2024 OCR A-Level Physics A - H556 模擬試題連答案詳解

The density of a uniform cylindrical wire is given by \(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 L}\).

The fractional uncertainty in \(\rho\) is:
\(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\)

Let's calculate each individual percentage uncertainty:
- Mass uncertainty: \(\frac{0.2}{24.6} \times 100\% \approx 0.81\%\)
- Diameter uncertainty: \(\frac{0.02}{1.25} \times 100\% = 1.60\%\)
- Length uncertainty: \(\frac{0.05}{2.50} \times 100\% = 2.00\%\)

Combining these into the total uncertainty equation:
Percentage uncertainty in \(\rho = 0.81\% + 2(1.60\%) + 2.00\% = 0.81\% + 3.20\% + 2.00\% = 6.01\%\), which rounds to \(6.0\%\).

1 mark for calculating the correct sum of the percentage uncertainties with the factor of 2 for diameter, leading to C.

First, convert the half-life to seconds:
\(t_{1/2} = 4.5 \times 10^9\text{ years} \times 3.16 \times 10^7\text{ s year}^{-1} = 1.422 \times 10^{17}\text{ s}\)

Next, calculate the decay constant \(\lambda\):
\(\lambda = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{1.422 \times 10^{17}\text{ s}} \approx 4.873 \times 10^{-18}\text{ s}^{-1}\)

Finally, calculate the activity \(A\):
\(A = \lambda N = (4.873 \times 10^{-18}\text{ s}^{-1}) \times (6.0 \times 10^{18}) \approx 29.2\text{ Bq}\)

This rounds to \(29\text{ Bq}\).

1 mark for correctly converting half-life to seconds, calculating the decay constant, and finding the activity as 29 Bq.

For the wire to remain suspended, the upward magnetic force must balance the downward weight of the wire:
\(F_B = W \implies B I L = m g\)

Rearranging to solve for current \(I\):
\(I = \frac{m g}{B L}\)

Convert mass from grams to kilograms: \(m = 12\text{ g} = 0.012\text{ kg}\).

Substitute the values into the equation:
\(I = \frac{0.012\text{ kg} \times 9.81\text{ m s}^{-2}}{0.40\text{ T} \times 0.15\text{ m}} = \frac{0.11772}{0.060} = 1.962\text{ A}\)

This rounds to \(2.0\text{ A}\).

1 mark for equating the magnetic force to the gravitational weight, converting units correctly, and solving for the current of 2.0 A.

Using the ideal gas relationship:
\(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)

First, convert the temperatures to Kelvin:
\(T_1 = 27 + 273 = 300\text{ K}\)
\(T_2 = 327 + 273 = 600\text{ K}\)

The volume is halved, so \(V_2 = 0.5 V_1\).

Rearranging for \(P_2\):
\(P_2 = P_1 \times \left(\frac{V_1}{V_2}\right) \times \left(\frac{T_2}{T_1}\right)\)
\(P_2 = (1.2 \times 10^5\text{ Pa}) \times 2 \times \left(\frac{600}{300}\right)\)
\(P_2 = (1.2 \times 10^5\text{ Pa}) \times 2 \times 2 = 4.8 \times 10^5\text{ Pa}\)

1 mark for converting temperatures to Kelvin and using the gas law relation to find the correct new pressure of 4.8 x 10^5 Pa.

The volume of a sphere is given by \(V = \frac{4}{3} \pi r^3\).

Substituting \(r = 4.20\text{ mm}\):
\(V = \frac{4}{3} \pi (4.20)^3 \approx 310.34\text{ mm}^3\)

The fractional uncertainty in volume \(\frac{\Delta V}{V}\) is three times the fractional uncertainty in the radius:
\(\frac{\Delta V}{V} = 3 \frac{\Delta r}{r}\)

Therefore, the absolute uncertainty in volume \(\Delta V\) is:
\(\Delta V = 3 \times \left(\frac{0.05}{4.20}\right) \times 310.34\text{ mm}^3 \approx 11.08\text{ mm}^3\)

Rounding to the appropriate number of significant figures, this gives \(\pm 11\text{ mm}^3\).

1 mark for calculating the volume, establishing the fractional uncertainty relationship, and calculating the absolute uncertainty of 11 mm^3.

First, calculate the mass defect \(\Delta m\) in the reaction:
\(\Delta m = m_{\text{U}} - (m_{\text{Th}} + m_{\alpha})\)
\(\Delta m = 238.0508\text{ u} - (234.0436\text{ u} + 4.0015\text{ u})\)
\(\Delta m = 238.0508\text{ u} - 238.0451\text{ u} = 0.0057\text{ u}\)

Now, convert this mass defect to energy using \(1\text{ u} = 931.5\text{ MeV}\):
\(E = 0.0057 \times 931.5\text{ MeV} \approx 5.31\text{ MeV}\)

Therefore, the energy released is approximately \(5.3\text{ MeV}\).

1 mark for calculating the mass defect and multiplying by 931.5 MeV to obtain 5.3 MeV.

The magnetic force on a charged particle provides the centripetal force keeping it in circular motion:
\(B q v = \frac{m v^2}{r}\)

Rearranging to solve for the radius \(r\):
\(r = \frac{m v}{B q}\)

Substitute the given values:
\(r = \frac{(1.67 \times 10^{-27}\text{ kg}) \times (2.4 \times 10^6\text{ m s}^{-1})}{0.15\text{ T} \times 1.60 \times 10^{-19}\text{ C}}\)
\(r = \frac{4.008 \times 10^{-21}}{2.40 \times 10^{-20}} \approx 0.167\text{ m}\)

This rounds to \(0.17\text{ m}\).

1 mark for using the formula for the radius of a charged particle's path in a magnetic field and calculating the correct radius of 0.17 m.

The root-mean-square speed of molecules in an ideal gas is given by:
\(c_{\text{rms}} = \sqrt{\frac{3RT}{M}}\)

In a mixture, both helium and oxygen are at the same temperature \(T\). Thus, \(c_{\text{rms}}\) is inversely proportional to the square root of the molar mass \(M\):
\(c_{\text{rms}} \propto \frac{1}{\sqrt{M}}\)

The ratio of the r.m.s. speed of helium to that of oxygen is:
\(\frac{c_{\text{rms, He}}}{c_{\text{rms, O}_2}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{He}}}} = \sqrt{\frac{32\text{ g mol}^{-1}}{4.0\text{ g mol}^{-1}}} = \sqrt{8} \approx 2.83\)

This rounds to \(2.8\).

1 mark for recognizing that r.m.s. speed is inversely proportional to the square root of the molar mass and calculating the correct ratio of 2.8.

The percentage uncertainty in resistivity is given by the sum of the percentage uncertainties of the independent variables, taking into account the power of each term: \(\% \Delta \rho = \% \Delta R + 2(\% \Delta d) + \% \Delta L\). Substituting the given values: \(\% \Delta \rho = 2.5\% + 2(4.0\%) + 2.0\% = 2.5\% + 8.0\% + 2.0\% = 12.5\%\).

1 mark: Correctly applies the uncertainty combination rule for powers and sums to calculate 12.5%.

First, subtract the background count rate to find the corrected count rates. Corrected initial rate = \(135 - 15 = 120\) counts per minute. Corrected final rate = \(45 - 15 = 30\) counts per minute. The ratio of final to initial corrected activity is \(\frac{30}{120} = \frac{1}{4} = (\frac{1}{2})^2\). This corresponds to exactly 2 half-lives. Since 2 half-lives elapsed in 6.0 hours, the half-life is \(\frac{6.0}{2} = 3.0\) hours.

1 mark: Correctly subtracts background, identifies that 2 half-lives have elapsed, and calculates 3.0 hours.

The radius of the path of a charged particle in a magnetic field is given by \(r = \frac{m v}{q B}\). In terms of kinetic energy \(E_k = \frac{1}{2} m v^2\), the momentum is \(m v = \sqrt{2 m E_k}\), so \(r = \frac{\sqrt{2 m E_k}}{q B}\). Since \(E_k\) and \(B\) are the same for both, \(r \propto \frac{\sqrt{m}}{q}\). For a proton, \(m_p = m\) and \(q_p = e\). For an alpha particle, \(m_{\alpha} = 4m\) and \(q_{\alpha} = 2e\). Therefore, \(r_p \propto \frac{\sqrt{m}}{e}\) and \(r_{\alpha} \propto \frac{\sqrt{4m}}{2e} = \frac{2\sqrt{m}}{2e} = \frac{\sqrt{m}}{e}\). The ratio \(\frac{r_p}{r_{\alpha}}\) is therefore 1.0.

1 mark: Derives the relationship between radius and kinetic energy, substitutes the mass and charge ratios, and correctly finds the ratio to be 1.0.

The root-mean-square speed is proportional to the square root of the absolute temperature: \(v_{\text{rms}} \propto \sqrt{T}\), or \(T \propto v_{\text{rms}}^2\). The initial temperature in Kelvin is \(T_1 = 27 + 273 = 300\,\text{K}\). Doubling the r.m.s. speed means the absolute temperature must increase by a factor of \(2^2 = 4\). Thus, the final temperature in Kelvin is \(T_2 = 4 \times 300\,\text{K} = 1200\,\text{K}\). Converting back to Celsius: \(T_2 = 1200 - 273 = 927\,^{\circ}\text{C}\).

1 mark: Converts temperature to Kelvin, relates r.m.s. speed to temperature squared, determines the final Kelvin temperature, and converts back to Celsius to get 927 °C.

Rearranging Stokes' Law for viscosity gives \(\eta = \frac{F}{6 \pi r v}\). The constant \(6 \pi\) is dimensionless. In terms of SI base units, the unit of force \(F\) is \(\text{kg}\,\text{m}\,\text{s}^{-2}\), the unit of radius \(r\) is \(\text{m}\), and the unit of speed \(v\) is \(\text{m}\,\text{s}^{-1}\). Substituting these gives: \([\eta] = \frac{\text{kg}\,\text{m}\,\text{s}^{-2}}{\text{m} \times \text{m}\,\text{s}^{-1}} = \frac{\text{kg}\,\text{m}\,\text{s}^{-2}}{\text{m}^2\,\text{s}^{-1}} = \text{kg}\,\text{m}^{-1}\,\text{s}^{-1}\).

1 mark: Correctly expresses the base units of force, radius, and speed, and simplifies to find kg m^-1 s^-1.

The energy released is the difference between the total binding energy of the product and the total binding energy of the reactants. Total binding energy of helium-4 (4 nucleons) = \(4 \times 7.07\,\text{MeV} = 28.28\,\text{MeV}\). Total binding energy of one deuterium nucleus (2 nucleons) = \(2 \times 1.11\,\text{MeV} = 2.22\,\text{MeV}\). Since there are two deuterium nuclei reacting, the total initial binding energy = \(2 \times 2.22\,\text{MeV} = 4.44\,\text{MeV}\). Therefore, the energy released = \(28.28 - 4.44 = 23.84\,\text{MeV}\).

1 mark: Correctly calculates total binding energy of reactants and product, and subtracts them to obtain 23.84 MeV.

According to Faraday's law of electromagnetic induction, the magnitude of the average induced e.m.f. is \(E = \frac{\Delta (N \Phi)}{\Delta t}\). The initial magnetic flux linkage is \(N \Phi_1 = N B A = 150 \times 0.20\,\text{T} \times 4.0 \times 10^{-3}\,\text{m}^2 = 0.12\,\text{Wb-turns}\). After a \(90^{\circ}\) rotation, the plane of the coil is parallel to the field lines, so the final magnetic flux linkage is \(N \Phi_2 = 0\). The change in flux linkage is \(\Delta (N \Phi) = 0.12\,\text{Wb-turns}\). The magnitude of the induced e.m.f. is \(E = \frac{0.12}{0.15} = 0.80\,\text{V}\).

