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The density of a cylinder is given by the formula:

\(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 L}\)

The expression for the fractional uncertainty in \(\rho\) is:

\(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\)

Now, calculate each percentage uncertainty:

- Percentage uncertainty in \(m = \frac{0.1}{25.0} \times 100\% = 0.40\%\)

- Percentage uncertainty in \(d = \frac{0.02}{1.20} \times 100\% = 1.67\%\)

- Percentage uncertainty in \(L = \frac{0.05}{5.00} \times 100\% = 1.00\%\)

Substitute these values back into the equation:

\(\frac{\Delta \rho}{\rho} \times 100\% = 0.40\% + 2(1.67\%) + 1.00\% = 0.40\% + 3.33\% + 1.00\% = 4.73\% \approx 4.7\%\)

1 mark for calculating individual percentage uncertainties, doubling the uncertainty of the diameter, and summing them to obtain 4.7% (accuracy).

The percentage uncertainty in \(E\) is given by summing the fractional uncertainties of the individual quantities, keeping in mind that the diameter \(d\) is squared:

\(\frac{\Delta E}{E} = \frac{\Delta F}{F} + \frac{\Delta L}{L} + 2 \frac{\Delta d}{d} + \frac{\Delta e}{e}\)

Substitute the given percentage values:

\(\frac{\Delta E}{E} = 1.5\% + 0.8\% + 2(1.2\%) + 2.5\%\)

\(\frac{\Delta E}{E} = 1.5\% + 0.8\% + 2.4\% + 2.5\% = 7.2\%\)

1 mark for doubling the percentage uncertainty of the diameter and summing all percentage values correctly to obtain 7.2%.

The period \(T\) is the total time divided by the number of oscillations \(n = 20\):

\(T = \frac{t}{n} = \frac{32.4}{20} = 1.62\text{ s}\)

The absolute uncertainty in the period, \(\Delta T\), is the absolute uncertainty in the total time, \(\Delta t\), divided by the number of oscillations:

\(\Delta T = \frac{\Delta t}{n} = \frac{0.2}{20} = 0.01\text{ s}\)

Therefore, the period of one oscillation is \((1.62 \pm 0.01)\text{ s}\).

1 mark for dividing both the measured time and the absolute uncertainty by 20 to yield 1.62 +/- 0.01 s.

- **Precision**: The readings are very close to one another, with a maximum variation of only \(\pm 0.01\text{ mm}\) from the average raw reading of \(1.25\text{ mm}\). This indicates high precision.
- **Accuracy**: The average of the raw readings is \(1.25\text{ mm}\), which is different from the actual value of \(1.20\text{ mm}\) by \(+0.05\text{ mm}\) (the systematic error). Because the raw readings are consistently offset from the true value, they have low accuracy.

1 mark for correctly identifying that the small spread means high precision and the systematic deviation from the actual value means low accuracy.

- **Quark charges**:
- up (\(u\)): \(+\frac{2}{3}e\)
- down (\(d\)): \(-\frac{1}{3}e\)
- strange (\(s\)): \(-\frac{1}{3}e\)

Total charge \(Q = +\frac{2}{3}e - \frac{1}{3}e - \frac{1}{3}e = 0\).

- **Strangeness**:
- The strange quark (\(s\)) has a strangeness of \(-1\).
- The other quarks have a strangeness of \(0\).

Total strangeness \(= -1\).

1 mark for summing the individual charges to 0 and identifying the strangeness as -1.

In \(\beta^+\) decay, a proton decays into a neutron.
- Proton quark composition is \(uud\) and neutron quark composition is \(udd\). Thus, an up quark (\(u\)) changes into a down quark (\(d\)), which is a \(u \rightarrow d\) conversion.
- To conserve charge and lepton number, the decay produces a positron (\(e^+\)) and an electron neutrino (\(\nu_e\)).

1 mark for identifying the correct quark transition (u to d) and the correct lepton pair (positron and electron neutrino).

According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. \(E\) is given by:

\(E = \frac{\Delta (N\Phi)}{\Delta t} = N \frac{\Delta B \cdot A}{\Delta t}\)

Given:
- \(N = 150\)
- \(A = 2.0 \times 10^{-3}\text{ m}^2\)
- \(\Delta B = 0.40\text{ T}\)
- \(\Delta t = 0.15\text{ s}\)

Substitute the values:

\(E = 150 \times \frac{0.40 \times 2.0 \times 10^{-3}}{0.15}\)

\(E = 150 \times \frac{8.0 \times 10^{-4}}{0.15} = \frac{0.12}{0.15} = 0.80\text{ V}\)

1 mark for using the correct formula for induced e.m.f. and performing the calculation to get 0.80 V.

Smoothing is determined by the time constant \(\tau = RC\) of the discharge circuit compared to the period of the input alternating current.
- To increase smoothing (and thus reduce the ripple voltage), the capacitor must discharge more slowly between peaks.
- This requires a larger time constant \(\tau = RC\).
- Increasing the resistance of the load resistor \(R\) increases the time constant, meaning the capacitor discharges slower, producing a smoother output.

1 mark for identifying that increasing the load resistance increases the time constant RC, reducing the discharge rate and therefore the ripple voltage.

The density \(\rho\) of a cylinder of mass \(m\), length \(l\), and diameter \(d\) is given by:

\(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 l}\)

The fractional uncertainty in \(\rho\) is determined by adding the fractional uncertainties of the independent variables, multiplying by their respective powers:

\(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2 \frac{\Delta d}{d} + \frac{\Delta l}{l}\)

Calculate each percentage uncertainty:
- For mass: \(\frac{\Delta m}{m} \times 100\% = \frac{3}{120} \times 100\% = 2.5\%\)
- For diameter: \(\frac{\Delta d}{d} \times 100\% = \frac{0.04}{2.00} \times 100\% = 2.0\%\)
- For length: \(\frac{\Delta l}{l} \times 100\% = \frac{0.1}{5.0} \times 100\% = 2.0\%\)

Substitute these into the equation:

\(\frac{\Delta \rho}{\rho} \times 100\% = 2.5\% + 2(2.0\%) + 2.0\% = 8.5\%\)

1 mark for the correct calculation of percentage uncertainty by summing individual percentage uncertainties with correct power factors:
- Mass percentage uncertainty = 2.5% (Method Mark)
- Diameter percentage uncertainty = 2.0% (multiplied by 2 due to exponent of 2) (Method Mark)
- Length percentage uncertainty = 2.0% (Method Mark)
- Total percentage uncertainty = 2.5% + 4.0% + 2.0% = 8.5% (Accuracy Mark)

The relationship for the fractional uncertainty in resistivity \(\rho\) is:

\(\frac{\Delta \rho}{\rho} = 2\frac{\Delta d}{d} + \frac{\Delta V}{V} + \frac{\Delta I}{I} + \frac{\Delta L}{L}\)

Using the given percentage uncertainties:
- Diameter contribution: \(2 \times 1.5\% = 3.0\%\)
- Potential difference contribution: \(2.0\%\)
- Current contribution: \(1.0\%\)
- Length contribution: \(0.5\%\)

Summing these contributions:

\(\text{Total percentage uncertainty} = 3.0\% + 2.0\% + 1.0\% + 0.5\% = 6.5\%\)

1 mark for correctly applying the uncertainty combination rule. Note that the uncertainty of the diameter must be multiplied by 2 because \(d\) is squared. Summing the terms yields: 2(1.5%) + 2.0% + 1.0% + 0.5% = 6.5%.

A zero error is a systematic error because it shifts all measurements by a constant value in the same direction.

- Accuracy is a measure of how close the experimental mean is to the true value. Because every individual reading is shifted by \(-0.02\text{ mm}\), the calculated mean will also be shifted from the true value, decreasing the accuracy.
- Precision is a measure of the agreement between the individual measurements (the spread of the values), which is determined by random errors. A constant systematic shift does not affect the spread of the data points relative to one another, so the precision remains unaffected.

1 mark for identifying that systematic errors (like zero error) decrease the accuracy of the final result but do not change the precision of the set of measurements.

In beta-minus (\(\beta^-\)) decay, a neutron within the nucleus decays into a proton, emitting an electron (\(e^-\)) and an electron antineutrino (\(\bar{
u}_e\)).

- The quark structure of a neutron is \(udd\).
- The quark structure of a proton is \(uud\).
- Therefore, during this decay, a down quark (\(d\)) is converted into an up quark (\(u\)).
- The emitted leptons are an electron and an electron antineutrino.

1 mark for identifying the correct quark transformation (down to up) and the correct pair of emitted leptons (electron and electron antineutrino) for beta-minus decay.

Let us check each reaction against conservation laws:

- Option A: \(p + p \rightarrow p + \pi^+\)
- Baryon number: LHS = \(1 + 1 = 2\); RHS = \(1 + 0 = 1\) (not conserved).

- Option B: \(\pi^+ + n \rightarrow p + e^+ + \nu_e\)
- Charge: LHS = \(+1 + 0 = +1\); RHS = \(+1 + 1 + 0 = +2\) (not conserved).

- Option C: \(\pi^- + p \rightarrow n + \pi^0\)
- Charge: LHS = \(-1 + 1 = 0\); RHS = \(0 + 0 = 0\) (conserved).
- Baryon number: LHS = \(0 + 1 = 1\); RHS = \(1 + 0 = 1\) (conserved).
- Lepton number: LHS = \(0\); RHS = \(0\) (conserved).
This reaction is possible.

- Option D: \(p + e^- \rightarrow n + \bar{
u}_e\)
- Lepton number: LHS = \(0 + 1 = 1\); RHS = \(0 + (-1) = -1\) (not conserved, as the electron antineutrino has lepton number \(-1\)).

1 mark for testing conservation of charge, baryon number, and lepton number for each reaction and identifying that only option C satisfies all conservation laws.

According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. \(E\) is given by the rate of change of magnetic flux linkage:

\(E = \frac{d(N\Phi)}{dt} = N \frac{d\Phi}{dt}\)

The magnetic flux \(\Phi\) through a single turn of the coil of area \(A\) inside the field is:

\(\Phi = B A\)

As the coil is pulled out of the field, the length of the square coil remaining inside the field decreases. Let the remaining length inside be \(x\). The area inside the field is \(A = s \cdot x\). Therefore:

\(\Phi = B s x\)

Taking the derivative with respect to time:

\(\frac{d\Phi}{dt} = B s \frac{dx}{dt} = B s v\)

Where \(v\) is the speed at which the coil is pulled out. Therefore, the total induced e.m.f. in the \(N\)-turn coil is:

\(E = N B s v\)

1 mark for correctly applying Faraday's law of electromagnetic induction to a moving square coil of N turns, obtaining the correct magnitude E = NBsv.

The ripple voltage represents the variation in the smoothed output voltage. It is determined by how much the capacitor discharges through the load resistor between successive peaks of the rectified input signal.

- The discharge is governed by the time constant \(\tau = R C\), where \(R\) is the resistance and \(C\) is the capacitance.
- To decrease the ripple (i.e., make the output voltage smoother), we want the capacitor to lose less charge between peaks.
- This can be achieved by:
1. Increasing the capacitance \(C\), which increases \(\tau\) and slows down the discharge rate.
2. Increasing the resistance \(R\), which also increases \(\tau\) and slows down the discharge rate.
3. Increasing the frequency \(f\) of the AC supply, which reduces the time interval between successive charging peaks, giving the capacitor less time to discharge.

Therfore, increasing the frequency of the alternating input voltage (Option D) decreases the ripple voltage.

1 mark for identifying that increasing the frequency of the alternating input reduces the time between charging peaks, resulting in less discharge and thus a decreased ripple voltage.

