2025 AQA IAS-Level Physics (9630) 模擬試題連答案詳解

The angular frequency is given by: \(\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{40}{0.45}} \approx 9.428\text{ rad s}^{-1}\). The maximum speed of the block is: \(v_{\max} = \omega A = 9.428 \times 0.080 \approx 0.754\text{ m s}^{-1}\).

[1 mark] for using \(v_{\max} = A\sqrt{\frac{k}{m}}\) or equivalent; [1 mark] for correct substitution of values; [1 mark] for correct final answer of \(0.75\text{ m s}^{-1}\) (accept \(0.75\) to \(0.76\text{ m s}^{-1}\)).

Using the equation of motion \(s = ut + \frac{1}{2}at^2\) where \(s = 15\text{ m}\), \(u = 25\text{ m s}^{-1}\), and \(a = -9.81\text{ m s}^{-2}\): \(15 = 25t - 4.905t^2\). Rearranging gives the quadratic equation: \(4.905t^2 - 25t + 15 = 0\). Solving this yields two times: \(t_1 \approx 0.69\text{ s}\) (on the way up) and \(t_2 \approx 4.40\text{ s}\) (on the way down). The total time interval spent above \(15\text{ m}\) is \(\Delta t = t_2 - t_1 = 4.40 - 0.69 = 3.71\text{ s}\).

[1 mark] for setting up the quadratic equation or an equivalent method to find times; [1 mark] for calculating the two times \(t_1 \approx 0.69\text{ s}\) and \(t_2 \approx 4.40\text{ s}\); [1 mark] for subtracting the times to find the time interval of \(3.7\text{ s}\) (accept \(3.7\) to \(3.71\text{ s}\)).

Using the definition of the Young modulus: \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{FL}{A\Delta L}\). Rearranging for extension gives: \(\Delta L = \frac{FL}{AE} = \frac{150 \times 2.5}{1.2 \times 10^{-6} \times 2.0 \times 10^{11}} = \frac{375}{2.4 \times 10^5} = 1.56 \times 10^{-3}\text{ m} = 1.56\text{ mm}\).

[1 mark] for rearranging the Young modulus formula to \(\Delta L = \frac{FL}{AE}\); [1 mark] for correct substitution of values; [1 mark] for final answer of \(1.6\text{ mm}\) (accept \(1.56\text{ mm}\) to \(1.6\text{ mm}\)).

The excitation energy required to raise the atom from \(n=1\) to \(n=3\) is: \(\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09\text{ eV}\). The remaining kinetic energy of the electron is: \(E_{\text{k}} = 12.5\text{ eV} - 12.09\text{ eV} = 0.41\text{ eV}\). Converting this energy to joules: \(0.41 \times 1.60 \times 10^{-19}\text{ J} = 6.56 \times 10^{-20}\text{ J}\) (or \(6.6 \times 10^{-20}\text{ J}\)).

[1 mark] for finding the excitation energy of \(12.09\text{ eV}\); [1 mark] for calculating the remaining kinetic energy as \(0.41\text{ eV}\); [1 mark] for converting the remaining energy to joules to give \(6.6 \times 10^{-20}\text{ J}\) (accept \(6.56 \times 10^{-20}\text{ J}\) to \(6.6 \times 10^{-20}\text{ J}\)).

Using the potential divider equation, the potential difference across the LDR is: \(V_{\text{LDR}} = V \times \frac{R_{\text{LDR}}}{R_{\text{fixed}} + R_{\text{LDR}}} = 9.0 \times \frac{12 \times 10^3}{4.0 \times 10^3 + 12 \times 10^3} = 9.0 \times \frac{12}{16} = 6.75\text{ V}\). Rounding to two significant figures gives \(6.8\text{ V}\).

[1 mark] for using the potential divider formula or equivalent multi-step method (e.g. finding current first); [1 mark] for correct substitution of values; [1 mark] for the final correct potential difference of \(6.8\text{ V}\) (accept \(6.75\text{ V}\)).

Part (a): Let \(v\) be the maximum speed. The acceleration phase has duration \(t_1 = v / 0.80 = 1.25 v\). The deceleration phase has duration \(t_3 = v / 1.20 = 0.833 v\). The constant speed phase has duration \(t_2 = 250 - t_1 - t_3 = 250 - 2.083 v\). The distance during acceleration is \(s_1 = 0.5 \times v \times t_1 = 0.625 v^2\). The distance during deceleration is \(s_3 = 0.5 \times v \times t_3 = 0.417 v^2\). The distance during constant speed is \(s_2 = v \times t_2 = 250 v - 2.083 v^2\). Total distance \(S = s_1 + s_2 + s_3 = 250 v - 1.042 v^2\). Setting this equal to \(6562.5\text{ m}\) gives: \(1.042 v^2 - 250 v + 6562.5 = 0\). Solving the quadratic equation gives \(v = 30\text{ m s}^{-1}\) (discarding the other root as physically impossible). Part (b): Distance while accelerating is \(s_1 = 0.5 \times v \times t_1 = 0.5 \times 30 \times (30 / 0.80) = 562.5\text{ m}\). Part (c): Average speed = total distance / total time = \(6562.5 / 250 = 26.25\text{ m s}^{-1}\).

Part (a) [4.2 marks]: [1] For expressing acceleration and deceleration times in terms of \(v\) as \(1.25v\) and \(0.833v\). [1] For writing a correct expression for the total distance in terms of \(v\) (e.g. \(250v - 1.042v^2 = 6562.5\)). [1] For setting up the quadratic equation correctly. [1.2] For solving and confirming \(v = 30\text{ m s}^{-1}\). Part (b) [3.0 marks]: [1] For calculating the acceleration time \(t_1 = 37.5\text{ s}\). [1] For using a correct equation for distance, e.g. \(s = 0.5 \times (u+v) \times t\) or \(s = v^2 / 2a\). [1] For obtaining \(562.5\text{ m}\) (accept \(563\text{ m}\) or \(560\text{ m}\)). Part (c) [3.0 marks]: [1] For stating average speed = total distance / total time. [1] For substituting values correctly: \(6562.5 / 250\). [1] For obtaining \(26.3\text{ m s}^{-1}\) (accept \(26\text{ m s}^{-1}\) or \(26.25\text{ m s}^{-1}\)).

Part (a): Cross-sectional area \(A = \pi \times (d/2)^2 = \pi \times (0.60 \times 10^{-3})^2 = 1.13 \times 10^{-6}\text{ m}^2\). Part (b): Extension \(\Delta L = F \times L / (A \times E) = 350 \times 2.5 / (1.13 \times 10^{-6} \times 2.0 \times 10^{11}) = 875 / (2.26 \times 10^5) = 3.87 \times 10^{-3}\text{ m}\). Part (c): Elastic strain energy \(E = 0.5 \times F \times \Delta L = 0.5 \times 350 \times 3.87 \times 10^{-3} = 0.677\text{ J}\).

Part (a) [2.0 marks]: [1] For using the correct formula \(A = \pi d^2 / 4\) or \(A = \pi r^2\). [1] For correct calculation to give \(1.13 \times 10^{-6}\text{ m}^2\) (accept \(1.1 \times 10^{-6}\text{ m}^2\)). Part (b) [4.2 marks]: [1] For recalling the Young modulus formula \(E = (F \times L) / (A \times \Delta L)\). [1] For rearranging the formula to make \(\Delta L\) the subject. [1] For substituting their area from (a) correctly. [1.2] For correct calculation to give \(3.9 \times 10^{-3}\text{ m}\) (accept \(3.87\text{ mm}\); allow error-carried-forward from part a). Part (c) [4.0 marks]: [2] For recalling \(E = 0.5 \times F \times \Delta L\). [2] For substituting values correctly and obtaining \(0.68\text{ J}\) (accept \(0.677\text{ J}\); allow error-carried-forward from part b).

Part (a): When light intensity decreases, fewer charge carriers are released in the LDR material, which causes the LDR's resistance to increase. Since the LDR and the fixed resistor are connected in series, the total resistance of the circuit increases, lowering the current. However, because the LDR's resistance is now a much larger fraction of the total resistance, it takes a larger share of the total input potential difference of \(12.0\text{ V}\), so the voltmeter reading increases. Part (b): Total circuit resistance in daylight = \(4000\ \Omega + 800\ \Omega = 4800\ \Omega\). Voltmeter reading \(V_{\text{out}} = 12.0 \times (800 / 4800) = 2.0\text{ V}\). Part (c): Since the voltmeter reads \(10.0\text{ V}\) across the LDR, the potential difference across the fixed resistor is \(12.0\text{ V} - 10.0\text{ V} = 2.0\text{ V}\). The current in the circuit is \(I = V / R = 2.0 / 4000 = 0.50\text{ mA}\). The resistance of the LDR in darkness is \(R = V / I = 10.0 / (0.50 \times 10^{-3}) = 20000\ \Omega = 20\text{ k}\Omega\).

Part (a) [3.2 marks]: [1] State that LDR resistance increases as light intensity decreases. [1] Explain that the total circuit current decreases. [1.2] Explain that the LDR takes a larger share of the total potential difference. Part (b) [3.0 marks]: [1] Use potential divider formula \(V_{\text{out}} = V_{\text{in}} \times R_{\text{LDR}} / (R_1 + R_{\text{LDR}})\). [1] Substitute values: \(12.0 \times 800 / 4800\). [1] Obtain \(2.0\text{ V}\). Part (c) [4.0 marks]: [1] State potential difference across the fixed resistor is \(2.0\text{ V}\). [1] Calculate current using \(I = 2.0 / 4000 = 0.50\text{ mA}\). [1] Apply Ohm's law for the LDR. [1] Obtain \(2.0 \times 10^4\ \Omega\) (or \(20\text{ k}\Omega\)).

