2023 OCR A-Level Physics A - H556 模擬試題連答案詳解

The formula for the density of a cylinder is \(\rho = \frac{4m}{\pi d^2 h}\). The percentage uncertainties in the measurements are: percentage uncertainty in \(m = \frac{0.5}{125.0} \times 100\% = 0.4\%\), percentage uncertainty in \(d = \frac{0.2}{20.0} \times 100\% = 1.0\%\), and percentage uncertainty in \(h = \frac{0.4}{80.0} \times 100\% = 0.5\%\). Combining these, the percentage uncertainty in \(\rho\) is \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta h}{h} = 0.4\% + 2(1.0\%) + 0.5\% = 2.9\%\).

1 mark for the correct calculation of percentage uncertainties and summing them appropriately with the correct factor of 2 for diameter.

The initial binding energy is \(2 \times (2 \times 1.11\text{ MeV}) = 4.44\text{ MeV}\). The final binding energy is \(3 \times 2.57\text{ MeV} = 7.71\text{ MeV}\) (since the free neutron has no binding energy). The energy released is the difference: \(7.71\text{ MeV} - 4.44\text{ MeV} = 3.27\text{ MeV}\).

1 mark for calculating the initial and final binding energies and subtracting them to obtain the correct energy released.

Using Stefan's Law, \(L = 4\pi R^2 \sigma T^4\). Comparing Star Y and Star X gives: \(\frac{L_Y}{L_X} = \left(\frac{R_Y}{R_X}\right)^2 \left(\frac{T_Y}{T_X}\right)^4 = (2)^2 \times \left(\frac{6000}{3000}\right)^4 = 4 \times (2)^4 = 4 \times 16 = 64\).

1 mark for utilizing Stefan's Law to calculate the correct ratio of 64.

The decay constant is \(\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{15 \times 3600\text{ s}} \approx 1.28 \times 10^{-5}\text{ s}^{-1}\). After \(30\text{ hours}\) (which is exactly two half-lives), the number of remaining nuclei \(N\) is \(8.0 \times 10^{18} \times (0.5)^2 = 2.0 \times 10^{18}\). The activity \(A\) is given by \(A = \lambda N = 1.28 \times 10^{-5}\text{ s}^{-1} \times 2.0 \times 10^{18} \approx 2.6 \times 10^{13}\text{ Bq}\).

1 mark for calculating the correct decay constant in seconds and multiplying by the remaining number of nuclei after 2 half-lives.

The distance between the 0th and 4th order bright fringes is \(4x = 15.0\text{ mm}\), so the fringe separation is \(x = 3.75\text{ mm} = 3.75 \times 10^{-3}\text{ m}\). Using the double-slit formula \(\lambda = \frac{ax}{D}\): \(\lambda = \frac{0.30 \times 10^{-3}\text{ m} \times 3.75 \times 10^{-3}\text{ m}}{1.8\text{ m}} = 6.25 \times 10^{-7}\text{ m} = 625\text{ nm}\).

1 mark for correctly determining the fringe spacing and calculating the wavelength using the double-slit formula.

Because the capacitor remains connected to the power supply, the potential difference remains constant at \(V\). The new capacitance is \(C' = 2C\). The charge on the plates becomes \(Q' = C'V = 2CV = 2Q\) (charge doubles). The stored energy becomes \(E' = \frac{1}{2}C'V^2 = \frac{1}{2}(2C)V^2 = 2E\) (energy doubles).

1 mark for concluding that both the charge and the stored energy double because the potential difference is kept constant.

The mean kinetic energy of gas molecules is directly proportional to absolute temperature: \(\frac{1}{2}m\langle c^2\rangle = \frac{3}{2}kT\). Since temperature \(T\) is doubled, the average kinetic energy doubles, and the mean-square speed \(\langle c^2\rangle\) doubles. Therefore, statement A is correct. The root-mean-square speed increases by \(\sqrt{2}\), so statement B is incorrect.

1 mark for correctly linking absolute temperature with mean-square speed.

At the highest point of the circular path, both the gravitational force \(mg\) and the normal force \(N\) act vertically downwards towards the center of the circle. The equation for centripetal force is \(N + mg = \frac{mv^2}{r}\). Solving for \(N\) yields \(N = \frac{mv^2}{r} - mg\).

1 mark for setting up the correct centripetal force equation at the top of the loop and solving for the normal force.

The formula for resistivity is \(\rho = \frac{R A}{L} = \frac{\pi R d^2}{4 L}\).

The percentage uncertainty in \(\rho\) is given by:

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

Calculate each term:
- \(\frac{\Delta R}{R} = \frac{0.3}{6.0} = 0.05\) (or \(5.0\%\))
- \(\frac{\Delta d}{d} = \frac{0.02}{0.40} = 0.05\) (or \(5.0\%\))
- \(\frac{\Delta L}{L} = \frac{0.03}{1.50} = 0.02\) (or \(2.0\%\))

Substituting these values:

\[ \frac{\Delta \rho}{\rho} \times 100 = 5.0\% + 2(5.0\%) + 2.0\% = 17\% \]

1 mark for correct calculation of fractional uncertainties and doubling the diameter fractional uncertainty leading to 17% (D).