In a full-wave bridge rectifier, at any given instant during either half-cycle of the AC input, current flows through exactly two diodes in series with the load resistor.

- Since each conducting diode has a forward potential drop of \(0.7\text{ V}\), the total potential drop across the two conducting diodes is:

\(V_{\text{diodes}} = 2 \times 0.7\text{ V} = 1.4\text{ V}\)

- The peak voltage \(V_{\text{peak, resistor}}\) across the load resistor is therefore:

\(V_{\text{peak, resistor}} = V_{\text{input, peak}} - V_{\text{diodes}} = 12.0\text{ V} - 1.4\text{ V} = 10.6\text{ V}\)

- The peak current \(I_{\text{peak}}\) through the resistor is calculated using Ohm's Law:

\(I_{\text{peak}} = \frac{V_{\text{peak, resistor}}}{R} = \frac{10.6\text{ V}}{150\ \Omega} \approx 0.0707\text{ A} = 70.7\text{ mA}\)

1 mark for calculating the combined forward voltage drop of two conducting diodes (1.4 V), subtracting this from the peak input voltage to find the peak voltage across the resistor (10.6 V), and then using Ohm's law to find the correct peak current (70.7 mA).

First, calculate the percentage uncertainty in each of the measured quantities: For resistance \(R\): \(\frac{0.15}{4.50} \times 100\% \approx 3.33\%\). For diameter \(d\): \(\frac{0.01}{0.50} \times 100\% = 2.00\%\). For length \(L\): \(\frac{0.006}{1.200} \times 100\% = 0.50\%\). Using the formula for resistivity \(\rho = \frac{R \pi d^2}{4 L}\), the percentage uncertainty in \(\rho\) is given by: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\). Substitute the calculated values: \(\frac{\Delta \rho}{\rho} = 3.33\% + 2(2.00\%) + 0.50\% = 3.33\% + 4.00\% + 0.50\% = 7.83\%\). Rounded to two significant figures, this is \(7.8\%\).

Correct answer: C. 1 mark for the correct calculation of the percentage uncertainty by combining the individual uncertainties with the correct powers.

The temperature rise is calculated as: \(\Delta T = T_2 - T_1 = 68.7\ ^\circ\text{C} - 23.4\ ^\circ\text{C} = 45.3\ ^\circ\text{C}\). For addition or subtraction, the absolute uncertainties of each quantity are added: \(\Delta (\Delta T) = \Delta T_1 + \Delta T_2 = 0.2\ ^\circ\text{C} + 0.3\ ^\circ\text{C} = 0.5\ ^\circ\text{C}\). Therefore, the temperature rise is \((45.3 \pm 0.5)\ ^\circ\text{C}\).

Correct answer: C. 1 mark for calculating the difference and adding the absolute uncertainties.

The relationship is \(h = \frac{1}{2}gt^2\), so the gradient \(m\) of the graph of \(h\) against \(t^2\) is \(m = \frac{1}{2}g\), which means \(g = 2m\). The best estimate for \(g\) is \(g_{\text{best}} = 2 \times 4.85 = 9.70\text{ m s}^{-2}\). The uncertainty in the gradient \(\Delta m\) is the difference between the best-fit gradient and the worst-fit gradient: \(\Delta m = 4.85 - 4.61 = 0.24\text{ m s}^{-2}\). Therefore, the absolute uncertainty in \(g\) is \(\Delta g = 2 \times \Delta m = 2 \times 0.24 = 0.48\text{ m s}^{-2}\). This gives \(g = (9.70 \pm 0.48)\text{ m s}^{-2}\).

Correct answer: D. 1 mark for correctly relating the gradient and its uncertainty to the value of g and its uncertainty.

A zero error in a measuring instrument (such as the closed reading of \(-0.02\text{ mm}\)) affects all measurements by the same constant amount, causing a systematic error. The slight variations in the readings at different positions of the wire are due to random fluctuations in the wire's diameter or the measurement process, which represent random errors.

Correct answer: A. 1 mark for identifying the zero reading as a systematic error and the variations as a random error.

An up quark \(u\) has an electric charge of \(+\frac{2}{3}e\) and a strange quark \(s\) has an electric charge of \(-\frac{1}{3}e\). The net charge \(Q\) is \(+\frac{2}{3}e + \frac{2}{3}e - \frac{1}{3}e = +1e\). Every quark has a baryon number of \(+\frac{1}{3}\), so the total baryon number \(B\) of the three-quark system is \(\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = +1\).

Correct answer: A. 1 mark for correctly calculating both the net charge and the baryon number.

Fundamental particles at A Level include leptons (such as electrons, muons, and neutrinos) and quarks. Protons, neutrons, and pions are hadrons, which are made of quarks and are therefore not fundamental. In group C, the electron, muon, and neutrino are all leptons, so they are all fundamental particles.

Correct answer: C. 1 mark for correctly identifying that all listed particles are fundamental leptons.

According to Faraday's law, the average induced e.m.f. \(E\) is given by: \(E = \frac{\Delta \Phi}{\Delta t} = N \frac{\Delta (B A)}{\Delta t}\). Substituting the given values: \(E = 200 \times \frac{(0.40 \times 1.5 \times 10^{-3})}{0.12} = 200 \times \frac{6.0 \times 10^{-4}}{0.12} = 1.0\text{ V}\).

Correct answer: C. 1 mark for calculating the magnetic flux linkage change and dividing by time to find the induced e.m.f.

The ripple voltage \(\Delta V\) of a smoothed rectifier output depends on how much the capacitor discharges between the peaks of the rectified input signal. The discharge rate is determined by the time constant \(\tau = RC\) where \(R\) is the load resistance and \(C\) is the capacitance. To increase the ripple voltage (making the output less smooth), the capacitor must discharge more rapidly, which requires a smaller time constant. Therefore, decreasing the capacitance \(C\) (or decreasing the load resistance \(R\)) will increase the ripple voltage. Decreasing the frequency of the input signal also increases the ripple voltage because there is a longer time between charge peaks.

Correct answer: D. 1 mark for identifying that a smaller capacitance increases the rate of discharge and hence increases the ripple voltage.

The volume is calculated as:

\( V = \pi r^2 h = \pi \times (2.00)^2 \times 10.0 \approx 125.66\text{ cm}^3 \)

To find the absolute uncertainty \( \Delta V \), we first calculate the fractional uncertainty:

\( \frac{\Delta V}{V} = 2 \left( \frac{\Delta r}{r} \right) + \frac{\Delta h}{h} \)

\( \frac{\Delta V}{V} = 2 \left( \frac{0.05}{2.00} \right) + \frac{0.2}{10.0} = 0.05 + 0.02 = 0.07 \)

This corresponds to a percentage uncertainty of \( 7\% \).

Now find \( \Delta V \):

\( \Delta V = 0.07 \times 125.66 = 8.796\text{ cm}^3 \)

In standard practice, experimental uncertainty is expressed to 1 significant figure:

\( \Delta V \approx 9\text{ cm}^3 \)

This means the volume should be rounded to match the precision of the uncertainty:

\( V = (126 \pm 9)\text{ cm}^3 \)

1 mark for the correct option C.
- Correctly calculates the value of volume \( V \approx 126\text{ cm}^3 \) (1 mark)
- Correctly combines fractional uncertainties to find \( \Delta V = 9\text{ cm}^3 \) (1 mark)

The fractional uncertainty in a calculated value \( g \) is found by adding the individual fractional uncertainties of the independent variables. For power terms, the fractional uncertainty is multiplied by the index:

\( \frac{\Delta g}{g} = \frac{\Delta h}{h} + 2 \frac{\Delta t}{t} \)

Given that the percentage uncertainty in \( h \) is \( 1.5\% \) and the percentage uncertainty in \( t \) is \( 2.0\% \):

\( \% \text{ uncertainty in } g = 1.5\% + 2 \times 2.0\% = 5.5\% \)

1 mark for the correct option B.
- Identifies the correct rule for combining uncertainties involving powers (1 mark).
- Correctly computes the final percentage uncertainty as \( 5.5\% \) (1 mark).

Accuracy refers to how close the average of the measurements is to the true value. Here, the average is around \( 356\text{ m s}^{-1} \), which is significantly higher than the true value of \( 343\text{ m s}^{-1} \). Hence, they are not accurate.

Precision refers to how close the independent measurements are to each other (i.e., the spread or range). Here, the range of measurements is very small (from \( 355\text{ m s}^{-1} \) to \( 357\text{ m s}^{-1} \)), meaning they are highly precise.

Thus, the measurements are precise but not accurate.

1 mark for the correct option B.
- Correctly identifies that the readings are precise because of their small spread (1 mark).
- Correctly identifies that the readings are inaccurate because they deviate significantly from the true value (1 mark).

Random errors are unpredictable and cause readings to scatter around the mean value. The main source of random error in timing a pendulum manually is human reaction time during starting and stopping the stopwatch.

Measuring the total time taken for a large number of oscillations (such as 20) and dividing by 20 distributes the random timing error (which remains roughly constant at about \( \pm 0.1 \text{ to } 0.2\text{ s} \) for human reactions) over 20 cycles. This reduces the percentage and absolute random error in the determined period of a single oscillation.

1 mark for the correct option B.
- Explains how dividing the total time of multiple oscillations reduces the effect of random timing error on the calculated period (1 mark).

In beta-minus (\( \beta^- \)) decay, a neutron (quark structure \( \text{udd} \)) changes into a proton (quark structure \( \text{uud} \)). This means a down quark (\( \text{d} \)) turns into an up quark (\( \text{u} \)).

The electron antineutrino (\( \bar{
u}_\text{e} \)) is an elementary particle that does not experience the strong nuclear force, so it is classified as a lepton.

1 mark for the correct option A.
- Identifies the quark transition as \( \text{d} \rightarrow \text{u} \) (1 mark).
- Identifies the antineutrino as a lepton (1 mark).

The total electric charge of the baryon is the sum of the charges of its constituent quarks:

\( Q = Q_\text{u} + Q_\text{s} + Q_\text{c} \)

\( Q = \left(+\frac{2}{3}e\right) + \left(-\frac{1}{3}e\right) + \left(+\frac{2}{3}e\right) = +\frac{3}{3}e = +e \)

1 mark for the correct option B.
- Correctly adds the individual quark charges (1 mark).

According to Faraday's law of electromagnetic induction, the magnitude of the induced e.m.f. \( E \) is given by:

\( E = N \frac{\Delta \Phi}{\Delta t} \)

where \( \Phi = B A \) is the magnetic flux since the coil is perpendicular to the field.

The change in magnetic flux is:

\( \Delta \Phi = \Delta B \times A = (0.40 - 0) \times 2.0 \times 10^{-3} = 8.0 \times 10^{-4}\text{ Wb} \)

Now, calculate the induced e.m.f.:

\( E = 150 \times \frac{8.0 \times 10^{-4}\text{ Wb}}{0.15\text{ s}} = 0.80\text{ V} \)

1 mark for the correct option C.
- Calculates change in magnetic flux \( \Delta \Phi = 8.0 \times 10^{-4}\text{ Wb} \) (1 mark).
- Applies Faraday's law with \( N = 150 \) and \( \Delta t = 0.15\text{ s} \) to find \( E = 0.80\text{ V} \) (1 mark).

To achieve a smooth output voltage with minimal ripple, the capacitor must discharge very slowly during the intervals between the peak voltages of the alternating supply.

The rate of discharge is characterized by the time constant \( \tau = RC \). A larger time constant means a slower discharge rate and hence less ripple.

