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To find the volume \(V\): \(V = \pi r^2 L = \pi \left(\frac{d}{2}\right)^2 L\). Substituting values: \(V = \pi \left(\frac{1.42 \times 10^{-3}\text{ m}}{2}\right)^2 \times (0.345\text{ m}) = 5.465 \times 10^{-7}\text{ m}^3\). To find the absolute uncertainty: Percentage uncertainty in \(d\) is: \(\frac{\Delta d}{d} \times 100\% = \frac{0.02}{1.42} \times 100\% \approx 1.408\%\). Percentage uncertainty in \(L\) is: \(\frac{\Delta L}{L} \times 100\% = \frac{0.1}{34.5} \times 100\% \approx 0.290\%\). Since \(V\) depends on \(d^2 L\), the percentage uncertainty in \(V\) is: \(\%\Delta V = 2 \times \%\Delta d + \%\Delta L = 2(1.408\%) + 0.290\% = 3.107\%\). Absolute uncertainty: \(\Delta V = 5.465 \times 10^{-7} \times 0.03107 = 1.70 \times 10^{-8}\text{ m}^3\).

Part (a): [3 marks] - Correct formula for volume [1 mark] - Correct substitution with SI conversions [1 mark] - Calculated value shown [1 mark]. Part (b): [3.6 marks] - Double the percentage uncertainty in diameter [1 mark] - Calculate percentage uncertainty in length [1 mark] - Add both to find total percentage uncertainty (approx. 3.1%) [1 mark] - Calculate absolute uncertainty (1.7 x 10^-8 m^3) [0.6 marks].

Using Newton's Second Law parallel to the slope (taking up the slope as positive): \(F_{\text{net}} = m a\) implies \(P - F_f - m g \sin\theta = m a\), where \(P = 240\text{ N}\), \(m = 35\text{ kg}\), \(g = 9.81\text{ m s}^{-2}\), \(\theta = 22^\circ\), and \(a = 1.2\text{ m s}^{-2}\). Substituting the values: \(240 - F_f - 35 \times 9.81 \times \sin(22^\circ) = 35 \times 1.2\) yields \(240 - F_f - 128.6 = 42\). Solving for \(F_f\) gives \(F_f = 240 - 128.6 - 42 = 69.4\text{ N}\).

- Resolving the weight component parallel to the slope (mg sin theta) [1.5 marks] - Formulating Newton's second law equation correctly [2 marks] - Correct substitution of values [1.5 marks] - Final correct answer of 69.4 N [1.6 marks]

The tension \(F = 150\text{ N}\) is the same in both sections of the wire. Using the formula for extension: \(\Delta L = \frac{F L}{A E}\). For the steel section: \(\Delta L_s = \frac{150 \times 1.60}{1.20 \times 10^{-6} \times 2.0 \times 10^{11}} = 1.00 \times 10^{-3}\text{ m}\). For the brass section: \(\Delta L_b = \frac{150 \times 1.20}{2.00 \times 10^{-6} \times 1.0 \times 10^{11}} = 0.90 \times 10^{-3}\text{ m}\). Total extension: \(\Delta L_{\text{total}} = \Delta L_s + \Delta L_b = 1.00 \times 10^{-3} + 0.90 \times 10^{-3} = 1.90 \times 10^{-3}\text{ m}\).

- Correct formula for extension used [1 mark] - Calculation of steel extension (1.00 mm) with working [2 marks] - Calculation of brass extension (0.90 mm) with working [2 marks] - Sum of both extensions to give 1.90 x 10^-3 m [1.6 marks]

First, find the initial velocity components: \(u_x = 18.0 \cos(30.0^\circ) = 15.59\text{ m s}^{-1}\) and \(u_y = 18.0 \sin(30.0^\circ) = 9.00\text{ m s}^{-1}\). Taking upwards as positive, the vertical motion equation is: \(y = u_y t - \frac{1}{2} g t^2\) which gives \(-35.0 = 9.00 t - 4.905 t^2\) or \(4.905 t^2 - 9.00 t - 35.0 = 0\). Using the quadratic formula: \(t = \frac{9.00 \pm \sqrt{(-9.00)^2 - 4(4.905)(-35.0)}}{2(4.905)} \approx 3.742\text{ s}\). The horizontal distance is: \(x = u_x t = 15.59 \times 3.742 \approx 58.3\text{ m}\).

- Resolve initial velocity into components [1.5 marks] - Set up the quadratic equation for vertical displacement [2 marks] - Solve for time t = 3.74 s [1.5 marks] - Calculate horizontal distance 58.3 m [1.6 marks]

Energy of the incident photons: \(E = \frac{h c}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{380 \times 10^{-9}} = 5.234 \times 10^{-19}\text{ J}\). In eV: \(E = \frac{5.234 \times 10^{-19}}{1.60 \times 10^{-19}} = 3.271\text{ eV}\). Using Einstein's photoelectric equation: \(E = \Phi + E_{k,\max}\). Since stopping potential \(V_s = 1.10\text{ V}\), the maximum kinetic energy is \(E_{k,\max} = 1.10\text{ eV}\). Thus, the work function is: \(\Phi = 3.271 - 1.10 = 2.171\text{ eV} = 3.474 \times 10^{-19}\text{ J}\). Threshold frequency is: \(f_0 = \frac{\Phi}{h} = \frac{3.474 \times 10^{-19}}{6.63 \times 10^{-34}} = 5.24 \times 10^{14}\text{ Hz}\).

- Calculation of incident photon energy [2 marks] - Relating stopping potential to max kinetic energy [1 mark] - Finding work function [1.6 marks] - Calculating threshold frequency [2 marks]

By conservation of momentum: \(m_A u_A + m_B u_B = (m_A + m_B) v\) yields \(0.80 \times 3.5 + 0 = (0.80 + 1.20) v\), so \(2.80 = 2.00 v\) and \(v = 1.40\text{ m s}^{-1}\). Initial Kinetic Energy: \(E_{k,i} = \frac{1}{2} m_A u_A^2 = 0.5 \times 0.80 \times 3.5^2 = 4.90\text{ J}\). Final Kinetic Energy: \(E_{k,f} = \frac{1}{2} (m_A + m_B) v^2 = 0.5 \times 2.00 \times 1.40^2 = 1.96\text{ J}\). Kinetic Energy Lost: \(\Delta E_k = 4.90 - 1.96 = 2.94\text{ J}\). Percentage of initial kinetic energy lost: \(\frac{2.94}{4.90} \times 100\% = 60\%\).

- Conservation of momentum used to find final velocity (1.4 m/s) [2.5 marks] - Calculation of initial kinetic energy (4.9 J) [1.5 marks] - Calculation of final kinetic energy (1.96 J) [1 mark] - Calculation of percentage loss (60%) [1.6 marks]

Let the weight of the beam be \(W = m g = 25 \times 9.81 = 245.25\text{ N}\). Let \(T_A = 160\text{ N}\) and \(T_B\) be the tension in the rope at B. For vertical equilibrium: \(T_A + T_B = W\) gives \(160 + T_B = 245.25\) so \(T_B = 85.25\text{ N}\). Let \(d\) be the distance from A to the centre of mass. Taking moments about end A: \(W \times d = T_B \times L\) gives \(245.25 \times d = 85.25 \times 4.0\). Thus, \(245.25 d = 341.0\) and \(d = \frac{341.0}{245.25} \approx 1.39\text{ m}\).

- Calculation of the weight of the beam [1 mark] - Determination of tension T_B from vertical equilibrium [2 marks] - Principle of moments applied about A [2 marks] - Calculation of distance d = 1.39 m [1.6 marks]

The kinetic energy \(E_k\) of the accelerated electron is: \(E_k = e V = 1.60 \times 10^{-19} \times 150 = 2.40 \times 10^{-17}\text{ J}\). The momentum \(p\) of the electron is: \(p = \sqrt{2 m_e E_k} = \sqrt{2 \times 9.11 \times 10^{-31} \times 2.40 \times 10^{-17}} \approx 6.613 \times 10^{-24}\text{ kg m s}^{-1}\). The de Broglie wavelength \(\lambda\) is: \(\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{6.613 \times 10^{-24}} \approx 1.00 \times 10^{-10}\text{ m}\).

- Use of E_k = eV to find kinetic energy [1.5 marks] - Use of p = sqrt(2 m E_k) to find momentum [2 marks] - Use of lambda = h/p [1.5 marks] - Correct calculation of lambda = 1.00 x 10^-10 m [1.6 marks]

(a)
Conservation of horizontal momentum:
\( p_{\text{initial}} = p_{\text{final, } x} \)
\( (0.45 \times 8.0) = (0.15 \times 15.0 \cos 30^\circ) + (0.30 \times v_{Bx}) \)
\( 3.60 = 2.25 \cos 30^\circ + 0.30 v_{Bx} \)
\( 3.60 = 1.949 + 0.30 v_{Bx} \)
\( v_{Bx} = \frac{3.60 - 1.949}{0.30} = 5.503 \approx 5.5\text{ m s}^{-1} \)

Conservation of vertical momentum:
\( 0 = m_A v_{Ay} + m_B v_{By} \)
\( 0 = (0.15 \times 15.0 \sin 30^\circ) + (0.30 \times v_{By}) \)
\( 0 = 1.125 + 0.30 v_{By} \)
\( v_{By} = -3.75\text{ m s}^{-1} \) (the minus sign indicates downwards direction)

(b)
Magnitude of velocity of fragment B:
\( v_B = \sqrt{v_{Bx}^2 + v_{By}^2} = \sqrt{5.503^2 + (-3.75)^2} = \sqrt{30.28 + 14.06} = 6.66\text{ m s}^{-1} \approx 6.7\text{ m s}^{-1} \)

Direction of velocity of fragment B:
\( \theta = \tan^{-1}\left(\frac{3.75}{5.503}\right) = 34.3^\circ \approx 34^\circ \) below the horizontal.

(c)
Initial kinetic energy:
\( E_{ki} = \frac{1}{2} M u^2 = \frac{1}{2} \times 0.45 \times 8.0^2 = 14.4\text{ J} \)

Final kinetic energy:
\( E_{kf} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = \frac{1}{2} \times 0.15 \times 15.0^2 + \frac{1}{2} \times 0.30 \times 6.66^2 = 16.88 + 6.65 = 23.53\text{ J} \approx 23.5\text{ J} \)

Since the total kinetic energy increases (by \( 9.1\text{ J} \)) due to the release of chemical energy, the explosion is inelastic (kinetic energy is not conserved).

(a) [3 marks]
- 1 mark: Correct expression for horizontal conservation of momentum with values substituted.
- 1 mark: Calculation showing \( v_{Bx} = 5.5\text{ m s}^{-1} \).
- 1 mark: Correct calculation of vertical velocity component as \( 3.75\text{ m s}^{-1} \) (ignore sign if direction indicated).

(b) [2 marks]
- 1 mark: Correct calculation of magnitude of \( v_B \) (\( 6.7\text{ m s}^{-1} \), accept range \( 6.6 \text{ to } 6.7 \)).
- 1 mark: Correct angle (\( 34^\circ \), accept range \( 34^\circ \text{ to } 34.5^\circ \)) and statement of direction (e.g., "below the horizontal" or "opposite vertical direction to A").

(c) [2 marks]
- 1 mark: Correct calculations of both initial (\( 14.4\text{ J} \)) and final (\( 23.5\text{ J} \)) kinetic energies.
- 1 mark: Correct conclusion that kinetic energy is not conserved, hence the explosion is inelastic.

(a)
Percentage uncertainty in diameter \( d \):
\( \%\Delta d = \frac{0.02}{2.00} \times 100\% = 1.0\% \)

Since volume \( V = \frac{\pi d^2 L}{4} \), the contribution from \( d^2 \) is:
\( 2 \times \%\Delta d = 2 \times 1.0\% = 2.0\% \)

Percentage uncertainty in length \( L \):
\( \%\Delta L = \frac{0.1}{5.0} \times 100\% = 2.0\% \)

Total percentage uncertainty in volume:
\( \%\Delta V = 2.0\% + 2.0\% = 4.0\% \)

(b)
Volume \( V \) of the cylinder:
\( V = \pi \left(\frac{d}{2}\right)^2 L = \pi \times (1.00\text{ cm})^2 \times 5.0\text{ cm} = 15.71\text{ cm}^3 \)

Density \( \rho \):
\( \rho = \frac{m}{V} = \frac{110.0\text{ g}}{15.71\text{ cm}^3} = 7.002\text{ g cm}^{-3} \approx 7.0\text{ g cm}^{-3} \)

(c)
Percentage uncertainty in mass \( m \):
\( \%\Delta m = \frac{0.5}{110.0} \times 100\% = 0.455\% \)

Total percentage uncertainty in density:
\( \%\Delta \rho = \%\Delta m + \%\Delta V = 0.455\% + 4.0\% = 4.455\% \)

Absolute uncertainty in density:
\( \Delta \rho = \rho \times \frac{\%\Delta \rho}{100} = 7.002 \times 0.04455 = 0.312\text{ g cm}^{-3} \)

Rounding the absolute uncertainty to 1 significant figure gives \( 0.3\text{ g cm}^{-3} \).
The density value must be quoted to the same decimal place (1 decimal place):
Density \( = 7.0 \pm 0.3\text{ g cm}^{-3} \)

(a) [3 marks]
- 1 mark: Calculation of percentage uncertainty in diameter as \( 1.0\% \).
- 1 mark: Calculation of percentage uncertainty in length as \( 2.0\% \).
- 1 mark: Summing \( 2.0\% \) (from diameter squared) and \( 2.0\% \) (from length) to get exactly \( 4.0\% \).

(b) [2 marks]
- 1 mark: Correct calculation of the volume \( V = 15.71\text{ cm}^3 \) (or \( 1.57 \times 10^{-5}\text{ m}^3 \)).
- 1 mark: Correct calculation of density \( 7.0\text{ g cm}^{-3} \) (or \( 7.0 \times 10^3\text{ kg m}^{-3} \)).

(c) [2 marks]
- 1 mark: Correct calculation of percentage uncertainty in density \( \approx 4.5\% \) or absolute uncertainty \( 0.31\text{ g cm}^{-3} \).
- 1 mark: Correct final representation \( 7.0 \pm 0.3\text{ g cm}^{-3} \) (both value and uncertainty rounded to 1 decimal place / appropriate significant figures).

To find the percentage uncertainty in the Young modulus \(E\), we sum the percentage uncertainties of the independent quantities, taking into account their powers: \(\%\Delta E = \%\Delta F + \%\Delta L + 2 \times \%\Delta d + \%\Delta e\). Substituting the given percentage uncertainties: \(\%\Delta E = 1.5\% + 0.5\% + 2 \times (2.0\%) + 3.0\% = 9.0\%\).

1 mark for the correct answer (9.0%).

The horizontal component of velocity is constant, \(v_x = u \cos\theta\), while the vertical component of velocity varies linearly with time, \(v_y = u \sin\theta - gt\). The kinetic energy of the ball is given by \(E_k = \frac{1}{2}m(v_x^2 + v_y^2) = \frac{1}{2}m(u^2\cos^2\theta + (u\sin\theta - gt)^2)\). Expanding this expression reveals a quadratic function in terms of time \(t\) with a positive \(t^2\) coefficient. Since \(v_x\) remains non-zero at the highest point of the trajectory, the kinetic energy decreases quadratically to a non-zero minimum, and then increases quadratically as the ball descends.

1 mark for the correct description of the quadratic variation to a non-zero minimum.

According to the impulse-momentum theorem, the change in momentum is equal to the total impulse. Since the body starts and ends at rest, the net change in momentum is zero: \(\Delta p = F t_1 - 2F t_2 = 0\). Rearranging this equation gives: \(F t_1 = 2F t_2\), which simplifies to \(\frac{t_1}{t_2} = 2.0\).

1 mark for the correct calculation of the ratio.

In the decay \(n \rightarrow p + e^- + \bar{\nu}_e\), baryon number is conserved (\(1 \rightarrow 1 + 0 + 0\)), electric charge is conserved (\(0 \rightarrow 1 - 1 + 0\)), and lepton number is conserved (\(0 \rightarrow 0 + 1 - 1 = 0\)), making this decay fully allowed. Option (a) violates lepton number conservation. Option (c) violates lepton number conservation (as both the positron and the electron antineutrino have lepton numbers of \(-1\)). Option (d) violates muon lepton number conservation.

1 mark for identifying the correct decay equation that conserves all quantum numbers.

At terminal velocity \(v_t\), the acceleration of the object is zero, which means the weight is balanced by the resistive force: \(mg = kv_t\). When the speed of the object is \(v = 0.5 v_t\), the net force acting on it is \(F_{\text{net}} = mg - kv = mg - k(0.5 v_t) = mg - 0.5(kv_t)\). Since \(kv_t = mg\), we have \(ma = mg - 0.5 mg = 0.5 mg\). Therefore, the acceleration is \(a = 0.50 g\).