To find the energy released, we calculate the difference between the total binding energy of the products and that of the reactants:

- Total binding energy of reactants (two \( ^{2}_{1}\text{H} \) nuclei):
\[ 2 \times (2 \times 1.11\text{ MeV}) = 4.44\text{ MeV} \]
- Total binding energy of products (one \( ^{3}_{2}\text{He} \) nucleus and one neutron):
\[ (3 \times 2.57\text{ MeV}) + 0 = 7.71\text{ MeV} \]
- Energy released:
\[ 7.71\text{ MeV} - 4.44\text{ MeV} = 3.27\text{ MeV} \]

1 mark for correct calculation of total binding energies of products and reactants, taking difference to yield 3.27 MeV (B).

According to Wien's displacement law:

\[ T \propto \frac{1}{\lambda_{\max}} \]

Thus, the ratio of temperatures is:

\[ \frac{T_X}{T_Y} = \frac{\lambda_Y}{\lambda_X} = \frac{600\text{ nm}}{400\text{ nm}} = 1.5 \]

According to Stefan-Boltzmann law, luminosity is:

\[ L = 4 \pi R^2 \sigma T^4 \implies L \propto R^2 T^4 \]

Therefore, the ratio of luminosities is:

\[ \frac{L_X}{L_Y} = \left( \frac{R_X}{R_Y} \right)^2 \left( \frac{T_X}{T_Y} \right)^4 = \left( \frac{1}{2} \right)^2 \left( 1.5 \right)^4 = 0.25 \times 5.0625 \approx 1.3 \]

1 mark for using Wien's law to find temperature ratio, and Stefan-Boltzmann's law to get luminosity ratio of 1.3 (C).

The activity \( A \) at time \( t \) is given by:

\[ A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \]

Given:
- Initial activity \( A_0 = 8.0 \times 10^7\text{ Bq} \)
- Half-life \( t_{1/2} = 15.0\text{ hours} \)
- Time \( t = 50.0\text{ hours} \)

Substituting the values:

\[ A = (8.0 \times 10^7) \times (0.5)^{\frac{50.0}{15.0}} \]
\[ A = (8.0 \times 10^7) \times (0.5)^{3.333} \]
\[ A = (8.0 \times 10^7) \times 0.09921 \approx 7.9 \times 10^6\text{ Bq} \]

1 mark for using the decay formula with the correct number of half-lives (3.33), giving 7.9 * 10^6 Bq (B).

The formula for fringe separation is:

\[ x = \frac{\lambda D}{a} \]

Since \( c = f \lambda \), the wavelength is inversely proportional to frequency:

\[ \lambda = \frac{c}{f} \]

When the frequency is multiplied by \( 1.5 \), the new wavelength \( \lambda' \) is:

\[ \lambda' = \frac{\lambda}{1.5} = \frac{2}{3}\lambda \]

Substituting the new parameters \( a' = 2a \) and \( D' = 0.5D = \frac{1}{2}D \) into the fringe separation formula:

\[ x' = \frac{\lambda' D'}{a'} = \frac{\left(\frac{2}{3}\lambda\right) \left(\frac{1}{2}D\right)}{2a} = \frac{\frac{1}{3}\lambda D}{2a} = \frac{1}{6}\frac{\lambda D}{a} = \frac{1}{6}x \]

1 mark for relating wavelength inversely to frequency and correctly substituting all changes into the double-slit equation to find 1/6 x (A).

First, we find the equivalent capacitance \( C_{eq1} \) of the first arrangement:
- Two capacitors in parallel: \( C_p = C + C = 2C \).
- In series with the third:
\[ \frac{1}{C_{eq1}} = \frac{1}{2C} + \frac{1}{C} = \frac{3}{2C} \implies C_{eq1} = \frac{2}{3}C \]

The energy stored is:
\[ E = \frac{1}{2} C_{eq1} V^2 = \frac{1}{3} C V^2 \]

Next, we find the equivalent capacitance \( C_{eq2} \) of the second arrangement (three in series):
\[ \frac{1}{C_{eq2}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} \implies C_{eq2} = \frac{1}{3}C \]

The energy stored in this case is:
\[ E' = \frac{1}{2} C_{eq2} V^2 = \frac{1}{6} C V^2 \]

Comparing the two energies:
\[ \frac{E'}{E} = \frac{\frac{1}{6} C V^2}{\frac{1}{3} C V^2} = \frac{1}{2} \implies E' = \frac{1}{2} E \]

1 mark for calculating both equivalent capacitances and finding the ratio of energies to be 1/2 (B).

From the kinetic theory of gases, the pressure \( p \), volume \( V \), and mean square speed \( \overline{c^2} = v_{\text{r.m.s.}}^2 \) of the molecules are related by:

\[ p V = \frac{1}{3} N m v_{\text{r.m.s.}}^2 \]

Since the vessel is sealed, the number of molecules \( N \) and the mass of each molecule \( m \) are constant. This gives:

\[ v_{\text{r.m.s.}} \propto \sqrt{p V} \]

If the pressure is doubled (\( 2p \)) and the volume is tripled (\( 3V \)), the product \( p V \) increases by a factor of:

\[ 2 \times 3 = 6 \]

Therefore, the root-mean-square speed increases by a factor of:

\[ \sqrt{6} \approx 2.45 \]

1 mark for identifying that v_r.m.s is proportional to the square root of pV and calculating sqrt(6) to get 2.45 (B).