1 mark: Calculates change in magnetic flux linkage and divides by time to determine the average induced e.m.f. of 0.80 V.

The formula for the density of a cylinder is:

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

Therefore, the fractional uncertainty in \(\rho\) is given by:

\[\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\]

Calculate each individual percentage uncertainty:

- Mass: \(\%\Delta m = \frac{0.1}{145.2} \times 100\% \approx 0.0689\%\)
- Diameter: \(\%\Delta d = \frac{0.02}{1.20} \times 100\% \approx 1.6667\%\)
- Length: \(\%\Delta L = \frac{0.05}{8.52} \times 100\% \approx 0.5869\%\)

Now, combine these to find the total percentage uncertainty:

\[\%\Delta \rho = 0.0689\% + 2(1.6667\%) + 0.5869\%\]
\[\%\Delta \rho = 0.0689\% + 3.3333\% + 0.5869\% = 3.989\% \approx 4.0\%\]

M1: Recall or use of \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\)
M1: Correct individual percentage uncertainty calculations for \(m\), \(d\), or \(L\)
A1: Correct final percentage uncertainty calculated to 2 s.f. as \(4.0\%\) (accept \(4\%\))

First, calculate the period \(T\):
\[T = \frac{39.2}{20} = 1.96\text{ s}\]

The absolute uncertainty in \(T\) is:
\[\Delta T = \frac{0.2}{20} = 0.01\text{ s}\]

Next, calculate the value of \(g\):
\[g = \frac{4\pi^2 \times 0.950}{1.96^2} \approx 9.762\text{ m s}^{-2}\]

Calculate the fractional uncertainty in \(g\):
\[\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}\]
\[\frac{\Delta g}{g} = \frac{0.5}{95.0} + 2\left(\frac{0.01}{1.96}\right) \approx 0.00526 + 0.01020 = 0.01546\]

Calculate the absolute uncertainty in \(g\):
\[\Delta g = 9.762 \times 0.01546 \approx 0.151\text{ m s}^{-2}\]

Rounding \(\Delta g\) to 1 significant figure gives \(0.2\text{ m s}^{-2}\), which means \(g\) must be rounded to 1 decimal place:
\[g = (9.8 \pm 0.2)\text{ m s}^{-2}\]
(Alternatively, writing as \(9.76 \pm 0.15\text{ m s}^{-2}\) is also acceptable as it preserves consistent precision).

M1: Correct calculation of \(g = 9.76\text{ m s}^{-2}\) and period \(T = 1.96\text{ s}\)
M1: Equation or substitution showing \(\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta T}{T}\) (gives \(1.55\%\))
A1: Correct final expression with matched precision: \((9.8 \pm 0.2)\text{ m s}^{-2}\) or \((9.76 \pm 0.15)\text{ m s}^{-2}\)

First, find the number of moles \(n\) of Uranium-235:
\[n = \frac{\text{mass}}{\text{molar mass}} = \frac{2.4 \times 10^{-9}\text{ kg}}{0.235\text{ kg mol}^{-1}} \approx 1.0213 \times 10^{-8}\text{ mol}\]

Now, calculate the number of nuclei \(N\):
\[N = n \times N_A = (1.0213 \times 10^{-8}\text{ mol}) \times (6.02 \times 10^{23}\text{ mol}^{-1}) \approx 6.148 \times 10^{15}\]

Finally, calculate the activity \(A\):
\[A = \lambda N = (3.1 \times 10^{-17}\text{ s}^{-1}) \times (6.148 \times 10^{15}) \approx 0.191\text{ Bq}\]

To 2 significant figures, the activity is \(0.19\text{ Bq}\).

M1: Correct conversion of mass to calculate moles of Uranium-235 (\(1.02 \times 10^{-8}\text{ mol}\))
M1: Correct calculation of the number of nuclei \(N\) (\(6.15 \times 10^{15}\))
A1: Correct final activity of \(0.19\text{ Bq}\) or \(0.191\text{ Bq}\)

We can determine the remaining activity using the decay equation or the number of half-lives:

Number of half-lives elapsed:
\[n = \frac{48.0}{15.0} = 3.2\]

The remaining activity \(A\) is:
\[A = A_0 \times (0.5)^n = 8.0 \times 10^7 \times (0.5)^{3.2}\]
\[A \approx 8.0 \times 10^7 \times 0.108819 = 8.706 \times 10^6\text{ Bq}\]

To 2 significant figures, this is \(8.7 \times 10^6\text{ Bq}\).

M1: Calculates decay constant \(\lambda = \frac{\ln 2}{15.0} = 0.0462\text{ h}^{-1}\) OR calculates number of half-lives as 3.2
M1: Uses \(A = A_0 e^{-\lambda t}\) or \(A = A_0 (0.5)^{3.2}\) with correct substitution
A1: Correct answer of \(8.7 \times 10^6\text{ Bq}\) (accept \(8.7 \times 10^6\) to \(8.8 \times 10^6\text{ Bq}\))

1. Calculate the magnetic force \(F_B\) acting on the wire:
\[F_B = B I L = 0.12\text{ T} \times 5.0\text{ A} \times 0.080\text{ m} = 0.048\text{ N}\text{ (upwards)}\]

2. Calculate the gravitational force (weight \(W\)) acting on the wire:
\[W = m g = (3.5 \times 10^{-3}\text{ kg}) \times 9.81\text{ m s}^{-2} = 0.034335\text{ N}\text{ (downwards)}\]

3. Calculate the net force \(F_{\text{net}}\):
\[F_{\text{net}} = F_B - W = 0.048 - 0.034335 = 0.013665\text{ N}\text{ (upwards since } F_B > W)\]

4. Calculate the vertical acceleration \(a\):
\[a = \frac{F_{\text{net}}}{m} = \frac{0.013665\text{ N}}{3.5 \times 10^{-3}\text{ kg}} \approx 3.90\text{ m s}^{-2}\]

Therefore, the acceleration is \(3.9\text{ m s}^{-2}\) in the upward direction.

M1: Calculates magnetic force \(F_B = 0.048\text{ N}\) and weight \(W = 0.0343\text{ N}\)
M1: Obtains net force \(F_{\text{net}} = 0.0137\text{ N}\) and correctly determines direction as upwards
A1: Calculates acceleration as \(3.9\text{ m s}^{-2}\) upwards

The magnetic force provides the required centripetal force for the circular motion:

\[B e v = \frac{m v^2}{r}\]

Rearranging for the radius \(r\):

\[r = \frac{m v}{B e}\]

Substitute the given values:

\[r = \frac{(9.11 \times 10^{-31}\text{ kg}) \times (3.0 \times 10^6\text{ m s}^{-1})}{(2.5 \times 10^{-3}\text{ T}) \times (1.60 \times 10^{-19}\text{ C})}\]
\[r = \frac{2.733 \times 10^{-24}}{4.00 \times 10^{-22}} = 6.8325 \times 10^{-3}\text{ m}\]

To 2 significant figures, the radius is \(6.8 \times 10^{-3}\text{ m}\) (or \(6.8\text{ mm}\)).

M1: Equates magnetic force to centripetal force: \(B e v = \frac{m v^2}{r}\)
M1: Rearranges formula to get \(r = \frac{m v}{B e}\) and substitutes values correctly
A1: Correctly calculates \(r = 6.8 \times 10^{-3}\text{ m}\) (accept \(6.8\text{ mm}\))

The mean kinetic energy of a molecule in an ideal gas depends only on the temperature \(T\):

\[E_k = \frac{1}{2} m \overline{c^2} = \frac{3}{2} k_B T\]

Since both gases are at the same temperature, their molecules have the same mean kinetic energy:

\[\frac{1}{2} m_{\text{He}} c_{\text{rms, He}}^2 = \frac{1}{2} m_{\text{Ar}} c_{\text{rms, Ar}}^2\]

Therefore, we can write:

\[c_{\text{rms, Ar}} = c_{\text{rms, He}} \sqrt{\frac{m_{\text{He}}}{m_{\text{Ar}}}}\]

The ratio of the molecular masses is equivalent to the ratio of their nucleon numbers (molar masses):

\[\frac{m_{\text{He}}}{m_{\text{Ar}}} = \frac{4}{40} = 0.1\]

Substitute the values:

\[c_{\text{rms, Ar}} = 1300 \times \sqrt{0.1} \approx 1300 \times 0.3162 = 411\text{ m s}^{-1}\]

To 2 significant figures, this is \(410\text{ m s}^{-1}\).

M1: Recalls that average kinetic energy is constant at the same temperature or states \(c_{\text{rms}} \propto \frac{1}{\sqrt{m}}\)
M1: Expresses the ratio of speeds correctly: \(c_{\text{rms, Ar}} = c_{\text{rms, He}} \sqrt{\frac{4}{40}}\)
A1: Correct calculation to give \(410\text{ m s}^{-1}\) or \(411\text{ m s}^{-1}\)

Using the ideal gas equation:

\[P V = n R T\]

Rearrange the equation to solve for the absolute temperature \(T\):

\[T = \frac{P V}{n R}\]

Substitute the given values:

\[T = \frac{(1.8 \times 10^5\text{ Pa}) \times 0.024\text{ m}^3}{1.2\text{ mol} \times 8.31\text{ J K}^{-1}\text{ mol}^{-1}}\]
\[T = \frac{4320}{9.972} \approx 433.2\text{ K}\]

Convert the temperature from Kelvin to Celsius:

\[\theta = 433.2 - 273.15 = 160.05\text{ }^\circ\text{C}\]

To 2 significant figures, this is \(160\text{ }^\circ\text{C}\).

M1: Recalls and rearranges the ideal gas equation: \(T = \frac{PV}{nR}\)
M1: Correctly calculates \(T \approx 433\text{ K}\) (or \(433.2\text{ K}\))
A1: Correctly converts Kelvin to Celsius to obtain \(160\text{ }^\circ\text{C}\) (accept \(160.1\text{ }^\circ\text{C}\))

1. Discard the anomaly: the remaining readings are \(0.41\text{ mm}\), \(0.42\text{ mm}\), \(0.42\text{ mm}\), and \(0.43\text{ mm}\).
2. Calculate the mean diameter \(d\):

\(d = \frac{0.41 + 0.42 + 0.42 + 0.43}{4} = 0.42\text{ mm}\)

3. Calculate the absolute uncertainty in \(d\) using half the range:

\(\Delta d = \frac{0.43 - 0.41}{2} = 0.01\text{ mm}\)

4. Calculate the percentage uncertainty in \(d\):

\(\%\Delta d = \left(\frac{0.01}{0.42}\right) \times 100 \approx 2.38\%\)

5. Since cross-sectional area \(A = \frac{\pi d^2}{4}\), the percentage uncertainty in \(A\) is twice the percentage uncertainty in \(d\):

\(\%\Delta A = 2 \times 2.38\% = 4.76\% \approx 4.8\%\)

- **C1 (Method Mark):** Determination of correct mean diameter of remaining readings (\(0.42\text{ mm}\)) and correct half-range uncertainty (\(0.01\text{ mm}\)).
- **C1 (Method Mark):** Formula for combining uncertainties for squared quantities: \(\%\Delta A = 2 \times \%\Delta d\).
- **A1.4 (Accuracy Mark):** Correct calculation of percentage uncertainty of cross-sectional area as \(4.8\%\) (or \(4.76\%\)).

1. Determine the number of half-lives that have elapsed:
The activity decreases from \(320\text{ Bq}\) to \(40\text{ Bq}\).

\(\frac{A}{A_0} = \frac{40}{320} = \frac{1}{8} = \left(\frac{1}{2}\right)^3\)

This corresponds to exactly 3 half-lives.

