2024 AQA AS-Level Physics 7407 Practice Paper with Answers

(a) Both the electron and electron antineutrino belong to the lepton class (specifically first-generation leptons). Differences include: the electron has charge $-1$ (or $-e$) while the antineutrino has zero charge; or the electron is a particle whereas the antineutrino is an antiparticle.

(b) The weak interaction is responsible, so the exchange particle is the $W^{-}$ boson (must specify negative charge).

(c) Initial Lepton Number (neutron): $L = 0$
Final Lepton Number: $L(p) = 0$, $L(e^{-}) = +1$, $L(\bar{\
u}_e) = -1$.
Total final lepton number $= 0 + 1 - 1 = 0$. Since initial $L$ = final $L$, lepton number is conserved.

(d) First, calculate the mass defect $\Delta m$:
$$\Delta m = m_n - (m_p + m_e) = 1.67493 \times 10^{-27}\text{ kg} - (1.67262 \times 10^{-27}\text{ kg} + 9.11 \times 10^{-31}\text{ kg})$$
$$\Delta m = 1.67493 \times 10^{-27} - 1.673531 \times 10^{-27} = 1.399 \times 10^{-30}\text{ kg}$$

Now calculate the equivalent energy release in joules:
$$E = \Delta m c^2 = 1.399 \times 10^{-30}\text{ kg} \times (3.00 \times 10^8\text{ m s}^{-1})^2 = 1.2591 \times 10^{-13}\text{ J}$$

Convert this energy to $\text{MeV}$:
$$E = \frac{1.2591 \times 10^{-13}\text{ J}}{1.60 \times 10^{-13}\text{ J MeV}^{-1}} \approx 0.787\text{ MeV}$$

Thus, the maximum kinetic energy of the electron is approximately $0.79\text{ MeV}$.

(a)
- 1 mark: Identifies both as leptons.
- 1 mark: Identifies a valid difference (e.g., charge, particle vs antiparticle).

(b)
- 1 mark: States $W$ boson.
- 1 mark: Specifies negative charge ($W^{-}$ boson).

(c)
- 1 mark: States initial lepton number is 0 and identifies individual lepton numbers for products ($L = +1$ for electron, $L = -1$ for antineutrino).
- 1 mark: Shows total final lepton number is 0, concluding it is conserved.

(d)
- 1 mark: Correct substitution to find mass defect $\Delta m = 1.399 \times 10^{-30}\text{ kg}$ (or equivalent in J: $1.26 \times 10^{-13}\text{ J}$).
- 1 mark: Converts energy to $\text{MeV}$ to yield a final value in the range $0.78\text{ MeV}$ to $0.80\text{ MeV}$.

(a) Elastic potential energy of the spring is transferred entirely to kinetic energy of the car because the track is frictionless:
$$E_{pe} = \frac{1}{2} k x^2 = \frac{1}{2} (120) (0.150)^2 = 1.35\text{ J}$$
$$E_k = \frac{1}{2} m_1 v_1^2 \implies 1.35 = \frac{1}{2} (0.250) v_1^2$$
$$v_1 = \sqrt{\frac{1.35}{0.125}} = 3.286\text{ m s}^{-1}$$

(b) Using conservation of linear momentum during the collision:
$$m_1 v_1 = (m_1 + m_2) v_2$$
$$0.250 \times 3.286 = (0.250 + 0.150) v_2$$
$$0.8215 = 0.400 v_2 \implies v_2 = 2.054\text{ m s}^{-1}$$

(c) The initial kinetic energy of the combined mass immediately after the collision is:
$$E_{k2} = \frac{1}{2} M v_2^2 = \frac{1}{2} (0.400) (2.054)^2 = 0.84375\text{ J}$$

As they move up the ramp, they gain gravitational potential energy ($Mgh$) and work is done against the average frictional force ($F_f d$):
$$E_{k2} = M g h + F_f d$$

where $h = d \sin(30.0^\circ) = 0.350 \sin(30.0^\circ) = 0.175\text{ m}$.

Calculate the gain in GPE:
$$\Delta E_p = M g h = 0.400\text{ kg} \times 9.81\text{ m s}^{-2} \times 0.175\text{ m} = 0.6867\text{ J}$$

Now apply conservation of energy:
$$0.84375\text{ J} = 0.6867\text{ J} + F_f \times 0.350\text{ m}$$
$$F_f \times 0.350 = 0.15705\text{ J}$$
$$F_f = \frac{0.15705}{0.350} \approx 0.4487\text{ N}$$

To 2 significant figures, the average frictional force is $0.45\text{ N}$.

(a)
- 1 mark: Equates $\frac{1}{2}kx^2$ to $\frac{1}{2}mv^2$ and calculates $E_{pe} = 1.35\text{ J}$.
- 1 mark: Correctly calculates speed of the car $v_1 = 3.29\text{ m s}^{-1}$ (or $3.3\text{ m s}^{-1}$).

(b)
- 1 mark: Uses conservation of momentum equation $m_1 v_1 = (m_1 + m_2) v_2$.
- 1 mark: Correctly calculates common speed $v_2 = 2.05\text{ m s}^{-1}$ (or $2.1\text{ m s}^{-1}$ using rounded $v_1$).

(c)
- 1 mark: Calculates initial kinetic energy of combined mass $E_{k2} \approx 0.844\text{ J}$.
- 1 mark: Identifies height gained $h = 0.350 \sin(30.0^\circ) = 0.175\text{ m}$ and calculates gain in GPE $= 0.687\text{ J}$.
- 1 mark: Uses energy conservation equation $E_{k2} = \Delta E_p + W_f$.
- 1 mark: Correctly calculates average frictional force as $0.45\text{ N}$ (allow $0.44\text{ N}$ to $0.46\text{ N}$ from rounding propagation).

(a) Using the equation for a complete circuit: $\varepsilon = I(R + r)$:

For the first set of values:
$$\varepsilon = 0.20(6.0 + r) \implies \varepsilon = 1.2 + 0.2r$$ [Equation 1]

For the second set of values:
$$\varepsilon = 0.50(2.0 + r) \implies \varepsilon = 1.0 + 0.5r$$ [Equation 2]

Equating both expressions for $\varepsilon$:
$$1.2 + 0.2r = 1.0 + 0.5r$$
$$0.2 = 0.3r \implies r = \frac{0.2}{0.3} = \frac{2}{3}\\ \Omega \approx 0.67\\ \Omega$$

Substitute $r = \frac{2}{3}\\ \Omega$ back into Equation 1 to find $\varepsilon$:
$$\varepsilon = 0.20 \left(6.0 + \frac{2}{3}\right) = 0.20 \left(\frac{20}{3}\right) = \frac{4}{3}\text{ V} \approx 1.33\text{ V}$$

(b) (i) For two identical resistors in parallel:
$$R_p = \frac{R_{\text{lamp}}}{2} = \frac{12.0}{2} = 6.0\\ \Omega$$

(ii) The total resistance of the new circuit is:
$$R_{\text{total}} = R_p + r = 6.0 + \frac{2}{3} = \frac{20}{3}\\ \Omega \approx 6.67\\ \Omega$$

The total current $I_{\text{new}}$ is:
$$I_{\text{new}} = \frac{\varepsilon}{R_{\text{total}}} = \frac{1.333}{6.667} = 0.20\text{ A}$$

The terminal potential difference $V$ across the cell is equal to the potential difference across the parallel network:
$$V = I_{\text{new}} \times R_p = 0.20\text{ A} \times 6.0\\ \Omega = 1.20\text{ V}$$

(a)
- 1 mark: Forms two correct equations based on $\varepsilon = I(R+r)$.
- 1 mark: Equates equations and solves to show $r = 0.67\\ \Omega$.
- 1 mark: Substitutes $r$ back to find $\varepsilon$.
- 1 mark: Obtains $\varepsilon = 1.3\text{ V}$ (accept $1.33\text{ V}$).