To increase \( \tau \), both the resistance \( R \) and the capacitance \( C \) should be increased.

1 mark for the correct option D.
- Relates the ripple reduction to a larger time constant \( \tau = RC \) (1 mark).
- Identifies that increasing both \( R \) and \( C \) increases \( \tau \) (1 mark).

The formula for density is \(\rho = \frac{4m}{\pi d^2 l}\). The fractional uncertainty in density is given by the sum of fractional uncertainties of each term (with the power of 2 for diameter): \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l}\). Substituting the values: \(\frac{\Delta m}{m} = \frac{0.02}{2.45} \approx 0.0082\), \(\frac{\Delta d}{d} = \frac{0.02}{1.24} \approx 0.0161\), and \(\frac{\Delta l}{l} = \frac{0.1}{15.0} \approx 0.0067\). Therefore, \(\frac{\Delta \rho}{\rho} = 0.0082 + 2(0.0161) + 0.0067 = 0.0471\). Expressed as a percentage, this is \(4.71\%\), which rounds to \(4.7\%\).

1 mark for the correct calculation of percentage uncertainty by summing the individual percentage uncertainties, doubling the uncertainty of diameter.

First, calculate the temperature rise: \(\Delta T = T_2 - T_1 = 58.7 - 23.4 = 35.3^\circ\text{C}\). When subtracting two quantities, their absolute uncertainties are added together: \(\Delta(\Delta T) = 0.2 + 0.3 = 0.5^\circ\text{C}\). The percentage uncertainty is \(\frac{0.5}{35.3} \times 100\% \approx 1.4\%\).

1 mark for calculating the absolute uncertainty by addition, and then expressing it as a percentage of the temperature difference.

A zero error is a constant offset in one direction, making it a systematic error. Parallax errors from randomly varying viewing positions are random in direction and magnitude, making them random errors.

1 mark for identifying the constant zero error as systematic and the varying parallax error as random.

The resistivity is given by \(\rho = \frac{R \pi d^2}{4 L} = \frac{4.0 \times \pi \times (0.50 \times 10^{-3})^2}{4 \times 1.00} = 7.854 \times 10^{-7}\ \Omega\text{ m}\). Percentage uncertainty in \(R\) is \(\frac{0.2}{4.0} \times 100\% = 5.0\%\). Percentage uncertainty in \(d\) is \(\frac{0.01}{0.50} \times 100\% = 2.0\%\). Percentage uncertainty in \(L\) is \(\frac{0.01}{1.00} \times 100\% = 1.0\%\). Total percentage uncertainty is \(5.0\% + 2(2.0\%) + 1.0\% = 10.0\%\). The absolute uncertainty is \(10.0\% \times 7.854 \times 10^{-7} \approx 0.8 \times 10^{-7}\ \Omega\text{ m}\). The final value is \((7.9 \pm 0.8) \times 10^{-7}\ \Omega\text{ m}\).

1 mark for calculating the correct value and total percentage uncertainty (10.0%), leading to the correct absolute uncertainty representation.

The up quark has a charge of \(+\frac{2}{3}e\). The strange quark has a charge of \(-\frac{1}{3}e\), so the strange antiquark \(\bar{\text{s}}\) has a charge of \(+\frac{1}{3}e\). The total charge is \(+\frac{2}{3}e + \frac{1}{3}e = +1e\). Since it consists of a quark-antiquark pair, it is classified as a meson.

1 mark for determining the correct charge (+1e) and identifying the quark-antiquark combination as a meson.

A proton has quark composition \(\text{uud}\) and a neutron has \(\text{udd}\). During \(\beta^+\) decay, one up quark (\(\text{u}\)) changes into a down quark (\(\text{d}\)). To conserve charge, lepton number, and baryon number, the decay is \(\text{u} \rightarrow \text{d} + \text{e}^+ + \nu_{\text{e}}\). The exchange boson carrying the positive charge in this weak interaction is the \(\text{W}^+\) boson.

1 mark for identifying the correct quark-level decay equation and the correct charge-carrying exchange boson.

By Faraday's law of electromagnetic induction, the average induced e.m.f. is \(E = \frac{\Delta \Phi}{\Delta t} = \frac{N A \Delta B}{\Delta t}\). Substituting the values: \(E = \frac{200 \times 1.5 \times 10^{-3} \times 0.40}{0.12} = \frac{0.12}{0.12} = 1.0\text{ V}\).

1 mark for calculating the correct change in flux linkage and dividing by the time interval to get 1.0 V.

The ripple voltage is decreased when the discharge of the capacitor between peaks is reduced. This occurs when the time constant \(\tau = RC\) is increased, or when the time between successive peaks of the rectified wave is decreased. Increasing the frequency of the alternating current supply decreases the period, meaning the capacitor has less time to discharge between cycles, thereby decreasing the ripple voltage.

1 mark for identifying that a higher frequency leads to shorter discharge times, resulting in a reduced ripple voltage.

**Part (a)**
- Systematic errors cause all readings to deviate from the true value by the same constant amount/in the same direction. They cannot be eliminated by averaging.
- Random errors cause readings to scatter around the true value due to unpredictable fluctuations. They can be reduced by taking repeated measurements and averaging.

**Part (b)(i)**
- Volume of the cylinder:
\(V = \frac{\pi d^2 L}{4} = \frac{\pi \times (1.86)^2 \times 6.42}{4} = 17.443 \text{ cm}^3\)
- Density:
\(\rho = \frac{m}{V} = \frac{145.2}{17.443} = 8.324 \text{ g cm}^{-3}\)

**Part (b)(ii)**
- The formula for density is:
\(\rho = \frac{4m}{\pi d^2 L}\)
- The fractional uncertainty in density is:
\(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\)
\(\frac{\Delta \rho}{\rho} = \frac{0.1}{145.2} + 2\left(\frac{0.02}{1.86}\right) + \frac{0.05}{6.42}\)
\(\frac{\Delta \rho}{\rho} = 0.000689 + 0.021505 + 0.007788 = 0.02998\)
- Percentage uncertainty:
\(\%\Delta \rho = 0.02998 \times 100\% \approx 3.00\%\) (which is approximately \(3\%\)).

**Part (b)(iii)**
- Absolute uncertainty:
\(\Delta \rho = 8.324 \times 0.02998 = 0.2496 \text{ g cm}^{-3}\)
- Rounded to 1 significant figure, \(\Delta \rho = 0.2 \text{ g cm}^{-3}\).
- Therefore, the density value must be quoted to the same decimal place (1 d.p.):
\(\rho = (8.3 \pm 0.2) \text{ g cm}^{-3}\)

**Part (a):**
- **C1**: Systematic error is a constant bias in one direction / cannot be eliminated by averaging. [1 mark]
- **C1**: Random error leads to scatter/fluctuations about a mean value / can be reduced by averaging. [1 mark]

**Part (b)(i):**
- **M1**: Calculation of volume: \(V = 17.4 \text{ cm}^3\) (or \(1.74 \times 10^{-5} \text{ m}^3\)). [1 mark]
- **A1**: Calculation of density: \(8.3 \text{ g cm}^{-3}\) (or \(8300 \text{ kg m}^{-3}\)). [1 mark]

**Part (b)(ii):**
- **C1**: Writes fractional uncertainty equation with factor of 2 for \(d\): \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\). [1 mark]
- **M1**: Correct substitution of values: \(\frac{0.1}{145.2} + 2\left(\frac{0.02}{1.86}\right) + \frac{0.05}{6.42}\). [1 mark]
- **A1**: Arrives at \(3.0\%\) or \(0.030\) with clear working. [0.57 marks]

**Part (b)(iii):**
- **M1**: Calculates absolute uncertainty as \(0.25\) or \(0.2\) and rounds to 1 s.f. [1 mark]
- **A1**: States final answer as \(8.3 \pm 0.2\) with correct unit \(\text{g cm}^{-3}\). [1 mark]

**Part (a)**
- Percentage uncertainty in \(L\):
\(\%\Delta L = \frac{0.003}{0.800} \times 100\% = 0.375\%\)
- Percentage uncertainty in \(t\) (and thus in \(T\)):
\(\%\Delta T = \frac{0.2}{35.8} \times 100\% = 0.559\%\)
- Because \(T\) is squared in the formula, its contribution to the uncertainty in \(g\) is \(2 \times \%\Delta T = 2 \times 0.559\% = 1.117\%\).
- Therefore, the measurement of time \(t\) contributes significantly more (nearly 3 times more) to the uncertainty of \(g\) than the measurement of \(L\).

**Part (b)(i)**
- The period is \(T = \frac{t}{20} = \frac{35.8}{20} = 1.79 \text{ s}\).
- \(g = \frac{4\pi^2 \times 0.800}{(1.79)^2} = 9.857 \text{ m s}^{-2}\).

**Part (b)(ii)**
- The total fractional uncertainty in \(g\) is:
\(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}\)
\(\frac{\Delta g}{g} = 0.00375 + 2 \times 0.00559 = 0.00375 + 0.01117 = 0.01492\)
- The absolute uncertainty in \(g\) is:
\(\Delta g = 9.857 \times 0.01492 = 0.147 \text{ m s}^{-2} \approx 0.1 \text{ m s}^{-2}\).

**Part (b)(iii)**
- Since \(\Delta g\) is rounded to 1 s.f. (\(0.1 \text{ m s}^{-2}\)), the value of \(g\) must be quoted to the same decimal place (1 d.p.):
\(g = (9.9 \pm 0.1) \text{ m s}^{-2}\)

**Part (a):**
- **M1**: Calculates percentage uncertainty in \(L\) as \(0.375\%\) (or fractional \(0.00375\)). [1 mark]
- **M1**: Calculates percentage uncertainty in \(t\) or \(T\) as \(0.559\%\) (or fractional \(0.00559\)). [1 mark]
- **A1**: Compares \(0.375\%\) with \(2 \times 0.559\% = 1.12\%\) and concludes \(t\) contributes more. [0.57 marks]

**Part (b)(i):**
- **M1**: Calculates period \(T = 1.79 \text{ s}\). [1 mark]
- **A1**: Evaluates \(g = 9.86\) or \(9.9 \text{ m s}^{-2}\). [1 mark]

**Part (b)(ii):**
- **C1**: Expresses fractional uncertainty equation: \(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}\). [1 mark]
- **M1**: Subsumes values: \(0.00375 + 2(0.00559) = 0.0149\). [1 mark]
- **A1**: Calculates \(\Delta g = 0.15\) or \(0.1 \text{ m s}^{-2}\). [1 mark]

**Part (b)(iii):**
- **A1**: Expresses final answer as \(g = (9.9 \pm 0.1) \text{ m s}^{-2}\) (consistent 1 d.p. for value and uncertainty). [1 mark]

**Part (a)**
- The wire may not be perfectly uniform in diameter along its entire length (non-uniformity), so taking measurements at different positions averages out this variation.
- The cross-section may not be perfectly circular (elliptical or out-of-round), so measuring at different orientations (orthogonal angles) averages out this non-circularity.

**Part (b)**
- Formula for resistivity: \(\rho = \frac{R A}{L} = \frac{\pi R d^2}{4 L}\)
- Fractional uncertainty: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\)
- Let's compute individual fractional uncertainties:
- \(\frac{\Delta R}{R} = \frac{0.2}{4.5} = 0.04444\) (\(4.44\%\))
- \(\frac{\Delta d}{d} = \frac{0.01}{0.38} = 0.02632\) (\(2.63\%\))
- \(\frac{\Delta L}{L} = \frac{0.002}{1.250} = 0.00160\) (\(0.16\%\))
- Total fractional uncertainty:
\(\frac{\Delta \rho}{\rho} = 0.04444 + 2(0.02632) + 0.00160 = 0.04444 + 0.05264 + 0.00160 = 0.09868\)
- Percentage uncertainty: \(0.09868 \times 100\% = 9.87\% \approx 9.9\%\).