2. Calculate the half-life \(T_{1/2}\):

\(3 \times T_{1/2} = 15.0\text{ hours} \implies T_{1/2} = 5.0\text{ hours}\)

Convert the half-life to seconds:

\(T_{1/2} = 5.0 \times 3600\text{ s} = 18,000\text{ s}\)

3. Calculate the decay constant \(\lambda\):

\(\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.6931}{18,000} \approx 3.85 \times 10^{-5}\text{ s}^{-1}\)

- **C1 (Method Mark):** Recognition that 3 half-lives have passed, giving a half-life of \(5.0\text{ hours}\) or conversion of time to seconds (\(54,000\text{ s}\)).
- **C1 (Method Mark):** Correct use of the exponential decay equation \(A = A_0 e^{-\lambda t}\) or relation \(\lambda = \frac{\ln(2)}{T_{1/2}}\).
- **A1.4 (Accuracy Mark):** Final answer of \(3.85 \times 10^{-5}\text{ s}^{-1}\) (or \(3.9 \times 10^{-5}\text{ s}^{-1}\)).

1. Find the initial magnetic flux linkage of the coil:

\(\Phi_{\text{initial}} = B A N = 0.15\text{ T} \times 3.5 \times 10^{-3}\text{ m}^2 \times 120 = 0.063\text{ Wb-turns}\)

2. Find the final magnetic flux linkage after a \(90^\circ\) rotation:
Since the plane of the coil is now parallel to the magnetic field, the normal to the area is perpendicular to the field lines.

\(\Phi_{\text{final}} = 0\text{ Wb-turns}\)

3. Calculate the magnitude of the average induced e.m.f. using Faraday's law:

\(V = \left|\frac{\Delta \Phi}{\Delta t}\right| = \frac{0.063}{0.24} = 0.2625\text{ V}\)

To 2 significant figures, the average induced e.m.f. is \(0.26\text{ V}\).

- **C1 (Method Mark):** Correct substitution into formula for initial flux linkage \(B A N\) to obtain \(0.063\text{ Wb-turns}\).
- **C1 (Method Mark):** Use of Faraday's law \(e.m.f. = \frac{\Delta (BAN)}{\Delta t}\).
- **A1.4 (Accuracy Mark):** Correct calculated average e.m.f. of \(0.26\text{ V}\) (or \(0.263\text{ V}\)).

1. Convert the initial temperature to Kelvin:

\(T_1 = 27 + 273 = 300\text{ K}\)

2. Since the volume of the rigid vessel is constant, the pressure is directly proportional to absolute temperature (\(P \propto T\)). Since pressure doubles:

\(T_2 = 2 \times T_1 = 600\text{ K}\)

Therefore, the change in temperature is:

\(\Delta T = T_2 - T_1 = 300\text{ K}\)

3. Since the volume is constant, no work is done on or by the gas. Thus, the thermal energy supplied equals the change in internal energy:

\(Q = \Delta U = \frac{3}{2}nR\Delta T\)

\(Q = 1.5 \times 0.12\text{ mol} \times 8.31\text{ J mol}^{-1}\text{ K}^{-1} \times 300\text{ K} = 448.74\text{ J}\)

Rounding to 2 significant figures gives \(450\text{ J}\).

- **C1 (Method Mark):** Conversion of temperature to Kelvin and recognition that \(T_2 = 600\text{ K}\) (or \(\Delta T = 300\text{ K}\)).
- **C1 (Method Mark):** Correct formulation of internal energy change using \(\Delta U = \frac{3}{2}nR\Delta T\).
- **A1.4 (Accuracy Mark):** Correct numerical result of \(450\text{ J}\) (or \(449\text{ J}\)).

1. Rearrange the formula to make \(g\) the subject:

\(g = \frac{2h}{t^2}\)

2. Calculate the central value of \(g\):

\(g = \frac{2 \times 1.20}{0.50^2} = 9.60\text{ m s}^{-2}\)

3. Calculate the fractional/percentage uncertainties:

\(\%\Delta h = \frac{0.02}{1.20} \times 100 \approx 1.67\%\)

\(\%\Delta t = \frac{0.02}{0.50} \times 100 = 4.00\%\)

4. Determine the percentage uncertainty in \(g\):

\(\%\Delta g = \%\Delta h + 2 \times \%\Delta t = 1.67\% + 2 \times 4.00\% = 9.67\%\)

5. Calculate the absolute uncertainty in \(g\):

\(\Delta g = 9.67\% \times 9.60\text{ m s}^{-2} = 0.0967 \times 9.60 \approx 0.928\text{ m s}^{-2}\)

To 1 significant figure (which is standard for absolute uncertainties), this is \(0.9\text{ m s}^{-2}\).

- **C1 (Method Mark):** Calculation of the central value of \(g = 9.6\text{ m s}^{-2}\).
- **C1 (Method Mark):** Correct addition of relative uncertainties to obtain \(\%\Delta g = \%\Delta h + 2\%\Delta t\) (leading to \(9.7\%\)).
- **A1.4 (Accuracy Mark):** Calculation of absolute uncertainty \(\Delta g\) as \(0.9\text{ m s}^{-2}\) (accept \(0.93\text{ m s}^{-2}\)).

1. Convert the photon energy from \(\text{MeV}\) to Joules:

\(E_{\text{photon}} = 1.25 \times 10^6 \times 1.60 \times 10^{-19}\text{ J} = 2.00 \times 10^{-13}\text{ J}\)

2. Calculate the total number of decays (and hence photons absorbed) in one minute (\(60\text{ s}\)):

\(N = \text{Activity} \times \text{time} = 1.8 \times 10^5\text{ Bq} \times 60\text{ s} = 1.08 \times 10^7\text{ decays}\)

3. Calculate the total energy absorbed:

\(E_{\text{total}} = N \times E_{\text{photon}} = 1.08 \times 10^7 \times 2.00 \times 10^{-13}\text{ J} = 2.16 \times 10^{-6}\text{ J}\)

4. Convert the energy into microjoules (\(\mu\text{J}\)):

\(E_{\text{total}} = 2.16\ \mu\text{J}\)

Rounding to 2 significant figures gives \(2.2\ \mu\text{J}\).

- **C1 (Method Mark):** Conversion of photon energy to Joules to obtain \(2.0 \times 10^{-13}\text{ J}\).
- **C1 (Method Mark):** Calculation of total energy emitted per second (power) or total decays in one minute (\(1.08 \times 10^7\)).
- **A1.4 (Accuracy Mark):** Correct calculation of total energy in microjoules as \(2.2\ \mu\text{J}\) (or \(2.16\ \mu\text{J}\)).

1. For the wire to be suspended, the upward magnetic force must equal the downward force of gravity:

\(F_{\text{magnetic}} = F_{\text{gravity}} \implies B I L = m g\)

2. Convert mass to kilograms:

\(m = 12\text{ g} = 0.012\text{ kg}\)

3. Substitute the values into the force equation:

\(B \times 2.5\text{ A} \times 0.45\text{ m} = 0.012\text{ kg} \times 9.81\text{ m s}^{-2}\)

\(1.125 \times B = 0.11772\)

\(B = \frac{0.11772}{1.125} \approx 0.1046\text{ T}\)

To 2 significant figures, the magnetic flux density is \(0.10\text{ T}\).

- **C1 (Method Mark):** Clear expression equating magnetic force to weight (\(B I L = m g\)).
- **C1 (Method Mark):** Correct substitution of values, including mass conversion to \(0.012\text{ kg}\).
- **A1.4 (Accuracy Mark):** Correct calculated magnetic flux density of \(0.10\text{ T}\) (or \(0.105\text{ T}\)).

1. State the ideal gas equation relating initial and final states:

\(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)

2. Convert the temperatures to Kelvin:

\(T_1 = 7.0 + 273 = 280\text{ K}\)

\(T_2 = 22.0 + 273 = 295\text{ K}\)

3. Rearrange the equation to solve for the final volume \(V_2\):

\(V_2 = V_1 \times \left(\frac{P_1}{P_2}\right) \times \left(\frac{T_2}{T_1}\right)\)

\(V_2 = (2.0 \times 10^{-6}\text{ m}^3) \times \left(\frac{2.4 \times 10^5}{1.0 \times 10^5}\right) \times \left(\frac{295}{280}\right)\)

\(V_2 = 2.0 \times 10^{-6} \times 2.4 \times 1.0536 \approx 5.057 \times 10^{-6}\text{ m}^3\)

To 2 significant figures, the final volume is \(5.1 \times 10^{-6}\text{ m}^3\).

- **C1 (Method Mark):** Conversion of temperatures to absolute scale (Kelvin), i.e., \(280\text{ K}\) and \(295\text{ K}\).
- **C1 (Method Mark):** Use of the combined gas law relationship \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\).
- **A1.4 (Accuracy Mark):** Final volume of \(5.1 \times 10^{-6}\text{ m}^3\) (or \(5.06 \times 10^{-6}\text{ m}^3\)).

First, calculate the percentage uncertainty in \(L\): \(\%\Delta L = \frac{0.2}{80.0} \times 100\% = 0.25\%\). Next, calculate the percentage uncertainty in \(T\): \(\%\Delta T = \frac{0.05}{1.80} \times 100\% = 2.78\%\). Since \(g \propto L / T^2\), the percentage uncertainty in \(g\) is given by: \(\%\Delta g = \%\Delta L + 2 \times \%\Delta T = 0.25\% + 2 \times 2.78\% = 5.81\%\). Calculating the value of \(g\): \(g = \frac{4\pi^2 \times 0.800}{1.80^2} = 9.75\text{ m s}^{-2}\). The absolute uncertainty in \(g\) is \(\Delta g = 9.75 \times 0.0581 = 0.57\text{ m s}^{-2}\), which rounds to \(0.6\text{ m s}^{-2}\) to 1 significant figure.

1 mark for calculating individual percentage uncertainties of 0.25% and 2.78%. 1 mark for summing uncertainties to find the percentage uncertainty in g is 5.81%. 1.4 marks for the final absolute uncertainty of 0.57 or 0.6 m s^-2.

The percentage uncertainty in mass \(M\) is: \(\%\Delta M = \frac{0.15}{4.50} \times 100\% = 3.33\%\). The percentage uncertainty in radius \(R\) is: \(\%\Delta R = \frac{0.2}{6.2} \times 100\% = 3.23\%\). Since density \(\rho = \frac{M}{V} = \frac{3 M}{4\pi R^3}\), the formula for percentage uncertainty in density is: \(\%\Delta\rho = \%\Delta M + 3 \times \%\Delta R = 3.33\% + 3 \times 3.23\% = 3.33\% + 9.68\% = 13.01\%\). This rounds to 13%.

1 mark for percentage uncertainty of mass (3.33%). 1 mark for percentage uncertainty of radius (3.23%) multiplied by 3 (9.68%). 1.4 marks for correct final sum to get 13%.

Using the relation \(A = \lambda N\), the decay constant is: \(\lambda = \frac{A}{N} = \frac{1.2 \times 10^{12}}{4.5 \times 10^{18}} = 2.67 \times 10^{-7}\text{ s}^{-1}\). The half-life \(T_{1/2}\) is: \(T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{2.67 \times 10^{-7}} = 2.60 \times 10^6\text{ s}\). Convert the half-life to days: \(T_{1/2} = \frac{2.60 \times 10^6}{3600 \times 24} = 30.1\text{ days}\), which rounds to 30 days.

1 mark for calculating the decay constant (2.67 x 10^-7 s^-1). 1 mark for calculating half-life in seconds (2.6 x 10^6 s). 1.4 marks for correct conversion to days to yield 30 days (or 30.1 days).

By the inverse square law, the intensity (and thus corrected count rate) is inversely proportional to the square of the distance: \(C \propto \frac{1}{d^2}\). The distance increases by a factor of \(\frac{0.45}{0.15} = 3\). Therefore, the new corrected count rate \(C_2\) is: \(C_2 = \frac{C_1}{3^2} = \frac{360}{9} = 40\text{ counts s}^{-1}\). The expected total measured count rate is the new corrected count rate plus the background count rate: \(40 + 25 = 65\text{ counts s}^{-1}\).