(b)(i)
- 1 mark: Correctly calculates parallel resistance $R_p = 6.0\\ \Omega$.

(b)(ii)
- 1 mark: Calculates total resistance $R_{\text{total}} = 6.67\\ \Omega$ (or uses potential divider formula).
- 1 mark: Calculates circuit current $I = 0.20\text{ A}$ (or uses correct potential divider ratio).
- 1 mark: Correctly calculates terminal potential difference as $1.2\text{ V}$.

(a) Progressive waves travel along the string from the generator and reflect at the fixed ends. These incident and reflected waves, which have the same frequency, amplitude, and speed but travel in opposite directions, superpose (interfere) with each other to produce a stationary wave.

(b) For the second harmonic (two loops):
- Number of nodes = 3 (one at each clamped end, one in the middle).
- Number of antinodes = 2.
Since the string length $L = 1.20\text{ m}$ contains exactly one full wavelength for the second harmonic:
$$\lambda = L = 1.20\text{ m}$$

(c) For a stretched string, wave speed $v$ is given by:
$$v = \sqrt{\frac{T}{\mu}}$$
where $T$ is the tension and $\mu$ is the mass per unit length.

Since $v = f \lambda$, the frequency of any harmonic $n$ is:
$$f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$$

Initially, the second harmonic ($n=2$) has frequency $f_2 = 150\text{ Hz}$. The initial fundamental frequency ($n=1$) is:
$$f_1 = \frac{f_2}{2} = 75\text{ Hz}$$

When the tension becomes $2.0T$, the new speed $v'$ becomes:
$$v' = \sqrt{\frac{2.0T}{\mu}} = \sqrt{2} \times v$$

Since the length $L$ of the string is constant, the new fundamental frequency $f'_1$ is:
$$f'_1 = \sqrt{2} \times f_1 = \sqrt{2} \times 75\text{ Hz} = 1.414 \times 75\text{ Hz} = 106.1\text{ Hz}$$

To 2 significant figures, the frequency of the fundamental mode is $110\text{ Hz}$ (or $106\text{ Hz}$ to 3 s.f.).

(a)
- 1 mark: Mentions reflection of progressive waves at the boundaries/ends (or waves travelling in opposite directions).
- 1 mark: Mentions superposition/interference of waves of the same frequency/amplitude.

(b)
- 1 mark: States 3 nodes and 2 antinodes.
- 1 mark: Obtains wavelength $\lambda = 1.20\text{ m}$.

(c)
- 1 mark: Recognizes that the initial fundamental frequency is $75\text{ Hz}$ (half of the second harmonic).
- 1 mark: Identifies the proportional relationship $f \propto \sqrt{T}$ (or uses $v = \sqrt{\frac{T}{\mu}}$).
- 1 mark: Sets up the ratio equation $\frac{f_1'}{f_1} = \sqrt{\frac{2T}{T}} = \sqrt{2}$.
- 1 mark: Calculates the final frequency as $106\text{ Hz}$ or $110\text{ Hz}$ (with correct unit).

(a) The energy of an incident photon is:
$$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{3.80 \times 10^{-7}\text{ m}}$$
$$E = 5.234 \times 10^{-19}\text{ J}$$

(b) First, convert the work function $\phi$ of sodium from $\text{eV}$ to joules:
$$\phi = 2.24 \times 1.60 \times 10^{-19}\text{ J} = 3.584 \times 10^{-19}\text{ J}$$

Using the photoelectric equation:
$$E_{k,\max} = E - \phi$$
$$E_{k,\max} = 5.234 \times 10^{-19}\text{ J} - 3.584 \times 10^{-19}\text{ J} = 1.650 \times 10^{-19}\text{ J}$$

(c) The momentum $p$ of the electron is related to its kinetic energy by:
$$E_k = \frac{p^2}{2m_e} \implies p = \sqrt{2 m_e E_{k,\max}}$$
$$p = \sqrt{2 \times 9.11 \times 10^{-31}\text{ kg} \times 1.650 \times 10^{-19}\text{ J}}$$
$$p = \sqrt{3.0063 \times 10^{-49}} = 5.483 \times 10^{-25}\text{ kg m s}^{-1}$$

Now, calculate the de Broglie wavelength $\lambda_{dB}$:
$$\lambda_{dB} = \frac{h}{p} = \frac{6.63 \times 10^{-34}\text{ J s}}{5.483 \times 10^{-25}\text{ kg m s}^{-1}} = 1.209 \times 10^{-9}\text{ m}$$

To 3 significant figures, the de Broglie wavelength is $1.21 \times 10^{-9}\text{ m}$ (or $1.21\text{ nm}$).

(a)
- 1 mark: Uses $E = \frac{hc}{\lambda}$ with correct substitutions.
- 1 mark: Obtains $5.2 \times 10^{-19}\text{ J}$ (or $5.23 \times 10^{-19}\text{ J}$).

(b)
- 1 mark: Converts work function to joules ($3.58 \times 10^{-19}\text{ J}$).
- 1 mark: Uses photoelectric equation to obtain $1.65 \times 10^{-19}\text{ J}$ (accept $1.6 \times 10^{-19}\text{ J}$ to $1.7 \times 10^{-19}\text{ J}$ depending on intermediate rounding).

(c)
- 1 mark: Uses $E_k = \frac{1}{2}m_e v^2$ (or $p^2/2m$) to write an expression for velocity or momentum.
- 1 mark: Calculates velocity $v \approx 6.0\\ \times 10^5\text{ m s}^{-1}$ or momentum $p \approx 5.5 \times 10^{-25}\text{ kg m s}^{-1}$.
- 1 mark: Uses $\lambda_{dB} = \frac{h}{p}$ (or $\lambda_{dB} = \frac{h}{m_e v}$) with correct substitutions.
- 1 mark: Obtains final answer in range $1.20 \times 10^{-9}\text{ m}$ to $1.22 \times 10^{-9}\text{ m}$ (must have correct units).

### Key Observations and Explanations:

1. **Threshold Frequency (\(f_0\))**
* **Observation**: No photoelectrons are emitted if the frequency of the incident light is below a minimum value (the threshold frequency), regardless of how intense the light is.
* **Wave Theory Failure**: According to wave theory, electromagnetic wave energy is delivered continuously. Therefore, light of any frequency should eventually deliver enough accumulated energy to emit electrons if shone for long enough.
* **Photon Model Explanation**: Light is packetized into discrete photons, where the energy of a single photon is \(E = hf\). The interaction is a one-to-one collision. If \(hf < \phi\) (where \(\phi\) is the work function), the electron does not gain enough energy to escape in a single interaction, and no emission occurs.

2. **Instantaneous Emission**
* **Observation**: If the frequency of light is above the threshold frequency, photoelectrons are emitted almost instantaneously (within nanoseconds) after the light source is turned on, even at extremely low intensities.
* **Wave Theory Failure**: Wave theory predicts that at low intensities, it would take a significant amount of time (seconds or hours) for the continuous wavefronts to transfer enough cumulative energy to a single electron to free it from the metal lattice.
* **Photon Model Explanation**: Because the collision between a single photon and a single electron is an instantaneous, all-or-nothing packet-based energy transfer, emission occurs immediately once a photon with \(hf \ge \phi\) strikes an electron.