(a) First, calculate the cross-sectional area \(A\):
\(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.38 \times 10^{-3}\ \text{m})^2}{4} = 1.1341 \times 10^{-7}\ \text{m}^2\)
Now, calculate the resistivity:
\(\rho = \frac{R A}{L} = \frac{4.20 \times 1.1341 \times 10^{-7}}{1.500} = 3.176 \times 10^{-7}\ \Omega\ \text{m} \approx 3.2 \times 10^{-7}\ \Omega\ \text{m}\).

(b) Next, determine the percentage uncertainty in each measured value:
- For \(R\): \(\frac{0.05}{4.20} \times 100\% \approx 1.190\%\)
- For \(L\): \(\frac{0.002}{1.500} \times 100\% \approx 0.133\%\)
- For \(d\): \(\frac{0.01}{0.38} \times 100\% \approx 2.632\%\)

Since \(\rho = \frac{\pi R d^2}{4 L}\), the percentage uncertainty in \(\rho\) is given by:
\(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\)
\(\frac{\Delta \rho}{\rho} = 1.190\% + 2(2.632\%) + 0.133\% = 6.587\%\)

Now, calculate the absolute uncertainty \(\Delta \rho\):
\(\Delta \rho = 6.587\% \times 3.176 \times 10^{-7}\ \Omega\ \text{m} = 0.2092 \times 10^{-7}\ \Omega\ \text{m} = 2.1 \times 10^{-8}\ \Omega\ \text{m}\) (or \(2 \times 10^{-8}\ \Omega\ \text{m}\) to 1 s.f.).

(a)
- C1: Area calculated correctly: \(A = 1.13 \times 10^{-7}\ \text{m}^2\) [1]
- A1: Value of \(\rho\) shown correctly as \(3.18 \times 10^{-7}\ \Omega\ \text{m}\) and rounded to \(3.2 \times 10^{-7}\ \Omega\ \text{m}\) [1]

(b)
- C1: Percentage uncertainties in \(R\) (\(1.19\%\)) AND \(L\) (\(0.13\%\)) AND \(d\) (\(2.63\%\)) calculated [1]
- C1: Triples/doubles diameter uncertainty correctly for \(d^2\) (yielding \(5.26\%\)) [1]
- C1: Sums all percentage uncertainties to obtain overall percentage uncertainty of \(6.59\%\) (accept \(6.6\%\)) [1]
- A1: Correct absolute uncertainty calculated: \(2.1 \times 10^{-8}\ \Omega\ \text{m}\) or \(2 \times 10^{-8}\ \Omega\ \text{m}\) [1]
- A1: Expresses absolute uncertainty to 1 or 2 significant figures with correct unit (\(\Omega\ \text{m}\)) [1]

(a) First, calculate the total binding energy (BE) of reactants and products:
- BE of deuterium (\(^{2}_{1}\text{H}\)): \(2 \times 1.11 = 2.22\ \text{MeV}\)
- BE of tritium (\(^{3}_{1}\text{H}\)): \(3 \times 2.83 = 8.49\ \text{MeV}\)
- Total BE of reactants = \(2.22 + 8.49 = 10.71\ \text{MeV}\)
- BE of helium (\(^{4}_{2}\text{He}\)): \(4 \times 7.07 = 28.28\ \text{MeV}\)
- BE of a neutron (\(^{1}_{0}\text{n}\)): \(0\ \text{MeV}\)
- Total BE of products = \(28.28\ \text{MeV}\)

Energy released \(= \text{BE(products)} - \text{BE(reactants)} = 28.28 - 10.71 = 17.57\ \text{MeV}\).

(b) Convert the power output to total thermal power required:
\(P_{\text{thermal}} = \frac{450 \times 10^6\ \text{W}}{0.35} = 1.286 \times 10^9\ \text{W}\)

Calculate thermal energy required per day (\(24 \times 3600\ \text{s} = 86400\ \text{s}\)):
\(E_{\text{day}} = 1.286 \times 10^9\ \text{W} \times 86400\ \text{s} = 1.111 \times 10^{14}\ \text{J}\)

Convert energy per single fusion reaction to Joules:
\(E_{\text{rxn}} = 17.57\ \text{MeV} \times 1.60 \times 10^{-13}\ \text{J/MeV} = 2.811 \times 10^{-12}\ \text{J}\)

Number of reactions (and hence deuterium atoms) per day:
\(N = \frac{E_{\text{day}}}{E_{\text{rxn}}} = \frac{1.111 \times 10^{14}}{2.811 \times 10^{-12}} = 3.952 \times 10^{25}\)

Calculate the mass of deuterium consumed:
\(m = N \times \text{mass of deuterium atom} = 3.952 \times 10^{25} \times (2.014 \times 1.66 \times 10^{-27}\ \text{kg}) = 0.132\ \text{kg}\) (or \(132\ \text{g}\)).

