2024 Edexcel A-Level Physics (9PH0) Practice Paper with Answers

The total resistance in the circuit is \(R + r\). The current is given by \(I = \frac{V_0}{R + r}\). As \(R\) decreases towards zero, the total resistance decreases, so the current \(I\) increases. The terminal potential difference is given by \(V = V_0 - Ir\). Since \(I\) increases, the lost volts \(Ir\) increase, causing the terminal potential difference \(V\) to decrease. Therefore, \(V\) decreases and \(I\) increases.

1 mark for identifying the correct relationship (Option A) where terminal potential difference decreases and current increases as external resistance decreases.

At the highest point, the net downward force provides the required centripetal acceleration: \(mg - N = \frac{m v^2}{r}\). Rearranging this equation for the normal contact force \(N\) yields \(N = mg - \frac{m v^2}{r}\).

1 mark for identifying that the centripetal force is the net force pointing towards the center of curvature, giving \(N = mg - \frac{mv^2}{r}\) (Option C).

The magnetic force provides the centripetal force: \(B e v = \frac{m_e v^2}{R}\). Solving for velocity gives \(v = \frac{B e R}{m_e}\). The kinetic energy is \(E_k = \frac{1}{2} m_e v^2 = \frac{1}{2} m_e \left(\frac{B e R}{m_e}\right)^2 = \frac{B^2 e^2 R^2}{2 m_e}\).

1 mark for equating magnetic force to centripetal force and substituting into the kinetic energy equation to find the correct expression (Option B).

A neutron has a quark composition of \(udd\). A proton has a quark composition of \(uud\). Therefore, during beta-minus decay, one of the down (\(d\)) quarks in the neutron changes into an up (\(u\)) quark: \(d \to u\).

1 mark for choosing Option B, corresponding to a down quark changing into an up quark.

The horizontal component of the velocity, \(u_x = u \cos\theta\), remains constant throughout the flight. The vertical component of velocity is zero at the highest point of the trajectory. Therefore, the minimum velocity is \(u \cos\theta\). The minimum kinetic energy is \(\frac{1}{2} m (u \cos\theta)^2 = \left(\frac{1}{2} m u^2\right) \cos^2\theta = E_k \cos^2\theta\).

1 mark for identifying that the minimum velocity occurs at the peak where only the horizontal component remains, resulting in Option C.

The equation for capacitor discharge is \(Q = Q_0 e^{-t/RC}\). Substituting \(t = 2RC\) gives \(Q = Q_0 e^{-2RC/RC} = Q_0 e^{-2}\). The fraction of the initial charge remaining is \(\frac{Q}{Q_0} = e^{-2}\).

1 mark for substituting the time of 2 time constants into the exponential decay formula to find Option A.

The useful work output is \(W = mgh = 250 \times 9.81 \times 12 = 29,430\text{ J}\). The useful power output is \(P_{\text{out}} = \frac{29430}{8.0} = 3678.75\text{ W}\). Given that the efficiency is \(60\%\), the input power is \(P_{\text{in}} = \frac{P_{\text{out}}}{0.60} = \frac{3678.75}{0.60} = 6131.25\text{ W} \approx 6.1\text{ kW}\).

1 mark for correctly calculating the useful work output, power output, and applying the efficiency formula to obtain the input power of 6.1 kW (Option C).

The resistance is given by \(R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}\). For wire X, \(R_X = \rho \frac{L}{\pi r^2}\). For wire Y, \(R_Y = \rho \frac{2L}{\pi (2r)^2} = \rho \frac{2L}{4\pi r^2} = \frac{1}{2} \rho \frac{L}{\pi r^2} = \frac{1}{2} R_X\). Therefore, \(\frac{R_X}{R_Y} = 2\), which is equivalent to a ratio of \(2:1\).

1 mark for calculating the correct ratio of resistances based on length and cross-sectional area variations to obtain 2 : 1 (Option C).

The charge \(Q\) on a discharging capacitor at time \(t\) is given by \(Q = Q_0 e^{-\frac{t}{\tau}}\), where \(\tau = RC\). If the charge falls to \(\frac{1}{e^2}\) of its initial value, then: \(Q_0 e^{-\frac{t}{\tau}} = Q_0 e^{-2} \implies \tau = \frac{t}{2}\). The energy stored in the capacitor is given by \(E = \frac{Q^2}{2C}\). Thus, the energy at time \(t'\) is: \(E = E_0 e^{-\frac{2t'}{\tau}}\). For the energy to fall to \(\frac{1}{e^2}\) of its initial value: \(E_0 e^{-\frac{2t'}{\tau}} = E_0 e^{-2} \implies \frac{2t'}{\tau} = 2 \implies t' = \tau\). Substituting \(\tau = \frac{t}{2}\) gives \(t' = \frac{t}{2}\).

1 mark for the correct option (B). Award 1 mark for showing that the time constant \(\tau = t/2\) and that the time taken for energy to fall to \(1/e^2\) of its initial value is equal to \(\tau\).

Since the two objects have identical masses and the collision is perfectly elastic, the two objects must move off at right angles to each other (which is confirmed by \(30^\circ + 60^\circ = 90^\circ\)). We can construct a right-angled vector triangle representing the conservation of momentum: \(\vec{p}_{initial} = \vec{p}_1 + \vec{p}_2\). Since mass \(m\) is constant for all, we can write the velocity vectors as a right-angled triangle with hypotenuse \(u\). The final velocity vector of the first object, \(v_1\), is adjacent to the angle of \(30^\circ\) relative to \(u\). Therefore: \(v_1 = u \cos(30^\circ) = \frac{\sqrt{3}u}{2}\).

1 mark for the correct option (B). Award 1 mark for identifying that the velocity vectors form a right-angled triangle with hypotenuse \(u\) and using trigonometry to find the component \(u \cos(30^\circ)\).

a) At the top of the loop, both the weight of the rider (\(mg\)) and the normal contact force (\(N\)) act downwards towards the center of the loop, providing the centripetal force: \(N + mg = \frac{mv^2}{r}\). "Weightlessness" is experienced when the normal contact force is zero (\(N = 0\)). Therefore, \(mg = \frac{mv^2}{r} \implies v = \sqrt{gr}\). Calculating this speed: \(v = \sqrt{9.81\text{ m s}^{-2} \times 12\text{ m}} = 10.85\text{ m s}^{-1} \approx 10.8\text{ m s}^{-1}\).

b) At the bottom of the loop, the normal contact force acts upwards and the weight acts downwards: \(N - mg = \frac{mu^2}{r} \implies N = m\left(g + \frac{u^2}{r}\right)\). Substituting values: \(N = 65\text{ kg} \times \left(9.81\text{ m s}^{-2} + \frac{(26\text{ m s}^{-1})^2}{12\text{ m}}\right) = 65 \times (9.81 + 56.33) = 65 \times 66.14 = 4299\text{ N} \approx 4.30 \times 10^3\text{ N}\).

c) At the bottom, the total mechanical energy of the cart (with the bottom as the zero-GPE reference) is: \(E_{\text{initial}} = E_k = \frac{1}{2} M u^2 = \frac{1}{2} \times 850\text{ kg} \times (26\text{ m s}^{-1})^2 = 287,300\text{ J}\). With \(15\%\) energy lost, the remaining energy is: \(E_{\text{remaining}} = 0.85 \times 287,300\text{ J} = 244,205\text{ J}\). At the top of the loop, the height is \(h = 2r = 24\text{ m}\). The gravitational potential energy at the top is: \(E_p = M g h = 850\text{ kg} \times 9.81\text{ m s}^{-2} \times 24\text{ m} = 200,124\text{ J}\). The remaining kinetic energy at the top is: \(E_k = E_{\text{remaining}} - E_p = 244,205\text{ J} - 200,124\text{ J} = 44,081\text{ J}\). Therefore, the speed at the top \(v\) is: \(v = \sqrt{\frac{2 E_k}{M}} = \sqrt{\frac{2 \times 44,081\text{ J}}{850\text{ kg}}} = 10.18\text{ m s}^{-1}\). Since \(10.18\text{ m s}^{-1} < 10.85\text{ m s}^{-1}\), the speed of the cart at the top is less than the minimum required speed to maintain contact. Thus, the cart would lose contact with the track.

a)
- States that \(N + mg = mv^2/r\) and identifies that weightlessness means \(N = 0\) [1]
- Derives expression \(v = \sqrt{gr}\) [1]
- Correct calculation to yield \(10.8\text{ m s}^{-1}\) [1]
- Qualitative explanation: the centripetal force is fully provided by the weight [1]

b)
- States correct equation at bottom: \(N - mg = mu^2/r\) [1]
- Correct substitution of values [1]
- Correct final answer with unit: \(4.30 \times 10^3\text{ N}\) (accept \(4300\text{ N}\)) [1]

c)
- Calculates initial energy or GPE at the top correctly [1]
- Calculates remaining kinetic energy at the top (\(44,081\text{ J}\)) and subsequent velocity (\(10.2\text{ m s}^{-1}\)) [1]
- Compares velocity to critical speed (\(10.8\text{ m s}^{-1}\)) and concludes that the cart does not safely complete the loop [1]

a) The electric field strength \(E\) is: \(E = \frac{V}{d} = \frac{450\text{ V}}{0.015\text{ m}} = 3.0 \times 10^4\text{ V m}^{-1}\).

b) The vertical force on the electron is \(F = eE\). Using Newton's Second Law, \(a = \frac{eE}{m_e} = \frac{1.60 \times 10^{-19}\text{ C} \times 3.0 \times 10^4\text{ V m}^{-1}}{9.11 \times 10^{-31}\text{ kg}} = 5.27 \times 10^{15}\text{ m s}^{-2} \approx 5.3 \times 10^{15}\text{ m s}^{-2}\).

c) Since the electron enters midway, the maximum vertical displacement before hitting the positive plate is \(s_y = \frac{d}{2} = 0.75\text{ cm} = 7.5 \times 10^{-3}\text{ m}\). Solving for the time taken to reach this displacement: \(s_y = \frac{1}{2} a t^2 \implies t = \sqrt{\frac{2 s_y}{a}} = \sqrt{\frac{2 \times 7.5 \times 10^{-3}\text{ m}}{5.27 \times 10^{15}\text{ m s}^{-2}}} = 1.69 \times 10^{-9}\text{ s}\). The horizontal distance traveled in this time is: \(x = v_x t = 2.5 \times 10^7\text{ m s}^{-1} \times 1.69 \times 10^{-9}\text{ s} = 0.042\text{ m} = 4.2\text{ cm}\). Since \(4.2\text{ cm} < 8.0\text{ cm}\) (the plate length), the electron collides with the positive plate at a horizontal distance of \(4.2\text{ cm}\) and does not exit.

d) After exiting the plates, the electric field is zero, so no electrostatic force acts on the electron. Neglecting gravity, there is no net force, so the electron travels in a straight line at a constant velocity, tangential to its exit trajectory.