1 mark for the correct derivation of the acceleration in terms of g.

The plank is uniform, so its weight of \(W = mg = 30\text{ kg} \times 9.8\text{ m s}^{-2} = 294\text{ N}\) acts at its midpoint, which is \(2.0\text{ m}\) from the left end. Taking moments about pivot A: \(\sum M_A = 0\), which gives \(W \times 2.0\text{ m} = F_B \times 3.0\text{ m}\). Solving for the force at pivot B: \(294 \times 2.0 = 3.0 F_B\), so \(F_B = 196\text{ N}\).

1 mark for calculating the correct vertical support force using the principle of moments.

The work done in stretching a wire elastically is given by \(W = \frac{1}{2}Fx\), where the extension is \(x = \frac{FL}{AE}\). For the second wire of length \(2L\) and area \(2A\), the extension \(x'\) under the same force \(F\) is \(x' = \frac{F(2L)}{(2A)E} = \frac{FL}{AE} = x\). Since both the force \(F\) and the resulting extension \(x'\) are identical to the first wire, the work done is \(W' = \frac{1}{2}Fx' = \frac{1}{2}Fx = W\).

1 mark for the correct deduction that the work done remains unchanged.

Each alpha decay decreases the nucleon number \(A\) by 4 and the proton number \(Z\) by 2. Each beta-minus decay leaves \(A\) unchanged and increases \(Z\) by 1. For three \(\alpha\) decays and two \(\beta^-\) decays: \(\Delta A = 3 \times (-4) + 2 \times 0 = -12\), so \(A_{\text{final}} = 226 - 12 = 214\). \(\Delta Z = 3 \times (-2) + 2 \times (+1) = -4\), so \(Z_{\text{final}} = 88 - 4 = 84\).

1 mark for the correct final proton and nucleon numbers.

The formula for the density of a cylinder is: \(\rho = \frac{m}{V} = \frac{m}{\pi (d/2)^2 h} = \frac{4m}{\pi d^2 h}\). The percentage uncertainty in each measurement is: \(\%\Delta m = \frac{0.1}{120.5} \times 100\% \approx 0.083\%\), \(\%\Delta h = \frac{0.05}{5.20} \times 100\% \approx 0.962\%\), and \(\%\Delta d = \frac{0.02}{1.42} \times 100\% \approx 1.408\%\). Summing the percentage uncertainties according to the rules for compounding: \(\%\Delta \rho = \%\Delta m + \%\Delta h + 2 \times (\%\Delta d) = 0.083\% + 0.962\% + 2 \times (1.408\%) \approx 3.86\%\). This rounds to \(3.9\%\).

1 mark for the correct option C.

By Newton's second law for the system, the net force causing motion is the weight of the hanging block minus the frictional force acting on the sliding block: \(F_{\text{net}} = mg - \mu Mg\). The total mass is \(M + m\). Therefore, acceleration \(a = \frac{g(m - \mu M)}{M + m} = \frac{9.81 \times (2.0 - 0.25 \times 3.0)}{3.0 + 2.0} = \frac{9.81 \times 1.25}{5.0} = 2.45\text{ m s}^{-2}\).

1 mark for the correct option B.

The extension is given by \(\Delta L = \frac{F L}{A E}\). For the second wire: the length is \(2L\), the cross-sectional area is \(0.25 A\) (since radius is halved), the force is \(2F\), and the Young modulus \(E\) is the same. Thus, \(\Delta L_2 = \frac{(2F)(2L)}{(0.25 A) E} = 16 \left(\frac{FL}{AE}\right) = 16 \Delta L\).

1 mark for the correct option D.

Since kinetic energy \(E_k \propto v^2\), the rebound speed is \(v = \sqrt{0.64} \times 8.0 = 6.4\text{ m s}^{-1}\). Taking the initial direction as positive, \(u = +8.0\text{ m s}^{-1}\) and \(v = -6.4\text{ m s}^{-1}\). The change in momentum is \(\Delta p = m(v - u) = 0.15 \times (-6.4 - 8.0) = -2.16\text{ N s}\). The average force magnitude is \(F = \frac{|\Delta p|}{\Delta t} = \frac{2.16}{0.050} = 43.2\text{ N}\).

1 mark for the correct option C.

Since \(X\) is a baryon with baryon number \(+1\), it consists of three quarks. Strangeness is \(-1\), indicating the presence of exactly one strange (\(s\)) quark (charge \(-\frac{1}{3}e\)). Since the particle is neutral (charge 0), the sum of the charges of the remaining two quarks must be \(+\frac{1}{3}e\). An up (\(u\), charge \(+\frac{2}{3}e\)) and a down (\(d\), charge \(-\frac{1}{3}e\)) quark sum to \(+\frac{1}{3}e\). Thus, the quark composition is \(uds\).

1 mark for the correct option C.

The useful work done is \(W = mgh = 120 \times 9.81 \times 15.0 = 17658\text{ J}\). The useful power output is \(P_{\text{out}} = \frac{W}{t} = \frac{17658}{12.0} = 1471.5\text{ W}\). The power input is \(P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{1471.5}{0.75} = 1962\text{ W} = 1.96\text{ kW}\).

1 mark for the correct option C.

(a) The energy of a photon is given by:
\( E = \frac{hc}{\lambda} \)
Substituting the values:
\( E = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{280 \times 10^{-9}\text{ m}} \)
\( E = 7.104 \times 10^{-19}\text{ J} \)
Converting this energy to electron-volts:
\( E = \frac{7.104 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} \approx 4.44\text{ eV} \)
This is approximately \( 4.4\text{ eV} \).

(b) Using Einstein's photoelectric equation:
\( E_{k,\text{max}} = hf - \Phi \)
\( E_{k,\text{max}} = 4.44\text{ eV} - 2.28\text{ eV} = 2.16\text{ eV} \)
Converting this kinetic energy to joules:
\( E_{k,\text{max}} = 2.16 \times 1.60 \times 10^{-19}\text{ J} = 3.46 \times 10^{-19}\text{ J} \).

(c) Doubling the intensity of the light means there are twice as many photons incident per second. This doubles the rate of electron emission (provided the frequency is above the threshold frequency). However, the energy of each photon remains unchanged, so the maximum kinetic energy of the emitted photoelectrons remains completely unaffected.

(a)
- 1 mark: Correct substitution into the photon energy formula: \( E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{280 \times 10^{-9}} = 7.10 \times 10^{-19}\text{ J} \).
- 1 mark: Correct conversion to eV to show \( E \approx 4.4\text{ eV} \).

(b)
- 1 mark: Calculation of maximum kinetic energy in eV using Einstein's equation: \( E_{k,\text{max}} = 4.44 - 2.28 = 2.16\text{ eV} \).
- 1 mark: Correct calculation of kinetic energy in Joules: \( 3.46 \times 10^{-19}\text{ J} \) (accept \( 3.5 \times 10^{-19}\text{ J} \)).

(c)
- 1 mark: Stating that maximum kinetic energy remains unchanged, with the reason that photon energy is dependent only on frequency/wavelength, not intensity.
- 1 mark: Stating that the rate of electron emission is doubled because the number of incident photons per second has doubled.

(a) Rearranging the resistivity formula \( R = \frac{\rho L}{A} \) to solve for the cross-sectional area \( A \):
\( A = \frac{\rho L}{R} = \frac{1.10 \times 10^{-6}\text{ }\Omega\text{ m} \times 1.50\text{ m}}{8.20\text{ }\Omega} = 2.012 \times 10^{-7}\text{ m}^2 \)
Using the area of a circle in terms of its diameter \( d \):
\( A = \frac{\pi d^2}{4} \Rightarrow d = \sqrt{\frac{4A}{\pi}} \)
\( d = \sqrt{\frac{4 \times 2.012 \times 10^{-7}\text{ m}^2}{\pi}} = 5.06 \times 10^{-4}\text{ m} = 0.506\text{ mm} \).

(b) The resistance of a wire is proportional to its length and inversely proportional to its cross-sectional area:
\( R \propto \frac{L}{A} \propto \frac{L}{d^2} \)
Letting the new values be primed, \( L' = 2L \) and \( d' = 0.5d \):
\( R' = \rho \frac{L'}{A'} = \rho \frac{2L}{\pi (0.5d)^2 / 4} = \rho \frac{2L}{0.25 A} = 8 \left(\rho \frac{L}{A}\right) = 8 R \)
\( R' = 8 \times 8.20\text{ }\Omega = 65.6\text{ }\Omega \).

(a)
- 1 mark: Rearranging formula to find cross-sectional area: \( A = \frac{\rho L}{R} = 2.01 \times 10^{-7}\text{ m}^2 \).
- 1 mark: Equating area to \( \frac{\pi d^2}{4} \) or finding radius \( r = 2.53 \times 10^{-4}\text{ m} \).
- 1 mark: Calculating correct diameter: \( d = 5.06 \times 10^{-4}\text{ m} \) (accept \( 5.1 \times 10^{-4}\text{ m} \) or \( 0.51\text{ mm} \)).

(b)
- 1 mark: Stating that halving the diameter reduces the cross-sectional area by a factor of 4 (i.e. \( A' = 0.25 A \)).
- 1 mark: Demonstrating that resistance increases by a factor of 8 because of the combined effect of twice the length and one-quarter the area (\( 2 / 0.25 = 8 \)).
- 1 mark: Correctly calculating the final resistance: \( R' = 65.6\text{ }\Omega \) (accept \( 66\text{ }\Omega \)).

(a) The distance from the central maximum (zero-order) to the fourth-order bright fringe corresponds to exactly 4 fringe spacings (\( 4w \)):
\( 4w = 32.0\text{ mm} \Rightarrow w = \frac{32.0\text{ mm}}{4} = 8.00\text{ mm} = 8.00 \times 10^{-3}\text{ m} \).

(b) Using the double-slit formula:
\( w = \frac{\lambda D}{d} \Rightarrow d = \frac{\lambda D}{w} \)
Substituting the values:
\( d = \frac{633 \times 10^{-9}\text{ m} \times 2.40\text{ m}}{8.00 \times 10^{-3}\text{ m}} = 1.899 \times 10^{-4}\text{ m} \approx 1.90 \times 10^{-4}\text{ m} \) (or \( 0.190\text{ mm} \)).

(c) Since \( w = \frac{\lambda D}{d} \), the fringe spacing is directly proportional to the wavelength of light used (\( w \propto \lambda \)) when \( D \) and \( d \) are kept constant. Therefore, replacing red light with blue light of a shorter wavelength will cause the fringe spacing \( w \) to decrease.

(a)
- 1 mark: Recognizing that \( 4w = 32.0\text{ mm} \).
- 1 mark: Correctly calculating the fringe spacing: \( w = 8.00\text{ mm} \) (or \( 8.00 \times 10^{-3}\text{ m} \)).

(b)
- 1 mark: Rearranging the double-slit formula to solve for \( d \) and substituting values correctly: \( d = \frac{633 \times 10^{-9} \times 2.40}{8.00 \times 10^{-3}} \).
- 1 mark: Calculating correct slit separation: \( d = 1.90 \times 10^{-4}\text{ m} \) (accept \( 1.9 \times 10^{-4}\text{ m} \) or \( 0.19\text{ mm} \)).

(c)
- 1 mark: Stating that the fringe spacing decreases.
- 1 mark: Explaining that fringe spacing is proportional to wavelength (\( w \propto \lambda \)), and since blue light has a shorter wavelength, the fringes will be closer together.

(a) A correct diagram must show a cell with an internal resistance \( r \) in series, connected in a single loop with a variable resistor. The voltmeter must be connected in parallel across the terminals of this real cell combination.

(b) Using the relationship: \( V = I R \)
First situation:
\( I_1 = \frac{V_1}{R_1} = \frac{1.6\text{ V}}{4.0\text{ }\Omega} = 0.40\text{ A} \)
Using \( \varepsilon = V + I r \):
\( \varepsilon = 1.6 + 0.40 r \) (Equation 1)

Second situation:
\( I_2 = \frac{V_2}{R_2} = \frac{1.8\text{ V}}{9.0\text{ }\Omega} = 0.20\text{ A} \)
Using \( \varepsilon = V + I r \):
\( \varepsilon = 1.8 + 0.20 r \) (Equation 2)

Since emf \( \varepsilon \) is constant, equate the two expressions:
\( 1.6 + 0.40 r = 1.8 + 0.20 r \)
\( 0.20 r = 0.20 \Rightarrow r = 1.0\text{ }\Omega \).

(c) Substitute \( r = 1.0\text{ }\Omega \) back into Equation 1:
\( \varepsilon = 1.6 + 0.40(1.0) = 2.0\text{ V} \).

(a)
- 1 mark: Circuit showing a cell and internal resistance in series connected with a variable resistor.
- 1 mark: Voltmeter connected in parallel across the cell-internal resistance combination (or across the variable resistor).

(b)
- 1 mark: Calculating currents for both states: \( I_1 = 0.40\text{ A} \) and \( I_2 = 0.20\text{ A} \).
- 1 mark: Setting up two equations using the terminal potential difference formula: \( \varepsilon = 1.6 + 0.40r \) and \( \varepsilon = 1.8 + 0.20r \).
- 1 mark: Solving the simultaneous equations correctly to show \( r = 1.0\text{ }\Omega \).

(c)
- 1 mark: Finding the correct emf: \( \varepsilon = 2.0\text{ V} \).

(a) From the de Broglie relationship:
\( p = \frac{h}{\lambda} \)
Substituting Planck's constant and the wavelength:
\( p = \frac{6.63 \times 10^{-34}\text{ J s}}{1.80 \times 10^{-10}\text{ m}} = 3.683 \times 10^{-24}\text{ kg m s}^{-1} \approx 3.68 \times 10^{-24}\text{ kg m s}^{-1} \).

(b) The kinetic energy \( E_k \) gained by the electron is given by the work done by the accelerating field:
\( E_k = eV = \frac{p^2}{2m_e} \)
Rearranging for \( V \):
\( V = \frac{p^2}{2 m_e e} \)
Substituting the values of mass of an electron \( m_e = 9.11 \times 10^{-31}\text{ kg} \) and electronic charge \( e = 1.60 \times 10^{-19}\text{ C} \):
\( V = \frac{(3.683 \times 10^{-24}\text{ kg m s}^{-1})^2}{2 \times (9.11 \times 10^{-31}\text{ kg}) \times (1.60 \times 10^{-19}\text{ C})} \)
\( V = \frac{1.3565 \times 10^{-47}}{2.9152 \times 10^{-49}} = 46.53\text{ V} \approx 46.5\text{ V} \).

(c) The relationship is \( \lambda = \frac{h}{\sqrt{2 m_e e V}} \), which shows that \( \lambda \propto \frac{1}{\sqrt{V}} \). If the accelerating potential difference is quadrupled, the de Broglie wavelength will be halved (reduced by a factor of 2, to \( 0.90 \times 10^{-10}\text{ m} \)).

(a)
- 1 mark: Correctly using \( p = \frac{h}{\lambda} \).
- 1 mark: Calculating correct momentum: \( p = 3.68 \times 10^{-24}\text{ kg m s}^{-1} \) (accept \( 3.7 \times 10^{-24} \)).

(b)
- 1 mark: Using kinetic energy equation linked to accelerating potential: \( E_k = eV \) and \( E_k = \frac{p^2}{2m_e} \).
- 1 mark: Substituting values with correct standard constants \( m_e \) and \( e \).
- 1 mark: Calculating correct voltage: \( V = 46.5\text{ V} \) (accept range \( 46.0\text{ V} \) to \( 47.0\text{ V} \)).

(c)
- 1 mark: Correctly identifying that the wavelength decreases by a factor of 2 (is halved) because \( \lambda \propto \frac{1}{\sqrt{V}} \).

(a) The third harmonic has three loops (loops/envelopes) within the length of the string. The fixed ends are nodes (N). The peak of each loop is an antinode (A). The sketch should show three clear loops with 4 nodes and 3 antinodes labeled.

(b) For the third harmonic on a string fixed at both ends, the length \( L \) contains 1.5 wavelengths:
\( L = \frac{3\lambda}{2} \Rightarrow \lambda = \frac{2L}{3} \)
Substituting the length:
\( \lambda = \frac{2 \times 1.20\text{ m}}{3} = 0.80\text{ m} \).

(c) Using the wave equation:
\( v = f \lambda \Rightarrow f = \frac{v}{\lambda} \)
Substituting the wave speed and wavelength:
\( f = \frac{48.0\text{ m s}^{-1}}{0.80\text{ m}} = 60.0\text{ Hz} \).

(a)
- 1 mark: Drawing a diagram with exactly three symmetrical loops/envelopes between fixed ends.
- 1 mark: Labelling all nodes (N) at the endpoints and crossover points (total 4) and antinodes (A) at maximum displacements (total 3).