Part (a): Using Snell's law: \(n_1 \sin\theta_1 = n_2 \sin\theta_2\) where \(1.00 \sin(45.0^\circ) = 1.52 \sin\theta_r\). \(\sin\theta_r = \sin(45.0^\circ) / 1.52 = 0.7071 / 1.52 = 0.4652\). Therefore, \(\theta_r = \sin^{-1}(0.4652) = 27.7^\circ\). Part (b): The critical angle is given by \(\sin\theta_c = n_2 / n_1 = 1.00 / 1.52 = 0.6579\). Therefore, \(\theta_c = \sin^{-1}(0.6579) = 41.1^\circ\). Part (c): The speed of light inside the glass block is \(v = c / n = 3.00 \times 10^8 / 1.52 = 1.97 \times 10^8\text{ m s}^{-1}\).

Part (a) [3.2 marks]: [1] Recall Snell's Law: \(n_1 \sin\theta_1 = n_2 \sin\theta_2\). [1] Substitute values: \(1.00 \sin(45.0^\circ) = 1.52 \sin\theta_r\). [1.2] Correct calculation to obtain \(27.7^\circ\) (accept \(28^\circ\)). Part (b) [3.0 marks]: [1] Recall formula for critical angle: \(\sin\theta_c = 1/n\). [1] Substitute values: \(\sin\theta_c = 1/1.52\). [1] Correct calculation to obtain \(41.1^\circ\) (accept \(41^\circ\)). Part (c) [4.0 marks]: [2] Recall relation \(n = c/v\) or \(v = c/n\). [2] Correct calculation to obtain \(1.97 \times 10^8\text{ m s}^{-1}\) (accept \(2.0 \times 10^8\text{ m s}^{-1}\)).

Part (a): Excitation is the process where an orbital electron in an atom absorbs a discrete amount of energy and transitions from a lower energy level (ground state) to a higher energy level. Part (b): The accelerated electron has a kinetic energy of \(12.5\text{ eV}\). The excitation energies required from the ground state are: \(E_2 - E_1 = -3.40 - (-13.6) = 10.2\text{ eV}\). \(E_3 - E_1 = -1.51 - (-13.6) = 12.09\text{ eV}\). \(E_4 - E_1 = -0.85 - (-13.6) = 12.75\text{ eV}\). Since \(12.09\text{ eV} < 12.5\text{ eV} < 12.75\text{ eV}\), the colliding electron can excite the atom up to \(n=3\), but not to \(n=4\). The remaining kinetic energy of the electron is \(12.5\text{ eV} - 12.09\text{ eV} = 0.41\text{ eV}\). Part (c): The energy of the emitted photon is \(\Delta E = 12.09\text{ eV} = 12.09 \times 1.60 \times 10^{-19}\text{ J} = 1.934 \times 10^{-18}\text{ J}\). Wavelength \(\lambda = hc / \Delta E = (6.63 \times 10^{-34} \times 3.00 \times 10^8) / (1.934 \times 10^{-18}) = 1.03 \times 10^{-7}\text{ m}\).

Part (a) [3.0 marks]: [1] Define excitation as an electron moving to a higher energy level. [1] State that this requires the absorption of energy. [1] Mention that the energy levels are discrete/quantised. Part (b) [4.2 marks]: [1] State the colliding electron energy is \(12.5\text{ eV}\). [1] Show calculations for excitation energies: \(n=2\) (\(10.2\text{ eV}\)) and \(n=3\) (\(12.09\text{ eV}\)). [1] Conclude \(n=3\) is the maximum level because \(12.5\text{ eV} > 12.09\text{ eV}\) but \(< 12.75\text{ eV}\). [1.2] Calculate remaining kinetic energy: \(12.5 - 12.09 = 0.41\text{ eV}\). Part (c) [3.0 marks]: [1] Convert photon energy from eV to Joules: \(12.09 \times 1.60 \times 10^{-19} = 1.93 \times 10^{-18}\text{ J}\). [1] Use \(\lambda = hc / E\) correctly. [1] Obtain \(\lambda = 1.03 \times 10^{-7}\text{ m}\) (accept \(1.0\) or \(1.03 \times 10^{-7}\text{ m}\) or \(103\text{ nm}\)).

Let upwards be the positive direction.

- The initial velocity is \(+u\).
- The acceleration is \(-g\).
- The displacement of the stone when it reaches the ground is \(-h\).

Using the equation of motion \(s = ut + \frac{1}{2}at^2\), we substitute these values:

\(-h = ut + \frac{1}{2}(-g)t^2\)

\(-h = ut - \frac{1}{2}gt^2\)

Multiplying both sides of the equation by \(-1\) gives:

\(h = \frac{1}{2}gt^2 - ut\)

This matches option C.

1 mark for the correct answer (C).

Inside a fluorescent tube, fast-moving electrons collide with mercury atoms, transferring energy and exciting the mercury atomic electrons to higher energy levels. When these excited electrons de-excite to lower energy levels, they emit ultraviolet (UV) photons.

These UV photons are then absorbed by the phosphor coating on the inner surface of the glass tube. This excites the phosphor, which subsequently de-excites in stages, emitting longer-wavelength visible light photons.

1 mark for the correct explanation of the excitation of mercury atoms, UV emission, and phosphor absorption (B).

The Young modulus \(E\) is given by:

\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{FL}{Ax}\)

Rearranging for extension \(x\):

\(x = \frac{FL}{AE}\)

For the second wire:
- The material is the same, so the Young modulus remains \(E\).
- The force applied is the same, \(F\).
- The length is \(L_2 = 2L\).
- The diameter is twice as large, which means the radius is twice as large. Since cross-sectional area \(A \propto r^2\), the new area is \(A_2 = 4A\).

Let \(x_2\) be the extension of the second wire:

\(x_2 = \frac{F(2L)}{(4A)E} = \frac{2}{4} \left(\frac{FL}{AE}\right) = 0.5x\)

Therefore, the extension is \(0.5x\).

1 mark for the correct answer (B).

First, calculate the output voltage in bright light:
- The resistance of the LDR is \(R_{\text{LDR}} = 4.0\,\Omega\).
- The total resistance of the circuit is \(R_{\text{total}} = 12.0 + 4.0 = 16.0\,\Omega\).
- The output voltage is:

\(V_{\text{out, bright}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R_{\text{total}}} = 6.0\,\text{V} \times \frac{4.0}{16.0} = 1.5\,\text{V}\)

Next, calculate the output voltage in darkness:
- The resistance of the LDR is \(R_{\text{LDR}} = 24.0\,\Omega\).
- The total resistance of the circuit is \(R_{\text{total}} = 12.0 + 24.0 = 36.0\,\Omega\).
- The output voltage is:

\(V_{\text{out, dark}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R_{\text{total}}} = 6.0\,\text{V} \times \frac{24.0}{36.0} = 4.0\,\text{V}\)

The change in the output voltage is:

\(\Delta V_{\text{out}} = V_{\text{out, dark}} - V_{\text{out, bright}} = 4.0\,\text{V} - 1.5\,\text{V} = 2.5\,\text{V}\)

1 mark for the correct answer (B).

The formula for the density \(\rho\) of a cylinder of diameter \(d\) and length \(h\) is:

\(\rho = \frac{\text{mass}}{\text{volume}} = \frac{m}{\pi \left(\frac{d}{2}\right)^2 h} = \frac{4m}{\pi d^2 h}\)

To find the fractional (or percentage) uncertainty in density, we sum the individual fractional uncertainties, multiplying by the powers of each variable:

\(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\left(\frac{\Delta d}{d}\right) + \frac{\Delta h}{h}\)

Substitute the given percentage uncertainties into this expression:

\(\% \text{ uncertainty in } \rho = 1.5\% + 2(1.0\%) + 2.0\%\)

\(\% \text{ uncertainty in } \rho = 1.5\% + 2.0\% + 2.0\% = 5.5\%\)

This corresponds to option B.

1 mark for the correct calculation of percentage uncertainty (B).

Let the mass on the table be \(m_1 = 3.0\,\text{kg}\) and the hanging mass be \(m_2 = 1.0\,\text{kg}\).

First, find the frictional force \(f\) acting on \(m_1\):
- Normal reaction force \(N = m_1 g = 3.0 \times 9.81 = 29.43\,\text{N}\).
- Frictional force \(f = \mu N = 0.20 \times 29.43 = 5.886\,\text{N}\).

The driving force of the system is the weight of the hanging block:
- \(F_{\text{drive}} = m_2 g = 1.0 \times 9.81 = 9.81\,\text{N}\).

Using Newton's second law for the combined system of mass \(m_1 + m_2 = 4.0\,\text{kg}\):

\((m_1 + m_2) a = F_{\text{drive}} - f\)

\(4.0 a = 9.81 - 5.886\)

\(4.0 a = 3.924\)

\(a = \frac{3.924}{4.0} = 0.981\,\text{m s}^{-2}\)

Rounding to two significant figures gives \(0.98\,\text{m s}^{-2}\).

1 mark for the correct answer (A).

Let the current in the circuit be \(I\).

- The power dissipated in the internal resistance is: \(P_{\text{internal}} = I^2 r\).
- The total electrical power produced by the cell is: \(P_{\text{total}} = I \varepsilon\).