1 mark for applying the inverse square law with distance ratio. 1 mark for finding the new corrected count rate of 40 counts s^-1. 1.4 marks for adding the background count rate to get 65 counts s^-1.

For the wire to be balanced, the magnetic force \(F = B I L\) must equal its weight \(W = m g\). This gives the relation: \(B I L = m g\). Solving for current: \(I = \frac{m g}{B L} = \frac{3.2 \times 10^{-3}\text{ kg} \times 9.81\text{ m s}^{-2}}{0.12\text{ T} \times 0.085\text{ m}} = \frac{0.031392}{0.0102} \approx 3.08\text{ A}\). Expressed to 2 significant figures, this is \(3.1\text{ A}\).

1 mark for equating BIL to mg. 1 mark for converting mass to kilograms and length to meters. 1.4 marks for calculating the current as 3.1 A.

The radius of the orbit of a charged particle in a magnetic field is given by \(r = \frac{p}{B q}\). First, determine the momentum \(p\) from the kinetic energy \(E_k\): \(E_k = \frac{p^2}{2m} \implies p = \sqrt{2 m E_k} = \sqrt{2 \times 6.64 \times 10^{-27} \times 3.2 \times 10^{-13}} = 6.52 \times 10^{-20}\text{ kg m s}^{-1}\). Now, calculate the radius: \(r = \frac{6.52 \times 10^{-20}}{0.45 \times 3.20 \times 10^{-19}} = 0.453\text{ m}\), which rounds to 0.45 m to 2 significant figures.

1 mark for linking kinetic energy to momentum and finding p = 6.52 x 10^-20 kg m s^-1. 1 mark for the magnetic centripetal force formula r = p/(Bq). 1.4 marks for the correct substitution and final answer of 0.45 m.

First, convert the initial temperature to Kelvin: \(T_1 = 22 + 273 = 295\text{ K}\). Since the volume of the canister is constant, the pressure law applies: \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\). Solve for the final temperature: \(T_2 = T_1 \times \frac{P_2}{P_1} = 295 \times \frac{2.1 \times 10^5}{1.5 \times 10^5} = 413\text{ K}\). Convert this back to Celsius: \(\theta_2 = 413 - 273 = 140^\circ\text{C}\).

1 mark for converting initial temperature to Kelvin (295 K). 1 mark for applying the pressure law to find T2 = 413 K. 1.4 marks for converting back to Celsius to get 140 °C.

The formula for the root-mean-square speed of gas molecules is \(v_{rms} = \sqrt{\frac{3 R T}{M}}\), where \(M\) is the molar mass in kilograms per mole. Substitute the given values into the formula: \(v_{rms} = \sqrt{\frac{3 \times 8.31 \times 350}{0.131}} = \sqrt{\frac{8725.5}{0.131}} = \sqrt{66606.87} \approx 258\text{ m s}^{-1}\). Rounded to 2 significant figures, this is \(260\text{ m s}^{-1}\).

1 mark for stating or using the correct r.m.s. equation. 1 mark for correct substitution. 1.4 marks for the final answer of 258 or 260 m s^-1.

1. Convert the temperature from Celsius to Kelvin: \(T = 120 + 273 = 393\text{ K}\). 2. Use the equation for the r.m.s. speed derived from kinetic theory: \(c_{\text{rms}} = \sqrt{\frac{3 R T}{M}}\). 3. Substitute the values: \(c_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \times 393}{4.0 \times 10^{-3}}} = \sqrt{2.449 \times 10^6} \approx 1565\text{ m s}^{-1}\). Rounded to an appropriate number of significant figures, this is \(1.56 \times 10^3\text{ m s}^{-1}\) (or \(1.6 \times 10^3\text{ m s}^{-1}\)).

[1 Mark] For converting the temperature to Kelvin: \(T = 393\text{ K}\). [1 Mark] For correct substitution of values into the r.m.s. speed equation: \(c_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \times 393}{4.0 \times 10^{-3}}}\) (or equivalent molecular calculation using \(k_B\) and atom mass \(m = 6.64 \times 10^{-27}\text{ kg}\)). [1 Mark] For the final correct value: \(1.56 \times 10^3\text{ m s}^{-1}\) or \(1560\text{ m s}^{-1}\) (allow \(1.6 \times 10^3\text{ m s}^{-1}\) or \(1600\text{ m s}^{-1}\)).

1. Convert the temperature from Celsius to Kelvin: \(T = 120 + 273 = 393\text{ K}\). 2. Use the equation for the r.m.s. speed derived from kinetic theory: \(c_{\text{rms}} = \sqrt{\frac{3 R T}{M}}\). 3. Substitute the values: \(c_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \times 393}{4.0 \times 10^{-3}}} = \sqrt{2.449 \times 10^6} \approx 1565\text{ m s}^{-1}\). Rounded to an appropriate number of significant figures, this is \(1.56 \times 10^3\text{ m s}^{-1}\) (or \(1.6 \times 10^3\text{ m s}^{-1}\)).

[1 Mark] For converting the temperature to Kelvin: \(T = 393\text{ K}\). [1 Mark] For correct substitution of values into the r.m.s. speed equation: \(c_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \times 393}{4.0 \times 10^{-3}}}\) (or equivalent molecular calculation using \(k_B\) and atom mass \(m = 6.64 \times 10^{-27}\text{ kg}\)). [1 Mark] For the final correct value: \(1.56 \times 10^3\text{ m s}^{-1}\) or \(1560\text{ m s}^{-1}\) (allow \(1.6 \times 10^3\text{ m s}^{-1}\) or \(1600\text{ m s}^{-1}\)).

The percentage uncertainty in current is \( \frac{0.05}{2.50} \times 100\% = 2.0\% \). The percentage uncertainty in diameter is \( \frac{0.02}{0.42} \times 100\% \approx 4.76\% \). Since \( J \propto \frac{I}{d^2} \), the percentage uncertainty in \( J \) is given by: \( \% \text{ uncertainty in } J = (\% \text{ uncertainty in } I) + 2 \times (\% \text{ uncertainty in } d) = 2.0\% + 2 \times 4.76\% = 11.5\% \).

[1 mark] Correctly calculates percentage uncertainty in both variables and doubles the uncertainty of diameter, leading to 11.5%.

After 48 days (which is exactly 2 half-lives), the number of active nuclei is \( N = \frac{4.0 \times 10^{18}}{4} = 1.0 \times 10^{18} \). The decay constant is \( \lambda = \frac{\ln(2)}{24 \times 24 \times 3600\text{ s}} \approx 3.34 \times 10^{-7}\text{ s}^{-1} \). The activity \( A \) is given by \( A = \lambda N = 3.34 \times 10^{-7} \times 1.0 \times 10^{18} \approx 3.3 \times 10^{11}\text{ Bq} \).

[1 mark] Correctly scales the number of nuclei over two half-lives and converts the half-life to seconds to calculate the correct activity.

The initial magnetic flux linkage is \( N\Phi_i = BAN = 0.080 \times (2.4 \times 10^{-3}) \times 150 = 0.0288\text{ Wb-turns} \). The final magnetic flux linkage is 0. According to Faraday\'s law, the induced e.m.f. is \( E = \frac{\Delta(N\Phi)}{\Delta t} = \frac{0.0288}{0.12} = 0.24\text{ V} \).

[1 mark] Correctly applies Faraday's Law using change in magnetic flux linkage divided by time, yielding 0.24 V.

The root-mean-square speed is proportional to the square root of the absolute temperature: \( c_{\text{rms}} \propto \sqrt{T} \). Doubling the speed requires the absolute temperature to quadruple. The initial temperature in Kelvin is \( T_1 = 27 + 273 = 300\text{ K} \). Therefore, the final temperature is \( T_2 = 4 \times 300 = 1200\text{ K} \). Converting back to Celsius gives \( 1200 - 273 = 927^\circ\text{C} \).

[1 mark] Converts initial temperature to Kelvin, quadruples the absolute temperature, and converts back to Celsius.

Resistivity is given by \( \rho = \frac{R A}{L} = \frac{V \pi d^2}{4 I L} \). The percentage uncertainty in \( \rho \) is the sum of the percentage uncertainties of each component, doubling the uncertainty of \( d \) because it is squared: \( \% \Delta \rho = \% \Delta V + 2(\% \Delta d) + \% \Delta I + \% \Delta L = 1.5\% + 2(2.0\%) + 1.0\% + 0.5\% = 7.0\% \).

[1 mark] Sums the percentage uncertainties correctly, including doubling the diameter contribution, to arrive at 7.0%.

The initial corrected count rate is \( 260 - 20 = 240 \) counts per minute. In 9.0 hours, 3.0 half-lives pass. The corrected count rate will halve three times: \( 240 \to 120 \to 60 \to 30 \) counts per minute. The recorded count rate is the corrected rate plus the background rate: \( 30 + 20 = 50 \) counts per minute.

[1 mark] Subtracts background radiation, halves the corrected rate three times, and adds the background back to get 50 counts per minute.

The area of the square loop is \( A = (0.050\text{ m})^2 = 2.5 \times 10^{-3}\text{ m}^2 \). The maximum torque \( \tau \) on a current-carrying loop is when the plane of the loop is parallel to the magnetic field lines (i.e. angle \( \theta = 90^\circ \) between the normal to the area and the field): \( \tau = B I A = 0.40 \times 2.5 \times 2.5 \times 10^{-3} = 2.5 \times 10^{-3}\text{ N m} \).

[1 mark] Correctly identifies that maximum torque occurs in this orientation, calculates area, and applies the formula to obtain 2.5 * 10^(-3) N m.

The mean translational kinetic energy of a molecule in an ideal gas depends only on the absolute temperature: \( E_{\text{mean}} = \frac{3}{2} k_B T \). Since both gases are at the same temperature, the mean kinetic energies of their molecules/atoms are equal, making the ratio exactly \( 1:1 \).

[1 mark] Correctly identifies that mean kinetic energy is independent of mass/molar mass and depends only on temperature, giving a 1:1 ratio.

First, calculate the percentage uncertainties of each individual quantity:
- For resistance \(R\): \(\frac{0.08}{4.00} \times 100 = 2.0\%\)
- For diameter \(d\): \(\frac{0.01}{0.50} \times 100 = 2.0\%\)
- For length \(L\): \(\frac{0.03}{1.20} \times 100 = 2.5\%\)
Now combine these according to the resistivity formula. Since \(d\) is squared, its percentage uncertainty must be multiplied by 2:
Percentage uncertainty in \(\rho = \%\Delta R + 2 \times (\%\Delta d) + \%\Delta L = 2.0\% + 2(2.0\%) + 2.5\% = 8.5\%\).

[1 mark] C is correct. Award 1 mark for the correct calculation of percentage uncertainties and correctly summing them with the appropriate power factor for diameter.

First, convert the initial temperature to kelvin: \(T_1 = 27 + 273.15 = 300\text{ K}\). The mean square speed of the gas molecules \(\langle c^2 \rangle\) is directly proportional to the absolute temperature \(T\), as shown by the relation \(\frac{1}{2}m\langle c^2 \rangle = \frac{3}{2}k_B T\). Doubling the mean square speed requires doubling the absolute temperature: \(T_2 = 2 \times T_1 = 2 \times 300\text{ K} = 600\text{ K}\). Convert this back to Celsius: \(t_2 = 600 - 273 = 327\ ^\circ\text{C}\).

[1 mark] B is correct. Award 1 mark for converting temperature to kelvin, doubling it, and correctly converting back to Celsius.

An activity of \(12.5\%\) corresponds to \(0.125 = (1/2)^3\) of the initial activity. This means exactly 3 half-lives have elapsed. Therefore:
3 \times T_{1/2} = 4.0\text{ hours} = 14400\text{ s}
This gives a half-life of \(T_{1/2} = 4800\text{ s}\).
Using the decay constant formula:
\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693}{4800} \approx 1.44 \times 10^{-4}\text{ s}^{-1}\).