(b)
- 1 mark: Stating relation \( L = 1.5\lambda \) or \( \lambda = \frac{2L}{3} \).
- 1 mark: Calculating wavelength correctly: \( \lambda = 0.80\text{ m} \).

(c)
- 1 mark: Using wave formula rearranged for frequency: \( f = \frac{v}{\lambda} \).
- 1 mark: Calculating frequency correctly: \( f = 60.0\text{ Hz} \) (or \( 60\text{ Hz} \)).

(a) Applying Snell's Law at the air-glass boundary (where the refractive index of air \( n_1 \approx 1.00 \)):
\( n_1 \sin \theta_1 = n_2 \sin \theta_2 \)
\( 1.00 \times \sin(40.0^\circ) = 1.52 \times \sin \theta_2 \)
\( \sin \theta_2 = \frac{\sin(40.0^\circ)}{1.52} = 0.4229 \)
\( \theta_2 = \arcsin(0.4229) = 25.0^\circ \).

(b) The speed of light in a medium is related to the refractive index by:
\( n = \frac{c}{v} \Rightarrow v = \frac{c}{n} \)
\( v = \frac{3.00 \times 10^8\text{ m s}^{-1}}{1.52} = 1.97 \times 10^8\text{ m s}^{-1} \).

(c) Total internal reflection occurs when light attempts to travel from the optically denser medium (glass, \( n_1 = 1.52 \)) to the less dense medium (water, \( n_2 = 1.33 \)) at an angle greater than the critical angle \( \theta_c \):
\( \sin \theta_c = \frac{n_{\text{water}}}{n_{\text{glass}}} \)
\( \sin \theta_c = \frac{1.33}{1.52} = 0.875 \)
\( \theta_c = \arcsin(0.875) = 61.0^\circ \).

(a)
- 1 mark: Correct substitution into Snell's law: \( \sin \theta_2 = \frac{\sin(40.0^\circ)}{1.52} \).
- 1 mark: Correctly calculating refraction angle: \( 25.0^\circ \) (accept \( 25^\circ \)).

(b)
- 1 mark: Correct use of relationship \( v = \frac{c}{n} \).
- 1 mark: Calculating speed correctly: \( 1.97 \times 10^8\text{ m s}^{-1} \) (accept \( 2.0 \times 10^8\text{ m s}^{-1} \)).

(c)
- 1 mark: Using critical angle relation: \( \sin \theta_c = \frac{n_2}{n_1} \) with correct assignments of \( n_1 = 1.52 \) and \( n_2 = 1.33 \).
- 1 mark: Calculating correct critical angle: \( 61.0^\circ \) (accept \( 61^\circ \)).

(a) In bright light, the total resistance of the circuit is:
\( R_{\text{total}} = 1200\text{ }\Omega + 300\text{ }\Omega = 1500\text{ }\Omega \)
Using the potential divider equation:
\( V_{\text{LDR}} = V_{\text{total}} \times \frac{R_{\text{LDR}}}{R_{\text{total}}} = 9.0\text{ V} \times \frac{300\text{ }\Omega}{1500\text{ }\Omega} = 1.8\text{ V} \).

(b) When \( V_{\text{LDR}} = 7.2\text{ V} \), the potential difference across the fixed resistor \( R \) is:
\( V_R = V_{\text{total}} - V_{\text{LDR}} = 9.0\text{ V} - 7.2\text{ V} = 1.8\text{ V} \)
The current in the circuit is:
\( I = \frac{V_R}{R} = \frac{1.8\text{ V}}{1200\text{ }\Omega} = 1.50 \times 10^{-3}\text{ A} \)
The resistance of the LDR under dark conditions is:
\( R_{\text{LDR}} = \frac{V_{\text{LDR}}}{I} = \frac{7.2\text{ V}}{1.50 \times 10^{-3}\text{ A}} = 4800\text{ }\Omega \) (or \( 4.8\text{ k}\Omega \)).

(c) As light intensity decreases, the resistance of the LDR increases. Consequently, the LDR takes a larger fraction of the total supply voltage, causing the potential difference across the fixed resistor to decrease.

(a)
- 1 mark: Setting up potential divider ratio: \( 9.0 \times \frac{300}{1500} \) or finding circuit current \( I = 6.0\text{ mA} \).
- 1 mark: Calculating correct voltage: \( 1.8\text{ V} \).

(b)
- 1 mark: Calculating voltage across the fixed resistor: \( V_R = 1.8\text{ V} \) (or setting up equation \( \frac{R_{\text{LDR}}}{1200 + R_{\text{LDR}}} = 0.80 \)).
- 1 mark: Calculating circuit current: \( I = 1.5\text{ mA} \) (or simplifying the ratio equation to \( R_{\text{LDR}} = 4R \)).
- 1 mark: Calculating correct resistance: \( 4800\text{ }\Omega \) (or \( 4.8\text{ k}\Omega \)).

(c)
- 1 mark: Stating that potential difference decreases because the resistance of the LDR increases, dropping a larger proportion of the total voltage.

(a) The energy of a photon is given by:
\(E = \frac{hc}{\lambda}\)
\(E = \frac{6.63 \times 10^{-34} \text{ J s} \times 3.00 \times 10^8 \text{ m s}^{-1}}{250 \times 10^{-9} \text{ m}} = 7.956 \times 10^{-19} \text{ J}\)

To convert this energy to electron-volts (\(\text{eV}\)):
\(E = \frac{7.956 \times 10^{-19} \text{ J}}{1.60 \times 10^{-19} \text{ J eV}^{-1}} \approx 4.97 \text{ eV}\)

(b) Using the photoelectric equation:
\(E_{\text{k,max}} = hf - \Phi = E - \Phi\)
\(E_{\text{k,max}} = 4.9725 \text{ eV} - 4.30 \text{ eV} = 0.6725 \text{ eV}\)

Converting this back to joules:
\(E_{\text{k,max}} = 0.6725 \text{ eV} \times 1.60 \times 10^{-19} \text{ J eV}^{-1} = 1.08 \times 10^{-19} \text{ J}\)

(c)
(i) The maximum kinetic energy of the photoelectrons remains unchanged because the photon energy (which depends solely on the constant wavelength) and the work function are both unchanged.
(ii) The number of photoelectrons emitted per second doubles. Doubling the intensity doubles the rate of incident photons. Since the interaction is 1-to-1, the emission rate of photoelectrons also doubles.

- (a) M1: Substitution of constants into \(E = \frac{hc}{\lambda}\) to obtain \(7.96 \times 10^{-19} \text{ J}\).
- (a) A1: Conversion of energy to obtain \(4.97 \text{ eV}\) (accept \(5.0 \text{ eV}\)).
- (b) M1: Use of \(E_{\text{k,max}} = E - \Phi\) with consistent units (either converting \(\Phi\) to joules or subtracting in eV first).
- (b) A1: Correct conversion to obtain \(1.08 \times 10^{-19} \text{ J}\) (accept range \(1.07 \times 10^{-19} \text{ J}\) to \(1.1 \times 10^{-19} \text{ J}\)).
- (c) B1: States that maximum kinetic energy is unchanged and explains that photon energy/frequency is constant.
- (c) B1: States that rate of emission doubles because double the number of photons are incident per second in a 1-to-1 process.

(a) First, determine the equivalent resistance of the two parallel resistors \(R_2\) and \(R_3\):
\(\frac{1}{R_{\text{p}}} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{12.0\ \Omega} + \frac{1}{12.0\ \Omega} = \frac{2}{12.0\ \Omega}\)
\(R_{\text{p}} = 6.0\ \Omega\)

Now, add the series resistor \(R_1\) to find the total external resistance \(R_{\text{ext}}\):
\(R_{\text{ext}} = R_1 + R_{\text{p}} = 2.0\ \Omega + 6.0\ \Omega = 8.0\ \Omega\) (as shown).

(b) The total resistance of the entire circuit, including the internal resistance \(r\), is:
\(R_{\text{total}} = R_{\text{ext}} + r = 8.0\ \Omega + 1.0\ \Omega = 9.0\ \Omega\)

The total current \(I\) in the circuit is:
\(I = \frac{\varepsilon}{R_{\text{total}}} = \frac{9.0\text{ V}}{9.0\ \Omega} = 1.0\text{ A}\)

The terminal potential difference \(V\) is:
\(V = \varepsilon - I r = 9.0\text{ V} - (1.0\text{ A} \times 1.0\ \Omega) = 8.0\text{ V}\)
(Alternatively: \(V = I R_{\text{ext}} = 1.0\text{ A} \times 8.0\ \Omega = 8.0\text{ V}\))

(c) The potential difference across the parallel network \(V_{\text{p}}\) is:
\(V_{\text{p}} = I R_{\text{p}} = 1.0\text{ A} \times 6.0\ \Omega = 6.0\text{ V}\)

Since \(R_2\) and \(R_3\) are in parallel, the potential difference across \(R_2\) is \(6.0\text{ V}\).
The electrical power dissipated in \(R_2\) is:
\(P = \frac{V_{\text{p}}^2}{R_2} = \frac{(6.0\text{ V})^2}{12.0\ \Omega} = \frac{36.0}{12.0} = 3.0\text{ W}\)
(Alternatively, the current through \(R_2\) is \(I_2 = 0.5\text{ A}\), yielding \(P = I_2^2 R_2 = 0.5^2 \times 12.0 = 3.0\text{ W}\))

- (a) M1: Calculates equivalent resistance of parallel combination to be \(6.0\ \Omega\).
- (a) A1: Adds series resistance to get \(8.0\ \Omega\) showing a clear, logical step.
- (b) M1: Calculates total circuit current as \(1.0\text{ A}\) by using total circuit resistance of \(9.0\ \Omega\).
- (b) A1: Calculates terminal potential difference as \(8.0\text{ V}\) using either formula.
- (c) M1: Determines either the voltage across \(R_2\) (\(6.0\text{ V}\)) or the branch current through \(R_2\) (\(0.5\text{ A}\)).
- (c) A1: Correctly calculates power as \(3.0\text{ W}\) (accept \(3\text{ W}\)).

(a) The grating spacing \(d\) is the distance between consecutive lines:
\(d = \frac{1\text{ mm}}{500} = \frac{1.00 \times 10^{-3} \text{ m}}{500} = 2.00 \times 10^{-6} \text{ m}\)

(b) Use the grating formula:
\(d \sin\theta = n\lambda\)
For the second-order maximum, \(n = 2\).
\(\sin\theta = \frac{2 \lambda}{d} = \frac{2 \times 632.8 \times 10^{-9} \text{ m}}{2.00 \times 10^{-6} \text{ m}} = 0.6328\)
\(\theta = \sin^{-1}(0.6328) \approx 39.3^\circ\) (or \(0.685\text{ rad}\))

(c) The maximum potential diffraction angle is \(\theta = 90^\circ\), which corresponds to \(\sin\theta = 1\).
\(n \le \frac{d}{\lambda} = \frac{2.00 \times 10^{-6} \text{ m}}{632.8 \times 10^{-9} \text{ m}} \approx 3.16\)
Since the order \(n\) must be an integer, the maximum observable order is \(n_{\text{max}} = 3\).
The observed orders are \(n = -3, -2, -1, 0, +1, +2, +3\).
Total number of maxima is \(2 n_{\text{max}} + 1 = 2(3) + 1 = 7\).

- (a) B1: Correct calculation of grating spacing: \(d = 2.00 \times 10^{-6} \text{ m}\).
- (b) M1: Rearrangement of grating formula to \(\sin\theta = \frac{n\lambda}{d}\) and substitution of values with \(n = 2\).
- (b) A1: Correct calculation of diffraction angle: \(39.3^\circ\) (accept range \(39.2^\circ\) to \(39.3^\circ\) or \(0.69\text{ rad}\)).
- (c) M1: Uses boundary condition \(\sin\theta \le 1\) to find limiting ratio \(n \approx 3.16\).
- (c) A1: Identifies maximum integer order of diffraction is \(n = 3\).
- (c) A1: Correctly calculates total number of maxima as \(7\) (taking negative orders and the central maximum into account).

The relationship between terminal potential difference \(V\), electromotive force \(\varepsilon\), current \(I\), and internal resistance \(r\) is given by:

\[V = \varepsilon - Ir\]

Rearranging this into the standard linear equation form \(y = mx + c\):

\[V = -rI + \varepsilon\]

- The dependent variable \(y\) is \(V\).
- The independent variable \(x\) is \(I\).
- The gradient \(m\) is \(-r\) (which is constant and negative).
- The y-intercept \(c\) is \(\varepsilon\) (which is constant and positive).

Thus, a graph of \(V\) against \(I\) is a straight line with a negative gradient and a positive y-intercept.

[1 mark] - B: Correct graph identified based on the linear relationship \(V = -rI + \varepsilon\).

The resistance of a wire is given by the formula:

\[R = \frac{\rho L}{A}\]

where \(\rho\) is the resistivity, \(L\) is the length, and \(A\) is the cross-sectional area. Since the cross-section is circular with diameter \(D\), the area is:

\[A = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4}\]

Substituting this into the resistance formula:

\[R = \frac{4\rho L}{\pi D^2}\]

For wire X (with length \(L\) and diameter \(d\)):

\[R_X = \frac{4\rho L}{\pi d^2}\]

For wire Y (with length \(2L\) and diameter \(2d\)):

\[R_Y = \frac{4\rho (2L)}{\pi (2d)^2} = \frac{8\rho L}{4\pi d^2} = \frac{2\rho L}{\pi d^2}\]

Taking the ratio of \(R_X\) to \(R_Y\):

\[\frac{R_X}{R_Y} = \frac{\frac{4\rho L}{\pi d^2}}{\frac{2\rho L}{\pi d^2}} = \frac{4}{2} = 2\]

Therefore, the ratio is \(2:1\).

[1 mark] - C: Correctly derived ratio of 2:1 by substituting dimensions into the resistivity formula.

We apply Snell's Law at the boundary between the two media:

\[n_1 \sin\theta_1 = n_2 \sin\theta_2\]

Given values:
- \(n_1 = 1.50\)
- \(\theta_1 = 40.0^\circ\)
- \(n_2 = 1.20\)

We rearrange to solve for \(\sin\theta_2\):

\[\sin\theta_2 = \frac{n_1 \sin\theta_1}{n_2}\]
\[\sin\theta_2 = \frac{1.50 \times \sin(40.0^\circ)}{1.20}\]
\[\sin\theta_2 = 1.25 \times 0.6428 = 0.8035\]

Taking the inverse sine:

\[\theta_2 = \arcsin(0.8035) \approx 53.5^\circ\]

[1 mark] - D: Correctly calculated angle of refraction using Snell's Law.

Einstein's photoelectric equation is given by:

\[hf = \Phi + E_k\]

where \(h\) is the Planck constant. Rearranging for the initial photon energy gives:

\[hf = E_k + \Phi\]

When the frequency is doubled, the energy of the incident photon becomes \(2hf\). The new maximum kinetic energy \(E_k'\) is:

\[E_k' = 2hf - \Phi\]

Substituting the expression for \(hf\) into this equation:

\[E_k' = 2(E_k + \Phi) - \Phi = 2E_k + 2\Phi - \Phi = 2E_k + \Phi\]

[1 mark] - B: Correct expression derived by applying Einstein's photoelectric equation to both cases.

For a stretched string fixed at both ends, the boundaries must be nodes. The permitted frequencies (harmonics) are given by integer multiples of the fundamental frequency \(f_1\):

\[f_n = n f_1\]

where \(n\) is the harmonic number. For the third harmonic (\(n = 3\)):

\[f_3 = 3 f_1 = 240\text{ Hz}\]

Solving for the fundamental frequency \(f_1\):

\[f_1 = \frac{240\text{ Hz}}{3} = 80\text{ Hz}\]

[1 mark] - A: Correctly computed fundamental frequency from the third harmonic frequency.

Let's check the conservation laws for option B, which represents beta-minus decay (\(n \rightarrow p + e^- + \bar{\nu}_e\)):
- **Charge (\(Q\))**: \(0 \rightarrow (+1) + (-1) + 0 = 0\) (conserved).
- **Baryon Number (\(B\))**: \(+1 \rightarrow (+1) + 0 + 0 = +1\) (conserved).
- **Lepton Number (\(L\))**: \(0 \rightarrow 0 + (+1) + (-1) = 0\) (conserved).

Therefore, option B is physically possible and satisfies all fundamental conservation laws for a weak decay.

Checking other options:
- **A**: Free proton decay \(p \rightarrow n + e^+ + \nu_e\) is forbidden because a free proton cannot decay into a heavier neutron due to conservation of energy.
- **C**: \(\pi^+ \rightarrow e^+ + \gamma\) violates lepton number: \(0 \rightarrow -1 + 0 = -1\).
- **D**: \(p \rightarrow \pi^+ + \pi^0\) violates baryon number: \(1 \rightarrow 0 + 0 = 0\).