(a)
- C1: Calculates total binding energy of reactants = \(10.71\ \text{MeV}\) [1]
- C1: Calculates total binding energy of products = \(28.28\ \text{MeV}\) [1]
- A1: Calculates energy released = \(17.57\ \text{MeV}\) (or \(17.6\ \text{MeV}\)) [1]

(b)
- C1: Calculates thermal power required (\(1.29 \times 10^9\ \text{W}\)) OR total thermal energy required per day (\(1.11 \times 10^{14}\ \text{J}\)) [1]
- C1: Converts single-reaction energy to Joules: \(2.81 \times 10^{-12}\ \text{J}\) [1]
- C1: Calculates the total number of reactions per day: \(3.95 \times 10^{25}\) [1]
- A1: Final mass of deuterium = \(0.132\ \text{kg}\) (accept range \(0.130\) to \(0.135\ \text{kg}\)) [1]

(a) Wien's displacement law states that the peak wavelength of emission is inversely proportional to the absolute temperature, \(\lambda_{\text{max}} T = \text{constant} \approx 2.90 \times 10^{-3}\ \text{m K}\).
Rearranging for temperature \(T\):
\(T = \frac{2.90 \times 10^{-3}\ \text{m K}}{193 \times 10^{-9}\ \text{m}} = 1.503 \times 10^4\ \text{K} \approx 1.5 \times 10^4\ \text{K}\).

(b) Using Stefan-Boltzmann law for stellar luminosity:
\(L = 4 \pi R^2 \sigma T^4\)
Rearranging for radius \(R\):
\(R = \sqrt{\frac{L}{4 \pi \sigma T^4}}\)
Using the calculated temperature \(15026\ \text{K}\):
\(R = \sqrt{\frac{1.92 \times 10^{29}}{4 \pi \times (5.67 \times 10^{-8}) \times (15026)^4}}\)
\(R = \sqrt{\frac{1.92 \times 10^{29}}{7.125 \times 10^{-7} \times 5.10 \times 10^{16}}} = \sqrt{\frac{1.92 \times 10^{29}}{3.634 \times 10^{10}}} = \sqrt{5.283 \times 10^{18}} = 2.30 \times 10^9\ \text{m}\).

(c) Radiant flux intensity (apparent brightness) is given by:
\(F = \frac{L}{4 \pi d^2}\)
First, find distance \(d\) in meters:
\(d = 860 \times 9.46 \times 10^{15}\ \text{m} = 8.136 \times 10^{18}\ \text{m}\)
Now compute \(F\):
\(F = \frac{1.92 \times 10^{29}}{4 \pi \times (8.136 \times 10^{18})^2} = \frac{1.92 \times 10^{29}}{8.318 \times 10^{38}} = 2.31 \times 10^{-10}\ \text{W m}^{-2}\).

(a)
- C1: Correct statement of Wien's law (\(\lambda_{\text{max}} \propto 1/T\) or formula with constant) [1]
- A1: Correct substitution showing \(T \approx 1.5 \times 10^4\ \text{K}\) [1]

(b)
- C1: Recalls and rearranges Stefan's law: \(R = \sqrt{\frac{L}{4\pi\sigma T^4}}\) [1]
- C1: Correct substitution of values (accept using either \(1.5 \times 10^4\ \text{K}\) or \(1.503 \times 10^4\ \text{K}\)) [1]
- A1: Correct radius \(R = 2.30 \times 10^9\ \text{m}\) (accept \(2.3 \times 10^9\ \text{m}\) to \(2.32 \times 10^9\ \text{m}\)) [1]

(c)
- C1: Calculates distance in meters (\(8.14 \times 10^{18}\ \text{m}\)) and uses formula \(F = \frac{L}{4\pi d^2}\) [1]
- A1: Correct radiant flux intensity = \(2.31 \times 10^{-10}\ \text{W m}^{-2}\) (accept range \(2.30 \times 10^{-10}\) to \(2.32 \times 10^{-10}\)) [1]

(a) Convert half-life from hours to seconds:
\(t_{1/2} = 6.0\ \text{hours} = 6.0 \times 3600\ \text{s} = 21600\ \text{s}\)
Now calculate the decay constant \(\lambda\):
\(\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.69315}{21600\ \text{s}} = 3.209 \times 10^{-5}\ \text{s}^{-1} \approx 3.2 \times 10^{-5}\ \text{s}^{-1}\).

(b) Calculate elapsed time from 08:00 AM to 08:00 PM:
\(t = 12.0\ \text{hours} = 12.0 \times 3600\ \text{s} = 43200\ \text{s}\)

Method 1: Find activity first
Since \(12.0\ \text{hours}\) is exactly two half-lives:
\(A = A_0 \times \left(\frac{1}{2}\right)^2 = 350\ \text{MBq} \times 0.25 = 87.5\ \text{MBq} = 8.75 \times 10^7\ \text{Bq}\)
Now use the relation \(A = \lambda N\):
\(N = \frac{A}{\lambda} = \frac{8.75 \times 10^7\ \text{Bq}}{3.209 \times 10^{-5}\ \text{s}^{-1}} = 2.726 \times 10^{12}\ \text{nuclei}\).

Method 2: Find initial number of nuclei \(N_0\) first
\(N_0 = \frac{A_0}{\lambda} = \frac{350 \times 10^6}{3.209 \times 10^{-5}} = 1.091 \times 10^{13}\ \text{nuclei}\)
Using decay equation:
\(N = N_0 e^{-\lambda t} = 1.091 \times 10^{13} \times e^{-(3.209 \times 10^{-5} \times 43200)} = 1.091 \times 10^{13} \times 0.25 = 2.73 \times 10^{12}\ \text{nuclei}\).