Thus, the ratio of these powers is:

\(\text{Ratio} = \frac{P_{\text{internal}}}{P_{\text{total}}} = \frac{I^2 r}{I \varepsilon} = \frac{Ir}{\varepsilon}\)

Since \(I = \frac{\varepsilon}{R + r}\), we can substitute this into the ratio:

\(\text{Ratio} = \frac{\left(\frac{\varepsilon}{R + r}\right) r}{\varepsilon} = \frac{r}{R + r}\)

This matches option B.

1 mark for the correct derivation of the power ratio (B).

When combining two forces of magnitudes \(F_1\) and \(F_2\), the magnitude of the resultant force \(R\) must lie within a specific range, depending on the angle between them:

- The minimum possible magnitude occurs when the forces act in opposite directions (an angle of \(180^\circ\)):
\(R_{\text{min}} = |F_2 - F_1| = 12.0\,\text{N} - 5.0\,\text{N} = 7.0\,\text{N}\)

- The maximum possible magnitude occurs when the forces act in the same direction (an angle of \(0^\circ\)):
\(R_{\text{max}} = F_1 + F_2 = 12.0\,\text{N} + 5.0\,\text{N} = 17.0\,\text{N}\)

Therefore, any possible resultant force must satisfy:

\(7.0\,\text{N} \le R \le 17.0\,\text{N}\)

A value of \(6.0\,\text{N}\) is less than the minimum possible magnitude of \(7.0\,\text{N}\), and therefore cannot be the magnitude of the resultant force.

1 mark for identifying that the resultant force must be in the range 7.0 N to 17.0 N, making A the correct answer.

Average velocity is defined as the total displacement divided by the total time. Since the ball returns to its starting point, its net displacement is zero, and thus its average velocity is zero. The total distance traveled is the actual path length, which is \(h\) upwards plus \(h\) downwards, giving a total distance of \(2h\).

1 mark for correct choice A. No partial marks.

Using the potential divider formula in bright light: \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\) which gives \(3.0 = 12.0 \times \frac{2.0}{R + 2.0}\). This simplifies to \(0.25 = \frac{2.0}{R + 2.0}\), meaning \(R + 2.0 = 8.0\), so \(R = 6.0\text{ k}\Omega\). In darkness, with \(R_{\text{LDR}} = 10.0\text{ k}\Omega\): \(V_{\text{out}} = 12.0 \times \frac{10.0}{6.0 + 10.0} = 12.0 \times \frac{10.0}{16.0} = 7.5\text{ V}\).

1 mark for correct choice B. No partial marks.

The Young modulus \(E\) is given by: \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L} = \frac{F L}{A \Delta L}\). Since \(A = \frac{\pi d^2}{4}\), we have: \(\Delta L = \frac{4 F L}{\pi d^2 E}\). Since \(F\), \(E\), and \(\pi\) are constant for both wires, \(\Delta L \propto \frac{L}{d^2}\). For wire X: \(\Delta L_X \propto \frac{L}{d^2}\). For wire Y: \(\Delta L_Y \propto \frac{2L}{(2d)^2} = \frac{L}{2d^2}\). Thus, the ratio of their extensions is: \(\frac{\Delta L_X}{\Delta L_Y} = \frac{L/d^2}{L/(2d^2)} = 2.0\).

1 mark for correct choice C. No partial marks.

In electron-atom collisions, a colliding electron can transfer a portion of its kinetic energy to excite an atomic electron to a higher energy level, provided its initial kinetic energy is greater than or equal to the excitation energy. Here, \(11.5\text{ eV} > 10.2\text{ eV}\). After transferring \(10.2\text{ eV}\) to excite the atom, the colliding electron retains the remaining kinetic energy: \(11.5\text{ eV} - 10.2\text{ eV} = 1.3\text{ eV}\).

1 mark for correct choice B. No partial marks.

Rearranging the equation for \(g\) gives: \(g = \frac{4\pi^2 L}{T^2}\). The fractional uncertainty in \(g\) is determined by adding the fractional uncertainty in \(L\) to twice the fractional uncertainty in \(T\): \(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}\). Converting to percentage uncertainties: \(\text{Percentage uncertainty in } g = 1.5\% + (2 \times 2.0\%) = 1.5\% + 4.0\% = 5.5\%\).

1 mark for correct choice B. No partial marks.

Specific charge is defined as the charge-to-mass ratio (\(q/m\)) of a particle. For a proton (\(_1^1\text{H}^+\)), charge \(q_{\text{p}} = e\) and mass \(m_{\text{p}} \approx 1u\), so its specific charge is \(\approx e/u\). For an alpha particle (\(_2^4\text{He}^{2+}\)), charge \(q_{\alpha} = 2e\) and mass \(m_{\alpha} \approx 4u\), so its specific charge is \(\approx 2e/4u = e/2u\). The ratio of their specific charges is: \(\frac{e/(2u)}{e/u} = 0.5\).

1 mark for correct choice B. No partial marks.

1. Two progressive waves of the same frequency and amplitude traveling in opposite directions meet and superpose. At positions where they are in phase, constructive interference occurs (antinodes), and where they are out of phase, destructive interference occurs (nodes).

2. For the third harmonic: \( L = \frac{3}{2}\lambda \Rightarrow \lambda = \frac{2}{3} L = \frac{2}{3} \times 1.20 = 0.80\text{ m} \). The wave speed is given by \( v = f\lambda = 120 \times 0.80 = 96.0\text{ m s}^{-1} \).

3. Using \( v = \sqrt{\frac{T}{\mu}} \Rightarrow \mu = \frac{T}{v^2} = \frac{18.0}{96.0^2} = 1.95 \times 10^{-3}\text{ kg m}^{-1} \).

1. M1: Progressive waves of same frequency/wavelength traveling in opposite directions superpose. (1) M2: Mention of constructive interference at antinodes / destructive interference at nodes. (1)
2. M1: Correct calculation of wavelength \( \lambda = 0.80\text{ m} \). (1) M2: Correct calculation of speed \( v = 96.0\text{ m s}^{-1} \). (1)
3. M1: Correct rearrangement or substitution into the tension-speed formula. (1) M2: Correct calculation of \( \mu = 1.95 \times 10^{-3}\text{ kg m}^{-1} \) (accept 2 or 3 SF). (1)

1. As light intensity increases, the resistance of the LDR decreases. Since the potential divider splits voltage in proportion to resistance, a smaller fraction of the total supply voltage is dropped across the LDR, causing \( V_{\text{out}} \) to decrease.

2. Using the potential divider equation: \( V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R_1 + R_{\text{LDR}}} = 12.0 \times \frac{2.5 \times 10^3}{4.0 \times 10^3 + 2.5 \times 10^3} = 12.0 \times \frac{2.5}{6.5} = 4.62\text{ V} \).

3. The voltmeter is in parallel with the LDR, which reduces the effective resistance of the parallel combination. Consequently, the fraction of voltage across this section decreases, leading to a lower measured output voltage.

1. M1: LDR resistance decreases as light intensity increases. (1) M2: Output voltage is proportional to fraction of total resistance. (1) M3: Therefore, \( V_{\text{out}} \) decreases. (1)
2. M1: Correct substitution into potential divider formula. (1) M2: Correct calculation of \( 4.62\text{ V} \) (or \( 4.6\text{ V} \)). (1)
3. M1: Identifies that parallel combination reduces resistance, hence reducing the measured output voltage. (1)

1. The equation is \( V = \varepsilon - Ir \). Comparing this to the equation of a straight line, \( y = mx + c \), a graph of \( V \) on the y-axis against \( I \) on the x-axis has a y-intercept equal to the emf \( \varepsilon \), and a gradient equal to \( -r \) (the magnitude of the gradient is the internal resistance).

2. Set up simultaneous equations:
\( 1.38 = \varepsilon - 0.40 r \) (1)
\( 1.10 = \varepsilon - 1.10 r \) (2)
Subtracting (2) from (1) gives:
\( 0.28 = 0.70 r \Rightarrow r = 0.40\ \Omega \).
Substitute \( r = 0.40\ \Omega \) back into (1):
\( \varepsilon = 1.38 + 0.40(0.40) = 1.54\text{ V} \).

1. M1: States correct equation \( V = \varepsilon - Ir \). (1) M2: Identifies that y-intercept is \( \varepsilon \). (1) M3: Identifies that gradient is equal to \( -r \) (or magnitude of gradient is \( r \)). (1)
2. M1: Sets up two simultaneous equations correctly. (1) M2: Solves for internal resistance \( r = 0.40\ \Omega \). (1) M3: Solves for emf \( \varepsilon = 1.54\text{ V} \). (1)

1. High-energy colliding electrons transfer a discrete amount of kinetic energy to orbital electrons in the mercury atoms. This raises the orbital electrons to a higher energy level (excitation). When these electrons fall back to lower energy levels (de-excitation), they emit a photon of discrete energy equal to the energy difference between the levels.

2. The energy difference \( \Delta E \) is:
\( \Delta E = -1.04\text{ eV} - (-5.52\text{ eV}) = 4.48\text{ eV} \).
In Joules: \( \Delta E = 4.48 \times 1.60 \times 10^{-19}\text{ C} = 7.168 \times 10^{-19}\text{ J} \).
Using \( \lambda = \frac{hc}{\Delta E} \):
\( \lambda = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{7.168 \times 10^{-19}} = 2.774 \times 10^{-7}\text{ m} \approx 2.77 \times 10^{-7}\text{ m} \).

1. M1: Free electrons transfer kinetic energy to atomic electrons during collisions. (1) M2: Atomic electrons jump to a higher energy level (excitation). (1) M3: When falling back, they emit a photon of energy equal to the energy gap. (1)
2. M1: Correct calculation of energy difference \( \Delta E = 4.48\text{ eV} \). (1) M2: Correct conversion to Joules \( 7.17 \times 10^{-19}\text{ J} \). (1) M3: Correct calculation of wavelength \( 2.77 \times 10^{-7}\text{ m} \). (1)

1. Refractive index is the ratio of the speed of light in a vacuum to the speed of light in the material (i.e., \( n = \frac{c}{v} \)).