a)
- Uses \(E = V/d\) [1]
- Correctly substitutes and gets \(3.0 \times 10^4\text{ V m}^{-1}\) [1]

b)
- Uses \(F = eE\) and \(F = ma\) [1]
- Correctly substitutes charge and mass of electron to show \(5.27 \times 10^{15}\text{ m s}^{-2}\) [1]

c)
- Identifies the maximum vertical displacement before collision is \(0.75\text{ cm}\) [1]
- Calculates time taken to reach this displacement as \(1.69 \times 10^{-9}\text{ s}\) [1]
- Calculates horizontal distance as \(4.2\text{ cm}\) [1]
- Explicitly compares \(4.2\text{ cm}\) with the plate length of \(8.0\text{ cm}\) and states it collides [1]

d)
- States that there is no net force/field outside the plates [1]
- Concludes the path is a straight line at a constant speed/velocity [1]

a) A cyclotron consists of two D-shaped hollow electrodes called "dees" with a small gap between them. A uniform magnetic field is applied perpendicular to the plane of the dees. The magnetic force (\(F = qvB\)) acts perpendicular to the velocity of the protons, providing the centripetal force (\(F = mv^2/r\)) to make them travel in circular paths inside the dees. An alternating electric field is applied across the gap. Every time protons cross the gap, the electric field accelerates them, increasing their speed and kinetic energy. Because the period of the circular path is independent of the speed (\(T = 2\pi m / qB\)), the frequency of the alternating electric field remains constant.

b) The magnetic force provides the centripetal force: \(q v B = \frac{m v^2}{r} \implies v = \frac{q B r}{m}\). Substituting proton properties (\(q = 1.60 \times 10^{-19}\text{ C}\), \(m = 1.67 \times 10^{-27}\text{ kg}\)): \(v = \frac{1.60 \times 10^{-19}\text{ C} \times 1.2\text{ T} \times 0.45\text{ m}}{1.67 \times 10^{-27}\text{ kg}} = 5.17 \times 10^7\text{ m s}^{-1} \approx 5.2 \times 10^7\text{ m s}^{-1}\).

c) The frequency \(f\) of the alternating potential difference must equal the orbital frequency of the protons: \(f = \frac{1}{T} = \frac{q B}{2 \pi m}\). Substituting values: \(f = \frac{1.60 \times 10^{-19}\text{ C} \times 1.2\text{ T}}{2 \pi \times 1.67 \times 10^{-27}\text{ kg}} = 1.83 \times 10^7\text{ Hz} \approx 1.8 \times 10^7\text{ Hz}\) (or \(18\text{ MHz}\)).

a)
- States that the magnetic field provides a force perpendicular to motion, causing circular orbits [1]
- States that the electric field accelerates protons across the gap, increasing their kinetic energy [1]
- States that the electric field must alternate in polarity to accelerate protons in both directions [1]
- Mentions that the orbital period is independent of radius/speed, allowing constant frequency [1]

b)
- Equates magnetic force to centripetal force: \(qvB = mv^2/r\) [1]
- Correct substitution of proton charge and mass [1]
- Correct calculation of velocity to \(5.2 \times 10^7\text{ m s}^{-1}\) [1]

c)
- Identifies frequency formula \(f = qB / (2 \pi m)\) [1]
- Substitutes values correctly [1]
- Correct calculation of frequency to \(1.8 \times 10^7\text{ Hz}\) [1]

a) The relationship between terminal potential difference and current is \(V = \varepsilon - I r\). Comparing this to the equation of a straight line, \(y = mx + c\), where \(V\) is on the y-axis and \(I\) is on the x-axis: the y-intercept of the graph represents the e.m.f. \(\varepsilon\), and the gradient of the graph represents the negative of the internal resistance, \(-r\).

b) The total resistance of the series circuit is the sum of all resistances: \(R_{\text{total}} = R_{\text{LDR}} + R_{\text{fixed}} + r = 350\ \Omega + 120\ \Omega + 1.5\ \Omega = 471.5\ \Omega\). The current \(I\) in the circuit is: \(I = \frac{\varepsilon}{R_{\text{total}}} = \frac{4.5\text{ V}}{471.5\ \Omega} = 9.544 \times 10^{-3}\text{ A}\). The potential difference across the LDR is: \(V_{\text{LDR}} = I \times R_{\text{LDR}} = 9.544 \times 10^{-3}\text{ A} \times 350\ \Omega = 3.34\text{ V}\).

c) When the light intensity on the LDR is increased: the resistance of the LDR decreases. This decreases the total resistance of the circuit. Since the e.m.f. is constant, the decrease in total resistance causes the current in the circuit to increase. The potential difference across the fixed resistor is given by \(V_{\text{fixed}} = I R_{\text{fixed}}\). Since \(R_{\text{fixed}}\) is constant and the current \(I\) has increased, the potential difference across the fixed resistor increases.

a)
- Recalls/uses \(V = \varepsilon - Ir\) [1]
- Identifies y-intercept as e.m.f. (\(\varepsilon\)) [1]
- Identifies gradient as negative internal resistance (\(-r\)) [1]

b)
- Calculates total resistance of the circuit (\(471.5\ \Omega\)) [1]
- Calculates circuit current (\(9.54\text{ mA}\)) [1]
- Calculates potential difference across the LDR as \(3.34\text{ V}\) [1]

c)
- States that LDR resistance decreases as light intensity increases [1]
- Explains that this leads to a decrease in total circuit resistance [1]
- Explains that the circuit current increases [1]
- Concludes that potential difference across the fixed resistor increases because \(V = IR\) [1]

a) The total linear momentum of a closed system remains constant, provided no external forces act on the system.

b) The impulse \(J\) is equal to the area under the force-time graph. Since the graph is a triangle: \(J = \frac{1}{2} \times \text{base} \times \text{height}\). Base \(t = 80\text{ ms} = 0.080\text{ s}\) and peak force \(F_{\text{max}} = 12\text{ N}\). Thus, \(J = 0.5 \times 0.080\text{ s} \times 12\text{ N} = 0.48\text{ N s}\) (or \(\text{kg m s}^{-1}\)).

c) The impulse delivered to glider \(B\) is equal to its change in momentum: \(J = \Delta p_B = m_B v_B - m_B u_B\). Since glider \(B\) was initially stationary (\(u_B = 0\)): \(0.48\text{ N s} = 0.60\text{ kg} \times v_B \implies v_B = \frac{0.48}{0.60} = 0.80\text{ m s}^{-1}\).

d) To determine if the collision is elastic, we compare the total kinetic energy before and after the collision. Initial kinetic energy: \(E_{k,i} = \frac{1}{2} m_A u_A^2 = 0.5 \times 0.40\text{ kg} \times (1.5\text{ m s}^{-1})^2 = 0.45\text{ J}\). To find the final velocity \(v_A\) of glider \(A\), we apply the conservation of momentum: \(m_A u_A = m_A v_A + m_B v_B \implies 0.40 \times 1.5 = 0.40 v_A + 0.60 \times 0.80 \implies 0.60 = 0.40 v_A + 0.48 \implies v_A = 0.30\text{ m s}^{-1}\). Final kinetic energy: \(E_{k,f} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = 0.5 \times 0.40 \times (0.30)^2 + 0.5 \times 0.60 \times (0.80)^2 = 0.018\text{ J} + 0.192\text{ J} = 0.21\text{ J}\). Since \(E_{k,f} < E_{k,i}\) (\(0.21\text{ J} < 0.45\text{ J}\)), kinetic energy is not conserved. Thus, the collision is inelastic.

a)
- States momentum is conserved (total initial momentum = total final momentum) [1]
- Specifies the condition: no external forces act / closed system [1]

b)
- States that impulse is the area under the force-time graph [1]
- Correctly substitutes values including millisecond-to-second conversion (\(0.080\text{ s}\)) [1]
- Obtains \(0.48\text{ N s}\) (or \(\text{kg m s}^{-1}\)) [1]

c)
- Equates impulse to change in momentum: \(J = m_B v_B\) [1]
- Calculates \(v_B = 0.80\text{ m s}^{-1}\) correctly [1]

d)
- Calculates initial kinetic energy correctly as \(0.45\text{ J}\) [1]
- Uses conservation of momentum to find final velocity of A as \(0.30\text{ m s}^{-1}\) and calculates final kinetic energy as \(0.21\text{ J}\) [1]
- Compares the two kinetic energies and correctly concludes that the collision is inelastic [1]

a) In a head-on collision, all kinetic energy is converted to electrostatic potential energy at the point of closest approach: \(E_k = E_p = \frac{Q_1 Q_2}{4 \pi \varepsilon_0 r}\). Here, \(Q_1 = 2e\) (alpha particle), \(Q_2 = 79e\) (gold nucleus). \(E_k = 7.7\text{ MeV} = 7.7 \times 10^6 \times 1.60 \times 10^{-19}\text{ J} = 1.232 \times 10^{-12}\text{ J}\). Rearranging for \(r\): \(r = \frac{2 \times 79 \times e^2}{4 \pi \varepsilon_0 E_k}\). Using \(\frac{1}{4 \pi \varepsilon_0} \approx 8.99 \times 10^9\text{ N m}^2\text{ C}^{-2}\): \(r = \frac{8.99 \times 10^9 \times 158 \times (1.60 \times 10^{-19})^2}{1.232 \times 10^{-12}} = 2.95 \times 10^{-14}\text{ m} \approx 3.0 \times 10^{-14}\text{ m}\).

b) Alpha particles are hadrons and therefore experience the strong nuclear force when they get very close to the nucleus, which perturbs the scattering pattern. High-energy electrons are leptons, so they do not feel the strong force; they only interact via the electromagnetic force. High-energy electrons have extremely small de Broglie wavelengths comparable to nuclear sizes, and their diffraction pattern (first minimum) directly yields the nuclear charge radius without needing to penetrate the nuclear potential barrier.

c) Assuming the nucleus is spherical, its volume is: \(V = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (r_0 A^{1/3})^3 = \frac{4}{3} \pi r_0^3 A\). The mass of the nucleus is given by \(M \approx A \times m_u\), where \(m_u\) is the average mass of a nucleon. The density of nuclear matter \(\rho\) is: \(\rho = \frac{M}{V} = \frac{A \cdot m_u}{\frac{4}{3} \pi r_0^3 A} = \frac{3 m_u}{4 \pi r_0^3}\). Since \(m_u\) and \(r_0\) are constants, the density \(\rho\) is constant and independent of the nucleon number \(A\).

a)
- Converts energy from MeV to Joules: \(1.23 \times 10^{-12}\text{ J}\) [1]
- Uses potential energy formula with charges \(2e\) and \(79e\) [1]
- Rearranges to solve for \(r\) [1]
- Correct calculation to yield \(2.95 \times 10^{-14}\text{ m}\) (or \(3.0 \times 10^{-14}\text{ m}\)) [1]

b)
- States that electrons are leptons and do not feel the strong force (unlike alpha particles) [1]
- Explains that high-energy electrons undergo diffraction (wave-like behavior) [1]
- Explains that the position of the first minimum in the diffraction pattern directly determines the nuclear radius [1]

c)
- Expresses volume in terms of \(A\): \(V = \frac{4}{3} \pi r_0^3 A\) [1]
- Expresses mass in terms of \(A\): \(M = A \times m_u\) [1]
- Shows that \(A\) cancels in \(\rho = M/V\), leaving only constants [1]

a) The circuit diagram should show: a power supply connected in series with an ammeter, a switch (or variable resistor), and the test wire. A voltmeter must be connected in parallel across the variable length of the wire using crocodile clips.

