71\text{ }\Omega\)). [2]
- Solves for \(\varepsilon\): \(8.6\text{ V}\) (accept range \(8.55 - 8.6\text{ V}\)). [1]

(b)(ii)
- Uses \(P = I^2 r\). [1]
- Obtains \(3.9\text{ W}\) (accept range \(3.8 - 3.9\text{ W}\) depending on rounding). [1]

(a) Specific latent heat of fusion is the amount of energy required to change unit mass (\(1\text{ kg}\)) of a substance from solid to liquid phase without any change in temperature.

(b) (i) \(Q_1 = m_{\text{ice}} L_f = 0.045 \times 3.3 \times 10^5 = 14850\text{ J} \approx 1.5 \times 10^4\text{ J}\).

(ii) Let \(T_f\) be the final equilibrium temperature in \(^\circ\text{C}\).
Heat lost by the warm water: \(Q_{\text{lost}} = m_{\text{water}} c_{\text{water}} (T_{\text{initial}} - T_f) = 0.25 \times 4200 \times (22 - T_f) = 1050(22 - T_f) = 23100 - 1050 T_f\).
Heat gained by the melting ice and the resulting ice-water: \(Q_{\text{gained}} = Q_1 + m_{\text{ice}} c_{\text{water}} (T_f - 0) = 14850 + 0.045 \times 4200 \times T_f = 14850 + 189 T_f\).
By conservation of energy, \(Q_{\text{lost}} = Q_{\text{gained}}\):
\(23100 - 1050 T_f = 14850 + 189 T_f\)
\(23100 - 14850 = (1050 + 189) T_f\)
\(8250 = 1239 T_f \implies T_f = 6.658^\circ\text{C} \approx 6.7^\circ\text{C}\).

(a)
- Energy per unit mass. [1]
- Required to change state from solid to liquid at constant temperature. [1]

(b)(i)
- Recalls \(Q = mL\). [1]
- Correct substitution of values. [1]
- Yields \(1.5 \times 10^4\text{ J}\) (or \(14850\text{ J}\)). [1]

(b)(ii)
- Expresses heat lost by warm water: \(0.25 \times 4200 \times (22 - T_f)\). [1]
- Expresses heat gained by melted ice water: \(0.045 \times 4200 \times T_f\). [1]
- Equates heat lost to heat gained: \(m_w c_w \Delta T_w = m_i L_f + m_i c_w T_f\). [1]
- Sets up the equation correctly: \(23100 - 1050 T_f = 14850 + 189 T_f\). [1]
- Yields \(6.7^\circ\text{C}\) (accept range \(6.6 - 6.7^\circ\text{C}\)). [1]

**(a)**
To maintain a stable circular orbit, the gravitational force acting on the probe provides the necessary centripetal force:
\( F_g = F_c \)

Substitute the expressions for gravitational and centripetal forces:
\( \frac{GMm}{r^2} = \frac{mv^2}{r} \)

where \( r = R + h \) is the orbital radius. Simplifying the equation:
\( \frac{GM}{r} = v^2 \Rightarrow v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{R+h}} \)

**(b)**
(i) First, find the orbital radius \( r \):
\( r = R + h = 9.50 \times 10^5 \text{ m} + 3.50 \times 10^5 \text{ m} = 1.30 \times 10^6 \text{ m} \)

Using Kepler's third law or the relation between orbital period \( T \) and orbital speed:
\( T^2 = \frac{4\pi^2 r^3}{GM} \)

Substitute the given values:
\( T^2 = \frac{4\pi^2 \times (1.30 \times 10^6)^3}{6.67 \times 10^{-11} \times 1.50 \times 10^{22}} \)
\( T^2 = \frac{39.478 \times 2.197 \times 10^{18}}{1.0005 \times 10^{12}} = 8.669 \times 10^7 \text{ s}^2 \)
\( T = \sqrt{8.669 \times 10^7} \approx 9.31 \times 10^3 \text{ s} \)

(ii) The gravitational potential energy \( E_p \) is given by:
\( E_p = -\frac{GMm}{r} \)

Substitute the values:
\( E_p = -\frac{6.67 \times 10^{-11} \times 1.50 \times 10^{22} \times 1200}{1.30 \times 10^6} \)
\( E_p = -9.24 \times 10^8 \text{ J} \) (or \( -9.2 \times 10^8 \text{ J} \))