(a)
- C1: Converts half-life into seconds correctly (\(21600\ \text{s}\)) [1]
- A1: Computes \(\lambda = \frac{\ln 2}{21600}\) to show \(3.2 \times 10^{-5}\ \text{s}^{-1}\) [1]

(b)
- C1: Identifies that time elapsed is \(12\ \text{hours}\) (or \(4.32 \times 10^4\ \text{s}\)) [1]
- C1: Uses the relation \(A = \lambda N\) or \(N = N_0 e^{-\lambda t}\) [1]
- C1: Calculates activity at 8:00 PM as \(87.5\ \text{MBq}\) OR calculates initial number of nuclei \(N_0 = 1.09 \times 10^{13}\) [1]
- C1: Substitutes values into final equation correctly [1]
- A1: Correct final answer for remaining nuclei: \(2.7 \times 10^{12}\) (accept range \(2.70 \times 10^{12}\) to \(2.74 \times 10^{12}\)) [1]

(a) Let \(y\) be the distance from the central maximum to the second-order maximum (\(1.65\ \text{m}\)) and \(D\) be the distance from the grating to the screen (\(1.80\ \text{m}\)).
Using trigonometry:
\(\tan \theta = \frac{y}{D} = \frac{1.65}{1.80} = 0.9167\)
\(\theta = \arctan(0.9167) = 42.51^\circ\) (or \(0.742\ \text{rad}\)).

(b) Calculate the grating spacing \(d\):
\(d = \frac{1}{600 \times 10^3\ \text{m}^{-1}} = 1.667 \times 10^{-6}\ \text{m}\)
Using the diffraction grating equation: \(d \sin \theta = n \lambda\)
For \(n = 2\):
\(\lambda = \frac{d \sin \theta}{2} = \frac{1.667 \times 10^{-6} \times \sin(42.51^\circ)}{2}\)
\(\lambda = \frac{1.667 \times 10^{-6} \times 0.6757}{2} = 5.631 \times 10^{-7}\ \text{m} = 563\ \text{nm}\).

(c) A higher number of lines per millimeter means that the grating spacing \(d\) decreases. According to \(d \sin \theta = n \lambda\), if \(d\) decreases, then \(\sin \theta\) must increase for the same wavelength and order. Consequently, the diffraction angles will be larger, and the maxima on the screen will be spaced further apart.

(a)
- C1: Uses correct trigonometric relation \(\tan \theta = 1.65 / 1.80\) [1]
- A1: Computes angle \(\theta = 42.5^\circ\) (or \(0.742\ \text{rad}\)) [1]

(b)
- C1: Calculates grating spacing \(d = 1.67 \times 10^{-6}\ \text{m}\) [1]
- C1: Recalls and substitutes correctly into \(d \sin \theta = n \lambda\) with \(n = 2\) [1]
- A1: Correctly calculates \(\lambda = 563\ \text{nm}\) (accept range \(560\ \text{nm}\) to \(565\ \text{nm}\)) [1]

(c)
- B1: States that the grating spacing \(d\) is smaller [1]
- B1: Explains that \(\theta\) is larger, hence the maxima are further apart / more widely separated on the screen [1]

(a) Time constant \(\tau\) is given by:
\(\tau = R C = 15.0 \times 10^3\ \Omega \times 470 \times 10^{-6}\ \text{F} = 7.05\ \text{s}\).

(b) Initial charge \(Q_0\) on the capacitor:
\(Q_0 = C V_0 = 470 \times 10^{-6}\ \text{F} \times 12.0\ \text{V} = 5.64 \times 10^{-3}\ \text{C}\)
The discharging equation for charge is:
\(Q = Q_0 e^{-t/\tau}\)
Since \(t = 2.5 \tau\), we have \(t/\tau = 2.5\):
\(Q = 5.64 \times 10^{-3} \times e^{-2.5} = 5.64 \times 10^{-3} \times 0.08208 = 4.63 \times 10^{-4}\ \text{C}\) (or \(0.463\ \text{mC}\)).

(c) Calculate the initial energy stored in the capacitor:
\(E_0 = \frac{1}{2} C V_0^2 = 0.5 \times 470 \times 10^{-6} \times (12.0)^2 = 0.03384\ \text{J}\)
Calculate the remaining energy in the capacitor after \(2.5 \tau\):
\(E = \frac{1}{2} \frac{Q^2}{C} = \frac{(4.63 \times 10^{-4})^2}{2 \times 470 \times 10^{-6}} = 2.28 \times 10^{-4}\ \text{J}\)
(Alternatively, using the fractional energy decay: \(E = E_0 e^{-2t/\tau} = E_0 e^{-5} = 0.03384 \times 0.006738 = 2.28 \times 10^{-4}\ \text{J}\)).

Energy dissipated in the resistor is the difference in stored energy:
\(E_{\text{dissipated}} = E_0 - E = 0.03384 - 2.28 \times 10^{-4} = 0.03361\ \text{J}\) (or \(33.6\ \text{mJ}\)).