3. **Maximum Kinetic Energy (\(E_{k,\text{max}}\))**
* **Observation**: The maximum kinetic energy of the emitted photoelectrons depends solely on the frequency of the incident light, not its intensity. Increasing the intensity increases the number of electrons emitted per second but not their maximum kinetic energy.
* **Wave Theory Failure**: Wave theory suggests that a higher intensity light wave has a larger amplitude, which should deliver more energy to individual electrons, leading to photoelectrons with greater kinetic energy.
* **Photon Model Explanation**: Increasing intensity increases the number of photons arriving per second, which increases the rate of emission (current). However, the maximum kinetic energy is governed by the conservation of energy: \(E_{k,\text{max}} = hf - \phi\). Because \(hf\) depends only on frequency, the maximum kinetic energy is independent of intensity.

This is a level-of-response question marked out of 6 marks:

**Level 3 (5–6 marks)**
- The candidate provides a detailed and coherent description of at least two of the key experimental observations.
- For each observation, they clearly explain why classical wave theory fails AND how the photon model successfully accounts for the observation.
- Precise physical terminology (e.g., work function, threshold frequency, intensity, photons, one-to-one interaction) is used correctly throughout.

**Level 2 (3–4 marks)**
- The candidate describes at least two observations and provides some explanation using either the failure of wave theory OR the success of the photon model, but not both in full depth.
- Alternatively, the candidate covers all three aspects but with minor errors, logical gaps, or omissions in explanations.
- The explanation is structured but may lack the rigorous detail of a Level 3 response.

**Level 1 (1–2 marks)**
- The candidate mentions some correct facts about the photoelectric effect (e.g., mentions threshold frequency or the photoelectric equation \(hf = \phi + E_{k,\text{max}}\)).
- However, they do not systematically link these facts to explain why wave theory fails or how photon theory is supported.
- The answer is fragmented or contains significant conceptual errors.

**Level 0 (0 marks)**
- No creditable physics is presented.

(a) Method 1 (Work-Energy):
Initial kinetic energy of block A: \(E_{k,i} = \frac{1}{2} m_1 u^2 = \frac{1}{2} \times 0.50 \times 5.0^2 = 6.25\text{ J}\).
Work done against friction: \(W = F_f \times s = 1.5 \times 1.5 = 2.25\text{ J}\).
Kinetic energy just before collision: \(E_{k,f} = E_{k,i} - W = 6.25 - 2.25 = 4.0\text{ J}\).
Using \(E_{k,f} = \frac{1}{2} m_1 v^2 \implies 4.0 = \frac{1}{2} \times 0.50 \times v^2 \implies v^2 = 16 \implies v = 4.0\text{ m s}^{-1}\).

Method 2 (Dynamics/SUVAT):
Deceleration due to friction: \(a = -\frac{F_f}{m_1} = -\frac{1.5}{0.50} = -3.0\text{ m s}^{-2}\).
Using \(v^2 = u^2 + 2as\):
\(v^2 = 5.0^2 + 2 \times (-3.0) \times 1.5 = 25 - 9.0 = 16\).
\(v = \sqrt{16} = 4.0\text{ m s}^{-1}\).

(b) Using conservation of linear momentum:
\(m_1 v_1 + m_2 v_2 = (m_1 + m_2) V\)
\(0.50 \times 4.0 + 0 = (0.50 + 0.30) V\)
\(2.0 = 0.80 V\)
\(V = 2.5\text{ m s}^{-1}\).

(c) The normal contact force is proportional to mass. Therefore, the frictional force on the combined mass is:
\(F_{f2} = F_{f1} \times \frac{m_1 + m_2}{m_1} = 1.5 \times \frac{0.80}{0.50} = 2.4\text{ N}\).
(Alternatively, \(\mu = \frac{1.5}{0.50 \times 9.81} = 0.306\), so \(F_{f2} = \mu (m_1 + m_2) g = 0.306 \times 0.80 \times 9.81 = 2.4\text{ N}\)).
Deceleration of the combined blocks: \(a_2 = -\frac{F_{f2}}{M} = -\frac{2.4}{0.80} = -3.0\text{ m s}^{-2}\) (Note: since \(a = \mu g\), deceleration remains constant regardless of mass).
Using \(v^2 = u^2 + 2as\) to find stopping distance \(d\):
\(0 = 2.5^2 + 2(-3.0)d\)
\(6.0 d = 6.25\)
\(d = 1.04\text{ m}\) (accept \(1.0\text{ m}\) to 2 s.f.).

(a) [3 marks total]
- 1 mark for calculating work done as \(2.25\text{ J}\) OR deceleration as \(-3.0\text{ m s}^{-2}\).
- 1 mark for correct substitution into \(E_{k,f} = E_{k,i} - W\) or into \(v^2 = u^2 + 2as\).
- 1 mark for showing clearly that \(v = 4.0\text{ m s}^{-1}\).

(b) [2 marks total]
- 1 mark for correct use of momentum conservation: \(0.50 \times 4.0 = 0.80 \times V\).
- 1 mark for correct calculation of \(V = 2.5\text{ m s}^{-1}\).

(c) [3 marks total]
- 1 mark for determining the new frictional force is \(2.4\text{ N}\) OR showing that deceleration remains \(3.0\text{ m s}^{-2}\).
- 1 mark for correct substitution of their velocity from (b) and deceleration into \(v^2 = u^2 + 2as\) (or equivalent work-energy equation).
- 1 mark for finding stopping distance \(d = 1.04\text{ m}\) (accept \(1.0\text{ m}\)).

(a) Two progressive waves of the same frequency (or wavelength) and similar amplitude, travelling in opposite directions, meet and superpose.

(b) First, find the mass per unit length, \(\mu\):
\(\mu = \frac{\text{mass}}{\text{length}} = \frac{4.50 \times 10^{-3}\text{ kg}}{1.20\text{ m}} = 3.75 \times 10^{-3}\text{ kg m}^{-1}\).
Next, find the tension in the string, \(T\):
\(T = mg = 2.50 \times 9.81 = 24.525\text{ N}\).
Use the wave speed formula:
\(v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{24.525}{3.75 \times 10^{-3}}} = \sqrt{6540} = 80.87\text{ m s}^{-1} \approx 80.9\text{ m s}^{-1}\).

(c) For the second harmonic on a string of vibrating length \(L = 0.80\text{ m}\), the pattern consists of a full wavelength within this length.
Therefore, \(\lambda = L = 0.80\text{ m}\).
Using the wave equation:
\(f = \frac{v}{\lambda} = \frac{80.87}{0.80} = 101.1\text{ Hz} \approx 101\text{ Hz}\) (accept answers in range \(101\text{ Hz}\) to \(102\text{ Hz}\) depending on rounding).

(a) [2 marks total]
- 1 mark for specifying two waves of the same frequency/wavelength travelling in opposite directions.
- 1 mark for stating that these waves superpose/interfere.

(b) [3 marks total]
- 1 mark for calculating \(\mu = 3.75 \times 10^{-3}\text{ kg m}^{-1}\).
- 1 mark for calculating tension \(T = 24.5\text{ N}\) (accept \(24.525\text{ N}\)).
- 1 mark for final calculated value of \(v = 80.9\text{ m s}^{-1}\) (accept range \(80.8\) to \(81.0\)).

(c) [3 marks total]
- 1 mark for stating/showing that for the second harmonic, \(\lambda = L = 0.80\text{ m}\).
- 1 mark for rearranging the wave equation to make \(f\) the subject: \(f = \frac{v}{\lambda}\).
- 1 mark for calculating frequency \(f = 101\text{ Hz}\) (accept range \(101\) to \(102\text{ Hz}\), ecf from b).

(a) Convert eV to J:
\(\Phi = 2.30 \times 1.60 \times 10^{-19}\text{ J} = 3.68 \times 10^{-19}\text{ J}\).

(b) First, find the energy \(E\) of the incident photon:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{280 \times 10^{-9}} = 7.104 \times 10^{-19}\text{ J}\).
Use Einstein's photoelectric equation:
\(E = \Phi + E_{k,max}\)
\(E_{k,max} = E - \Phi = 7.104 \times 10^{-19} - 3.68 \times 10^{-19} = 3.424 \times 10^{-19}\text{ J}\).
Rounding to 3 significant figures gives \(3.42 \times 10^{-19}\text{ J}\).