[1 mark] - B: Correct decay identified by validating all relevant quantum number conservation laws.

The potential divider equation is given by:

\[V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{LDR}}}{R_{\text{fixed}} + R_{\text{LDR}}}\]

Given:
- \(V_{\text{in}} = 12.0\text{ V}\]
- \)R_{\text{fixed}} = 3.0\text{ k}\Omega = 3000\text{ }\Omega\]
- \(R_{\text{LDR}} = 1.0\text{ k}\Omega = 1000\text{ }\Omega\]

Substituting these values:

\[V_{\text{out}} = 12.0\text{ V} \times \frac{1.0\text{ k}\Omega}{3.0\text{ k}\Omega + 1.0\text{ k}\Omega}\]
\[V_{\text{out}} = 12.0 \times \frac{1.0}{4.0} = 3.0\text{ V}\]

[1 mark] - A: Correct potential difference calculated using the potential divider formula.

The formula for the fringe separation in a double-slit setup is:

\[w = \frac{\lambda D}{d}\]

Let the new fringe separation be \(w'\). Substituting the new parameters:
- New wavelength \(\lambda' = 2\lambda\)
- New slit spacing \(d' = 0.5d\)

\[w' = \frac{\lambda' D}{d'} = \frac{(2\lambda) D}{0.5d} = 4 \left(\frac{\lambda D}{d}\right) = 4w\]

Therefore, the new fringe separation is \(4w\).

[1 mark] - D: Correctly determined the factor of 4 change in fringe spacing.

In a series circuit, the current \(I\) is constant throughout the wires. The formula for electrical current is given by:

\(I = n A v q\)

where:
- \(n\) is the number density of charge carriers
- \(A\) is the cross-sectional area of the conductor
- \(v\) is the drift speed of the charge carriers
- \(q\) is the charge of a carrier

Since both wires are made of copper, the values of \(n\) and \(q\) are identical. Since they are connected in series:

\(I_X = I_Y\)

\(n A_X v_X q = n A_Y v_Y q \implies A_X v_X = A_Y v_Y\)

The cross-sectional area of a wire of diameter \(d\) is proportional to \(d^2\).
Therefore:

\(A_Y = 4 A_X\)

Substituting this into the ratio:

\(A_X v_X = 4 A_X v_Y \implies v_Y = \frac{v_X}{4} = \frac{v}{4}\)

1 mark: Correctly identifies that current \(I\) is constant in series and uses \(I = nAvq\) to show that drift speed is inversely proportional to cross-sectional area, leading to \(v_Y = \frac{v}{4}\).

At the critical angle \(\theta_c\), refraction occurs at \(90^\circ\). Applying Snell's Law:

\(n_{\text{glass}} \sin(\theta_c) = n_{\text{liquid}} \sin(90^\circ)\)

Since \(\sin(90^\circ) = 1\), this simplifies to:

\(n_{\text{liquid}} = n_{\text{glass}} \sin(\theta_c)\)

Substitute the given values:

\(n_{\text{liquid}} = 1.52 \times \sin(58.0^\circ)\)

\(n_{\text{liquid}} = 1.52 \times 0.84805 = 1.29\)

1 mark: Recalls and applies \(\sin(\theta_c) = \frac{n_2}{n_1}\) with correct substitution to find \(n_{\text{liquid}} = 1.29\).

The formula for the fringe spacing in a double-slit interference pattern is:

\(w = \frac{\lambda D}{s}\)

where \(s\) is the slit separation.

For the first arrangement:
\(w_1 = \frac{\lambda_1 D}{s_1}\)

For the second arrangement:
\(w_2 = \frac{\lambda_2 D}{s_2}\)

We are given that \(s_2 = 2s_1\) and \(w_2 = \frac{2}{3}w_1\). Substituting these relations:

\(\frac{2}{3}w_1 = \frac{\lambda_2 D}{2s_1}\)

Now substitute \(w_1 = \frac{\lambda_1 D}{s_1}\) into this expression:

\(\frac{2}{3} \left(\frac{\lambda_1 D}{s_1}\right) = \frac{\lambda_2 D}{2s_1}\)

Dividing both sides by \(\frac{D}{s_1}\):

\(\frac{2}{3} \lambda_1 = \frac{\lambda_2}{2} \implies \lambda_2 = \frac{4}{3} \lambda_1\)

Substitute \(\lambda_1 = 600\text{ nm}\):

\(\lambda_2 = \frac{4}{3} \times 600\text{ nm} = 800\text{ nm}\)

1 mark: Relates the two fringe spacing equations to find the correct ratio and calculates the new wavelength as \(800\text{ nm}\).

The terminal potential difference \(V\) is related to the emf \(\varepsilon\) and internal resistance \(r\) by:

\(V = \varepsilon - I r\)

Using Ohm's law, \(I = \frac{V}{R}\), so this can be written as:

\(\varepsilon = V + \left(\frac{V}{R}\right)r = V\left(1 + \frac{r}{R}\right)\)

Let's apply this equation to both scenarios:

1) For \(R = 4.0\ \Omega\) and \(V = 6.0\text{ V}\):
\(\varepsilon = 6.0\left(1 + \frac{r}{4.0}\right) = 6.0 + 1.5r\) (Equation 1)

2) For \(R = 8.0\ \Omega\) and \(V = 7.2\text{ V}\):
\(\varepsilon = 7.2\left(1 + \frac{r}{8.0}\right) = 7.2 + 0.9r\) (Equation 2)

Since \(\varepsilon\) is constant, equate Equation 1 and Equation 2:

\(6.0 + 1.5r = 7.2 + 0.9r\)

\(0.6r = 1.2 \implies r = 2.0\ \Omega\)

Now substitute \(r = 2.0\ \Omega\) back into Equation 1 to find \(\varepsilon\):

\(\varepsilon = 6.0 + 1.5(2.0) = 9.0\text{ V\)}

1 mark: Sets up simultaneous equations for the two cases and solves for \(\varepsilon = 9.0\text{ V}\) and \(r = 2.0\ \Omega\).

Einstein's photoelectric equation is given by:

\(hf = \Phi + E_{k,\max}\)

First, calculate the energy of the incident photons in joules:

\(E = hf = (6.63 \times 10^{-34}\text{ J s}) \times (1.20 \times 10^{15}\text{ Hz}) = 7.956 \times 10^{-19}\text{ J}\)

Convert this photon energy into electronvolts (\(eV\)):

\(E = \frac{7.956 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} = 4.9725\text{ eV}\)

Now, find the maximum kinetic energy of the emitted electrons:

\(E_{k,\max} = hf - \Phi = 4.9725\text{ eV} - 3.40\text{ eV} = 1.5725\text{ eV}\)

The stopping potential \(V_s\) is related to \(E_{k,\max}\) by \(e V_s = E_{k,\max}\). Since the energy is in electronvolts, the magnitude of the stopping potential is simply equal to the maximum kinetic energy in electronvolts:

\(V_s = 1.57\text{ V}\)

1 mark: Calculates photon energy in eV, subtracts work function to find maximum kinetic energy, and identifies stopping potential as \(1.57\text{ V}\).

Let's test each reaction against key conservation laws:

**Option A: \(\pi^- + p \rightarrow n +
\pi^0\)**
- **Charge \(Q\)**: \((-1) + (+1) = 0 \rightarrow 0 + 0 = 0\) (Conserved)
- **Baryon number \(B\)**: \(0 + 1 = 1 \rightarrow 1 + 0 = 1\) (Conserved)
- **Lepton number \(L\)**: \(0 + 0 = 0 \rightarrow 0 + 0 = 0\) (Conserved)
This reaction is allowed.

**Option B: \(p + p \rightarrow p + \pi^+\)**
- **Baryon number \(B\)**: \(1 + 1 = 2 \rightarrow 1 + 0 = 1\) (Violated!)
This reaction is prohibited.

**Option C: \(\pi^+ \rightarrow e^+ + \pi^0\)**
- **Lepton number \(L\)**: \(0 \rightarrow (-1) + 0 = -1\) (Violated!)
This reaction is prohibited.

**Option D: \(n \rightarrow p + e^-\)**
- **Lepton number \(L\)**: \(0 \rightarrow 0 + 1 = 1\) (Violated! It requires an electron antineutrino \(\bar{
u}_e\) to conserve lepton number).
This reaction is prohibited.

1 mark: Systematically applies conservation laws to identify that only reaction A is allowed, while others violate baryon number or lepton number conservation.

For part (a):
From Faraday's law of electromagnetic induction, the average induced emf \(E\) is given by:
\(E = N \frac{\Delta \Phi}{\Delta t} = N \frac{B A}{\Delta t}\)
Rearranging for \(B\):
\(B = \frac{E \Delta t}{N A}\)
\(B = \frac{0.12 \times 0.045}{250 \times 3.5 \times 10^{-4}}\).
\(B = \frac{0.0054}{0.0875} = 0.0617 \text{ T} \approx 0.062 \text{ T}\).

For part (b):
When the coil is rotated by \(90^\circ\), the final flux through the coil is zero (as its plane becomes parallel to the magnetic field). The initial flux linkage is \(N B A\) and the final flux linkage is 0. This results in the same change in flux linkage (\(\Delta \Phi N = N B A\)) over the same time interval of \(0.045 \text{ s}\). Therefore, the mean induced emf will remain exactly the same, which is \(0.12 \text{ V}\).

(a)
- Method Mark: Correctly using Faraday's law equation rearranged for \(B\) [1 mark].
- Calculation Mark: Correct substitution of values into the formula [1 mark].
- Accuracy Mark: Correct final answer to 2 significant figures with correct unit (\(0.062 \text{ T}\) or \(6.2 \times 10^{-2} \text{ T}\)) [1.3 marks].

(b)
- Explanation Mark: Explaining that the final flux linkage is zero when rotated by \(90^\circ\) [2 marks].
- Reasoning Mark: Showing that the change in flux linkage is the same as withdrawing the coil [2 marks].
- Conclusion Mark: Concluding that the induced emf is unchanged (\(0.12 \text{ V}\)) [2 marks].

For part (a):
For sphere A in equilibrium, the forces are tension \(T_A\), electrostatic repulsion \(F\), and weight \(m_A g\).
Resolving forces:
\(T_A \sin \theta_A = F\)
\(T_A \cos \theta_A = m_A g\)
Dividing the equations:
\(F = m_A g \tan \theta_A = 1.5 \times 10^{-4} \times 9.81 \times \tan(6.0^\circ)\)
\(F = 1.5 \times 10^{-4} \times 9.81 \times 0.1051 = 1.547 \times 10^{-4} \text{ N} \approx 1.5 \times 10^{-4} \text{ N}\).

For part (b):
By Newton's third law, sphere B experiences the same magnitude of electrostatic force \(F\). Thus, for sphere B:
\(F = m_B g \tan \theta_B\)
\(\tan \theta_B = \frac{F}{m_B g} = \frac{1.547 \times 10^{-4}}{3.0 \times 10^{-4} \times 9.81} = 0.05256\).
\(\theta_B = \arctan(0.05256) \approx 3.01^\circ\).
The horizontal separation \(r\) between the spheres is:
\(r = L \sin \theta_A + L \sin \theta_B = 0.80 \sin(6.0^\circ) + 0.80 \sin(3.01^\circ)\)
\(r = 0.80 \times 0.1045 + 0.80 \times 0.0525 = 0.0836 + 0.0420 = 0.1256 \text{ m} \approx 0.13 \text{ m}\).

For part (c):
Using Coulomb's law:
\(F = \frac{q^2}{4\pi \epsilon_0 r^2}\)
\(q^2 = 4\pi \epsilon_0 r^2 F = \frac{r^2 F}{k}\) where \(k = 8.99 \times 10^9 \text{ N m}^2 \text{ C}^{-2}\).
\(q^2 = \frac{(0.1256)^2 \times 1.547 \times 10^{-4}}{8.99 \times 10^9} = 2.715 \times 10^{-16} \text{ C}^2\)
\(q = \sqrt{2.715 \times 10^{-16}} = 1.65 \times 10^{-8} \text{ C} \approx 1.7 \times 10^{-8} \text{ C}\).

(a)
- Method Mark: Using force resolution to show \(F = m g \tan \theta\) [1 mark].
- Calculation Mark: Substitution of values to get \(1.55 \times 10^{-4} \text{ N}\) [1 mark].

(b)
- Analysis Mark: Realising that sphere B hangs at a different angle and setting up \(\tan \theta_B = \frac{m_A}{m_B} \tan \theta_A\) [1.3 marks].
- Calculation Mark: Correct angle for B (\(3.0^\circ\)) [1 mark].
- Geometry Mark: Using \(r = L \sin \theta_A + L \sin \theta_B\) with correct substitution [1 mark].
- Accuracy Mark: Finding \(0.126 \text{ m}\) or \(0.13 \text{ m}\) [1 mark].

(c)
- Formula Mark: Stating Coulomb's law [1 mark].
- Calculation Mark: Rearranging and substitution of calculated values [1 mark].
- Accuracy Mark: Final answer in range \(1.6 \times 10^{-8} \text{ C}\) to \(1.7 \times 10^{-8} \text{ C}\) [1 mark].

For part (a):
The time constant \(\tau\) is:
\(\tau = R C = (15 \times 10^3 \text{ }\Omega) \times (470 \times 10^{-6} \text{ F}) = 7.05 \text{ s} \approx 7.1 \text{ s}\).

For part (b):
The discharge formula is:
\(V = V_0 e^{-t/\tau}\)
\(3.0 = 12.0 e^{-t/7.05}\)
\(0.25 = e^{-t/7.05}\)
Taking the natural logarithm of both sides:
\(\ln(0.25) = -\frac{t}{7.05}\)
\(t = -7.05 \times \ln(0.25) = 7.05 \times 1.386 = 9.77 \text{ s} \approx 9.8 \text{ s}\).

For part (c):
The energy \(E\) stored in a capacitor is:
\(E = \frac{1}{2} C V^2\)
Initial energy:
\(E_i = 0.5 \times 470 \times 10^{-6} \times 12.0^2 = 0.03384 \text{ J}\).
Final energy:
\(E_f = 0.5 \times 470 \times 10^{-6} \times 3.0^2 = 0.002115 \text{ J}\).
Energy dissipated in the resistor:
\(\Delta E = E_i - E_f = 0.03384 - 0.002115 = 0.031725 \text{ J} \approx 0.032 \text{ J}\).

(a)
- Formula Mark: \(\tau = R C\) [1 mark].
- Accuracy Mark: Correct calculated time constant with unit (\(7.1 \text{ s}\) or \(7.05 \text{ s}\)) [1.3 marks].

(b)
- Formula Mark: Correct exponential decay relation used [1 mark].
- Method Mark: Rearranging equation with logs correctly [1 mark].
- Accuracy Mark: Correct answer of \(9.8 \text{ s}\) (allow 9.7s to 9.8s) [1 mark].

(c)
- Formula Mark: Using \(E = \frac{1}{2} C V^2\) [1 mark].
- Calculation Mark: Correctly calculated initial and final energy values [1 mark].
- Method Mark: Subtracting final energy from initial energy [1 mark].
- Accuracy Mark: Correct energy loss with appropriate precision (\(0.032 \text{ J}\) or \(3.2 \times 10^{-2} \text{ J}\)) [1 mark].

For part (a):
Conservation of energy during acceleration:
\(\frac{1}{2} m v^2 = q V \implies v = \sqrt{\frac{2 q V}{m}}\).
Mass of \(^{12}\text{C}^+\) ion:
\(m = 12 \times 1.66 \times 10^{-27} \text{ kg} = 1.992 \times 10^{-26} \text{ kg}\).
\(v_{12} = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 2500}{1.992 \times 10^{-26}}} = \sqrt{4.016 \times 10^{10}} = 2.004 \times 10^5 \text{ m s}^{-1} \approx 2.0 \times 10^5 \text{ m s}^{-1}\). (Shown)

For part (b):
In a magnetic field, the centripetal force is provided by the magnetic force:
\(B q v = \frac{m v^2}{r} \implies r = \frac{m v}{B q}\).
Substituting the speed formula into the radius formula:
\(r = \frac{1}{B} \sqrt{\frac{2 m V}{q}}\).
For \(^{12}\text{C}^+\):
\(r_{12} = \frac{1}{0.18} \sqrt{\frac{2 \times 1.992 \times 10^{-26} \times 2500}{1.60 \times 10^{-19}}} = 0.1386 \text{ m}\).
For \(^{14}\text{C}^+\):
\(m_{14} = 14 \times 1.66 \times 10^{-27} \text{ kg} = 2.324 \times 10^{-26} \text{ kg}\).
\(r_{14} = \frac{1}{0.18} \sqrt{\frac{2 \times 2.324 \times 10^{-26} \times 2500}{1.60 \times 10^{-19}}} = 0.1497 \text{ m}\).
After completing a semi-circle, the ions travel in opposite directions, and their separation is the difference in diameters of the circular orbits:
\(\Delta D = 2(r_{14} - r_{12}) = 2(0.1497 - 0.1386) = 2 \times 0.0111 = 0.0222 \text{ m} \approx 0.022 \text{ m}\) (or \(2.2 \text{ cm}\)).

