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First, calculate the percentage uncertainty for each measured quantity:

Percentage uncertainty in current \( I \) is:

\( \frac{0.04}{2.00} \times 100\% = 2.0\% \)

Percentage uncertainty in resistance \( R \) is:

\( \frac{1.5}{50.0} \times 100\% = 3.0\% \)

Percentage uncertainty in time \( t \) is:

\( \frac{0.1}{10.0} \times 100\% = 1.0\% \)

Using the rules for combining independent uncertainties, the total percentage uncertainty in \( P = \frac{I^2 R}{t} \) is given by:

\( \frac{\Delta P}{P} = 2 \left( \frac{\Delta I}{I} \right) + \frac{\Delta R}{R} + \frac{\Delta t}{t} \)

\( \frac{\Delta P}{P} = 2(2.0\%) + 3.0\% + 1.0\% = 8.0\% \)

C is the correct answer. 1 mark is awarded for the correct calculation of individual percentage uncertainties (current = 2.0%, resistance = 3.0%, time = 1.0%), correctly doubling the percentage uncertainty of current due to the squared term, summing the values to obtain 8.0%, and selecting option C.

Using the ideal gas equation, \( pV = nRT \).

Initially, the state of the gas is described by:

\( P_0 V = n_0 R T_0 \)

where \( T_0 = 27^\circ\text{C} = 27 + 273 = 300\text{ K} \).

When half of the mass escapes, the number of moles is halved, so \( n_1 = 0.5 n_0 \).

The gas is heated to \( T_1 = 327^\circ\text{C} = 327 + 273 = 600\text{ K} \).

Let the final pressure be \( P_1 \). Since the volume \( V \) is constant:

\( P_1 V = n_1 R T_1 = (0.5 n_0) R (600) = 300 n_0 R \)

Comparing this to the initial state:

\( P_0 V = n_0 R (300) = 300 n_0 R \)

Therefore, \( P_1 V = P_0 V \), which means \( P_1 = P_0 \).

C is the correct answer. 1 mark is awarded for correctly converting temperatures to Kelvin (300 K and 600 K), recognizing that the number of moles is halved, setting up the ratio of the ideal gas state equations, and finding that the final pressure is equal to the initial pressure.

According to Hubble's Law, the recession velocity \( v \) of the galaxy is given by:

\( v = H_0 d \)

\( v = (2.3 \times 10^{-18}\text{ s}^{-1}) \times (4.0 \times 10^{24}\text{ m}) = 9.2 \times 10^6\text{ m s}^{-1} \)

The Doppler redshift relation is:

\( z = \frac{\Delta \lambda}{\lambda} \approx \frac{v}{c} \)

Substitute the values to find the change in wavelength \( \Delta \lambda \):

\( \Delta \lambda = \lambda \left( \frac{v}{c} \right) = 656.3\text{ nm} \times \left( \frac{9.2 \times 10^6}{3.0 \times 10^8} \right) \approx 20.1\text{ nm} \)

The observed wavelength \( \lambda_{\text{obs}} \) is:

\( \lambda_{\text{obs}} = \lambda + \Delta \lambda = 656.3\text{ nm} + 20.1\text{ nm} = 676.4\text{ nm} \approx 676\text{ nm} \)

C is the correct answer. 1 mark is awarded for calculating the recession velocity of the galaxy using Hubble's law, calculating the wavelength change using the Doppler redshift equation, and adding this change to the laboratory wavelength to find the correct observed wavelength of 676 nm.

1. Maximum equivalent capacitance is achieved when all three capacitors are connected in parallel:

\( C_{\text{max}} = C_1 + C_2 + C_3 = 3.0\ \mu\text{F} + 6.0\ \mu\text{F} + 9.0\ \mu\text{F} = 18.0\ \mu\text{F} \)

2. Minimum equivalent capacitance is achieved when all three capacitors are connected in series:

\( \frac{1}{C_{\text{min}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{3.0} + \frac{1}{6.0} + \frac{1}{9.0} = \frac{6 + 3 + 2}{18} = \frac{11}{18.0} \)

\( C_{\text{min}} = \frac{18.0}{11} \approx 1.6\ \mu\text{F} \)

A is the correct answer. 1 mark is awarded for correctly identifying that parallel connection yields maximum capacitance and series connection yields minimum capacitance, and accurately calculating both values.

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

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

where \( \Delta \Phi \) is the change in magnetic flux linkage:

\( \Delta \Phi = |\Phi_f - \Phi_i| \)

Initially, the plane of the coil is perpendicular to the magnetic field, so the flux linkage is:

\( \Phi_i = N B A = 200 \times 0.40\text{ T} \times (1.5 \times 10^{-3}\text{ m}^2) = 0.12\text{ Wb} \)

After rotating by \( 90^\circ \), the plane of the coil is parallel to the magnetic field, so the magnetic flux linkage is:

\( \Phi_f = 0 \)

Therefore, the change in flux linkage is:

\( \Delta \Phi = 0.12\text{ Wb} - 0 = 0.12\text{ Wb} \)

The average induced e.m.f. is:

\( E = \frac{0.12\text{ Wb}}{0.15\text{ s}} = 0.80\text{ V} \)

C is the correct answer. 1 mark is awarded for calculating the initial magnetic flux linkage of the coil, recognizing that the final flux linkage is zero, and dividing the change in flux linkage by the time interval to get 0.80 V.

The displacement \( x \) of a particle in simple harmonic motion starting from equilibrium at \( t = 0 \) is given by:

\( x = A \sin(\omega t) \)

where \( \omega = \frac{2\pi}{T} \).

We set \( x = 0.5A \):

\( 0.5A = A \sin(\omega t) \implies \sin(\omega t) = 0.5 \)

For the minimum positive time \( t \):

\( \omega t = \frac{\pi}{6} \)

Substitute \( \omega = \frac{2\pi}{T} \):

\( \left(\frac{2\pi}{T}\right) t = \frac{\pi}{6} \implies t = \frac{T}{12} \)

A is the correct answer. 1 mark is awarded for using the displacement equation for SHM starting from equilibrium, solving for the phase angle corresponding to half of the amplitude, and correctly expressing the minimum time in terms of the period T.

The total energy released in the reaction is the difference between the total binding energy of the products and the total binding energy of the reactants.

1. Calculate total binding energy of reactants:

- For \( ^{2}\text{H} \) (2 nucleons): \( 2 \times 1.11\text{ MeV} = 2.22\text{ MeV} \)

- For \( ^{3}\text{H} \) (3 nucleons): \( 3 \times 2.83\text{ MeV} = 8.49\text{ MeV} \)

- Total initial binding energy = \( 2.22 + 8.49 = 10.71\text{ MeV} \)

2. Calculate total binding energy of products:

- For \( ^{4}\text{He} \) (4 nucleons): \( 4 \times 7.07\text{ MeV} = 28.28\text{ MeV} \)

- For a single neutron \( ^{1}\text{n} \): \( 0\text{ MeV} \)

- Total final binding energy = \( 28.28\text{ MeV} \)

3. Calculate energy released:

- \( E = \text{Total final binding energy} - \text{Total initial binding energy} \)

- \( E = 28.28\text{ MeV} - 10.71\text{ MeV} = 17.57\text{ MeV} \approx 17.6\text{ MeV} \)

C is the correct answer. 1 mark is awarded for calculating the total binding energy of the reactants (10.71 MeV) and product (28.28 MeV) by multiplying each given value by its respective nucleon number, and then finding the difference to arrive at 17.6 MeV.