2. The critical angle \( \theta_c \) is given by:
\( \sin(\theta_c) = \frac{1}{n} = \frac{1}{1.52} \Rightarrow \theta_c = \sin^{-1}(0.6579) = 41.14^\circ \approx 41.1^\circ \).

3. Using Snell's law: \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \).
Here, \( n_1 = 1.52 \), \( \theta_1 = 35.0^\circ \), and \( n_2 = 1.00 \) (air).
\( 1.52 \times \sin(35.0^\circ) = 1.00 \times \sin(\theta_2) \)
\( \sin(\theta_2) = 1.52 \times 0.5736 = 0.8718 \Rightarrow \theta_2 = 60.67^\circ \approx 60.7^\circ \).

1. M1: Ratio of speed of light in vacuum to speed of light in medium. (1)
2. M1: Correct formula used \( \sin(\theta_c) = 1/n \). (1) M2: Correct calculation of \( 41.1^\circ \). (1)
3. M1: Use of Snell's law. (1) M2: Correct substitution of values. (1) M3: Correct calculation of angle of refraction \( 60.7^\circ \) (or \( 61^\circ \)). (1)

1. Coherent sources are those that maintain a constant phase relationship/difference over time and have the same frequency/wavelength.

2. The distance of \( 3.80\text{ cm} \) corresponds to 10 fringe spacings:
\( w = \frac{3.80 \times 10^{-2}\text{ m}}{10} = 3.80 \times 10^{-3}\text{ m} \) (or \( 3.80\text{ mm} \)).

3. Using the double-slit formula \( w = \frac{\lambda D}{s} \):
\( s = \frac{\lambda D}{w} = \frac{633 \times 10^{-9} \times 2.40}{3.80 \times 10^{-3}} = 3.998 \times 10^{-4}\text{ m} \approx 4.00 \times 10^{-4}\text{ m} \) (or \( 0.400\text{ mm} \)).

1. M1: Constant phase difference. (1) M2: Same frequency or wavelength. (1)
2. M1: Correctly calculates fringe spacing \( w = 3.80 \times 10^{-3}\text{ m} \). (1)
3. M1: Correct rearrangement of formula to make \( s \) the subject. (1) M2: Correct substitution with consistent units. (1) M3: Correct calculation of \( 4.00 \times 10^{-4}\text{ m} \). (1)

1. In reverse bias, the diode has very high resistance and conducts negligible current. In forward bias, current remains negligible until the threshold voltage of approximately \( 0.7\text{ V} \) is reached. Above this voltage, the resistance drops significantly, and current increases rapidly with any further increase in voltage.

2. In a filament lamp, an increase in voltage leads to an increase in current, which raises the temperature of the filament. This causes the metal ions in the lattice to vibrate with greater amplitude, increasing the frequency of collisions with flowing electrons, thereby increasing the resistance. In contrast, in an NTC thermistor, an increase in temperature releases additional charge carriers from the valence band, which dramatically decreases its resistance.

1. M1: Reverse bias: negligible current / very high resistance. (1) M2: Forward bias: threshold/turn-on voltage of around \( 0.6 - 0.7\text{ V} \). (1) M3: Sharp increase in current after threshold voltage. (1)
2. M1: Filament: Higher current increases temperature, causing greater lattice vibration. (1) M2: Increased vibrations increase electron collisions, increasing resistance. (1) M3: Thermistor: Increased temperature releases more charge carriers, decreasing resistance. (1)

1. Resistivity is a property of a material defined by \( \rho = \frac{RA}{L} \), where \( R \) is the resistance, \( A \) is the cross-sectional area, and \( L \) is the length (i.e., resistance per unit length for a unit cross-section).

2. Rearranging \( R = \rho \frac{L}{A} \):
\( A = \rho \frac{L}{R} = 1.72 \times 10^{-8} \times \frac{2.50}{1.60} = 2.6875 \times 10^{-8}\text{ m}^2 \approx 2.69 \times 10^{-8}\text{ m}^2 \).

3. Since volume \( V = A \times L \) is constant, if length is doubled (\( L' = 2L \)), the cross-sectional area must be halved (\( A' = \frac{A}{2} \)).
The new resistance is:
\( R' = \rho \frac{L'}{A'} = \rho \frac{2L}{A/2} = 4 \left(\rho \frac{L}{A}\right) = 4R \).
\( R' = 4 \times 1.60 = 6.40\ \Omega \).

1. M1: Correct definition of resistivity or formula defined with all terms. (1)
2. M1: Correct rearrangement to find \( A \). (1) M2: Correct calculation of area \( 2.69 \times 10^{-8}\text{ m}^2 \). (1)
3. M1: Realises that area is halved when length is doubled for constant volume. (1) M2: Shows that new resistance \( R' = 4R \). (1) M3: Calculates final value of \( 6.40\ \Omega \). (1)

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

(b)
Using the relationship between the Young modulus \(E\), force \(F\), extension \(\Delta L\), original length \(L\), and area \(A\):
\[E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{\Delta L / L} = \left(\frac{F}{\Delta L}\right) \frac{L}{A}\]
Since the gradient of the force-extension graph is \(k = \frac{F}{\Delta L} = 5.6 \times 10^3\text{ N m}^{-1}\):
\[E = k \frac{L}{A} = (5.6 \times 10^3\text{ N m}^{-1}) \times \frac{2.45\text{ m}}{1.134 \times 10^{-7}\text{ m}^2} = 1.21 \times 10^{11}\text{ Pa}\]
Accept \(1.2 \times 10^{11}\text{ Pa}\) (or \(\text{N m}^{-2}\)).

(c)
Since \(A = \frac{\pi d^2}{4}\), doubling the diameter \(d\) increases the cross-sectional area \(A\) by a factor of \(2^2 = 4\).
From the equation:
\[\text{gradient } k = \frac{E A}{L}\]
Because the material (Young modulus \(E\)) and the length \(L\) remain unchanged, the gradient is directly proportional to the area \(A\). Therefore, the gradient increases by a factor of 4.

(a)
- Correct substitution of diameter into \(A = \frac{\pi d^2}{4}\) (including unit conversion from mm to m) [1 mark]

(b)
- Rearranges Young modulus equation to use gradient: \(E = \text{gradient} \times \frac{L}{A}\) [1 mark]
- Calculates \(E = 1.2 \times 10^{11}\) (accept range \(1.2 \times 10^{11}\) to \(1.21 \times 10^{11}\)) [1 mark]
- Correct unit: \(\text{Pa}\) or \(\text{N m}^{-2}\) [1 mark]

(c)
- States that area increases by a factor of 4 (or \(A \propto d^2\)) [1 mark]
- States that gradient is directly proportional to area and therefore increases by a factor of 4 [1 mark]

(a)
At the equilibrium position, the tension in the spring equals the weight of the mass:
\[mg = k \Delta x\]
\[k = \frac{mg}{\Delta x} = \frac{0.15\text{ kg} \times 9.81\text{ m s}^{-2}}{0.082\text{ m}} = 17.94\text{ N m}^{-1}\]
This is approximately \(18\text{ N m}^{-1}\).

(b)
First, find the angular frequency \(\omega\):
\[\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{17.94\text{ N m}^{-1}}{0.15\text{ kg}}} = 10.94\text{ rad s}^{-1}\]
The amplitude of the oscillation is the initial displacement from equilibrium, \(A = 0.040\text{ m}\).
The maximum speed is given by:
\[v_{\text{max}} = \omega A = 10.94\text{ rad s}^{-1} \times 0.040\text{ m} = 0.438\text{ m s}^{-1} \approx 0.44\text{ m s}^{-1}\]

(c)
The acceleration is given by \(a = -\omega^2 x\), where \(x\) is the displacement from the equilibrium position.
- At the lowest point (maximum positive displacement downwards), the acceleration is at its maximum value and directed upwards (towards equilibrium).
- As the mass moves through the equilibrium position (\(x = 0\)), the acceleration is zero.
- At the highest point (maximum negative displacement upwards), the acceleration is at its maximum value and directed downwards (towards equilibrium).

(a)
- Equates gravitational force to spring force: \(mg = k \Delta x\) [1 mark]
- Substitutes values and calculates \(k \approx 17.9\text{ N m}^{-1}\) or \(18\text{ N m}^{-1}\) [1 mark]

(b)
- Calculates angular frequency: \(\omega = \sqrt{\frac{k}{m}} \approx 10.9\text{ rad s}^{-1}\) [1 mark]
- Calculates maximum speed: \(v_{\text{max}} = \omega A = 0.44\text{ m s}^{-1}\) (accept range \(0.438\) to \(0.44\)) [1 mark]

(c)
- States that acceleration is proportional to displacement but in the opposite direction, and is zero at the equilibrium position [1 mark]
- States that acceleration is maximum at the extreme positions (lowest and highest points) [1 mark]

(a)
Using the time constant formula:
\[\tau = RC = (15 \times 10^3\text{ }\Omega) \times (220 \times 10^{-6}\text{ F}) = 3.3\text{ s}\]

(b)
The discharging equation for potential difference is:
\[V = V_0 e^{-\frac{t}{RC}}\]
Substituting the values:
\[V = 9.0\text{ V} \times e^{-\frac{4.5}{3.3}}\]
\[V = 9.0\text{ V} \times e^{-1.364} = 9.0 \times 0.2557 = 2.30\text{ V}\]
Accept \(2.3\text{ V}\).