(a)
- Formula Mark: Equating kinetic energy to electrical potential energy [1 mark].
- Mass Calculation: Calculation of carbon-12 ion mass [1 mark].
- Speed Calculation: Correctly showing value around \(2.0 \times 10^5 \text{ m s}^{-1}\) [1 mark].

(b)
- Formula Mark: Equating centripetal force to magnetic force to find \(r\) [1 mark].
- Radius Method: Substituting speed or finding individual radii correctly [1 mark].
- Diameter Link: Showing that distance is the difference in diameters (\(2 \Delta r\)) [1.3 marks].
- Calculations: Correct values of both radii or diameters [2 marks].
- Accuracy Mark: Final separation value of \(0.022 \text{ m}\) or \(2.2 \text{ cm}\) (allow \(2.1-2.3 \text{ cm}\)) [1 mark].

For part (a):
Gravitational potential at surface:
\(V_s = -\frac{G M_E}{R_E} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^6} = -6.251 \times 10^7 \text{ J kg}^{-1}\).
At altitude \(h = 1400 \text{ km} = 1.40 \times 10^6 \text{ m}\):
Orbital radius \(r = R_E + h = 6.37 \times 10^6 + 1.40 \times 10^6 = 7.77 \times 10^6 \text{ m}\).
Gravitational potential at orbital radius:
\(V_o = -\frac{G M_E}{r} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{7.77 \times 10^6} = -5.125 \times 10^7 \text{ J kg}^{-1}\).

For part (b):
Work done against gravity to change potential energy:
\(W = \Delta E_p = m(V_o - V_s)\)
\(W = 450 \times (-5.125 \times 10^7 - (-6.251 \times 10^7))\)
\(W = 450 \times 1.126 \times 10^7 = 5.067 \times 10^9 \text{ J} \approx 5.07 \times 10^9 \text{ J}\).

For part (c):
In orbit, circular motion requires:
\(\frac{G M_E m}{r^2} = \frac{m v^2}{r} \implies E_k = \frac{1}{2} m v^2 = \frac{G M_E m}{2 r}\).
This can be written as:
\(E_k = -\frac{1}{2} m V_o = -0.5 \times 450 \times (-5.125 \times 10^7) = 1.153 \times 10^{10} \text{ J} \approx 1.15 \times 10^{10} \text{ J}\).

(a)
- Formula Mark: \(V = -\frac{GM}{r}\) identified [1 mark].
- Calculation Mark: Correct potential at the Earth's surface with negative sign [1 mark].
- Altitude Calculation: Combining radius and altitude correctly to get \(r = 7.77 \times 10^6 \text{ m}\) [1 mark].
- Accuracy Mark: Correct potential at orbit [1.3 marks].

(b)
- Method Mark: Using \(W = m \Delta V\) [1 mark].
- Accuracy Mark: Final work value in range \(5.0 \times 10^9 \text{ J}\) to \(5.1 \times 10^9 \text{ J}\) [1 mark].

(c)
- Formula Mark: Deriving/using \(E_k = \frac{G M m}{2r}\) or equivalent [1 mark].
- Calculation Mark: Substitution of orbital radius and mass [1 mark].
- Accuracy Mark: Final kinetic energy value of \(1.15 \times 10^{10} \text{ J}\) [1 mark].

For part (a):
The area \(A\) of the coil is:
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.064)^2}{4} = 3.217 \times 10^{-3} \text{ m}^2\).
Flux linkage is \(\Phi N = N B(t) A = 120 \times (0.45 - 0.080 t^2) \times 3.217 \times 10^{-3}\).
\(\Phi N = 0.386 \times (0.45 - 0.080 t^2) = 0.1737 - 0.03088 t^2 \approx 0.17 - 0.031 t^2 \text{ Wb-turns}\).

For part (b):
Using Faraday's Law, the magnitude of the induced emf \(E\) is:
\(E = \left| \frac{d(\Phi N)}{dt} \right| = \left| \frac{d}{dt}(0.1737 - 0.03088 t^2) \right| = 2 \times 0.03088 t = 0.06176 t\).
At \(t = 2.5 \text{ s}\):
\(E = 0.06176 \times 2.5 = 0.1544 \text{ V} \approx 0.15 \text{ V}\).

For part (c):
According to Lenz's law, the induced current creates a magnetic field that opposes the change in magnetic flux. Since the external magnetic field is decreasing over time (as shown by the negative coefficient of \(t^2\)), the induced current must flow in a direction that creates a magnetic field in the same direction as the external field to maintain the declining flux.

(a)
- Area calculation: finding \(A = 3.22 \times 10^{-3} \text{ m}^2\) [1 mark].
- Expression formulation: multiplying by \(N = 120\) and the expression for \(B\) [1 mark].
- Accuracy Mark: simplified correct equation (e.g., \(0.17 - 0.031 t^2\)) [1.3 marks].

(b)
- Calculus/Method Mark: differentiating the expression with respect to time [2 marks].
- Accuracy Mark: correct value of \(0.15 \text{ V}\) (allow \(0.15-0.16 \text{ V}\)) [2 marks].

(c)
- State Lenz's Law: reference to opposing the change in flux [1 mark].
- Physics Logic: identifying that field is decreasing and therefore the induced field must reinforce the external field [1 mark].

For part (a):
For static equilibrium, the electric force upwards must balance the downward weight:
\(F_e = W \implies q E = m g\).
Since \(E = \frac{V}{d}\):
\(q \frac{V}{d} = m g \implies V = \frac{m g d}{q}\).
\(V = \frac{4.2 \times 10^{-5} \times 9.81 \times 0.0150}{6.4 \times 10^{-9}} = \frac{6.1803 \times 10^{-6}}{6.4 \times 10^{-9}} = 965.7 \text{ V} \approx 970 \text{ V}\).

For part (b):
The sphere is negatively charged. To experience an upward force, it must be attracted to the upper plate and repelled by the lower plate. Therefore, the upper plate must be positively charged (higher potential) and the lower plate negatively charged.

For part (c):
If potential difference increases by \(20\%\), the electric force increases to:
\(F_{e,\text{new}} = 1.20 \times F_e = 1.20 \times m g\).
The net upward force is:
\(F_{\text{net}} = F_{e,\text{new}} - m g = 0.20 m g\).
The acceleration \(a\) is:
\(a = \frac{F_{\text{net}}}{m} = 0.20 g = 0.20 \times 9.81 = 1.96 \text{ m s}^{-2}\).
The sphere will accelerate vertically upwards at \(1.96 \text{ m s}^{-2}\).

(a)
- Formula Mark: Equating electric force to gravitational force, \(q E = m g\) [1 mark].
- Method Mark: Correct substitution of \(E = V / d\) [1 mark].
- Accuracy Mark: Final potential difference calculated to 2 or 3 sig figs (\(970 \text{ V}\) or \(966 \text{ V}\)) with correct unit [1.3 marks].

(b)
- Identification: Stating that the upper plate must be at the higher potential [1 mark].
- Explanation: Linking the negative charge of the sphere to the direction of attraction/electric field [1 mark].

(c)
- Description Mark: Correctly stating that the sphere will accelerate upwards [1 mark].
- Force calculation: Identifying that net force is \(0.20 \times \text{weight}\) [2 marks].
- Accuracy Mark: Calculating correct acceleration of \(1.96 \text{ m s}^{-2}\) (or \(2.0 \text{ m s}^{-2}\)) [1 mark].

The mass of a uniform sphere of radius \(r\) and density \(\rho\) is given by:
\(M = \rho \times V = \rho \frac{4}{3}\pi r^3\)

Thus, mass is proportional to the cube of the radius: \(M \propto r^3\).
When the radius is doubled from \(R\) to \(2R\), the mass of each sphere increases by a factor of \(2^3 = 8\):
\(M' = 8M\)

When the spheres are in contact, the distance between their centers is equal to the sum of their radii:
Initially, \(d = R + R = 2R\).
With the new spheres, \(d' = 2R + 2R = 4R = 2d\).

Using Newton's law of gravitation, the initial force is:
\(F = G \frac{M^2}{d^2}\)

The new gravitational force \(F'\) is:
\(F' = G \frac{M'^2}{d'^2} = G \frac{(8M)^2}{(2d)^2} = G \frac{64 M^2}{4 d^2} = 16 \left(G \frac{M^2}{d^2}\right) = 16F\)

1 mark for the correct option D.

Incorrect options breakdown:
- A: Assumes the force scales linearly with radius.
- B: Only accounts for the change in distance or mass incorrectly.
- C: Fails to cube the radius to find the change in mass.

The launch platform is at a height of \(2R\) above the surface, so the initial distance of the spacecraft from the center of the planet is:
\(r_i = R + 2R = 3R\)

The initial gravitational potential energy of the spacecraft is:
\(E_i = -\frac{GMm}{r_i} = -\frac{GMm}{3R\)}

To completely escape the gravitational field, the spacecraft must reach infinity (where the potential energy \(E_f = 0\)).
The minimum energy required to escape is the change in gravitational potential energy:
\(\Delta E = E_f - E_i = 0 - \left(-\frac{GMm}{3R}\right) = \frac{GMm}{3R\)}

Since \(g = \frac{GM}{R^2}\) at the surface of the planet, we can substitute \(GM = gR^2\):
\(\Delta E = \frac{(gR^2)m}{3R} = \frac{1}{3}mgR\)

1 mark for the correct option B.

Incorrect options breakdown:
- A: Incorrectly uses the distance as \(9R\).
- C: Uses the height above the surface of \(2R\) directly as the orbital radius from the center, leading to \(\frac{1}{2}mgR\) modified incorrectly.
- D: Incorrectly calculates the required energy using the surface potential.

According to Kepler's third law of planetary motion, the square of the orbital period is directly proportional to the cube of the orbital radius:
\(T^2 \propto r^3\)

Therefore, we can write:
\(\frac{T_B^2}{T_A^2} = \left(\frac{r_B}{r_A}\right)^3\)

Substitute the given values into the equation:
\(\frac{T_B^2}{T^2} = \left(\frac{3R}{R}\right)^3 = 3^3 = 27\)

Taking the square root of both sides:
\(T_B = \sqrt{27} T = 3\sqrt{3} T\)

1 mark for the correct option B.

Incorrect options breakdown:
- A: Direct linear relationship assumed, \(3T\).
- C: Squares the scale factor, \(9T\).
- D: Forgets to take the square root of the ratio, \(27T\).

Let the three charges be located at the vertices \(A, B,\) and \(C\). Let's calculate the net force on the charge at \(C\).

The electrostatic force between any pair of charges is given by Coulomb's law:
\(F = \frac{q^2}{4\pi\varepsilon_0 a^2}\)

The force on \(C\) due to \(A\) is \(F_{AC} = F\), directed along the line \(AC\) away from \(A\).
The force on \(C\) due to \(B\) is \(F_{BC} = F\), directed along the line \(BC\) away from \(B\).

Since the triangle is equilateral, the angle between the two forces \(F_{AC}\) and \(F_{BC}\) is \(60^\circ\).

Using the vector addition formula, the magnitude of the resultant force \(F_{\text{net}}\) is:
\(F_{\text{net}} = \sqrt{F^2 + F^2 + 2F^2 \cos(60^\circ)}\)

Since \(\cos(60^\circ) = 0.5\):
\(F_{\text{net}} = \sqrt{F^2 + F^2 + F^2} = F\sqrt{3}\)

Substituting the expression for \(F\):
\(F_{\text{net}} = \frac{\sqrt{3} q^2}{4\pi\varepsilon_0 a^2}\)

1 mark for the correct option C.

Incorrect options breakdown:
- A: Shows only the magnitude of a single interaction.
- B: Uses an angle of \(90^\circ\) instead of \(60^\circ\) for vector addition.
- D: Simply adds the magnitudes of the forces without accounting for their different directions.

Let the charge \(+Q\) be at position \(x = 0\) and the charge \(-3Q\) be at \(x = d\).
Let \(x\) be the distance from the charge \(+Q\) where the electric potential is zero, such that \(0 < x < d\).

The electric potential \(V_1\) due to charge \(+Q\) is:
\(V_1 = \frac{1}{4\pi\varepsilon_0} \frac{Q}{x}\)

The electric potential \(V_2\) due to charge \(-3Q\) is:
\(V_2 = \frac{1}{4\pi\varepsilon_0} \frac{-3Q}{d-x}\)

Set the total potential to zero:
\(V_1 + V_2 = 0\)
\(\frac{1}{4\pi\varepsilon_0} \frac{Q}{x} - \frac{1}{4\pi\varepsilon_0} \frac{3Q}{d-x} = 0\)

Simplify by dividing out common terms:
\(\frac{1}{x} = \frac{3}{d-x}\)

Cross-multiplying yields:
\(d - x = 3x \implies 4x = d \implies x = \frac{d}{4}\)

1 mark for the correct option C.

Incorrect options breakdown:
- A: Incorrectly assumes the potential is zero midway between them.
- B: Assumes the potential scales inversely with the square of distance.
- D: Incorrect algebraic manipulation resulting in \(\frac{d}{5}\).

The potential difference across a discharging capacitor decreases exponentially with time according to:
\(V = V_0 e^{-t / RC}\)

Substituting the given time \(t = RC \ln 2\):
\(V = V_0 e^{-(RC \ln 2) / RC} = V_0 e^{-\ln 2} = \frac{V_0}{2}\)

The energy stored in a capacitor is proportional to the square of its potential difference:
\(E = \frac{1}{2} C V^2\)

Therefore, the ratio of the energy stored at time \(t\) to the initial energy is:
\(\frac{E}{E_0} = \left(\frac{V}{V_0}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} = 0.25\)

1 mark for the correct option D.

Incorrect options breakdown:
- A: Incorrectly calculated using the complement of the ratio \(1 - 0.25 = 0.75\).
- B: Shows the ratio of the potential difference (or charge) rather than the energy.
- C: Approximates \(1/e = 0.37\), which would occur at the time constant \(t = RC\).

The kinetic energy gained by a charge \(q\) accelerated through potential difference \(V\) is:
\(\frac{1}{2}mv^2 = qV \implies v = \sqrt{\frac{2qV}{m}}\)

When entering the magnetic field, the magnetic force provides the necessary centripetal force:
\(qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB}\)

Substitute the expression for velocity \(v\) into the radius formula:
\(r = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}\)

Since the potential difference \(V\) and magnetic field \(B\) are the same for both particles, the radius of the path is proportional to:
\(r \propto \sqrt{\frac{m}{q}}\)

For the proton: \(r_p \propto \sqrt{\frac{m}{e}}\).
For the alpha particle: \(r_{\alpha} \propto \sqrt{\frac{4m}{2e}} = \sqrt{\frac{2m}{e}}\).

The ratio is:
\(\frac{r_p}{r_{\alpha}} = \frac{\sqrt{m/e}}{\sqrt{2m/e}} = \frac{1}{\sqrt{2}}\)

1 mark for the correct option B.

Incorrect options breakdown:
- A: Compares charges and masses incorrectly without taking the square root.
- C: Calculates the inverse ratio \(\frac{r_{\alpha}}{r_p}\).
- D: Fails to incorporate the square root of the mass-to-charge ratios correctly.

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

Initially, the plane of the coil is perpendicular to the field lines, meaning the normal to the plane of the coil is parallel to the magnetic field. Thus, the initial magnetic flux linkage is:
\(\Phi_1 = N B A \cos(0^\circ) = N B A\)

After rotating through \(90^\circ\), the plane of the coil is parallel to the field lines. Thus, the final magnetic flux linkage is:
\(\Phi_2 = N B A \cos(90^\circ) = 0\)

The change in magnetic flux linkage over the time interval \(\Delta t\) is:
\(\Delta \Phi = |\Phi_2 - \Phi_1| = N B A\)

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

1 mark for the correct option C.

Incorrect options breakdown:
- A: Incorrectly assumes no emf is induced since the field is uniform.
- B: Mistakenly divides by 2, possibly due to confusion with averaging peak and rms values.
- D: Incorrectly doubles the change in flux.

The kinetic energy in a circular orbit of radius \(R\) is given by:
\(E_k = \frac{GMm}{2R}\)

The gravitational potential energy of the satellite in the orbit of radius \(R\) is:
\(E_{p1} = -\frac{GMm}{R} = -2E_k\)

In the new orbit of radius \(2R\), the gravitational potential energy is:
\(E_{p2} = -\frac{GMm}{2R} = -E_k\)

The change in the gravitational potential energy is:
\(\Delta E_p = E_{p2} - E_{p1} = -E_k - (-2E_k) = +E_k\)

Award 1 mark for the correct option C.