The Doppler effect formula for a moving source is:

\( f_o = f_s \left( \frac{v}{v \mp v_s} \right) \)

1. As the ambulance approaches the observer, the observed frequency \( f_{\text{app}} \) is:

\( f_{\text{app}} = f_s \left( \frac{v}{v - v_s} \right) = 900 \left( \frac{340}{340 - 25} \right) = 900 \left( \frac{340}{315} \right) \approx 971.4\text{ Hz} \)

2. As the ambulance recedes from the observer, the observed frequency \( f_{\text{rec}} \) is:

\( f_{\text{rec}} = f_s \left( \frac{v}{v + v_s} \right) = 900 \left( \frac{340}{340 + 25} \right) = 900 \left( \frac{340}{365} \right) \approx 838.4\text{ Hz} \)

3. The change in the observed frequency \( \Delta f \) is:

\( \Delta f = f_{\text{app}} - f_{\text{rec}} = 971.4\text{ Hz} - 838.4\text{ Hz} \approx 133\text{ Hz} \)

B is the correct answer. 1 mark is awarded for correctly applying the Doppler effect formulas for an approaching source and a receding source to find the respective observed frequencies, and calculating the difference between them to get 133 Hz.

To find the percentage uncertainty in the resistivity \(\rho\), we first find the percentage uncertainty in each measured value:

\(\frac{\Delta R}{R} = \frac{0.05}{2.50} \times 100 = 2.0\%\)

\(\frac{\Delta d}{d} = \frac{0.02}{0.40} \times 100 = 5.0\%\)

\(\frac{\Delta L}{L} = \frac{0.005}{1.000} \times 100 = 0.5\%\)

Using the rule for combining uncertainties, since \(d\) is squared, its fractional uncertainty is multiplied by 2:

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

\(\frac{\Delta \rho}{\rho} = 2.0\% + 2(5.0\%) + 0.5\% = 12.5\%\)

Therefore, the correct option is B.

1 mark for the correct calculation of percentage uncertainty (12.5%) showing correct summation with twice the uncertainty of the diameter.

The root-mean-square speed of molecules in an ideal gas depends only on the absolute temperature \(T\) and the molecular mass \(m\), according to the formula:

\(c_{\text{rms}} = \sqrt{\frac{3 k T}{m}}\)

Since the mass \(m\) and Boltzmann's constant \(k\) are constant, the r.m.s. speed is proportional to \(\sqrt{T}\). Changing the volume of the gas does not directly alter this relationship.

When the temperature increases from \(T\) to \(2T\):

\(c_{\text{rms}}' = \sqrt{2} \cdot c_{\text{rms}} \approx 1.41 c_{\text{rms}}\)

Therefore, the correct option is C.

1 mark for identifying that r.m.s. speed is proportional to the square root of absolute temperature and is unaffected by volume change.

The potential difference \(V\) across a discharging capacitor decreases exponentially with time \(t\) according to:

\(V = V_0 e^{-t / RC}\)

At a time equal to the time constant, \(t = RC\):

\(V = V_0 e^{-1}\)

The energy \(E\) stored in the capacitor is proportional to the square of the potential difference:

\(E = \frac{1}{2} C V^2\)

Therefore, the energy remaining is:

\(E = \frac{1}{2} C (V_0 e^{-1})^2 = E_0 e^{-2}\)

Where \(E_0 = \frac{1}{2} C V_0^2\) is the initial energy.

Calculating the percentage remaining:

\(\frac{E}{E_0} = e^{-2} \approx 0.135 = 13.5\% \approx 14\%\)

Thus, the correct option is A.

1 mark for establishing that stored energy is proportional to the square of voltage, and calculating the remaining energy percentage as 14%.

The recession speed \(v\) of the galaxy is related to the redshift \(z\) by:

\(v = z c = 0.050 \times 3.00 \times 10^8\text{ m s}^{-1} = 1.50 \times 10^7\text{ m s}^{-1}\)

Using Hubble's law, \(v = H_0 d\):

\(d = \frac{v}{H_0} = \frac{1.50 \times 10^7\text{ m s}^{-1}}{2.0 \times 10^{-18}\text{ s}^{-1}} = 7.5 \times 10^{24}\text{ m}\)

Therefore, the correct option is B.

1 mark for the correct calculation of recession velocity and subsequent calculation of distance using Hubble's law.

The observed frequency when the source moves towards a stationary observer is:

\(f_{\text{toward}} = f_s \left( \frac{v}{v - v_s} \right) = 600 \left( \frac{340}{340 - 30} \right) = 600 \left( \frac{340}{310} \right) \approx 658.1\text{ Hz}\)

The observed frequency when the source moves away from the stationary observer is:

\(f_{\text{away}} = f_s \left( \frac{v}{v + v_s} \right) = 600 \left( \frac{340}{340 + 30} \right) = 600 \left( \frac{340}{370} \right) \approx 551.4\text{ Hz}\)

The change in frequency is:

\(\Delta f = f_{\text{toward}} - f_{\text{away}} = 658.1 - 551.4 = 106.7\text{ Hz} \approx 107\text{ Hz}\)

Therefore, the correct option is C.

1 mark for the correct calculation of both Doppler-shifted frequencies and subtracting them to find the change (107 Hz).

The extension \(x\) of a wire under load \(F\) is given by:

\(x = \frac{F L}{A E}\)

where \(A\) is the cross-sectional area and \(E\) is the Young modulus of the material. Since \(A = \frac{\pi d^2}{4}\), we have:

\(x = \frac{4 F L}{\pi d^2 E}\)

For the second wire, the length is \(2L\) and the diameter is \(2d\). The new extension \(x'\) is:

\(x' = \frac{4 F (2L)}{\pi (2d)^2 E} = \frac{8 F L}{4 \pi d^2 E} = \frac{1}{2} \left( \frac{4 F L}{\pi d^2 E} \right) = 0.50 x\)

Therefore, the correct option is B.

1 mark for correctly formulating extension in terms of length and diameter, and substituting the new values to get 0.50x.

The relation between current \(I\) and drift speed \(v\) is given by:

\(I = n A v q\)

where \(n\) is the number density of conduction electrons and \(q\) is the elementary charge. For wires of the same material, \(n\) is constant.

Rearranging for drift speed:

\(v = \frac{I}{n A q}\)

For the second wire:

\(v' = \frac{3I}{n (2A) q} = 1.5 \left( \frac{I}{n A q} \right) = 1.5 v\)

Therefore, the correct option is B.

1 mark for using the drift velocity equation to correctly deduce that the new drift speed is 1.5v.

The total binding energy of a nucleus is equal to the binding energy per nucleon multiplied by the number of nucleons (mass number).

Total binding energy of the reactants:
- For \(_{1}^{2}\text{H}\): \(2 \times 1.11\text{ MeV} = 2.22\text{ MeV}\)
- For \(_{1}^{3}\text{H}\): \(3 \times 2.57\text{ MeV} = 7.71\text{ MeV}\)
- Total initial binding energy \(= 2.22 + 7.71 = 9.93\text{ MeV}\)

Total binding energy of the products:
- For \(_{2}^{4}\text{He}\): \(4 \times 7.07\text{ MeV} = 28.28\text{ MeV}\)
- For the free neutron \(_{0}^{1}\text{n}\): \(0\text{ MeV}\) (since it is a single nucleon)
- Total final binding energy \(= 28.28\text{ MeV}\)

Energy released in the reaction is the difference in total binding energy between products and reactants:

\(Q = 28.28\text{ MeV} - 9.93\text{ MeV} = 18.35\text{ MeV} \approx 18.4\text{ MeV}\)

Therefore, the correct option is C.

1 mark for calculating the total binding energy of reactants (9.93 MeV) and products (28.28 MeV), and finding the difference to get 18.4 MeV.