**(c)**
- In a circular orbit, the total energy is \( E_{\text{total}} = E_p + E_k = -\frac{GMm}{2r} \).
- When moving to a larger orbital radius \( r \), the gravitational potential energy increases (becomes less negative/more positive) because work is done against the gravitational field:
\( \Delta E_p = -\frac{GMm}{2r_1} - \left(-\frac{GMm}{r_1}\right) = +\frac{GMm}{2r_1} \).
- At the same time, the orbital speed decreases, so kinetic energy decreases by:
\( \Delta E_k = -\frac{GMm}{4r_1} \).
- Since the increase in potential energy is twice the magnitude of the decrease in kinetic energy, the total mechanical energy of the satellite increases: \( \Delta E_{\text{total}} = +\frac{GMm}{4r_1} > 0 \).
- Therefore, positive work must be done by the engines to increase the total energy of the orbital system.

**(a)**
- Equates gravitational force to centripetal force: \( \frac{GMm}{r^2} = \frac{mv^2}{r} \) OR \( g = \frac{v^2}{r} \) [1]
- States that \( r = R+h \) and correctly rearranges to show \( v = \sqrt{\frac{GM}{R+h}} \) [1]

**(b)(i)**
- Calculates orbital radius: \( r = 1.30 \times 10^6 \text{ m} \) [1]
- Correctly uses a formula linking period, radius, and mass, e.g., \( T = \sqrt{\frac{4\pi^2 r^3}{GM}} \) OR finds \( v \approx 877 \text{ m s}^{-1} \) and uses \( T = \frac{2\pi r}{v} \) [1]
- Final answer: \( 9.31 \times 10^3 \text{ s} \) (accept \( 9.3 \times 10^3 \text{ s} \) to \( 9.4 \times 10^3 \text{ s} \)) [1]

**(b)(ii)**
- Correct formula choice: \( E_p = -\frac{GMm}{r} \) [1]
- Correct calculation with negative sign: \( -9.24 \times 10^8 \text{ J} \) (accept \( -9.2 \times 10^8 \text{ J} \)) [1]
*Award [1 max] if negative sign is omitted.*

**(c)**
- Realizes that potential energy increases (becomes less negative) as radius increases [1]
- Realizes that kinetic energy decreases as radius increases (due to lower orbital speed) [1]
- Explains that the increase in gravitational potential energy is greater than the decrease in kinetic energy, so the overall total energy of the orbit increases, meaning positive work must be done [1]

(a) Measuring 20 oscillations spreads the human reaction time (which is a constant systematic/random error of about \(0.1\text{ s}\) to \(0.2\text{ s}\) at both start and stop) over a much longer total duration. The absolute uncertainty in the total time \(\Delta t = 0.4\text{ s}\) translates to an absolute uncertainty in the period \(T\) of \(\Delta T = \frac{\Delta t}{20} = 0.02\text{ s}\), which is 20 times smaller than timing a single oscillation directly.

(b) (i) The period is calculated as:
\(T = \frac{36.0\text{ s}}{20} = 1.80\text{ s}\)

The absolute uncertainty in the period is:
\(\Delta T = \frac{0.4\text{ s}}{20} = 0.02\text{ s}\)

Thus, \(T = 1.80 \pm 0.02\text{ s}\).

(ii) The percentage uncertainty in \(T\) is:
\(\frac{\Delta T}{T} \times 100\% = \frac{0.02}{1.80} \times 100\% \approx 1.11\%\)

Since \(T^2\) is proportional to the square of \(T\), its percentage uncertainty is doubled:
\(2 \times 1.11\% = 2.22\%\) (accept \(2.2\%\)).

(c) (i) Squaring both sides of the pendulum equation gives:
\(T^2 = \left(\frac{4\pi^2}{g}\right) L\)

Comparing this to the equation of a straight line \(y = mx\), the gradient is:
\(\text{gradient} = \frac{4\pi^2}{g}\)

(ii) Rearranging for \(g\):
\(g = \frac{4\pi^2}{\text{gradient}} = \frac{4\pi^2}{4.02} \approx 9.82\text{ m s}^{-2}\).