- Correct identification of potential energy in terms of kinetic energy (1 mark).

Initially, the charge stored is \(Q = C V\), and the energy stored is:
\(W_0 = \frac{1}{2} C V^2 = \frac{Q^2}{2C}\)

When the capacitor is isolated, the charge \(Q\) remains constant.
Inserting a dielectric with \(\varepsilon_r = 3\) increases the capacitance to \(C' = 3C\).

The new energy stored in the capacitor is:
\(W' = \frac{Q^2}{2C'} = \frac{Q^2}{2(3C)} = \frac{1}{3} \left(\frac{Q^2}{2C}\right) = \frac{1}{3} W_0 = \frac{1}{6} C V^2\)

Award 1 mark for the correct option A.

- Correct application of constant charge and change in capacitance (1 mark).

Since the two charges have opposite signs, the position where the net electric field (and thus the net force on any charge \(Q\)) is zero must lie outside the region between the charges, on the side of the smaller magnitude charge (i.e., to the left of \(+q\)).

Let this position be at a distance \(x_0\) from \(+q\). The electric field at this point is:
\(E = \frac{q}{4\pi\varepsilon_0 x_0^2} - \frac{4q}{4\pi\varepsilon_0 (d + x_0)^2} = 0\)

This simplifies to:
\(\frac{1}{x_0^2} = \frac{4}{(d + x_0)^2}\)

Taking the square root of both sides:
\(\frac{1}{x_0} = \frac{2}{d + x_0} \implies d + x_0 = 2x_0 \implies x_0 = d\)

Therefore, the charge \(Q\) must be placed at a distance \(d\) from \(+q\) on the side opposite to \(-4q\).

Award 1 mark for the correct option C.

- Correct formulation of Coulomb's law or electric field balance equation (1 mark).

The kinetic energy gained by a charge \(q\) accelerated through a potential difference \(V\) is:
\(E_k = qV = \frac{p^2}{2m} \implies p = \sqrt{2mqV}\)

The radius \(r\) of the circular path in a magnetic field is given by:
\(r = \frac{p}{qB} = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B} \sqrt{\frac{2mV}{q}}\)

Thus, \(r \propto \sqrt{\frac{m}{q}}\).

For the proton: \(r_p \propto \sqrt{\frac{m_p}{e}}\)

For the alpha particle: \(r_{\alpha} \propto \sqrt{\frac{4m_p}{2e}} = \sqrt{2} \sqrt{\frac{m_p}{e}}\)

Therefore, the ratio is:
\(\frac{r_p}{r_{\alpha}} = \frac{1}{\sqrt{2}}\)

Award 1 mark for the correct option B.

- Correct derivation of radius in terms of potential difference and charge-to-mass ratio (1 mark).

The initial magnetic flux linkage of the coil is:
\(\Phi_1 = N B A\)

When the coil is rotated by \(180^\circ\), its orientation relative to the field is reversed, so the final magnetic flux linkage is:
\(\Phi_2 = -N B A\)

The magnitude of the change in magnetic flux linkage is:
\(|\Delta \Phi| = |\Phi_2 - \Phi_1| = |-N B A - N B A| = 2 N B A\)

According to Faraday's law, the magnitude of the average induced emf is:
\(E = \frac{|\Delta \Phi|}{\Delta t} = \frac{2 N B A}{\Delta t}\)

Award 1 mark for the correct option C.

- Correct calculation of the change in magnetic flux linkage for a \(180^\circ\) rotation (1 mark).

The equation for the discharge of a capacitor is:
\(Q = Q_0 e^{-\frac{t}{RC}}\)

We are given that at \(t = T\), the charge decreases to half of its initial value, so:
\(\frac{1}{2} Q_0 = Q_0 e^{-\frac{T}{RC}}\)

Dividing both sides by \(Q_0\):
\(\frac{1}{2} = e^{-\frac{T}{RC}}\)

Taking the natural logarithm of both sides:
\(\ln\left(\frac{1}{2}\right) = -\frac{T}{RC} \implies -\ln(2) = -\frac{T}{RC}\)

This simplifies to:
\(T = RC \ln(2)\)

Award 1 mark for the correct option A.

- Correct application of the logarithmic transformation of the decay equation (1 mark).

The gravitational field strength \(g\) at the surface of a planet (at distance \(R\) from the center) is:
\(g = \frac{GM}{R^2}\)

At a height of \(2R\) above the surface, the distance from the center of the planet is:
\(r = R + 2R = 3R\)

The gravitational field strength \(g'\) at this height is:
\(g' = \frac{GM}{r^2} = \frac{GM}{(3R)^2} = \frac{GM}{9R^2} = \frac{g}{9}\)

Award 1 mark for the correct option D.

- Correct identification of the total distance from the center of the planet as \(3R\) and inverse-square relation (1 mark).

(a) In 1 second, the volume of air passing through the swept area \(A\) is \(V = A v\).
Since the swept area is circular, \(A = \pi R^2\), so the volume per second is \(\pi R^2 v\).
The mass of air passing through this area per second is \(m = \rho V = \rho \pi R^2 v\).
The kinetic energy of this mass of air per second is:
\(P = \frac{1}{2} m v^2 = \frac{1}{2} (\rho \pi R^2 v) v^2 = \frac{1}{2} \pi \rho R^2 v^3\).

(b)(i) Using the power formula:
\(P_{\text{available}} = \frac{1}{2} \pi \rho R^2 v^3\)
\(P_{\text{available}} = 0.5 \times \pi \times 1.22 \times 45^2 \times 12^3\)
\(P_{\text{available}} = 0.5 \times 3.14159 \times 1.22 \times 2025 \times 1728 \approx 6.70 \times 10^6\text{ W}\)
Since the efficiency \(\eta = 0.38\):
\(P_{\text{electrical}} = 0.38 \times 6.70 \times 10^6\text{ W} \approx 2.54 \times 10^6\text{ W} = 2.5\text{ MW}\) (to 2 significant figures, matching the efficiency and wind speed data).

(b)(ii) Mass flow rate \(\frac{\Delta m}{\Delta t} = \rho A v\)
\(\frac{\Delta m}{\Delta t} = 1.22 \times \pi \times 45^2 \times 12 \approx 9.31 \times 10^4\text{ kg s}^{-1}\).

(c) Reasons why actual power is lower:
1. Betz's law limits the maximum extractable kinetic energy to about \(59.3\\%\), because air must retain some kinetic energy to leave the turbine area, otherwise the air flow would stall.
2. Frictional losses in the turbine bearings, gearbox, and generator dissipate energy as thermal energy.
3. Aerodynamic drag on the blades and turbulence reduce efficiency.

(a)
- Correct expression for volume of air passing per second: \(V = A v\) or mass per second: \(m = \rho A v\) [1 mark]
- Substitutes area \(A = \pi R^2\) to get mass per second \(m = \rho \pi R^2 v\) [1 mark]
- Equates power to kinetic energy per second: \(P = \frac{1}{2} m v^2\) and completes derivation to show \(P = \frac{1}{2} \pi \rho R^2 v^3\) [1 mark]

(b)(i)
- Correct calculation of available power: \(P_{\text{available}} = 6.70 \times 10^6\text{ W}\) (or \(6.7\text{ MW}\)) [1 mark]
- Multiplies available power by 0.38 to find electrical power: \(2.54 \times 10^6\text{ W}\) [1 mark]
- Final answer given to 2 s.f.: \(2.5\text{ MW}\) (or \(2.5 \times 10^6\text{ W}\)) with appropriate units [1 mark]

(b)(ii)
- Correct formula for mass per second: \(\rho A v\) [1 mark]
- Correct calculation: \(9.3 \times 10^4\text{ kg s}^{-1}\) (accept \(9.31 \times 10^4\)) [1 mark]

(c)
- Mentions that the wind cannot be slowed to a complete stop / Betz limit [1 mark]
- Mentions mechanical friction / heat dissipation in bearings or generator [1 mark]
- Mentions aerodynamic losses / turbulence [1 mark]

(a) The solar constant is the mean solar electromagnetic radiation (power) received per unit area perpendicular to the rays, outside the Earth's atmosphere, at the mean distance of the Earth from the Sun. Its typical value is approximately \(1.35\text{ to }1.40\text{ kW m}^{-2}\).

(b)(i) Daily energy required = \(18\text{ kWh}\).
This can be converted to Joules: \(18 \times 10^3\text{ Wh} \times 3600\text{ s h}^{-1} = 6.48 \times 10^7\text{ J}\).
Alternatively, work in power:
Average daily peak sunlight duration = \(6.0\text{ hours}\).
Average electrical power required during these 6 hours to accumulate 18 kWh:
\(P_{\text{elec}} = \frac{18\text{ kWh}}{6.0\text{ h}} = 3.0\text{ kW} = 3000\text{ W}\).
The electrical power output per unit area of the panel is:
\(P_{\text{unit}} = \text{Insolation} \times \text{efficiency} = 850\text{ W m}^{-2} \times 0.18 = 153\text{ W m}^{-2}\).
The required panel area is:
\(A = \frac{P_{\text{elec}}}{P_{\text{unit}}} = \frac{3000\text{ W}}{153\text{ W m}^{-2}} \approx 19.6\text{ m}^2\) (or \(20\text{ m}^2\) to 2 s.f.).

(b)(ii) The peak electrical power is achieved during peak insolation:
\(P_{\text{peak}} = A \times 850 \times 0.18 = 19.6 \times 153 = 3000\text{ W} = 3.0\text{ kW}\).

(c) The daily savings on the grid electricity bill is:
\(\text{Daily Savings} = 18\text{ kWh} \times \\$0.25/\text{kWh} = \\$4.50\).
Annual savings:
\(\text{Annual Savings} = \\$4.50 \times 365 = \\$1642.50\).
Payback period:
\(\text{Payback Period} = \frac{\\$12,000}{\\$1642.50} \approx 7.31\text{ years}\).

(a)
- Defines solar constant as power per unit area normal to the radiation outside atmosphere [1 mark]
- Correct typical value: \(1.3\text{ to }1.4\text{ kW m}^{-2}\) [1 mark]

(b)(i)
- Converts daily energy to Joules (\(6.48 \times 10^7\text{ J}\)) OR calculates electrical power requirement (3000 W) [1 mark]
- Calculates energy/power delivered per unit area: \(153\text{ W m}^{-2}\) OR \(3.3 \times 10^6\text{ J m}^{-2}\) [1 mark]
- Correct method of dividing total power/energy by power/energy per unit area [1 mark]
- Correct area: \(19.6\text{ m}^2\) (accept \(20\text{ m}^2\)) [1 mark]

(b)(ii)
- Identifies peak power is \(3.0\text{ kW}\) (or \(3000\text{ W}\)) [1 mark]
- Shows calculation using their calculated area from (b)(i) [1 mark]

(c)
- Calculates daily savings (\$4.50) OR annual savings (\$1642.50) [1 mark]
- Divides installation cost by annual savings [1 mark]
- Correct final payback period of \(7.3\text{ years}\) (accept \(7.31\text{ years}\)) [1 mark]

(a)(i) Electrical power output, \(P_{\text{elec}} = 1.20 \times 10^8\text{ W}\).
Since \(\eta = 0.85\), the mechanical power input from the falling water is:
\(P_{\text{input}} = \frac{P_{\text{elec}}}{\eta} = \frac{1.20 \times 10^8}{0.85} \approx 1.412 \times 10^8\text{ W}\).
This power is provided by the rate of loss of gravitational potential energy:
\(P_{\text{input}} = \frac{\Delta E_p}{\Delta t} = \frac{\Delta m}{\Delta t} g h\)
Where \(\frac{\Delta m}{\Delta t}\) is the mass flow rate.
\(\frac{\Delta m}{\Delta t} = \frac{1.412 \times 10^8}{9.81 \times 240} = 6.00 \times 10^4\text{ kg s}^{-1}\).
Since \(\text{volume} = \frac{\text{mass}}{\rho}\), the volume flow rate \(\frac{\Delta V}{\Delta t}\) is:
\(\frac{\Delta V}{\Delta t} = \frac{6.00 \times 10^4\text{ kg s}^{-1}}{1000\text{ kg m}^{-3}} = 60.0\text{ m}^3\text{ s}^{-1}\).

(a)(ii) Total time \(t = 4.0\text{ hours} = 4.0 \times 3600\text{ s} = 1.44 \times 10^4\text{ s}\).
Total mass \(M = \text{mass flow rate} \times t = (6.00 \times 10^4\text{ kg s}^{-1}) \times 1.44 \times 10^4\text{ s} = 8.64 \times 10^8\text{ kg}\).

(b)(i) Pumped-storage systems can respond extremely quickly (within seconds/minutes) to sudden changes in demand, whereas coal/fossil fuel plants take hours to heat up. It also allows excess wind/nuclear/solar energy generated at night to be stored rather than wasted.

(b)(ii) The gravitational potential energy gained by the water when pumped back up:
\(E_p = M g h = 8.64 \times 10^8\text{ kg} \times 9.81\text{ m s}^{-2} \times 240\text{ m} = 2.034 \times 10^{12}\text{ J}\).
Since pumping efficiency is \(78\\%\), the electrical energy required is:
\(E_{\text{input}} = \frac{E_p}{0.78} = \frac{2.034 \times 10^{12}}{0.78} \approx 2.61 \times 10^{12}\text{ J} = 2610\text{ GJ}\).

(a)(i)
- Uses \(P_{\text{input}} = \frac{P_{\text{elec}}}{\eta}\) to find input power of \(1.41 \times 10^8\text{ W}\) [1 mark]
- Equates input power to \(\frac{\Delta m}{\Delta t} g h\) [1 mark]
- Correctly calculates mass flow rate of \(6.0 \times 10^4\text{ kg s}^{-1}\) [1 mark]
- Divides by density to find volume flow rate of \(60\text{ m}^3\text{ s}^{-1}\) (or \(60.0\text{ m}^3\text{ s}^{-1}\)) [1 mark]

(a)(ii)
- Multiplies mass flow rate by time in seconds (14,400 s) [1 mark]
- Correct final mass of \(8.6 \times 10^8\text{ kg}\) (accept \(8.64 \times 10^8\text{ kg}\)) [1 mark]

(b)(i)
- Mentions fast start-up time / rapid response to peak demand [1 mark]
- Mentions storage of surplus energy that would otherwise be wasted during low-demand periods [1 mark]

(b)(ii)
- Calculates potential energy gained: \(2.03 \times 10^{12}\text{ J}\) [1 mark]
- Divides energy by efficiency (0.78) [1 mark]
- Correct final energy in GJ: \(2600\text{ GJ}\) (or \(2.6 \times 10^3\text{ GJ}\); accept \(2610\text{ GJ}\)) [1 mark]

(a) A self-sustaining chain reaction occurs when the neutrons released in one fission reaction go on to cause at least one further fission reaction, maintaining the reaction rate without requiring external neutron sources. Since each fission event here produces 3 neutrons, only one needs to successfully induce another fission (while the others are absorbed or escape) to keep the reaction going.

(b)(i) Total mass of reactants:
\(m_{\text{reactants}} = 235.0439\text{ u} + 1.0087\text{ u} = 236.0526\text{ u}\).
Total mass of products:
\(m_{\text{products}} = 140.9144\text{ u} + 91.9262\text{ u} + 3 \times 1.0087\text{ u} = 235.8667\text{ u}\).
Mass defect:
\(\Delta m = 236.0526\text{ u} - 235.8667\text{ u} = 0.1859\text{ u}\).

(b)(ii) Energy released:
\(E = 0.1859 \times 931.5\text{ MeV} = 173.17\text{ MeV} \approx 173\text{ MeV}\).

(c) Electrical power output \(P_{\text{elec}} = 8.0 \times 10^8\text{ W}\).
Thermal power required:
\(P_{\text{thermal}} = \frac{8.0 \times 10^8}{0.32} = 2.5 \times 10^9\text{ W}\).
Total thermal energy required in 24 hours:
\(E_{\text{thermal}} = 2.5 \times 10^9\text{ W} \times 24 \times 3600\text{ s} = 2.16 \times 10^{14}\text{ J}\).
Energy released per fission event in Joules:
\(E_{\text{fission}} = 173.17\text{ MeV} = 173.17 \times 10^6 \times 1.602 \times 10^{-19}\text{ J} \approx 2.774 \times 10^{-11}\text{ J}\).
Number of fission events required:
\(N = \frac{2.16 \times 10^{14}\text{ J}}{2.774 \times 10^{-11}\text{ J}} \approx 7.787 \times 10^{24}\text{ fissions}\).
Since 1 atom of U-235 is consumed per fission, the number of moles of U-235 consumed is:
\(n = \frac{7.787 \times 10^{24}}{6.022 \times 10^{23}\text{ mol}^{-1}} \approx 12.93\text{ mol}\).
Mass of U-235 consumed:
\(m = n \times 235\text{ g mol}^{-1} \approx 3039\text{ g} \approx 3.04\text{ kg}\) (accept \(3.0\text{ kg}\)).