[1 mark] B is correct. Award 1 mark for finding the number of half-lives, calculating the half-life in seconds, and finding the correct value of the decay constant.

The force on a current-carrying wire in a magnetic field is given by \(F = B I L \sin(\theta)\). When plotting \(F\) against \(\sin(\theta)\), the equation represents a straight line \(y = mx\) where the gradient \(m = B I L\). Thus:
\text{Gradient} = B I L = 0.18\text{ N}
We are given \(I = 4.0\text{ A}\) and \(L = 0.15\text{ m}\). Hence:
B \times 4.0 \times 0.15 = 0.18
0.60 B = 0.18
B = 0.30\text{ T}\).

[1 mark] C is correct. Award 1 mark for identifying the gradient as equal to BIL and calculating B correctly.

First, find the mean of the five readings:
\text{Mean} = \frac{0.38 + 0.39 + 0.37 + 0.38 + 0.38}{5} = 0.38\text{ mm}
Next, determine the uncertainty using half the range of the readings:
\text{Range} = \text{Maximum value} - \text{Minimum value} = 0.39 - 0.37 = 0.02\text{ mm}
\text{Uncertainty} = \frac{\text{Range}}{2} = 0.01\text{ mm}
Combining these gives \((0.38 \pm 0.01)\text{ mm}\).

[1 mark] B is correct. Award 1 mark for correctly calculating the mean and half the range to represent the absolute uncertainty.

According to Faraday's law, the magnitude of the induced e.m.f. is \(E = N \frac{\Delta \Phi}{\Delta t} = N A \frac{\Delta B}{\Delta t}\) (since the field is perpendicular to the area of the coil).
Calculating the values:
\Delta B = 0.40 - 0.10 = 0.30\text{ T}
E = 200 \times (1.5 \times 10^{-3}\text{ m}^2) \times \frac{0.30\text{ T}}{0.12\text{ s}} = 0.30 \times 2.5 = 0.75\text{ V}\).

[1 mark] C is correct. Award 1 mark for using Faraday's law to correctly compute the magnitude of the induced e.m.f.

The energy released is equal to the total binding energy of the products minus the total binding energy of the reactants:
- Reactants: There are two deuterium nuclei, each having 2 nucleons. Total binding energy of reactants = \(2 \times (2 \times 1.11\text{ MeV}) = 4.44\text{ MeV}\).
- Products: Helium-3 has 3 nucleons. Total binding energy of helium-3 = \(3 \times 2.57\text{ MeV} = 7.71\text{ MeV}\). The neutron is a free nucleon and has zero binding energy.
- Energy released = \(7.71\text{ MeV} - 4.44\text{ MeV} = 3.27\text{ MeV}\).

[1 mark] B is correct. Award 1 mark for multiplying the binding energies per nucleon by their respective number of nucleons and subtracting reactants from products.

First, calculate the percentage uncertainty in \(L\): \(\%\Delta L = \frac{0.004}{0.800} \times 100\% = 0.50\%\). Next, calculate the percentage uncertainty in \(T\): \(\%\Delta T = \frac{0.03}{1.80} \times 100\% \approx 1.67\%\). Since \(g \propto \frac{L}{T^2}\), the total percentage uncertainty in \(g\) is \(\%\Delta g = \%\Delta L + 2 \times (\%\Delta T) = 0.50\% + 2 \times 1.67\% = 3.84\%\). Rounded to 2 significant figures, this is \(3.8\%\).

C1: for calculating the percentage uncertainty of \(L\) as \(0.50\%\) or \(T\) as \(1.67\%\). C1: for adding the percentage uncertainty of \(L\) to twice the percentage uncertainty of \(T\). A1: for the correct final answer of \(3.8\%\) (accept \(3.8\) or \(3.8\%\)).

Percentage uncertainty in \(V\) is \(\%\Delta V = \frac{0.02}{1.80} \times 100\% = 1.11\%\). Percentage uncertainty in \(I\) is \(\%\Delta I = \frac{0.05}{2.50} \times 100\% = 2.00\%\). Percentage uncertainty in \(d\) is \(\%\Delta d = \frac{0.01}{0.38} \times 100\% \approx 2.63\%\). The total percentage uncertainty in \(\rho\) is \(\%\Delta \rho = \%\Delta V + \%\Delta I + 2 \times \%\Delta d = 1.11\% + 2.00\% + 2 \times 2.63\% = 8.37\%\). To 2 significant figures, this is \(8.4\%\).

C1: for calculating the percentage uncertainty of at least two variables correctly. C1: for doubling the percentage uncertainty of \(d\) and adding to those of \(V\) and \(I\). A1: for correct final answer of \(8.4\%\) (accept \(8.4\) or \(8.4\%\); accept answers in the range \(8.3\%\) to \(8.4\%\)).

Using the radioactive decay law \(A = A_0 e^{-\lambda t}\), we get \(1.2 \times 10^5 = \theta.4 \times 10^5 e^{-\lambda \times 4.5}\). This simplifies to \(e^{-4.5\lambda} = \frac{1}{7}\), so \(4.5\lambda = \ln(7)\), giving \(\lambda \approx 0.4324\text{ hours}^{-1}\). The half-life in hours is \(t_{1/2} = \frac{\ln(2)}{\lambda} = \frac{0.69315}{0.4324} \approx 1.603\text{ hours}\). Converting this to minutes gives \(t_{1/2} = 1.603 \times 60 = 96.18\text{ minutes}\). To 2 significant figures, the half-life is \(96\text{ minutes}\).

C1: for using the radioactive decay equation to set up a correct relation or find \(\lambda \approx 0.43\text{ h}^{-1}\). C1: for calculating the half-life in hours as \(1.6\text{ h}\) or using ratio equations. A1: for the correct final answer of \(96\text{ minutes}\) (accept \(96\)).

The initial activity is \(A_0 = \lambda N_0 = 1.8 \times 10^{-6} \times 3.5 \times 10^{18} = 6.30 \times 10^{12}\text{ Bq}\). The elapsed time in seconds is \(t = 5.0 \times 24 \times 3600 = 4.32 \times 10^5\text{ s}\). The activity after this time is \(A = A_0 e^{-\lambda t} = 6.30 \times 10^{12} \times e^{-(1.8 \times 10^{-6} \times 4.32 \times 10^5)} = 6.30 \times 10^{12} \times e^{-0.7776} = 2.895 \times 10^{12}\text{ Bq}\). Converting to gigabecquerels (\(1\text{ GBq} = 10^9\text{ Bq}\)), we get \(2895\text{ GBq}\), which rounds to \(2900\text{ GBq}\) to 2 significant figures.

C1: for calculating the initial activity as \(6.3 \times 10^{12}\text{ Bq}\) or converting time to \(4.32 \times 10^5\text{ s}\). C1: for substituting values correctly into \(A = A_0 e^{-\lambda t}\) or \(N = N_0 e^{-\lambda t}\). A1: for the correct final activity of \(2900\text{ GBq}\) (accept range \(2890\) to \(2900\)).

The weight of the wire is \(W = m g = 18 \times 10^{-3}\text{ kg} \times 9.81\text{ m s}^{-2} = 0.17658\text{ N}\). The magnetic force is given by \(F = B I L = 0.45 \times I \times 0.15 = 0.0675 I\). For the forces to balance, \(B I L = m g\), so \(0.0675 I = 0.17658\), which gives \(I \approx 2.616\text{ A}\). To 2 significant figures, the current is \(2.6\text{ A}\).

C1: for calculating the weight of the wire as \(0.177\text{ N}\). C1: for equating the weight to \(B I L\) and rearranging for \(I\). A1: for the correct current of \(2.6\text{ A}\) (accept \(2.6\)).

The initial magnetic flux linkage of the coil is \(N \Phi = N B A = 400 \times 0.18 \times 2.5 \times 10^{-4} = 0.018\text{ Wb-turns}\). Since the coil is completely removed, the final flux linkage is zero. According to Faraday's law of electromagnetic induction, the average induced e.m.f. is \(E = \frac{\Delta (N \Phi)}{\Delta t} = \frac{0.018}{0.060} = 0.30\text{ V}\).

C1: for calculating the initial magnetic flux linkage as \(0.018\text{ Wb-turns}\) (or magnetic flux \(\Phi = 4.5 \times 10^{-5}\text{ Wb}\)). C1: for using Faraday's law equation \(E = \frac{\Delta (N \Phi)}{\Delta t}\). A1: for obtaining \(0.30\text{ V}\) (accept \(0.3\) or \(0.30\)).

First, convert the initial temperature to Kelvin: \(T_1 = 27 + 273 = 300\text{ K}\). Since the volume \(V\) and the amount of gas \(n\) are kept constant, the pressure is directly proportional to the absolute temperature (\(P \propto T\)). Since the pressure is doubled (\(P_2 = 2 P_1\)), the absolute temperature is also doubled: \(T_2 = 2 \times T_1 = 2 \times 300 = 600\text{ K}\). Converting back to degrees Celsius: \(t_2 = 600 - 273 = 327\text{ }^\circ\text{C}\).

C1: for converting the initial temperature to Kelvin (\(300\text{ K}\)). C1: for recognizing that \(P \propto T\) and finding the final temperature in Kelvin (\(600\text{ K}\)). A1: for the correct final temperature of \(327\text{ }^\circ\text{C}\) (accept \(327\)).

The mean kinetic energy of a single atom of an ideal gas is given by \(E_k = \frac{3}{2} k T\). The total translational kinetic energy for \(N\) atoms is \(E_{\text{total}} = N \times \frac{3}{2} k T = \frac{3}{2} (N k T)\). Since \(P V = N k T\), the total kinetic energy is simply \(E_{\text{total}} = \frac{3}{2} P V\). Substituting the values, we get: \(E_{\text{total}} = 1.5 \times 1.5 \times 10^5 \times 0.045 = 10125\text{ J}\). To 3 significant figures, this is \(10100\text{ J}\) (or \(1.01 \times 10^4\text{ J}\)).

C1: for stating or using the equation \(E = \frac{3}{2} N k T\) or \(E = \frac{3}{2} P V\). C1: for correct substitution of pressure and volume (or number of atoms/moles and temperature). A1: for the correct total kinetic energy of \(10100\text{ J}\) (accept \(1.01 \times 10^4\text{ J}\) or \(10125\text{ J}\)).

First, calculate the value of \(g\) using the formula rearranged as \(g = \frac{2h}{t^2}\):
\(g = \frac{2 \times 1.50}{0.55^2} = 9.917 \text{ m s}^{-2}\).

Next, calculate the percentage uncertainty in each measurement:
\% \text{ uncertainty in } h = \frac{0.02}{1.50} \times 100 \approx 1.33\%
\% \text{ uncertainty in } t = \frac{0.03}{0.55} \times 100 \approx 5.45\%

Combine these to find the percentage uncertainty in \(g\):
\% \text{ uncertainty in } g = \% \text{ uncertainty in } h + 2 \times (\% \text{ uncertainty in } t)
\% \text{ uncertainty in } g = 1.33\% + 2 \times 5.45\% = 12.23\%

Calculate the absolute uncertainty in \(g\):
\Delta g = 12.23\% \times 9.917 \approx 1.21 \text{ m s}^{-2}.

Expressing the final answer with appropriate significant figures gives \(g = (9.9 \pm 1.2) \text{ m s}^{-2}\).

M1: Calculation of the value of \(g = 9.9 \text{ m s}^{-2}\) (or \(9.92 \text{ m s}^{-2}\))
M1: Correct determination of the total percentage uncertainty in \(g\) as approximately \(12.2\%\)
A1.1: Correct calculation of the absolute uncertainty to give \(\pm 1.2 \text{ m s}^{-2}\) (accept \(\pm 1.2\) or \(\pm 1\)) to write the final format \(9.9 \pm 1.2 \text{ m s}^{-2}\)

The radioactive decay law states that \(A = A_0 e^{-\lambda t}\).

