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(a) The absolute uncertainty in length is directly given as \(\Delta L = \pm 0.002\text{ m}\).
The percentage uncertainty is:
\(\%\Delta L = \frac{0.002}{0.850} \times 100\% \approx 0.235\% \approx 0.24\%\).

(b) The time period \(T\) is the total time divided by the number of oscillations:
\(T = \frac{37.1}{20} = 1.855\text{ s} \approx 1.86\text{ s}\).
The absolute uncertainty in the total time is \(\Delta t = \pm 0.2\text{ s}\). Since the number of oscillations is a constant count with no uncertainty, the absolute uncertainty in the period is:
\(\Delta T = \frac{\Delta t}{20} = \frac{0.2}{20} = \pm 0.01\text{ s}\).

(c) Rearranging \(T = 2\pi\sqrt{\frac{L}{g}}\) for \(g\):
\(g = \frac{4\pi^2 L}{T^2}\)
Using the calculated values:
\(g = \frac{4 \times \pi^2 \times 0.850}{1.855^2} = 9.753\text{ m s}^{-2} \approx 9.75\text{ m s}^{-2}\).

For percentage uncertainty in \(g\):
\(\%\Delta g = \%\Delta L + 2 \times \%\Delta T\)
\(\%\Delta T = \frac{0.01}{1.855} \times 100\% \approx 0.539\%\)
\(\%\Delta g = 0.235\% + 2 \times 0.539\% = 1.313\% \approx 1.3\%\).

(d) To reduce the percentage uncertainty in \(T\), the student should record the time for a greater number of oscillations (e.g., 50 oscillations). Since the human reaction time (the primary source of absolute uncertainty in manual timing) remains constant at about \(\pm 0.2\text{ s}\), dividing this fixed uncertainty by a larger total time reduces the percentage uncertainty in the measured time and therefore reduces the percentage uncertainty in the calculated period \(T\).

- **(a)**
- 1 mark for stating the absolute uncertainty is \(\pm 0.002\text{ m}\).
- 1 mark for calculating the percentage uncertainty as \(0.24\%\) (accept \(0.235\%\) or \(0.2\%\)).
- **(b)**
- 1 mark for \(T = 1.855\text{ s}\) or \(1.86\text{ s}\).
- 1 mark for dividing the absolute uncertainty of total time by 20: \(\Delta T = \frac{0.2}{20}\).
- 1 mark for final absolute uncertainty of \(\pm 0.01\text{ s}\).
- **(c)**
- 1 mark for correct rearranged formula and calculation of \(g = 9.75\text{ m s}^{-2}\) (accept \(9.75 - 9.8\)).
- 1 mark for identifying that percentage uncertainty of \(g\) is \(\%\Delta L + 2 \times \%\Delta T\).
- 1 mark for correct final percentage uncertainty of \(1.3\%\) (accept \(1.31\%\)).
- **(d)**
- 1 mark for identifying that measuring more oscillations increases the total recorded time.
- 1 mark for explaining that because the absolute timing uncertainty (reaction time) remains constant, the percentage uncertainty is reduced.

(a) A neutral kaon \(K^0\) is a meson with strangeness \(+1\), meaning it contains an anti-strange quark (\(\bar{s}\)) and a down quark (\(d\)): \(d\bar{s}\). (If \(\bar{K}^0\) is assumed, the composition is \(\bar{d}s\), which is also accepted).
A positive pion \(\pi^+\) is a meson with charge \(+1\), consisting of an up quark and an anti-down quark: \(u\bar{d}\).

(b) Conservation checks:
- Baryon number \(B\): Mesons have \(B = 0\). \(B_{\text{initial}} (K^0) = 0\). \(B_{\text{final}} (\pi^+ + \pi^-) = 0 + 0 = 0\). (Conserved)
- Lepton number \(L\): None of these particles are leptons, so \(L = 0\) throughout. \(L_{\text{initial}} = 0\). \(L_{\text{final}} = 0 + 0 = 0\). (Conserved)
- Charge \(Q\): \(Q_{\text{initial}} = 0\). \(Q_{\text{final}} = (+1) + (-1) = 0\). (Conserved)

(c) Strangeness \(S\):
- Before decay: \(S(K^0) = +1\) (since it contains \(\bar{s}\) which has \(S = +1\)).
- After decay: Both \(\pi^+\) (\(u\bar{d}\)) and \(\pi^-\) (\(\bar{u}d\)) have no strange quarks, so \(S = 0\).
Total final strangeness = \(0\).
The change in strangeness is \(\Delta S = -1\), so strangeness is NOT conserved. The student's claim is incorrect. This decay is allowed because it proceeds via the weak interaction, which is the only force capable of changing quark flavour and is allowed to violate strangeness conservation by \(\pm 1\).

(d) Reaction: \(p + \pi^- \rightarrow K^0 + X\)
This proceeds via the strong interaction, meaning all quantum numbers, including strangeness, must be conserved.
- Baryon number \(B\): \(1 + 0 \rightarrow 0 + B_X \implies B_X = 1\) (so \(X\) is a baryon, consisting of 3 quarks).
- Charge \(Q\): \(+1 + (-1) \rightarrow 0 + Q_X \implies Q_X = 0\) (neutral baryon).
- Strangeness \(S\): \(0 + 0 \rightarrow +1 + S_X \implies S_X = -1\) (contains one strange quark \(s\)).
To form a neutral baryon (\(Q=0\), \(B=1\)) with strangeness \(-1\), the quarks must be \(uds\). This particle is the Lambda baryon (\(\Lambda^0\)) or the Sigma-zero baryon (\(\Sigma^0\)).

- **(a)**
- 1 mark for correct quark structure of \(K^0\): \(d\bar{s}\) (accept \(\bar{d}s\)).
- 1 mark for correct quark structure of \(\pi^+\): \(u\bar{d}\).
- **(b)**
- 1 mark for correctly showing conservation of charge: \(0 = +1 - 1\).
- 1 mark for correctly showing conservation of baryon number: \(0 = 0 + 0\).
- 1 mark for stating that lepton number is zero on both sides, hence conserved.
- **(c)**
- 1 mark for stating strangeness before is \(\pm 1\) and after is \(0\), proving the student is incorrect.
- 1 mark for stating that the decay is mediated by the weak interaction/force.
- 1 mark for explaining that the weak interaction can change strangeness by \(\pm 1\).
- **(d)**
- 1 mark for deducing the quark composition is \(uds\).
- 1 mark for identifying the particle as the Lambda (\(\Lambda\)) baryon or Sigma-zero (\(\Sigma^0\)) baryon.

(a) Let the direction to the right be positive.
Initial values:
\(m_A = 0.35\text{ kg}\), \(u_A = +2.4\text{ m s}^{-1}\)
\(m_B = 0.15\text{ kg}\), \(u_B = -1.2\text{ m s}^{-1}\)
Final values:
\(v_A = +0.80\text{ m s}^{-1}\)
\(v_B = ?\)

By conservation of linear momentum:
\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)
\((0.35 \times 2.4) + (0.15 \times -1.2) = (0.35 \times 0.80) + 0.15 v_B\)
\(0.84 - 0.18 = 0.28 + 0.15 v_B\)
\(0.66 = 0.28 + 0.15 v_B\)
\(0.15 v_B = 0.38\)
\(v_B = \frac{0.38}{0.15} = 2.533\text{ m s}^{-1} \approx 2.5\text{ m s}^{-1}\) to the right (since positive).

(b) Total initial kinetic energy (\(E_{k,i}\)):
\(E_{k,i} = \frac{1}{2} m_A u_A^2 + \frac{1}{2} m_B u_B^2\)
\(E_{k,i} = \frac{1}{2}(0.35)(2.4)^2 + \frac{1}{2}(0.15)(-1.2)^2 = 1.008 + 0.108 = 1.116\text{ J} \approx 1.12\text{ J}\).

Total final kinetic energy (\(E_{k,f}\)):
\(E_{k,f} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2\)
\(E_{k,f} = \frac{1}{2}(0.35)(0.80)^2 + \frac{1}{2}(0.15)(2.533)^2 = 0.112 + 0.481 = 0.593\text{ J} \approx 0.59\text{ J}\).

Since \(E_{k,f} < E_{k,i}\), kinetic energy is lost to other forms (e.g., thermal energy and sound), meaning the collision is inelastic.

(c) Impulse is equal to the change in momentum:
For glider \(B\):
\(\text{Impulse} = \Delta p_B = m_B (v_B - u_B)\)
\(\text{Impulse} = 0.15 \times (2.533 - (-1.2)) = 0.15 \times 3.733 = 0.560\text{ N s}\) (or \(\text{kg m s}^{-1}\)).

(d) Average force is given by:
\(F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{0.560}{0.045} = 12.44\text{ N} \approx 12\text{ N}\).
Since glider \(B\)'s momentum changed from leftwards to rightwards, the direction of the force acting on it must be to the right.

- **(a)**
- 1 mark for correct conservation of momentum equation setup with signs: \(0.35(2.4) + 0.15(-1.2) = 0.35(0.80) + 0.15 v_B\).
- 1 mark for correct calculation of \(v_B = 2.5\text{ m s}^{-1}\) (or \(2.53\text{ m s}^{-1}\)).
- 1 mark for stating the direction is "to the right".
- **(b)**
- 1 mark for calculating correct initial kinetic energy as \(1.12\text{ J}\).
- 1 mark for calculating correct final kinetic energy as \(0.59\text{ J}\).
- 1 mark for concluding that the collision is inelastic because kinetic energy is not conserved.
- **(c)**
- 1 mark for using \(\text{Impulse} = \Delta p = m \Delta v\).
- 1 mark for correct value \(0.56\text{ N s}\) (or \(\text{kg m s}^{-1}\)).
- **(d)**
- 1 mark for using \(F = \frac{\Delta p}{\Delta t}\) and calculating \(12\text{ N}\) (or \(12.4\text{ N}\)).
- 1 mark for identifying the direction of the force is to the right.

(a) Total resistance in the circuit is:
\(R_{\text{total}} = R_1 + R_T + r = 12.0 + 6.5 + 1.5 = 20.0\,\Omega\).
The current \(I\) is:
\(I = \frac{\varepsilon}{R_{\text{total}}} = \frac{9.0}{20.0} = 0.45\text{ A}\).

The terminal potential difference \(V\) is:
\(V = \varepsilon - Ir = 9.0 - (0.45 \times 1.5) = 9.0 - 0.675 = 8.325\text{ V} \approx 8.3\text{ V}\).

(b) The power dissipated in the external circuit is:
\(P_{\text{ext}} = I^2 \times R_{\text{ext}} = (0.45)^2 \times (12.0 + 6.5) = 0.2025 \times 18.5 = 3.746\text{ W} \approx 3.7\text{ W}\) (as shown).

The efficiency \(\eta\) of the electrical energy transfer is:
\(\eta = \frac{P_{\text{ext}}}{I \varepsilon} = \frac{V}{\varepsilon} = \frac{8.325}{9.0} \times 100\% = 92.5\%\).

(c) An NTC thermistor's resistance decreases as temperature increases. Therefore, the total circuit resistance \(R_{\text{total}}\) decreases. Since \(I = \frac{\varepsilon}{R_{\text{total}}}\) and \(\varepsilon\) is constant, the circuit current \(I\) increases.
The terminal pd is given by \(V = \varepsilon - Ir\). With an increased current, the "lost volts" \(Ir\) across the internal resistance increases, which causes the terminal pd \(V\) to decrease.