(a)
- A1: Correct time constant \(7.05\ \text{s}\) (accept \(7.1\ \text{s}\)) [1]

(b)
- C1: Calculates initial charge \(Q_0 = 5.64 \times 10^{-3}\ \text{C}\) [1]
- C1: Uses correct decay expression \(Q = Q_0 e^{-2.5}\) [1]
- A1: Correct remaining charge \(4.6 \times 10^{-4}\ \text{C}\) (accept range \(4.6 \times 10^{-4}\) to \(4.7 \times 10^{-4}\ \text{C}\)) [1]

(c)
- C1: Calculates initial energy stored: \(E_0 = 0.0338\ \text{J}\) [1]
- C1: Calculates remaining energy: \(E = 2.28 \times 10^{-4}\ \text{J}\) OR uses formula \(\Delta E = E_0 (1 - e^{-5})\) [1]
- A1: Calculates correct energy difference: \(0.0336\ \text{J}\) or \(33.6\ \text{mJ}\) (accept range \(0.0335\) to \(0.0340\ \text{J}\)) [1]

(a) Convert temperature to Kelvin:
\(T = 27 + 273.15 = 300.15\ \text{K}\) (accept \(300\ \text{K}\)).
Use the ideal gas equation:
\(p V = N k_{\text{B}} T \Rightarrow N = \frac{p V}{k_{\text{B}} T}\)
\(N = \frac{1.8 \times 10^5\ \text{Pa} \times 0.035\ \text{m}^3}{1.38 \times 10^{-23}\ \text{J K}^{-1} \times 300\ \text{K}} = \frac{6300}{4.14 \times 10^{-21}} = 1.522 \times 10^{24}\ \text{atoms}\).

(b) Method 1: Using molecular mass
Mass of one helium atom:
\(m = \frac{4.0 \times 10^{-3}\ \text{kg mol}^{-1}}{6.02 \times 10^{23}\ \text{mol}^{-1}} = 6.645 \times 10^{-27}\ \text{kg}\)
Use the kinetic theory equation:
\(\frac{1}{2} m c_{\text{r.m.s.}}^2 = \frac{3}{2} k_{\text{B}} T \Rightarrow c_{\text{r.m.s.}} = \sqrt{\frac{3 k_{\text{B}} T}{m}}\)
\(c_{\text{r.m.s.}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times 300}{6.645 \times 10^{-27}}} = \sqrt{1.869 \times 10^6} = 1.367 \times 10^3\ \text{m s}^{-1}\).

Method 2: Using molar gas formula
\(c_{\text{r.m.s.}} = \sqrt{\frac{3 R T}{M}}\)
\(c_{\text{r.m.s.}} = \sqrt{\frac{3 \times 8.31 \times 300}{4.0 \times 10^{-3}}} = \sqrt{1.86975 \times 10^6} = 1.367 \times 10^3\ \text{m s}^{-1}\).

(a)
- C1: Converts temperature to Kelvin: \(T = 300\ \text{K}\) [1]
- C1: Recalls and rearranges \(p V = N k_{\text{B}} T\) or uses \(pV = nRT\) with \(N = n N_{\text{A}}\) [1]
- A1: Calculates number of atoms correctly: \(1.5 \times 10^{24}\) (accept \(1.50 \times 10^{24}\) to \(1.53 \times 10^{24}\)) [1]

(b)
- C1: Converts helium molar mass to kg or determines mass of a helium atom (\(6.64 \times 10^{-27}\ \text{kg}\)) [1]
- C1: Equates mean kinetic energy of a molecule to \(\frac{3}{2} k_{\text{B}} T\) (or uses \(c_{\text{r.m.s.}} = \sqrt{3RT/M}\)) [1]
- C1: Substitutes values into expression correctly [1]
- A1: Obtains \(c_{\text{r.m.s.}} = 1.37 \times 10^3\ \text{m s}^{-1}\) (accept range \(1.36 \times 10^3\) to \(1.38 \times 10^3\ \text{m s}^{-1}\)) [1]

(a) Centripetal acceleration \(a_c\) is given by:
\(a_c = \frac{v^2}{r} = \frac{(14.5\ \text{m s}^{-1})^2}{12.0\ \text{m}} = 17.52\ \text{m s}^{-2}\).

(b) At the top of the loop, both the weight \(W = mg\) and the normal contact force \(N\) act vertically downwards, pointing toward the center of the circular path. Therefore, they combine to provide the centripetal force:
\(F_c = N + mg = \frac{m v^2}{r}\)
Rearranging for the normal contact force \(N\):
\(N = \frac{m v^2}{r} - mg = m(a_c - g)\)
\(N = 450\ \text{kg} \times (17.52\ \text{m s}^{-2} - 9.81\ \text{m s}^{-2}) = 450 \times 7.71 = 3469.5\ \text{N} \approx 3.47 \times 10^3\ \text{N}\).

(c) To maintain contact with the track, the normal contact force must be greater than or equal to zero (\(N \ge 0\)).
The minimum speed \(v_{\text{min}}\) occurs when \(N = 0\):
\(\frac{m v_{\text{min}}^2}{r} = mg \Rightarrow v_{\text{min}} = \sqrt{r g}\)
\(v_{\text{min}} = \sqrt{12.0\ \text{m} \times 9.81\ \text{m s}^{-2}} = \sqrt{117.72} = 10.85\ \text{m s}^{-1} \approx 10.9\ \text{m s}^{-1}\).