b) The two variables that must be kept constant are: the temperature of the wire, and the cross-sectional area (uniformity) of the wire. To control the temperature, the student should use a low current to avoid heating effects, and open the switch between readings to let the wire cool down.

c) The resistance is given by \(R = \frac{\rho L}{A}\), which can be written as \(R = \left(\frac{\rho}{A}\right) L\). Therefore, the gradient \(m\) of a graph of \(R\) against \(L\) is \(m = \frac{\rho}{A}\). The cross-sectional area \(A\) of the wire of diameter \(d = 0.38 \times 10^{-3}\text{ m}\) is: \(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.38 \times 10^{-3}\text{ m})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\). Thus, resistivity \(\rho = m \times A = 4.31\ \Omega\text{ m}^{-1} \times 1.134 \times 10^{-7}\text{ m}^2 = 4.89 \times 10^{-7}\ \Omega\text{ m}\).

d) A source of systematic error is the contact resistance at the connections (e.g., crocodile clips). This adds a constant additional resistance to all measurements. On the graph of \(R\) against \(L\), this shifts the line of best fit upwards, creating a positive y-intercept instead of passing through the origin, but it does not change the gradient.

a)
- Draw ammeter in series and voltmeter in parallel across the wire [1]
- Includes a power supply and a method to vary current/turn off circuit (switch or variable resistor) [1]

b)
- Identifies temperature and cross-sectional area/diameter [1]
- Explains that resistance increases with temperature, which changes resistivity [1]
- Explains how to control: use low currents / switch off circuit between readings to minimize heating [1]

c)
- Calculates cross-sectional area correctly: \(A = 1.13 \times 10^{-7}\text{ m}^2\) [1]
- Uses relationship \(\rho = \text{gradient} \times A\) [1]
- Obtains resistivity of \(4.89 \times 10^{-7}\ \Omega\text{ m}\) with correct unit [1]

d)
- Identifies a systematic error (e.g., contact resistance / zero error on micrometer) [1]
- Explains that this causes a constant shift, changing the y-intercept but leaving the gradient unchanged [1]

a) The induced e.m.f. \(\varepsilon\) is given by: \(\varepsilon = B l v = 0.65\text{ T} \times 0.15\text{ m} \times 8.0\text{ m s}^{-1} = 0.78\text{ V}\).

b) The induced current \(I\) is: \(I = \frac{\varepsilon}{R} = \frac{0.78\text{ V}}{0.30\ \Omega} = 2.6\text{ A}\). The magnetic force \(F_{\text{mag}}\) acting on the bar is: \(F_{\text{mag}} = B I l = 0.65\text{ T} \times 2.6\text{ A} \times 0.15\text{ m} = 0.2535\text{ N} \approx 0.25\text{ N}\).

c) According to Lenz's law, the direction of the induced current is such that it opposes the change in magnetic flux that produces it. The motion of the bar cuts magnetic field lines, causing an increase in magnetic flux linkage. The resulting magnetic force on the induced current acts in the opposite direction to the velocity of the bar, attempting to slow it down. To keep the bar moving at a constant velocity (zero acceleration), the net force must be zero, so an external force equal in magnitude and opposite in direction to this magnetic force must be continuously applied.

d) Electrical power dissipated in the resistor is: \(P_{\text{elec}} = I^2 R = (2.6\text{ A})^2 \times 0.30\ \Omega = 2.03\text{ W}\). The mechanical power (rate of work done by the external force) is: \(P_{\text{mech}} = F_{\text{ext}} \times v = 0.2535\text{ N} \times 8.0\text{ m s}^{-1} = 2.03\text{ W}\). Thus, \(P_{\text{elec}} = P_{\text{mech}} = 2.0\text{ W}\) (to 2 s.f.), which demonstrates conservation of energy.

a)
- Uses \(\varepsilon = Blv\) [1]
- Correctly calculates \(0.78\text{ V}\) [1]

b)
- Uses Ohm's Law to find current: \(I = 2.6\text{ A}\) [1]
- Uses magnetic force formula: \(F = BIl\) [1]
- Calculates force as \(0.25\text{ N}\) [1]

c)
- States Lenz's law (induced current opposes the motion producing it) [1]
- Explains that the magnetic force on the induced current opposes the direction of velocity [1]
- Concludes that an external force of equal magnitude is required to balance this force and maintain constant speed [1]

d)
- Calculates electrical power using \(P = I^2 R\) or \(P = IV\) to show \(2.03\text{ W}\) [1]
- Calculates mechanical power using \(P = Fv\) to show \(2.03\text{ W}\) and concludes they are equal [1]

The mean kinetic energy of the molecules of an ideal gas is directly proportional to its absolute temperature: \(\frac{1}{2} m \langle c^2 \rangle = \frac{3}{2} k T\). This implies that the mean square speed \(\langle c^2 \rangle\) is directly proportional to the absolute temperature \(T\). Since the temperature is doubled, the mean square speed of the molecules is also doubled, becoming \(2.0 \langle c^2 \rangle\). The change in volume does not affect the mean square speed of the molecules, as the temperature remains the sole determinant of their average kinetic energy.

1 mark for the correct answer C. No partial marks.

From Wien's displacement law: \(\lambda_{\text{max}} T = \text{constant}\). Therefore, \(\frac{T_X}{T_Y} = \frac{\lambda_Y}{\lambda_X} = \frac{800\text{ nm}}{400\text{ nm}} = 2\). Using Stefan-Boltzmann law: \(L = 4 \pi R^2 \sigma T^4\). The ratio of luminosities is: \(\frac{L_X}{L_Y} = \left(\frac{R_X}{R_Y}\right)^2 \left(\frac{T_X}{T_Y}\right)^4 = \left(\frac{R}{3R}\right)^2 \left(2\right)^4 = \frac{1}{9} \times 16 = \frac{16}{9}\).

1 mark for the correct answer C. No partial marks.

The total energy of the oscillator is \(E_{\text{total}} = \frac{1}{2} k A^2\). The potential energy at displacement \(x\) is \(E_p = \frac{1}{2} k x^2\). The kinetic energy is \(E_k = E_{\text{total}} - E_p = \frac{1}{2} k (A^2 - x^2)\). We require \(E_k = 3 E_p\), which gives: \(\frac{1}{2} k (A^2 - x^2) = 3 \left(\frac{1}{2} k x^2\right) \Rightarrow A^2 - x^2 = 3 x^2 \Rightarrow A^2 = 4 x^2 \Rightarrow x = \pm \frac{A}{2}\).

1 mark for the correct answer B. No partial marks.

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

1 mark for the correct answer C. No partial marks.

The gravitational potential energy of a satellite in orbit is \(E_p = -\frac{GMm}{r}\). If the radius increases to \(2r\), the potential energy becomes less negative (closer to zero), meaning GPE increases. The kinetic energy of the satellite in a stable orbit is given by \(E_k = \frac{GMm}{2r}\). If the orbit radius increases to \(2r\), the kinetic energy decreases to half of its initial value. Therefore, GPE increases and KE decreases.

1 mark for the correct answer B. No partial marks.

The energy supplied by the heater is \(Q = P \times t = 100\text{ W} \times 200\text{ s} = 20000\text{ J}\). The energy required to melt a mass \(m\) is \(Q = m L_f\), where \(L_f\) is the specific latent heat of fusion. Therefore, \(L_f = \frac{Q}{m} = \frac{20000\text{ J}}{0.50\text{ kg}} = 40000\text{ J kg}^{-1} = 4.0 \times 10^4\text{ J kg}^{-1}\).

1 mark for the correct answer D. No partial marks.

The distance in parsecs is given by \(d = \frac{1}{p}\), where \(p\) is the parallax angle in arcseconds. Thus, \(d = \frac{1}{0.125\text{ arcseconds}} = 8\text{ parsecs}\). To convert parsecs into light-years: \(d = 8 \times 3.26 = 26.08\text{ light-years} \approx 26.1\text{ light-years}\).

1 mark for the correct answer D. No partial marks.

As damping increases, the peak of the resonance curve becomes broader and the maximum amplitude of oscillation decreases. Furthermore, the peak shifts slightly to a lower frequency (the resonant frequency decreases). Therefore, the peak amplitude decreases and the resonant frequency shifts to a lower value.

1 mark for the correct answer B. No partial marks.

The mean kinetic energy of a molecule in an ideal gas is given by \(\frac{1}{2} m \langle c^2 \rangle = \frac{3}{2} k T\) where \(T\) is the absolute temperature in Kelvin. Therefore, the mean square speed \(\langle c^2 \rangle\) is directly proportional to the absolute temperature \(T\). Converting the temperatures from Celsius to Kelvin gives: \(T_{\text{initial}} = 20 + 273 = 293\text{ K}\) and \(T_{\text{final}} = 120 + 273 = 393\text{ K}\). The ratio of the final mean square speed to the initial mean square speed is \(\frac{393\text{ K}}{293\text{ K}} \approx 1.34\). Option A is the ratio of the root-mean-square speeds. Option C is the ratio of the root-mean-square speeds in Celsius. Option D is the ratio of Celsius temperatures.

B is the correct answer. 1 mark for correctly converting temperatures to Kelvin and finding the ratio of absolute temperatures: 393 K / 293 K = 1.34.

From Wien's displacement law, the absolute temperature \(T\) of a star is inversely proportional to its peak wavelength \(\lambda_{\text{max}}\), so \(T \propto \frac{1}{\lambda_{\text{max}}}\). Since Star Y has twice the peak wavelength of Star X (\(800\text{ nm}\) compared to \(400\text{ nm}\)), its absolute temperature is half that of Star X: \(T_{\text{Y}} = 0.5 T_{\text{X}}\). From the Stefan-Boltzmann law, the luminosity of a star is given by \(L = 4 \pi r^2 \sigma T^4\). Since the stars have the same radius, \(L \propto T^4\). Therefore, the luminosity of Star Y is \(L_{\text{Y}} = L_{\text{X}} \times (0.5)^4 = L \times 0.0625 = 0.0625 L\).

D is the correct answer. 1 mark for relating peak wavelength to temperature to find \(T_{\text{Y}} = 0.5 T_{\text{X}}\), and then applying Stefan-Boltzmann law to show \(L \propto T^4\) to obtain \(0.0625 L\).