(d) To obtain an output voltage that increases with temperature, we should monitor the voltage across the fixed resistor \(R_1\). As the temperature increases and the thermistor's resistance decreases, the current in the circuit increases. According to \(V_1 = I R_1\), because \(R_1\) is constant, the potential difference across it increases.

- **(a)**
- 1 mark for calculating the current: \(I = 0.45\text{ A}\).
- 1 mark for calculating the terminal pd: \(V = 8.3\text{ V}\).
- **(b)**
- 1 mark for showing external power: \(P = I^2 R = 3.75\text{ W}\).
- 1 mark for formula for efficiency: \(\eta = \frac{V}{\varepsilon}\).
- 1 mark for correct efficiency calculation of \(92.5\%\).
- **(c)**
- 1 mark for stating that resistance of the thermistor decreases as temperature increases, which increases circuit current.
- 1 mark for referencing \(V = \varepsilon - Ir\) or stating "lost volts increase".
- 1 mark for concluding that terminal pd decreases.
- **(d)**
- 1 mark for suggesting that the output voltage is taken across the fixed resistor \(R_1\).
- 1 mark for explaining that as current increases, the potential difference across the fixed resistor increases.

(a) The force acting on the wire is:
\(F = m g = 6.0 \times 9.81 = 58.86\text{ N}\).

Using the formula for Young modulus \(E = \frac{F L}{A \Delta L}\):
\(\Delta L = \frac{F L}{A E}\)
\(\Delta L = \frac{58.86 \times 2.2}{(4.5 \times 10^{-7}) \times (2.0 \times 10^{11})} = \frac{129.492}{9.0 \times 10^4} = 1.439 \times 10^{-3}\text{ m} \approx 1.4\text{ mm}\) (or \(1.44 \times 10^{-3}\text{ m}\)).

(b) Assuming the limit of proportionality has not been exceeded, the elastic strain energy \(E_{\text{str}}\) is given by:
\(E_{\text{str}} = \frac{1}{2} F \Delta L = \frac{1}{2} \times 58.86 \times 1.439 \times 10^{-3} = 0.0423\text{ J} \approx 4.2 \times 10^{-2}\text{ J}\).

(c)
- **Elastic behavior:** Under low stress, the applied force pulls the metal atoms slightly further apart from their equilibrium positions. The metallic bonds act like springs, and when the stress is removed, the interatomic forces pull the atoms back to their original positions.
- **Plastic behavior:** Under stress beyond the elastic limit, the planes of atoms slide past each other. This involves the movement of dislocations through the crystal lattice. When the stress is removed, the atoms do not return to their original positions, resulting in permanent deformation.

(d) The Young modulus is a measure of stiffness. From \(\Delta L = \frac{F L}{A E}\), for wires of identical length \(L\) and area \(A\) under the same load \(F\), the extension is inversely proportional to the Young modulus:
\(\Delta L \propto \frac{1}{E}\).
Since the Young modulus of copper (\(1.1 \times 10^{11}\text{ Pa}\)) is lower than that of steel (\(2.0 \times 10^{11}\text{ Pa}\)), copper is less stiff. Therefore, the copper wire will experience a larger extension under the same force.

- **(a)**
- 1 mark for calculating the force \(F = 58.86\text{ N}\).
- 1 mark for correct rearrangement: \(\Delta L = \frac{FL}{AE}\).
- 1 mark for final calculated value of \(\Delta L = 1.4\text{ mm}\).
- **(b)**
- 1 mark for using \(E_{\text{str}} = \frac{1}{2} F \Delta L\).
- 1 mark for correct energy value of \(4.2 \times 10^{-2}\text{ J}\).
- **(c)**
- 1 mark for describing elastic behavior in terms of temporary atomic displacement.
- 1 mark for describing plastic behavior in terms of sliding atomic planes / dislocation movement.
- 1 mark for stating that plastic deformation results in permanent shift of atomic positions.
- **(d)**
- 1 mark for showing that extension is inversely proportional to Young modulus (\(\Delta L \propto \frac{1}{E}\)).
- 1 mark for stating that because copper has a smaller Young modulus (is less stiff), it extends more under the same force.

(a) The time period \(T\) of a mass-spring system in simple harmonic motion is given by:
\(T = 2\pi \sqrt{\frac{m}{k}}\)
\(T = 2\pi \sqrt{\frac{0.40}{25}} = 2\pi \sqrt{0.016} \approx 0.795\text{ s} \approx 0.79\text{ s}\).

(b) Since the mass is released from its maximum positive displacement (\(A = 8.0\text{ cm} = 0.080\text{ m}\)) at \(t = 0\), its displacement can be described by a cosine function:
\(x(t) = A \cos(\omega t)\)

Where the angular frequency \(\omega\) is:
\(\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{25}{0.40}} = \sqrt{62.5} \approx 7.91\text{ rad s}^{-1}\).
Thus, the expression is:
\(x(t) = 0.080 \cos(7.91 t)\).

At \(t = 0.25\text{ s}\) (with calculator in radian mode):
\(\omega t = 7.9057 \times 0.25 = 1.9764\text{ rad}\).
\(x(0.25) = 0.080 \cos(1.9764) = 0.080 \times (-0.3923) = -0.0314\text{ m} \approx -3.1\text{ cm}\).
The negative sign indicates the mass is \(3.1\text{ cm}\) above the equilibrium position.

(c) The maximum speed \(v_{\text{max}}\) is:
\(v_{\text{max}} = \omega A = 7.906 \times 0.080 = 0.632\text{ m s}^{-1} \approx 0.63\text{ m s}^{-1}\).

The maximum acceleration \(a_{\text{max}}\) is:
\(a_{\text{max}} = \omega^2 A = 62.5 \times 0.080 = 5.0\text{ m s}^{-2}\).

(d)
- **Maximum speed** occurs as the mass passes through the equilibrium position (\(x = 0\)), where all potential energy is kinetic.
- **Maximum acceleration** occurs at the positions of maximum displacement (\(x = \pm 8.0\text{ cm}\)), where the restoring force is greatest.

- **(a)**
- 1 mark for using \(T = 2\pi \sqrt{\frac{m}{k}}\).
- 1 mark for calculating \(T = 0.79\text{ s}\).
- **(b)**
- 1 mark for writing the expression \(x = 0.080 \cos(7.9 t)\).
- 1 mark for evaluating the displacement with calculator in radian mode.
- 1 mark for final displacement of \(-3.1\text{ cm}\) (accept range \(-3.1\text{ cm}\) to \(-3.2\text{ cm}\)).
- **(c)**
- 1 mark for calculating maximum speed: \(v_{\text{max}} = 0.63\text{ m s}^{-1}\).
- 1 mark for formula: \(a_{\text{max}} = \omega^2 A\).
- 1 mark for calculating maximum acceleration: \(5.0\text{ m s}^{-2}\).
- **(d)**
- 1 mark for identifying maximum speed is at the equilibrium position.
- 1 mark for identifying maximum acceleration is at the amplitudes (\(x = \pm 8.0\text{ cm}\)).

The power dissipated in the load resistor \(R\) is given by \(P = I^2 R = \left(\frac{\varepsilon}{R+r}\right)^2 R = \frac{\varepsilon^2 R}{(R+r)^2}\). We are given that \(P(2.0\ \Omega) = P(8.0\ \Omega)\). Therefore: \(\frac{\varepsilon^2 (2.0)}{(2.0+r)^2} = \frac{\varepsilon^2 (8.0)}{(8.0+r)^2}\). Simplifying this gives: \(\frac{1}{(2.0+r)^2} = \frac{4}{(8.0+r)^2}\). Taking the square root of both sides (since both sides are positive): \(\frac{1}{2.0+r} = \frac{2}{8.0+r}\). Cross-multiplying gives: \(8.0 + r = 2(2.0 + r) = 4.0 + 2r\). Solving for \(r\) yields: \(r = 4.0\ \Omega\). This is also the geometric mean of the two resistances, \(r = \sqrt{R_1 R_2} = \sqrt{2.0 \times 8.0} = 4.0\ \Omega\).

1 mark for the correct answer B. (Method: Set up power equations for both cases, equate them, and solve for internal resistance r.)

Let the direction to the right be positive. The initial velocities are \(u_P = 3u\) and \(u_Q = -u\). In a perfectly elastic collision, the relative velocity of separation equals the relative velocity of approach: \(v_Q - v_P = u_P - u_Q = 3u - (-u) = 4u \implies v_P = v_Q - 4u\). By conservation of linear momentum: \(m_P u_P + m_Q u_Q = m_P v_P + m_Q v_Q\). Substituting the masses and initial velocities: \(m(3u) + 2m(-u) = m(v_Q - 4u) + 2m v_Q \implies mu = 3m v_Q - 4mu \implies 5mu = 3m v_Q \implies v_Q = \frac{5}{3}u\) to the right.

1 mark for the correct answer C. (Method: Combine conservation of momentum and the elastic collision relative velocity relationship to find the final velocity of Q.)

The total mechanical energy \(E\) of a satellite of mass \(m\) in a circular orbit of radius \(r\) is given by \(E = -\frac{GMm}{2r}\). The initial energy at orbit radius \(r_1 = 3R\) is \(E_1 = -\frac{GMm}{6R}\). The final energy at orbit radius \(r_2 = 4R\) is \(E_2 = -\frac{GMm}{8R}\). The increase in total mechanical energy is \(\Delta E = E_2 - E_1 = -\frac{GMm}{8R} - \left(-\frac{GMm}{6R}\right) = \left(\frac{1}{6} - \frac{1}{8}\right)\frac{GMm}{R} = \frac{1}{24}\frac{GMm}{R}\). Since the gravitational field strength at the planet's surface is \(g = \frac{GM}{R^2}\), we have \(GM = gR^2\). Substituting this in: \(\Delta E = \frac{m(gR^2)}{24R} = \frac{mgR}{24}\).

1 mark for the correct answer C. (Method: Calculate initial and final orbital mechanical energy, find the difference, and express in terms of surface gravity g.)

Let the displacement of the particle be given by \(x(t) = A \cos(\omega t)\), since it starts at its maximum displacement \(x(0) = A\). To travel a total distance of \(1.5 A\), the particle first moves from \(x = A\) to \(x = 0\) (distance of \(A\)), and then continues to move to \(x = -0.5 A\) (additional distance of \(0.5 A\)). Therefore, we need to solve for \(t\) when \(x(t) = -0.5 A\). This gives \(A \cos(\omega t) = -0.5 A \implies \cos(\omega t) = -0.5\). The smallest positive solution is \(\omega t = \frac{2\pi}{3}\). Since \(\omega = \frac{2\pi}{T}\), we have \(\frac{2\pi}{T} t = \frac{2\pi}{3} \implies t = \frac{T}{3}\).

1 mark for the correct answer C. (Method: Identify the final displacement required, use the cosine displacement equation, and solve for the time t.)

The reactant \(\Sigma^-\) has quark composition \(dds\). The products are a neutron (\(udd\)) and a pion \(\pi^-\) (\(d\bar{u}\)). Looking at the quark transition, an \(s\) quark decays into a \(u\) quark by emitting a virtual \(W^-\in\) boson (since charge changes from \(-1/3\) to \(+2/3\), which requires a change of \(-1\) carried away by the \(W^-\)). The \(W^-\in\) then decays into a \(d\bar{u}\) pair, forming the \(\pi^-\). Thus, the correct transition is \(s \rightarrow u\) via a \(W^-\) boson.