(c) (i) No effect on maximum kinetic energy. This is because the maximum kinetic energy depends only on the frequency (or wavelength) of individual incident photons and the work function, both of which are unchanged.

(ii) The rate of emission of photoelectrons is doubled. Doubling the intensity of the radiation doubles the number of photons incident on the surface per second. Since there is a one-to-one interaction between a photon and a conduction electron, the rate of emission doubles.

(a) [1 mark total]
- 1 mark for \(3.68 \times 10^{-19}\text{ J}\) (accept \(3.7 \times 10^{-19}\text{ J}\)).

(b) [4 marks total]
- 1 mark for recalling the formula for photon energy: \(E = \frac{hc}{\lambda}\).
- 1 mark for calculating photon energy as \(7.10 \times 10^{-19}\text{ J}\) (or \(7.1 \times 10^{-19}\text{ J}\)).
- 1 mark for subtracting their work function from the calculated photon energy.
- 1 mark for the final answer of \(3.42 \times 10^{-19}\text{ J}\) (accept range \(3.41 \times 10^{-19}\) to \(3.43 \times 10^{-19}\text{ J}\)).

(c) [3 marks total]
- 1 mark for stating that there is no effect on the maximum kinetic energy of the photoelectrons, because photon energy is unchanged.
- 1 mark for stating that the rate of emission of photoelectrons doubles (or increases proportionally).
- 1 mark for explaining that higher intensity means more photons per second, leading to more one-to-one interactions with electrons.

**1.1**
Plotting \(v\) against \(r^2\) yields a linear (straight-line) graph through the origin if the theory is correct. A straight line is easier to verify visually and its gradient can be calculated accurately. Plotting \(v\) against \(r\) would yield a curve, which is much harder to verify and analyze quantitatively.

**1.2**
Comparing the given formula \(v = \left[ \frac{2g(\rho_s - \rho_l)}{9\eta} \right] r^2\) with the equation of a straight line through the origin, \(y = mx\) where \(y = v\) and \(x = r^2\):
\[ m = \frac{2g(\rho_s - \rho_l)}{9\eta} \]
Rearranging for \(\eta\):
\[ \eta = \frac{2g(\rho_s - \rho_l)}{9m} \]

**1.3**
Substitute the values into the equation:
\[ \eta = \frac{2 \times 9.81 \times (7.80 \times 10^3 - 9.50 \times 10^2)}{9 \times 4.50 \times 10^3} \]
\[ \eta = \frac{2 \times 9.81 \times 6850}{40500} = \frac{134397}{40500} \approx 3.32\text{ Pa s} \]
Rounding to 3 significant figures (consistent with the provided data):
\(\eta = 3.32\text{ Pa s}\) (or \(\text{kg m}^{-1}\text{ s}^{-1}\) or \(\text{N s m}^{-2}\)).

**1.4**
Percentage uncertainty in the gradient \(m\):
\[ \% \text{ uncertainty in } m = \frac{0.30 \times 10^3}{4.50 \times 10^3} \times 100\% = 6.67\% \]
Since \(\eta \propto \frac{1}{m}\) and all other values have negligible uncertainty, the percentage uncertainty in \(\eta\) is also \(6.67\%\).
Absolute uncertainty in \(\eta\):
\[ \Delta \eta = 3.3184 \times 0.0667 \approx 0.22\text{ Pa s} \]
This is stated as \(\pm 0.22\text{ Pa s}\) (or \(\pm 0.2\text{ Pa s}\)).

**1.1**
- Award 1 mark for stating that a straight-line graph through the origin confirms the relationship (or is easier to analyze than a curve).
- Award 1 mark for explaining that the gradient of a straight line is constant and can be used directly to calculate \(\eta\).

**1.2**
- Award 1 mark for equating the gradient \(m\) to \(\frac{2g(\rho_s - \rho_l)}{9\eta}\).
- Award 1 mark for showing correct algebraic steps to rearrange the equation to make \(\eta\) the subject.

**1.3**
- Award 1 mark for correct substitution of values: \(\eta = \frac{2 \times 9.81 \times 6850}{9 \times 4500}\).
- Award 1 mark for the correct numerical value: \(3.32\) (accept \(3.3\) to \(3.32\)).
- Award 1 mark for correct SI unit: \(\text{Pa s}\) or \(\text{kg m}^{-1}\text{ s}^{-1}\) or \(\text{N s m}^{-2}\).

**1.4**
- Award 1 mark for calculating the percentage uncertainty of the gradient: \(\frac{0.30}{4.50} \times 100 = 6.67\%\).
- Award 1 mark for equating the percentage uncertainty of \(\eta\) to that of the gradient (or showing \(\frac{\Delta \eta}{\eta} = \frac{\Delta m}{m}\)).
- Award 1 mark for the correct absolute uncertainty: \(\pm 0.22\) or \(\pm 0.2\) with the correct unit.

**2.1**
To ensure the thermometer reading represents the wire's temperature:
1. The water must be stirred thoroughly before taking a reading to ensure a uniform temperature throughout the water bath.
2. The thermometer bulb should be placed in close proximity to the wire (without touching the container walls) so they are in the same thermal environment.

**2.2**
Expanding the theoretical equation gives:
\[ R = R_0 \alpha \theta + R_0 \]
Comparing this to \(y = mx + c\) where \(y = R\) and \(x = \theta\):
- The y-intercept \(c = R_0\).
- The gradient \(m = R_0 \alpha\).
Thus, \(R_0\) is the y-intercept, and \(\alpha = \frac{\text{gradient}}{\text{y-intercept}} = \frac{m}{R_0}\).

**2.3**
From the line of best fit parameters:
- \(R_0 = 12.0\ \Omega\)
- Gradient \(m = 0.048\ \Omega\ ^{\circ}\text{C}^{-1}\)

Calculate \(\alpha\):
\[ \alpha = \frac{0.048}{12.0} = 4.0 \times 10^{-3}\ ^{\circ}\text{C}^{-1} \]
(Accept \(\text{K}^{-1}\) as temperature differences/coefficients are equivalent in Celsius and Kelvin).

**2.4**
(a) Percentage uncertainty in temperature:
\[ \% \text{ uncertainty} = \frac{\text{uncertainty}}{\text{value}} \times 100\% = \frac{0.5}{20.0} \times 100\% = 2.5\% \]

(b) If the water bath is heated rapidly:
- Thermal lag exists between the water, the wire, and the thermometer bulb (they do not reach thermal equilibrium instantly).
- Temperature gradients will exist in the water, meaning different parts of the bath are at different temperatures, increasing the systematic error beyond the reading resolution.

**2.1**
- Award 1 mark for mentioning stirring the water bath (to ensure uniform temperature distribution).
- Award 1 mark for placing the thermometer close to the wire / avoiding contact with the container walls (which may be at a different temperature).

**2.2**
- Award 1 mark for stating that \(R_0\) is directly equal to the y-intercept.
- Award 1 mark for stating that \(\alpha\) is equal to the gradient divided by the y-intercept (or \(\alpha = m / R_0\)).

**2.3**
- Award 1 mark for \(R_0 = 12.0\ \Omega\) (must include unit).
- Award 1 mark for calculating \(\alpha = 4.0 \times 10^{-3}\) (or \(0.0040\)).
- Award 1 mark for the correct unit of \(\alpha\): \(^{\circ}\text{C}^{-1}\) or \(\text{K}^{-1}\).