Divide by \(A_0\) and take the natural logarithm of both sides:
\(\frac{1.2 \times 10^5}{8.4 \times 10^5} = e^{-\lambda \times 45\)}
\(\frac{1}{7} = e^{-45\lambda\)}
\(\ln\left(\frac{1}{7}\right) = -45\lambda\)
\(-1.9459 = -45\lambda\)
\(\lambda = 0.04324 \text{ h}^{-1}\)

Now, calculate the half-life \(t_{1/2}\):
\(t_{1/2} = \frac{\ln(2)}{\lambda} = \frac{0.69315}{0.04324} \approx 16.03 \text{ hours}\).

Rounded to 2 significant figures, the half-life is \(16 \text{ hours}\).

M1: Correct substitution of values into \(A = A_0 e^{-\lambda t}\) or \(A = A_0 (0.5)^{t/t_{1/2}}\)
M1: Correct determination of decay constant \(\lambda = 0.043 \text{ h}^{-1}\) (or setting up \(\frac{45}{t_{1/2}} = \frac{\ln(7)}{\ln(2)}\))
A1.1: Final answer for half-life evaluated to \(16 \text{ hours}\) (accept range \(16.0\) to \(16.1\))

According to Faraday's law of electromagnetic induction, the average induced e.m.f. \(E\) is given by:
\(E = \frac{\Delta(N\Phi)}{\Delta t}\)

Since the plane of the coil is initially perpendicular to the field:
\(\Phi_{\text{initial}} = B A = 0.15 \times 3.5 \times 10^{-4} = 5.25 \times 10^{-5} \text{ Wb}\)

Therefore, the initial flux linkage is:
\(N\Phi_{\text{initial}} = 120 \times 5.25 \times 10^{-5} = 6.30 \times 10^{-3} \text{ Wb-turns}\)

When the coil is completely removed from the field, the final flux linkage is zero. Thus, the magnitude of the change in flux linkage is:
\(\Delta(N\Phi) = 6.30 \times 10^{-3} \text{ Wb-turns}\)

Calculate the magnitude of the induced e.m.f.:
\(E = \frac{6.30 \times 10^{-3}}{0.080} = 0.07875 \text{ V}\)

Rounded to 2 significant figures, this is \(0.079 \text{ V}\).

M1: Recalls and uses Faraday's law equation \(E = \frac{\Delta(N\Phi)}{\Delta t}\)
M1: Correctly calculates initial flux linkage as \(6.30 \times 10^{-3} \text{ Wb-turns}\) (or \(6.3 \times 10^{-3} \text{ Wb}\))
A1.1: Final evaluated answer of \(0.079 \text{ V}\) (or \(0.0788 \text{ V}\))

Use the ideal gas equation:
\(P V = n R T\)

Rearranging to solve for the final temperature \(T_f\) in Kelvin:
\(T_f = \frac{P_f V}{n R}\)

Substitute the given values into the formula:
\(P_f = 2.5 \times 10^5 \text{ Pa}\)
\(V = 0.025 \text{ m}^3\)
\(n = 1.8 \text{ mol}\)
\(R = 8.31 \text{ J mol}^{-1} \text{ K}^{-1}\)

\(T_f = \frac{2.5 \times 10^5 \times 0.025}{1.8 \times 8.31} = \frac{6250}{14.958} \approx 417.84 \text{ K}\)

Convert the temperature to Celsius:
\(T_f (\text{ }^\circ\text{C}) = 417.84 - 273.15 = 144.69 \text{ }^\circ\text{C}\)

Rounding to 3 significant figures, the final temperature is \(145 \text{ }^\circ\text{C}\).

M1: Recalls and correctly rearranges \(PV = nRT\) to \(T = \frac{PV}{nR}\)
M1: Correct substitution of values to calculate temperature in Kelvin as \(418 \text{ K}\) (or \(417.8 \text{ K}\))
A1.1: Correct subtraction of 273 (or 273.15) to obtain a Celsius temperature of \(145 \text{ }^\circ\text{C}\) (accept range \(144\) to \(145\))

The resistivity \(\rho\) is calculated using the formula:
\(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\)

Therefore, the fractional uncertainty in \(\rho\) is given by the sum of the fractional uncertainties of its independent components (with the factor of 2 for \(d\) due to the square term):
\(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\)

Calculate individual percentage uncertainties:
\% \text{ uncertainty in } R = \frac{0.1}{3.6} \times 100\% \approx 2.78\%
\% \text{ uncertainty in } d = \frac{0.02}{0.42} \times 100\% \approx 4.76\%
\% \text{ uncertainty in } L = \frac{0.01}{1.20} \times 100\% \approx 0.83\%

Sum the values to find the total percentage uncertainty:
\% \text{ uncertainty in } \rho = 2.78\% + 2 \times 4.76\% + 0.83\% = 13.13\%

To 3 significant figures, this is \(13.1\%\).

M1: Identifies the formula for uncertainty summing components, correctly incorporating the coefficient of 2 for diameter: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\)
M1: Calculates individual percentage uncertainties correctly (\(\%R = 2.78\%\), \(\%d = 4.76\%\), \(\%L = 0.83\%\))
A1.1: Sums components correctly to yield \(13\%\) or \(13.1\%\)

Let \(x\) represent the number of \(\alpha\) particles (\(^{4}_{2}\text{He}\)) and \(y\) represent the number of \(\beta^-\) particles (\(^{\ 0}_{-1}\text{e}\)) emitted.

The overall decay equation is:
\(^{238}_{\ 92}\text{U} \rightarrow\ ^{206}_{\ 82}\text{Pb} + x \ ^{4}_{2}\text{He} + y \ ^{\ 0}_{-1}\text{e} + y \bar{\nu}_e\)

Apply the conservation of nucleon number (mass number):
\(238 = 206 + 4x + 0y\)
\(32 = 4x\)
\(x = 8\)

Apply the conservation of proton number (atomic number):
\(92 = 82 + 2x - y\)

Substitute \(x = 8\) into the equation:
\(92 = 82 + 16 - y\)
\(92 = 98 - y\)
\(y = 6\)

Thus, 8 \(\alpha\) particles and 6 \(\beta^-\) particles are emitted.

M1: Uses conservation of mass number to determine the number of \(\alpha\) particles: \(238 - 206 = 4x \implies x = 8\)
M1: Sets up the equation for proton number conservation: \(92 = 82 + 2x - y\)
A1.1: Solves correctly to determine the number of \(\beta^-\) particles as \(6\) and states both values clearly

First, calculate the cross-sectional area \(A\) of the square coil:
\(A = (0.050 \text{ m})^2 = 2.5 \times 10^{-3} \text{ m}^2\).

Next, determine the change in magnetic flux linkage \(\Delta(N\Phi)\) of the coil:
\(\Delta(N\Phi) = N \Delta B A = 80 \times 0.45 \times 2.5 \times 10^{-3} = 0.090 \text{ Wb-turns}\).

Using Faraday's law, calculate the magnitude of the average induced e.m.f. \(E\):
\(E = \frac{\Delta(N\Phi)}{\Delta t} = \frac{0.090}{0.12} = 0.75 \text{ V}\).

Using Ohm's law, find the induced current \(I\):
\(I = \frac{E}{R} = \frac{0.75}{3.0} = 0.25 \text{ A}\).

M1: Calculates correct coil area \(A = 2.5 \times 10^{-3} \text{ m}^2\) and flux linkage change \(\Delta(N\Phi) = 0.090 \text{ Wb-turns}\)
M1: Calculates average induced e.m.f. of \(0.75 \text{ V}\) using Faraday's law
A1.1: Obtains the correct induced current of \(0.25 \text{ A}\)

First, find the number of moles \(n\) of helium gas in the cylinder:
\(n = \frac{\text{mass}}{\text{molar mass}} = \frac{0.40 \text{ kg}}{4.0 \times 10^{-3} \text{ kg mol}^{-1}} = 100 \text{ mol}\).

The total internal kinetic energy \(E_{\text{total}}\) of a monatomic ideal gas is given by:
\(E_{\text{total}} = \frac{3}{2} n R T\)

Substitute the values:
\(E_{\text{total}} = 1.5 \times 100 \times 8.31 \times 300\)
\(E_{\text{total}} = 373,950 \text{ J} \approx 3.7 \times 10^5 \text{ J}\).

Alternatively, calculating using number of atoms \(N = n \times N_A = 6.02 \times 10^{25}\) atoms and \(E_{\text{total}} = N \left( \frac{3}{2} k_B T \right)\) gives identical results.

M1: Calculates number of moles \(n = 100 \text{ mol}\) (or number of molecules \(N = 6.02 \times 10^{25}\))
M1: Recalls and uses \(E_{\text{total}} = \frac{3}{2} n R T\) (or \(E_{\text{total}} = \frac{3}{2} N k_B T\))
A1.1: Obtains the correct final answer of \(3.7 \times 10^5 \text{ J}\) (accept range \(3.7 \times 10^5\) to \(3.74 \times 10^5\))

The volume \( V \) of a cylinder of diameter \( d \) and length \( L \) is given by:

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

The percentage uncertainty in the volume is given by:

\[ \frac{\Delta V}{V} \times 100 = 2 \left( \frac{\Delta d}{d} \times 100 \right) + \left( \frac{\Delta L}{L} \times 100 \right) \]

Calculate individual percentage uncertainties:
- For diameter \( d \): \( \frac{0.02}{1.24} \times 100 \approx 1.613\% \)
- For length \( L \): \( \frac{0.2}{85.0} \times 100 \approx 0.235\% \)

Combine the uncertainties:

\[ \frac{\Delta V}{V} \times 100 = 2(1.613\%) + 0.235\% = 3.226\% + 0.235\% \approx 3.46\% \]

Rounding to 2 significant figures gives \( 3.5\% \).

M1: Identifies the percentage uncertainty equation for volume: % uncertainty in V = 2 * (% uncertainty in d) + (% uncertainty in L)
A1: Correctly calculates individual percentage uncertainties: d = 1.61% and L = 0.24%
A1: Adds the components correctly to obtain a total of 3.5% (allow 3.46% or 3.4%)

First, calculate the percentage uncertainty in the acceleration of free fall \( g \):

\[ \frac{\Delta g}{g} \times 100 = \frac{\Delta h}{h} \times 100 + 2 \left( \frac{\Delta t}{t} \times 100 \right) \]

Substitute the given values:

\[ \text{Percentage uncertainty in } g = 1.5\% + 2(2.5\%) = 1.5\% + 5.0\% = 6.5\% \]

Next, calculate the absolute uncertainty \( \Delta g \) using the calculated value of \( g = 9.81 \text{ m s}^{-2} \):

\[ \Delta g = g \times \frac{6.5}{100} = 9.81 \times 0.065 \approx 0.638 \text{ m s}^{-2} \]

Rounding to 2 significant figures, the absolute uncertainty is \( 0.64 \text{ m s}^{-2} \) (or \( 0.6 \text{ m s}^{-2} \) to 1 significant figure).