1 mark for the correct answer B. (Method: Dedicate quark transition from reactant to product and use charge conservation to identify the W- boson.)

Using Einstein's photoelectric equation: \(E_k = hf - \Phi\), where \(\Phi\) is the work function. For the second frequency: \(2.5E_k = h(1.5f) - \Phi\). Substituting the first expression into the second: \(2.5(hf - \Phi) = 1.5hf - \Phi \implies 2.5hf - 2.5\Phi = 1.5hf - \Phi \implies hf = 1.5\Phi \implies \Phi = \frac{2}{3}hf\). Since the threshold frequency \(f_0\) is defined by \(\Phi = hf_0\), we get \(hf_0 = \frac{2}{3}hf \implies f_0 = \frac{2}{3}f \approx 0.67f\).

1 mark for the correct answer C. (Method: Write the photoelectric equations for both cases, eliminate Ek to find the work function in terms of hf, and deduce the threshold frequency.)

The energy stored in a capacitor is \(E = \frac{1}{2} C V^2\). During discharge, the potential difference decreases according to \(V = V_0 e^{-t/RC}\). Substituting this into the energy formula gives: \(E = \frac{1}{2} C \left(V_0 e^{-t/RC}\right)^2 = E_0 e^{-2t/RC}\), where \(E_0 = \frac{1}{2} C V_0^2\). We are given that \(E = 0.10 E_0\), so: \(0.10 E_0 = E_0 e^{-2t/RC} \implies e^{-2t/RC} = 0.10 \implies -2t/RC = \ln(0.10) = -\ln(10)\). Therefore, \(2t/RC = \ln 10 \implies t = 0.5 RC \ln 10\).

1 mark for the correct answer B. (Method: Formulate energy as a function of time, equate to 0.10 of the initial energy, and solve the exponential equation for t.)

The elastic strain energy stored in a wire under load \(F\) is \(U = \frac{1}{2} F \Delta L\). The extension is given by \(\Delta L = \frac{F L}{A E_Y}\), where \(A\) is the cross-sectional area and \(E_Y\) is the Young modulus. This gives \(U = \frac{F^2 L}{2 A E_Y}\). Since the wires are of the same material, \(E_Y\) is identical. The loads \(F\) are also identical. The cross-sectional area \(A\) is proportional to \(d^2\). For wire X: \(U_X \propto \frac{L}{d^2}\). For wire Y: \(U_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\). Therefore, the ratio of the energies is: \(\frac{U_X}{U_Y} = \frac{L/d^2}{L/(2d^2)} = 2.0\).

1 mark for the correct answer C. (Method: Use the formula for elastic strain energy in terms of tension, length, and cross-sectional area, and apply the scaling factors for X and Y.)

First, calculate the percentage uncertainty in the length \( L \):
\[ \% \Delta L = \frac{2 \text{ mm}}{730 \text{ mm}} \times 100\% \approx 0.274\% \]

Next, calculate the percentage uncertainty in the total measured time \( t \). Since \( T = \frac{t}{20} \), the percentage uncertainty in \( T \) is equal to the percentage uncertainty in \( t \):
\[ \% \Delta T = \% \Delta t = \frac{0.2 \text{ s}}{34.2 \text{ s}} \times 100\% \approx 0.585\% \]

Since \( g \propto \frac{L}{T^2} \), the percentage uncertainty in \( g \) is:
\[ \% \Delta g = \% \Delta L + 2 \times \% \Delta T \]
\[ \% \Delta g = 0.274\% + 2 \times 0.585\% = 0.274\% + 1.17\% \approx 1.44\% \]

Rounding to 2 significant figures gives \( 1.4\% \).

1 mark for the correct calculation of percentage uncertainties in L and t, doubling the uncertainty of t, adding them, and choosing option D.
- Correct answer is D.
- Reject other choices due to incorrect propagation of uncertainty.

The mean square speed of the molecules in an ideal gas is directly proportional to its absolute temperature \( T \) in Kelvin:
\[ c_{\text{rms}}^2 \propto T \]

Convert the initial temperature to Kelvin:
\[ T_1 = 27 + 273 = 300 \text{ K} \]

Since the mean square speed is doubled, the absolute temperature must also double:
\[ T_2 = 2 \times T_1 = 2 \times 300 \text{ K} = 600 \text{ K} \]

Convert this back to degrees Celsius:
\[ T_2 = 600 - 273 = 327^\circ\text{C} \]

1 mark for converting Celsius to Kelvin, realizing that doubling the mean square speed doubles the absolute temperature, and converting back to Celsius.
- Correct answer is C.
- Choice A is incorrect (simply doubling the Celsius value: 2 x 27 = 54).
- Choice B is incorrect (miscalculation or taking square root of temperature instead of direct proportion).
- Choice D is incorrect (retaining the temperature in Kelvin).

Let the initial number of nuclei be \( N_0 \).
After three half-lives, the fraction of remaining undecayed nuclei is:
\[ \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]
So, the number of remaining undecayed nuclei is \( N = \frac{1}{8} N_0 \).

The number of decayed nuclei is:
\[ N_{\text{decayed}} = N_0 - N = N_0 - \frac{1}{8} N_0 = \frac{7}{8} N_0 \]

The ratio of decayed nuclei to remaining undecayed nuclei is:
\[ \frac{N_{\text{decayed}}}{N} = \frac{\frac{7}{8} N_0}{\frac{1}{8} N_0} = \frac{7}{1} \]
Therefore, the ratio is \( 7 : 1 \).

1 mark for calculating the fraction of undecayed nuclei remaining after 3 half-lives, determining the fraction of decayed nuclei, and finding the correct ratio.
- Correct answer is C.
- Choice A is the inverse ratio.
- Choice B is the ratio of remaining to initial nuclei.
- Choice D is the ratio of initial to remaining nuclei.

Because the voltmeter is of high resistance, it draws negligible current.
When the resistor \( R \) is connected across the cell, the circuit is a simple series loop consisting of the cell's internal resistance \( r \) and the external resistor \( R \).

The terminal potential difference \( V \), which is measured by the voltmeter, is given by the potential divider formula:
\[ V = \varepsilon \frac{R}{R + r} \]

We are given that \( V = \frac{2}{3}\varepsilon \), so:
\[ \frac{2}{3}\varepsilon = \varepsilon \frac{R}{R + r} \]
\[ \frac{2}{3} = \frac{R}{R + r} \]
\[ 2(R + r) = 3R \]
\[ 2R + 2r = 3R \]
\[ R = 2r \]

Therefore, the ratio \( \frac{R}{r} = 2.0 \).

1 mark for using the terminal potential difference formula with the given values to solve for the ratio of R to r.
- Correct answer is C.
- Other options correspond to algebraic errors in rearranging the potential divider formula.

By Kepler's Third Law, the square of the orbital period \( T \) is directly proportional to the cube of the orbital radius \( R \):
\[ T^2 \propto R^3 \]

Therefore:
\[ \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{R_2}{R_1}\right)^3 \]

Given \( R_1 = R \), \( R_2 = 4R \), and \( T_1 = T \):
\[ \left(\frac{T_2}{T}\right)^2 = \left(\frac{4R}{R}\right)^3 = 4^3 = 64 \]

Taking the square root of both sides:
\[ \frac{T_2}{T} = \sqrt{64} = 8 \]
\[ T_2 = 8T \]

1 mark for applying Kepler's Third Law, substituting the change in radius, and solving for the new period.
- Correct answer is C.
- Other options arise from incorrect relationships (such as linear or inverse proportionality).

We can find the distance \( s \) using the work-energy theorem. The work done by the braking force \( F \) equals the loss of kinetic energy:
\[ W = F s = \frac{1}{2}mv^2 \implies s = \frac{mv^2}{2F} \]

We can find the time \( t \) using the impulse-momentum theorem. The impulse of the braking force equals the change in momentum:
\[ F t = m v \implies t = \frac{mv}{F} \]

Therefore, the correct pair of (distance, time) is \( \left(\frac{mv^2}{2F}, \frac{mv}{F}\right) \).

1 mark for deriving correct expressions for both distance and time using work-energy and momentum principles, and selecting option A.
- Correct answer is A.
- Other options confuse the factor of 2 in either the work-energy or the momentum equation.

Initially, the plane of the coil is perpendicular to the magnetic field, so the magnetic flux linkage through the coil is at its maximum value:
\[ \Phi_{\text{initial}} = N B A = N B (\pi r^2) \]

When the coil is rotated by \( 90^\circ \), its plane is parallel to the magnetic field, meaning no flux lines pass through it:
\[ \Phi_{\text{final}} = 0 \]

The change in magnetic flux linkage is:
\[ \Delta \Phi = \Phi_{\text{initial}} - \Phi_{\text{final}} = \pi N B r^2 \]

By Faraday's Law, the magnitude of the average induced emf \( \varepsilon \) is given by:
\[ \varepsilon = \frac{\Delta \Phi}{\Delta t} = \frac{\pi N B r^2}{\Delta t} \]

1 mark for calculating the initial magnetic flux linkage using the area of a circle, finding the change in flux linkage, and applying Faraday's Law.
- Correct answer is B.
- Choice A is incorrect (omits the factor of pi in the area formula).
- Choice C is incorrect (double the correct value).
- Choice D is incorrect (no net induced emf is false; average emf is non-zero).

The initial charge on the first capacitor is:
\[ Q_1 = C V \]
The initial electrostatic energy stored is:
\[ E_{\text{initial}} = \frac{1}{2} C V^2 \]

When connected in parallel to the uncharged capacitor of capacitance \( 2C \), the total capacitance of the system is:
\[ C_{\text{total}} = C + 2C = 3C \]

By the conservation of charge, the total charge in the parallel system remains \( Q = Q_1 = C V \).
The final potential difference \( V_{\text{final}} \) across the combination is:
\[ V_{\text{final}} = \frac{Q}{C_{\text{total}}} = \frac{C V}{3C} = \frac{V}{3} \]

The total energy stored in the two capacitors after the connection is:
\[ E_{\text{final}} = \frac{1}{2} C_{\text{total}} V_{\text{final}}^2 = \frac{1}{2} (3C) \left(\frac{V}{3}\right)^2 = \frac{1}{2} (3C) \frac{V^2}{9} = \frac{1}{6} C V^2 \]

Comparing the final energy to the initial energy:
\[ E_{\text{final}} = \frac{1}{3} \left(\frac{1}{2} C V^2\right) = \frac{1}{3} E_{\text{initial}} \]

The energy lost as heat and radiation is:
\[ \Delta E = E_{\text{initial}} - E_{\text{final}} = E_{\text{initial}} - \frac{1}{3} E_{\text{initial}} = \frac{2}{3} E_{\text{initial}} \]

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

1 mark for applying the conservation of charge to find the final potential difference, calculating the final energy of the system, and determining the fraction of energy lost.
- Correct answer is C.
- Choice A is incorrect (this is the fraction of energy remaining).
- Choice B is incorrect (intuitive guess that doesn't account for the quadratic relationship of energy with voltage).
- Choice D is incorrect (represents the fraction of energy lost if the final voltage was squared without adjusting the capacitance factor properly).