**2.4**
- Award 1 mark for (a) showing \(\frac{0.5}{20.0} \times 100 = 2.5\%\).
- Award 1 mark for (b) identifying thermal lag (temperature of the wire lags behind that of the water/thermometer or vice versa).
- Award 1 mark for (b) identifying temperature gradients or lack of thermal equilibrium within the water bath.

(a) The cross-sectional area of the brass wire is calculated using:
\( A_2 = \pi r^2 = \pi \left(\frac{0.60 \times 10^{-3} \text{ m}}{2}\right)^2 = 2.827 \times 10^{-7} \text{ m}^2 \approx 2.83 \times 10^{-7} \text{ m}^2 \)

(b) The tension in both wires is equal to the applied load of \( 45 \text{ N} \).
Extension of the steel wire (1):
\( A_1 = \pi \left(0.40 \times 10^{-3}\right)^2 = 5.027 \times 10^{-7} \text{ m}^2 \)
\( \Delta L_1 = \frac{F L_1}{A_1 E_1} = \frac{45 \times 1.8}{5.027 \times 10^{-7} \times 2.0 \times 10^{11}} = 8.06 \times 10^{-4} \text{ m} \)

Extension of the brass wire (2):
\( \Delta L_2 = \frac{F L_2}{A_2 E_2} = \frac{45 \times 1.2}{2.827 \times 10^{-7} \times 1.0 \times 10^{11}} = 1.91 \times 10^{-3} \text{ m} \)

Total extension:
\( \Delta L_{\text{total}} = \Delta L_1 + \Delta L_2 = 8.06 \times 10^{-4} \text{ m} + 1.91 \times 10^{-3} \text{ m} = 2.72 \times 10^{-3} \text{ m} \) (or \( 2.7 \text{ mm} \))

(c) Since both wires behave elastically under the initial load, the total elastic strain energy stored is:
\( E_{\text{total}} = \frac{1}{2} F \Delta L_{\text{total}} = 0.5 \times 45 \times 2.72 \times 10^{-3} = 0.0612 \text{ J} \approx 0.061 \text{ J} \) (or \( 6.1 \times 10^{-2} \text{ J} \))

(d) When the load is removed:
- The steel wire underwent only elastic deformation, meaning Hooke's law was obeyed/elastic limit was not exceeded. It will fully recover and return to its original length of \( 1.8 \text{ m} \).
- The brass wire underwent plastic deformation, meaning its elastic limit was exceeded. It will experience permanent strain (permanent set) and will be permanently longer than its original length of \( 1.2 \text{ m} \).
- The unloading curve for the brass wire would run parallel to its initial loading line.

(a)
- 1 Mark (AO1): Correct calculation of cross-sectional area: \( 2.8 \times 10^{-7} \text{ m}^2 \) or \( 2.83 \times 10^{-7} \text{ m}^2 \).

(b)
- 1 Mark (AO2): Correct calculation of the steel wire's cross-sectional area: \( 5.0 \times 10^{-7} \text{ m}^2 \).
- 1 Mark (AO2): Correct calculation of individual extension of steel: \( 8.06 \times 10^{-4} \text{ m} \) or \( 0.81 \text{ mm} \).
- 1 Mark (AO2): Correct calculation of individual extension of brass: \( 1.91 \times 10^{-3} \text{ m} \) or \( 1.9 \text{ mm} \).
- 1 Mark (AO1): Adding both extensions to get \( 2.7 \times 10^{-3} \text{ m} \) (accept range \( 2.70 \times 10^{-3} \) to \( 2.73 \times 10^{-3} \text{ m} \)).

(c)
- 1 Mark (AO2): Use of \( E = \frac{1}{2} F \Delta L \) (either calculating individually and adding or using total extension).
- 1 Mark (AO1): Correct calculation of energy: \( 0.061 \text{ J} \) (accept \( 0.0612 \text{ J} \)). Allow ECF from (b).

(d)
- 1 Mark (AO1): Steel wire returns to its original length / has no permanent extension because it only deformed elastically.
- 1 Mark (AO1): Brass wire has permanent deformation / permanent set / does not return to its original length because it deformed plastically.
- 1 Mark (AO2): Clear description of the physical state, e.g., the steel atoms return to equilibrium positions while brass planes of atoms have slid past one another permanently.

(a) The energy of a photon in joules is:
\( E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \text{ J s} \times 3.00 \times 10^8 \text{ m s}^{-1}}{240 \times 10^{-9} \text{ m}} = 8.2875 \times 10^{-19} \text{ J} \)

Converting to electron-volts:
\( E_{\text{eV}} = \frac{8.2875 \times 10^{-19} \text{ J}}{1.60 \times 10^{-19} \text{ J eV}^{-1}} \approx 5.18 \text{ eV} \)

(b) For \( \lambda = 320 \text{ nm} \):
\( E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{320 \times 10^{-9}} = 6.216 \times 10^{-19} \text{ J} \)
In electron-volts:
\( E_{\text{eV}} = \frac{6.216 \times 10^{-19}}{1.60 \times 10^{-19}} \approx 3.89 \text{ eV} \)

Since the photon energy \( 3.89 \text{ eV} \) is less than the work function of zinc (\( 4.3 \text{ eV} \)), a single photon does not have enough energy to liberate an electron from the surface. Therefore, no photoelectric emission occurs.
(Alternatively, threshold wavelength can be calculated as \( \lambda_{\text{th}} = \frac{hc}{\Phi} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{4.3 \times 1.60 \times 10^{-19}} = 2.89 \times 10^{-7} \text{ m} = 289 \text{ nm} \). Since \( 320 \text{ nm} > 289 \text{ nm} \), the incident wavelength is greater than the threshold wavelength, so no emission occurs.)

(c) According to Einstein's photoelectric equation:
\( E_{\text{k,max}} = hf - \Phi \)
In eV:
\( E_{\text{k,max}} = 5.18 \text{ eV} - 4.3 \text{ eV} = 0.88 \text{ eV} \)

Converting back to Joules:
\( E_{\text{k,max}} = 0.88 \times 1.60 \times 10^{-19} \text{ J} \approx 1.41 \times 10^{-19} \text{ J} \) (or \( 1.4 \times 10^{-19} \text{ J} \))

(d) The stopping potential \( V_{\text{s}} \) is related to the maximum kinetic energy by:
\( e V_{\text{s}} = E_{\text{k,max}} \)
Since \( E_{\text{k,max}} = 0.88 \text{ eV} \), the stopping potential is simply \( 0.88 \text{ V} \).

(a)
- 1 Mark (AO1): Use of \( E = \frac{hc}{\lambda} \) with correct values substituted.
- 1 Mark (AO2): Convert Joules to eV by dividing by \( 1.60 \times 10^{-19} \).
- 1 Mark (AO1): Correct final value: \( 5.18 \text{ eV} \) (accept \( 5.2 \text{ eV} \)).

(b)
- 1 Mark (AO2): Calculation of new photon energy: \( 3.89 \text{ eV} \) (or \( 6.22 \times 10^{-19} \text{ J} \)) OR threshold wavelength calculation: \( 2.89 \times 10^{-7} \text{ m} \).
- 1 Mark (AO1): Explicit comparison showing \( 3.89 \text{ eV} < 4.3 \text{ eV} \) (or energy in Joules is less than work function in Joules, or wavelength is greater than threshold wavelength).
- 1 Mark (AO1): Explanation that a single photon interacts with a single electron, meaning energy cannot accumulate over time, hence no emission.

(c)
- 1 Mark (AO1): Recall and use of \( E_{\text{k,max}} = hf - \Phi \) (or equivalent in Joules).
- 1 Mark (AO2): Correct calculation of kinetic energy: \( 1.41 \times 10^{-19} \text{ J} \) (accept \( 1.4 \times 10^{-19} \text{ J} \) to \( 1.42 \times 10^{-19} \text{ J} \)).