(a)
- Explains that neutrons from one fission induce subsequent fissions [1 mark]
- Explains "self-sustaining" as maintaining a constant rate of reaction without external inputs [1 mark]
- Mentions that because multiple (3) neutrons are released, there is a high probability that at least one initiates another fission [1 mark]

(b)(i)
- Correct calculation of reactant mass (236.0526 u) and product mass (235.8667 u) [1 mark]
- Correct mass defect: \(0.1859\text{ u}\) [1 mark]

(b)(ii)
- Multiplies mass defect by 931.5 to get \(173\text{ MeV}\) (accept \(173.2\text{ MeV}\)) [2 marks]

(c)
- Calculates total thermal energy required in 24 hours: \(2.16 \times 10^{14}\text{ J}\) [1 mark]
- Converts fission energy to Joules: \(2.77 \times 10^{-11}\text{ J}\) [1 mark]
- Calculates total number of fissions: \(7.8 \times 10^{24}\) [1 mark]
- Correctly finds mass of U-235: \(3.0\text{ kg}\) (accept \(3.04\text{ kg}\)) [1 mark]

(a)(i) The binding energy of a nucleus is the work that must be done / energy required to separate the nucleus completely into its constituent protons and neutrons (nucleons).

(a)(ii) Total binding energy of reactants:
- Deuterium (\({}^2_1\text{H}\)): \(2 \text{ nucleons} \times 1.11\text{ MeV/nucleon} = 2.22\text{ MeV}\)
- Tritium (\({}^3_1\text{H}\)): \(3 \text{ nucleons} \times 2.83\text{ MeV/nucleon} = 8.49\text{ MeV}\)
- Total reactant BE \(= 2.22 + 8.49 = 10.71\text{ MeV}\).
Total binding energy of products:
- Helium-4 (\({}^4_2\text{He}\)): \(4 \text{ nucleons} \times 7.07\text{ MeV/nucleon} = 28.28\text{ MeV}\)
- Neutron: \(0\text{ MeV}\)
- Total product BE \(= 28.28\text{ MeV}\).
Energy released \(Q = \text{Total BE of products} - \text{Total BE of reactants} = 28.28 - 10.71 = 17.57\text{ MeV}\).

(b)(i) Both deuterium and tritium nuclei have a charge of \(q = +1e = 1.60 \times 10^{-19}\text{ C}\).
Electrostatic potential energy:
\(E_p = \frac{q_1 q_2}{4 \pi \varepsilon_0 r}\)
\(E_p = \frac{(1.60 \times 10^{-19})^2}{4 \pi \times 8.85 \times 10^{-12} \times 2.5 \times 10^{-15}}\)
\(E_p = \frac{2.56 \times 10^{-38}}{2.780 \times 10^{-25}} \approx 9.21 \times 10^{-14}\text{ J}\).

(b)(ii) The average kinetic energy of a gas particle at temperature \(T\) is given by:
\(E_k = \frac{3}{2} k T\)
Setting \(E_k = E_p\):
\(\frac{3}{2} k T = 9.21 \times 10^{-14}\text{ J}\)
\(T = \frac{2 \times 9.21 \times 10^{-14}}{3 \times 1.38 \times 10^{-23}} = \frac{1.842 \times 10^{-13}}{4.14 \times 10^{-23}} \approx 4.45 \times 10^9\text{ K}\).
(Accept answers using \(E_k = k T\) yielding \(T \approx 6.7 \times 10^9\text{ K}\) as a simplified estimation).

(a)(i)
- Correct definition: energy required to separate a nucleus into individual/constituent nucleons [1 mark]
- Mention of 'completely' or 'to infinity' or 'individual protons and neutrons' [1 mark]

(a)(ii)
- Calculates total binding energy of reactants: \(10.71\text{ MeV}\) [1 mark]
- Calculates total binding energy of products: \(28.28\text{ MeV}\) [1 mark]
- Correctly finds difference: \(17.57\text{ MeV}\) (accept \(17.6\text{ MeV}\)) [1 mark]

(b)(i)
- Selects correct formula \(E_p = \frac{q_1 q_2}{4\pi\varepsilon_0 r}\) [1 mark]
- Identifies charges as \(1.60 \times 10^{-19}\text{ C}\) each [1 mark]
- Correct calculation of \(E_p = 9.2 \times 10^{-14}\text{ J}\) (accept \(9.21 \times 10^{-14}\)) [1 mark]

(b)(ii)
- Recalls \(E_k = \frac{3}{2} k T\) [1 mark]
- Rearranges and substitutes values correctly [1 mark]
- Correct temperature of \(4.4 \times 10^9\text{ K}\) (or \(4.5 \times 10^9\text{ K}\); accept \(6.7 \times 10^9\text{ K}\) if \(E_k = k T\) is used) [1 mark]

(a)(i) The role of the moderator is to slow down the fast-moving neutrons produced by fission to "thermal" speeds (kinetic energy of about 0.025 eV). This is necessary because slow neutrons have a much higher probability of being captured by Uranium-235 nuclei to cause further fission. The moderator achieves this as neutrons undergo elastic collisions with the moderator's nuclei, transferring kinetic energy to them.

(a)(ii) The moderator nuclei should have a mass close to the mass of a neutron (e.g., hydrogen or carbon) to maximize the energy transfer per collision, and they must have a low neutron absorption cross-section.

(b)(i) Control rods absorb neutrons. When control rods are lowered further into the reactor core, they absorb more neutrons, reducing the number of neutrons available to cause fission, thereby slowing down or stopping the chain reaction. If they are raised, fewer neutrons are absorbed, increasing the rate of fission and the power output.

(b)(ii) Control rods: Boron (or Cadmium/Indium). Coolant: Water (or Carbon Dioxide gas / Liquid Sodium).

(c) Unused fuel contains mostly Uranium-235 and Uranium-238, which have extremely long half-lives and therefore very low activity. Spent fuel rods contain fission fragments (daughter nuclei) and neutron-activation products which are highly unstable, neutron-rich, and have very short half-lives, resulting in extremely high activity. They are initially stored under water in deep cooling ponds on-site to absorb the radiation and dissipate the decay heat.

(a)(i)
- Explains that the moderator slows down fast neutrons to thermal speeds [1 mark]
- Explains that slow/thermal neutrons are more likely to cause fission in U-235 [1 mark]
- Mentions that this occurs through elastic collisions with moderator nuclei [1 mark]

(a)(ii)
- States that the nuclei of the moderator should have a mass similar to that of a neutron [1 mark]

(b)(i)
- States that control rods are made of neutron-absorbing materials [1 mark]
- Explains that lowering them decreases the neutron population / rate of fission [1 mark]
- Explains that raising them increases the rate of fission / power output [1 mark]

(b)(ii)
- Correctly names a control rod material: Boron / Cadmium / Indium [1 mark]
- Correctly names a coolant: Water / Carbon Dioxide / Liquid Sodium [1 mark]

(c)
- Mentions that spent fuel contains fission fragments with short half-lives / high activity [1 mark]
- States they are initially stored in cooling ponds filled with water on-site [1 mark]

The theoretical power available from the wind is given by the formula \(P = \frac{1}{2} \rho A v^3\), where \(\rho\) is the air density, \(A\) is the swept area of the turbine blades, and \(v\) is the wind speed. Thus, power is proportional to the cube of the wind speed: \(P \propto v^3\). If the wind speed increases by 25%, the new speed is \(1.25v\). The new power \(P'\) is proportional to \((1.25v)^3 = 1.953125v^3\). The percentage increase in power is \((1.953125 - 1) \times 100\% \approx 95.3\%\).

1 mark for the correct calculation and selecting option C.

The rate of loss of gravitational potential energy represents the input power: \(P_{\text{in}} = \frac{\Delta m}{\Delta t} g h = (4.2 \times 10^4\text{ kg s}^{-1}) \times 9.81\text{ m s}^{-2} \times 350\text{ m} = 1.442 \times 10^8\text{ W}\). Since the overall efficiency of the power station is \(82\%\), the electrical power output is \(P_{\text{out}} = 0.82 \times 1.442 \times 10^8\text{ W} \approx 1.2 \times 10^8\text{ W}\).

1 mark for the correct power calculation and choosing option B.

The energy required to heat the water is: \(Q = mc\Delta T = 150\text{ kg} \times 4200\text{ J kg}^{-1}\text{ K}^{-1} \times (55 - 15)\text{ K} = 2.52 \times 10^7\text{ J}\). The useful power required over \(2.0\text{ hours}\) (\(7200\text{ s}\)) is: \(P_{\text{useful}} = \frac{Q}{t} = \frac{2.52 \times 10^7\text{ J}}{7200\text{ s}} = 3500\text{ W}\). With an efficiency of \(18\%\), the total incident solar power must be: \(P_{\text{solar}} = \frac{P_{\text{useful}}}{0.18} = \frac{3500\text{ W}}{0.18} \approx 19444\text{ W}\). Given a solar intensity \(I = 1000\text{ W m}^{-2}\), the area \(A\) is: \(A = \frac{P_{\text{solar}}}{I} = \frac{19444\text{ W}}{1000\text{ W m}^{-2}} \approx 19.4\text{ m}^2\), which rounds to \(19\text{ m}^2\).

1 mark for the correct required area of 19 m² (Option C).

The rotational kinetic energy of the flywheel at maximum speed is given by: \(E_k = \frac{1}{2} I \omega^2 = \frac{1}{2} \times 12\text{ kg m}^2 \times (150\text{ rad s}^{-1})^2 = 1.35 \times 10^5\text{ J}\). Since power is the rate of energy transfer and there are no losses, the time \(t\) taken is: \(t = \frac{E_k}{P} = \frac{1.35 \times 10^5\text{ J}}{4500\text{ W}} = 30\text{ s}\).

1 mark for the correct calculated time of 30 s (Option C).

The theoretical maximum power of a wind turbine is given by \(P = \frac{1}{2} \rho A v^3\). The swept area \(A\) is \(\pi R^2\), so \(P \propto R^2 v^3\). For the second turbine: \(P' \propto (1.5R)^2 (0.8v)^3 = 2.25 R^2 \times 0.512 v^3 = 1.152 (R^2 v^3)\). Thus, the power output is approximately \(1.15P\).

1 mark for identifying the correct power relationship and finding 1.15P (Option B).

First, calculate the gravitational potential energy of the stored water: \(E_p = mgh = (2.0 \times 10^6\text{ kg}) \times 9.81\text{ m s}^{-2} \times 180\text{ m} = 3.53 \times 10^9\text{ J} = 3.53\text{ GJ}\). To pump the water, the electrical energy input must be greater than \(E_p\): \(E_{\text{input}} = \frac{E_p}{0.75} = \frac{3.53\text{ GJ}}{0.75} \approx 4.7\text{ GJ}\). Upon release, the generated electrical energy is less than \(E_p\): \(E_{\text{output}} = E_p \times 0.85 = 3.53\text{ GJ} \times 0.85 \approx 3.0\text{ GJ}\).

1 mark for correctly dividing by 0.75 for input and multiplying by 0.85 for output (Option B).

For each branch with 9 cells in series: \(\text{e.m.f.}_{\text{branch}} = 9 \times 0.60\text{ V} = 5.4\text{ V}\) and \(r_{\text{branch}} = 9 \times 0.20\ \Omega = 1.8\ \Omega\). For 4 of these branches connected in parallel, the total e.m.f. is the same as that of a single branch: \(\text{e.m.f.}_{\text{total}} = 5.4\text{ V}\). The total internal resistance is: \(\frac{1}{r_{\text{total}}} = 4 \times \frac{1}{r_{\text{branch}}} \implies r_{\text{total}} = \frac{1.8\ \Omega}{4} = 0.45\ \Omega\).

1 mark for the correct combination of total e.m.f. and total internal resistance (Option C).

First, calculate the angular acceleration \(\alpha\): \(\alpha = \frac{\tau}{I} = \frac{1.2 \times 10^4\text{ N m}}{2.5 \times 10^5\text{ kg m}^2} = 0.048\text{ rad s}^{-2}\). The angular velocity \(\omega\) after \(15\text{ s}\) is: \(\omega = \alpha t = 0.048\text{ rad s}^{-2} \times 15\text{ s} = 0.72\text{ rad s}^{-1}\). The rotational kinetic energy \(E_k\) is: \(E_k = \frac{1}{2} I \omega^2 = \frac{1}{2} \times (2.5 \times 10^5\text{ kg m}^2) \times (0.72\text{ rad s}^{-1})^2 = 6.48 \times 10^4\text{ J} \approx 65\text{ kJ}\).

1 mark for the correct calculation of rotational kinetic energy to be 65 kJ (Option B).

First, calculate the useful electrical power generated by the solar panel array: \(P = I \times A \times \text{efficiency} = 1000\text{ W m}^{-2} \times 4.0\text{ m}^2 \times 0.18 = 720\text{ W}\). Next, calculate the thermal energy required to heat the water: \(Q = m c \Delta T = 150\text{ kg} \times 4200\text{ J kg}^{-1}\text{ K}^{-1} \times 15\text{ K} = 9.45 \times 10^6\text{ J}\). Finally, calculate the time required: \(t = \frac{Q}{P} = \frac{9.45 \times 10^6\text{ J}}{720\text{ W}} = 13125\text{ s}\). Converting to hours gives \(13125 / 3600 = 3.65\text{ hours} \approx 3.6\text{ hours}\).

B is the correct option. 1 mark for the correct answer.

The area swept out by the blades is \(A = \pi L^2\), so power \(P \propto L^2 v^3\). When the blade length is \(1.5L\) and the speed is \(1.2v\), the new power is \(P' \propto (1.5L)^2 (1.2v)^3 = 2.25 L^2 \times 1.728 v^3 = 3.888 L^2 v^3\). Therefore, the new power is \(3.888 P \approx 3.9 P\).

C is the correct option. 1 mark for the correct answer.

The gravitational potential energy stored in the water is \(E_p = m g h = 2.0 \times 10^6\text{ kg} \times 9.81\text{ m s}^{-2} \times 150\text{ m} = 2.943 \times 10^9\text{ J}\). The electrical energy input required to pump this water up is \(E_{\text{in}} = \frac{E_p}{0.80} = 3.679 \times 10^9\text{ J}\). The electrical energy generated during the day is \(E_{\text{out}} = E_p \times 0.85 = 2.502 \times 10^9\text{ J}\). The total wasted electrical energy is \(E_{\text{wasted}} = E_{\text{in}} - E_{\text{out}} = 3.679 \times 10^9\text{ J} - 2.502 \times 10^9\text{ J} = 1.177 \times 10^9\text{ J} \approx 1.2\text{ GJ}\).

B is the correct option. 1 mark for the correct answer.

First, calculate the moment of inertia of the flywheel: \(I = \frac{1}{2} M R^2 = 0.5 \times 40\text{ kg} \times (0.25\text{ m})^2 = 1.25\text{ kg m}^2\). Then, calculate the rotational kinetic energy: \(E_k = \frac{1}{2} I \omega^2 = 0.5 \times 1.25\text{ kg m}^2 \times (1200\text{ rad s}^{-1})^2 = 9.0 \times 10^5\text{ J} = 900\text{ kJ}\

B is the correct option. 1 mark for the correct answer.

The solar intensity \(I\) decreases with distance \(d\) according to the inverse-square law: \(I \propto \frac{1}{d^2}\). Since the area and efficiency of the solar panels are constant, the electrical power generated \(P\) is directly proportional to the intensity, hence \(P \propto \frac{1}{d^2}\). At a distance of \(3.0\text{ AU}\) (which is \(3\) times the initial distance), the power generated will be \(P' = \frac{P}{3^2} = \frac{450\text{ W}}{9} = 50\text{ W}\).

A is the correct option. 1 mark for the correct answer.

The area swept out by the turbine blades is \(A = \pi r^2 = \pi \times 25^2 \approx 1963.5\text{ m}^2\). The total kinetic power in the wind is \(P_{\text{wind}} = \frac{1}{2} \rho A v^3 = 0.5 \times 1.2\text{ kg m}^{-3} \times 1963.5\text{ m}^2 \times (8.0\text{ m s}^{-1})^3 \approx 603186\text{ W}\). The electrical power output is \(P_{\text{elec}} = 0.35 \times 603186\text{ W} = 211115\text{ W} \approx 210\text{ kW}\).

C is the correct option. 1 mark for the correct answer.

The mass flow rate of water is \(\frac{\Delta m}{\Delta t} = \text{volume flow rate} \times \text{density} = 12\text{ m}^3\text{ s}^{-1} \times 1000\text{ kg m}^{-3} = 12000\text{ kg s}^{-1}\). The input power from the falling water is \(P_{\text{in}} = \frac{\Delta m}{\Delta t} g h = 12000\text{ kg s}^{-1} \times 9.81\text{ m s}^{-2} \times 85\text{ m} = 1.00 \times 10^7\text{ W}\). The electrical power output is \(P_{\text{out}} = P_{\text{in}} \times 0.78 = 1.00 \times 10^7\text{ W} \times 0.78 = 7.8 \times 10^6\text{ W} = 7.8\text{ MW}\).