M1: Computes total percentage uncertainty in g: 1.5% + 2 * 2.5% = 6.5%
A1: Sets up calculation for absolute uncertainty: 9.81 * 0.065
A1: Obtains 0.64 m s^-2 (accept 0.6 m s^-2)

First, calculate the initial activity \( A_0 \) using:

\[ A_0 = \lambda N_0 = (1.2 \times 10^{-5} \text{ s}^{-1}) \times (4.5 \times 10^{18}) = 5.4 \times 10^{13} \text{ Bq} \]

Next, use the decay equation to find the time \( t \) for the activity to fall to \( 10\% \) of its initial value (i.e., \( A = 0.10 A_0 \)):

\[ A = A_0 e^{-\lambda t} \implies 0.10 = e^{-\lambda t} \]

Taking the natural logarithm of both sides:

\[ \ln(0.10) = -\lambda t \implies t = \frac{-\ln(0.10)}{\lambda} = \frac{2.3026}{1.2 \times 10^{-5} \text{ s}^{-1}} \approx 191882 \text{ s} \]

Convert this time into hours:

\[ t = \frac{191882 \text{ s}}{3600 \text{ s hour}^{-1}} \approx 53.3 \text{ hours} \]

To 2 significant figures, this is \( 53 \text{ hours} \).

M1: Calculates initial activity correctly using A = \lambda N to get 5.4 * 10^13 Bq
A1: Applies the exponential decay equation to establish ln(0.10) = -\lambda t
A1: Converts time from seconds to hours correctly to obtain 53 hours (accept 53.3 hours)

First, determine the decay constant \( \lambda \) of Sodium-24:

\[ T_{1/2} = 15 \text{ hours} = 15 \times 3600 \text{ s} = 54000 \text{ s} \]
\[ \lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.69315}{54000 \text{ s}} \approx 1.2836 \times 10^{-5} \text{ s}^{-1} \]

After \( 45 \text{ hours} \), which corresponds to exactly \( 3 \) half-lives (\( 45 / 15 = 3 \)), the activity \( A \) will have halved three times:

\[ A = \frac{A_0}{2^3} = \frac{3.2 \text{ MBq}}{8} = 0.40 \text{ MBq} = 4.0 \times 10^5 \text{ Bq} \]

Finally, calculate the number of undecayed nuclei remaining \( N \) using \( A = \lambda N \):

\[ N = \frac{A}{\lambda} = \frac{4.0 \times 10^5 \text{ Bq}}{1.2836 \times 10^{-5} \text{ s}^{-1}} \approx 3.12 \times 10^{10} \]

To 2 significant figures, this is \( 3.1 \times 10^{10} \).

M1: Calculates decay constant \lambda correctly (1.28 * 10^-5 s^-1) or identifies remaining activity after 3 half-lives is 0.40 MBq
A1: Links current activity and number of remaining nuclei via N = A / \lambda
A1: Correctly evaluates remaining nuclei as 3.1 * 10^10 (accept 3.12 * 10^10)

First, calculate the area \( A \) of the square coil:

\[ A = (0.040 \text{ m})^2 = 1.6 \times 10^{-3} \text{ m}^2 \]

The initial magnetic flux linkage of the coil is:

\[ N\Phi_{\text{initial}} = N B A = 50 \times 0.15 \text{ T} \times (1.6 \times 10^{-3} \text{ m}^2) = 0.012 \text{ Wb-turns} \]

Since the field is reduced to zero, the final flux linkage is zero, giving a change in flux linkage of \( \Delta(N\Phi) = 0.012 \text{ Wb-turns} \).

Using Faraday's law of electromagnetic induction, the average induced e.m.f. \( E \) is:

\[ E = \frac{\Delta(N\Phi)}{\Delta t} = \frac{0.012 \text{ Wb-turns}}{0.12 \text{ s}} = 0.10 \text{ V} \]

M1: Calculates the cross-sectional area of the coil correctly as 1.6 * 10^-3 m^2
A1: Determines the change in magnetic flux linkage as 0.012 Wb-turns
A1: Applies Faraday's law to find the induced e.m.f. of 0.10 V

The force \( F \) on a current-carrying conductor in a magnetic field is given by:

\[ F = B I L \sin\theta \]

Since the wire is perpendicular to the field, \( \theta = 90^\circ \) and \( \sin(90^\circ) = 1 \), simplifying the formula to:

\[ F = B I L \]

Rearrange the equation to find the magnetic flux density \( B \):

\[ B = \frac{F}{I L} \]

Substitute the given values:

\[ B = \frac{0.61 \text{ N}}{4.2 \text{ A} \times 0.85 \text{ m}} = \frac{0.61}{3.57} \approx 0.171 \text{ T} \]

To 2 significant figures, the magnetic flux density is \( 0.17 \text{ T} \).

M1: Recalls and rearranges the magnetic force formula: B = F / (I * L)
A1: Correctly substitutes the values into the rearranged equation
A1: Obtains 0.17 T (accept 0.171 T)

For an ideal gas in a rigid container, the volume \( V \) and number of moles \( n \) remain constant.

From the equation of state \( P V = n R T \), pressure is directly proportional to absolute temperature \( T \) (Charles' law / Pressure law):

\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]

Convert the initial temperature from Celsius to Kelvin:

\[ T_1 = 20^\circ\text{C} + 273.15 = 293.15 \text{ K} \]

Since the pressure doubles, \( P_2 = 2 P_1 \):

\[ T_2 = 2 T_1 = 2 \times 293.15 \text{ K} = 586.3 \text{ K} \]

Convert the final temperature back to Celsius:

\[ T_2 = 586.3 - 273.15 = 313.15^\circ\text{C} \]

To the nearest degree, the final temperature is \( 313^\circ\text{C} \).

M1: Converts the initial temperature of 20°C to 293 K
A1: Identifies and applies the pressure law relation (P is proportional to T) to determine final temperature in Kelvin as 586 K
A1: Subtracts 273 to give the final temperature of 313°C (accept 313.15°C)

The pressure of an ideal gas can be expressed in terms of its total mass \( M \) (where \( M = N m \), total number of molecules times the mass of one molecule), volume \( V \), and mean square speed \( \overline{c^2} \) by the equation:

\[ P = \frac{1}{3} \frac{M}{V} \overline{c^2} \]

Rearranging this equation to solve for the total mass \( M \):

\[ M = \frac{3 P V}{\overline{c^2}} \]

Substitute the given values into the equation:

\[ M = \frac{3 \times (1.8 \times 10^5 \text{ Pa}) \times (3.5 \times 10^{-3} \text{ m}^3)}{2.4 \times 10^5 \text{ m}^2\text{ s}^{-2}} \]
\[ M = \frac{1890}{2.4 \times 10^5} \approx 0.007875 \text{ kg} \]

To 2 significant figures, the total mass is \( 7.9 \times 10^{-3} \text{ kg} \) (or \( 0.0079 \text{ kg} \)).

M1: Recalls and rearranges the kinetic theory of gases pressure equation to obtain M = 3PV /
A1: Substitutes values correctly: (3 * 1.8*10^5 * 3.5*10^-3) / 2.4*10^5
A1: Calculates final mass correctly as 7.9 * 10^-3 kg (accept 0.0079 kg)

Resistivity is given by \(\rho = \frac{R A}{L} = \frac{\pi R d^2}{4 L}\). The percentage uncertainty in resistivity is \(\frac{\Delta \rho}{\rho} \times 100 = \left( \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L} \right) \times 100\). Calculating each component: \(\frac{\Delta R}{R} \times 100 = \frac{0.2}{4.5} \times 100 \approx 4.44\%\), \(\frac{\Delta d}{d} \times 100 = \frac{0.01}{0.38} \times 100 \approx 2.63\%\) (which is doubled to \(5.26\%\)), and \(\frac{\Delta L}{L} \times 100 = \frac{0.02}{1.20} \times 100 \approx 1.67\%\). Summing these gives: \(4.44\% + 5.26\% + 1.67\% = 11.37\%\). To an appropriate number of significant figures, this is \(11\%\).

1 mark for identifying the correct uncertainty combination formula with diameter squared contributing double, 1 mark for calculating correct individual percentage uncertainties (R = 4.44%, d = 2.63%, L = 1.67%), 1 mark for the final percentage uncertainty of 11% (accept 11.4% or 11%).

The induced e.m.f. is given by Faraday's law: \(V = N \frac{\Delta \Phi}{\Delta t}\). The cross-sectional area of the coil is \(A = \pi r^2 = \pi (0.025)^2 \approx 1.96 \times 10^{-3}\text{ m}^2\). The change in magnetic flux density is \(\Delta B = 0.40 - 0.10 = 0.30\text{ T}\). The change in magnetic flux is \(\Delta \Phi = A \Delta B = (1.96 \times 10^{-3}) \times 0.30 \approx 5.89 \times 10^{-4}\text{ Wb}\). The total induced e.m.f. is \(V = 150 \times \frac{5.89 \times 10^{-4}}{0.12} \approx 0.736\text{ V}\), which rounds to \(0.74\text{ V}\).

1 mark for calculating cross-sectional area as \(1.96 \times 10^{-3}\text{ m}^2\), 1 mark for substitution into Faraday's law formula, 1 mark for final answer of 0.74 V (accept 0.736 V to 0.74 V).

From the kinetic theory of gases, pressure is \(p = \frac{1}{3} \rho \overline{c^2} = \frac{1}{3} \frac{N m}{V} \overline{c^2}\). Rearranging for root-mean-square speed: \(c_{\text{rms}} = \sqrt{\overline{c^2}} = \sqrt{\frac{3 p V}{N m}}\). First, calculate the total mass of the helium gas: \(M = N m = (3.2 \times 10^{23}) \times (6.64 \times 10^{-27}) = 2.1248 \times 10^{-3}\text{ kg}\). Now, substitute the values: \(c_{\text{rms}} = \sqrt{\frac{3 \times (1.2 \times 10^5) \times 0.045}{2.1248 \times 10^{-3}}} = \sqrt{\frac{16200}{2.1248 \times 10^{-3}}} = \sqrt{7.624 \times 10^6} \approx 2761\text{ m s}^{-1}\). Rounding to 2 significant figures gives \(2800\text{ m s}^{-1}\).

1 mark for calculating total gas mass or density of the gas, 1 mark for correctly rearranging the kinetic theory equation for r.m.s. speed, 1 mark for final correct value of 2800 m s^-1 (accept 2760 m s^-1 to 2800 m s^-1).

Part (a): Resistivity \(\rho\) is defined by \(R = \frac{\rho L}{A}\), and resistance \(R = \frac{V}{I}\). Therefore, \(\rho = \frac{V A}{I L}\) where \(A = \frac{\pi d^2}{4}\). Combining these gives \(\rho = \frac{\pi V d^2}{4 I L}\). Part (b): First calculate cross-sectional area \(A = \frac{\pi (0.45 \times 10^{-3})^2}{4} = 1.5904 \times 10^{-7}\text{ m}^2\). Then calculate resistance \(R = \frac{V}{I} = \frac{3.20}{1.45} = 2.2069\ \Omega\). Substituting these values into \(\rho = \frac{R A}{L}\) gives \(\rho = \frac{2.2069 \times 1.5904 \times 10^{-7}}{1.25} = 2.808 \times 10^{-7}\ \Omega\text{ m}\), which is approximately \(2.8 \times 10^{-7}\ \Omega\text{ m}\). Part (c): The percentage uncertainties are: \(\%\Delta V = \frac{0.05}{3.20} \times 100\% = 1.56\%\); \(\%\Delta I = \frac{0.02}{1.45} \times 100\% = 1.38\%\); \(\%\Delta L = \frac{0.01}{1.25} \times 100\% = 0.80\%\); \(\%\Delta d = \frac{0.02}{0.45} \times 100\% = 4.44\%\). Since \(\rho \propto \frac{V d^2}{I L}\), the percentage uncertainty in resistivity is: \(\%\Delta \rho = \%\Delta V + 2(\%\Delta d) + \%\Delta I + \%\Delta L = 1.56\% + 2(4.44\%) + 1.38\% + 0.80\% = 12.62\% \approx 12.6\%\). Part (d): The absolute uncertainty in \(\rho = 2.808 \times 10^{-7} \times 0.1262 = 3.54 \times 10^{-8}\ \Omega\text{ m}\) (or \(0.4 \times 10^{-7}\ \Omega\text{ m}\)).