(d)
- 1 Mark (AO2): Use of \( e V_{\text{s}} = E_{\text{k,max}} \).
- 1 Mark (AO1): Correct stopping potential of \( 0.88 \text{ V} \) (accept range \( 0.87 \text{ V} \) to \( 0.90 \text{ V} \), depending on intermediate rounding; must have unit \( \text{V} \)).

First, analyze the conservation laws for the strong interaction (where charge \(Q\), baryon number \(B\), and strangeness \(S\) must all be conserved):

1. **Charge (\(Q\)) conservation**:
\(Q(\pi^-) + Q(p) = Q(K^0) + Q(X)\)
\(-1 + 1 = 0 + Q(X) \implies Q(X) = 0\)

2. **Baryon number (\(B\)) conservation**:
\(B(\pi^-) + B(p) = B(K^0) + B(X)\)
\(0 + 1 = 0 + B(X) \implies B(X) = 1\)

3. **Strangeness (\(S\)) conservation** (conserved in strong interactions):
\(S(\pi^-) + S(p) = S(K^0) + S(X)\)
\(0 + 0 = +1 + S(X) \implies S(X) = -1\)

We need a baryon (three quarks) with a total charge of \(0\) and strangeness of \(-1\) (meaning it contains exactly one strange quark, \(s\), which has charge \(-1/3\) and strangeness \(-1\)).

Let's test the options:
- **A**: \(uds\) has charge \(+2/3 - 1/3 - 1/3 = 0\), baryon number \(1/3 + 1/3 + 1/3 = 1\), and strangeness \(-1\). This is correct.
- **B**: \(uss\) has strangeness \(-2\).
- **C**: \(udd\) has strangeness \(0\).
- **D**: \(uus\) has charge \(+2/3 + 2/3 - 1/3 = +1\).

1 mark for selecting the correct option (A).

- Correct deduction of conservation properties: baryon number \(B = 1\), charge \(Q = 0\), and strangeness \(S = -1\).
- Identification of \(uds\) as the only option matching these properties.

Use the principle of conservation of energy:

Total work done by the applied force (\(W_{\text{in}}\)) goes into increasing the gravitational potential energy (\(\Delta E_p\)) and overcoming friction (\(W_f\)), since kinetic energy remains constant (constant speed).

1. Calculate total work input:
\(W_{\text{in}} = F \times d = 30\text{ N} \times 5.0\text{ m} = 150\text{ J}\)

2. Calculate change in gravitational potential energy:
\(\Delta h = d \sin(30^\circ) = 5.0 \times 0.5 = 2.5\text{ m}\)
\(\Delta E_p = m g \Delta h = 4.0\text{ kg} \times 9.81\text{ m s}^{-2} \times 2.5\text{ m} = 98.1\text{ J}\)

3. Calculate work done against friction:
\(W_f = W_{\text{in}} - \Delta E_p = 150\text{ J} - 98.1\text{ J} = 51.9\text{ J}\)

To two significant figures, this is \(52\text{ J}\).

1 mark for the correct option (A).

- Method mark for calculating the total work input (\(150\text{ J}\)) and change in gravitational potential energy (\(98.1\text{ J}\)).
- Accuracy mark for calculating the difference to yield \(52\text{ J}\).

We can set up two simultaneous equations using the formula for terminal potential difference:
\(\varepsilon = V + I r\)

For the first scenario:
- \(R_1 = 3.0\,\Omega\)
- \(V_1 = 6.0\text{ V}\)
- Current \(I_1 = \frac{V_1}{R_1} = \frac{6.0}{3.0} = 2.0\text{ A}\)
- Equation 1: \(\varepsilon = 6.0 + 2.0 r\)

For the second scenario:
- \(R_2 = 7.0\,\Omega\)
- \(V_2 = 8.4\text{ V}\)
- Current \(I_2 = \frac{V_2}{R_2} = \frac{8.4}{7.0} = 1.2\text{ A}\)
- Equation 2: \(\varepsilon = 8.4 + 1.2 r\)

Equating the two expressions for \(\varepsilon\):
\(6.0 + 2.0 r = 8.4 + 1.2 r\)
\(0.8 r = 2.4\)
\(r = 3.0\,\Omega\)

1 mark for the correct option (C).

- Method mark for calculating the currents (\(2.0\text{ A}\) and \(1.2\text{ A}\)) and setting up the simultaneous emf equations.
- Accuracy mark for solving to find \(r = 3.0\,\Omega\).

1. For the original string of length \(L\), the frequency of the third harmonic is \(f_3 = 3 f_1 = 360\text{ Hz}\).
Therefore, the fundamental frequency (first harmonic) of the original string is:
\(f_1 = \frac{360}{3} = 120\text{ Hz}\).

2. The formula for the fundamental frequency is:
\(f = \frac{v}{2l}\)
where \(v\) is the wave speed (which remains constant because the tension and mass per unit length are unchanged) and \(l\) is the length of the string.

3. For a new string of length \(l' = 2L\):
\(f'_1 = \frac{v}{2(2L)} = \frac{1}{2} \left( \frac{v}{2L} \right) = \frac{f_1}{2}\)
\(f'_1 = \frac{120\text{ Hz}}{2} = 60\text{ Hz}\).

1 mark for the correct option (A).

- Method mark for identifying the original fundamental frequency (\(120\text{ Hz}\)).
- Accuracy mark for correctly applying the inverse relationship between frequency and length to find \(60\text{ Hz}\).

Since the wires are connected in series, the tension (force \(F\)) in both wires is identical. Since they are made of the same material, they have the same Young modulus \(E\).

Using the Young modulus formula:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L} \implies \Delta L = \frac{F L}{A E}\)

Since area \(A = \frac{\pi d^2}{4}\), we have:
\(\Delta L \propto \frac{L}{d^2}\)

Let the length and diameter of wire Y be \(L_Y = L\) and \(d_Y = d\).
For wire X:
- \(L_X = 2L\)
- \(d_X = 2d\)

Now find the ratio of their extensions:
\(\frac{\Delta L_X}{\Delta L_Y} = \frac{L_X / d_X^2}{L_Y / d_Y^2} = \frac{2L / (2d)^2}{L / d^2} = \frac{2L / 4d^2}{L / d^2} = \frac{2}{4} = 0.5\)

1 mark for the correct option (B).

- Method mark for establishing that \(F\) and \(E\) are constant, and showing the proportionality \(\Delta L \propto \frac{L}{d^2}\).
- Accuracy mark for evaluating the ratio as \(0.5\).

1. Find the grating spacing \(d\):
\(d = \frac{1}{400\text{ lines/mm}} = \frac{1}{400 \times 10^3\text{ lines/m}} = 2.5 \times 10^{-6}\text{ m}\)

2. Use the grating equation \(d \sin\theta = n \lambda\), where \(n = 2\) and \(\theta = 30^\circ\):
\(2.5 \times 10^{-6} \times \sin(30^\circ) = 2 \lambda\)
\(2.5 \times 10^{-6} \times 0.5 = 2 \lambda\)
\(1.25 \times 10^{-6} = 2 \lambda\)
\(\lambda = 6.25 \times 10^{-7}\text{ m} = 625\text{ nm}\)

1 mark for the correct option (C).

- Method mark for calculating the slit spacing \(d = 2.5 \times 10^{-6}\text{ m}\).
- Accuracy mark for applying the diffraction grating formula to find \(\lambda = 625\text{ nm}\).