By Coulomb's Law, the electrostatic force is given by \(F = \frac{k Q_1 Q_2}{r^2}\). The new force is \(F' = \frac{k (2Q_1) (0.5Q_2)}{(3r)^2} = \frac{k Q_1 Q_2}{9r^2} = \frac{1}{9}F\). Thus, the ratio of the new force to the original force is 1/9.

1 mark for identifying Coulomb's Law, applying the multipliers correctly, and arriving at 1/9.

The total energy supplied by the power supply is given by \(E_{\text{total}} = Q V = C V^2\). The energy stored in the fully charged capacitor is \(E_{\text{stored}} = \frac{1}{2} C V^2\). The ratio of the energy stored to the total energy supplied is therefore \(\frac{0.5 C V^2}{C V^2} = 0.50\).

1 mark for correctly determining the total energy supplied and the energy stored, then finding the correct ratio.

The mean square speed of ideal gas molecules is proportional to the absolute temperature, \(\langle c^2 \rangle \propto T\). The root-mean-square speed is therefore proportional to the square root of the absolute temperature, \(c_{\text{rms}} \propto \sqrt{T}\). Changing the volume has no effect on the speed distribution of the molecules at a constant temperature. Since the absolute temperature doubles, the root-mean-square speed increases by a factor of \(\sqrt{2} \approx 1.41\).

1 mark for establishing the relationship between root-mean-square speed and absolute temperature, and correctly calculating the ratio.

The volume \(V = A \times L\) is constant. Under stretching, the length increases to \(L' = 1.10 L\), which means the area must decrease to \(A' = A / 1.10\). Using the resistance formula \(R = \rho \frac{L}{A}\), the new resistance is \(R' = \rho \frac{1.10 L}{A / 1.10} = 1.21 \rho \frac{L}{A} = 1.21 R\). The fractional increase in resistance is \(0.21\), representing a \(21\%\) increase.

1 mark for applying constant volume to find the new area and calculating the final resistance to show a 21% increase.

The radius of the circular path is \(r = \frac{mv}{Bq} = \frac{\sqrt{2m E_k}}{Bq}\). Since \(B\) and \(E_k\) are constant, \(r \propto \frac{\sqrt{m}}{q}\). The alpha particle has mass \(4m_p\) and charge \(2e\), while the proton has mass \(m_p\) and charge \(e\). Thus, \(\frac{r_{\alpha}}{r_p} = \frac{\sqrt{4m_p} / 2e}{\sqrt{m_p} / e} = 1.00\).

1 mark for relating radius to mass and charge under constant kinetic energy, and calculating the correct ratio.

Total energy is given by \(E = E_k + E_p = \frac{1}{2} k A^2\). Since we are given \(E_k = 3 E_p\), the expression becomes \(3 E_p + E_p = 4 E_p = \frac{1}{2} k A^2\). Substituting the formula for potential energy \(E_p = \frac{1}{2} k x^2\) gives \(4 \left( \frac{1}{2} k x^2 \right) = \frac{1}{2} k A^2\), which simplifies to \(4 x^2 = A^2\), leading to \(x = 0.50 A\).

1 mark for linking kinetic and potential energies to total energy and finding the correct displacement in terms of amplitude.

The extension of a wire is given by \(\Delta L = \frac{F L}{A E} = \frac{4 F L}{\pi d^2 E}\). Since \(F\) and \(E\) are the same for both wires, the extension is proportional to \(\frac{L}{d^2}\). Thus, the ratio of the extensions is \(\frac{\Delta L_X}{\Delta L_Y} = \frac{L_X}{L_Y} \times \left( \frac{d_Y}{d_X} \right)^2 = 2 \times 2^2 = 8\).

1 mark for establishing the correct relationship between extension, length, and diameter, and computing the correct ratio.

After three half-lives, the remaining fraction of active nuclei is \((1/2)^3 = 1/8\). The fraction of active nuclei that has decayed is the total minus the remaining fraction, which is \(1 - 1/8 = 7/8\).

1 mark for calculating the fraction remaining after three half-lives and subtracting it from 1 to find the decayed fraction.

The initial energy stored in the capacitor is given by \(E_0 = \frac{1}{2} C V_0^2\). When the capacitor discharges, the potential difference across it at any time \(t\) is given by \(V = V_0 e^{-\frac{t}{RC}}\). At time \(t = 2RC\), the potential difference is \(V = V_0 e^{-2}\). The energy remaining in the capacitor at this time is \(E_{\text{remaining}} = \frac{1}{2} C V^2 = \frac{1}{2} C (V_0 e^{-2})^2 = \frac{1}{2} C V_0^2 e^{-4} = E_0 e^{-4}\). By conservation of energy, the energy dissipated in the resistor is \(E_{\text{dissipated}} = E_0 - E_{\text{remaining}} = E_0 (1 - e^{-4})\). The ratio of the energy dissipated to the remaining energy is therefore \(\frac{E_0 (1 - e^{-4})}{E_0 e^{-4}} = \frac{1 - e^{-4}}{e^{-4}} = e^4 - 1\).

1 mark for the correct answer B.

**(a)** Using the ideal gas equation:
\(p V = n R T\)
\(p = \frac{n R T}{V} = \frac{0.12 \times 8.31 \times 290}{2.4 \times 10^{-3}} = 1.205 \times 10^5\text{ Pa} \approx 1.2 \times 10^5\text{ Pa}\).

**(b)** For constant volume, pressure is directly proportional to temperature:
\(\frac{p_1}{T_1} = \frac{p_2}{T_2} \Rightarrow T_2 = T_1 \times \frac{p_2}{p_1}\)
Using the exact initial pressure:
\(T_2 = 290 \times \frac{2.8 \times 10^5}{1.205 \times 10^5} = 674\text{ K}\) (using the rounded value of \(1.2 \times 10^5\text{ Pa}\) yields \(677\text{ K}\)).

**(c)**
- As temperature increases, the average kinetic energy of the helium atoms increases, so they move with higher mean speed.
- This means there is a greater change in momentum during each collision with the container walls.
- In addition, collisions with the walls occur more frequently because of the higher speed and constant volume.
- Since force is the rate of change of momentum, the average force on the walls increases, leading to an increase in pressure (\(p = F/A\)).

**(d)** The internal energy \(U\) of a monatomic ideal gas is given by:
\(U = \frac{3}{2} n R T\)
\(\Delta U = \frac{3}{2} n R \Delta T\)
\(\Delta T = 674 - 290 = 384\text{ K}\)
\(\Delta U = 1.5 \times 0.12 \times 8.31 \times 384 = 574\text{ J}\) (or \(579\text{ J}\) if using \(T_2 = 677\text{ K}\)).
Since temperature increased, this is an **increase** in internal energy.

**(a) [2 Marks]**
- 1 mark: Correct substitution into \(p = \frac{n R T}{V}\)
- 1 mark: Calculation showing \(1.205 \times 10^5\text{ Pa}\) and concluding to \(\approx 1.2 \times 10^5\text{ Pa}\).

**(b) [2 Marks]**
- 1 mark: Correct use of gas law relation \(\frac{p_1}{T_1} = \frac{p_2}{T_2}\) or equivalent substitution into \(pV = nRT\).
- 1 mark: Correct final temperature of \(674\text{ K}\) or \(677\text{ K}\).

**(c) [3 Marks]**
- 1 mark: Higher temperature means higher average speed/kinetic energy of molecules.
- 1 mark: Change in momentum per collision is greater OR collisions are more frequent.
- 1 mark: Linking rate of change of momentum to average force per unit area (\(p = F/A\)) to explain the pressure increase.

**(d) [3 Marks]**
- 1 mark: Correct formula for internal energy change of a monatomic gas, \(\Delta U = \frac{3}{2} n R \Delta T\).
- 1 mark: Correct calculation of energy change, \(574\text{ J}\) (or \(579\text{ J}\)).
- 1 mark: Stating clearly that this is an **increase** (consistent with their calculation).

**(a)** In \(\beta^-\) decay, a neutron decays into a proton, emitting an electron and an electron antineutrino:
\(^{24}_{11}\text{Na} \rightarrow ^{24}_{12}\text{Mg} + ^{0}_{-1}\beta + \bar{\nu}_e\)
(Note: \(\beta^-\) can also be written as \(e^-\) or \(\text{e}\)).

**(b)** The half-life \(T_{1/2} = 15.0 \times 3600\text{ s} = 54000\text{ s}\).
\(\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.69315}{54000} = 1.28 \times 10^{-5}\text{ s}^{-1}\).

**(c)** Time elapsed \(t = 45.0\text{ hours}\), which corresponds to exactly \(\frac{45.0}{15.0} = 3.0\) half-lives.
Activity \(A = \frac{A_0}{2^3} = \frac{3.6 \times 10^7}{8} = 4.5 \times 10^6\text{ Bq}\).
Using \(A = \lambda N\):
\(N = \frac{A}{\lambda} = \frac{4.5 \times 10^6}{1.283 \times 10^{-5}} = 3.51 \times 10^{11}\) nuclei (accept \(3.5 \times 10^{11}\)).

**(d)** Gamma radiation follows the inverse square law because it spreads out uniformly in all directions from a point source:
\(I \propto \frac{1}{r^2} \Rightarrow C_1 r_1^2 = C_2 r_2^2\)
\(C_2 = C_1 \left(\frac{r_1}{r_2}\right)^2 = 2400 \times \left(\frac{0.50}{1.50}\right)^2 = 2400 \times \left(\frac{1}{3}\right)^2 = 2400 \times \frac{1}{9} = 267\text{ s}^{-1}\) (or \(270\text{ s}^{-1}\) to 2 s.f.).

**(a) [2 Marks]**
- 1 mark: Correct magnesium symbol with mass number 24 and proton number 12 (\(^{24}_{12}\text{Mg}\)) and \(\beta^-\) particle (\(^{0}_{-1}\beta\)).
- 1 mark: Inclusion of electron antineutrino (\(\bar{\nu}_e\)) with balanced conservation laws.

**(b) [2 Marks]**
- 1 mark: Correct conversion of hours to seconds (\(15.0 \times 3600 = 54000\text{ s}\)).
- 1 mark: Calculation of decay constant \(\lambda = 1.28 \times 10^{-5}\text{ s}^{-1}\) (accept \(1.3 \times 10^{-5}\text{ s}^{-1}\)).

**(c) [3 Marks]**
- 1 mark: Determination that 45 hours is 3 half-lives, giving \(A = 4.5 \times 10^6\text{ Bq}\).
- 1 mark: Correct rearranged equation \(N = A / \lambda\).
- 1 mark: Correct final answer for number of nuclei \(N = 3.5 \times 10^{11}\) (allow ecf from b).

**(d) [3 Marks]**
- 1 mark: Stating and explaining the inverse square law for gamma radiation (\(I \propto r^{-2}\) due to point source and isotropic spherical spreading).
- 1 mark: Correct calculation of count rate ratio (factor of 9 decrease).
- 1 mark: Correct final count rate of \(267\text{ s}^{-1}\) or \(270\text{ s}^{-1}\).