**Part (c)**
- Value of resistivity:
\(\rho = \frac{\pi \times 4.5 \times (3.8 \times 10^{-4})^2}{4 \times 1.250}\)
\(\rho = \frac{\pi \times 4.5 \times 1.444 \times 10^{-7}}{5.00} = 4.083 \times 10^{-7} \ \Omega \text{ m}\)
- Absolute uncertainty:
\(\Delta \rho = 4.083 \times 10^{-7} \times 0.09868 = 4.029 \times 10^{-8} \ \Omega \text{ m} \approx 0.4 \times 10^{-7} \ \Omega \text{ m}\)
- Writing to 1 s.f. for uncertainty, the value is:
\(\rho = (4.1 \pm 0.4) \times 10^{-7} \ \Omega \text{ m}\)

**Part (a):**
- **C1**: Mentions averaging out variations in diameter along the wire (non-uniformity). [1 mark]
- **C1**: Mentions averaging out non-circularity / ensuring cross-section is represented accurately. [1 mark]

**Part (b):**
- **C1**: Recognizes the formula is proportional to \(d^2\) and correctly doubles the fractional uncertainty of \(d\). [1 mark]
- **M1**: Computes individual fractional uncertainties: \(0.0444\) for \(R\), \(0.0263\) for \(d\), \(0.0016\) for \(L\). [1 mark]
- **M1**: Correct substitution: \(0.0444 + 2(0.0263) + 0.0016 = 0.0987\). [1 mark]
- **A1**: Obtains \(9.9\%\) (or \(10\%\)). [0.57 marks]

**Part (c):**
- **M1**: Calculates resistivity value: \(4.1 \times 10^{-7} \ \Omega \text{ m}\). [1 mark]
- **M1**: Calculates absolute uncertainty: \(0.4 \times 10^{-7} \ \Omega \text{ m}\). [1 mark]
- **A1**: Expresses final answer clearly with correct unit: \((4.1 \pm 0.4) \times 10^{-7} \ \Omega \text{ m}\). [1 mark]

**Part (a)**
- Precision refers to how close a set of repeated measurements are to one another (it depends on the size of random errors).
- Accuracy refers to how close the measured average value is to the true value of the quantity (it depends on the size of systematic errors).

**Part (b)**
- Kinetic energy is given by: \(E_k = \frac{1}{2} m v^2\)
- Fractional uncertainty: \(\frac{\Delta E_k}{E_k} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v}\)
- Let's compute the components:
- \(\frac{\Delta m}{m} = \frac{5}{340} = 0.0147\) (or \(1.47\%\))
- \(\frac{\Delta v}{v} = \frac{0.04}{1.24} = 0.03226\) (or \(3.23\%\))
- \(2\frac{\Delta v}{v} = 2 \times 0.03226 = 0.06452\) (or \(6.45\%\))
- Total fractional uncertainty:
\(\frac{\Delta E_k}{E_k} = 0.0147 + 0.06452 = 0.0792\)
- Percentage uncertainty: \(0.0792 \times 100\% = 7.92\% \approx 7.9\%\).

**Part (c)**
- Mass \(m = 0.340 \text{ kg}\)
- Kinetic energy:
\(E_k = \frac{1}{2} \times 0.340 \times (1.24)^2 = 0.26146 \text{ J}\)
- Absolute uncertainty:
\(\Delta E_k = 0.26146 \times 0.0792 = 0.0207 \text{ J} \approx 0.02 \text{ J}\)
- Rounding the uncertainty to 1 s.f. gives \(0.02 \text{ J}\). The value is therefore rounded to 2 decimal places:
\(E_k = (0.26 \pm 0.02) \text{ J}\)

**Part (a):**
- **C1**: Precision: closeness of agreement between independent measurements / determined by random error. [1 mark]
- **C1**: Accuracy: closeness of the agreement between the result of a measurement and a true value / determined by systematic error. [1 mark]

**Part (b):**
- **C1**: Correct formula for combining uncertainties: \(\frac{\Delta E_k}{E_k} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v}\). [1 mark]
- **M1**: Substitution of values: \(\frac{5}{340} + 2\left(\frac{0.04}{1.24}\right)\). [1 mark]
- **M1**: Correct calculations of components: \(1.47\%\) and \(6.45\%\). [1 mark]
- **A1**: Correct total percentage uncertainty of \(7.9\%\) (or \(8\%\)). [0.57 marks]

**Part (c):**
- **M1**: Calculates kinetic energy: \(E_k = 0.261 \text{ J}\) or \(0.26 \text{ J}\). [1 mark]
- **M1**: Calculates absolute uncertainty: \(\Delta E_k = 0.02 \text{ J}\). [1 mark]
- **A1**: Expresses final answer as \((0.26 \pm 0.02) \text{ J}\). [1 mark]

**Part (a)**
(i) Neutron quark composition: up, down, down (\(udd\)).
(ii) Proton quark composition: up, up, down (\(uud\)).

**Part (b)(i)**
- During \(\beta^-\)-decay, a down quark decays into an up quark, emitting an electron and an electron antineutrino:
\(d \rightarrow u + e^- + \bar{\nu}_e\)

**Part (b)(ii)**
- The particle is the **electron antineutrino** (symbol \(\bar{\nu}_e\)).
- Since it is a lepton and not a baryon, its baryon number is **0**.

**Part (c)**
- The fundamental force responsible is the **weak nuclear force** (or weak interaction).
- The exchange particle is the **\(W^-\) boson**.

**Part (a):**
- **A1**: Neutron: \(udd\). [1 mark]
- **A1**: Proton: \(uud\). [1 mark]

**Part (b)(i):**
- **M1**: Shows down quark turning into up quark: \(d \rightarrow u\). [1 mark]
- **A1**: Full correct quark equation including electron and electron antineutrino: \(d \rightarrow u + e^- + \bar{\nu}_e\). [0.57 marks]

**Part (b)(ii):**
- **A1**: Electron antineutrino. [1 mark]
- **A1**: Baryon number is 0. [1 mark]

**Part (c):**
- **C1**: Weak nuclear force / weak interaction. [1 mark]
- **C1**: \(W^-\) boson. [1 mark]
- **A1**: Specifies the correct charge on the boson (negative / \(W^-\) boson, NOT \(W^+\) or \(Z^0\)). [1 mark]

**Part (a)**
- Magnetic flux linkage is defined as the product of the magnetic flux and the number of turns of the coil:
\(\Phi = N B A \cos\theta\)
where \(N\) is the number of turns, \(B\) is the magnetic flux density, \(A\) is the cross-sectional area, and \(\theta\) is the angle between the normal to the plane of the coil and the magnetic field.

**Part (b)**
- Faraday's law states that the magnitude of the induced electromotive force (e.m.f.) is directly proportional to the rate of change of magnetic flux linkage.

**Part (c)**
- The initial magnetic flux linkage is:
\(\Phi_{\text{initial}} = N B_{\text{initial}} A = 150 \times 0.45 \text{ T} \times 8.4 \times 10^{-4} \text{ m}^2 = 0.0567 \text{ Wb}\)
- The final magnetic flux linkage is:
\(\Phi_{\text{final}} = N B_{\text{final}} A = 150 \times 0.15 \text{ T} \times 8.4 \times 10^{-4} \text{ m}^2 = 0.0189 \text{ Wb}\)
- The change in magnetic flux linkage is:
\(\Delta \Phi = \Phi_{\text{initial}} - \Phi_{\text{final}} = 0.0567 - 0.0189 = 0.0378 \text{ Wb}\)
- The magnitude of the induced e.m.f. is given by Faraday's law:
\(E = \frac{\Delta \Phi}{\Delta t} = \frac{0.0378}{0.12} = 0.315 \text{ V}\)
- Rounding to 2 significant figures gives \(0.32 \text{ V}\).

**Part (a):**
- **C1**: Product of magnetic flux and number of turns (or \(N\Phi\) where \(\Phi = BA\)). [1 mark]
- **C1**: Specifies that the magnetic field is perpendicular to the area / mentions \(\cos \theta\) term. [1 mark]

**Part (b):**
- **C1**: Induced e.m.f. is proportional to / equal to the rate of change of... [1 mark]
- **C1**: ...magnetic flux linkage. [1 mark]

**Part (c):**
- **M1**: Calculates initial flux linkage: \(0.0567 \text{ Wb}\) (or final: \(0.0189 \text{ Wb}\)). [1 mark]
- **M1**: Calculates change in magnetic flux linkage \(\Delta (NBA) = 150 \times (0.45 - 0.15) \times 8.4 \times 10^{-4} = 0.0378 \text{ Wb}\). [1 mark]
- **M1**: Uses Faraday's equation \(E = \frac{\Delta (NBA)}{\Delta t}\). [1 mark]
- **A1**: Evaluates \(E = 0.315 \text{ V}\) or \(0.32 \text{ V}\). [1.57 marks]

**Part (a)**
- During the positive half-cycle of the a.c. input, two diagonally opposite diodes conduct while the other two are reverse-biased (non-conducting).
- During the negative half-cycle of the a.c. input, the other pair of diagonally opposite diodes conduct while the first pair is reverse-biased.
- In both half-cycles, the current flows in the same direction through the load resistor \(R\). This achieves full-wave rectification.

**Part (b)**
- When the output voltage from the rectifier rises, the capacitor charges up and stores electrical energy.
- When the output voltage from the rectifier begins to fall, the capacitor discharges energy slowly through the load resistor \(R\).
- This prevents the potential difference across \(R\) from falling to zero, reducing the fluctuation (ripple) in the output voltage.

**Part (c)**
- The time constant \(\tau\) is given by:
\(\tau = R C\)
- Substituting the given values:
\(\tau = 1.2 \times 10^3 \ \Omega \times 470 \times 10^{-6} \text{ F}\)
\(\tau = 0.564 \text{ s}\)
- Rounding to 2 significant figures gives \(0.56 \text{ s}\).