(a)
- Explains that the constant absolute reaction time error is divided/distributed over 20 cycles. [1]
- Connects this to a 20-fold reduction in the absolute uncertainty of the period (\(\Delta T = \frac{\Delta t}{20}\)). [1]

(b) (i)
- Correctly calculates \(T = 1.80\text{ s}\). [1]
- Correctly calculates \(\Delta T = 0.02\text{ s}\). [1]

(b) (ii)
- Correctly determines the fractional or percentage uncertainty in \(T\) as \(1.11\%\) (or \(0.0111\)). [1]
- Doubles the uncertainty for the squared quantity to get \(2.22\%\) or \(2.2\%\). [1]

(c) (i)
- Correctly identifies the gradient relation: \(\text{gradient} = \frac{4\pi^2}{g}\). [1]

(c) (ii)
- Correctly calculates \(g = 9.82\text{ m s}^{-2}\) (accept 9.8). [1]

(a) The circuit diagram should feature a series loop containing:
- The cell to be tested
- An ammeter connected in series to measure the circuit current
- A variable resistor (rheostat) to alter the load and vary the current
Additionally, a voltmeter must be connected in parallel across the terminals of the cell (or across the variable resistor) to measure the terminal potential difference.

(b) Comparing the equation \(V = -r I + \varepsilon\) to the equation of a straight line \(y = mx + c\):
- When \(I = 0\), \(V = \varepsilon\). Therefore, the vertical (y) intercept of the graph represents the electromotive force \(\varepsilon\).
- The slope/gradient of the graph is equal to \(-r\). Therefore, the internal resistance \(r\) is equal to the negative gradient (\(r = -\text{gradient}\)).

(c) The power delivered to the external circuit is:
\(P = V \times I = 1.35\text{ V} \times 0.40\text{ A} = 0.54\text{ W}\)

To calculate the uncertainty in the power, we add the relative (fractional) uncertainties of the multiplied quantities:
\(\frac{\Delta P}{P} = \frac{\Delta I}{I} + \frac{\Delta V}{V}\)

\(\frac{\Delta I}{I} = \frac{0.02}{0.40} = 0.050\) (or \(5.0\%\))

\(\frac{\Delta V}{V} = \frac{0.05}{1.35} \approx 0.037\) (or \(3.7\%\))

\(\frac{\Delta P}{P} = 0.050 + 0.037 = 0.087\) (or \(8.7\%\))

Now, find the absolute uncertainty in power:
\(\Delta P = P \times 0.087 = 0.54\text{ W} \times 0.087 \approx 0.047\text{ W}\)

Rounding to an appropriate single significant figure for absolute uncertainty yields \(0.05\text{ W}\).

Hence, \(P = 0.54 \pm 0.05\text{ W}\).

(a)
- Shows ammeter, cell, and variable resistor all connected in a single series loop. [1]
- Voltmeter connected correctly in parallel across the cell or across the load resistor. [1]

(b)
- Explains that the vertical intercept equals the emf \(\varepsilon\). [1]
- Explains that the internal resistance \(r\) is found from the negative gradient (or \(r = -\text{gradient}\)). [1]

(c)
- Correctly calculates power \(P = 0.54\text{ W}\). [1]
- Correctly calculates the sum of fractional/percentage uncertainties (\(8.7\%\) or \(0.087\)). [1]
- Correctly determines the absolute uncertainty \(\Delta P = 0.05\text{ W}\) (accept \(0.047\text{ W}\)). [1]

(a) A diagram showing Earth, Sun, and the star forming a right-angled triangle, where the base is \(1\text{ AU}\) (Earth-Sun distance) and the parallax angle at the star is \(1\text{ arcsecond}\). The parsec is the distance from the Sun to the star when this angle is exactly 1 arcsecond.

(b) (i) Distance is given by:
\(d = \frac{1}{p} = \frac{1}{0.125} = 8.0\text{ pc}\)

(ii) Convert parsecs to meters:
\(d = 8.0 \times 3.09 \times 10^{16}\text{ m} = 2.47 \times 10^{17}\text{ m}\)
Using the brightness formula \(b = \frac{L}{4\pi d^2}\):
\(L = 4\pi d^2 b = 4\pi \times (2.47 \times 10^{17})^2 \times (2.4 \times 10^{-10}) = 1.84 \times 10^{26}\text{ W}\)

(a)
- Award [1] for a clear diagram showing a right-angled triangle with a baseline of 1 AU (or Earth-Sun distance) and a parallax angle of 1 arcsecond.
- Award [1] for stating that 1 parsec is the distance at which 1 AU subtends an angle of 1 arcsecond.