(i) Using the ideal gas equation: \( p_1 = \frac{n R T_1}{V} = \frac{1.2 \times 8.31 \times 290}{0.025} = 1.16 \times 10^5\text{ Pa} \) (or \( 1.2 \times 10^5\text{ Pa} \) to 2 s.f.). (ii) Since all energy increases internal kinetic energy: \( \Delta U = \frac{3}{2} n R \Delta T \implies 4500 = 1.5 \times 1.2 \times 8.31 \times \Delta T \implies \Delta T = \frac{4500}{14.958} \approx 300.8\text{ K} \). The final temperature is \( T_2 = T_1 + \Delta T = 290 + 300.8 = 590.8\text{ K} \approx 591\text{ K} \). (iii) The formula for r.m.s. speed is \( v_{\text{rms}} = \sqrt{\frac{3 R T}{M}} \). Initial r.m.s. speed: \( v_1 = \sqrt{\frac{3 \times 8.31 \times 290}{4.0 \times 10^{-3}}} = 1344.4\text{ m s}^{-1} \). Final r.m.s. speed: \( v_2 = \sqrt{\frac{3 \times 8.31 \times 590.8}{4.0 \times 10^{-3}}} = 1918.9\text{ m s}^{-1} \). The change in r.m.s. speed is \( \Delta v = 1918.9 - 1344.4 = 574.5\text{ m s}^{-1} \approx 575\text{ m s}^{-1} \).

(i) [2 marks] 1 mark for rearranging the ideal gas law, 1 mark for correct value of initial pressure with units. (ii) [3 marks] 1 mark for substituting values into the internal energy equation, 1 mark for calculating temperature change, 1 mark for correct final temperature in Kelvin. (iii) [4 marks] 1 mark for r.m.s. speed formula, 1 mark for correct initial speed, 1 mark for correct final speed, 1 mark for correct speed change with units (accept \( 570\text{ m s}^{-1} \) to \( 580\text{ m s}^{-1} \) depending on rounding steps).

(i) Luminosity is the total electromagnetic energy radiated per second by a star (total power output). (ii) First find the absolute luminosity in Watts: \( L = 1.2 \times 10^5 \times 3.85 \times 10^{26}\text{ W} = 4.62 \times 10^{31}\text{ W} \). Stefan-Boltzmann's law is given by \( L = 4\pi r^2 \sigma T^4 \). Rearranging for radius \( r \): \( r = \sqrt{\frac{L}{4\pi \sigma T^4}} = \sqrt{\frac{4.62 \times 10^{31}}{4 \times \pi \times 5.67 \times 10^{-8} \times 3500^4}} = \sqrt{\frac{4.62 \times 10^{31}}{1.0702 \times 10^8}} = 6.57 \times 10^{11}\text{ m} \). (iii) Light from stars moving away from Earth has its spectral lines shifted towards longer wavelengths (redshift). By measuring the shift in wavelength \( \Delta \lambda \) and the rest wavelength \( \lambda \), astronomers calculate the redshift parameter \( z = \frac{\Delta \lambda}{\lambda} \). Since \( z = \frac{v}{c} \) for non-relativistic speeds, the recession speed is \( v = z c \), where \( c \) is the speed of light.

(i) [1 mark] Correct definition of luminosity as total power / total energy radiated per second. (ii) [4 marks] 1 mark for converting solar luminosity to Watts, 1 mark for rearranging Stefan's law for r, 1 mark for correct substitution including Stefan-Boltzmann constant, 1 mark for correct final value of radius in metres. (iii) [4 marks] 1 mark for identifying redshift / lines shifting to longer wavelengths, 1 mark for explaining that redshift indicates recession, 1 mark for Doppler relation formula, 1 mark for identifying variables and speed of light.

(i) Time period is given by \( T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{15}{240}} = 2\pi \sqrt{0.0625} = 2\pi \times 0.25 = 1.57\text{ s} \). (ii) First, calculate the angular frequency: \( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{240}{15}} = 4.0\text{ rad s}^{-1} \). Maximum acceleration is \( a_{\text{max}} = \omega^2 A = (4.0)^2 \times 0.12 = 1.92\text{ m s}^{-2} \). Maximum kinetic energy is equal to the total elastic potential energy at maximum amplitude: \( E_{k,\text{max}} = \frac{1}{2} k A^2 = 0.5 \times 240 \times (0.12)^2 = 120 \times 0.0144 = 1.73\text{ J} \). (iii) Critical damping is the name given to the damping condition where the system returns to its equilibrium position in the minimum possible time without overshooting or oscillating.

(i) [2 marks] 1 mark for formula of time period, 1 mark for correct calculation with units. (ii) [4 marks] 1 mark for calculating angular frequency \( \omega \), 1 mark for max acceleration calculation, 1 mark for max kinetic energy formula, 1 mark for correct value of max kinetic energy. (iii) [3 marks] 1 mark for naming 'critical damping', 2 marks for explaining its characteristics (fastest return, no oscillation/overshoot).

(i) Decay constant \( \lambda = \frac{\ln(2)}{T_{1/2}} \). Half-life in seconds: \( T_{1/2} = 6.0 \times 3600\text{ s} = 21600\text{ s} \). Therefore, \( \lambda = \frac{0.69315}{21600} = 3.21 \times 10^{-5}\text{ s}^{-1} \), which is approximately \( 3.2 \times 10^{-5}\text{ s}^{-1} \). (ii) \( 18\text{ hours} \) is exactly \( 3 \) half-lives (\( 18 / 6 = 3 \)). Activity after \( 18\text{ hours} \) is \( A = A_0 \left(\frac{1}{2}\right)^3 = 4.5 \times 10^8 \times 0.125 = 5.625 \times 10^7\text{ Bq} \). Using \( A = \lambda N \), the number of nuclei remaining is \( N = \frac{A}{\lambda} = \frac{5.625 \times 10^7}{3.21 \times 10^{-5}} = 1.75 \times 10^{12} \) nuclei. (iii) A short half-life ensures that the isotope decays rapidly inside the body, reducing prolonged exposure and minimizing the radiation dose received by the patient. However, the half-life of several hours is long enough to prepare the tracer, administer it, and complete the diagnostic scan before the activity drops too low.

(i) [2 marks] 1 mark for converting half-life to seconds, 1 mark for calculating decay constant showing at least 3 significant figures. (ii) [4 marks] 1 mark for identifying 3 half-lives or using decay equation, 1 mark for calculating final activity, 1 mark for using \( A = \lambda N \), 1 mark for correct number of nuclei. (iii) [3 marks] 1 mark for patient safety (rapid drop in activity minimizes dose), 1 mark for practical scan duration (lasts long enough to record data), 1 mark for linking these two factors together.

(i) The gravitational force provides the centripetal force: \( \frac{G M m}{r^2} = \frac{m v^2}{r} \implies v^2 = \frac{GM}{r} \). Since \( v = \frac{2\pi r}{T} \), we substitute to get \( \frac{4\pi^2 r^2}{T^2} = \frac{GM}{r} \). Rearranging gives \( T^2 = \frac{4\pi^2 r^3}{GM} \). (ii) Convert period to seconds: \( T = 130 \times 24 \times 3600 = 1.123 \times 10^7\text{ s} \). Rearranging for radius: \( r^3 = \frac{GM T^2}{4\pi^2} = \frac{(6.67 \times 10^{-11}) \times (9.5 \times 10^{29}) \times (1.123 \times 10^7)^2}{4\pi^2} = 2.025 \times 10^{32}\text{ m}^3 \). Taking the cube root: \( r = (2.025 \times 10^{32})^{1/3} = 5.87 \times 10^{10}\text{ m} \). (iii) Gravitational potential at a point is the work done per unit mass to bring a small test mass from infinity to that point in the gravitational field.

(i) [3 marks] 1 mark for equating gravitational force and centripetal force, 1 mark for substituting linear speed using period, 1 mark for correct algebraic manipulation leading to the given formula. (ii) [4 marks] 1 mark for converting orbital period from days to seconds, 1 mark for rearranging formula for \( r^3 \), 1 mark for substitution of values, 1 mark for correct orbital radius in metres. (iii) [2 marks] 1 mark for 'work done per unit mass', 1 mark for 'from infinity (to the point)'.

(i) Convert work function to joules: \( \phi = 4.3 \times 1.60 \times 10^{-19}\text{ J} = 6.88 \times 10^{-19}\text{ J} \). Since \( \phi = h f_0 \), the threshold frequency is \( f_0 = \frac{6.88 \times 10^{-19}}{6.63 \times 10^{-34}} = 1.04 \times 10^{15}\text{ Hz} \). (ii) Calculate photon energy: \( E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{220 \times 10^{-9}} = 9.04 \times 10^{-19}\text{ J} \). Using Einstein's equation: \( E_{k,\text{max}} = E - \phi = 9.04 \times 10^{-19} - 6.88 \times 10^{-19} = 2.16 \times 10^{-19}\text{ J} \). (iii) In wave theory, light is a continuous wave of energy. Energy would build up on the electrons over time, regardless of the light's frequency. Thus, even low-frequency light should eventually deliver enough cumulative energy to eject an electron. In reality, emission is instantaneous and occurs only if individual photon energy is above the work function, showing light behaves as packets of energy (photons).

(i) [3 marks] 1 mark for converting electronvolts to joules, 1 mark for using threshold relation, 1 mark for correct frequency with units. (ii) [4 marks] 1 mark for calculating energy of incident photon, 1 mark for stating/using Einstein's equation, 1 mark for subtraction, 1 mark for correct maximum kinetic energy in joules. (iii) [2 marks] 1 mark for stating wave theory suggests energy accumulates over time, 1 mark for explaining that wave theory predicts emission at any frequency given sufficient time/intensity.

(i) Convert temperature to Kelvin: \( T_1 = 20 + 273.15 = 293.15\text{ K} \). Ideal gas law: \( p V = n R T \implies n = \frac{p V}{R T} = \frac{1.5 \times 10^6 \times 0.040}{8.31 \times 293.15} = 24.63\text{ moles} \). Mass: \( m = n \times M = 24.63 \times 4.0 \times 10^{-3} = 0.0985\text{ kg} \) (accept \( 0.0986\text{ kg} \) if using 293 K). (ii) For a constant volume, \( \frac{p_1}{T_1} = \frac{p_2}{T_2} \). Convert new temperature: \( T_2 = 80 + 273.15 = 353.15\text{ K} \). Therefore: \( p_2 = p_1 \times \frac{T_2}{T_1} = 1.5 \times 10^6 \times \frac{353.15}{293.15} = 1.81 \times 10^6\text{ Pa} \). (iii) An increase in temperature increases the average kinetic energy of the gas molecules, meaning they move with a higher average speed. As a result, they collide with the walls of the container more frequently and experience a larger change in momentum during each collision. This increases the average rate of change of momentum (force) exerted on the walls, leading to an increased pressure.

(i) [3 marks] 1 mark for converting temperature to Kelvin, 1 mark for calculating the number of moles, 1 mark for multiplying moles by molar mass. (ii) [2 marks] 1 mark for converting both temperatures to Kelvin, 1 mark for correct pressure value. (iii) [4 marks] 1 mark for stating that higher temperature means higher molecular speed/kinetic energy, 1 mark for stating collisions with the walls are more frequent, 1 mark for stating that collisions involve a greater change in momentum, 1 mark for concluding that the average force per unit area increases.