Let's determine the energy states of the system during each step:

1. Initially, connected to battery \(V\) with capacitance \(C\), the stored energy is \(E = \frac{1}{2} C V^2\).
2. When the dielectric of \(\varepsilon_r = 3.0\) is inserted while remaining connected to the battery, the capacitance becomes \(C' = 3.0 C\). The potential difference remains \(V\). The charge on the plates becomes \(Q' = C' V = 3 C V\).
3. The capacitor is now disconnected from the battery, so the charge is isolated and must remain constant at \(Q' = 3 C V\).
4. The dielectric is slowly removed, returning the capacitance back to \(C'' = C\).
5. The energy stored in this final state is:
\(E'' = \frac{Q'^2}{2 C''} = \frac{(3 C V)^2}{2 C} = \frac{9 C^2 V^2}{2 C} = 9 \left(\frac{1}{2} C V^2\right) = 9 E\).
6. The energy stored in the capacitor immediately before the dielectric was removed (after disconnection) was:
\(E' = \frac{Q'^2}{2 C'} = \frac{(3 C V)^2}{2 (3 C)} = 3 \left(\frac{1}{2} C V^2\right) = 3 E\).
7. Since the system is isolated from the battery, the work done \(W\) by the external agent is simply equal to the increase in stored electrostatic potential energy:
\(W = E'' - E' = 9 E - 3 E = 6 E\).

C is the correct answer. 1 mark is awarded for obtaining the correct work done of 6E.

1. Find the redshift \(z\):
\(z = \frac{\Delta\lambda}{\lambda_0} = \frac{15.3\text{ nm}}{656.3\text{ nm}} \approx 0.02331\).

2. Use the Doppler relation to find the recession speed \(v\):
\(v = z c = 0.02331 \times 3.00 \times 10^8\text{ m s}^{-1} \approx 6.99 \times 10^6\text{ m s}^{-1} = 6990\text{ km s}^{-1}\).

3. Use Hubble's law, \(v = H_0 d\), to solve for distance \(d\):
\(d = \frac{v}{H_0} = \frac{6990\text{ km s}^{-1}}{70\text{ km s}^{-1}\text{Mpc}^{-1}} \approx 100\text{ Mpc\}.

B is the correct answer. 1 mark is awarded for calculating the correct distance of 100 Mpc.

1. Write the Doppler equations for a moving source approaching and receding from a stationary observer:
- Approaching: \(f_{\text{approach}} = f_s \left(\frac{v}{v - v_s}\right) = 480\text{ Hz}\)
- Receding: \(f_{\text{recede}} = f_s \left(\frac{v}{v + v_s}\right) = 400\text{ Hz}\)

2. Divide the first equation by the second:
\(\frac{f_{\text{approach}}}{f_{\text{recede}}} = \frac{v + v_s}{v - v_s} = \frac{480}{400} = 1.20\)

3. Solve for \(v_s\):
\(v + v_s = 1.20(v - v_s)\)
\(v + v_s = 1.20v - 1.20v_s\)
\(2.20v_s = 0.20v\)
\(v_s = \frac{0.20}{2.20} v = \frac{1}{11} v\)

4. Calculate the value of \(v_s\) with \(v = 340\text{ m s}^{-1}\):
\(v_s = \frac{340}{11} \approx 30.9\text{ m s}^{-1} \approx 31\text{ m s}^{-1}\).

B is the correct answer. 1 mark is awarded for using the Doppler formulas for a moving source and correctly determining the speed to be approximately 31 m/s.

1. Express the fractional uncertainty equation from the resistivity formula \(\rho = \frac{R \pi d^2}{4 L}\):
\(\frac{\Delta\rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\)

2. Calculate each term:
- \(\frac{\Delta R}{R} = \frac{0.05}{4.20} \approx 0.0119\)
- \(\frac{\Delta d}{d} = \frac{0.01}{0.38} \approx 0.0263\)
- \(\frac{\Delta L}{L} = \frac{0.002}{0.824} \approx 0.0024\)

3. Combine these values, remembering to multiply the diameter uncertainty term by 2 because diameter is squared:
\(\frac{\Delta\rho}{\rho} = 0.0119 + 2(0.0263) + 0.0024 = 0.0119 + 0.0526 + 0.0024 = 0.0669\)

4. Convert to a percentage:
\(\frac{\Delta\rho}{\rho} \times 100\% \approx 6.7\%\).

D is the correct answer. 1 mark is awarded for correctly scaling the diameter fractional uncertainty by 2 and summing the terms to get 6.7%.

In electron-positron annihilation, the total initial momentum of the system is nearly zero.
To conserve momentum, at least two photons must be produced, and they must travel in opposite directions.
To find the energy of each photon, we use the mass-energy conservation. The rest mass of both the electron and positron is converted into two photons:
\(2 E_{\text{photon}} = 2 m_e c^2\)
\(E_{\text{photon}} = m_e c^2 \approx 0.51\text{ MeV}\).
Thus, two photons of approximately \(0.51\text{ MeV}\) each are produced, traveling in opposite directions.

B is the correct answer. 1 mark is awarded for selecting the correct explanation of PET photon production.

1. Calculate the total initial binding energy of the reactants:
- For \({}_{1}^{2}\text{H}\) (2 nucleons): \(2 \times 1.11\text{ MeV} = 2.22\text{ MeV}\).
- For \({}_{1}^{3}\text{H}\) (3 nucleons): \(3 \times 2.83\text{ MeV} = 8.49\text{ MeV}\).
- Total initial binding energy = \(2.22 + 8.49 = 10.71\text{ MeV}\).

2. Calculate the total final binding energy of the products:
- For \({}_{2}^{4}\text{He}\) (4 nucleons): \(4 \times 7.07\text{ MeV} = 28.28\text{ MeV}\).
- The single neutron \({}_{0}^{1}\text{n}\) has zero binding energy.
- Total final binding energy = \(28.28\text{ MeV}\).

3. Calculate the energy released \(Q\):
\(Q = \text{BE}_{\text{final}} - \text{BE}_{\text{initial}} = 28.28 - 10.71 = 17.57\text{ MeV} \approx 17.6\text{ MeV}\).

C is the correct answer. 1 mark is awarded for correctly determining the total binding energies of products and reactants and taking the difference.

1. Use Gay-Lussac's Law for an ideal gas at constant volume:
\(\frac{P_1}{T_1} = \frac{P_2}{T_2}\), where temperature must be in Kelvin.

2. Convert the initial temperature to Kelvin:
\(T_1 = 27 + 273 = 300\text{ K}\).

3. Calculate the final temperature \(T_2\):
\(T_2 = T_1 \times \frac{P_2}{P_1} = 300\text{ K} \times \frac{3.5 \times 10^5\text{ Pa}}{2.0 \times 10^5\text{ Pa}} = 525\text{ K}\).

4. Convert the final temperature back to degrees Celsius:
\(\theta_2 = 525 - 273 = 252^\circ\text{C}\).

B is the correct answer. 1 mark is awarded for converting the temperatures to Kelvin, applying the ideal gas equation of state, and correctly converting back to Celsius.

1. The general sinusoidal alternating voltage equation is given by:
\(V = V_0 \sin(\omega t)\)

2. Calculate the peak voltage \(V_0\):
\(V_0 = \sqrt{2} \times V_{\text{rms}} = \sqrt{2} \times 12.0\text{ V} \approx 16.97\text{ V} \approx 17.0\text{ V}\).

3. Calculate the angular frequency \(\omega\):
\(\omega = 2\pi f = 2\pi \times 50.0\text{ Hz} = 100\pi \approx 314\text{ rad s}^{-1}\).

4. Write the complete expression:
\(V = 17.0 \sin(314 t)\).

D is the correct answer. 1 mark is awarded for obtaining both correct values of peak voltage and angular frequency.