(a)
- C1: Recalls formula \(a_c = \frac{v^2}{r}\) and substitutes values [1]
- A1: Correct calculation: \(17.5\ \text{m s}^{-2}\) (accept \(17.52\ \text{m s}^{-2}\)) [1]

(b)
- C1: Identifies the dynamic equation at the top: \(N + mg = F_c\) [1]
- C1: Substitutes numbers correctly: \(N = 450(17.52 - 9.81)\) [1]
- A1: Correct contact force: \(3.47 \times 10^3\ \text{N}\) (accept \(3460\ \text{N}\) to \(3500\ \text{N}\)) [1]

(c)
- C1: Recognizes condition for maintaining contact as \(N = 0\) leading to \(v = \sqrt{rg}\) [1]
- A1: Correct calculation: \(10.9\ \text{m s}^{-1}\) (accept \(10.8\ \text{m s}^{-1}\) to \(10.9\ \text{m s}^{-1}\)) [1]

(i) The time constant is given by \( \tau = RC \). Substituting the values: \( \tau = 47 \times 10^3\text{ }\Omega \times 330 \times 10^{-6}\text{ F} = 15.51\text{ s} \approx 15.5\text{ s} \). (ii) The discharge equation is \( V = V_0 e^{-t/\tau} \). Substituting the values: \( V = 12.0 \times e^{-20.0 / 15.51} = 12.0 \times e^{-1.289} = 12.0 \times 0.2754 = 3.30\text{ V} \approx 3.3\text{ V} \). (iii) The initial energy stored in the capacitor is \( E_i = \frac{1}{2} C V_0^2 = 0.5 \times 330 \times 10^{-6} \times (12.0)^2 = 0.02376\text{ J} \). The final energy remaining at \( 20.0\text{ s} \) is \( E_f = \frac{1}{2} C V_f^2 = 0.5 \times 330 \times 10^{-6} \times (3.30)^2 = 0.00180\text{ J} \). The energy delivered to the resistor is the difference: \( \Delta E = E_i - E_f = 0.02376 - 0.00180 = 0.02196\text{ J} \approx 0.022\text{ J} \) (or \( 2.2 \times 10^{-2}\text{ J} \)).

(i) [2 marks] C1: \( \tau = RC = 47 \times 10^3 \times 330 \times 10^{-6} \) | A1: \( 15.5\text{ s} \) (accept 15 s or 16 s if 2 s.f. used). (ii) [2 marks] C1: \( V = 12.0 \times e^{-20.0/15.5} \) (allow ecf from (i)) | A1: \( 3.3\text{ V} \) (accept 3.30 V). (iii) [3 marks] C1: \( E = \frac{1}{2}CV^2 \) used for at least one energy calculation (either initial or final) | C1: \( E_i = 0.0238\text{ J} \) AND \( E_f = 0.0018\text{ J} \) (or correct subtraction structure) | A1: \( 0.022\text{ J} \) (accept \( 2.2 \times 10^{-2}\text{ J} \), allow ecf from (ii)).

(i) Using the wave equation: \( \lambda = \frac{c}{f} = \frac{3.00 \times 10^8\text{ m s}^{-1}}{1.5 \times 10^9\text{ Hz}} = 0.20\text{ m} \). (ii) The path difference \( \Delta x = 3.70 - 3.20 = 0.50\text{ m} \). Comparing this to the wavelength: \( \frac{\Delta x}{\lambda} = \frac{0.50}{0.20} = 2.5 \). Since the path difference is a half-integer number of wavelengths (\( 2.5\lambda \)), the waves arrive completely out of phase, resulting in destructive interference. (iii) If transmitter B is adjusted to be \( 180^\circ \) out of phase with A, this introduces an additional phase difference of \( 180^\circ \) (or \( 0.5\lambda \)). The total effective phase difference at P becomes equivalent to an integer number of wavelengths (\( 2.5\lambda - 0.5\lambda = 2.0\lambda \)), leading to constructive interference and a maximum signal intensity at P.

(i) [2 marks] C1: \( \lambda = \frac{c}{f} \) or \( \lambda = \frac{3.00 \times 10^8}{1.5 \times 10^9} \) | A1: \( 0.20\text{ m} \). (ii) [3 marks] C1: \( \text{Path difference} = 3.70 - 3.20 = 0.50\text{ m} \) | B1: Compares path difference to wavelength, e.g. path difference is \( 2.5\lambda \) or \( (2n+1)\frac{\lambda}{2} \) | A1: States 'destructive interference' (with valid reasoning based on phase or path difference). (iii) [2 marks] B1: Explains that the source phase difference of \( 180^\circ \) compensates for the half-wavelength path difference | B1: Concludes that interference becomes constructive / signal becomes a maximum.

(i) Mass of reactants = \( 2.01355 + 3.01550 = 5.02905\text{ u} \). Mass of products = \( 4.00150 + 1.00867 = 5.01017\text{ u} \). Mass defect \( \Delta m = 5.02905 - 5.01017 = 0.01888\text{ u} \). (ii) Mass defect in kg: \( \Delta m = 0.01888 \times 1.661 \times 10^{-27}\text{ kg} = 3.1360 \times 10^{-29}\text{ kg} \). Energy released: \( E = \Delta m c^2 = 3.1360 \times 10^{-29}\text{ kg} \times (3.00 \times 10^8\text{ m s}^{-1})^2 = 2.8224 \times 10^{-12}\text{ J} \approx 2.82 \times 10^{-12}\text{ J} \). (iii) Required rate of fusion: \( \text{Rate} = \frac{\text{Power}}{\text{Energy per reaction}} = \frac{50 \times 10^6\text{ W}}{2.8224 \times 10^{-12}\text{ J}} = 1.772 \times 10^{19}\text{ s}^{-1} \approx 1.8 \times 10^{19}\text{ s}^{-1} \) (or \( 1.77 \times 10^{19}\text{ s}^{-1} \)).