Use Einstein's photoelectric equation:
\(E_{k(\text{max})} = hf - \Phi\)

1. Calculate the energy of the incident photon (\(hf\)):
\(E = (6.63 \times 10^{-34}\text{ J s}) \times (8.0 \times 10^{14}\text{ Hz}) = 5.304 \times 10^{-19}\text{ J}\)

2. Convert the work function (\(\Phi\)) from \(\text{eV}\) to Joules:
\(\Phi = 2.4\text{ eV} \times (1.60 \times 10^{-19}\text{ J/eV}) = 3.84 \times 10^{-19}\text{ J}\)

3. Calculate the maximum kinetic energy:
\(E_{k(\text{max})} = 5.304 \times 10^{-19}\text{ J} - 3.84 \times 10^{-19}\text{ J} = 1.464 \times 10^{-19}\text{ J}\)

This rounds to \(1.5 \times 10^{-19}\text{ J}\) to two significant figures.

1 mark for the correct option (A).

- Method mark for converting either the work function to Joules (\(3.84 \times 10^{-19}\text{ J}\)) or photon energy to eV (\(3.32\text{ eV}\)).
- Accuracy mark for computing the difference to get \(1.5 \times 10^{-19}\text{ J}\).

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

To find the overall percentage uncertainty, sum the individual percentage uncertainties, multiplying by the power of each term:
\(\%\text{ uncertainty in } \rho = \%\text{ uncertainty in } R + 2 \times (\%\text{ uncertainty in } d) + \%\text{ uncertainty in } L\)

Substitute the given values:
\(\%\text{ uncertainty in } \rho = 2\% + 2(3\%) + 1\%\)
\(\%\text{ uncertainty in } \rho = 2\% + 6\% + 1\% = 9\%\)

1 mark for the correct option (D).

- Method mark for identifying that the percentage uncertainty of the diameter must be multiplied by 2 because it is squared in the area formula.
- Accuracy mark for summing the uncertainties correctly to yield \(9\%\).

Using conservation laws: Charge: \(0 + 0 = +1 + Q_X \implies Q_X = -1\). Baryon number: \(0 + 1 = 1 + B_X \implies B_X = 0\). Lepton number: \(1 + 0 = 0 + L_{e,X} \implies L_{e,X} = 1\). The only particle matching these is the electron \(e^-\).

1 mark for the correct answer (B). Accept any equivalent derivation showing conservation laws.

The change in momentum is \(\Delta p = m \Delta v = 0.15 \times (36 - (-24)) = 9.0\text{ kg m s}^{-1}\). The average force is \(F = \frac{\Delta p}{\Delta t} = \frac{9.0}{8.0 \times 10^{-3}} = 1125\text{ N}\).

1 mark for the correct answer (C).

The equivalent resistance of the two parallel resistors is \(R_p = \frac{10}{2} = 5\ \Omega\). The total resistance of the circuit is \(R_{total} = R_p + 15 = 20\ \Omega\). The total current is \(I = \frac{V}{R_{total}} = \frac{6.0}{20} = 0.30\text{ A}\). The potential difference across the \(15\ \Omega\) resistor is \(V = I \times 15 = 0.30 \times 15 = 4.5\text{ V}\).

1 mark for the correct answer (D).

The grating spacing \(d\) is \(\frac{1}{5.00 \times 10^5} = 2.00 \times 10^{-6}\text{ m}\). Using \(d \sin\theta = n \lambda\) with \(n = 3\), we find \(\lambda = \frac{2.00 \times 10^{-6} \times \sin(50.0^\circ)}{3} = 5.11 \times 10^{-7}\text{ m} = 511\text{ nm}\).

1 mark for the correct answer (B).

The percentage uncertainty in \(\rho\) is given by the sum of the percentage uncertainties of individual terms, multiplying by the powers: \(\% \text{ uncertainty in } \rho = \% \Delta R + 2 \times (\% \Delta d) + \% \Delta L = 2\% + 2(1.5\%) + 1\% = 6.0\%\).

1 mark for the correct answer (C).

Energy of the incident photon: \(E = hf = 6.63 \times 10^{-34} \times 1.20 \times 10^{15} = 7.956 \times 10^{-19}\text{ J}\). Work function in Joules: \(\Phi = 3.40 \times 1.60 \times 10^{-19} = 5.44 \times 10^{-19}\text{ J}\). Maximum kinetic energy: \(E_k = E - \Phi = 7.956 \times 10^{-19} - 5.44 \times 10^{-19} = 2.52 \times 10^{-19}\text{ J} \approx 2.5 \times 10^{-19}\text{ J}\).

1 mark for the correct answer (B).

For the second harmonic, the wavelength of the stationary wave is equal to the length of the string: \(\lambda = L = 0.80\text{ m}\). Using \(v = f \lambda\), the frequency is \(f = \frac{v}{\lambda} = \frac{120}{0.80} = 150\text{ Hz}\).

1 mark for the correct answer (B).

The tension force is \(F = mg = 15 \times 9.81 = 147.15\text{ N}\). Young's modulus is \(E = \frac{FL}{A\Delta L} \implies \Delta L = \frac{FL}{AE} = \frac{147.15 \times 2.0}{1.5 \times 10^{-6} \times 2.0 \times 10^{11}} = 9.81 \times 10^{-4}\text{ m} \approx 0.98\text{ mm}\).

1 mark for the correct answer (B).

In electron capture, a proton interacts with an electron to produce a neutron and an electron neutrino: \(p + e^- \rightarrow n + \nu_e\). Let us check the conservation laws for this interaction: 1. Charge: \(+1 + (-1) = 0 + 0\) (conserved). 2. Baryon number: \(+1 + 0 = +1 + 0\) (conserved). 3. Lepton number: \(0 + 1 = 0 + 1\) (conserved). Therefore, this interaction is fully permitted. Option B violates lepton number (lepton number is +2 on the right-hand side, but 0 on the left-hand side because a antineutrino should be produced instead). Option C is a pion decay that violates lepton family number. Option D violates muon lepton number.

1 mark for the correct choice of Option A. Reject other options due to violations of conservation laws.

The kinetic energy \(E_k\) of the electron accelerated through a potential difference \(V\) is given by \(E_k = eV\). The momentum \(p\) of the electron is \(p = \sqrt{2m_e E_k} = \sqrt{2m_e eV}\). The de Broglie wavelength is \(\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m_e eV}}\). If the potential difference becomes \(3V\), the new wavelength is \(\lambda' = \frac{h}{\sqrt{2m_e e(3V)}} = \frac{1}{\sqrt{3}} \times \frac{h}{\sqrt{2m_e eV}} = \frac{\lambda}{\sqrt{3}}\).

1 mark for identifying the inverse square root relationship between wavelength and voltage and selecting Option C.

For the third harmonic on a string of length \(L\) fixed at both ends, there are three loops (half-wavelengths). Thus, the length of the string is equal to \(L = 3 \times \frac{\lambda}{2}\), which gives the wavelength \(\lambda = \frac{2}{3}L\). The distance between a node and its adjacent antinode is a quarter of a wavelength: \(\text{Distance} = \frac{\lambda}{4} = \frac{1}{4} \times \left(\frac{2}{3}L\right) = \frac{L}{6}\).

1 mark for finding the wavelength in terms of length \(L\) and calculating a quarter of that wavelength to select Option C.

Using the diffraction grating equation: \(d \sin\theta = n \lambda\). For the third-order maximum, \(n = 3\), so \(d \sin\theta = 3\lambda \implies d = \frac{3\lambda}{\sin\theta}\). The slit spacing \(d\) in metres is related to the number of lines per millimetre \(N\) by \(d = \frac{10^{-3}}{N}\) because there are \(1000N\) lines per metre. Therefore, \(\frac{1}{1000N} = \frac{3\lambda}{\sin\theta}\), which rearranges to \(N = \frac{\sin\theta}{3000\lambda}\).

1 mark for correctly applying the grating formula, substituting \(n = 3\), and converting to lines per millimetre to select Option A.