The fractional uncertainty in resistivity \(\rho\) is given by:

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

First, calculate the individual fractional uncertainties:
- For resistance: \(\frac{\Delta R}{R} = \frac{0.05}{4.50} \approx 0.0111\) (or \(1.11\%\))
- For diameter: \(\frac{\Delta d}{d} = \frac{0.01}{0.35} \approx 0.0286\) (or \(2.86\%\))
- For length: \(\frac{\Delta L}{L} = \frac{0.002}{1.200} \approx 0.0017\) (or \(0.17\%\))

Now substitute these values into the uncertainty equation:

\(\frac{\Delta \rho}{\rho} = 0.0111 + 2(0.0286) + 0.0017 = 0.0111 + 0.0571 + 0.0017 = 0.0699\)

As a percentage uncertainty, this is \(0.0699 \times 100\% \approx 7.0\%\).

1 mark for the correct answer C.
- Correctly identifies that the fractional uncertainty in diameter must be doubled because of the \(d^2\) term.
- Correctly calculates and sums the individual percentage uncertainties.

The Young modulus \(E\) of the material is given by:

\(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{e/L} = \frac{FL}{Ae}\)

Rearranging this for the extension \(e\):

\(e = \frac{FL}{AE}\)

Since the cross-sectional area \(A\) is related to the diameter \(d\) by \(A = \frac{\pi d^2}{4}\), we have:

\(e = \frac{4FL}{\pi d^2 E}\)

For wires made of the same material (same \(E\)) and subjected to the same force (same \(F\)):

\(e \propto \frac{L}{d^2}\)

Let the second wire have length \(L_2 = 2L\) and diameter \(d_2 = 0.5d\). Its extension \(e_2\) is:

\(e_2 \propto \frac{2L}{(0.5d)^2} = \frac{2L}{0.25d^2} = 8\left(\frac{L}{d^2}\right)\)

Thus, \(e_2 = 8e\).

1 mark for the correct answer C.
- Recognizes the relationship between extension, length, and diameter: \(e \propto L/d^2\).
- Applies the ratios to find the new extension of \(8e\).

According to the ideal gas equation, at constant volume, pressure is directly proportional to absolute temperature (Kelvin temperature):

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

First, convert the initial temperature to Kelvin:

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

Using the proportion to find the final temperature in Kelvin:

\(T_2 = T_1 \times \frac{P_2}{P_1} = 300\ \text{K} \times \frac{2.0 \times 10^5\ \text{Pa}}{1.2 \times 10^5\ \text{Pa}} = 500\ \text{K}\)

Convert the temperature back to degrees Celsius:

\(t_2 = 500 - 273 = 227^\circ\text{C}\).

1 mark for the correct answer C.
- Converts Celsius to Kelvin before using the gas laws.
- Correctly calculates the final temperature in Kelvin as 500 K and converts back to 227 degrees Celsius.

The gravitational potential \(\phi\) at a distance \(r\) from the center of a planet of mass \(M\) is given by:

\(\phi = -\frac{GM}{r}\)

At the surface (where \(r = R\)):

\(\phi_{\text{surface}} = -\frac{GM}{R} = -6.4 \times 10^7\ \text{J}\ \text{kg}^{-1}\)

At the orbit radius (where \(r = 4R\)):

\(\phi_{\text{orbit}} = -\frac{GM}{4R} = \frac{\phi_{\text{surface}}}{4} = -1.6 \times 10^7\ \text{J}\ \text{kg}^{-1}\)

The change in gravitational potential \(\Delta \phi\) is:

\(\Delta \phi = \phi_{\text{orbit}} - \phi_{\text{surface}} = -1.6 \times 10^7 - (-6.4 \times 10^7) = +4.8 \times 10^7\ \text{J}\ \text{kg}^{-1}\)

The gain in gravitational potential energy \(\Delta U\) is:

\(\Delta U = m \Delta \phi = 500\ \text{kg} \times 4.8 \times 10^7\ \text{J}\ \text{kg}^{-1} = 2.4 \times 10^{10}\ \text{J}\).

1 mark for the correct answer C.
- Calculates the potential at 4R correctly as one quarter of the surface potential.
- Finds the change in potential and multiplies by the mass of the satellite to obtain 2.4 x 10^10 J.

The potential divider equation for the output voltage across the fixed resistor \(R\) is:

\(V_{\text{out}} = E \times \frac{R}{R_T + R}\)

Initially, the thermistor resistance \(R_{T1} = 8.0\ \text{k}\Omega\):

\(V_{\text{out, initial}} = 12.0 \times \frac{4.0}{8.0 + 4.0} = 12.0 \times \frac{4.0}{12.0} = 4.0\ \text{V}\)

Finally, the thermistor resistance \(R_{T2} = 2.0\ \text{k}\Omega\):

\(V_{\text{out, final}} = 12.0 \times \frac{4.0}{2.0 + 4.0} = 12.0 \times \frac{4.0}{6.0} = 8.0\ \text{V}\)

Therefore, the change in output voltage is:

\(\Delta V_{\text{out}} = V_{\text{out, final}} - V_{\text{out, initial}} = 8.0\ \text{V} - 4.0\ \text{V} = +4.0\ \text{V}\) (an increase of \(4.0\ \text{V}\)).

1 mark for the correct answer B.
- Calculates the initial output voltage correctly as 4.0 V.
- Calculates the final output voltage correctly as 8.0 V.
- Calculates the correct change (increase of 4.0 V).

For an observer at rest and a source in motion, the observed frequency is given by:

- When approaching:
\(f_{\text{approach}} = f_s \left( \frac{v}{v - v_s} \right)\)

- When receding:
\(f_{\text{recede}} = f_s \left( \frac{v}{v + v_s} \right)\)

Using the given values (\(f_s = 800\ \text{Hz}\), \(v = 340\ \text{m}\ \text{s}^{-1}\), \(v_s = 30.0\ \text{m}\ \text{s}^{-1}\)):

\(f_{\text{approach}} = 800 \times \frac{340}{340 - 30} = 800 \times \frac{340}{310} \approx 877.4\ \text{Hz}\)

\(f_{\text{recede}} = 800 \times \frac{340}{340 + 30} = 800 \times \frac{340}{370} \approx 735.1\ \text{Hz}\)

Now find the difference between these two frequencies:

\(\Delta f = f_{\text{approach}} - f_{\text{recede}} = 877.4\ \text{Hz} - 735.1\ \text{Hz} = 142.3\ \text{Hz} \approx 142\ \text{Hz}\).

1 mark for the correct answer B.
- Correctly calculates both the approaching frequency (877.4 Hz) and receding frequency (735.1 Hz).
- Correctly calculates the difference as 142 Hz.

First, calculate the total binding energy of the reactants:
- Each lithium-7 nucleus contains 7 nucleons, and there are 2 lithium-7 nuclei.
- \(\text{Total binding energy of reactants} = 2 \times (7 \times 5.60\ \text{MeV}) = 78.40\ \text{MeV}\).

Next, calculate the total binding energy of the products:
- Each helium-4 nucleus contains 4 nucleons, and there are 3 helium-4 nuclei.
- Individual protons (\(^{1}_{1}\text{p}\)) have a binding energy of 0 MeV.
- \(\text{Total binding energy of products} = 3 \times (4 \times 7.07\ \text{MeV}) + 2 \times 0 = 84.84\ \text{MeV}\).

The net energy released in the reaction is the difference in binding energy between products and reactants:

\(\text{Energy released} = \text{Binding Energy of Products} - \text{Binding Energy of Reactants}\)
\(\text{Energy released} = 84.84\ \text{MeV} - 78.40\ \text{MeV} = 6.44\ \text{MeV}\).

1 mark for the correct answer B.
- Correctly calculates total reactant binding energy (78.40 MeV) and product binding energy (84.84 MeV).
- Correctly subtracts reactant binding energy from product binding energy.