Part (a): [2 marks] 1 mark for identifying \(R = \rho L / A\) and \(R = V/I\), 1 mark for correct combination to obtain \(\rho = \pi V d^2 / (4 I L)\). Part (b): [3 marks] 1 mark for correct cross-sectional area \(1.59 \times 10^{-7}\text{ m}^2\), 1 mark for resistance \(2.21\ \Omega\), 1 mark for final calculated resistivity of \(2.81 \times 10^{-7}\ \Omega\text{ m}\). Part (c): [4 marks] 1 mark for individual uncertainties of V, I, L, 1 mark for uncertainty of d (4.44%), 1 mark for doubling the uncertainty of d (8.89%), 1 mark for summing all terms to get 12.6%. Part (d): [2.6 marks] 1.6 marks for multiplying percentage uncertainty by the calculated value, 1 mark for absolute uncertainty value: \(3.5 \times 10^{-8}\ \Omega\text{ m}\) (accept range \(3.5 \times 10^{-8}\) to \(4.0 \times 10^{-8}\)).

Part (a): Absolute zero is the temperature at which the pressure of an ideal gas is zero. Setting \(p = 0\) in the equation of the line \(p = m\theta + c\) gives: \(0 = m T_0 + c\). Rearranging this equation yields \(m T_0 = -c \implies T_0 = -\frac{c}{m}\). Part (b): Substituting the given values: \(T_0 = -\frac{9.83 \times 10^4}{360} = -273.06\ ^\circ\text{C}\), which rounds to \(-273\ ^\circ\text{C}\). Part (c): Since \(T_0 = -\frac{c}{m}\), the percentage uncertainty in \(T_0\) is the sum of the percentage uncertainties in \(c\) and \(m\). \(\%\Delta c = \frac{0.12 \times 10^4}{9.83 \times 10^4} \times 100\% = 1.22\%\). \(\%\Delta m = \frac{15}{360} \times 100\% = 4.17\%\). Total percentage uncertainty \(= 1.22\% + 4.17\% = 5.39\% \approx 5.4\%\). Part (d): Systematic errors can be reduced by ensuring that the volume of the connection tube between the flask and pressure gauge is as small as possible (reducing dead space), or by ensuring the thermometer is fully immersed and calibrated in an ice bath and boiling water bath.

Part (a): [3 marks] 1 mark for stating that pressure is zero at absolute zero, 1 mark for setting up \(0 = m T_0 + c\), 1 mark for algebraic progression to \(T_0 = -c/m\). Part (b): [2 marks] 1 mark for substitution, 1 mark for correct calculation of \(-273\ ^\circ\text{C}\) (or \(-273.1\)). Part (c): [4 marks] 1 mark for \(\%\Delta c = 1.22\%\), 1 mark for \(\%\Delta m = 4.17\%\), 1 mark for adding the component percentage uncertainties, 1 mark for final result \(5.4\%\). Part (d): [2.6 marks] 1.6 marks for identifying a practical source of systematic error (e.g. dead space, poorly calibrated thermometer), 1 mark for explaining how to minimize it (e.g. using thin capillary tube, calibrating thermometer).

Part (a): The decay is represented by \(^{220}_{86}\text{Rn} \rightarrow ^{216}_{84}\text{Po} + ^{4}_{2}\text{He}\) (or \(\alpha\)). Part (b): Mass defect \(\Delta m = m_{\text{Rn}} - (m_{\text{Po}} + m_{\alpha}) = 219.9678 - (215.9557 + 4.0015) = 0.0106\text{ u}\). Converting this mass defect to energy using \(1\text{ u} = 931.5\text{ MeV}\): \(E = 0.0106 \times 931.5 = 9.87\text{ MeV}\). Alternatively, using \(E = \Delta m c^2\) with \(1\text{ u} = 1.661 \times 10^{-27}\text{ kg}\) yields \(9.90\text{ MeV}\). Part (c): The decay constant is \(\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{55.6} = 0.01247\text{ s}^{-1}\). The initial activity is \(A_0 = \lambda N_0 = 0.01247 \times (4.5 \times 10^{18}) = 5.61 \times 10^{16}\text{ Bq}\). Part (d): Using the decay law \(A = A_0 e^{-\lambda t}\), for \(A = 0.10 A_0\): \(0.10 = e^{-\lambda t} \implies \ln(10) = \lambda t \implies t = \frac{\ln(10)}{0.01247} = 184.7\text{ s} \approx 185\text{ s}\).

Part (a): [2 marks] 1 mark for correct mass numbers and atomic numbers, 1 mark for correct elements (Po and He/alpha). Part (b): [4 marks] 1 mark for finding \(\Delta m = 0.0106\text{ u}\), 1 mark for correct method of conversion to energy, 2 marks for final value of \(9.87\text{ MeV}\) (or \(9.90\text{ MeV}\) if using \(E=\Delta mc^2\)). Part (c): [3 marks] 1 mark for \(\lambda = 0.0125\text{ s}^{-1}\), 1 mark for using \(A_0 = \lambda N_0\), 1 mark for calculating \(5.6 \times 10^{16}\text{ Bq}\). Part (d): [2.6 marks] 1.6 marks for correct exponential decay formula setup, 1 mark for correct time calculation of \(185\text{ s}\) (accept \(184\text{ s}\) to \(185\text{ s}\)).

Part (a): The ion experiences an electric force \(F_E = qE\) and a magnetic force \(F_B = qvB_1\) in opposite directions. For the ion to travel in a straight, undeflected line, these forces must be equal in magnitude: \(qE = qvB_1\), which simplifies to \(v = E/B_1\). Ions with larger or smaller velocities will experience a net force and be deflected away from the exit slit. Part (b): \(v = \frac{1.20 \times 10^5}{0.40} = 3.00 \times 10^5\text{ m s}^{-1}\). Part (c): In the second magnetic field, the magnetic force acts as the centripetal force: \(q v B_2 = \frac{m v^2}{R} \implies R = \frac{m v}{q B_2}\). Substituting the values: \(R = \frac{(6.64 \times 10^{-27}) \times (3.00 \times 10^5)}{(3.20 \times 10^{-19}) \times 0.60} = 1.0375 \times 10^{-2}\text{ m} \approx 1.04\text{ cm}\). Part (d): Since \(R = \frac{m v}{q B_2}\), the radius is inversely proportional to the magnetic flux density (\(R \propto B_2^{-1}\)). Therefore, a \(\pm 5\%\) percentage uncertainty in \(B_2\) will directly propagate to a \(\pm 5\%\) uncertainty in the calculated radius \(R\), assuming all other quantities are known precisely.

Part (a): [3 marks] 1 mark for identifying both forces and their expressions, 1 mark for stating that forces must balance for zero deflection, 1 mark for deriving \(v = E/B\) and explaining why other velocities are deflected. Part (b): [2 marks] 1 mark for formula \(v = E/B_1\), 1 mark for \(3.00 \times 10^5\text{ m s}^{-1}\). Part (c): [4 marks] 1 mark for equating centripetal force to magnetic force, 1 mark for rearranging to find \(R = mv / (qB_2)\), 1 mark for correct substitution, 1 mark for final calculated value of \(1.04\text{ cm}\) (or \(1.04 \times 10^{-2}\text{ m}\)). Part (d): [2.6 marks] 1.6 marks for expressing \(R \propto B_2^{-1}\), 1 mark for concluding that the uncertainty in \(R\) is also \(\pm 5\%\).

Part (a): The discharging equation is \(V = V_0 e^{-t / RC}\). Taking the natural logarithm of both sides: \(\ln(V) = \ln(V_0 e^{-t / RC}) = \ln(V_0) - \frac{1}{RC} t\). This matches the straight-line equation \(y = mx + c\) where \(y = \ln(V)\), \(x = t\), and gradient \(m = -\frac{1}{RC}\). Part (b): Rearranging the gradient equation for \(C\) gives \(C = -\frac{1}{m R}\). Using the magnitude of \(m\): \(C = \frac{1}{0.0141 \times 47.0 \times 10^3} = 1.509 \times 10^{-3}\text{ F} \approx 1.51\text{ mF}\). Part (c): Since \(C = \frac{1}{m R}\), the percentage uncertainty in \(C\) is given by: \(\%\Delta C = \%\Delta m + \%\Delta R = \frac{0.0005}{0.0141} \times 100\% + \frac{1.0}{47.0} \times 100\% = 3.55\% + 2.13\% = 5.68\% \approx 5.7\%\). Part (d): To improve accuracy and reduce timing errors, the student could use a voltage sensor connected to a computer data logger to record potential difference and time automatically, eliminating human reaction time errors in stopwatch measurements.

Part (a): [3 marks] 1 mark for stating \(V = V_0 e^{-t/RC}\), 1 mark for correct log algebra, 1 mark for identifying the gradient \(m = -1/RC\). Part (b): [3 marks] 1 mark for formula \(C = -1/(mR)\), 1 mark for substituting values, 1 mark for calculating \(1.51\text{ mF}\) (or \(1.51 \times 10^{-3}\text{ F}\)). Part (c): [3.6 marks] 1.6 marks for calculating both component percentage uncertainties, 1 mark for adding them, 1 mark for final percentage uncertainty of \(5.7\%\). Part (d): [2 marks] 1 mark for recommending a data logger / digital sensor, 1 mark for explaining that it removes human reaction time.

Part (a): Using the ideal gas equation \(pV = NkT\): \(N = \frac{pV}{kT} = \frac{1.20 \times 10^4 \times 2.50 \times 10^{-4}}{1.38 \times 10^{-23} \times 293} = \frac{3.00}{4.0434 \times 10^{-21}} = 7.42 \times 10^{20}\) helium atoms. Part (b): Since each alpha decay yields one helium atom, the total number of decays is equal to the number of helium atoms: \(7.42 \times 10^{20}\). Time period \(t = 3.0\text{ years} = 3.0 \times 365.25 \times 24 \times 3600 = 9.47 \times 10^7\text{ s}\). Average activity \(A = \frac{N}{t} = \frac{7.42 \times 10^{20}}{9.47 \times 10^7} = 7.84 \times 10^{12}\text{ Bq}\). Part (c): The half-life of Radium-226 (\(1600\text{ years}\)) is much larger than the observational period (\(3.0\text{ years}\)). The fraction of radium nuclei that decay is approximately \(\lambda t = \frac{\ln(2) \times 3.0}{1600} \approx 0.0013\) (or \(0.13\%\)), which is extremely small. Therefore, the activity is virtually constant. Part (d): The mean kinetic energy is \(\frac{1}{2} m c_{\text{rms}}^2 = \frac{3}{2} k T \implies c_{\text{rms}} =
\sqrt{\frac{3 k T}{m}}\). The mass of a helium-4 atom is \(m = 4.00 \times 1.661 \times 10^{-27}\text{ kg} = 6.644 \times 10^{-27}\text{ kg}\). \(c_{\text{rms}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 293}{6.644 \times 10^{-27}}} = 1351\text{ m s}^{-1} \approx 1350\text{ m s}^{-1}\).

Part (a): [3 marks] 1 mark for stating \(pV = NkT\), 1 mark for substitution, 1 mark for correct calculation of \(7.42 \times 10^{20}\). Part (b): [3 marks] 1 mark for converting 3 years to seconds (around \(9.5 \times 10^7\text{ s}\)), 1 mark for dividing number of atoms by time, 1 mark for calculating \(7.8 \times 10^{12}\text{ Bq}\). Part (c): [3 marks] 1 mark for stating that \(3.0\text{ years} \ll 1600\text{ years}\), 1 mark for calculating the fraction decayed is very small (around 0.13%), 1 mark for concluding the activity change is negligible. Part (d): [2.6 marks] 1.6 marks for formula \(c_{\text{rms}} = \sqrt{3kT/m}\) and finding mass of helium atom, 1 mark for correct final speed of \(1350\text{ m s}^{-1}\).