**(a)** For a circular orbit, the gravitational force provides the centripetal force:
\(\frac{G M m}{r^2} = \frac{m v^2}{r} = m \omega^2 r = m \left(\frac{2\pi}{T}\right)^2 r\)
Rearranging for \(r^3\):
\(r^3 = \frac{G M T^2}{4\pi^2}\)
Substitute the values:
\(r^3 = \frac{(6.67 \times 10^{-11}) \times (6.00 \times 10^{24}) \times (7200)^2}{4\pi^2}
= \frac{4.002 \times 10^{14} \times 5.184 \times 10^7}{39.478} = 5.255 \times 10^{20}\text{ m}^3\)
\(r = (5.255 \times 10^{20})^{1/3} = 8.07 \times 10^6\text{ m} \approx 8.1 \times 10^6\text{ m}\).

**(b)** Gravitational potential energy:
\(E_p = -\frac{G M m}{r} = -\frac{(6.67 \times 10^{-11}) \times (6.00 \times 10^{24}) \times 1200}{8.07 \times 10^6} = -5.95 \times 10^{10}\text{ J}\)
(If using \(8.1 \times 10^6\text{ m}\), \(E_p = -5.93 \times 10^{10}\text{ J}\)).

**(c)** For a stable circular orbit:
\(E_k = \frac{1}{2} m v^2 = \frac{G M m}{2r} = -\frac{E_p}{2}\)
\(E_k = \frac{5.95 \times 10^{10}}{2} = 2.98 \times 10^{10}\text{ J}\).
Total mechanical energy:
\(E_{\text{total}} = E_p + E_k = -5.95 \times 10^{10} + 2.98 \times 10^{10} = -2.97 \times 10^{10}\text{ J}\) (or \(-2.98 \times 10^{10}\text{ J}\) depending on rounding).

**(d)**
- As the satellite loses energy due to drag, it moves to a lower orbit (radius \(r\) decreases).
- As the orbit radius decreases, the potential energy decreases (becomes more negative) and the kinetic energy increases, which means the orbital speed \(v\) actually increases (\(v = \sqrt{\frac{GM}{r}}\)).

**(a) [3 Marks]**
- 1 mark: Equating gravitational force to centripetal force (\frac{GMm}{r^2} = m\omega^2 r or equivalent).
- 1 mark: Correctly rearranging to get Keplers third law \(r^3 = \frac{GMT^2}{4\pi^2}\).
- 1 mark: Substitution of correct values to show \(r = 8.07 \times 10^6\text{ m}\).

**(b) [2 Marks]**
- 1 mark: Recall and substitution of \(E_p = -\frac{GMm}{r}\) (must include the negative sign).
- 1 mark: Correct value \(-5.95 \times 10^{10}\text{ J}\) (accept \(-5.9 \times 10^{10}\text{ J}\) to \(-6.0 \times 10^{10}\text{ J}\)).

**(c) [3 Marks]**
- 1 mark: Recall that \(E_k = \frac{GMm}{2r}\) or \(E_k = -\frac{1}{2} E_p\) and calculating \(2.98 \times 10^{10}\text{ J}\).
- 1 mark: Correct equation for total energy \(E = E_p + E_k\).
- 1 mark: Correct total energy value of \(-2.97 \times 10^{10}\text{ J}\) or \(-2.98 \times 10^{10}\text{ J}\) (must be negative).

**(d) [2 Marks]**
- 1 mark: Stating that orbit radius decreases (satellite spiralling inwards).
- 1 mark: Stating that orbital speed increases because kinetic energy increases as it falls.

**(a)**
(i) \(\%\Delta L = \frac{0.002}{1.240} \times 100\% = 0.16\%\)
(ii) \(\%\Delta d = \frac{0.01}{0.38} \times 100\% = 2.63\%\)
(iii) Area \(A = \frac{\pi d^2}{4}\). Since \(A \propto d^2\), the percentage uncertainty in \(A\) is twice that of \(d\):
\(\%\Delta A = 2 \times 2.63\% = 5.26\%\) (accept \(5.3\%\)).

**(b)**
\(A = \frac{\pi (0.38 \times 10^{-3})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\)
Resistivity \(\rho = \frac{R A}{L} = \frac{4.82 \times 1.134 \times 10^{-7}}{1.240} = 4.408 \times 10^{-7}\ \Omega\text{ m}\).

**(c)** The percentage uncertainty in \(\rho\) is the sum of percentage uncertainties of the terms in the formula:
\(\%\Delta \rho = \%\Delta R + \%\Delta A + \%\Delta L\)
\(\%\Delta R = \frac{0.05}{4.82} \times 100\% = 1.04\%\)
\(\%\Delta \rho = 1.04\% + 5.26\% + 0.16\% = 6.46\%\)
Absolute uncertainty in \(\rho\):
\(\Delta \rho = 4.408 \times 10^{-7} \times 0.0646 = 0.285 \times 10^{-7}\ \Omega\text{ m} \approx 0.3 \times 10^{-7}\ \Omega\text{ m}\) (rounded to 1 s.f.).
Therefore, the resistivity is \(\rho = (4.4 \pm 0.3) \times 10^{-7}\ \Omega\text{ m}\).

**(d)** To reduce percentage uncertainty in \(d\):
- Measure diameter at multiple positions and angles along the wire and compute the mean. This minimizes random errors.
- Or, use a wire of larger diameter (which increases the measured value \(d\) while keeping absolute uncertainty constant, reducing % uncertainty).

**(a) [3 Marks]**
- 1 mark: Correct % uncertainty for length: \(0.16\%\).
- 1 mark: Correct % uncertainty for diameter: \(2.6\%\) or \(2.63\%\).
- 1 mark: Correct % uncertainty for area: \(5.3\%\) or \(5.26\%\) (must be twice the % uncertainty in \(d\)).

**(b) [3 Marks]**
- 1 mark: Correct calculation of cross-sectional area \(A = 1.13 \times 10^{-7}\text{ m}^2\).
- 1 mark: Correct rearrangement of resistivity formula \(\rho = \frac{RA}{L}\).
- 1 mark: Correct value for resistivity \(\rho = 4.4 \times 10^{-7}\ \Omega\text{ m}\).

**(c) [3 Marks]**
- 1 mark: Correct addition of percentage uncertainties to find total % uncertainty \(\approx 6.5\%\).
- 1 mark: Correct calculation of absolute uncertainty \(\Delta \rho = 0.3 \times 10^{-7}\ \Omega\text{ m}\).
- 1 mark: Expressing final answer correctly with matching precision: \(\rho = (4.4 \pm 0.3) \times 10^{-7}\ \Omega\text{ m}\).

**(d) [1 Mark]**
- 1 mark: Suggesting taking multiple readings of diameter along the wire and finding the mean (or equivalent valid method).

**(a)** Electric potential energy lost = kinetic energy gained:
\(q V = \frac{1}{2} m v^2\)
\(v = \sqrt{\frac{2 q V}{m}}\)
\(v = \sqrt{\frac{2 \times (3.20 \times 10^{-19}) \times (120 \times 10^3)}{6.64 \times 10^{-27}}} = \sqrt{\frac{7.68 \times 10^{-14}}{6.64 \times 10^{-27}}} = \sqrt{1.1566 \times 10^{13}} = 3.401 \times 10^6\text{ m s}^{-1} \approx 3.4 \times 10^6\text{ m s}^{-1}\).

**(b)** The path is a circle (or circular arc).
Explanation: The magnetic force \(F = B q v\) is always perpendicular to the velocity of the particle. This constant-magnitude perpendicular force provides the centripetal acceleration, causing circular motion with constant speed.

**(c)** Equating magnetic force to centripetal force:
\(B q v = \frac{m v^2}{r} \Rightarrow r = \frac{m v}{B q}\)
\(r = \frac{(6.64 \times 10^{-27}) \times (3.40 \times 10^6)}{0.45 \times (3.20 \times 10^{-19})} = \frac{2.2576 \times 10^{-20}}{1.44 \times 10^{-19}} = 0.157\text{ m}\) (or \(0.16\text{ m}\)).

**(d)**
1. The beta-minus particle has an opposite charge, so it deflects in the opposite direction (e.g., curves clockwise instead of counter-clockwise).
2. The beta-minus particle has a much smaller mass than the alpha particle, so the radius of its path will be much smaller (\(r = \frac{mv}{Bq}\) is directly proportional to mass, which is approx. 7300 times smaller).

**(a) [3 Marks]**
- 1 mark: Equating work done \(qV\) to kinetic energy \(\frac{1}{2}mv^2\).
- 1 mark: Correct substitution of charge \(2e\) and voltage \(120\text{ kV}\).
- 1 mark: Correct calculation showing speed is \(3.4 \times 10^6\text{ m s}^{-1}\).

**(b) [2 Marks]**
- 1 mark: Identifies the path as circular / circular arc.
- 1 mark: Explains that the magnetic force is perpendicular to velocity, acting as centripetal force.

**(c) [3 Marks]**
- 1 mark: Recalls or derives \(r = \frac{mv}{Bq}\).
- 1 mark: Correct substitution of mass, charge, velocity, and field strength.
- 1 mark: Correct calculation of radius as \(0.16\text{ m}\) (or \(0.157\text{ m}\)).

**(d) [2 Marks]**
- 1 mark: Stating the opposite direction of curvature.
- 1 mark: Stating a much smaller radius of curvature (due to much smaller mass).

**(a)**
Initial charge: \(Q_0 = C V = (470 \times 10^{-6}\text{ F}) \times 12.0\text{ V} = 5.64 \times 10^{-3}\text{ C}\) (or \(5.6\text{ mC}\)).
Initial energy stored: \(E_0 = \frac{1}{2} C V^2 = 0.5 \times (470 \times 10^{-6}) \times (12.0)^2 = 0.0338\text{ J}\) (or \(33.8\text{ mJ}\)).

**(b)** The time constant \(\tau\) is:
\(\tau = R C = (15 \times 10^3\ \Omega) \times (470 \times 10^{-6}\text{ F}) = 7.05\text{ s} \approx 7.0\text{ s}\).

**(c)** During discharge:
\(V = V_0 e^{-t/RC}\)
At \(t = 10.0\text{ s}\):
\(V = 12.0 \times e^{-10.0 / 7.05} = 12.0 \times e^{-1.418} = 12.0 \times 0.242 = 2.91\text{ V}\).
The current \(I\) in the circuit is:
\(I = \frac{V}{R} = \frac{2.91}{15 \times 10^3\ \Omega} = 1.94 \times 10^{-4}\text{ A}\) (or \(0.19\text{ mA}\)).

**(d)**
- Sketch: The graph of Energy \(E\) against time \(t\) starts at \(E_0\) and decays exponentially towards zero. The curve is significantly steeper than the \(V-t\) curve.
- Explanation: Since energy \(E = \frac{1}{2}CV^2\), energy is proportional to the square of voltage. Thus, \(E = E_0 e^{-2t/RC}\). The decay constant for energy is twice that of voltage (or the time constant for energy decay is \(\frac{RC}{2} \approx 3.5\text{ s}\)), resulting in a much steeper decay.

**(a) [2 Marks]**
- 1 mark: Correct calculation of initial charge \(Q_0 = 5.6 \times 10^{-3}\text{ C}\).
- 1 mark: Correct calculation of energy stored \(E_0 = 3.4 \times 10^{-2}\text{ J}\) or \(33.8\text{ mJ}\).

**(b) [2 Marks]**
- 1 mark: State formula \(\tau = RC\).
- 1 mark: Substitute values and calculate \(7.05\text{ s}\) to show it is \(\approx 7.0\text{ s}\).