**Part (a):**
- **C1**: Mentions that two diagonally opposite diodes conduct during one half-cycle. [1 mark]
- **C1**: Mentions that the other two diodes conduct during the opposite half-cycle. [1 mark]
- **C1**: Explains that current flows in the same direction through the load resistor in both half-cycles. [1.57 marks]

**Part (b):**
- **C1**: States that the capacitor charges up (stores energy) when the supply voltage is near its peak. [1 mark]
- **C1**: States that the capacitor discharges through the load resistor when the supply voltage falls. [1 mark]
- **C1**: Mentions reduction of ripple / keeping the output voltage high / preventing it from falling to zero. [1 mark]

**Part (c):**
- **M1**: Recalls and uses \(\tau = RC\) with correct powers of ten: \((1.2 \times 10^3) \times (470 \times 10^{-6})\). [1 mark]
- **A1**: Evaluates \(\tau = 0.56 \text{ s}\) (accept \(0.564 \text{ s}\)). [1 mark]

**Sample Experimental Data:**

* Diameter of wire, \(D = 0.38 \times 10^{-3}\text{ m}\)
* Cross-sectional area, \(A = \frac{\pi D^2}{4} = \frac{\pi (0.38 \times 10^{-3})^2}{4} = 1.13 \times 10^{-7}\text{ m}^2\)

| \(L\text{ / m}\) | \(I\text{ / A}\) | \(\frac{1}{I}\text{ / A}^{-1}\) |
|---|---|---|
| 0.200 | 0.211 | 4.74 |
| 0.350 | 0.175 | 5.71 |
| 0.500 | 0.150 | 6.67 |
| 0.650 | 0.131 | 7.63 |
| 0.800 | 0.116 | 8.62 |
| 0.900 | 0.108 | 9.26 |

**Graph Analysis:**
* The graph of \(\frac{1}{I}\) against \(L\) is a straight line.
* Gradient, \(P = \frac{9.26 - 4.74}{0.900 - 0.200} = \frac{4.52}{0.700} = 6.46\text{ A}^{-1}\text{ m}^{-1}\)
* y-intercept, \(Q = 3.45\text{ A}^{-1}\)

**Resistivity Calculation:**
* \(P = \frac{\rho}{A \cdot E} \implies \rho = P \cdot A \cdot E\)
* \(\rho = 6.46 \times 1.13 \times 10^{-7} \times 1.50 = 1.10 \times 10^{-6}\ \Omega\text{ m}\)

**Marking Scheme (Total: 20 marks)**

* **Successful collection of raw data (4 marks):**
* [1] Six sets of values of \(L\) and \(I\) recorded.
* [1] Range of \(L\) is at least \(0.600\text{ m}\).
* [1] Column headings: Each column in the table must have a proper heading with the unit (e.g., \(L\text{ / m}\), \(I\text{ / A}\), \(\frac{1}{I}\text{ / A}^{-1}\)).
* [1] All raw values of \(L\) recorded to the nearest millimeter (\(0.001\text{ m}\)).

* **Data consistency and calculated values (2 marks):**
* [1] Precision: Raw values of \(I\) recorded to a consistent number of decimal places (e.g. 3 d.p. for ammeter reading in A).
* [1] Significant figures: The number of significant figures for \(\frac{1}{I}\) must be the same as, or one more than, the number of significant figures in the raw current \(I\).

* **Graph (4 marks):**
* [1] Axes: Sensible scales must be chosen so that the plotted points occupy more than half the graph grid in both directions. Axes must be clearly labelled with quantity and unit.
* [1] Plotting: Check that at least two points are plotted correctly (within half a small square). Diameter of plot points must be \(\le 1\text{ mm}\).
* [1] Line of best fit: Must be a single straight line with a balanced distribution of points about it.
* [1] Quality: Little or no scatter of points around the line of best fit.

* **Calculations and Analysis (5 marks):**
* [1] Gradient: Calculated using a triangle whose hypotenuse is at least half the length of the drawn line. Correct read-offs from the line must be used.
* [1] y-intercept: Calculated using a point on the line substituted into \(y = mx + c\) or read directly from the y-axis if \(x=0\).
* [1] Value of \(P\) set equal to gradient and \(Q\) set equal to y-intercept, with correct values from graph.
* [2] Units: Correct units for \(P\) (\(\text{A}^{-1}\text{ m}^{-1}\) or \(\Omega\text{ m}^{-1}\text{ V}^{-1}\)) and \(Q\) (\(\text{A}^{-1}\) or \(\Omega\text{ V}^{-1}\)).

* **Resistivity Determination (5 marks):**
* [1] Calculation of cross-sectional area \(A = \frac{\pi D^2}{4}\) correctly with correct substitution of \(D = 0.38 \times 10^{-3}\text{ m}\) to obtain \(A \approx 1.13 \times 10^{-7}\text{ m}^2\).
* [1] Formula \(\rho = P \cdot A \cdot E\) used correctly.
* [1] Substitution of candidate's value of \(P\), \(A\), and \(E = 1.50\text{ V}\) to find \(\rho\).
* [1] Final value of \(\rho\) in the range \(1.0 \times 10^{-6}\) to \(1.5 \times 10^{-6}\ \Omega\text{ m}\) (or consistent with candidate's experimental slope).
* [1] Unit of \(\rho\) given as \(\Omega\text{ m}\).

**Sample Experimental Data:**

* **First measurement:**
* \(d_1 = 0.300\text{ m} = 30.0\text{ cm}\)
* Uncertainty in \(d_1\): \(\Delta d = 2\text{ mm} = 0.2\text{ cm}\) (due to the difficulty in measuring to the exact centers of the clamp attachments).
* Percentage uncertainty in \(d_1\): \(\frac{0.2}{30.0} \times 100\% \approx 0.67\%\).
* Time for 20 oscillations: \(t_{1,1} = 26.24\text{ s}\), \(t_{1,2} = 26.30\text{ s}\). Average \(t_1 = 26.27\text{ s}\).
* Period \(T_1 = \frac{26.27}{20} = 1.314\text{ s}\).

* **Second measurement:**
* \(d_2 = 0.450\text{ m} = 45.0\text{ cm}\)
* Time for 20 oscillations: \(t_{2,1} = 24.12\text{ s}\), \(t_{2,2} = 24.18\text{ s}\). Average \(t_2 = 24.15\text{ s}\).
* Period \(T_2 = \frac{24.15}{20} = 1.208\text{ s}\).

**Calculation of constant \(k\):**
* Since \(L = 0.500\text{ m}\), \(4L^2 = 4 \times (0.500)^2 = 1.000\text{ m}^2\).
* Therefore: \(k = \frac{T^2}{1 - d^2}\).

* **For \(d_1 = 0.300\text{ m}\):**
* \(k_1 = \frac{1.314^2}{1 - (0.300)^2} = \frac{1.727}{0.91} = 1.898\text{ s}^2\)

* **For \(d_2 = 0.450\text{ m}\):**
* \(k_2 = \frac{1.208^2}{1 - (0.450)^2} = \frac{1.459}{0.7975} = 1.830\text{ s}^2\)

**Conclusion and Comparison:**
* Percentage difference in \(k\): \(\frac{|1.898 - 1.830|}{1.898} \times 100\% \approx 3.6\%\).
* Since this percentage difference is less than the typical experimental threshold of \(10\%\) (or the uncertainty in the period determination), the suggested relationship is supported.

**Marking Scheme (Total: 20 marks)**

* **Measurements & Raw Data (5 marks):**
* [1] Measurement of \(d_1\) to the nearest millimetre with correct unit.
* [1] Measurement of \(t_1\) for at least 20 oscillations, repeated and averaged, with correct decimal places.
* [1] Correct calculation of period \(T_1\).
* [1] Measurement of \(d_2\) and \(t_2\) for a second distance where \(d_2 > d_1\).
* [1] Values of \(T_1\) and \(T_2\) recorded with correct units (\(\text{s}\)).

* **Uncertainty Analysis (2 marks):**
* [1] Absolute uncertainty in \(d_1\) estimated as \(1\text{ mm}\) to \(3\text{ mm}\) with clear justification (e.g. alignment with clamp centres).
* [1] Percentage uncertainty calculated correctly using: \(\frac{\text{absolute uncertainty}}{d_1} \times 100\%\).

* **Constant \(k\) and Comparison (5 marks):**
* [1] Correct calculation of first value of \(k\) with appropriate significant figures (typically 3 s.f.).
* [1] Correct calculation of second value of \(k\).
* [1] Consistent units for \(k\) (\(\text{s}^2\)).
* [1] Calculation of percentage difference between the two \(k\) values: \(\frac{|k_1 - k_2|}{\text{mean } k} \times 100\%\).
* [1] Conclusion: A clear statement of whether the results support the relationship, comparing the percentage difference in \(k\) directly to a specified limit (e.g. \(10\%\) or candidate's estimate of uncertainty).

* **Limitations & Improvements (8 marks - 1 mark for each of 4 limitations, and 1 mark for each of 4 matching improvements):**
* **Limitation 1:** Two readings of \(d\) are not enough to confirm a relationship.
* **Improvement 1:** Take more sets of readings for different values of \(d\) and plot a graph of \(T^2\) against \(1 - d^2/4L^2\).
* **Limitation 2:** Difficult to measure the exact distance \(d\) between the two suspension points because the string attachment points are inside the clamps.
* **Improvement 2:** Mark the exact point of suspension on the clamps, or use a fiducial marker directly below the clamp.
* **Limitation 3:** Difficult to ensure the two strings are of exactly equal length \(L\).
* **Improvement 3:** Measure the length of both strings under tension from the clamp to the mass using a vertical rule and plumb-line, or use a single continuous string passing through a smooth ring at the mass.
* **Limitation 4:** Oscillations are not perfectly planar / the mass undergoes circular or twisting motion.
* **Improvement 4:** Use a guide or release mechanism (e.g., burning a thread) to ensure purely perpendicular release.

(a) Systematic errors cause measurements to be consistently larger or consistently smaller than the true value. They cannot be eliminated by repeating and averaging, but can be reduced by calibrating instruments. Random errors cause measurements to fluctuate around the true value due to unpredictable variations. They can be reduced by taking repeated measurements and calculating an average.

(b)
(i) The period \(T\) is the time for one oscillation:
\(T = \frac{t}{50} = \frac{89.4}{50} = 1.788\text{ s}\)
The absolute uncertainty in \(T\) is:
\(\Delta T = \frac{\Delta t}{50} = \frac{0.4}{50} = 0.008\text{ s}\)
So, \(T = (1.788 \pm 0.008)\text{ s}\).

(ii) Rearranging the pendulum equation for \(g\):
\(g = \frac{4\pi^2 L}{T^2}\)
Using \(L = 0.800\text{ m}\) and \(T = 1.788\text{ s}\):
\(g = \frac{4\pi^2 \times 0.800}{1.788^2} \approx 9.879\text{ m s}^{-2}\)

(iii) The percentage uncertainty in \(g\) is given by:
\(\frac{\Delta g}{g} \times 100\% = \left(\frac{\Delta L}{L} + 2 \frac{\Delta T}{T}\right) \times 100\%\)
\(\frac{\Delta L}{L} = \frac{0.2}{80.0} = 0.0025\)
\(\frac{\Delta T}{T} = \frac{0.008}{1.788} \approx 0.00447\)
\(\frac{\Delta g}{g} = 0.0025 + 2(0.00447) = 0.0025 + 0.00894 = 0.0114\)
Percentage uncertainty in \(g = 0.0114 \times 100\% \approx 1.1\%\) (or \(1.14\%\)).

(iv) The absolute uncertainty in \(g\) is:
\(\Delta g = 9.879 \times 0.0114 \approx 0.11\text{ m s}^{-2}\).
To 1 significant figure, \(\Delta g = 0.1\text{ m s}^{-2}\).
Thus, the value of \(g\) is expressed as:
\(g = (9.9 \pm 0.1)\text{ m s}^{-2}\).

(a)
- Systematic error: constant bias / deviation in one direction from true value [1]
- Random error: scatter / spread of values about the mean [1]

(b)(i)
- \(T = 1.788\text{ s}\) [1]
- \(\Delta T = \frac{0.4}{50} = 0.008\text{ s}\) [1]
- Final statement: \(T = (1.788 \pm 0.008)\text{ s}\) [1]

(b)(ii)
- Correct rearrangement: \(g = \frac{4\pi^2 L}{T^2}\) [1]
- \(g = 9.88\text{ m s}^{-2}\) (accept \(9.9\text{ m s}^{-2}\)) [1]

(b)(iii)
- \(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}\) [1]
- \(\% \text{ uncertainty} = 1.1\%\) (or \(1.14\%\)) [1]

(b)(iv)
- \(g = (9.9 \pm 0.1)\text{ m s}^{-2}\) [1] (consistent decimal places for value and uncertainty)

(a) The resistance is given by \(R = \frac{\rho L}{A}\), where the cross-sectional area of a cylinder is \(A = \frac{\pi d^2}{4}\).
Substituting \(A\) gives:
\(\rho = \frac{\pi R d^2}{4 L}\)

(b)
(i) Substituting the values into the equation:
\(R = 2.45\ \Omega\)
\(L = 1.542\text{ m}\)
\(d = 0.38 \times 10^{-3}\text{ m}\)
\(\rho = \frac{\pi \times 2.45 \times (0.38 \times 10^{-3})^2}{4 \times 1.542}\)
\(\rho = \frac{\pi \times 2.45 \times 1.444 \times 10^{-7}}{6.168} \approx 1.802 \times 10^{-7}\ \Omega\text{ m}\).