Two identical cells connected in parallel behave as a single equivalent cell with an emf of \(\varepsilon\) and an equivalent internal resistance of \(\frac{r}{2}\). The total resistance of the circuit is therefore \(R + \frac{r}{2}\). The total current \(I\) in the external resistor is given by: \(I = \frac{\varepsilon}{R + \frac{r}{2}} = \frac{2\varepsilon}{2R + r}\).

1 mark for calculating the parallel equivalent emf and internal resistance to find the correct current expression, selecting Option B.

First, find the final velocity \(v_Q\) of trolley Q using conservation of momentum: \(2mu = 2m\left(\frac{1}{3}u\right) + mv_Q \implies v_Q = \frac{4}{3}u\). The total initial kinetic energy of the system is \(E_{k,i} = \frac{1}{2}(2m)u^2 = mu^2\). The total final kinetic energy of the system is \(E_{k,f} = \frac{1}{2}(2m)\left(\frac{1}{3}u\right)^2 + \frac{1}{2}m\left(\frac{4}{3}u\right)^2 = \frac{1}{9}mu^2 + \frac{8}{9}mu^2 = mu^2\). Since \(E_{k,i} = E_{k,f}\), the total kinetic energy does not change (the collision is perfectly elastic).

1 mark for determining the final velocity of trolley Q, computing both initial and final kinetic energies, and selecting Option C.

The elastic strain energy stored is given by \(U = \frac{1}{2}F \Delta L\). Since the extension is \(\Delta L = \frac{FL}{AE}\), the stored energy is \(U = \frac{F^2 L}{2AE}\). The wires are of the same material, so the Young's modulus \(E\) is the same, and they experience the same force \(F\). This means that \(U \propto \frac{L}{A} \propto \frac{L}{d^2}\). For wire X: \(U_X \propto \frac{L}{d^2}\). For wire Y: \(U_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\). The ratio \(\frac{U_X}{U_Y} = 2\), which is a ratio of \(2:1\).

1 mark for expressing the elastic strain energy in terms of length and diameter squared, finding the correct ratio, and selecting Option C.

The relationship for \(g\) is given by \(g = \frac{4\pi^2 L}{T^2}\). The percentage uncertainty in the total measured time is \(\frac{0.3}{30.0} \times 100\% = 1\%\). Since the period \(T\) is the total time divided by 20, its percentage uncertainty is also \(1\%\). The percentage uncertainty in \(T^2\) is twice the percentage uncertainty in \(T\), which is \(2 \times 1\% = 2\%\). Combining these, the percentage uncertainty in \(g\) is \(\% \text{ uncertainty in } L + \% \text{ uncertainty in } T^2 = 1\% + 2\% = 3\%\).

1 mark for calculating the percentage uncertainty in the period, doubling it for \(T^2\), adding the percentage uncertainty in \(L\), and selecting Option C.

In \(\beta^-\)\ decay, a neutron (\(udd\)) converts into a proton (\(uud\)). This means a down quark (\(d\)) is converted into an up quark (\(u\)). To conserve charge at the quark-lepton vertex, a \(W^-\)\ boson is emitted: \(d \rightarrow u + W^-\). The \(W^-\)\ boson then decays into an electron and an electron antineutrino (\(W^- \rightarrow e^- + \bar{\nu}_e\)). Hence, the exchange particle is the \(W^-\)\ boson, and the quark flavour change is down to up.

[1 mark] A: Correctly identifies both the exchange particle as \(W^-\)\ and the quark flavour change as down to up.

First, apply conservation of linear momentum to find the common velocity \(v\) after the collision: \(p_i = p_f\) which gives \((2.0 \times 6.0) + (4.0 \times 0) = (2.0 + 4.0) v\). Solving for \(v\): \(12.0 = 6.0 v \implies v = 2.0\text{ m s}^{-1}\). Next, calculate the initial kinetic energy: \(E_{ki} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 2.0 \times 6.0^2 = 36.0\text{ J}\). Calculate the final kinetic energy: \(E_{kf} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times 6.0 \times 2.0^2 = 12.0\text{ J}\). The loss in kinetic energy is: \(\Delta E_k = E_{ki} - E_{kf} = 36.0 - 12.0 = 24.0\text{ J}\).

[1 mark] B: Correctly calculates the common final velocity as \(2.0\text{ m s}^{-1}\) and the loss in kinetic energy as \(24\text{ J}\).

For a string fixed at both ends, the wavelength of the \(n\)-th harmonic is given by: \(L = n \frac{\lambda}{2}\). For the third harmonic (\(n = 3\)): \(1.2 = 3 \frac{\lambda}{2} \implies \lambda = \frac{2 \times 1.2}{3} = 0.80\text{ m}\). Using the wave equation \(v = f \lambda\): \(f = \frac{v}{\lambda} = \frac{180}{0.80} = 225\text{ Hz}\).

[1 mark] C: Correctly determines the wavelength for the third harmonic as \(0.80\text{ m}\) and calculates the frequency as \(225\text{ Hz}\).

Using the diffraction grating equation: \(d \sin\theta = n \lambda\), where \(n = 3\), \(\lambda = 632.8 \times 10^{-9}\text{ m}\), and \(\theta = 51.4^\circ\). Calculate grating spacing \(d\): \(d = \frac{3 \times 632.8 \times 10^{-9}}{\sin(51.4^\circ)} = \frac{1.8984 \times 10^{-6}}{0.7815} \approx 2.429 \times 10^{-6}\text{ m}\). The number of lines per metre \(N\) is: \(N = \frac{1}{d} = \frac{1}{2.429 \times 10^{-6}} \approx 4.117 \times 10^5\text{ lines/m}\). Converting to lines per millimetre: \(N_{\text{mm}} = \frac{4.117 \times 10^5}{1000} \approx 412\text{ lines/mm}\).

[1 mark] C: Correctly applies the grating formula to find \(d \approx 2.43 \times 10^{-6}\text{ m}\) and converts to obtain \(412\text{ lines/mm}\).

First, calculate the initial output voltage using the potential divider formula: \(V_{\text{out}} = V_{\text{supply}} \times \frac{R_{\text{fixed}}}{R_{\text{thermistor}} + R_{\text{fixed}}}\), which gives \(V_{\text{out}} = 12.0 \times \frac{3.0\text{ k}\Omega}{6.0\text{ k}\Omega + 3.0\text{ k}\Omega} = 12.0 \times \frac{3.0}{9.0} = 4.0\text{ V}\). For a negative temperature coefficient (NTC) thermistor, as the temperature increases, its resistance decreases. When \(R_{\text{thermistor}}\) decreases, the total resistance of the series circuit decreases, increasing the circuit current. This increases the potential difference across the fixed resistor, so \(V_{\text{out}}\) increases.

[1 mark] A: Correctly calculates \(V_{\text{out}} = 4.0\text{ V}\) and explains that \(V_{\text{out}}\) increases as the NTC thermistor's resistance decreases with temperature.

First, calculate the extension \(\Delta L\) of the wire using the Young modulus formula: \(E_Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L} = \frac{FL}{A \Delta L}\). This gives \(\Delta L = \frac{FL}{A E_Y} = \frac{36 \times 2.5}{1.2 \times 10^{-7} \times 2.0 \times 10^{11}} = \frac{90}{2.4 \times 10^4} = 3.75 \times 10^{-3}\text{ m}\). Now, calculate the stored elastic strain energy \(E_s\): \(E_s = \frac{1}{2} F \Delta L = \frac{1}{2} \times 36 \times 3.75 \times 10^{-3} = 18 \times 3.75 \times 10^{-3} = 0.0675\text{ J}\). Rounding to 2 significant figures gives \(6.8 \times 10^{-2}\text{ J}\).

[1 mark] C: Correctly determines the extension to be \(3.75\text{ mm}\) and calculates the stored energy to be \(6.8 \times 10^{-2}\text{ J}\).