**(c) [4 Marks]**
- 1 mark: Correct formula for voltage decay \(V = V_0 e^{-t/RC}\).
- 1 mark: Correct calculation of voltage at \(10\text{ s}\) as \(2.91\text{ V}\).
- 1 mark: State and use Ohm's Law \(I = V/R\) or current decay equation \(I = I_0 e^{-t/RC}\).
- 1 mark: Correct calculation of current as \(1.94 \times 10^{-4}\text{ A}\) or \(0.19\text{ mA}\).

**(d) [2 Marks]**
- 1 mark: Correctly sketched exponential decay curve for energy starting on y-axis and asymptoting to zero.
- 1 mark: Explaining that \(E \propto V^2\), so it decays as \(e^{-2t/RC}\), which has a shorter time constant (\(\tau/2\)), making it steeper.

From the kinetic theory of gases, the pressure of an ideal gas is given by \(p = \frac{1}{3} \frac{N m c^2}{V}\). Rearranging this formula for the root-mean-square speed squared gives \(c^2 = \frac{3 p V}{N m}\). Since pressure \(p\), the total number of molecules \(N\), and the mass of a single molecule \(m\) remain constant, the mean square speed is directly proportional to the volume: \(c^2 \propto V\). Therefore, when the volume is reduced to \(0.75V\), the new mean square speed becomes \(0.75 c^2\). The new root-mean-square speed is \(c' = \sqrt{0.75 c^2} = \sqrt{0.75} c \approx 0.866 c\), which rounds to \(0.87 c\).

1 mark for selecting the correct option (C). Correct mathematical relationship established showing \(c^2 \propto V\) at constant pressure, followed by calculating the square root of the scale factor 0.75.

Let the initial activity of isotope Y be \(A_Y(0) = A_0\). The initial activity of isotope X is \(A_X(0) = 2 A_0\). The activities at time \(t\) are given by \(A_X(t) = 2 A_0 \cdot 2^{-t/3.0}\) and \(A_Y(t) = A_0 \cdot 2^{-t/6.0}\). Equating the two activities: \(2 A_0 \cdot 2^{-t/3.0} = A_0 \cdot 2^{-t/6.0}\). Dividing both sides by \(A_0\) gives \(2^1 \cdot 2^{-t/3.0} = 2^{-t/6.0}\). Combining exponents on the left hand side: \(2^{1 - t/3.0} = 2^{-t/6.0}\). Equating the powers yields: \(1 - \frac{t}{3.0} = -\frac{t}{6.0}\). Solving for \(t\): \(1 = \frac{t}{3.0} - \frac{t}{6.0} = \frac{t}{6.0}\), which gives \(t = 6.0\text{ hours}\).

1 mark for selecting the correct option (B). Correctly equating the activity decay equations for both isotopes and solving the exponential equation for the time variable.

The total energy in simple harmonic motion is constant and is given by \(E_{\text{total}} = \frac{1}{2} k A^2\). The potential energy at displacement \(x\) is \(E_p = \frac{1}{2} k x^2\). Substituting \(x = \frac{1}{3} A\), we get \(E_p = \frac{1}{2} k \left(\frac{A}{3}\right)^2 = \frac{1}{9} \left(\frac{1}{2} k A^2\right) = \frac{1}{9} E_{\text{total}}\). The kinetic energy of the system is \(E_k = E_{\text{total}} - E_p = E_{\text{total}} - \frac{1}{9} E_{\text{total}} = \frac{8}{9} E_{\text{total}}\). The ratio of kinetic energy to potential energy is \(\frac{E_k}{E_p} = \frac{\frac{8}{9} E_{\text{total}}}{\frac{1}{9} E_{\text{total}}} = 8\).

1 mark for selecting the correct option (C). Accurate application of SHM energy formulas to determine the fraction of total energy represented by potential and kinetic energy, followed by calculating their ratio.

For a satellite in a circular orbit of radius \(r\), the gravitational potential energy is \(E_p = -\frac{GMm}{r}\) and its kinetic energy is \(E_k = \frac{GMm}{2r}\). The total mechanical energy is \(E = E_k + E_p = -\frac{GMm}{2r}\). In the new orbit of radius \(2r\), the total mechanical energy is \(E' = -\frac{GMm}{2(2r)} = -\frac{GMm}{4r}\). The change in the total mechanical energy is \(\Delta E = E' - E = -\frac{GMm}{4r} - \left(-\frac{GMm}{2r}\right) = +\frac{GMm}{4r}\).

1 mark for selecting the correct option (B). Identifying the correct expression for total orbital energy as a function of orbital radius, and evaluating the final minus initial energy difference.

The initial energy stored in the capacitor is \(E_i = \frac{1}{2} C V^2\). The charge on the capacitor is \(Q = C V\). When connected in parallel with an uncharged capacitor of capacitance \(2C\), the total capacitance of the combination is \(C_{\text{total}} = C + 2C = 3C\). Since the total charge is conserved, the final total energy stored in the two capacitors is \(E_f = \frac{Q^2}{2 C_{\text{total}}} = \frac{(CV)^2}{2(3C)} = \frac{1}{6} C V^2\). The energy dissipated is \(\Delta E = E_i - E_f = \frac{1}{2} C V^2 - \frac{1}{6} C V^2 = \frac{1}{3} C V^2\). The fraction of energy dissipated is \(\frac{\Delta E}{E_i} = \frac{\frac{1}{3} C V^2}{\frac{1}{2} C V^2} = \frac{2}{3}\).

1 mark for selecting the correct option (C). Correct calculation of initial energy and final system energy using charge conservation, followed by finding the ratio of lost energy to initial energy.

The radius of a circular path in a magnetic field is given by \(R = \frac{m v}{B q}\). Since kinetic energy is \(E_k = \frac{1}{2} m v^2\), the momentum can be written as \(m v = \sqrt{2 m E_k}\), which gives \(R = \frac{\sqrt{2 m E_k}}{B q}\). For the first particle, \(R_1 = \frac{\sqrt{2 m E_k}}{B q} = R\). For the second particle, the mass is \(2m\), the charge is \(2q\), and the kinetic energy is \(2E_k\). The new radius is \(R_2 = \frac{\sqrt{2 (2m) (2E_k)}}{B (2q)} = \frac{\sqrt{8 m E_k}}{2 B q} = \frac{2 \sqrt{2 m E_k}}{2 B q} = \frac{\sqrt{2 m E_k}}{B q} = R\). Thus, the radius of the path remains unchanged.

1 mark for selecting the correct option (B). Expressing path radius in terms of kinetic energy and showing that doubling the mass, doubling the charge, and doubling the kinetic energy leaves the radius unchanged.

For the net electric field to be zero, the individual fields from the two charges must be equal in magnitude and opposite in direction. Because the charges have opposite signs, the field vectors only point in opposite directions outside the region between the charges. Since \(|+4Q| > |-Q|\), the zero-field point must lie closer to the smaller charge, which means \(x > d\). Let the point be at a distance \(r\) to the right of the charge \(-Q\), so its position is \(x = d + r\). The condition for equal field magnitudes is: \(\frac{4Q}{4\pi\varepsilon_0 (d+r)^2} = \frac{Q}{4\pi\varepsilon_0 r^2}\). Simplifying this gives \(\frac{4}{(d+r)^2} = \frac{1}{r^2}\). Taking the square root of both sides gives \(\frac{2}{d+r} = \frac{1}{r}\). Multiplying out gives \(2r = d + r \implies r = d\). Therefore, the position is \(x = d + d = 2d\).

1 mark for selecting the correct option (C). Establishing the correct region for zero electric field and solving the inverse square relation to determine the distance from the origin.

The percentage uncertainty in each measurement is: For resistance \(R\): \(\frac{0.1}{4.0} \times 100\% = 2.5\%\). For diameter \(d\): \(\frac{0.01}{0.50} \times 100\% = 2.0\%\). For length \(L\): \(\frac{0.02}{2.00} \times 100\% = 1.0\%\). The formula for resistivity is \(\rho = \frac{R \pi d^2}{4 L}\). The total percentage uncertainty is the sum of the percentage uncertainties of each term, with the percentage uncertainty of \(d\) multiplied by 2 because it is raised to the second power: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L} = 2.5\% + 2(2.0\%) + 1.0\% = 7.5\%\).

1 mark for selecting the correct option (B). Calculating the individual percentage uncertainties correctly and adding them with the appropriate weighting factor of 2 for the squared diameter term.

First, the mean of the raw measurements is calculated: \(\text{mean raw} = \frac{0.42 + 0.44 + 0.41 + 0.43 + 0.45}{5} = 0.43\text{ mm}\). Next, the zero error of \(+0.02\text{ mm}\) is subtracted to find the corrected mean: \(\text{corrected mean} = 0.43\text{ mm} - 0.02\text{ mm} = 0.41\text{ mm}\). The absolute uncertainty is estimated as half the range of the repeat readings: \(\text{uncertainty} = \frac{0.45 - 0.41}{2} = 0.02\text{ mm}\). Finally, the percentage uncertainty is: \(\frac{0.02\text{ mm}}{0.41\text{ mm}} \times 100\% \approx 4.88\%\), which rounds to \(4.9\%\).

1 mark for the correct answer C. Method involves subtracting the zero error to find the corrected mean, determining the absolute uncertainty as half the range, and correctly calculating the percentage uncertainty.

From the ideal gas equation, \(pV = N k_B T\), meaning the absolute temperature \(T\) is proportional to the product \(pV\). Since \(p_2 = 3 p_1\) and \(V_2 = 0.5 V_1\), the final temperature is \(T_2 \propto (3 p_1)(0.5 V_1) = 1.5 p_1 V_1\), so \(T_2 = 1.5 T_1\). The mean kinetic energy of the gas molecules is proportional to the temperature, meaning that \(\frac{1}{2} m c^2 \propto T\), hence the rms speed \(c \propto \sqrt{T}\). The final rms speed is therefore \(c_2 = c \sqrt{1.5} \approx 1.22 c\).

1 mark for the correct answer B. Correctly identifying that absolute temperature increases by a factor of 1.5 and that the rms speed is proportional to the square root of temperature.

Using the radioactive decay law \(N = N_0 e^{-\lambda t}\), the number of parent nuclei remaining at \(t = \frac{1}{\lambda}\) is \(N_1 = N_0 e^{-\lambda (1/\lambda)} = N_0 e^{-1}\). The number of parent nuclei remaining at \(t = \frac{2}{\lambda}\) is \(N_2 = N_0 e^{-\lambda (2/\lambda)} = N_0 e^{-2}\). The number of daughter nuclei formed in this time interval is equal to the number of parent nuclei that decayed: \(\Delta N = N_1 - N_2 = N_0 (e^{-1} - e^{-2}) \approx N_0 (0.368 - 0.135) = 0.233 N_0 \approx 0.23 N_0\).

1 mark for the correct answer B. Method involves using the decay equation to find the parent populations at both times and calculating the difference to determine the number of decayed nuclei.

Initially, the total external resistance is \(R_{\text{ext}} = 3R\) and the total circuit resistance is \(3.5R\). The current is \(I_1 = \frac{\varepsilon}{3.5R}\), giving a power dissipation of \(P_1 = I_1^2 R_{\text{ext}} = \left(\frac{\varepsilon}{3.5R}\right)^2 (3R) = \frac{3}{12.25}\frac{\varepsilon^2}{R}\). After one resistor is short-circuited, the external resistance is \(R_{\text{ext}}' = 2R\) and the total circuit resistance is \(2.5R\). The new current is \(I_2 = \frac{\varepsilon}{2.5R}\), giving a power dissipation of \(P_2 = I_2^2 R_{\text{ext}}' = \left(\frac{\varepsilon}{2.5R}\right)^2 (2R) = \frac{2}{6.25}\frac{\varepsilon^2}{R}\). The ratio is \(\frac{P_2}{P_1} = \frac{2/6.25}{3/12.25} = \frac{0.32}{0.2449} \approx 1.31\).