B is the correct option. 1 mark for the correct answer.

(a)
Mean diameter:
\(d_{\text{mean}} = \frac{0.32 + 0.34 + 0.31 + 0.33 + 0.32}{5} = 0.324\text{ mm}\)

Range of values:
\(\text{Range} = 0.34 - 0.31 = 0.03\text{ mm}\)

Absolute uncertainty:
\(\Delta d = \frac{\text{Range}}{2} = \frac{0.03}{2} = 0.015\text{ mm}\)
(Or if using 0.02 mm due to resolution/rounding: \(0.32 \pm 0.02\text{ mm}\))

(b)
Area \(A = \frac{\pi d^2}{4}\)
Percentage uncertainty in \(d\):
\(\%\Delta d = \frac{0.015}{0.324} \times 100\% \approx 4.63\%\) (or \(\frac{0.02}{0.32} \times 100\% = 6.25\%\))

Percentage uncertainty in \(A\):
\(\%\Delta A = 2 \times \%\Delta d = 2 \times 4.63\% = 9.26\%\) (accept \(9.3\%\) or \(12.5\%\))

(c)
1. To account for any variation in the wire's diameter along its length (non-uniformity).
2. To check if the cross-section is circular or oval (non-circularity).

(d)
By definition, \(E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{e / L} = \frac{F L}{A e}\).
The gradient of the graph of \(e\) against \(F\) is \(m = \frac{e}{F}\).
Therefore, \(E = \frac{L}{A m}\).

(e)
1. Close the micrometer jaws completely using the ratchet until it clicks.
2. Note the scale reading; if it is not zero, record this value as the zero error (noting whether it is positive or negative).
3. Subtract this zero error from all subsequent experimental measurements of the diameter.

(a)
- 1 mark for correct mean diameter (0.324 mm or 0.32 mm).
- 1 mark for calculating the range / 2 as 0.015 mm.
- 1 mark for expressing final value with consistent precision (e.g., 0.324 +/- 0.015 mm or 0.32 +/- 0.02 mm).

(b)
- 1 mark for identifying that percentage uncertainty in area is twice that of diameter.
- 1 mark for calculating percentage uncertainty in diameter correctly (4.6% or 6.3%).
- 1 mark for final percentage uncertainty in area (9.3% or 13% depending on rounding/approach).

(c)
- 1 mark for identifying non-uniformity along the length.
- 1 mark for identifying non-circularity / ovality.

(d)
- 1 mark for starting formula \(E = \frac{F L}{A e}\).
- 1 mark for identifying gradient \(m = \frac{e}{F}\).
- 0.3 marks for correct final formula \(E = \frac{L}{A m}\).

(e)
- 1 mark for closing jaws using ratchet without any object.
- 1 mark for reading the non-zero scale division to identify the magnitude and sign of zero error.
- 1 mark for stating that the zero error must be subtracted from all subsequent readings.

(a)
1. Adjust the Y-sensitivity (Y-gain / volts per division) until the sine wave trace covers a significant portion of the screen.
2. Count the vertical grid divisions from the center line to the peak of the wave.
3. Multiply this number of divisions by the Y-sensitivity setting (V/div) to find \(V_0\).

(b)
1. Plot a graph of \(V_0\) (y-axis) against \(f\) (x-axis).
2. The relationship is linear and passes through the origin, so the line of best fit will be straight.
3. The gradient \(m\) is given by \(m = 2\pi N A B_0\).
4. Rearranging this gives: \(B_0 = \frac{m}{2\pi N A}\).

(c)
1. Misalignment reduces the peak emf.
2. The magnetic flux linkage depends on the angle \(\theta\) between the magnetic field lines and the normal to the search coil's face: \(\Phi = B A \cos\theta\).
3. When tilted, \(\theta > 0\) hence \(\cos\theta < 1\), meaning less flux links the search coil, reducing the rate of change of flux linkage and the induced emf.

(d)
1. Use shielded / coaxial cables or tightly twisted wires to connect the search coil to the oscilloscope (prevents induction of noise in lead wires).
2. Move the apparatus away from high-current mains appliances, power supplies, or transformers that generate stray magnetic fields.

(a)
- 1 mark for adjusting Y-gain/sensitivity to ensure a clear trace.
- 1 mark for measuring the peak-to-peak divisions and halving (or measuring peak from central line).
- 1 mark for multiplying the vertical divisions by the volts-per-division setting.

(b)
- 1 mark for specifying a plot of \(V_0\) against \(f\).
- 1 mark for stating that the gradient of this graph represents \(2\pi N A B_0\).
- 1 mark for stating the equation to find \(B_0\) from the gradient.
- 1 mark for noting the graph should be a straight line passing through the origin.

(c)
- 1 mark for stating peak emf decreases.
- 1 mark for referencing \(\Phi = B A \cos\theta\) (or equivalent projection expression).
- 1.3 marks for explaining that the normal component of magnetic flux through the coil is reduced when tilted.

(d)
- 1 mark for choosing shielded/coaxial/twisted leads.
- 1 mark for keeping the setup away from other power supplies/magnetic fields.
- 1 mark for general neatness of electrical setup (e.g. short leads).

(a)
Advantage: A stopwatch is simple, highly portable, and does not require complex alignment. Additionally, light gates can fail if the liquid is highly opaque or if the ball does not fall exactly along the path of the light beam.
Disadvantage: Human reaction time at start and stop introduces a large random uncertainty (typically \(\sim 0.1-0.2\text{ s}\)), reducing accuracy.

(b)
(i)
\(v = \frac{d}{t} = \frac{0.450}{1.84} \approx 0.24457\text{ m s}^{-1}\)
Rounded to 3 sig figs: \(v = 0.245\text{ m s}^{-1}\)

(ii)
Percentage uncertainty in \(d\):
\(\%\Delta d = \frac{0.002}{0.450} \times 100\% \approx 0.44\%\)

Percentage uncertainty in \(t\):
\(\%\Delta t = \frac{0.12}{1.84} \times 100\% \approx 6.52\%\)

Total percentage uncertainty in \(v\):
\(\%\Delta v = \%\Delta d + \%\Delta t = 0.44\% + 6.52\% = 6.96\%\)

Absolute uncertainty in \(v\):
\(\Delta v = 6.96\% \times 0.24457 = 0.0170\text{ m s}^{-1}\)
Therefore, \(v = 0.245 \pm 0.017\text{ m s}^{-1}\) (or \(0.24 \pm 0.02\text{ m s}^{-1}\))

(c)
When the ball is released, it accelerates from rest under gravity while drag increases. It must travel some distance before the drag force plus upthrust balance the weight. If the first marker is too close to the top, the ball will still be accelerating when timing begins, leading to an underestimation of the true terminal velocity.

(d)
As the temperature of the liquid increases, its viscosity decreases. A lower viscosity reduces the drag force on the ball at any given speed. As a result, the ball must reach a higher speed before the drag force matches the net downward force, meaning the terminal velocity increases.

(a)
- 1 mark for a valid advantage (e.g. alignment issues avoided, easier setup).
- 1 mark for a valid disadvantage (e.g. human reaction time error).
- 1 mark for contextualizing why this matters (e.g. reaction time is a significant portion of total travel time).

(b)
- 1 mark for formula \(v = d/t\) and value 0.245 (or 0.24).
- 1 mark for correct calculation of individual percentage uncertainties.
- 1 mark for summing the percentage uncertainties (approx 6.96% or 7.0%).
- 0.3 marks for correct absolute uncertainty (0.017 or 0.02 m/s) with consistent sig figs.

(c)
- 1 mark for stating that the ball bearing starts from rest and accelerates.
- 1 mark for stating that terminal velocity requires forces to balance (weight = drag + upthrust).
- 1 mark for explaining that placing the marker lower ensures constant speed has been reached before measurements begin.

(d)
- 1 mark for identifying that viscosity decreases with increasing temperature.
- 1 mark for stating this reduces drag force at a given speed.
- 1 mark for concluding that terminal velocity increases.

(a)
1. Measure the background counts in the laboratory over a long time period (e.g. 10 minutes) with the radioactive source removed from the room.
2. Calculate the average background count rate (counts per second).
3. Subtract this background count rate from each of the raw count rates measured with the source present to obtain the corrected count rates.

(b)
Take the natural logarithm of both sides of \(R = R_0 e^{-\lambda t}\):
\(\ln(R) = \ln(R_0 e^{-\lambda t})\)
\(\ln(R) = \ln(R_0) - \lambda t\)

Comparing this to the equation of a straight line, \(y = mx + c\):
- Plot \(\ln(R)\) on the vertical axis (y-axis).
- Plot \(t\) on the horizontal axis (x-axis).
- The gradient of the line is \(m = -\lambda\). Therefore, \(\lambda = -\text{gradient}\).

(c)
\(\lambda = -\text{gradient} = 0.0125\text{ s}^{-1}\).
Using the half-life equation:
\(T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.6931}{0.0125} = 55.45\text{ s}\).
Rounding to 3 sig figs gives \(55.4\text{ s}\) (or \(55\text{ s}\) to 2 sig figs).

(d)
Alpha particles have low penetrating power and are absorbed by very thin barriers, including standard glass/metal tube walls. A thin mica window allows alpha particles to enter the tube.
Precaution: Handle the source using long tongs to maximize distance from the body (reducing dose rate due to inverse square law / range limits), or keep the source in a lead-lined container when not in use.

(a)
- 1 mark for stating that background radiation must be measured with the source absent / far away.
- 1 mark for measuring over a long time to reduce statistical uncertainty.
- 1 mark for subtracting background count rate from raw count rate.

(b)
- 1 mark for applying natural logs correctly to show linear form: \(\ln R = -\lambda t + \ln R_0\).
- 1 mark for plotting \(\ln R\) on the vertical axis.
- 1 mark for plotting \(t\) on the horizontal axis.
- 1 mark for showing \(\lambda = -\text{gradient}\).

(c)
- 1 mark for identifying \(\lambda = 0.0125\text{ s}^{-1}\).
- 1 mark for using \(T_{1/2} = \ln 2 / \lambda\).
- 1.3 marks for correct value (55.4 s or 55 s) with correct units (seconds / s).

(d)
- 1 mark for explaining that alpha particles are highly ionizing/weakly penetrating and would otherwise be absorbed by the container walls.
- 1 mark for stating a correct precaution (use tongs / store in lead box / point away from eyes/body).
- 1 mark for explaining the physical basis of the precaution (e.g. keeping distance is highly effective as alpha has short range in air).

(a)
Using the coordinates \((0.200, 3.1)\) and \((0.600, 9.4)\):
\(\text{gradient } m = \frac{9.4 - 3.1}{0.600 - 0.200} = \frac{6.3}{0.400} = 15.75\ \Omega\text{ m}^{-1}\)
(Accept values in the range \(15.5 - 15.8\ \Omega\text{ m}^{-1}\) depending on the choice of points or best-fit line approximation).

(b)
(i)
Cross-sectional area:
\(A = \frac{\pi d^2}{4} = \frac{\pi (0.24 \times 10^{-3}\text{ m})^2}{4} = 4.524 \times 10^{-8}\text{ m}^2\)

From resistivity definition:
\(R = \frac{\rho L}{A} \implies \frac{R}{L} = \frac{\rho}{A} \implies m = \frac{\rho}{A}\)
\(\rho = m \times A = 15.75 \times 4.524 \times 10^{-8}\text{ m}^2 \approx 7.12 \times 10^{-7}\ \Omega\text{ m}\)
(Using \(15.75\): \(7.1 \times 10^{-7}\ \Omega\text{ m}\))

(ii)
Percentage uncertainty in \(d\):
\(\%\Delta d = \frac{0.01}{0.24} \times 100\% \approx 4.17\%\)

Percentage uncertainty in \(A\):
\(\%\Delta A = 2 \times 4.17\% = 8.34\%\)

Assuming the uncertainty in the gradient is negligible (or considering only \(d\)'s uncertainty contribution):
\(\Delta \rho = 8.34\% \times 7.12 \times 10^{-7} \approx 0.59 \times 10^{-7}\ \Omega\text{ m}\)
Rounding: \(\pm 0.6 \times 10^{-7}\ \Omega\text{ m}\) (accept range \(0.5 - 0.7 \times 10^{-7}\ \Omega\text{ m}\)).

(c)
A constant systematic error shifts the entire line upwards vertically (increases the y-intercept by \(0.2\ \Omega\)). However, the slope of the line (gradient) is unchanged. Since resistivity \(\rho\) is determined solely from the gradient \(m\), the calculated value of resistivity remains unchanged.

(d)
A high current causes electrical energy to be dissipated as heat, raising the temperature of the wire. The resistivity of metals increases with temperature. Keeping the current low ensures the wire's temperature remains constant, ensuring a fair and consistent resistance measurement.

(a)
- 1 mark for correct selection of points from the table.
- 1 mark for calculating rise over run correctly.
- 1 mark for obtaining gradient in range \(15.5\) to \(15.8\ \Omega\text{ m}^{-1}\).

(b)
- 1 mark for correct calculation of cross-sectional area \(A = 4.52 \times 10^{-8}\text{ m}^2\).
- 1 mark for relationship \(\rho = m \times A\).
- 1 mark for correct resistivity calculation (approx \(7.1 \times 10^{-7}\ \Omega\text{ m}\)).
- 1 mark for calculating the percentage uncertainty in area as twice the percentage uncertainty in diameter (approx 8.3%).
- 1.3 marks for calculating absolute uncertainty in resistivity correctly as approx \(0.6 \times 10^{-7}\ \Omega\text{ m}\).

(c)
- 1 mark for stating that the line shifts vertically / the y-intercept increases.
- 1 mark for stating that the gradient remains unchanged.
- 1 mark for concluding that the calculated resistivity is unaffected.

(d)
- 1 mark for identifying that current causes heating.
- 1 mark for explaining that a rise in temperature would change (increase) the resistance, violating the assumption of constant resistivity.

(a)
(i)
\(T = \frac{\text{Total Time}}{N_{\text{oscillations}}} = \frac{30.4}{20} = 1.52\text{ s}\)

(ii)
Percentage uncertainty in total time is:
\(\%\Delta t = \frac{0.2\text{ s}}{30.4\text{ s}} \times 100\% \approx 0.658\%\)
Since the dividing factor (20) has no uncertainty, the percentage uncertainty in period \(T\) is identical to that of total time:
\(\%\Delta T \approx 0.66\%\).

(b)
Squaring the equation \(T = 2\pi\sqrt{\frac{L}{g}}\):
\(T^2 = \frac{4\pi^2}{g} L\)

Comparing this with \(y = mx + c\):
- Plot \(T^2\) on the vertical axis (y-axis).
- Plot \(L\) on the horizontal axis (x-axis).
- The graph is a straight line through the origin with gradient \(m = \frac{4\pi^2}{g}\).
- To determine \(g\), calculate: \(g = \frac{4\pi^2}{m}\).

(c)
\(g = \frac{4\pi^2}{m} = \frac{4\pi^2}{4.02} \approx \frac{39.478}{4.02} \approx 9.820\text{ m s}^{-2}\)
Rounding to 3 sig figs: \(9.82\text{ m s}^{-2}\).

(d)
- Place the fiducial mark (e.g. a pointer or card with a vertical line) at the center of the swing (the equilibrium position).
- Start and stop the stopwatch exactly as the pendulum bob passes this mark.
- Ensure the bob is moving in the same direction each time a complete oscillation is counted (or count passes at the equilibrium position where speed is maximum, minimizing human reaction time error).

(a)
- 1 mark for correct period (1.52 s).
- 1 mark for identifying that percentage uncertainty of \(T\) equals that of total time.
- 1.3 marks for correct percentage calculation (0.66%).

(b)
- 1 mark for correct rearrangement of the formula to \(T^2 = \frac{4\pi^2}{g} L\).
- 1 mark for specifying \(T^2\) on the y-axis and \(L\) on the x-axis.
- 1 mark for stating that the gradient is equal to \(\frac{4\pi^2}{g}\).
- 1 mark for stating \(g = \frac{4\pi^2}{\text{gradient}}\).

(c)
- 1 mark for substituting the gradient value into the rearranged formula.
- 2 marks for correct calculation of \(g = 9.82\text{ m s}^{-2}\) (accept 9.8 or 9.82, unit required for full marks).

(d)
- 1 mark for placing the mark at the center of the oscillation (equilibrium position).
- 1 mark for explaining that the bob moves fastest at this point, which minimizes the timing error.
- 1 mark for ensuring counts are taken as the bob passes in the same direction.