(i) Find redshift parameter: \( z = \frac{\Delta \lambda}{\lambda_0} = \frac{678.1 - 656.3}{656.3} = \frac{21.8}{656.3} = 0.0332 \). Recession velocity: \( v = z c = 0.0332 \times 3.00 \times 10^8\text{ m s}^{-1} = 9.96 \times 10^6\text{ m s}^{-1} \) (or \( 9960\text{ km s}^{-1} \)). (ii) Use Hubble's law: \( v = H_0 d \implies d = \frac{v}{H_0} \). Since \( v = 9960\text{ km s}^{-1} \): \( d = \frac{9960}{68} = 146.5\text{ Mpc} \approx 146\text{ Mpc} \). (iii) To convert \( H_0 \) to SI units: \( 68\text{ km s}^{-1}\text{ Mpc}^{-1} = \frac{68 \times 10^3\text{ m s}^{-1}}{10^6 \times 3.09 \times 10^{16}\text{ m}} = 2.20 \times 10^{-18}\text{ s}^{-1} \). The age of the universe is approximately \( T \approx \frac{1}{H_0} = \frac{1}{2.20 \times 10^{-18}} = 4.545 \times 10^{17}\text{ s} \). In years: \( T = \frac{4.545 \times 10^{17}}{3.156 \times 10^7} \approx 1.44 \times 10^{10}\text{ years} \).

(i) [3 marks] 1 mark for finding the change in wavelength, 1 mark for calculating redshift parameter, 1 mark for correct speed with unit. (ii) [3 marks] 1 mark for converting velocity to km/s, 1 mark for applying Hubble's Law, 1 mark for correct distance in Mpc. (iii) [3 marks] 1 mark for converting Hubble constant to SI units, 1 mark for relating age to the reciprocal of the Hubble constant, 1 mark for correct age in years.

**(a)**
Assumptions of the kinetic theory of gases include:
* Gas molecules are in rapid, continuous, random motion.
* The volume of the molecules is negligible compared to the total volume occupied by the gas.
* All collisions between molecules and the container walls are perfectly elastic.
* The duration of collisions is negligible compared to the time between collisions.
* There are no intermolecular forces acting between the molecules except during collisions.

**(b)**
1. Convert temperature to Kelvin:
\(T = 20.0 + 273.15 = 293.15 \text{ K}\) (or \(293 \text{ K}\))

2. Use the ideal gas equation \(pV = nRT\) to find the number of moles \(n\):
\(n = \frac{pV}{RT} = \frac{1.20 \times 10^5 \text{ Pa} \times 2.50 \times 10^{-2} \text{ m}^3}{8.31 \text{ J mol}^{-1}\text{ K}^{-1} \times 293.15 \text{ K}} = 1.232 \text{ mol}\)

3. Calculate the mass \(m\):
\(m = n \times \text{molar mass} = 1.232 \text{ mol} \times 4.00 \times 10^{-3} \text{ kg mol}^{-1} = 4.93 \times 10^{-3} \text{ kg}\) (or \(4.9 \text{ g}\))

**(c)**
1. Since the volume is constant, the pressure is directly proportional to absolute temperature (\(p \propto T\)):
\(T_2 = T_1 \times \frac{p_2}{p_1} = 293.15 \text{ K} \times \frac{1.80 \times 10^5 \text{ Pa}}{1.20 \times 10^5 \text{ Pa}} = 439.7 \text{ K}\)

2. Find the temperature change \(\Delta T\):
\(\Delta T = T_2 - T_1 = 439.7 \text{ K} - 293.15 \text{ K} = 146.6 \text{ K}\)

3. Use the equation for the mean kinetic energy of a molecule, \(E_k = \frac{3}{2}kT\):
\(\Delta E_k = \frac{3}{2} k \Delta T\)
\(\Delta E_k = 1.5 \times (1.38 \times 10^{-23} \text{ J K}^{-1}) \times 146.6 \text{ K}\)
\(\Delta E_k = 3.03 \times 10^{-21} \text{ J}\)

**Part (a):**
* **MP1:** Any one valid assumption of the kinetic theory of gases. (1)
* **MP2:** Any second valid assumption of the kinetic theory of gases. (1)
*(Do not accept "molecules are small" without "compared to the distance between them" or "volume of container".)*

**Part (b):**
* **MP3:** Converts temperature to Kelvin: \(293 \text{ K}\) seen or used. (1)
* **MP4:** Applies \(pV = nRT\) to find the number of moles (\(n = 1.23 \text{ mol}\)). (1)
* **MP5:** Calculates mass to give \(4.93 \times 10^{-3} \text{ kg}\) (accept range \(4.9 \times 10^{-3} \text{ kg}\) to \(5.0 \times 10^{-3} \text{ kg}\)). (1)

**Part (c):**
* **MP6:** Applies pressure law \(p_1/T_1 = p_2/T_2\) to determine new temperature \(T_2 = 440 \text{ K}\) or temperature difference \(\Delta T = 147 \text{ K}\). (1)
* **MP7:** Recalls or uses the expression for mean kinetic energy of an ideal gas particle, \(E_k = \frac{3}{2}kT\). (1)
* **MP8:** Correct substitution of Boltzmann constant \(k = 1.38 \times 10^{-23} \text{ J K}^{-1}\) and temperature change. (1)
* **MP9:** Calculates correct increase in kinetic energy to give \(3.0 \times 10^{-21} \text{ J}\) (accept range \(3.0 \times 10^{-21} \text{ J}\) to \(3.1 \times 10^{-21} \text{ J}\)). (1)

Step 1: Correct the diameter measurements for the zero error of \(+0.01\text{ mm}\) by subtracting \(0.01\text{ mm}\) from each value:
Corrected values: \(0.41\text{ mm}\), \(0.43\text{ mm}\), \(0.40\text{ mm}\), \(0.42\text{ mm}\), \(0.41\text{ mm}\).

Step 2: Calculate the mean diameter:
\(d_{\text{mean}} = \frac{0.41 + 0.43 + 0.40 + 0.42 + 0.41}{5} = 0.414\text{ mm}\).

Step 3: Calculate the uncertainty in diameter using half-range:
\(\text{Range} = 0.43 - 0.40 = 0.03\text{ mm}\).
\(\Delta d = \frac{0.03}{2} = 0.015\text{ mm}\).
\(\%\text{ uncertainty in } d = \left(\frac{0.015}{0.414}\right) \times 100\% = 3.62\%\).

Step 4: Calculate the cross-sectional area:
\(A = \frac{\pi d^2}{4} = \frac{\pi (0.414 \times 10^{-3})^2}{4} = 1.346 \times 10^{-7}\text{ m}^2\).
\(\%\text{ uncertainty in } A = 2 \times \%\text{ uncertainty in } d = 2 \times 3.62\% = 7.24\%\).

Step 5: Determine the Young Modulus \(E\):
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta x / L} = \left(\frac{F}{\Delta x}\right) \frac{L}{A} = \text{Gradient} \times \frac{L}{A}\).
\(E = (1.25 \times 10^4) \times \frac{1.540}{1.346 \times 10^{-7}} = 1.430 \times 10^{11}\text{ Pa}\).

Step 6: Calculate the percentage and absolute uncertainty in \(E\):
\(\%\text{ uncertainty in Gradient} = \left(\frac{0.05}{1.25}\right) \times 100\% = 4.0\%\).
\(\%\text{ uncertainty in } L = \left(\frac{0.002}{1.540}\right) \times 100\% = 0.13\%\).
\(\text{Total } \%\text{ uncertainty in } E = 4.0\% + 0.13\% + 7.24\% = 11.37\%\).
\(\text{Absolute uncertainty in } E = 11.37\% \times 1.430 \times 10^{11}\text{ Pa} = 0.163 \times 10^{11}\text{ Pa}\).

Thus, \(E = (1.4 \pm 0.2) \times 10^{11}\text{ Pa}\) (or \((1.43 \pm 0.16) \times 10^{11}\text{ Pa}\)).

(a)
- 1 mark: Long wire (approx 1.5 m) suspended from a fixed point with a marker attached to the wire near a scale/vernier, or use of a travelling microscope to measure small extensions.
- 1 mark: Use of a digital micrometer at multiple different orientations and positions to measure diameter.
- 1 mark: Reference to taking a reference reading and adding standard masses to vary load, measuring extension each time.

(b)
- 1 mark: Correct correction of zero error and calculation of mean diameter \(d_{\text{mean}} = 0.414\text{ mm}\).
- 1 mark: Correct area calculation \(A = 1.35 \times 10^{-7}\text{ m}^2\).
- 1 mark: Correct percentage uncertainty in area (7.2% or 7%).

(c)
- 1 mark: Correct formula for \(E\) linking gradient, original length, and area.
- 1 mark: Correct substitution giving \(E = 1.4 \times 10^{11}\text{ Pa}\).
- 1 mark: Summing individual percentage uncertainties correctly (approx 11.4%).
- 1 mark: Correct absolute uncertainty to 1 or 2 s.f. with units.

Step 1: Identify \(\varepsilon\) and \(r\) from the graph parameters.
Since \(V = \varepsilon - Ir\), the y-intercept is equal to \(\varepsilon\) and the gradient is equal to \(-r\).
Thus, \(\varepsilon = 1.65 \pm 0.05\text{ V}\) and \(r = 4.50 \pm 0.30\text{ }\Omega\).

Step 2: Recall the condition for maximum power transfer.
Maximum power occurs when the load resistance \(R\) is matched to the internal resistance of the cell, \(R = r\).
At this condition, \(V = \frac{\varepsilon}{2}\) and \(I = \frac{\varepsilon}{2r}\).
Thus, the maximum power \(P_{\text{max}} = V \times I = \frac{\varepsilon^2}{4r}\).

Step 3: Calculate the maximum power:
\(P_{\text{max}} = \frac{1.65^2}{4 \times 4.50} = \frac{2.7225}{18} = 0.15125\text{ W}\).

Step 4: Propagate uncertainties to find the absolute uncertainty in \(P_{\text{max}}\):
\(\%\text{ uncertainty in } \varepsilon = \left(\frac{0.05}{1.65}\right) \times 100\% = 3.03\%\).
\(\%\text{ uncertainty in } \varepsilon^2 = 2 \times 3.03\% = 6.06\%\).
\(\%\text{ uncertainty in } r = \left(\frac{0.30}{4.50}\right) \times 100\% = 6.67\%\).
\(\%\text{ uncertainty in } P_{\text{max}} = 6.06\% + 6.67\% = 12.73\%\).
\(\text{Absolute uncertainty in } P_{\text{max}} = 12.73\% \times 0.15125\text{ W} = 0.01925\text{ W}\).

Rounding to appropriate significant figures:
\(P_{\text{max}} = (0.15 \pm 0.02)\text{ W}\).

(a)
- 1 mark: Valid circuit diagram showing the solar cell connected in a closed loop containing a variable resistor (or potentiometer) and an ammeter, with a voltmeter connected in parallel across the solar cell.
- 1 mark: Explicit statement on how to vary the resistance of the variable resistor to obtain a range of values for current and potential difference.

(b)
- 1 mark: Ensure that the ambient room light is kept constant (e.g. conduct the experiment in a darkened room) to avoid external light interference.
- 1 mark: Keep the illumination period short or use a heat-absorbing glass filter to prevent the temperature of the solar cell from rising, as internal resistance is temperature-dependent.