(b) (i)
- Award [1] for \(8.0\text{ pc}\).

(b) (ii)
- Award [1] for converting parsecs to meters: \(d = 2.47 \times 10^{17}\text{ m}\) (accept \(2.5 \times 10^{17}\text{ m}\)).
- Award [1] for stating the formula \(L = 4\pi d^2 b\).
- Award [1] for the final calculation of \(1.8 \times 10^{26}\text{ W}\) (accept range \(1.8 \times 10^{26}\text{ W}\) to \(1.9 \times 10^{26}\text{ W}\)).

(a) A standard candle is an astronomical object with a known intrinsic luminosity. By measuring its apparent brightness from Earth, astronomers can determine its distance using the inverse square law of radiation.

(b) (i) Substitute \(T = 25\) days into the formula:
\(\frac{L}{L_{\odot}} = 100 \times (25)^{1.2} = 100 \times 47.59 \approx 4.8 \times 10^3\)
So, \(L \approx 4.8 \times 10^3 L_{\odot}\) (or \(4759 L_{\odot}\)).

(ii) First, calculate the absolute luminosity of the star in Watts:
\(L = 4759 \times 3.8 \times 10^{26}\text{ W} = 1.81 \times 10^{30}\text{ W}\)
Now, use the brightness-distance relation:
\(b = \frac{L}{4\pi d^2} \implies d = \sqrt{\frac{L}{4\pi b}}\)
\(d = \sqrt{\frac{1.81 \times 10^{30}}{4\pi \times 5.2 \times 10^{-16}}} = \sqrt{2.77 \times 10^{44}} = 1.66 \times 10^{22}\text{ m} \approx 1.7 \times 10^{22}\text{ m}\)

(a)
- Award [1] for defining standard candle as an object of known luminosity.
- Award [1] for explaining that measuring its apparent brightness allows the determination of its distance.

(b) (i)
- Award [1] for \(4.8 \times 10^3 L_{\odot}\) (accept range \(4.7 \times 10^3 L_{\odot}\) to \(4.8 \times 10^3 L_{\odot}\)).

(b) (ii)
- Award [1] for calculating absolute luminosity in Watts: \(1.8 \times 10^{30}\text{ W}\).
- Award [1] for correct substitution into \(d = \sqrt{\frac{L}{4\pi b}}\).
- Award [1] for obtaining \(1.7 \times 10^{22}\text{ m}\) (accept range \(1.6 \times 10^{22}\text{ m}\) to \(1.7 \times 10^{22}\text{ m}\)).

(a) When hydrogen runs out in the core of a Sun-like star, the core contracts and heats up while the outer envelope expands and cools, turning the star into a Red Giant. During this phase, helium fusion occurs in the core (producing carbon and oxygen) and hydrogen fusion occurs in a shell around the core. Eventually, fusion stops in the core, and the outer layers are ejected to form a Planetary Nebula. The remaining hot, dense core contracts to form a White Dwarf, where no nuclear fusion takes place.

(b) The Chandrasekhar limit is the maximum mass at which a white dwarf can remain stable, supported by electron degeneracy pressure. This limit is approximately \(1.4 M_{\odot}\). For a Sun-like star, the mass of the remaining core after the planetary nebula phase is less than the Chandrasekhar limit. Therefore, the electron degeneracy pressure is strong enough to resist gravitational collapse, and the star remains stable as a white dwarf instead of collapsing further to form a neutron star.

(a)
- Award [1] for describing the transition to a Red Giant with helium fusion in the core and/or hydrogen fusion in a surrounding shell.
- Award [1] for describing the planetary nebula phase where the outer layers of the star are ejected.
- Award [1] for identifying that the remaining core becomes a white dwarf where all nuclear fusion has ceased.