(i) [2 marks] C1: Sums reactant masses and product masses | A1: Mass defect \( = 0.01888\text{ u} \). (ii) [3 marks] C1: Multiplies mass defect by \( 1.661 \times 10^{-27} \) | C1: Uses \( E = \Delta m c^2 \) | A1: \( 2.82 \times 10^{-12}\text{ J} \) (accept \( 2.8 \times 10^{-12}\text{ J} \)). (iii) [2 marks] C1: Uses \( \text{Rate} = \frac{50 \times 10^6}{\text{Energy}} \) | A1: \( 1.8 \times 10^{19}\text{ s}^{-1} \) (or \( 1.77 \times 10^{19}\text{ s}^{-1} \), allow ecf from (ii)).

### 1. Circuit Diagram & Experimental Procedure
* **Circuit Diagram:** Draw a DC power supply in series with a single-pole double-throw (SPDT) switch, which can connect the capacitor either to the power supply (to charge it) or across a resistor \(R\) in parallel with a digital voltmeter (to discharge it).
* **Procedure:**
1. Close the switch to the power supply to fully charge the capacitor to initial potential difference \(V_0\).
2. Flip the switch to disconnect the power supply and connect the capacitor to the resistor \(R\), simultaneously starting a stopwatch.
3. Record the potential difference \(V\) at regular, predefined time intervals \(t\) (e.g., every 5 seconds) as the capacitor discharges.
4. Continue taking readings until \(V\) falls to a small fraction of \(V_0\) (e.g., at least \(3\) to \(5\) time constants).
5. Repeat the entire process at least twice more to obtain average values of \(V\) for each corresponding time \(t\) to reduce random errors.

### 2. Graphical Analysis
* The potential difference \(V\) during discharge is given by: \(V = V_0 e^{-\frac{t}{RC}}\)
* Taking natural logarithms of both sides yields: \(\ln(V) = \ln(V_0) - \frac{t}{RC}\)
* This matches the equation of a straight line, \(y = mx + c\), where:
* \(y = \ln(V)\)
* \(x = t\)
* \(m = -\frac{1}{RC}\) (gradient)
* \(c = \ln(V_0)\) (y-intercept)
* Plot a graph of \(\ln(V)\) on the y-axis against \(t\) on the x-axis.
* Draw a straight line of best fit through the data points.
* Measure the gradient \(m\) of this line. The capacitance is then calculated as: \(C = -\frac{1}{m R}\)

### 3. Estimating Uncertainty in \(C\)
* **Uncertainty in the Gradient (\(\Delta m\)):**
* Plot error bars on the \(\ln(V)\) vs \(t\) graph based on the spread of the repeated trials.
* Draw the line of best fit (gradient \(m_{\text{best}}\)) and the worst acceptable line of fit (gradient \(m_{\text{worst}}\)), which is the steepest or shallowest possible straight line that still passes through all error bars.
* Find the absolute uncertainty in the gradient: \(\Delta m = |m_{\text{best}} - m_{\text{worst}}|\)
* **Combining Uncertainties:**
* Since \(C = \frac{1}{|m| R}\), the fractional uncertainty in \(C\) is the sum of the fractional uncertainties in the gradient and the resistance:
\(\frac{\Delta C}{C} = \frac{\Delta m}{|m_{\text{best}}|} + \frac{\Delta R}{R}\)
* Calculate the percentage uncertainty in \(R\): \(\%\Delta R = \frac{1 \text{ k}\Omega}{47 \text{ k}\Omega} \times 100\% \approx 2.13\%\).
* Find the absolute uncertainty in \(C\) using: \(\Delta C = C \times \left( \frac{\Delta m}{|m_{\text{best}}|} + \frac{\Delta R}{R} \right)\)

**[8 Marks - Level of Response]**

* **Level 3 (7-8 marks):**
- Detailed, correct circuit diagram and procedure with multiple repeats described.
- Full mathematical justification showing how \(\ln(V)\) vs \(t\) produces a linear relationship, identifying the gradient as \(-\frac{1}{RC}\) and correctly expressing \(C\).
- Excellent description of determining gradient uncertainty via the worst acceptable line of fit, and correctly combining this with the uncertainty of \(R\) (using fractional/percentage uncertainties) to find \(\Delta C\).

* **Level 2 (4-6 marks):**
- Good circuit and procedure described, although some minor details like repeats or diagrammatic precision might be missing.
- Correctly identifies that a plot of \(\ln(V)\) vs \(t\) is required and relates the gradient to \(C\), but may have minor sign errors or lack full algebraic proof.
- Mentions using a worst-fit line for gradient uncertainty or combining percentage uncertainties, but does not provide a complete explanation of both steps.

* **Level 1 (1-3 marks):**
- Simple circuit diagram or basic procedural steps given.
- Understands that logarithmic treatment of data is needed or mentions a graph of \(V\) vs \(t\), but is unable to link gradient directly to \(-\frac{1}{RC}\).
- Limited or incorrect discussion of uncertainties.

* **0 marks:** No response or no physics of any merit.