(c)
- 1 mark: Identify emf and internal resistance with correct values and uncertainties: \(\varepsilon = 1.65 \pm 0.05\text{ V}\) and \(r = 4.50 \pm 0.30\text{ }\Omega\).
- 1 mark: Use correct formula for maximum power \(P_{\text{max}} = \frac{\varepsilon^2}{4r}\) and calculate \(0.15\text{ W}\).
- 1 mark: Sum percentage uncertainties correctly (approx 12.7%).
- 1 mark: Calculate correct absolute uncertainty of \(0.02\text{ W}\) with correct formatting.

Step 1: Calculate the grating spacing \(d\):
\(d = \frac{1}{600\text{ lines mm}^{-1}} = 1.667 \times 10^{-3}\text{ mm} = 1.667 \times 10^{-6}\text{ m}\).
Uncertainty in \(d\) is \(1\%\) of \(1.667 \times 10^{-6}\text{ m} = 0.017 \times 10^{-6}\text{ m}\).
Thus, \(d = (1.67 \pm 0.02) \times 10^{-6}\text{ m}\).

Step 2: Show that the angle of diffraction \(\theta\) is approximately \(49.4^\circ\):
\(\tan\theta = \frac{y}{D} = \frac{2.859}{2.450} = 1.1669\).
\(\theta = \arctan(1.1669) = 49.405^\circ \approx 49.4^\circ\).

Step 3: Calculate the wavelength using the grating equation \(d\sin\theta = n\lambda\):
\(\lambda = \frac{d\sin\theta}{n} = \frac{1.667 \times 10^{-6} \times \sin(49.405^\circ)}{2} = \frac{1.667 \times 10^{-6} \times 0.7593}{2} = 6.328 \times 10^{-7}\text{ m} \approx 633\text{ nm}\).

Step 4: Propagate percentage uncertainties:
To find the percentage uncertainty in \(\sin\theta\), calculate maximum and minimum values:
\(y_{\text{max}} = 2.869\text{ m}\), \(D_{\text{min}} = 2.445\text{ m}\)
\(\theta_{\text{max}} = \arctan\left(\frac{2.869}{2.445}\right) = 49.564^\circ\).
\(y_{\text{min}} = 2.849\text{ m}\), \(D_{\text{max}} = 2.455\text{ m}\)
\(\theta_{\text{min}} = \arctan\left(\frac{2.849}{2.455}\right) = 49.246^\circ\).
\(\sin\theta_{\text{max}} = \sin(49.564^\circ) = 0.7611\).
\(\sin\theta_{\text{min}} = \sin(49.246^\circ) = 0.7575\).
Uncertainty in \(\sin\theta = \frac{0.7611 - 0.7575}{2} = 0.0018\).
\(\%\text{ uncertainty in } \sin\theta = \left(\frac{0.0018}{0.7593}\right) \times 100\% = 0.24\%\).

Step 5: Total percentage uncertainty in \(\lambda\):
\(\%\text{ uncertainty in } \lambda = \%\text{ uncertainty in } d + \%\text{ uncertainty in } \sin\theta = 1.00\% + 0.24\% = 1.24\% \approx 1.2\%\).

(a)
- 1 mark: Ensure the laser beam is incident perpendicularly to the face of the diffraction grating.
- 1 mark: Ensure the screen is parallel to the grating so that the distances from the grating to left and right maxima of the same order are symmetrical.
- 1 mark: Position the screen at a large distance \(D\) to maximize \(y\), thereby minimizing percentage uncertainty in distance measurements.

(b)
- 1 mark: Correct conversion and calculation of \(d = 1.67 \times 10^{-6}\text{ m}\).
- 1 mark: Accurate calculation of uncertainty in \(d\) as \(1.7 \times 10^{-8}\text{ m}\).

(c)
- 1 mark: Correct use of trigonometry to show \(\theta \approx 49.4^\circ\).
- 1 mark: Correct application of the diffraction grating equation \(d\sin\theta = n\lambda\).
- 1 mark: Correct value for wavelength \(\lambda = 6.33 \times 10^{-7}\text{ m}\) (or \(633\text{ nm}\)).
- 2 marks: Correct calculation of the percentage uncertainty in \(\sin\theta\) (approx 0.2%) and combination to find the final percentage uncertainty of approx 1.2% (accept 1% to 1.3%). Deduct 1 mark if maximum/minimum method is not attempted or done incorrectly.

Step 1: Calculate the total electrical energy input \(E\):
\(E = V I t = 12.0\text{ V} \times 4.50\text{ A} \times 600\text{ s} = 32,400\text{ J}\).

Step 2: Calculate specific heat capacity \(c\):
\(Q = mc\Delta\theta \Rightarrow c = \frac{E}{m\Delta\theta}\).
\(c = \frac{32,400}{1.20\text{ kg} \times 22.5\text{ }^\circ\text{C}} = 1200\text{ J kg}^{-1}\text{ K}^{-1}\).

Step 3: Calculate percentage uncertainties in each variable:
\(\%\Delta V = \frac{0.2}{12.0} \times 100\% = 1.67\%\).
\(\%\Delta I = \frac{0.05}{4.50} \times 100\% = 1.11\%\).
\(\%\Delta t = \frac{1}{600} \times 100\% = 0.17\%\).
\(\%\Delta m = \frac{0.01}{1.20} \times 100\% = 0.83\%\).
\(\%\Delta(\Delta\theta) = \frac{0.5}{22.5} \times 100\% = 2.22\%\).

Step 4: Sum individual percentage uncertainties:
Total percentage uncertainty in \(c\) is:
\(\%\Delta c = 1.67\% + 1.11\% + 0.17\% + 0.83\% + 2.22\% = 6.00\%\).

Step 5: Calculate absolute uncertainty in \(c\):
\(\text{Absolute uncertainty} = 6.00\% \times 1200\text{ J kg}^{-1}\text{ K}^{-1} = 72\text{ J kg}^{-1}\text{ K}^{-1} \approx 70\text{ J kg}^{-1}\text{ K}^{-1}\).
Therefore, \(c = (1200 \pm 70)\text{ J kg}^{-1}\text{ K}^{-1}\).

Step 6: Analysis of thermal energy loss:
If heat is lost to the surroundings, the measured temperature rise \(\Delta\theta\) is smaller than it would be under ideal conditions. Looking at the equation \(c = \frac{E}{m\Delta\theta}\), a smaller \(\Delta\theta\) leads to a larger calculated value of \(c\). Hence, the calculated value is higher than the true value.

(a)
- 1 mark: Wrap the metal block in a thick layer of high-quality thermal insulator (e.g., polystyrene or glass wool).
- 1 mark: Put a small amount of oil or water in the thermometer hole to ensure good thermal contact between the block and the thermometer.
- 1 mark: Start the experiment with the block below room temperature (by approximately half the expected temperature rise) and heat it until it is the same amount above room temperature so that heat gains from surroundings in the first half cancel heat losses in the second half.

(b)
- 1 mark: Correct calculation of energy input \(E = 32,400\text{ J}\).
- 1 mark: Correct substitution into specific heat capacity formula.
- 1 mark: Correct value for specific heat capacity \(1200\text{ J kg}^{-1}\text{ K}^{-1}\) (or \(\text{J kg}^{-1}\text{ }^\circ\text{C}^{-1}\)).

(c)
- 1 mark: Correct calculations of individual percentage uncertainties.
- 1 mark: Correct final percentage uncertainty of \(6.0\%\).
- 1 mark: Correct absolute uncertainty (\(70\text{ J kg}^{-1}\text{ K}^{-1}\)) and presentation with unit.
- 1 mark: Clear explanation that thermal energy loss results in a smaller measured temperature rise, which makes the calculated value of \(c\) systematically higher than the true value.

Step 1: Describe fiducial marker use:
Place the marker at the equilibrium position (where the mass moves at maximum speed). Time the system as it crosses the marker in the same direction. Time 20 full oscillations and divide the total time by 20 to reduce the fractional uncertainty due to human reaction time.

Step 2: Linearize the equation:
Squaring both sides:
\(T^2 = 4\pi^2 \frac{m + m_{\text{eff}}}{k} = \left(\frac{4\pi^2}{k}\right)m + \frac{4\pi^2 m_{\text{eff}}}{k}\).
This fits \(y = mx + c\), where:
- \(y = T^2\)
- \(x = m\)
- \(\text{Gradient } G = \frac{4\pi^2}{k} \Rightarrow k = \frac{4\pi^2}{G}\)
- \(\text{y-intercept } C = \frac{4\pi^2 m_{\text{eff}}}{k} = G \times m_{\text{eff}} \Rightarrow m_{\text{eff}} = \frac{C}{G}\).

Step 3: Calculate spring constant \(k\) and its uncertainty:
\(k = \frac{4\pi^2}{1.58} = 24.986\text{ N m}^{-1} \approx 25.0\text{ N m}^{-1}\).
Since \(k \propto G^{-1}\), the percentage uncertainty in \(k\) is the same as the percentage uncertainty in the gradient \(G\):
\(\%\text{ uncertainty in } k = \left(\frac{0.04}{1.58}\right) \times 100\% = 2.53\% \approx 2.5\%\).
Absolute uncertainty in \(k = 2.53\% \times 24.986 = 0.63\text{ N m}^{-1}\).

Step 4: Calculate effective mass \(m_{\text{eff}}\) and its uncertainty:
\(m_{\text{eff}} = \frac{C}{G} = \frac{0.038}{1.58} = 0.02405\text{ kg}\).
Since \(m_{\text{eff}} = C \times G^{-1}\), the percentage uncertainty is:
\(\%\text{ uncertainty in } m_{\text{eff}} = \%\text{ uncertainty in } C + \%\text{ uncertainty in } G\).
\(\%\text{ uncertainty in } C = \left(\frac{0.002}{0.038}\right) \times 100\% = 5.26\%\).
\(\%\text{ uncertainty in } m_{\text{eff}} = 5.26\% + 2.53\% = 7.79\% \approx 7.8\%\).
Absolute uncertainty in \(m_{\text{eff}} = 7.79\% \times 0.02405\text{ kg} = 0.00187\text{ kg}\).

Final values:
\(k = (25.0 \pm 0.6)\text{ N m}^{-1}\)
\(m_{\text{eff}} = (0.024 \pm 0.002)\text{ kg}\) (or \((24.1 \pm 1.9)\text{ g}\)).

(a)
- 1 mark: Place the fiducial marker at the centre of the oscillation / equilibrium position.
- 1 mark: Time multiple (e.g. 10 to 20) oscillations.
- 1 mark: Divide the total time by the number of oscillations, reducing the absolute impact of human reaction time on the final period \(T\).

(b)
- 1 mark: Correctly square the original equation to find \(T^2 = \left(\frac{4\pi^2}{k}\right)m + \frac{4\pi^2 m_{\text{eff}}}{k}\).
- 1 mark: Identify that the spring constant is calculated from the gradient: \(k = \frac{4\pi^2}{\text{Gradient}}\).
- 1 mark: Identify that the effective mass is calculated from the ratio of the intercept to the gradient: \(m_{\text{eff}} = \frac{\text{y-intercept}}{\text{Gradient}}\).