(c)
The energy \(E\) stored in a capacitor is:
\[E = \frac{1}{2} C V^2\]
Using \(C = 220 \times 10^{-6}\text{ F}\) and the calculated value of \(V = 2.30\text{ V}\):
\[E = 0.5 \times (220 \times 10^{-6}\text{ F}) \times (2.30\text{ V})^2\]
\[E = 1.10 \times 10^{-4} \times 5.29 = 5.82 \times 10^{-4}\text{ J}\]
Accept \(5.8 \times 10^{-4}\text{ J}\) or \(0.58\text{ mJ}\) (allow ecf from part b).

(a)
- Correct calculation of \(RC\) showing \(15 \times 10^3 \times 220 \times 10^{-6} = 3.3\text{ s}\) [1 mark]

(b)
- Uses the exponential discharge formula \(V = V_0 e^{-t/RC}\) [1 mark]
- Obtains a potential difference of \(2.3\text{ V}\) (accept \(2.3\) to \(2.31\)) [1 mark]

(c)
- Recalls and writes down the equation \(E = \frac{1}{2}CV^2\) (or equivalent) [1 mark]
- Substitutes their value of \(V\) from part (b) along with \(C = 220 \times 10^{-6}\text{ F}\) [1 mark]
- Obtains a correct final energy value with unit: \(5.8 \times 10^{-4}\text{ J}\) (or \(5.82 \times 10^{-4}\text{ J}\)) [1 mark]

For total internal reflection to just occur, the angle of incidence must equal the critical angle: \(\theta_i = \theta_c = 38.0^\circ\).
Using the formula for the critical angle at a boundary with air (where \(n_{\text{air}} = 1.00\)):
\(\sin\theta_c = \frac{1}{n}\)
\(n = \frac{1}{\sin(38.0^\circ)}\)
\(n = \frac{1}{0.61566} \approx 1.62\).

1 mark for the correct calculation of the refractive index, leading to option D.

For a string fixed at both ends, the length \(L\) in terms of wavelength \(\lambda\) for the third harmonic is:
\(3 \left(\frac{\lambda}{2}\right) = L \implies \lambda = \frac{2L}{3}\).

The distance from a node (fixed end) to the adjacent antinode is a quarter of a wavelength (\(\frac{\lambda}{4}\)):
\(d = \frac{\lambda}{4} = \frac{1}{4} \left(\frac{2L}{3}\right) = \frac{2L}{12} = \frac{L}{6}\).

1 mark for relating wavelength to string length and finding the correct node-to-antinode distance, leading to option B.

The energy levels of a hydrogen atom are:
- \(E_1 = -13.6\text{ eV}\) (ground state)
- \(E_2 = -3.4\text{ eV}\) (first excited state; \(\Delta E_1 = 10.2\text{ eV}\))
- \(E_3 = -1.51\text{ eV}\) (second excited state; \(\Delta E_2 = 12.09\text{ eV}\))
- \(E_4 = -0.85\text{ eV}\) (third excited state; \(\Delta E_3 = 12.75\text{ eV}\))

Since the incident electron has \(12.5\text{ eV}\) of energy, it cannot excite the atom to \(n=4\) (which requires \(12.75\text{ eV}\)). The highest reachable level is \(n=3\), requiring \(12.09\text{ eV}\).

After exciting the atom to \(n=3\), the kinetic energy remaining is:
\(12.5\text{ eV} - 12.09\text{ eV} = 0.41\text{ eV}\).

1 mark for calculating energy transitions and subtracting the correct excitation energy, leading to option A.

Using the potential divider formula for the voltage across the LDR:
\(V_{\text{LDR}} = V_s \times \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}}\)
\(3.0 = 12.0 \times \frac{400}{R + 400}\)
\(0.25 = \frac{400}{R + 400}\)
\(R + 400 = \frac{400}{0.25} = 1600\ \Omega\)
\(R = 1200\ \Omega\).

1 mark for using the potential divider relation to solve for the fixed resistance, leading to option C.

We set up two equations using \(\varepsilon = I(R + r)\):
1) \(\varepsilon = 1.2(3.0 + r) = 3.6 + 1.2r\)
2) \(\varepsilon = 0.60(8.0 + r) = 4.8 + 0.60r\)

Equating the two expressions for \(\varepsilon\):
\(3.6 + 1.2r = 4.8 + 0.60r\)
\(0.60r = 1.2\)
\(r = 2.0\ \Omega\).

1 mark for constructing the simultaneous equations and solving for the internal resistance, leading to option B.

Einstein's photoelectric equation states:
\(h f = \Phi + E_{k,\text{max}} \implies E_{k,\text{max}} = h f - \Phi\)

If the frequency is doubled, the new equation is:
\(h (2f) = \Phi + E'_{k,\text{max}} \implies E'_{k,\text{max}} = 2hf - \Phi\)

Substitute \(hf = E_{k,\text{max}} + \Phi\) into this equation:
\(E'_{k,\text{max}} = 2(E_{k,\text{max}} + \Phi) - \Phi = 2 E_{k,\text{max}} + \Phi\)

Since the work function \(\Phi\) is a positive physical value (\(\Phi > 0\)), the new maximum kinetic energy must be greater than \(2 E_{k,\text{max}}\).

1 mark for using the photoelectric equation to deduce the relationship between old and new maximum kinetic energies, leading to option C.

The fringe spacing is given by \(w = \frac{\lambda D}{d}\).
With the changes:
\(\lambda' = 1.2\lambda\)
\(d' = 0.5d\)

The new fringe spacing is:
\(w' = \frac{\lambda' D}{d'} = \frac{1.2\lambda D}{0.5d} = 2.4 \frac{\lambda D}{d} = 2.4 w\).

1 mark for correctly applying the double-slit scaling to find the ratio, leading to option D.

By definition, in a longitudinal wave (such as a sound wave in air), the particles of the medium oscillate back and forth parallel to the direction in which the wave travels (energy transfer). Transverse waves oscillate perpendicular to this direction. Additionally, only transverse waves can be polarized.

1 mark for identifying the correct characteristic definition of longitudinal waves, leading to option C.

According to Snell's Law:
\(n_1 \sin\theta_1 = n_2 \sin\theta_2\)

where:
- \(n_1 = 1.40\) (refractive index of liquid)
- \(n_2 = 1.60\) (refractive index of glass)
- \(\theta_2 = 32.0^\circ\) (angle of refraction in glass)
- \(\theta_1\) is the angle of incidence in the liquid.

Rearranging to solve for \(\sin\theta_1\):
\(\sin\theta_1 = \frac{n_2}{n_1} \sin\theta_2\)
\(\sin\theta_1 = \frac{1.60}{1.40} \sin(32.0^\circ)\)
\(\sin\theta_1 \approx 1.1429 \times 0.5299 \approx 0.6056\)

Taking the inverse sine:
\(\theta_1 = \arcsin(0.6056) \approx 37.3^\circ\)

1 mark for the correct calculation of the angle of incidence in the liquid.

The terminal potential difference \(V\) of a cell is related to the electromotive force \(\varepsilon\) and internal resistance \(r\) by the equation:
\(V = \varepsilon - Ir\)

Rearranging this into the standard straight-line equation form \(y = mx + c\):
\(V = -rI + \varepsilon\)

Comparing variables:
- Vertical axis (y) is \(V\)
- Horizontal axis (x) is \(I\)
- Gradient (m) is \(-r\), so the magnitude of the gradient is \(r\) (internal resistance).
- y-intercept (c) is \(\varepsilon\) (emf).

Therefore, the y-intercept represents the emf, and the magnitude of the gradient represents the internal resistance.

1 mark for identifying both the y-intercept and gradient correctly.

The potential divider formula for the output voltage across the fixed resistor is:
\(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{fixed}}}{R_{\text{fixed}} + R_{\text{LDR}}}\)

1. In the dark:
\(R_{\text{LDR}} = 10.0\text{ k}\Omega = 10\,000\ \Omega\)
\(V_{\text{out, dark}} = 12.0\text{ V} \times \frac{2.0\text{ k}\Omega}{2.0\text{ k}\Omega + 10.0\text{ k}\Omega} = 12.0 \times \frac{2}{12} = 2.0\text{ V}\)

2. In bright light:
\(R_{\text{LDR}} = 400\ \Omega = 0.4\text{ k}\Omega\)
\(V_{\text{out, light}} = 12.0\text{ V} \times \frac{2.0\text{ k}\Omega}{2.0\text{ k}\Omega + 0.4\text{ k}\Omega} = 12.0 \times \frac{2.0}{2.4} = 10.0\text{ V}\)

3. Calculate the change in output voltage:
\(\Delta V_{\text{out}} = V_{\text{out, light}} - V_{\text{out, dark}} = 10.0\text{ V} - 2.0\text{ V} = 8.0\text{ V}\)

1 mark for calculating both output voltages and finding the difference of 8.0 V.

For an air column closed at one end, the boundary conditions require a node at the closed end and an antinode at the open end.

This geometry only supports odd harmonics:
\(f_n = n f_1\) where \(n = 1, 3, 5, \dots\)

- The first harmonic (fundamental) has a frequency of \(f_1\).
- The next possible harmonic is the third harmonic (first overtone), which has a frequency of \(3.0 f_1\).

1 mark for identifying that only odd harmonics are possible in a closed-open tube, yielding 3.0 f1.