(b)
- Award [1] for defining the Chandrasekhar limit as the maximum mass of a stable white dwarf (approx. \(1.4\) solar masses).
- Award [1] for stating that the remaining core mass of a Sun-like star is below this limit.
- Award [1] for explaining that electron degeneracy pressure is sufficient to halt gravitational collapse (preventing the transition to a neutron star).

(a) The cosmic microwave background (CMB) radiation is isotropic (highly uniform across the sky) and matches a perfect blackbody spectrum at a temperature of approximately \(2.7\text{ K}\). This matches the prediction of the Big Bang model that the early universe was extremely hot and dense. As the universe expanded, the short-wavelength radiation emitted during that hot phase was stretched (redshifted) over time, cooling to the microwave wavelengths observed today.

(b) (i) The recessional speed \(v\) is given by:
\(v = z c = 0.045 \times 3.0 \times 10^8\text{ m s}^{-1} = 1.35 \times 10^7\text{ m s}^{-1}\) (or \(13500\text{ km s}^{-1}\))

(ii) Using Hubble's law, \(v = H_0 d\):
\(d = \frac{v}{H_0} = \frac{13500\text{ km s}^{-1}}{70\text{ km s}^{-1}\text{ Mpc}^{-1}} = 192.85\text{ Mpc} \approx 193\text{ Mpc}\)

(a)
- Award [1] for stating that CMB is isotropic / highly uniform throughout space.
- Award [1] for stating that CMB has a blackbody spectrum corresponding to a temperature of about \(2.7\text{ K}\).
- Award [1] for explaining that this represents the redshifted thermal radiation of the early, hot, dense universe that cooled as space expanded.

(b) (i)
- Award [1] for \(1.35 \times 10^7\text{ m s}^{-1}\) (or \(1.4 \times 10^7\text{ m s}^{-1}\) or \(13500\text{ km s}^{-1}\)).

(b) (ii)
- Award [1] for using \(d = \frac{v}{H_0}\) with consistent units.
- Award [1] for obtaining \(193\text{ Mpc}\) (accept \(190\text{ Mpc}\) if using rounded velocity of \(1.4 \times 10^7\text{ m s}^{-1}\)).

(a) (i) Using Wien's displacement law:
\(\lambda_{\text{max}} T = 2.90 \times 10^{-3}\text{ m K}\)
\(T = \frac{2.90 \times 10^{-3}\text{ m K}}{410 \times 10^{-9}\text{ m}} = 7073\text{ K}\)
This is approximately \(7100\text{ K}\).

(ii) Luminosity is proportional to \(R^2 T^4\):
\(\frac{L_{\text{star}}}{L_{\odot}} = \left(\frac{R_{\text{star}}}{R_{\odot}}\right)^2 \times \left(\frac{T_{\text{star}}}{T_{\odot}}\right)^4\)
Using \(T_{\text{star}} = 7073\text{ K}\):
\(\frac{L_{\text{star}}}{L_{\odot}} = (1.8)^2 \times \left(\frac{7073}{5800}\right)^4 = 3.24 \times (1.2195)^4 = 3.24 \times 2.21 = 7.16 \approx 7.2\)
(If using \(7100\text{ K}\): \(\frac{L_{\text{star}}}{L_{\odot}} = (1.8)^2 \times \left(\frac{7100}{5800}\right)^4 = 3.24 \times 2.24 = 7.26 \approx 7.3\)).

(b) Dark lines in the absorption spectrum represent specific wavelengths of light absorbed by elements in the star's outer atmosphere. Comparing these lines to known laboratory spectral lines allows identification of elements present in the star.

(a) (i)
- Award [1] for showing Wien's displacement law formula: \(\lambda_{\text{max}} T = 2.90 \times 10^{-3}\text{ m K}\).
- Award [1] for calculating \(7073\text{ K}\) and showing it is close to \(7100\text{ K}\).

(a) (ii)
- Award [1] for stating the relationship \(L \propto R^2 T^4\).
- Award [1] for correct substitution: \((1.8)^2 \times \left(\frac{7073}{5800}\right)^4\) or \((1.8)^2 \times \left(\frac{7100}{5800}\right)^4\).
- Award [1] for the final answer of \(7.2\) or \(7.3\).

(b)
- Award [1] for stating that the wavelengths of dark lines in the absorption spectrum are matched/compared to known laboratory spectral lines of specific elements.