(c)
- 1 mark: Correct calculation of \(k = 25.0\text{ N m}^{-1}\) and \(\%\text{ uncertainty} = 2.5\%\).
- 1 mark: Correct calculation of \(m_{\text{eff}} = 0.024\text{ kg}\) (or \(24\text{ g}\)).
- 1 mark: Correct calculation of percentage uncertainty for \(m_{\text{eff}}\) as \(7.8\%\).
- 1 mark: Correct units for both values (\(\text{N m}^{-1}\) or \(\text{kg s}^{-2}\) for \(k\), and \(\text{kg}\) or \(\text{g}\) for \(m_{\text{eff}}\)).

Step 1: Background correction:
Measure the background count rate over a long time (e.g., 10 minutes) with the source removed from the room, to find the background count rate in counts per second (s\(^{-1}\)). Subtract this value from each measured count rate to obtain the corrected count rate \(C\).
Safety: Handle the source using long tongs; keep the source pointed away from body; store it in a lead-lined container when not in use.

Step 2: Analysis of the systematic error in distance:
The actual distance between the source active centre and the GM tube's active volume is \(r = d + x\), where \(d\) is the measured distance from the casing and \(x\) is a systematic error/offset due to internal positioning of components.
According to the inverse-square law:
\(C = \frac{k}{r^2} = \frac{k}{(d + x)^2}\).
Taking the square root and reciprocal of both sides:
\(C^{-1/2} = \frac{1}{\sqrt{k}}(d + x) = \frac{1}{\sqrt{k}}d + \frac{x}{\sqrt{k}}\).
Thus, a plot of \(C^{-1/2}\) against \(d\) is a straight line.
When \(C^{-1/2} = 0\), \(d = -x\). Therefore, the magnitude of the negative x-intercept of this linear plot yields the systematic offset \(x\).

Step 3: Compare GM tube and scintillation counter:
- Detection efficiency: Scintillation counters are much more efficient at detecting gamma rays (typically 10-50%) than GM tubes (<1%), because the solid crystal scintillator has a higher density and stopping power for high-energy photons.
- Dead time: Scintillation counters have a much shorter dead time (typically a few nanoseconds to microseconds) than GM tubes (typically 100-200 \(\mu\)s). Thus, the scintillation counter is less prone to under-counting at high intensities (small distances), making it highly suitable for verifying the inverse-square law.

(a)
- 1 mark: Measure counts for a long duration without the source present, calculate the background count rate, and subtract this from all measured counts.
- 1 mark: Use tongs to maintain distance / point source window away from people / minimize exposure time.

(b)
- 1 mark: State that the systematic error \(x\) arises because the active centers of the source/detector are inside their respective protective housings.
- 1 mark: Show mathematically that \(C^{-1/2}\) is linearly proportional to \(d\).
- 1 mark: State that plotting \(C^{-1/2}\) against \(d\) yields a straight line.
- 1 mark: Explain that the magnitude of the x-intercept of this graph gives the systematic correction factor \(x\).

(c)
- 1 mark: Scintillation counters have much higher efficiency for gamma detection due to solid scintillating medium compared to low-density gas in GM tubes.
- 1 mark: Scintillation counters have significantly shorter dead times than GM tubes.
- 1 mark: Shorter dead time reduces systematic under-counting (coincidence loss) at small distances where intensity is high, making the scintillation counter superior.

Step 1: Calibration and alignment of the Hall probe:
Calibrate the probe by placing it in a known magnetic field (e.g. at the center of a standard solenoid). Orient the flat face of the Hall probe sensor perpendicular to the axis of the coil to ensure it measures only the axial component of the magnetic field.

Step 2: Linear graphical analysis at \(x=0\):
Substituting \(x=0\) into the given equation yields:
\(B = \frac{\mu_0 N I R^2}{2(R^2)^{3/2}} = \frac{\mu_0 N I}{2R}\).
Rearranging this:
\(B = \left(\frac{\mu_0 N}{2R}\right) I\).
Plot a graph of magnetic flux density \(B\) on the y-axis against current \(I\) on the x-axis.
The graph is a straight line through the origin with gradient \(m = \frac{\mu_0 N}{2R}\).
Calculate \(\mu_0\) using the gradient: \(\mu_0 = \frac{2 R m}{N}\).

Step 3: Calculate the value of \(B\) at the center of the coil:
\(B = \frac{(4\pi \times 10^{-7}\text{ H m}^{-1}) \times 150 \times 2.40\text{ A}}{2 \times 0.085\text{ m}}\).
\(B = \frac{4.524 \times 10^{-4}}{0.170} = 2.661 \times 10^{-3}\text{ T} = 2.66\text{ mT}\).

Step 4: Percentage uncertainty calculation:
Since \(B = \frac{\mu_0 N I}{2R}\), and assuming only \(R\) has significant experimental uncertainty:
\(\%\text{ uncertainty in } B = \%\text{ uncertainty in } R\).
\(\%\text{ uncertainty in } R = \left(\frac{0.1\text{ cm}}{8.5\text{ cm}}\right) \times 100\% = 1.18\% \approx 1.2\%\).

Step 5: Calculate absolute uncertainty:
\(\Delta B = 1.18\% \times 2.661 \times 10^{-3}\text{ T} = 0.031 \times 10^{-3}\text{ T}\).
Thus, \(B = (2.66 \pm 0.03) \times 10^{-3}\text{ T}\).

(a)
- 1 mark: Calibrate the Hall probe in a magnetic field of known strength (solenoid).
- 1 mark: Align the flat plate of the Hall probe perpendicularly to the central axis of the circular coil.
- 1 mark: Mention resetting the probe to zero far from any magnetic fields to account for the Earth's magnetic field.

(b)
- 1 mark: Simplify the formula for \(x = 0\) to get \(B = \frac{\mu_0 N I}{2R}\).
- 1 mark: State that plotting \(B\) against \(I\) gives a straight line through the origin.
- 1 mark: Identify that the gradient is \(\frac{\mu_0 N}{2R}\) and show how \(\mu_0\) is found from it.

(c)
- 1 mark: Correct substitution into the simplified formula at \(x=0\).
- 1 mark: Correct value calculated as \(2.66 \times 10^{-3}\text{ T}\).
- 1 mark: Identify that percentage uncertainty in \(B\) is equal to percentage uncertainty in \(R\) (1.18% or 1.2%).
- 1 mark: Express absolute uncertainty correctly as \(0.03 \times 10^{-3}\text{ T}\) (or \(0.03\text{ mT}\)).

Step 1: Derivation:
Draw a diagram showing tension \(T_s\) acting along the string at an angle \(\theta\) to the vertical, and weight \(W = mg\) acting vertically downwards.
Horizontal component provides centripetal force:
\(T_s \sin\theta = m\omega^2 r\).
Vertical component balances weight:
\(T_s \cos\theta = mg\).
Dividing the two equations gives:
\(\tan\theta = \frac{\omega^2 r}{g}\).
Since the radius of the circle is \(r = L \sin\theta\):
\(\tan\theta = \frac{\omega^2 L \sin\theta}{g} \Rightarrow \cos\theta = \frac{g}{\omega^2 L} \Rightarrow \omega = \sqrt{\frac{g}{L \cos\theta}}\).
Using \(T = \frac{2\pi}{\omega}\):
\(T = 2\pi\sqrt{\frac{L\cos\theta}{g}}\).

Step 2: Calculate the experimental value of \(g\):
Time period \(T = \frac{36.64}{20} = 1.832\text{ s}\).
Rearranging for \(g\):
\(g = \frac{4\pi^2 L \cos\theta}{T^2}\).
Substitute the given values:
\(g = \frac{4 \times \pi^2 \times 0.920 \times \cos(25.0^\circ)}{1.832^2} = \frac{32.915}{3.3562} = 9.807\text{ m s}^{-2}\).

Step 3: Propagate uncertainties:
- \(\%\text{ uncertainty in } L = \left(\frac{0.005}{0.920}\right) \times 100\% = 0.54\%\).
- \(\%\text{ uncertainty in } T = \left(\frac{0.10}{36.64}\right) \times 100\% = 0.27\%\).
- \(\%\text{ uncertainty in } T^2 = 2 \times 0.27\% = 0.54\%\).
- Uncertainty in \(\cos\theta\):
\(\cos(24^\circ) = 0.9135\)
\(\cos(26^\circ) = 0.8988\)
\(\text{Uncertainty } \Delta(\cos\theta) = \frac{0.9135 - 0.8988}{2} = 0.00735\).
\(\%\text{ uncertainty in } \cos\theta = \left(\frac{0.00735}{\cos(25^\circ)}\right) \times 100\% = \left(\frac{0.00735}{0.9063}\right) \times 100\% = 0.81\%\).
- Total percentage uncertainty in \(g\):
\(\%\Delta g = \%\Delta L + \%\Delta(\cos\theta) + \%\Delta(T^2) = 0.54\% + 0.81\% + 0.54\% = 1.89\%\).

Step 4: Calculate absolute uncertainty:
\(\Delta g = 1.89\% \times 9.807\text{ m s}^{-2} = 0.185\text{ m s}^{-2} \approx 0.19\text{ m s}^{-2}\).
So, \(g = (9.81 \pm 0.19)\text{ m s}^{-2}\).

Step 5: Consistency:
The accepted value of \(9.81\text{ m s}^{-2}\) lies within the experimental range of \(9.62\text{ m s}^{-2}\) to \(10.00\text{ m s}^{-2}\). Therefore, the student's result is highly consistent with the accepted value.

(a)
- 1 mark: Correct free-body force diagram showing tension along the string and weight downwards.
- 1 mark: Setting up correct simultaneous equations for horizontal (centripetal) and vertical forces.
- 1 mark: Correct substitution of \(r = L \sin\theta\) to complete the derivation.

(b)
- 1 mark: Calculation of period \(T = 1.832\text{ s}\).
- 1 mark: Correct transposition of formula to solve for \(g\).
- 1 mark: Correct calculation of \(g = 9.81\text{ m s}^{-2}\) (accept 9.80 - 9.81).

(c)
- 1 mark: Correct calculation of the percentage uncertainty in \(\cos\theta\) using max/min bounds (approx 0.8%).
- 1 mark: Correctly adding percentage uncertainties for \(L\), \(\cos\theta\), and \(T^2\) to get approx 1.9%.
- 1 mark: Correct absolute uncertainty of \(0.19\text{ m s}^{-2}\) (or \(0.2\text{ m s}^{-2}\)).
- 1 mark: Concluding that the value is consistent because the accepted value lies within the range of uncertainty.

(a) The circuit must include a power supply, an ammeter in series, and a voltmeter in parallel across the portion of the wire under test. A flying lead or crocodile clip is used to alter the length of wire in the circuit.

(b) To determine the cross-sectional area of the wire:
1. Check the micrometer screw gauge for zero error and subtract/add this correction value from all future diameter readings to eliminate systematic error.
2. Measure the diameter at several different points along the length of the wire, and at different orientations at each point, to calculate an average diameter and minimize random error.
3. Use the formula \(A = \frac{\pi d^2}{4}\) to find the cross-sectional area.