Einstein's photoelectric equation is:
\(hf = \Phi + E_{\text{k,max}}\)

The relationship between the speed of light \(c\), frequency \(f\), and wavelength \(\lambda\) is:
\(c = f\lambda \Rightarrow f = \frac{c}{\lambda}\)

Substituting this frequency into the photoelectric equation:
\(h\left(\frac{c}{\lambda}\right) = \Phi + E_{\text{k,max}}\)

Rearranging to make \(E_{\text{k,max}}\) the subject:
\(E_{\text{k,max}} = \frac{hc}{\lambda} - \Phi\)

1 mark for deriving the expression correctly by substituting f = c / lambda.

First, calculate the excitation energy required to raise the hydrogen atom from the ground state (\(E_1 = -13.6\text{ eV}\)) to the higher energy levels:

- To \(E_2\): \(-3.40 - (-13.6) = 10.2\text{ eV}\)
- To \(E_3\): \(-1.51 - (-13.6) = 12.09\text{ eV}\)
- To \(E_4\): \(-0.85 - (-13.6) = 12.75\text{ eV}\)

Since the colliding electron has \(12.5\text{ eV}\) of kinetic energy:
- It can excite the atom to \(n = 2\) (requires \(10.2\text{ eV}\))
- It can excite the atom to \(n = 3\) (requires \(12.09\text{ eV}\))
- It cannot excite the atom to \(n = 4\) (requires \(12.75\text{ eV}\))

The maximum excited state reached is \(n = 3\).

From \(n = 3\), the possible de-excitation transitions to lower levels are:
1. Transition from \(3 \rightarrow 1\)
2. Transition from \(3 \rightarrow 2\)
3. Transition from \(2 \rightarrow 1\)

Each of these 3 transitions involves a different energy change, producing 3 different photon frequencies.

1 mark for identifying n = 3 as the highest reachable state and listing the 3 unique transitions.

For part (a):
Phase 1 (acceleration):
\(u = 0\text{ m s}^{-1}\), \(a = 0.45\text{ m s}^{-2}\), \(t_1 = 40\text{ s}\).
Final velocity reached: \(v = u + a t_1 = 0 + 0.45 \times 40 = 18\text{ m s}^{-1}\).
Distance traveled in Phase 1: \(s_1 = \frac{1}{2}(u + v)t_1 = \frac{1}{2} \times 18 \times 40 = 360\text{ m}\).

Phase 2 (constant velocity):
\(v = 18\text{ m s}^{-1}\), \(t_2 = 120\text{ s}\).
Distance traveled in Phase 2: \(s_2 = v \times t_2 = 18 \times 120 = 2160\text{ m}\).

Phase 3 (deceleration to rest):
\(u = 18\text{ m s}^{-1}\), \(v = 0\text{ m s}^{-1}\), \(t_3 = 30\text{ s}\).
Distance traveled in Phase 3: \(s_3 = \frac{1}{2}(u + v)t_3 = \frac{1}{2} \times 18 \times 30 = 270\text{ m}\).

Total distance: \(s_{\text{total}} = s_1 + s_2 + s_3 = 360 + 2160 + 270 = 2790\text{ m}\).

For part (b):
In both phases, the displacement is increasing, so the gradient is always positive. During acceleration, the velocity is increasing, so the gradient of the displacement-time graph increases (curves upwards). During deceleration, the velocity is decreasing, so the gradient decreases (curves downwards), eventually becoming flat (zero gradient) when the train stops.

Part (a) [6 marks total]:
- \(1\) mark for calculating the maximum velocity reached: \(18\text{ m s}^{-1}\).
- \(1\) mark for calculating distance \(s_1 = 360\text{ m}\).
- \(1\) mark for calculating distance \(s_2 = 2160\text{ m}\).
- \(1\) mark for calculating distance \(s_3 = 270\text{ m}\).
- \(1\) mark for summing the three distances: \(s_{\text{total}} = 2790\text{ m}\).
- \(1\) mark for correct unit (\(\text{m}\)) and appropriate significant figures.

Part (b) [3.28 marks total]:
- \(1\) mark for stating that both have positive gradients (displacement always increases).
- \(1\) mark for explaining that the acceleration phase curve gets steeper (gradient increases).
- \(1.28\) marks for explaining that the deceleration phase curve flattens out / gradient decreases to zero.

For part (a):
The period of vertical oscillation for a mass-spring system is given by:
\(T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.15}{24}} = 2\pi \sqrt{0.00625} = 2\pi \times 0.07906 \approx 0.497\text{ s}\), which rounds to \(0.50\text{ s}\) (to 2 significant figures).

For part (b):
The maximum kinetic energy is equal to the total mechanical energy of the oscillator, given by:
\(E_k = \frac{1}{2} k A^2\), where \(A = 5.0\text{ cm} = 0.050\text{ m}\).
\(E_{\text{max}} = 0.5 \times 24 \times (0.050)^2 = 12 \times 0.0025 = 0.030\text{ J}\).

For part (c):
The magnitude of acceleration in SHM is given by:
\(|a| = \omega^2 x\), where \(\omega^2 = \frac{k}{m} = \frac{24}{0.15} = 160\text{ rad}^2\text{ s}^{-2}\).
At \(x = 3.0\text{ cm} = 0.030\text{ m}\):
\(|a| = 160 \times 0.030 = 4.8\text{ m s}^{-2}\).

Part (a) [3 marks total]:
- \(1\) mark for selecting the correct formula: \(T = 2\pi \sqrt{\frac{m}{k}}\).
- \(1\) mark for correct substitution: \(2\pi \sqrt{\frac{0.15}{24}}\).
- \(1\) mark for final correct value with unit: \(0.50\text{ s}\) (or \(0.497\text{ s}\)).

Part (b) [3 marks total]:
- \(1\) mark for formula for maximum energy: \(\frac{1}{2} k A^2\) or \(\frac{1}{2} m \omega^2 A^2\).
- \(1\) mark for converting amplitude to meters: \(0.050\text{ m}\).
- \(1\) mark for correct final value with unit: \(0.030\text{ J}\).

Part (c) [3.28 marks total]:
- \(1\) mark for calculating the angular frequency squared: \(\omega^2 = 160\text{ rad}^2\text{ s}^{-2}\).
- \(1\) mark for substitution into \(|a| = \omega^2 x\) using meters: \(160 \times 0.030\).
- \(1.28\) marks for the correct value of acceleration: \(4.8\text{ m s}^{-2}\).

For part (a):
The normal contact force \(N\) acts perpendicular to the slope. Resolving forces perpendicular to the slope gives:
\(N = m g \cos\theta\)
\(N = 8.0 \times 9.81 \times \cos(25^\circ) = 78.48 \times 0.9063 \approx 71.13\text{ N} \approx 71\text{ N}\) (to 2 s.f.).

For part (b):
The kinetic frictional force \(F_f\) is given by:
\(F_f = \mu N\)
\(F_f = 0.35 \times 71.13 = 24.90\text{ N} \approx 25\text{ N}\) (to 2 s.f.).

For part (c):
Resolving forces parallel to the slope (taking up the slope as positive):
\(F_{\text{net}} = T - F_f - m g \sin\theta\)
where \(T = 75\text{ N}\) and the weight component down the slope is:
\(m g \sin\theta = 8.0 \times 9.81 \times \sin(25^\circ) = 78.48 \times 0.4226 \approx 33.17\text{ N}\).
\(F_{\text{net}} = 75 - 24.90 - 33.17 = 16.93\text{ N}\).
Applying Newton's second law:
\(a = \frac{F_{\text{net}}}{m} = \frac{16.93}{8.0} \approx 2.12\text{ m s}^{-2} \approx 2.1\text{ m s}^{-2}\) (to 2 s.f.).

Part (a) [3 marks total]:
- \(1\) mark for resolving perpendicular to slope: \(N = m g \cos\theta\).
- \(1\) mark for substituting values correctly: \(8.0 \times 9.81 \times \cos(25^\circ)\).
- \(1\) mark for final correct value: \(71\text{ N}\) (accept \(71.1\text{ N}\)).

Part (b) [2 marks total]:
- \(1\) mark for using \(F_f = \mu N\) with their normal force.
- \(1\) mark for correct final value: \(25\text{ N}\) (accept \(24.9\text{ N}\)).

Part (c) [4.28 marks total]:
- \(1\) mark for calculating the parallel component of gravity: \(m g \sin(25^\circ) \approx 33\text{ N}\).
- \(1\) mark for setting up the motion equation: \(F_{\text{net}} = T - F_f - m g \sin\theta\).
- \(1\) mark for calculating the net force: \(16.9\text{ N}\).
- \(1.28\) marks for the final acceleration value with unit: \(2.1\text{ m s}^{-2}\) (allow ECF from previous parts).

For part (a):
The time constant \(\tau\) is given by:
\(\tau = R C = (15.0 \times 10^3\ \Omega) \times (470 \times 10^{-6}\text{ F}) = 7.05\text{ s}\).

For part (b):
The initial charge stored is:
\(Q_0 = C V_0 = (470 \times 10^{-6}\text{ F}) \times 12.0\text{ V} = 5.64 \times 10^{-3}\text{ C}\).
The charge \(Q\) at time \(t\) during discharge is:
\(Q = Q_0 e^{-t/RC} = 5.64 \times 10^{-3} \times e^{-5.0 / 7.05}\).
\(Q = 5.64 \times 10^{-3} \times e^{-0.7092} = 5.64 \times 10^{-3} \times 0.4920 \approx 2.77 \times 10^{-3}\text{ C} \approx 2.8 \times 10^{-3}\text{ C}\).