(ii) The fractional uncertainty in \(\rho\) is:
\(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\)
\(\frac{\Delta R}{R} = \frac{0.05}{2.45} \approx 0.0204\)
\(\frac{\Delta d}{d} = \frac{0.01}{0.38} \approx 0.0263\)
\(\frac{\Delta L}{L} = \frac{0.002}{1.542} \approx 0.0013\)
\(\frac{\Delta \rho}{\rho} = 0.0204 + 2(0.0263) + 0.0013 = 0.0204 + 0.0526 + 0.0013 = 0.0743\)

(iii) The absolute uncertainty in \(\rho\) is:
\(\Delta \rho = 1.802 \times 10^{-7} \times 0.0743 = 0.134 \times 10^{-7}\ \Omega\text{ m}\)
Rounding the uncertainty to 1 significant figure gives \(\Delta \rho = 0.1 \times 10^{-7}\ \Omega\text{ m}\).
Thus, the value of \(\rho\) must be quoted to the same decimal position:
\(\rho = (1.8 \pm 0.1) \times 10^{-7}\ \Omega\text{ m}\).

(a)
- \(\rho = \frac{\pi R d^2}{4 L}\) [1]

(b)(i)
- Area \(A = 1.13 \times 10^{-7}\text{ m}^2\) [1]
- Substitute values into the correct resistivity formula [1]
- \(\rho = 1.8 \times 10^{-7}\ \Omega\text{ m}\) [1]

(b)(ii)
- Fractional uncertainty equation: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\) [1]
- Calculations of individual fractional terms (e.g., \(2 \times 0.0263\)) [1]
- Final value: \(0.074\) (or \(7.4\%\)) [1]

(b)(iii)
- Calculate \(\Delta \rho = 0.13 \times 10^{-7}\ \Omega\text{ m}\) [1]
- Round uncertainty to 1 s.f.: \(0.1 \times 10^{-7}\ \Omega\text{ m}\) [1]
- Final expression with matching s.f. and unit: \(\rho = (1.8 \pm 0.1) \times 10^{-7}\ \Omega\text{ m}\) [1]

(a) Precision is a measure of the agreement between repeated independent measurements (i.e., how close they are to each other; related to the size of random errors). Accuracy is how close a measured value is to the true value of the quantity being measured (related to systematic errors).

(b)
(i) Volume \(V = x y z = 5.20 \times 3.80 \times 2.40 = 47.424\text{ cm}^3\).
Density \(D = \frac{m}{V} = \frac{124.6}{47.424} \approx 2.6273\text{ g cm}^{-3}\).

(ii) The volume uncertainty is given by:
\(\frac{\Delta V}{V} = \frac{\Delta x}{x} + \frac{\Delta y}{y} + \frac{\Delta z}{z}\)
\(\frac{\Delta V}{V} = \frac{0.05}{5.20} + \frac{0.05}{3.80} + \frac{0.05}{2.40}\)
\(\frac{\Delta V}{V} \approx 0.00962 + 0.01316 + 0.02083 = 0.0436\)
Percentage uncertainty in \(V = 0.0436 \times 100\% \approx 4.36\% \approx 4.4\%\).

(iii) The fractional uncertainty in density \(D\) is:
\(\frac{\Delta D}{D} = \frac{\Delta m}{m} + \frac{\Delta V}{V}\)
\(\frac{\Delta m}{m} = \frac{0.1}{124.6} \approx 0.00080\)
\(\frac{\Delta D}{D} = 0.00080 + 0.0436 = 0.0444\) (or 4.44%)
The absolute uncertainty in \(D\) is:
\(\Delta D = 0.0444 \times 2.6273 = 0.117\text{ g cm}^{-3}\).
Rounding to a suitable number of decimal places:
\(\Delta D \approx 0.12\text{ g cm}^{-3}\) (or \(0.1\text{ g cm}^{-3}\)).

(a)
- Precision: closeness of agreement between independent measurements / small random error [1]
- Accuracy: closeness of measurement to the true value / small systematic error [1]

(b)(i)
- Volume \(V = 47.4\text{ cm}^3\) [1]
- Density \(D = 2.63\text{ g cm}^{-3}\) [1]

(b)(ii)
- Fractional terms: \(\frac{0.05}{5.20}\), \(\frac{0.05}{3.80}\), \(\frac{0.05}{2.40}\) [1]
- Sum of fractional terms: \(0.0436\) [1]
- Multiplied by 100% to show \(4.4\%\) [1]

(b)(iii)
- \(\frac{\Delta m}{m} \approx 0.0008\) [1]
- \(\frac{\Delta D}{D} \approx 0.0444\) [1]
- \(\Delta D \approx 0.12\text{ g cm}^{-3}\) (accept \(0.1\text{ g cm}^{-3}\)) [1]

(a) Velocity is defined as the rate of change of displacement. It is a vector quantity.

(b)
(i) Speed \(v = \frac{\text{distance}}{\text{time}} = \frac{140}{5.6} = 25.0\text{ m s}^{-1}\).

(ii) The percentage uncertainty in speed \(v\) is:
\(\frac{\Delta v}{v} = \frac{\Delta d}{d} + \frac{\Delta t}{t}\)
\(\frac{\Delta v}{v} = \frac{2}{140} + \frac{0.2}{5.6} = 0.0143 + 0.0357 = 0.0500\)
Percentage uncertainty in \(v = 5.0\%\).

(iii) Kinetic energy:
\(E_k = \frac{1}{2} M v^2 = 0.5 \times 1200 \times (25.0)^2 = 3.75 \times 10^5\text{ J}\) (or \(375\text{ kJ}\)).
The fractional uncertainty in \(E_k\) is:
\(\frac{\Delta E_k}{E_k} = \frac{\Delta M}{M} + 2 \frac{\Delta v}{v}\)
\(\frac{\Delta M}{M} = \frac{50}{1200} \approx 0.0417\)
\(\frac{\Delta E_k}{E_k} = 0.0417 + 2(0.0500) = 0.1417\) (or \(14.2\%\)).
The absolute uncertainty in \(E_k\) is:
\(\Delta E_k = 3.75 \times 10^5 \times 0.1417 \approx 5.3 \times 10^4\text{ J} \approx 0.5 \times 10^5\text{ J}\) (or \(50\text{ kJ}\)).
Expressing \(E_k\) with its absolute uncertainty:
\(E_k = (3.8 \pm 0.5) \times 10^5\text{ J}\) (or \((380 \pm 50)\text{ kJ}\)).

(a)
- Rate of change of displacement [1]
- Vector [1]

(b)(i)
- Formula \(v = \frac{d}{t}\) used [1]
- \(v = 25\text{ m s}^{-1}\) [1]

(b)(ii)
- Sum of fractional uncertainties: \(\frac{2}{140} + \frac{0.2}{5.6}\) [1]
- Percentage uncertainty in \(v = 5.0\%\) [1]

(b)(iii)
- \(E_k = 3.75 \times 10^5\text{ J}\) (or \(375\text{ kJ}\)) [1]
- Fractional uncertainty equation: \(\frac{\Delta E_k}{E_k} = \frac{\Delta M}{M} + 2\frac{\Delta v}{v}\) [1]
- Fractional uncertainty in \(E_k = 0.142\) (or \(14\%\)) [1]
- \(\Delta E_k = 0.5 \times 10^5\text{ J}\) and final answer \((3.8 \pm 0.5) \times 10^5\text{ J}\) [1]

(a) A fundamental particle is a particle with no internal structure that cannot be broken down into simpler constituents. From the list, the fundamental particles are the electron, neutrino, and up quark. (Protons and neutrons are composite particles made of quarks).

(b)
(i) In \(\beta^+\) decay, a proton (quark structure \(uud\)) decays into a neutron (quark structure \(udd\)). One up quark (\(u\)) changes into a down quark (\(d\)):
\(u \rightarrow d + e^+ + \nu_e\)
(ii) The complete nuclear decay equation is:
\(p \rightarrow n + e^+ + \nu_e\)

(c)
For charge conservation:
- Before decay: \(p\) has charge \(+1e\) (or \(+1.6 \times 10^{-19}\text{ C}\)).
- After decay: \(n\) has charge \(0\), \(e^+\) has charge \(+1e\), and \(\nu_e\) has charge \(0\). Total final charge \(= 0 + 1 + 0 = +1e\).
Since total charge before equals total charge after, charge is conserved.

For lepton number conservation:
- Before decay: \(p\) is a baryon, so lepton number \(L = 0\).
- After decay: \(n\) is a baryon (\(L = 0\)), \(e^+\) is an anti-lepton (\(L = -1\)), and \(\nu_e\) is a lepton (\(L = +1\)). Total final lepton number \(= 0 + (-1) + 1 = 0\).
Since total lepton number before equals total lepton number after, lepton number is conserved.

(a)
- Fundamental particle definition: a particle that cannot be split / has no internal structure [1]
- Correct identification of: electron, neutrino, and up quark [2] (lose 1 mark for each incorrect or missing item, minimum 0)

(b)(i)
- Proton changes to a neutron / \(uud \rightarrow udd\) [1]
- Specifically, an up quark changes to a down quark (\(u \rightarrow d\)) [1]

(b)(ii)
- \(p \rightarrow n + e^+\) [1]
- \(+ \nu_e\) (electron neutrino included to balance lepton number) [1]

(c)
- Charge calculation: Initial \(+1e\) and Final \((0 + 1 + 0) = +1e\), hence conserved [1]
- Lepton number calculation: Initial \(0\) and Final \((0 - 1 + 1) = 0\), hence conserved [2]

(a)
(i) Hadrons experience the strong nuclear interaction (and other forces), whereas leptons do not experience the strong nuclear interaction.
(ii) Leptons are fundamental particles, while hadrons are composite particles made up of quarks.

(b)
(i) A baryon consists of three quarks (\(qqq\)). A meson consists of a quark-antiquark pair (\(q\bar{q}\)).
(ii) The baryon number of a meson is \(0\). The baryon number of an anti-baryon is \(-1\) (since each antiquark has a baryon number of \(-\frac{1}{3}\)).

(c)
(i) The quark structure is \(dds\).
Total charge \(= Q(d) + Q(d) + Q(s) = -\frac{1}{3}e - \frac{1}{3}e - \frac{1}{3}e = -1e\).
(ii) The decay is:
\(\Sigma^- (dds) \rightarrow n (udd) + \pi^- (d\bar{u})\)
In this process, the strange quark (\(s\)) decays into an up quark (\(u\)), releasing a \(W^-\) boson. The \(W^-\) boson then decays into a down quark and an anti-up quark (\(d\bar{u}\)).
- The exchange boson is the \(W^-\) boson.
- Strangeness:
- The \(\Sigma^-\) particle contains one strange quark (\(s\)), so its strangeness \(S = -1\).
- Neither the neutron nor the \(\pi^-\) meson contains any strange quarks, so their strangeness \(S = 0\).
- Therefore, strangeness is not conserved; it changes from \(-1\) to \(0\) (a change of \(\Delta S = +1\)) due to the weak interaction.