(c) Calculations:
1. Cross-sectional area:
\(A = \frac{\pi (0.38 \times 10^{-3}\text{ m})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\)

2. Resistivity \(\rho\):
\(\rho = \frac{R A}{L} = \frac{4.25 \times 1.134 \times 10^{-7}}{0.800} = 6.02 \times 10^{-7}\ \Omega\text{ m}\)

3. Percentage Uncertainties:
- Percentage uncertainty in \(L = \frac{0.002}{0.800} \times 100\% = 0.25\%\)
- Percentage uncertainty in \(R = \frac{0.05}{4.25} \times 100\% = 1.18\%\)
- Percentage uncertainty in \(d = \frac{0.01}{0.38} \times 100\% = 2.63\%\)
- Since \(A \propto d^2\), percentage uncertainty in \(A = 2 \times 2.63\% = 5.26\%\)
- Total percentage uncertainty in \(\rho = \%\Delta R + \%\Delta A + \%\Delta L = 1.18\% + 5.26\% + 0.25\% = 6.69\% \approx 6.7\%\)

(a)
- MP1: Ammeter in series, voltmeter in parallel with the wire under test. (1)
- MP2: Complete, functional circuit containing power source and showing a clear method to vary the length of the wire connected (e.g., sliding contact / flying lead). (1)

(b)
- MP1: Checking and correcting for zero error (to reduce systematic uncertainty). (1)
- MP2: Measuring the diameter in multiple positions and orientations, and calculating the mean (to reduce random uncertainty). (1)
- MP3: Use of \(A = \frac{\pi d^2}{4}\) to determine area. (1)

(c)
- MP1: Calculation of cross-sectional area as \(1.13 \times 10^{-7}\text{ m}^2\). (1)
- MP2: Calculation of resistivity \(\rho = 6.0 \times 10^{-7}\ \Omega\text{ m}\) (accept range \(6.0 \times 10^{-7}\) to \(6.03 \times 10^{-7}\)). (1)
- MP3: Correct percentage uncertainty in \(R\) (1.18%) and \(L\) (0.25%). (1)
- MP4: Correct percentage uncertainty in \(A\) as double the uncertainty of \(d\) (5.26% or 5.3%). (1)
- MP5: Correct total percentage uncertainty of 6.7% (accept 6.69% or 7%). (1)

(a)
- Insulating material reduces the rate of thermal energy transfer from the block to the surroundings, minimizing systematic error.
- Oil in the thermometer hole ensures good thermal contact between the block and the thermometer bulb, reducing the time lag in temperature measurement.

(b)
- Electrical energy supplied is \(E = V I t\).
- Temperature rise equation: \(V I t = m c (\theta - \theta_0)\), where \(\theta_0\) is the initial temperature.
- Rearranging gives \(\theta = \left(\frac{VI}{mc}\right) t + \theta_0\).
- This is in the form \(y = mx + c\), where gradient \(m = \frac{VI}{mc}\).

(c)
- Since \(\text{gradient} = \frac{VI}{mc}\), we have:
\(c = \frac{VI}{m \times \text{gradient}} = \frac{11.8\text{ V} \times 4.15\text{ A}}{1.00\text{ kg} \times 0.0524\text{ }^\circ\text{C s}^{-1}} = 934.5\text{ J kg}^{-1}\text{ K}^{-1} \approx 935\text{ J kg}^{-1}\text{ K}^{-1}\).

(d)
- The experimental value is larger than the accepted value because thermal energy is lost to the environment despite the insulation.
- Since some supplied energy is lost, more electrical energy is needed to raise the temperature of the block, which is recorded as a smaller gradient and hence a larger calculated specific heat capacity.
- To reduce this discrepancy, a cooling correction curve could be used, or the block could be cooled below room temperature by a certain amount before heating it to the same amount above room temperature (equalizing energy gained and lost).

(a)
- MP1: Insulation reduces heat loss to the surroundings. (1)
- MP2: Oil ensures good thermal contact (reducing measurement lag). (1)

(b)
- MP1: States \(E = V I t = m c \Delta \theta\) or \(m c (\theta - \theta_0)\). (1)
- MP2: Clearly rearranges to the form \(\theta = \left(\frac{VI}{mc}\right) t + \theta_0\) and identifies the gradient. (1)

(c)
- MP1: Rearranges gradient formula to make \(c\) the subject: \(c = \frac{VI}{m \cdot \text{gradient}}\). (1)
- MP2: Correct substitution and calculation to give \(935\text{ J kg}^{-1}\text{ K}^{-1}\) (or \(934\text{ J kg}^{-1}\text{ K}^{-1}\)). (1)

(d)
- MP1: Identifies that heat loss to the surroundings is the primary systematic error. (1)
- MP2: Explains that this means a lower temperature rise is observed (or more electrical energy is required), yielding a smaller gradient and therefore a larger calculated \(c\). (1)
- MP3: Suggests a practical improvement, such as cooling the block below room temperature first so that net heat exchange with surroundings is zero. (1)
- MP4: Or suggests using a cooling correction curve / plot temperature against time after switching off the heater. (1)

(a)
- Position the laser so that the beam is incident normally (at 90 degrees) to the diffraction grating. Place the screen parallel to the grating at a distance of 1.5 - 2.0 meters to ensure the maxima are widely separated and easy to measure.
- Safety: Avoid looking directly into the laser beam or its specular reflections. Use a black non-reflective backstop behind the screen to safely absorb the beam. Put warning signs outside the laboratory.

(b)
- From geometry: \(\tan \theta = \frac{y_2}{D} = \frac{0.442}{1.50} = 0.29467\)
- \(\theta = \arctan(0.29467) = 16.42^\circ\)
- To find the uncertainty in \(\theta\):
- \(\theta_{\max} = \arctan\left(\frac{0.442 + 0.005}{1.50 - 0.01}\right) = \arctan\left(\frac{0.447}{1.49}\right) = 16.70^\circ\)
- \(\theta_{\min} = \arctan\left(\frac{0.442 - 0.005}{1.50 + 0.01}\right) = \arctan\left(\frac{0.437}{1.51}\right) = 16.14^\circ\)
- Absolute uncertainty \(\Delta \theta = \frac{16.70^\circ - 16.14^\circ}{2} = 0.28^\circ\)
- Percentage uncertainty in \(\theta = \frac{0.28^\circ}{16.42^\circ} \times 100\% \approx 1.70\%\)
- (Alternative calculation using small-angle approximation: \(\%\Delta \theta \approx \%\Delta y_2 + \%\Delta D = \frac{0.005}{0.442} \times 100\% + \frac{0.01}{1.50} \times 100\% = 1.13\% + 0.67\% = 1.80\%\). Both methods are fully acceptable).

(c)
- Use the grating equation: \(d \sin \theta = n \lambda\)
- For \(n = 2\):
- \(d = \frac{2 \lambda}{\sin \theta} = \frac{2 \times 632.8 \times 10^{-9}\text{ m}}{\sin(16.42^\circ)} = \frac{1.2656 \times 10^{-6}\text{ m}}{0.2827} = 4.477 \times 10^{-6}\text{ m}\)
- Number of lines per millimetre \(N_{\text{mm}}\):
- \(N_{\text{mm}} = \frac{10^{-3}\text{ m}}{d} = \frac{10^{-3}}{4.477 \times 10^{-6}} = 223.4\text{ lines mm}^{-1} \approx 223\text{ lines mm}^{-1}\)

(a)
- MP1: Aligns grating perpendicular to laser beam and screen parallel to grating. (1)
- MP2: Measures distance from grating to screen using a metre rule (and keeps it >1 m to reduce uncertainty). (1)
- MP3: Mentions safety precaution: avoid direct eye exposure, or use a black backstop behind the screen. (1)

(b)
- MP1: Calculates \(\theta = 16.42^\circ\) (or 0.287 rad). (1)
- MP2: Shows clear method to calculate the maximum and minimum angles, or sums fractional uncertainties. (1)
- MP3: Obtains percentage uncertainty of 1.7% to 1.8%. (1)

(c)
- MP1: Uses \(d \sin \theta = n \lambda\) with \(n = 2\). (1)
- MP2: Correctly substitutes values to find \(d = 4.48 \times 10^{-6}\text{ m}\) (or \(4.5 \times 10^{-6}\text{ m}\)). (1)
- MP3: Relates line density \(N\) to grating spacing as \(N = \frac{1}{d}\). (1)
- MP4: Correctly calculates \(N = 223\text{ lines mm}^{-1}\) (allow 220 - 225 depending on rounding). (1)

(a)
- Diagram should show: a vertical rule, an electromagnet holding a steel ball bearing at the top, a trapdoor connected to a timer circuit at the bottom, and an electronic timer.
- The timer is started when the switch to the electromagnet is opened, breaking the circuit. The timer is stopped when the falling ball bearing hits the trapdoor, breaking (or closing) the electronic connection.

(b)
- A steel ball bearing is much denser and heavier than a table tennis ball of the same size, which means the air resistance is negligible compared to its weight.
- This ensures the motion is a very close approximation to true free fall (constant acceleration equal to \(g\)).

(c)
- Using \(s = ut + \frac{1}{2}at^2\) with \(u = 0\), the equation becomes \(h = \frac{1}{2}gt^2\).
- Plot a graph of height \(h\) on the y-axis against the square of time \(t^2\) on the x-axis.
- The graph should be a straight line passing through the origin. Since the gradient \(m = \frac{1}{2}g\), the acceleration of free fall is obtained by \(g = 2 \times \text{gradient}\).

(d)
- A delay in releasing the ball bearing after the timer starts means the recorded time \(t\) is larger than the actual fall time.
- If a single data point is used, \(g = \frac{2h}{t^2}\), so an overestimate of \(t\) leads to an underestimated value of \(g\).
- In the equation \(\sqrt{h} = \sqrt{\frac{g}{2}} (t - t_{\text{delay}})\), plotting \(\sqrt{h}\) against \(t\) yields a straight line where the gradient is still \(\sqrt{\frac{g}{2}}\). The delay only shifts the intercept on the time axis (by \(t_{\text{delay}}\)), leaving the gradient (and thus the calculated value of \(g\)) completely unaffected by the systematic error.

(a)
- MP1: Draw a clear, labelled diagram showing electromagnet, ball, scale, and trapdoor. (1)
- MP2: Explains that opening the switch cuts the current to the electromagnet, starting the timer. (1)
- MP3: Explains that the ball hitting the trapdoor breaks/closes the electrical connection, stopping the timer. (1)

(b)
- MP1: Explains that the steel ball has a higher mass/density (or weight-to-drag ratio) than a table tennis ball. (1)
- MP2: Concludes that air resistance is negligible, so the acceleration is constant and equal to \(g\). (1)

(c)
- MP1: States the equation \(h = \frac{1}{2}gt^2\). (1)
- MP2: Identifies that \(h\) is plotted on the vertical axis and \(t^2\) on the horizontal axis (or vice versa). (1)
- MP3: Explains that \(g = 2 \times \text{gradient}\) (or \(g = \frac{2}{\text{gradient}}\) if \(t^2\) is on the vertical axis). (1)

(d)
- MP1: Explains that the delay increases the recorded time \(t\), leading to a lower calculated \(g\) for a single point. (1)
- MP2: Explains that plotting \(\sqrt{h}\) against \(t\) results in a linear relationship where the systematic error only affects the intercept, leaving the gradient unaffected. (1)