For part (c):
The current in the circuit at time \(t\) is:
\(I = \frac{V}{R} = \frac{Q}{RC}\) or \(I = I_0 e^{-t/RC}\).
Using \(I = \frac{Q}{\tau}\):
\(I = \frac{2.775 \times 10^{-3}}{7.05} = 3.94 \times 10^{-4}\text{ A} \approx 3.9 \times 10^{-4}\text{ A}\) (or \(0.39\text{ mA}\)).

Part (a) [2 marks total]:
- \(1\) mark for formula: \(\tau = RC\).
- \(1\) mark for correct calculation: \(7.05\text{ s}\) (accept \(7.1\text{ s}\)).

Part (b) [4 marks total]:
- \(1\) mark for initial charge calculation: \(Q_0 = 5.64 \times 10^{-3}\text{ C}\).
- \(1\) mark for selecting exponential decay formula: \(Q = Q_0 e^{-t/RC}\).
- \(1\) mark for correct substitution of values.
- \(1\) mark for final correct value with unit: \(2.8 \times 10^{-3}\text{ C}\) (or \(2.77 \times 10^{-3}\text{ C}\)).

Part (c) [3.28 marks total]:
- \(1\) mark for formula: \(I = I_0 e^{-t/RC}\) or \(I = Q/RC\).
- \(1\) mark for correct calculation of \(I_0 = 8.0 \times 10^{-4}\text{ A}\) (if using exponential formula).
- \(1.28\) marks for correct final value with unit: \(3.9 \times 10^{-4}\text{ A}\) (accept \(0.39\text{ mA}\)).

For part (a):
The tension force \(F\) on the wire is due to gravity on the mass:
\(F = m g = 15.0 \times 9.81 = 147.15\text{ N}\).
The stress \(\sigma\) is:
\(\sigma = \frac{F}{A} = \frac{147.15}{1.20 \times 10^{-6}} = 1.226 \times 10^8\text{ Pa} \approx 1.23 \times 10^8\text{ Pa}\).

For part (b):
The Young modulus is given by:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{\sigma}{\epsilon}\) where strain \(\epsilon = \frac{\Delta L}{L}\).
Therefore, \(\Delta L = \frac{\sigma L}{E}\).
\[\Delta L = \frac{(1.226 \times 10^8) \times 2.50}{2.00 \times 10^{11}} = 1.53 \times 10^{-3}\text{ m}\]
This corresponds to \(1.53\text{ mm}\).

For part (c):
The elastic strain energy \(E_{\text{str}}\) stored in the wire is:
\(E_{\text{str}} = \frac{1}{2} F \Delta L = 0.5 \times 147.15 \times (1.53 \times 10^{-3}) = 0.1126\text{ J} \approx 0.113\text{ J}\).

Part (a) [3 marks total]:
- \(1\) mark for calculating load force: \(F = 147.15\text{ N}\).
- \(1\) mark for formula for stress: \(\sigma = F / A\).
- \(1\) mark for correct final value: \(1.23 \times 10^8\text{ Pa}\) (accept \(\text{N m}^{-2}\)).

Part (b) [3 marks total]:
- \(1\) mark for the relation of Young modulus, stress, and strain.
- \(1\) mark for correct substitution of values.
- \(1\) mark for final value of extension: \(1.53 \times 10^{-3}\text{ m}\) (or \(1.53\text{ mm}\)).

Part (c) [3.28 marks total]:
- \(1\) mark for energy formula: \(E_{\text{str}} = \frac{1}{2} F \Delta L\).
- \(1\) mark for substitution of values.
- \(1.28\) marks for correct final energy value with unit: \(0.113\text{ J}\) (accept \(0.11\text{ J}\)).

For part (a):
Using Snell's Law:
\(n_l \sin\theta_l = n_p \sin\theta_p\)
\(1.38 \sin(32.0^\circ) = 1.56 \sin\theta_p\)
\(1.38 \times 0.5299 = 1.56 \sin\theta_p\)
\(\sin\theta_p = \frac{0.7313}{1.56} \approx 0.4688\)
\(\theta_p = \arcsin(0.4688) \approx 27.96^\circ \approx 28.0^\circ\).

For part (b):
Critical angle \(\theta_c\) occurs for light going from high refractive index (plastic, \(1.56\)) to low refractive index (liquid, \(1.38\)):
\(\sin\theta_c = \frac{n_l}{n_p} = \frac{1.38}{1.56} \approx 0.8846\)
\(\theta_c = \arcsin(0.8846) = 62.2^
\circ\).

For part (c):
Since the ray is traveling from a less optically dense medium (liquid, \(1.38\)) to a more optically dense medium (plastic, \(1.56\)), refraction always occurs. Total internal reflection is physically impossible in this direction, regardless of the angle of incidence.

Part (a) [3 marks total]:
- \(1\) mark for Snell's Law equation.
- \(1\) mark for correct substitution: \(1.38 \sin(32.0^\circ) = 1.56 \sin\theta_p\).
- \(1\) mark for correct angle: \(28.0^\circ\) (accept \(28^\circ\)).

Part (b) [3 marks total]:
- \(1\) mark for using critical angle formula: \(\sin\theta_c = n_2 / n_1\).
- \(1\) mark for substituting correct indices: \(1.38 / 1.56\).
- \(1\) mark for correct critical angle: \(62.2^\circ\).

Part (c) [3.28 marks total]:
- \(1\) mark for stating clearly that TIR will not occur.
- \(1\) mark for stating the physical rule: TIR can only occur when going from a high refractive index to a lower refractive index.
- \(1.28\) marks for applying it to this scenario: liquid has a lower refractive index than plastic (\(1.38 < 1.56\)).

For part (a):
The fraction of remaining activity is:
\(\frac{A(t)}{A_0} = \frac{40}{320} = \frac{1}{8} = \left(\frac{1}{2}\right)^3\).
This corresponds to exactly 3 half-lives.
\(3 T_{1/2} = 24.0\text{ hours} \implies T_{1/2} = 8.0\text{ hours}\).

For part (b):
The decay constant \(\lambda\) is linked to the half-life by:
\(\lambda = \frac{\ln 2}{T_{1/2}}\).
First, convert the half-life to seconds:
\(T_{1/2} = 8.0\text{ hours} = 8.0 \times 3600\text{ s} = 28,800\text{ s}\).
Now calculate \(\lambda\):
\(\lambda = \frac{0.69315}{28800} \approx 2.41 \times 10^{-5}\text{ s}^{-1} \approx 2.4 \times 10^{-5}\text{ s}^{-1}\).

For part (c):
The activity is given by \(A = \lambda N\).
Initially, \(A_0 = \lambda N_0\).
Therefore, \(N_0 = \frac{A_0}{\lambda} = \frac{320}{2.41 \times 10^{-5}} \approx 1.33 \times 10^7\text{ nuclei} \approx 1.3 \times 10^7\text{ nuclei}\).

Part (a) [3 marks total]:
- \(1\) mark for showing that activity has halved 3 times (factor of 1/8).
- \(1\) mark for relation \(3 T_{1/2} = 24.0\text{ hours}\).
- \(1\) mark for correct final value with unit: \(8.0\text{ hours}\).

Part (b) [3 marks total]:
- \(1\) mark for converting half-life into seconds: \(28,800\text{ s}\).
- \(1\) mark for equation \(\lambda = \ln 2 / T_{1/2}\).
- \(1\) mark for correct final value with unit: \(2.4 \times 10^{-5}\text{ s}^{-1}\) (accept \(2.41 \times 10^{-5}\text{ s}^{-1}\)).

Part (c) [3.28 marks total]:
- \(1\) mark for formula linking activity and number of nuclei: \(A_0 = \lambda N_0\).
- \(1\) mark for correct substitution of values.
- \(1.28\) marks for correct final value of nuclei (no unit needed, or "nuclei"): \(1.3 \times 10^7\) (accept \(1.33 \times 10^7\)).

For the first phase, the distance is \(d\) and the average speed is \(v/2\), so the time taken is \(t_1 = \frac{d}{v/2} = \frac{2d}{v}\). For the second phase, the distance is \(2d\) and the speed is constant at \(v\), so the time taken is \(t_2 = \frac{2d}{v}\). For the third phase, the distance is \(3d\) and the average speed is \(v/2\), so the time taken is \(t_3 = \frac{3d}{v/2} = \frac{6d}{v}\). The total distance is \(S = d + 2d + 3d = 6d\), and the total time is \(T = t_1 + t_2 + t_3 = \frac{10d}{v}\). Therefore, the average speed of the car is \(\frac{S}{T} = \frac{6d}{10d/v} = 0.60v\).

1 mark for the correct option B.

Since the total energy \(E\) is the sum of the kinetic energy \(E_k\) and potential energy \(E_p\), we have \(E = E_k + E_p\). Given \(E_k = 3E_p\), it follows that \(E = 4E_p\). Using the expressions for total and potential energy, \(\frac{1}{2} m \omega^2 A^2 = 4 \left(\frac{1}{2} m \omega^2 x^2\right)\), which simplifies to \(x^2 = \frac{A^2}{4}\) or \(x = \pm \frac{A}{2}\). The particle starts at maximum displacement, so its displacement is given by \(x(t) = A \cos(\omega t)\). To find the first time it reaches \(x = A/2\), we solve \(A \cos(\omega t) = \frac{A}{2}\), which gives \(\omega t = \frac{\pi}{3}\). Substituting \(\omega = \frac{2\pi}{T}\), we obtain \(\frac{2\pi}{T} t = \frac{\pi}{3}\), which yields \(t = \frac{T}{6}\).

1 mark for the correct option C.