The potential difference \(V\) across a discharging capacitor at any time \(t\) is given by:

\(V = V_0 e^{-t/RC}\)

The energy \(W\) stored in a capacitor is given by:

\(W = \frac{1}{2} C V^2\)

Substituting \(V\) into the energy equation:

\(W(t) = \frac{1}{2} C \left(V_0 e^{-t/RC}\right)^2 = \left(\frac{1}{2} C V_0^2\right) e^{-2t/RC} = W_0 e^{-2t/RC}\)

where \(W_0\) is the initial energy stored in the capacitor.

At \(t = 2RC\), the stored energy is \(E\):

\(E = W_0 e^{-2(2RC)/RC} = W_0 e^{-4}\)

Solving for the initial energy \(W_0\):

\(W_0 = e^4 E\).

1 mark for the correct answer B.
- Relates energy to the square of voltage, giving \(W \propto e^{-2t/RC}\).
- Correctly evaluates \(W(2RC) = W_0 e^{-4}\) and rearranges to obtain \(W_0 = e^4 E\).

First, calculate the percentage uncertainty for each of the measured quantities:

Percentage uncertainty in \(R = \frac{0.05}{2.50} \times 100\% = 2.0\%\)

Percentage uncertainty in \(d = \frac{0.01}{0.40} \times 100\% = 2.5\%\)

Percentage uncertainty in \(L = \frac{0.005}{1.250} \times 100\% = 0.4\%\)

Using the formula \(\rho = \frac{R \pi d^2}{4 L}\), the percentage uncertainty in \(\rho\) is calculated by adding the individual percentage uncertainties, multiplying the percentage uncertainty of \(d\) by 2 because it is squared in the formula:

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

\frac{\Delta \rho}{\rho} \times 100\% = 2.0\% + 2(2.5\%) + 0.4\% = 7.4\%

C is the correct answer. 1 mark for calculating the individual percentage uncertainties and summing them correctly with the factor of 2 for the diameter.

The equation of state of an ideal gas can be written as \(pV = N k T\), where \(V\) is the volume, \(k\) is the Boltzmann constant, and \(N\) is the number of molecules. Rearranging for pressure gives:

p = \frac{N k T}{V}

Since the volume \(V\) is fixed, the new pressure \(p'\) with the new number of molecules \(N' = 0.5 N\) and new temperature \(T' = 1.5 T\) is:

p' = \frac{(0.5 N) k (1.5 T)}{V} = 0.75 \frac{N k T}{V} = 0.75 p

B is the correct answer. 1 mark for applying the ideal gas equation and correctly scaling the variables.

First, calculate the redshift \(z\) of the galaxy:

z = \frac{\Delta \lambda}{\lambda_0} = \frac{672\text{ nm} - 656\text{ nm}}{656\text{ nm}} = \frac{16}{656} \approx 0.0244

The recessional velocity \(v\) is related to redshift by:

v = z c = 0.0244 \times (3.00 \times 10^8\text{ m s}^{-1}) = 7.32 \times 10^6\text{ m s}^{-1} = 7320\text{ km s}^{-1}

Using Hubble's Law, \(v = H_0 d\):

d = \frac{v}{H_0} = \frac{7320\text{ km s}^{-1}}{70\text{ km s}^{-1}\text{ Mpc}^{-1}} \approx 105\text{ Mpc}

This is approximately \(1.0 \times 10^2\text{ Mpc}\) to two significant figures.

C is the correct answer. 1 mark for finding the recessional speed from the redshift and dividing by the Hubble constant to obtain the distance.

Let's find the equivalent capacitance for all possible arrangements of three identical capacitors:
1. Three in series: \(\frac{1}{C_{\text{eq}}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} \implies C_{\text{eq}} = \frac{1}{3}C\)
2. Three in parallel: \(C_{\text{eq}} = C + C + C = 3C\)
3. Two in parallel, connected in series with the third: The parallel pair has \(2C\). In series with \(C\), we have \(\frac{1}{C_{\text{eq}}} = \frac{1}{2C} + \frac{1}{C} = \frac{3}{2C} \implies C_{\text{eq}} = \frac{2}{3}C\)
4. Two in series, connected in parallel with the third: The series pair has \(\frac{1}{2}C\). In parallel with \(C\), we have \(C_{\text{eq}} = \frac{1}{2}C + C = \frac{3}{2}C\).

Therefore, \(\frac{3}{4}C\) is not a possible value.

C is the correct answer. 1 mark for identifying that 3/4 C is not a possible equivalent capacitance for any network of three identical capacitors.

The intensity reflection coefficient \(\alpha\) at a boundary between two media with acoustic impedances \(Z_1\) and \(Z_2\) is given by:

\alpha = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2

Substituting the given values:

\alpha = \left(\frac{1.60 \times 10^6 - 1.50 \times 10^6}{1.60 \times 10^6 + 1.50 \times 10^6}\right)^2 = \left(\frac{0.10}{3.10}\right)^2 \approx 0.00104

The percentage of intensity reflected is \(0.00104 \times 100\% \approx 0.10\%\).

Since there is no absorption, the percentage of intensity transmitted is:

100\% - 0.10\% = 99.9\%

D is the correct answer. 1 mark for finding the reflection coefficient and subtracting from 100% to find the transmission percentage.

The Doppler effect formula for a moving source approaching a stationary observer is:

f_{\text{obs}} = f_s \left( \frac{v}{v - v_s} \right)

where \(f_s = 440\text{ Hz}\), \(v = 340\text{ m s}^{-1}\), and \(v_s = 25.0\text{ m s}^{-1}\).

f_{\text{obs, approaching}} = 440 \left( \frac{340}{340 - 25.0} \right) = 440 \left( \frac{340}{315} \right) \approx 475\text{ Hz}

When the car has passed and is moving away from the observer, the formula is:

f_{\text{obs, receding}} = f_s \left( \frac{v}{v + v_s} \right) = 440 \left( \frac{340}{340 + 25.0} \right) = 440 \left( \frac{340}{365} \right) \approx 410\text{ Hz}

B is the correct answer. 1 mark for correctly calculating both frequencies using the correct Doppler effect formula versions for source approaching and receding.

The strain energy \(E\) stored in a wire under elastic deformation is given by:

E = \frac{1}{2} F e

where \(F\) is the applied force and \(e\) is the extension. Since the load \(F\) is identical for both wires, the ratio of strain energies is:

\frac{E_X}{E_Y} = \frac{e_X}{e_Y}

The extension \(e\) is related to Young's modulus \(E_{\text{Young}}\) by:

e = \frac{F L}{A E_{\text{Young}}}

Since the load \(F\), initial length \(L\), and Young's modulus \(E_{\text{Young}}\) (same metal) are identical for both wires, the extension is inversely proportional to the cross-sectional area \(A\):

e \propto \frac{1}{A} \propto \frac{1}{d^2}

Thus, the ratio of the extensions is:

\frac{e_X}{e_Y} = \frac{d_Y^2}{d_X^2} = \frac{(2d)^2}{d^2} = 4

Therefore, \frac{E_X}{E_Y} = 4.

D is the correct answer. 1 mark for recognizing that strain energy is proportional to extension, and extension is inversely proportional to the square of the diameter, leading to a ratio of 4.

The peak power dissipated in the resistor is \(P_{\text{peak}} = 80.0\text{ W}\).

The peak power is given by:

P_{\text{peak}} = I_0^2 R

where \(I_0\) is the peak current. Thus:

I_0 = \sqrt{\frac{P_{\text{peak}}}{R}} = \sqrt{\frac{80.0}{20.0}} = 2.00\text{ A}

For a sinusoidal alternating current, the root-mean-square current \(I_{\text{rms}}\) is:

I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{2.00}{\sqrt{2}} \approx 1.41\text{ A}

B is the correct answer. 1 mark for finding the peak current from the peak power and then using the r.m.s. relationship.