(a)(i)
- Hadrons feel the strong force / strong interaction, leptons do not [1]

(a)(ii)
- Leptons are fundamental, hadrons are composite / made of quarks [1]

(b)(i)
- Baryon: three quarks [1]
- Meson: quark-antiquark [1]

(b)(ii)
- Meson baryon number \(= 0\) [1]
- Anti-baryon baryon number \(= -1\) [1]

(c)(i)
- Sum of charges: \(-\frac{1}{3} - \frac{1}{3} - \frac{1}{3} = -1\) [1]

(c)(ii)
- Exchange boson: \(W^-\) (or \(W\) boson) [1]
- Initial strangeness of \(\Sigma^- = -1\) and final strangeness \(= 0\) [1]
- State that strangeness is not conserved / changes by \(+1\) in weak interaction [1]

(a)
(i) Magnetic flux \(\Phi\) is the product of the magnetic flux density and the area perpendicular to the magnetic field: \(\Phi = BA\).
(ii) Magnetic flux linkage is the product of the magnetic flux and the number of turns of the coil: \(\text{Flux linkage} = N\Phi = NBA\).

(b) Faraday's law states that the magnitude of the induced electromotive force (e.m.f.) is directly proportional to the rate of change of magnetic flux linkage.

(c)
(i) The maximum magnetic flux density is \(B_0 = 0.40\text{ T}\).
The maximum magnetic flux linkage is:
\(\text{Flux linkage}_{\text{max}} = N B_0 A\)
\(\text{Flux linkage}_{\text{max}} = 250 \times 0.40\text{ T} \times 1.5 \times 10^{-3}\text{ m}^2 = 0.15\text{ Wb}\).

(ii) By Faraday's law, the induced e.m.f. \(E\) is:
\(E = -\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t} = -N A \frac{\mathrm{d}B}{\mathrm{d}t}\)
Given \(B = B_0 \sin(\omega t)\):
\(\frac{\mathrm{d}B}{\mathrm{d}t} = B_0 \omega \cos(\omega t)\)
Therefore:
\(E = -N A B_0 \omega \cos(\omega t)\)
The maximum induced e.m.f. \(E_{\text{max}}\) is:
\(E_{\text{max}} = N B_0 A \omega = (\text{Flux linkage}_{\text{max}}) \times \omega\)
\(E_{\text{max}} = 0.15\text{ Wb} \times 50\text{ rad s}^{-1} = 7.5\text{ V}\).

(iii) Since \(B \propto \sin(\omega t)\) and \(E \propto \cos(\omega t)\), the phase difference between \(B\) and \(E\) is \(\frac{\pi}{2}\text{ rad}\) (or \(90^\circ\)).

(a)(i)
- Product of magnetic flux density and area perpendicular to the field [1]

(a)(ii)
- Product of magnetic flux and number of turns [1]

(b)
- E.m.f. induced is proportional to [1]
- rate of change of magnetic flux linkage [1]

(c)(i)
- Formula \(\text{Flux linkage} = NBA\) used [1]
- Calculation: \(250 \times 0.40 \times 1.5 \times 10^{-3} = 0.15\text{ Wb}\) [1]

(c)(ii)
- Recall of \(E_{\text{max}} = N B_0 A \omega\) (or understanding that e.m.f. is derivative of flux linkage) [1]
- Calculation: \(0.15 \times 50\) [1]
- \(E_{\text{max}} = 7.5\text{ V}\) [1]

(c)(iii)
- \(\frac{\pi}{2}\text{ rad}\) or \(90^\circ\) [1]

(a) The bridge rectifier consists of four diodes arranged in a loop.
- The diodes must point in the same direction around the loop (e.g., clockwise or counterclockwise).
- The AC input is connected to the two nodes where the cathode of one diode meets the anode of another.
- The DC output (across the load resistor \(R\)) is taken from the remaining two nodes: the positive output is at the junction of the two cathodes, and the negative output is at the junction of the two anodes.
[Diagram Description: A diamond/bridge of 4 diodes pointing such that current flows to the top node (positive output) and returns to the bottom node (negative output). The AC input is connected to the left and right nodes.]

(b)
(i) For a \(50\text{ Hz}\) input, the period is \(T = \frac{1}{f} = \frac{1}{50} = 0.02\text{ s} = 20\text{ ms}\).
Full-wave rectification converts both half-cycles into positive output pulses, so the rectified output has peaks every \(\frac{T}{2} = 10\text{ ms}\).
In \(40\text{ ms}\), there will be exactly 4 positive half-sine-wave peaks.
- Peak voltage on the y-axis should be labelled as \(12\text{ V}\).
- Time axis on the x-axis should have markings at \(10\text{ ms}\), \(20\text{ ms}\), \(30\text{ ms}\), and \(40\text{ ms}\) corresponding to the peaks/valleys of the rectified wave.

(b)(ii)
1. When the input voltage exceeds the capacitor voltage, the capacitor charges up rapidly to the peak value of the input voltage. When the input voltage drops below the peak, the diodes become reverse-biased, and the capacitor discharges slowly through the load resistor \(R\). This maintains a higher output voltage between the peaks, reducing the 'ripple' voltage.

2. Time constant \(\tau = RC = (1.5 \times 10^3\ \Omega) \times (22 \times 10^{-6}\text{ F}) = 0.033\text{ s} = 33\text{ ms}\).
For effective smoothing, the time constant \(\tau\) must be much greater than the time interval between peaks of the rectified output. The time between peaks is \(10\text{ ms}\). Since \(\tau = 33\text{ ms} > 10\text{ ms}\), the capacitor only discharges slightly between peaks, which means it provides effective smoothing.

(a)
- 4 diodes in correct loop configuration (pointing in same circular direction) [1]
- Input AC connected to opposite corners (where anode meets cathode) [1]
- Output load resistor connected to other opposite corners [1]
- Polarity of DC output marked correctly (cathodes junction is positive, anodes junction is negative) [1]

(b)(i)
- 4 positive half-cycles of equal amplitude shown from \(t=0\) to \(t=40\text{ ms}\) [1]
- Peak voltage marked as \(12\text{ V}\) and correct time scale (peaks every 10 ms) [1]

(b)(ii)1.
- Capacitor charges to peak voltage (when input voltage is high) [1]
- Capacitor discharges through resistor \(R\) (when input voltage falls), preventing voltage falling to zero / reducing fluctuation [1]

(b)(ii)2.
- \(\tau = RC = 33\text{ ms}\) [1]
- Comparison with time between peaks (\(10\text{ ms}\)): since \(\tau > 10\text{ ms}\), discharge is slow enough to ensure effective smoothing [1]

(a) Magnetic flux linkage is defined as the product of the magnetic flux and the number of turns of the coil, \(N\Phi\). Magnetic flux is the product of the magnetic flux density and the area perpendicular to the magnetic field, \(\Phi = BA\).

(b)(i) The change in magnetic flux linkage is given by:
\(\Delta(N\Phi) = N A \Delta B\)
\(\Delta(N\Phi) = 650 \times (4.2 \times 10^{-4}\text{ m}^2) \times 0.15\text{ T} = 0.04095\text{ Wb} \approx 0.041\text{ Wb}\)

(ii) The induced electromotive force (e.m.f.) is given by Faraday's law of electromagnetic induction:
\(E = \frac{\Delta(N\Phi)}{\Delta t}
\)E = \frac{0.04095\text{ Wb}}{0.18\text{ s}} = 0.2275\text{ V} \approx 0.23\text{ V}\)

(iii) Lenz's law states that the direction of the induced e.m.f. or current is such as to oppose the change in magnetic flux that produces it. Since the external magnetic field is decreasing, the induced current in the coil will flow in a direction that creates its own magnetic field in the same direction as the external field, attempting to maintain the magnetic flux and oppose the decrease.

(c) The magnetic flux density changes from zero to \(0.20\text{ T}\) in the opposite direction. The change in magnetic flux linkage is:
\(\Delta(N\Phi) = N A \Delta B = 650 \times (4.2 \times 10^{-4}\text{ m}^2) \times 0.20\text{ T} = 0.0546\text{ Wb}\)
Using Faraday's law:
\(E = \frac{\Delta(N\Phi)}{\Delta t} = \frac{0.0546\text{ Wb}}{0.12\text{ s}} = 0.455\text{ V} \approx 0.46\text{ V}\)

(a)
- M1: Product of magnetic flux and number of turns (or \(N\Phi\)). [1]
- A1: Magnetic flux is defined as product of magnetic flux density and area perpendicular to the field (or terms in \(NBA\) defined with \(B\) perpendicular to \(A\)). [1]

(b)(i)
- C1: Substitution into \(\Delta(N\Phi) = N A \Delta B\) (e.g., \(650 \times 4.2 \times 10^{-4} \times 0.15\)). [1]
- A1: \(0.041\text{ Wb}\) (or \(0.0410\text{ Wb}\)). [1]

(b)(ii)
- C1: \(E = \frac{\text{change in flux linkage}}{\Delta t}\). [1]
- A1: \(0.23\text{ V}\) (or \(0.228\text{ V}\)). [1]

(b)(iii)
- B1: Statement of Lenz's law: The direction of the induced e.m.f./current opposes the change in magnetic flux that produces it. [1]
- B1: Explanation: As the external field decreases, the induced field is in the same direction to oppose this decrease. [1]

(c)
- C1: Calculation of change in flux linkage: \(650 \times 4.2 \times 10^{-4} \times 0.20 = 0.0546\text{ Wb}\) OR \(E = \frac{650 \times 4.2 \times 10^{-4} \times 0.20}{0.12}\). [1]
- A1: \(0.46\text{ V}\) (or \(0.455\text{ V}\)). [1]

(a)(i) Diodes \(D_1\) and \(D_4\) conduct when terminal A is positive with respect to terminal B.

(ii) Diodes \(D_2\) and \(D_3\) conduct when terminal B is positive with respect to terminal A.

(iii) During both half-cycles, current flows through the load resistor \(R\) in the same direction, from X to Y.

(b)(i) The capacitor stores charge/energy when the alternating input voltage is greater than the potential difference across the capacitor (near the peak values). When the input voltage drops below the capacitor's voltage, the capacitor discharges through the load resistor \(R\), preventing the output potential difference from dropping to zero and maintaining a more constant output voltage.

(ii) The time constant \(\tau\) of the discharge circuit is given by:
\(\tau = RC\)
\(\tau = (2.4 \times 10^3\ \Omega) \times (47 \times 10^{-6}\text{ F}) = 0.1128\text{ s} \approx 0.11\text{ s}\)

(iii) 1. Increasing the value of the capacitance \(C\) increases the time constant \(\tau\). The capacitor discharges more slowly, leading to a smaller decrease in potential difference between peaks, which reduces the ripple.
2. Decreasing the frequency of the alternating supply increases the period (time between peaks). This gives the capacitor more time to discharge before being recharged by the next cycle, resulting in a larger drop in voltage and hence an increased ripple.