Because the ray is incident normally on the first face, it enters without refracting and meets the second face at an angle of incidence equal to the prism angle \(\theta\). For total internal reflection to just occur at this face, the angle of incidence must be equal to the critical angle \(\theta_c\). According to Snell's law, \(\sin \theta_c = \frac{n_{\text{air}}}{n_{\text{prism}}} = \frac{1.00}{1.50} = \frac{2}{3}\). Solving for \(\theta_c\) gives \(\theta_c = \sin^{-1}(2/3) \approx 41.8^\circ\), which rounds to \(42^\circ\).

1 mark for the correct option B.

At the highest point of the circular bridge, the forces acting on the car are its weight \(mg\) acting downwards and the normal contact force \(N\) acting upwards. The net force directing the car towards the center of the circular path (downwards) provides the necessary centripetal force: \(mg - N = \frac{mv^2}{R}\). Rearranging this equation for \(N\) gives \(N = mg - \frac{mv^2}{R}\).

1 mark for the correct option C.

The extension \(\Delta L\) of a wire is given by the formula \(\Delta L = \frac{F L}{A E}\). Since both wires are made of the same material (same Young modulus \(E\)) and support identical loads (same force \(F = W\)), the extension is proportional to \(\frac{L}{A}\). Because the cross-sectional area \(A\) is proportional to the square of the diameter \(d^2\), the extension is proportional to \(\frac{L}{d^2}\). For wire X, \(\Delta L_X \propto \frac{L}{d^2}\). For wire Y, \(\Delta L_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = 0.5\frac{L}{d^2}\). Therefore, the ratio of their extensions is \(\frac{\Delta L_X}{\Delta L_Y} = \frac{1}{0.5} = 2.0\).

1 mark for the correct option C.

An increase in light intensity on an LDR causes its resistance to decrease. Because the LDR is connected in series with a fixed resistor, a decrease in the LDR's resistance means it forms a smaller fraction of the total series resistance. Consequently, the potential difference across the LDR (which is measured by the voltmeter) decreases.

1 mark for the correct option A.

The energy levels of hydrogen are: \(E_1 = -13.60\text{ eV}\), \(E_2 = -3.40\text{ eV}\), \(E_3 = -1.51\text{ eV}\), and \(E_4 = -0.85\text{ eV}\). The excitation energies required to raise the atom from the ground state \(n=1\) are: to \(n=2\): \(10.20\text{ eV}\); to \(n=3\): \(12.09\text{ eV}\); and to \(n=4\): \(12.75\text{ eV}\). Because the colliding electron has a kinetic energy of \(12.5\text{ eV}\), it has enough energy to excite the atom to the \(n=3\) state (requiring \(12.09\text{ eV}\)) but not the \(n=4\) state (which requires \(12.75\text{ eV}\)). After exciting the atom to \(n=3\), the remaining kinetic energy of the electron is \(12.5\text{ eV} - 12.09\text{ eV} = 0.41\text{ eV}\).

1 mark for the correct option B.

Using the rules for combining uncertainties, for a relationship of the form \(g = \frac{2h}{t^2}\), the percentage uncertainty in \(g\) is equal to the percentage uncertainty in \(h\) plus twice the percentage uncertainty in \(t\). This gives: \(\%\Delta g = \%\Delta h + 2(\%\Delta t) = 1.5\% + 2(2.0\%) = 5.5\%\).

1 mark for the correct option B.

Let the upward direction be positive.

The displacement \(s\) of the ball when it reaches the ground at the base of the cliff is \(-H\).

Using the kinematic equation:
\[s = u t + \frac{1}{2} a t^2\]
where:
- \(s = -H\)
- \(t = T\)
- \(a = -g\)

Substituting these values:
\[-H = u T - \frac{1}{2} g T^2\]

Multiplying both sides by \(-1\) to solve for \(H\):
\[H = \frac{1}{2} g T^2 - u T\]

1 mark for the correct algebraic derivation of the height of the cliff.

When a free electron collides with a gas atom, it can transfer a portion of its kinetic energy equal to the excitation energy of the atom.

1. **Elastic Collision:** The electron loses negligible kinetic energy, so its remaining kinetic energy is \(12.5\text{ eV}\).
2. **Inelastic Collision (to \(n=2\)):** The electron excites the atom by transferring \(10.2\text{ eV}\). Its remaining kinetic energy is:
\[12.5\text{ eV} - 10.2\text{ eV} = 2.3\text{ eV}\]
3. **Inelastic Collision (to \(n=3\)):** The electron excites the atom by transferring \(12.1\text{ eV}\). Its remaining kinetic energy is:
\[12.5\text{ eV} - 12.1\text{ eV} = 0.4\text{ eV}\]

Note: The electron cannot excite the atom to \(n=4\) (excitation energy \(12.75\text{ eV}\)) because this exceeds the electron's initial kinetic energy of \(12.5\text{ eV}\).

1 mark for identifying all three possible remaining kinetic energies (elastic and both inelastic states).

By the definition of the critical angle for total internal reflection:
\[\sin\theta_c = \frac{n_l}{n_g}\]

The refractive index of a medium is defined by \(n = \frac{c}{v}\). Thus, we can express the refractive indices of the liquid and glass in terms of the speeds of light:
\[n_l = \frac{c}{v_l} \quad \text{and} \quad n_g = \frac{c}{v_g}\]

Substituting these into the critical angle equation:
\[\sin\theta_c = \frac{c/v_l}{c/v_g} = \frac{v_g}{v_l}\]

Rearranging the equation to solve for the speed of light in the liquid, \(v_l\):
\[v_l = \frac{v_g}{\sin\theta_c}\]

1 mark for correctly linking the critical angle to the ratio of refractive indices and translating this to the speeds of light.

The elastic strain energy \(E_s\) stored in a wire under load \(F\) is:
\[E_s = \frac{1}{2} F \Delta L\]

Since the same force \(F\) is applied to both wires, the ratio of the elastic strain energies is:
\[\frac{E_{s, P}}{E_{s, Q}} = \frac{\Delta L_P}{\Delta L_Q}\]

The extension \(\Delta L\) is given by:
\[\Delta L = \frac{F L}{A E} = \frac{4 F L}{\pi d^2 E}\]

Because the wires are made of the same material, the Young modulus \(E\) is identical. Thus:
\[\Delta L \propto \frac{L}{d^2}\]

For wire P:
\[\Delta L_P \propto \frac{L}{d^2}\]

For wire Q:
\[\Delta L_Q \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2} \frac{L}{d^2}\]

Therefore, \(\Delta L_P = 2 \Delta L_Q\), which gives:
\[\frac{E_{s, P}}{E_{s, Q}} = 2\]

1 mark for the correct ratio of 2:1.

At temperature \(T_1\), the potential difference across the thermistor is \(V_{th} = 8.0\text{ V}\).

Since the supply is \(12\text{ V}\), the potential difference across the fixed resistor is:
\[V_R = 12\text{ V} - 8.0\text{ V} = 4.0\text{ V}\]

Using the potential divider ratio:
\[\frac{R_{th}}{R} = \frac{V_{th}}{V_R} = \frac{8.0\text{ V}}{4.0\text{ V}} = 2\]

Since \(R = 3.0\text{ k}\Omega\), the resistance of the thermistor at temperature \(T_1\) is:
\[R_{th} = 2 \times 3.0\text{ k}\Omega = 6.0\text{ k}\Omega\]

At temperature \(T_2\), the resistance of the thermistor is halved:
\[R_{th}' = \frac{6.0\text{ k}\Omega}{2} = 3.0\text{ k}\Omega\]

Since both the thermistor and the fixed resistor now have equal resistance (\(3.0\text{ k}\Omega\)), the supply voltage of \(12\text{ V}\) is divided equally between them.

Thus, the new voltmeter reading is:
\[V_{th}' = \frac{12\text{ V}}{2} = 6.0\text{ V}\]

1 mark for the correct determination of the thermistor's initial and halved resistance, leading to the new voltmeter reading.

According to Einstein's photoelectric equation:
\[E_{photon} = \Phi + E_{k,\max}\]

For the first case (wavelength \(\lambda\)):
\[\frac{hc}{\lambda} = E_{k,\max} + \Phi \quad \text{--- (Equation 1)}\]

For the second case (wavelength \(\frac{\lambda}{2}\)):
\[\frac{hc}{\lambda / 2} = E' + \Phi \implies \frac{2hc}{\lambda} = E' + \Phi \quad \text{--- (Equation 2)}\]

Substitute Equation 1 into Equation 2:
\[2(E_{k,\max} + \Phi) = E' + \Phi\]
\[2E_{k,\max} + 2\Phi = E' + \Phi\]
\[E' = 2E_{k,\max} + \Phi\]

1 mark for utilizing Einstein's photoelectric equation to set up simultaneous equations and obtaining the correct expression.

The formula for the resistivity \(\rho\) of a cylindrical wire is:
\[\rho = \frac{R A}{L} = \frac{\pi R d^2}{4 L}\]

The fractional uncertainty in \(\rho\) is given by:
\[\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\]

Let us calculate the percentage uncertainty for each measured variable:
- For resistance \(R\):
\[\frac{0.2}{4.0} \times 100\% = 5.0\%\]
- For diameter \(d\):
\[\frac{0.02}{0.80} \times 100\% = 2.5\%\]
- For length \(L\):
\[\frac{0.05}{1.25} \times 100\% = 4.0\%\]

Substituting these percentages into the uncertainty combination equation:
\[\text{Percentage uncertainty in } \rho = 5.0\% + 2(2.5\%) + 4.0\% = 14.0\%\]

1 mark for calculating the percentage uncertainties of each quantity and correctly doubling the diameter uncertainty to obtain 14.0%.