(a) Velocity is defined as the rate of change of displacement with respect to time. Velocity has both magnitude and direction (it is a vector), whereas speed has only magnitude (it is a scalar). (b)(i) The vertical component of the initial velocity is given by \(u_y = u \sin(\theta) = 24 \sin(30^\circ) = 12.0\text{ m s}^{-1}\). (b)(ii) Using \(s = ut + \frac{1}{2}at^2\) for vertical motion, with downwards as negative: \(-15 = 12t - 4.905t^2\), which simplifies to \(4.905t^2 - 12t - 15 = 0\). Solving the quadratic equation gives \(t = \frac{12 \pm \sqrt{144 - 4(4.905)(-15)}}{2(4.905)} = \frac{12 + \sqrt{438.3}}{9.81} = 3.36\text{ s}\). (b)(iii) The horizontal velocity \(u_x = 24 \cos(30^\circ) = 20.78\text{ m s}^{-1}\). The horizontal distance is \(x = u_x \times t = 20.78 \times 3.357 = 69.8\text{ m}\). (b)(iv) Using conservation of energy, \(\frac{1}{2} m u^2 + m g h = \frac{1}{2} m v^2\). Thus, \(v = \sqrt{u^2 + 2gh} = \sqrt{24^2 + 2(9.81)(15)} = \sqrt{576 + 294.3} = 29.5\text{ m s}^{-1}\).

(a) 1 mark for defining velocity as rate of change of displacement. 1 mark for stating velocity is a vector / has direction whereas speed is a scalar / has no direction. (b)(i) 1 mark for showing \(24 \sin(30^\circ) = 12\text{ m s}^{-1}\). (b)(ii) 1 mark for setting up the equation \(-15 = 12t - 0.5(9.81)t^2\), 1 mark for solving the quadratic formula, 1 mark for correct final answer \(3.36\text{ s}\) (accept 3.4 s). (b)(iii) 1 mark for calculating \(u_x = 20.8\text{ m s}^{-1}\), 1 mark for final answer \(69.8\text{ m}\) (accept \(70\text{ m}\) for 2 s.f.). (b)(iv) 1 mark for vertical velocity component at impact \(21.0\text{ m s}^{-1}\) or using energy conservation, 1 mark for final answer \(29.5\text{ m s}^{-1}\) (accept range 29 to 30).

(a)(i) Stress is defined as force applied per unit cross-sectional area: \(\sigma = F / A\). (ii) Strain is the ratio of extension to original length: \(\varepsilon = x / L\). (b)(i) Area \(A = \pi d^2 / 4 = \pi (0.80 \times 10^{-3})^2 / 4 = 5.03 \times 10^{-7}\text{ m}^2\). (b)(ii) Tension \(T = mg = 6.5 \times 9.81 = 63.8\text{ N}\) (or \(64\text{ N}\)). (b)(iii) Young modulus \(E = \frac{\text{stress}}{\text{strain}} = \frac{T L}{A x}\). Extension \(x = \frac{T L}{A E} = \frac{63.8 \times 2.4}{(5.03 \times 10^{-7}) \times (2.0 \times 10^{11})} = 1.52 \times 10^{-3}\text{ m}\) (or \(1.5\text{ mm}\)). (b)(iv) The stress on the wire is \(\sigma = \frac{T}{A} = \frac{63.8}{5.03 \times 10^{-7}} = 1.27 \times 10^8\text{ Pa}\). Since this value is less than the yield strength of \(2.5 \times 10^8\text{ Pa}\), the deformation is within the elastic limit. Therefore, the wire will return to its original length when the mass is removed.

(a)(i) 1 mark for force per unit cross-sectional area. (a)(ii) 1 mark for change in length / extension per unit original length. (b)(i) 1 mark for formula \(A = \pi d^2/4\), 1 mark for correct value \(5.03 \times 10^{-7}\text{ m}^2\). (b)(ii) 1 mark for \(T = 63.8\text{ N}\) (or \(64\text{ N}\)). (b)(iii) 1 mark for formula rearranging for \(x\), 1 mark for substitution of values, 1 mark for final answer \(1.5\text{ mm}\) (or \(1.52 \times 10^{-3}\text{ m}\)). (b)(iv) 1 mark for calculating the stress \(1.27 \times 10^8\text{ Pa}\), 1 mark for stating that because stress is below the yield strength, the deformation is elastic and it returns to its original length.

(a) The Doppler effect is the change in the observed frequency (or wavelength) of a wave when there is relative motion between the source of the waves and the observer. (b)(i) For an approaching source, the observed frequency \(f_o\) is given by \(f_o = f_s \left(\frac{v}{v - v_s}\right)\). Substituting the values: \(710 = 650 \left(\frac{340}{340 - v_s}\right)\). Rearranging: \(340 - v_s = \frac{650 \times 340}{710} = 311.27\text{ m s}^{-1}\). Therefore, \(v_s = 340 - 311.27 = 28.7\text{ m s}^{-1}\). (b)(ii) For a receding source, \(f_o = f_s \left(\frac{v}{v + v_s}\right)\). Substituting the values: \(f_o = 650 \left(\frac{340}{340 + 28.73}\right) = 650 \left(\frac{340}{368.73}\right) = 599.4\text{ Hz}\), which rounds to \(599\text{ Hz}\). (b)(iii) The loudness of the sound decreases. This is because the sound intensity decreases as the distance between the source and the observer increases, since the energy of the sound waves spreads over an increasingly large spherical surface area.

(a) 1 mark for stating that there is a change in observed frequency / wavelength, 1 mark for mentioning this is due to the relative motion between source and observer. (b)(i) 1 mark for using the correct Doppler formula for an approaching source, 1 mark for substituting values, 1 mark for correct final answer \(28.7\text{ m s}^{-1}\) (or \(29\text{ m s}^{-1}\)). (b)(ii) 1 mark for the correct Doppler formula for a receding source, 1 mark for substituting values, 1 mark for correct final answer \(599\text{ Hz}\) (or \(600\text{ Hz}\)). (b)(iii) 1 mark for stating that the loudness decreases, 1 mark for explaining that intensity decreases as the wave energy spreads out over a larger distance.

(a)(i) Kirchhoff's first law states that the algebraic sum of currents entering a junction is equal to the algebraic sum of currents leaving that junction. This is a consequence of the conservation of electric charge. (a)(ii) Kirchhoff's second law states that in any closed loop of a circuit, the sum of the electromotive forces (e.m.f.s) is equal to the sum of the potential differences (p.d.s). This is a consequence of the conservation of energy. (b)(i) Applying Kirchhoff's first law at junction A: \(I_1 + I_2 = I_3\). (b)(ii) For Loop 1 (containing \(E_1\), \(R_1\), and \(R_3\)): \(E_1 = I_1 R_1 + I_3 R_3\), which gives \(12.0 = 3.0 I_1 + 2.0 I_3\). For Loop 2 (containing \(E_2\), \(R_2\), and \(R_3\)): \(E_2 = I_2 R_2 + I_3 R_3\), which gives \(6.0 = 6.0 I_2 + 2.0 I_3\). (b)(iii) Express \(I_1\) and \(I_2\) in terms of \(I_3\): From Loop 1, \(3.0 I_1 = 12.0 - 2.0 I_3 \implies I_1 = 4.0 - \frac{2}{3}I_3\). From Loop 2, \(6.0 I_2 = 6.0 - 2.0 I_3 \implies I_2 = 1.0 - \frac{1}{3}I_3\). Substitute these into the junction equation: \(I_3 = (4.0 - \frac{2}{3}I_3) + (1.0 - \frac{1}{3}I_3) = 5.0 - I_3\). This gives \(2 I_3 = 5.0\), so \(I_3 = 2.5\text{ A}\).