1 mark for the correct answer C. Method involves formulating expressions for external power in both configurations using total loop resistance including internal resistance, then dividing the values.

The gravitational potential energy is given by \(E_p = -\frac{GMm}{r}\). Thus, \(\Delta E_p = -\frac{GMm}{2R} - \left(-\frac{GMm}{R}\right) = \frac{GMm}{2R}\). The kinetic energy of a satellite in a circular orbit is given by \(E_k = \frac{GMm}{2r}\). Thus, \(\Delta E_k = \frac{GMm}{4R} - \frac{GMm}{2R} = -\frac{GMm}{4R}\). Comparing the two changes, we find the ratio is \(\frac{\Delta E_p}{\Delta E_k} = \frac{GMm / (2R)}{-GMm / (4R)} = -2\).

1 mark for the correct answer A. Recognition of standard orbital mechanical energy formulas and finding the ratio of their differences.

The total work done by the force \(F\) is \(W = F d\). According to the conservation of energy, this work equals the gain in kinetic energy \(E_k\) plus the work done against resistive forces \(W_r\), so \(W_r = F d - E_k\). For uniform acceleration from rest, the final velocity is \(v = \frac{2d}{t}\). This means the final kinetic energy is \(E_k = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{2d}{t}\right)^2 = \frac{2 m d^2}{t^2}\). Thus, the work done against resistive forces is \(W_r = F d - \frac{2 m d^2}{t^2}\).

1 mark for the correct answer C. Method requires using SUVAT to link distance, time, and final velocity to find kinetic energy, and then subtracting kinetic energy from the work done by the force.

The initial magnetic flux linkage through the coil is \(\Phi_1 = N B A = N B L^2\) because the magnetic field is perpendicular to the plane of the coil. After rotating by \(90^\circ\), the plane of the coil is parallel to the magnetic field lines, so the flux linkage is \(\Phi_2 = 0\). The change in magnetic flux linkage is \(\Delta \Phi = N B L^2\). According to Faraday's law, the average induced emf is \(\varepsilon = \frac{\Delta \Phi}{\Delta t} = \frac{N B L^2}{\Delta t}\).

1 mark for the correct answer C. Method involves identifying the initial flux linkage, identifying the final flux linkage as zero, and using Faraday's law to get average emf.

The initial charge on the first capacitor is \(Q = C V\), and its initial energy is \(E_i = \frac{Q^2}{2C}\). When connected to the uncharged capacitor, charge is shared such that total charge remains \(Q\). The total capacitance of the parallel combination is \(C_{\text{total}} = C + 2C = 3C\). The final stored energy is \(E_f = \frac{Q^2}{2(3C)} = \frac{Q^2}{6C}\). The energy dissipated is \(\Delta E = E_i - E_f = \frac{Q^2}{2C} - \frac{Q^2}{6C} = \frac{Q^2}{3C}\). The fraction of initial energy dissipated is \(\frac{\Delta E}{E_i} = \frac{Q^2 / (3C)}{Q^2 / (2C)} = \frac{2}{3}\).

1 mark for the correct answer C. Method involves using charge conservation, calculating final equivalent capacitance, comparing initial and final electrostatic energies, and finding the difference as a fraction.

The root-mean-square speed of an ideal gas molecule is given by \(c_{\text{rms}} = \sqrt{\frac{3kT}{m}}\), which means \(c_{\text{rms}} \propto \sqrt{T}\) where \(T\) is the absolute temperature in kelvin. Converting the temperatures: \(T_1 = 27 + 273.15 = 300.15\text{ K}\) and \(T_2 = 327 + 273.15 = 600.15\text{ K}\). The ratio is therefore \(\frac{v_2}{v_1} = \sqrt{\frac{600.15}{300.15}} = \sqrt{2} \approx 1.41\).

1 mark for the correct option B.

At \(t = T_{1/2}\), the number of remaining active nuclei is \(N(T_{1/2}) = 0.5N_0\). At \(t = 3T_{1/2}\), the number of remaining active nuclei is \(N(3T_{1/2}) = N_0 \left(\frac{1}{2}\right)^3 = 0.125N_0\). The number of nuclei that decay in this time interval is the difference: \(0.5N_0 - 0.125N_0 = 0.375N_0\). Thus, the fraction of the initial nuclei that decay is \(0.375\).

1 mark for the correct option C.

For a satellite of mass \(m\) in a circular orbit of radius \(r\) around a planet of mass \(M\), equating gravitational force and centripetal force gives \(\frac{GMm}{r^2} = \frac{mv^2}{r}\), which simplifies to \(v = \sqrt{\frac{GM}{r}}\). Therefore, the orbital speed is inversely proportional to the square root of the orbital radius: \(v \propto \frac{1}{\sqrt{r}}\). Since \(r_{\text{Q}} = 2r_{\text{P}}\), we have \(v_{\text{Q}} = \frac{v}{\sqrt{2}} \approx 0.71 v\).

1 mark for the correct option C.

The kinetic energy gained by a charged particle is \(E_k = qV = \frac{p^2}{2m}\), so its momentum is \(p = \sqrt{2mqV}\). The radius of the path in the magnetic field is \(r = \frac{p}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}\). Thus, \(r \propto \sqrt{\frac{m}{q}}\). For a proton, \(m_p = m\) and \(q_p = e\). For an \(\alpha\) particle, \(m_{\alpha} = 4m\) and \(q_{\alpha} = 2e\). The ratio is \(\frac{r_{\alpha}}{r_p} = \sqrt{\frac{m_{\alpha}}{q_{\alpha}} \cdot \frac{q_p}{m_p}} = \sqrt{\frac{4}{2} \cdot \frac{1}{1}} = \sqrt{2} \approx 1.41\).

1 mark for the correct option B.

The initial energy stored is \(E_i = \frac{1}{2}CV^2\) and the initial charge is \(Q_i = CV\). When connected in parallel to the \(3C\) capacitor, the total capacitance becomes \(C_{\text{total}} = C + 3C = 4C\). By conservation of charge, the total charge is still \(Q_f = CV\). The final potential difference is \(V_f = \frac{Q_f}{C_{\text{total}}} = \frac{V}{4\). The final energy stored in the system is \(E_f = \frac{1}{2}(4C)\left(\frac{V}{4}\right)^2 = \frac{1}{8}CV^2\). The fraction of the initial energy remaining is \(\frac{E_f}{E_i} = \frac{1/8}{1/2} = 0.25\). Thus, the fraction of energy lost is \(1 - 0.25 = 0.75\).

1 mark for the correct option D.

Let \(x\) be the distance in meters from the \(+2.0\,\mu\text{C}\) charge. The electric potential is \(V = \frac{1}{4\pi\varepsilon_0} \left( \frac{q_1}{x} + \frac{q_2}{0.12 - x} \right)\). Setting \(V = 0\) gives \(\frac{2.0 \times 10^{-6}}{x} + \frac{-8.0 \times 10^{-6}}{0.12 - x} = 0\), which simplifies to \(\frac{2}{x} = \frac{8}{0.12 - x}\). Solving for \(x\) yields \(2(0.12 - x) = 8x \implies 0.24 = 10x \implies x = 0.024\text{ m} = 2.4\text{ cm}\).

1 mark for the correct option B.

The displacement of the mass is described by \(x(t) = A \cos(\omega t)\), starting at \(x = +A\) at \(t = 0\). We seek the minimum time \(t\) such that \(A \cos(\omega t) = -A/2 \implies \cos(\omega t) = -1/2\). The smallest positive angle satisfying this is \(\omega t = \frac{2\pi}{3}\). Since \(\omega = \frac{2\pi}{T}\), we have \(\frac{2\pi}{T} t = \frac{2\pi}{3}\), which yields \(t = \frac{T}{3}\).

1 mark for the correct option C.

The percentage uncertainty in the resistivity is found by adding the individual percentage uncertainties, doubling the contribution of diameter because it is raised to the power of 2: \(\%\Delta \rho = \%\Delta R + 2(\%\Delta d) + \%\Delta L = \left(\frac{0.05}{4.20}\right)\times 100\% + 2\left(\frac{0.01}{0.38}\right)\times 100\% + \left(\frac{0.002}{0.824}\right)\times 100\% \approx 1.19\% + 5.26\% + 0.24\% = 6.69\% \approx 6.7\%\).

1 mark for the correct option D.

The initial energy stored in the capacitor of capacitance \(C\) with charge \(Q\) is given by \(E_i = \frac{Q^2}{2C}\). When the capacitor is disconnected from the power supply and connected in parallel to an uncharged capacitor of capacitance \(3C\), the total charge \(Q\) is conserved and distributed across both capacitors. The combined capacitance of the parallel combination is \(C_{\text{total}} = C + 3C = 4C\). The final total energy stored in the system is \(E_f = \frac{Q^2}{2C_{\text{total}}} = \frac{Q^2}{2(4C)} = \frac{Q^2}{8C}\). The ratio of the final total energy to the initial energy is therefore \(\frac{E_f}{E_i} = \frac{Q^2 / 8C}{Q^2 / 2C} = \frac{2}{8} = 0.25\).

B is the correct answer (1 mark). Award 1 mark for the correct identification of the conservation of charge, the parallel capacitance calculation, and the resulting energy ratio of 0.25.

(a) Measuring 20 oscillations reduces the impact of human reaction time on the final calculated time period, thereby reducing the percentage uncertainty in the measurement. It also averages out minor random fluctuations in timing.

(b) Mean time for 20 oscillations:
\(t_{\text{mean}} = \frac{28.2 + 28.5 + 28.1}{3} = 28.27\text{ s}\)
Mean time period for one oscillation:
\(T = \frac{28.27}{20} = 1.413\text{ s}\) (accept \(1.41\text{ s}\))
Range of the 20-oscillation times: \(28.5 - 28.1 = 0.4\text{ s}\)
Absolute uncertainty in time for 20 oscillations: \(\Delta t = \frac{\text{range}}{2} = \frac{0.4}{2} = 0.2\text{ s}\)
Absolute uncertainty in the mean time period \(T\):
\(\Delta T = \frac{0.2}{20} = 0.01\text{ s}\)

(c) A plot of \(T^2\) on the vertical y-axis against \(L\) on the horizontal x-axis yields a straight line through the origin. Since \(T^2 = \left(\frac{4\pi^2}{g}\right)L\), the gradient \(m\) of this line is given by \(m = \frac{4\pi^2}{g}\). Therefore, \(g\) can be determined from the gradient using: \(g = \frac{4\pi^2}{m}\).

(d) Percentage uncertainty in \(T\):
\(\frac{\Delta T}{T} \times 100\% = \frac{0.01}{1.413} \times 100\% \approx 0.708\%\)
Since \(T^2\) involves raising \(T\) to the power of 2, the percentage uncertainty in \(T^2\) is twice the percentage uncertainty in \(T\):
Percentage uncertainty in \(T^2\) = \(2 \times 0.708\% = 1.416\%\) (accept \(1.4\%\) to \(1.42\%\)).