(a)(i)
- B1: \(D_1\) and \(D_4\) (both required). [1]

(a)(ii)
- B1: \(D_2\) and \(D_3\) (both required). [1]

(a)(iii)
- B1: From X to Y (accept: same direction / unidirectional). [1]

(b)(i)
- B1: Capacitor charges up / stores energy near peak values of input voltage. [1]
- B1: Capacitor discharges through the load resistor when input voltage falls, preventing the output voltage from dropping to zero. [1]

(b)(ii)
- C1: Use of \(\tau = RC\) with correct substitution: \(2.4 \times 10^3 \times 47 \times 10^{-6}\). [1]
- A1: \(0.11\text{ s}\) (or \(0.113\text{ s}\)). [1]

(b)(iii)
1.
- B1: Ripple is reduced / smaller because the time constant increases, leading to a slower rate of discharge. [1]

2.
- B1: Ripple is increased / larger. [1]
- B1: Lower frequency means a longer period (or longer time between peaks), allowing more time for the capacitor to discharge. [1]

### 1. Variables
- **Independent variable:** Frequency \(f\) of the alternating current.
- **Dependent variable:** Peak induced e.m.f. \(V_0\) in the search coil.
- **Control variables:** Peak current \(I_0\) in the solenoid, number of turns of the solenoid \(N_s\), length of the solenoid \(L\), number of turns of the search coil \(N_c\), and cross-sectional area of the search coil \(A\).

### 2. Experimental Setup and Procedure
- Connect the signal generator in series with the long solenoid and a standard low-resistance resistor \(R\) (e.g., \(1.0\ \Omega\) or \(10\ \Omega\)) to monitor the current.
- Rigidly mount the small search coil at the geometric centre of the solenoid using a clamp stand, ensuring its longitudinal axis is perfectly parallel to the solenoid’s axis.
- Connect **Channel 1** of a dual-channel oscilloscope across the standard resistor \(R\) to display the current profile. Since \(I_0 = \frac{V_{R,0}}{R}\), keeping the peak voltage \(V_{R,0}\) on Channel 1 constant ensures \(I_0\) remains constant while \(f\) is varied.
- Connect **Channel 2** of the oscilloscope directly across the terminals of the search coil to display the induced e.m.f. waveform and measure its peak value \(V_0\).

### 3. Measurements and Calibration
- Set the signal generator to a specific frequency. Adjust the output amplitude so that the peak voltage across resistor \(R\) is at a predetermined constant value (meaning constant \(I_0\)).
- Measure \(V_0\) as half of the peak-to-peak voltage displayed on Channel 2 of the calibrated oscilloscope screen.
- Determine the frequency \(f\) by measuring the period \(T\) from the horizontal time-base of the oscilloscope (\(f = 1/T\)) to achieve higher accuracy than the signal generator dial.

### 4. Analysis of Data
- The relationship predicts \(V_0 = k f\), where \(k = \frac{2\pi^2 \mu_0 N_s N_c A I_0}{L}\).
- A graph of \(V_0\) on the y-axis against \(f\) on the x-axis should yield a straight line passing through the origin.
- The gradient of the graph is equal to \(\frac{2\pi^2 \mu_0 N_s N_c A I_0}{L}\).

### 5. Safety and Reliability
- **Overheating Prevention:** Use a relatively low peak current to avoid heating the solenoid, and turn off the power supply between taking readings.
- **Solenoid Field Uniformity:** The search coil must be much smaller in diameter than the solenoid to ensure it is subjected to a uniform, axial magnetic field.
- **Eliminating Interference:** Keep the search coil away from external alternating fields, such as mains transformers or other electronics, to avoid stray induced voltages.

Defining the Problem (3 marks):
- [1] Identify f as the independent variable and V_0 as the dependent variable.
- [1] Identify I_0 (peak current) as a key control variable that must be kept constant.
- [1] State that the dimensions of the search coil (A, N_c) and solenoid (L, N_s) must be kept constant.

Methods of Data Collection (5 marks):
- [1] Draw a clear circuit diagram showing the signal generator, solenoid, a standard series resistor, and connections to the oscilloscope.
- [1] Describe how the search coil is positioned at the center of the solenoid and coaxial with it.
- [1] Describe how peak current I_0 is monitored and maintained constant by measuring the peak voltage across the series resistor using Channel 1 of the oscilloscope.
- [1] Describe how the peak induced e.m.f. V_0 is measured using Channel 2 of the oscilloscope.
- [1] Describe how the frequency f is determined accurately by measuring the period T from the oscilloscope time-base (f = 1/T).

Method of Analysis (2 marks):
- [1] State that a graph of V_0 against f should be plotted.
- [1] State that a straight line passing through the origin confirms the relationship, with gradient equal to 2*pi^2 * mu_0 * N_s * N_c * A * I_0 / L.

Additional Details / Safety / Reliability (5 marks) - any 5 from:
- [1] Avoid overheating by keeping current low and turning off the power supply between readings.
- [1] Use a search coil with a small diameter relative to the solenoid's diameter to ensure the magnetic field is uniform across its area.
- [1] Use a clamp and stand to rigidly fix the position and orientation of the search coil.
- [1] Perform the experiment away from other magnetic fields or sources of electromagnetic noise (e.g., mains power supplies).
- [1] Use a high-resistance digital voltmeter or oscilloscope to ensure negligible current is drawn from the search coil.
- [1] Detail on measuring diameter of the search coil in perpendicular directions with calipers to find area A.

### (a) Linearised Equation
Taking the natural logarithm of both sides of \(V = V_0 e^{-\frac{t}{RC}}\):
\[\ln(V / \text{V}) = -\frac{1}{RC} t + \ln(V_0 / \text{V})\]
This is of the form \(y = m x + c\), where \(y = \ln(V / \text{V})\), \(x = t\), the gradient is \(m = -\frac{1}{RC}\), and the y-intercept is \(c = \ln(V_0 / \text{V})\).

### (b) Calculated Table of Values
The absolute uncertainty in \(\ln(V)\) is calculated using \(\Delta [\ln(V)] \approx \frac{\Delta V}{V}\):
- For \(t = 1.0\ \text{s}\): \(\ln(9.7) = 2.272\); \(\Delta[\ln(V)] = \frac{0.2}{9.7} = 0.021 \approx 0.02\) \(\implies 2.27 \pm 0.02\)
- For \(t = 2.0\ \text{s}\): \(\ln(7.8) = 2.054\); \(\Delta[\ln(V)] = \frac{0.2}{7.8} = 0.026 \approx 0.03\) \(\implies 2.05 \pm 0.03\)
- For \(t = 3.0\ \text{s}\): \(\ln(6.3) = 1.841\); \(\Delta[\ln(V)] = \frac{0.2}{6.3} = 0.032 \approx 0.03\) \(\implies 1.84 \pm 0.03\)
- For \(t = 4.0\ \text{s}\): \(\ln(5.1) = 1.629\); \(\Delta[\ln(V)] = \frac{0.2}{5.1} = 0.039 \approx 0.04\) \(\implies 1.63 \pm 0.04\)
- For \(t = 5.0\ \text{s}\): \(\ln(4.1) = 1.411\); \(\Delta[\ln(V)] = \frac{0.2}{4.1} = 0.049 \approx 0.05\) \(\implies 1.41 \pm 0.05\)
- For \(t = 6.0\ \text{s}\): \(\ln(3.3) = 1.194\); \(\Delta[\ln(V)] = \frac{0.2}{3.3} = 0.061 \approx 0.06\) \(\implies 1.19 \pm 0.06\)

### (c) Plotting the Graph
Plot \(\ln(V / \text{V})\) on the y-axis against \(t / \text{s}\) on the x-axis. Error bars are plotted vertically for each point.
- **Line of Best Fit (LOBF):** A line drawn through the center of gravity of the data points, balanced evenly.
- **Worst Acceptable Line (WAL):** A line with the steepest or shallowest possible gradient that still passes through all error bars (e.g., from the top of the error bar of the first point to the bottom of the error bar of the last point).

### (d) Determination of Gradient and Intercept
- **Best Gradient (m):** From linear regression or best fit line, \(m = -0.215\ \text{s}^{-1}\).
- **Worst Gradient (m_w):** From WAL, e.g., passing through \((1.0, 2.25)\) and \((6.0, 1.25)\):
\[m_w = \frac{1.25 - 2.25}{6.0 - 1.0} = -0.200\ \text{s}^{-1}\]
- **Uncertainty in Gradient (\(\Delta m\)):** \(\Delta m = |m_{\text{best}} - m_{\text{worst}}| = 0.015\ \text{s}^{-1}\) (or up to \(0.016\ \text{s}^{-1}\)).
So, \(m = -0.215 \pm 0.016\ \text{s}^{-1}\).
- **Best Y-intercept (c):** \(c = 2.48\).
- **Worst Y-intercept (c_w):** \(c_w = 2.45\) (or \(2.52\)).
- **Uncertainty in Intercept (\(\Delta c\)):** \(|2.48 - 2.44| = 0.04\).
So, \(c = 2.48 \pm 0.04\).

### (e) Calculation of Capacitance \(C\) and \(V_0\)
**(i) Capacitance \(C\):**
\[m = -\frac{1}{RC} \implies C = -\frac{1}{m R}\]
\[C = \frac{1}{0.215 \times 47.0 \times 10^3\ \Omega} = 9.896 \times 10^{-5}\ \text{F} \approx 99.0\ \mu\text{F}\]

Uncertainty in \(C\):
\[\frac{\Delta C}{C} = \frac{\Delta |m|}{|m|} + \frac{\Delta R}{R} = \frac{0.016}{0.215} + \frac{1.5}{47.0} = 0.0744 + 0.0319 = 0.1063\ (10.6\%)\]
\[\Delta C = 0.1063 \times 99.0\ \mu\text{F} \approx 11\ \mu\text{F}\]
\[C = (99 \pm 11)\ \mu\text{F}\]

**(ii) Peak Initial Voltage \(V_0\):**
\[c = \ln(V_0 / \text{V}) \implies V_0 = e^{c} = e^{2.48} = 11.94\ \text{V} \approx 11.9\ \text{V}\]
\[\Delta V_0 = V_{0,\text{max}} - V_0 = e^{2.52} - 11.94 = 12.43 - 11.94 = 0.49\ \text{V} \approx 0.5\ \text{V}\]
\[V_0 = (11.9 \pm 0.5)\ \text{V}\]

Part (a) [1 Mark]:
- [1] Correct equation: ln(V) = -t / (RC) + ln(V_0) or equivalent linear format.

Part (b) [3 Marks]:
- [1] All ln(V) calculated correctly to 2 decimal places (2.27, 2.05, 1.84, 1.63, 1.41, 1.19).
- [2] Correctly calculated absolute uncertainties for ln(V) (all 6 correct for 2 marks, 4-5 correct for 1 mark): 0.02, 0.03, 0.03, 0.04, 0.05, 0.06.

Part (c) [3 Marks]:
- [1] Points plotted correctly with y-error bars correctly shown according to calculated uncertainties.
- [1] Line of best fit drawn as a single, clean, balanced straight line.
- [1] Worst acceptable line drawn with steepest/shallowest possible slope passing through all error bars and clearly labelled.

Part (d) [4 Marks]:
- [1] Correct calculation of the gradient of the line of best fit (yields -0.215 +- 0.010 s^-1 depending on student graph drawing).
- [1] Correct calculation of the uncertainty in the gradient using the worst acceptable line.
- [1] Correct calculation of the y-intercept of the line of best fit (yields 2.48 +- 0.05).
- [1] Correct calculation of the uncertainty in the y-intercept using the worst acceptable line.

Part (e) [4 Marks]:
- [1] Value of C calculated in the range 94 to 104 uF.
- [1] Uncertainty in C calculated using fractional/percentage uncertainties of both R and m (yielding ~11 uF).
- [1] Value of V_0 calculated using exponential of y-intercept (yielding 11.5 V to 12.5 V).
- [1] Uncertainty in V_0 calculated correctly using extreme intercept value or standard propagation (yielding ~0.5 V).