(a)(i) 1 mark for correct statement of first law, 1 mark for stating conservation of charge. (a)(ii) 1 mark for correct statement of second law, 1 mark for stating conservation of energy. (b)(i) 1 mark for correct junction equation \(I_1 + I_2 = I_3\). (b)(ii) 1 mark for Loop 1 equation: \(12 = 3I_1 + 2I_3\), 1 mark for Loop 2 equation: \(6 = 6I_2 + 2I_3\), 1 mark for attempting to solve by substitution. (b)(iii) 1 mark for correct algebraic manipulation to eliminate \(I_1\) and \(I_2\), 1 mark for correct final value \(I_3 = 2.5\text{ A}\).

(a)(i) The quark composition of a proton is \(uud\) (up, up, down). (a)(ii) The quark composition of a neutron is \(udd\) (up, down, down). (b)(i) In beta-minus decay, a down quark inside the neutron decays into an up quark. The decay equation in terms of quarks is: \(d \rightarrow u + e^{-} + \bar{\nu}_e\). (b)(ii) The fundamental force responsible for beta decay is the weak nuclear force (or weak interaction). (b)(iii) The leptons emitted are: 1. An electron (charge \(-1e\) or \(-1\), lepton number \(+1\)). 2. An electron antineutrino (charge \(0\), lepton number \(-1\)). (c)(i) An electron is classified as a lepton. (c)(ii) A neutrino is classified as a lepton.

(a)(i) 1 mark for \(uud\). (a)(ii) 1 mark for \(udd\). (b)(i) 1 mark for showing \(d \rightarrow u\), 1 mark for showing \(+ e^{-} + \bar{\nu}_e\). (b)(ii) 1 mark for naming weak nuclear force. (b)(iii) 1 mark for identifying electron and electron antineutrino, 1 mark for correct charges (\(-1\) and \(0\)), 1 mark for correct lepton numbers (\(+1\) and \(-1\)). (c)(i) 1 mark for classifying electron as a lepton. (c)(ii) 1 mark for classifying neutrino as a lepton.

(a) The principle of conservation of energy states that the total energy of an isolated system remains constant; energy cannot be created or destroyed, it can only be converted from one form to another. (b)(i) First, find the vertical height of the slope: \(h = L \sin(\theta) = 6.0 \sin(25^\circ) = 2.536\text{ m}\). The gravitational potential energy is: \(E_p = mgh = 3.5 \times 9.81 \times 2.536 = 87.1\text{ J}\). (b)(ii) The kinetic energy at the bottom is: \(E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 3.5 \times (4.2)^2 = 1.75 \times 17.64 = 30.9\text{ J}\). (b)(iii) The work done against resistive forces is the loss of mechanical energy: \(W = E_p - E_k = 87.1 - 30.9 = 56.2\text{ J}\). (b)(iv) The average resistive force \(F\) is related to the work done against it and the distance traveled along the slope \(L\): \(W = F \times L \implies F = W / L = 56.2 / 6.0 = 9.37\text{ N}\).

(a) 2 marks for stating that energy cannot be created or destroyed, only transformed from one form to another (or total energy of a closed system remains constant). (b)(i) 1 mark for finding the height \(h = 2.54\text{ m}\), 1 mark for correct potential energy calculation \(87.1\text{ J}\) (accept \(87\text{ J}\)). (b)(ii) 1 mark for kinetic energy formula, 1 mark for correct final answer \(30.9\text{ J}\) (accept \(31\text{ J}\)). (b)(iii) 1 mark for recognizing that work done is \(E_p - E_k\), 1 mark for final answer \(56.2\text{ J}\) (or \(56\text{ J}\)). (b)(iv) 1 mark for formula \(W = F \times d\), 1 mark for final answer \(9.37\text{ N}\) (accept range 9.3 to 9.4).

### **Step-by-Step Solution**

1. **Data Collection and Table construction:**
* For a rule suspended by two vertical threads, as the thread spacing \(d\) increases, the restoring torque increases, which leads to a decrease in the period of oscillation \(T\).
* Example measurements:
* \(d = 0.100\text{ m} \implies 1/d = 10.0\text{ m}^{-1} \implies t_{10} = 17.0\text{ s} \implies T = 1.70\text{ s}\)
* \(d = 0.150\text{ m} \implies 1/d = 6.67\text{ m}^{-1} \implies t_{10} = 12.0\text{ s} \implies T = 1.20\text{ s}\)
* \(d = 0.200\text{ m} \implies 1/d = 5.00\text{ m}^{-1} \implies t_{10} = 9.5\text{ s} \implies T = 0.95\text{ s}\)
* \(d = 0.300\text{ m} \implies 1/d = 3.33\text{ m}^{-1} \implies t_{10} = 7.0\text{ s} \implies T = 0.70\text{ s}\)
* \(d = 0.400\text{ m} \implies 1/d = 2.50\text{ m}^{-1} \implies t_{10} = 5.8\text{ s} \implies T = 0.58\text{ s}\)
* \(d = 0.450\text{ m} \implies 1/d = 2.22\text{ m}^{-1} \implies t_{10} = 5.3\text{ s} \implies T = 0.53\text{ s}\)

2. **Graph Plotting:**
* Plot \(T / \text{s}\) on the vertical axis against \((1/d) / \text{m}^{-1}\) on the horizontal axis.
* The points will form a clear linear trend.
* Draw a straight line of best fit through the points.

3. **Gradient and Intercept Calculations:**
* **Gradient:** Choose two points on the line that are far apart (e.g., \((2.50, 0.58)\) and \((10.0, 1.70)\)).
\[\text{Gradient} = \frac{1.70 - 0.58}{10.0 - 2.50} = \frac{1.12}{7.50} \approx 0.149\text{ s m}\]
* **y-intercept:** Since the horizontal axis does not start at \(0\), calculate \(Q\) using the equation \(T = m(1/d) + c\):
\[0.58 = 0.149 \times 2.50 + Q \implies Q = 0.58 - 0.3725 \approx 0.21\text{ s}\]

4. **Determination of Constants:**
* \(P = \text{gradient} \approx 0.15\text{ s m}\)
* \(Q = \text{y-intercept} \approx 0.21\text{ s}\)

**Data Collection (4 Marks):**
* **[1M]** At least 6 sets of readings of \(d\) and \(t\) recorded.
* **[1M]** Range of \(d\) includes a small value (\(\le 12.0\text{ cm}\)) and a large value (\(\ge 40.0\text{ cm}\)).
* **[1M]** Correct trend: \(T\) decreases as \(d\) increases.
* **[1M]** Quality of data: scatter of points about the line of best fit is minimal.

**Table Quality (4 Marks):**
* **[1M]** Column headings are complete with appropriate units: \(d/\text{m}\), \((1/d)/\text{m}^{-1}\), \(t/\text{s}\), and \(T/\text{s}\).
* **[1M]** Raw values of \(d\) recorded to the nearest millimeter.
* **[1M]** Stopwatch readings recorded to 0.1 s or 0.01 s.
* **[1M]** Calculated column \(1/d\) is given to the same or one more significant figure as the raw \(d\).

**Graph and Line of Best Fit (4 Marks):**
* **[1M]** Sensible scales used (points cover more than half the grid, no inconvenient fractions).
* **[1M]** All points plotted correctly to within half a small square.
* **[1M]** Line of best fit is drawn with a balanced distribution of points above and below the line.
* **[1M]** Points are clearly identified (e.g., small crosses or circled dots).