Part (a): [2 marks]
- 1 mark for stating that timing multiple oscillations reduces the percentage uncertainty arising from human reaction time.
- 1 mark for stating that it helps average out random timing errors.

Part (b): [3 marks]
- 1 mark for finding the mean time period \(T = 1.41\text{ s}\) (or \(1.413\text{ s}\)).
- 1 mark for calculating the range of 20-oscillation times as \(0.4\text{ s}\) or the absolute uncertainty in 20 oscillations as \(\pm 0.2\text{ s}\).
- 1 mark for dividing the timing uncertainty by 20 to obtain \(\Delta T = \pm 0.01\text{ s}\).

Part (c): [2 marks]
- 1 mark for identifying that the gradient \(m\) of the plot of \(T^2\) against \(L\) is equal to \(\frac{4\pi^2}{g}\).
- 1 mark for explicitly stating how to compute \(g\): \(g = \frac{4\pi^2}{m}\).

Part (d): [4.25 marks]
- 1 mark for calculating the percentage uncertainty in \(T\) (approx. \(0.71\%\)).
- 2 marks for stating that the percentage uncertainty in \(T^2\) is twice the percentage uncertainty in \(T\).
- 1.25 marks for calculating the correct final percentage uncertainty of \(1.4\%\) (accept \(1.4\%\) to \(1.42\%\)).

(a) An electrical heater is placed inside a central hole in the aluminum block and connected in series with a low-voltage DC power supply, an ammeter, and a variable resistor. A voltmeter is connected in parallel directly across the terminals of the heater to monitor the voltage accurately. This setup allows the student to measure current \(I\) and potential difference \(V\) to calculate the electrical power delivered, \(P = VI\).

(b) Sources of heat loss to the surroundings:
1. Convection and radiation from the exposed outer surfaces of the aluminum block to the surrounding air.
2. Conduction from the base of the block to the table or bench surface.
Methods to minimize heat loss:
- Wrap the outer surface of the block with a high-quality thermal insulator (such as cotton wool or bubble wrap).
- Place the block on a poor conductor, such as a polystyrene tile, to reduce heat conduction to the bench.
- Add a small drop of thermal paste or oil in the thermometer hole to improve thermal contact and prevent air gaps.

(c) The electrical energy supplied is \(E = V I t\).
Assuming no heat is lost, this energy is completely absorbed by the aluminum block: \(Q = m c \Delta \theta\).
Therefore:
\(V I t = m c \Delta \theta \implies c = \frac{V I t}{m \Delta \theta}\)
\(c = \frac{12.0 \times 4.10 \times 300}{1.00 \times 12.4} = \frac{14760}{12.4} \approx 1190\text{ J kg}^{-1}\,^{\circ}\text{C}^{-1}\) (or \(\text{J kg}^{-1}\text{ K}^{-1}\)).

(d) The fractional uncertainty in \(c\) is the sum of the fractional uncertainties of the constituent independent measurements:
\(\frac{\Delta c}{c} = \frac{\Delta V}{V} + \frac{\Delta I}{I} + \frac{\Delta t}{t} + \frac{\Delta m}{m} + \frac{\Delta (\Delta \theta)}{\Delta \theta}\)
Calculating each individual percentage uncertainty:
- \(\%\Delta V = \frac{0.2}{12.0} \times 100\% \approx 1.67\%\)
- \(\%\Delta I = \frac{0.05}{4.10} \times 100\% \approx 1.22\%\)
- \(\%\Delta t = \frac{1}{300} \times 100\% \approx 0.33\%\)
- \(\%\Delta m = \frac{0.01}{1.00} \times 100\% = 1.00\%\)
- \(\%\Delta (\Delta \theta) = \frac{0.5}{12.4} \times 100\% \approx 4.03\%\)

Sum of percentage uncertainties:
\(\%\Delta c = 1.67\% + 1.22\% + 0.33\% + 1.00\% + 4.03\% = 8.25\% \approx 8.3\%\) (accept \(8.2\%\) to \(8.3\%\)).

Part (a): [3 marks]
- 1 mark for stating that the heater is connected in series with the power supply and an ammeter.
- 1 mark for stating that a voltmeter is connected in parallel across the heater terminals.
- 1 mark for mentioning the use of a variable resistor (or variable voltage supply) to keep current/voltage stable during the heating time.

Part (b): [4 marks]
- 1 mark each for identifying any two separate sources of heat loss (e.g., surface convection, conduction to support, heat capacity of the heater itself, air gaps around thermometer).
- 2 marks for proposing a single valid, practical minimization method with brief reasoning (e.g., 1 mark for wrapping with insulation, 1 mark for explaining that it slows down heat transfer to the air).

Part (c): [2 marks]
- 1 mark for rearranging the energy balance equation to make \(c\) the subject: \(c = \frac{VIt}{m\Delta\theta}\) (or showing correct substitution).
- 1 mark for final calculated specific heat capacity value of \(1190\text{ J kg}^{-1}\text{ K}^{-1}\) (accept range \(1190\text{ to } 1200\text{ J kg}^{-1}\text{ K}^{-1}\)).

Part (d): [2.25 marks]
- 1 mark for showing that the percentage uncertainties of all five independent variables must be added together.
- 1.25 marks for the correct summation, leading to \(8.3\%\) (accept range \(8.2\%\) to \(8.3\%\)).

(a) The student measures the potential difference \(V\) across the capacitor at regular time intervals \(t\) as it discharges.
Starting with the discharge equation:
\(V = V_0 e^{-t/RC}\)
Taking natural logarithms of both sides:
\(\ln V = \ln V_0 - \frac{1}{RC} t\)
This is of the form \(y = mx + c\). The student plots \(\ln(V/\text{V})\) on the vertical y-axis against time \(t\) on the horizontal x-axis.
The resulting graph is a straight line with a negative gradient. Since the gradient \(m = -\frac{1}{RC} = -\frac{1}{\tau}\), the experimental time constant is determined as:
\(\tau = -\frac{1}{\text{gradient}}\).

(b) Nominal theoretical time constant:
\(\tau = RC = (150 \times 10^3\text{ }\Omega) \times (470 \times 10^{-6}\text{ F}) = 70.5\text{ s}\).

(c) Using the discharge equation:
\(V = V_0 e^{-t/\tau}\)
\(3.30 = 9.00 e^{-70.0/\tau}\)
\(\frac{3.30}{9.00} = e^{-70.0/\tau}\)
Taking natural logarithms on both sides:
\(\ln\left(\frac{3.30}{9.00}\right) = -\frac{70.0}{\tau}\)
\(-1.0033 = -\frac{70.0}{\tau}\)
\(\tau = \frac{70.0}{1.0033} \approx 69.8\text{ s}\).

(d) If the discharge happens too quickly for manual measurements with a stopwatch:
1. Use a storage oscilloscope connected in parallel with the capacitor. Set the trigger to single-sweep mode to capture the voltage decay curve, allowing coordinates to be read from the frozen display.
2. Alternatively, connect a digital data logger equipped with a high-sampling-frequency voltage sensor to a computer. The data logging software can capture multiple readings per millisecond and automatically generate the \(V\) vs \(t\) curve.

Part (a): [4 marks]
- 1 mark for stating that potential difference \(V\) is measured against time \(t\).
- 1 mark for showing the correct derivation of the logarithmic relationship: \(\ln V = \ln V_0 - \frac{t}{RC}\).
- 1 mark for specifying that \(\ln(V)\) is plotted on the y-axis and \(t\) on the x-axis.
- 1 mark for stating that the time constant \(\tau = -\frac{1}{\text{gradient}}\).

Part (b): [1.5 marks]
- 1 mark for substitute values of \(R\) and \(C\) correctly with correct power-of-ten multipliers.
- 0.5 marks for correct final calculation of \(70.5\text{ s}\) (accept \(71\text{ s}\)).

Part (c): [2.75 marks]
- 1 mark for setting up the decay equation: \(3.30 = 9.00 e^{-70.0/\tau}\).
- 1 mark for taking logs correctly to obtain \(\ln(0.367) = -70.0/\tau\) (or equivalent form).
- 0.75 marks for final calculated experimental value of \(\tau = 69.8\text{ s}\) (accept \(69.7\text{ s}\) to \(70.0\text{ s}\)).

Part (d): [3 marks]
- 1 mark for suggesting a storage oscilloscope OR a digital data logger with a voltage sensor.
- 1 mark for explaining the setup to capture fast-changing signals (e.g., using a high sampling rate or setting a single-trigger sweep).
- 1 mark for describing how the raw data is extracted from the instrument (e.g., reading coordinates from frozen grid, or transferring plotted values to software).

(a) The circuit diagram should contain the following:
- A single loop containing the cell (represented as a cell symbol, optionally with a series resistor labeled \(r\) enclosed in a dashed box), an ammeter, a switch, and a variable resistor connected in series.
- A voltmeter connected in parallel across the terminals of the cell (or, equivalently, across the outer combination of the variable resistor and ammeter).

(b) By rearranging \(V = \varepsilon - Ir\) into the form \(y = mx + c\):
\(V = -r I + \varepsilon\)
The student should plot terminal potential difference \(V\) on the vertical y-axis against current \(I\) on the horizontal x-axis.
- The y-intercept of the resulting straight-line graph corresponds directly to the electromotive force \(\varepsilon\).
- The gradient of the graph is equal to \(-r\), meaning the internal resistance is \(r = -\text{gradient}\).

(c) Matching the experimental line equation \(V = -1.45 I + 1.54\) with the linear model \(V = -r I + \varepsilon\):
- The y-intercept represents the emf: \(\varepsilon = 1.54\text{ V}\).
- The negative of the gradient represents the internal resistance: \(r = 1.45\text{ }\Omega\).

(d) If the cell warms up, its chemical reaction rates and physical properties change, causing its internal resistance \(r\) to increase. Consequently, the relationship between \(V\) and \(I\) becomes non-linear as \(r\) changes across different current ranges, causing scatter in the graph and making the line-of-best-fit less reliable.
Practical precaution:
- Keep the switch open between individual trials and only close it briefly to note down the steady ammeter and voltmeter readings. Open the switch immediately after each reading to prevent continuous discharge and heating.

Part (a): [3 marks]
- 1 mark for drawing a cell, an ammeter, a switch, and a variable resistor in a single series loop with standard symbols.
- 1 mark for connecting the voltmeter in parallel across either the cell or the variable resistor.
- 1 mark for utilizing completely correct, standard electrical symbols with no gaps or short-circuits.

Part (b): [3 marks]
- 1 mark for specifying that \(V\) is plotted on the vertical axis and \(I\) on the horizontal axis.
- 1 mark for stating that the y-intercept of the straight line equals the emf \(\varepsilon\).
- 1 mark for stating that the internal resistance \(r\) is found from the magnitude of the gradient (or \(r = -\text{gradient}\)).

Part (c): [2.25 marks]
- 1 mark for \(\varepsilon = 1.54\text{ V}\) (both numeric value and unit needed).
- 1.25 marks for \(r = 1.45\text{ }\Omega\) (both numeric value and unit needed).

Part (d): [3 marks]
- 1 mark for explaining that temperature changes alter internal resistance, causing it to vary rather than remain constant (resulting in non-linear data points).
- 2 marks for stating a clear practical precaution: 1 mark for opening the switch between readings to prevent heating; 1 mark for explaining that this limits continuous current flow.