**Gradient and Intercept (4 Marks):**
* **[1M]** Large triangle used for gradient, hypotenuse length must be greater than half the length of the line of best fit.
* **[1M]** Values read correctly from the graph and gradient calculated correctly.
* **[1M]** Intercept calculated using a point on the line substituted into \(y = mx + c\) (unless the x-axis starts at zero).
* **[1M]** Precision of gradient and intercept matches the graphical data (typically 2 or 3 s.f.).

**Analysis and Constants (4 Marks):**
* **[1M]** Value of \(P\) equated to the gradient.
* **[1M]** Value of \(Q\) equated to the y-intercept.
* **[1M]** Correct units stated: \(P\) has unit \(\text{s m}\) and \(Q\) has unit \(\text{s}\).
* **[1M]** Final values of \(P\) and \(Q\) given with appropriate significant figures (2 or 3 s.f.).

### **Step-by-Step Solution**

1. **Diameter and Area Calculation:**
* External diameter of the boiling tube: \(D = 2.80\text{ cm}\).
* Cross-sectional area: \(A = \frac{\pi D^2}{4} = \frac{\pi (2.80)^2}{4} = 6.16\text{ cm}^2\).

2. **Initial Depth and Uncertainty:**
* Depth of immersion: \(h_1 = 8.5\text{ cm}\).
* Absolute uncertainty in reading depth against a scale through a curved glass beaker is typically \(1\text{ mm to }2\text{ mm} = 0.1\text{ to }0.2\text{ cm}\).
* Percentage uncertainty:
\[\frac{0.1\text{ cm}}{8.5\text{ cm}} \times 100\% \approx 1.2\%\]

3. **First Measurement with \(m = 20\text{ g}\):**
* \(h_2 = 11.7\text{ cm}\).
* \(\Delta h = 11.7 - 8.5 = 3.2\text{ cm}\).
* First value of \(k\):
\[k_1 = \frac{3.2\text{ cm} \times 6.16\text{ cm}^2}{20\text{ g}} = 0.986\text{ cm}^3 \text{g}^{-1}\]

4. **Second Measurement with \(m = 40\text{ g}\):**
* \(h_2 = 14.9\text{ cm}\).
* \(\Delta h = 14.9 - 8.5 = 6.4\text{ cm}\).
* Second value of \(k\):
\[k_2 = \frac{6.4\text{ cm} \times 6.16\text{ cm}^2}{40\text{ g}} = 0.986\text{ cm}^3 \text{g}^{-1}\]

5. **Comparing \(k_1\) and \(k_2\):**
* Percentage difference:
\[\text{Percentage Difference} = \frac{|0.986 - 0.986|}{0.986} \times 100\% = 0\%\]
* Criterion: 10%.
* Since \(0\% < 10\%\), the experimental results strongly support the suggestion that \(k\) is constant.

**Measurements and Uncertainty (5 Marks):**
* **[1M]** Value of \(D\) recorded to at least 0.1 mm (or 0.01 cm) with appropriate unit.
* **[1M]** Value of \(h_1\) recorded to the nearest millimeter with unit.
* **[1M]** Correct calculation of percentage uncertainty in \(h_1\) using an absolute uncertainty of \(1\text{ mm}\) or \(2\text{ mm}\).
* **[1M]** Area \(A\) calculated correctly with consistent units.
* **[1M]** Raw depth \(h_2\) and calculated \(\Delta h\) recorded correctly with consistent units.

**Constant Determination and Comparison (7 Marks):**
* **[1M]** First value of \(k\) calculated correctly with unit (e.g., \(\text{cm}^3 \text{g}^{-1}\) or \(\text{m}^3 \text{kg}^{-1}\)).
* **[1M]** Second mass \(m = 40\text{ g}\) used and corresponding second depth \(h_2\) recorded.
* **[1M]** Second change in depth \(\Delta h\) calculated correctly.
* **[1M]** Second value of \(k\) calculated correctly.
* **[1M]** Percentage difference between \(k_1\) and \(k_2\) calculated correctly using: \(|k_1 - k_2| / k_1 \times 100\%\).
* **[1M]** Sensible criterion specified (e.g., 10% or 15%).
* **[1M]** Clear conclusion comparing the percentage difference to the criterion and stating whether the suggestion is supported.

**Limitations and Improvements (8 Marks):**
* **[4M]** Four distinct difficulties/limitations identified:
* 1. Parallax error when reading ruler scale against the water meniscus/tube bottom.
* 2. Meniscus of water makes it hard to locate the exact water surface line.
* 3. Boiling tube does not float perfectly vertically (it tilts slightly).
* 4. Friction/surface tension forces when the tube touches the beaker walls.
* **[4M]** Four corresponding improvements described:
* 1. Use a travelling microscope or a height gauge to align readings precisely.
* 2. Add a surfactant (e.g., dishwashing liquid) to minimize meniscus curvature or use dye.
* 3. Place slotted masses symmetrically inside the tube, or use a guide to maintain vertical alignment without friction.
* 4. Use a wider beaker or container to prevent the tube from touching the walls.

(a) Gravitational potential at a point is defined as the work done per unit mass in bringing a small test mass from infinity to that point.

(b) The formula for gravitational potential is \(\phi = -\frac{GM}{R}\).
Substituting the values:
\(\phi = -\frac{6.67 \times 10^{-11} \times 4.8 \times 10^{16}}{1.2 \times 10^4}\)
\(\phi = -2.67 \times 10^2\text{ J kg}^{-1}\) (or \(-2.7 \times 10^2\text{ J kg}^{-1}\)).

(c) To escape to infinity (where kinetic energy and potential energy are zero):
Total energy at surface = Total energy at infinity
\(\frac{1}{2} m v^2 + m \phi = 0\)
\(\frac{1}{2} v^2 = -\phi\)
\(v = \sqrt{-2\phi}\).

(d) Using the expression from (c):
\(v = \sqrt{-2 \times (-2.67 \times 10^2)} = \sqrt{534} \approx 23.1\text{ m s}^{-1}\) (or \(23\text{ m s}^{-1}\)).

(e) Due to the low escape velocity of the asteroid (approx. \(23\text{ m s}^{-1}\)), the average (r.m.s.) speed of gas molecules at even very low temperatures is significantly higher than this value. Hence, most gas molecules will exceed the escape velocity and escape into space.

(a)
- Work done per unit mass [1]
- In bringing a test mass from infinity to the point [1]

(b)
- Formula \(\phi = -GM/R\) used with correct values [1]
- Value: \(-2.7 \times 10^2\text{ J kg}^{-1}\) (must include negative sign and correct unit) [1]

(c)
- Equating gain in gravitational potential energy to loss in kinetic energy: \(\Delta E_k + \Delta E_p = 0\) or equivalent [1]
- Clear algebraic manipulation showing \(v = \sqrt{-2\phi}\) [1]

(d)
- Substitution of \(\phi\) from (b) into formula [1]
- Value: \(23\text{ m s}^{-1}\) (allow \(23.1\text{ m s}^{-1}\)) [1]

(e)
- Statement that the r.m.s. speed of gas molecules is higher than the escape velocity [1]
- Therefore, gas molecules will overcome the weak gravitational pull and escape [1]

(a) Internal energy is the sum of the random distribution of kinetic and potential energies associated with the molecules of a system.

(b) The first law of thermodynamics is written as \(\Delta U = q + w\), where:
- \(\Delta U\) is the change in internal energy
- \(q\) is the thermal energy supplied to the system
- \(w\) is the work done on the system.

(c)(i) Since the gas absorbs heat, \(q = +450\text{ J}\).
Since the gas expands and does work on the surroundings, work is done *by* the gas, so \(w = -180\text{ J}\).
\(\Delta U = q + w = 450 - 180 = +270\text{ J}\).

(c)(ii) For an ideal gas, there are no intermolecular forces, which means the potential energy of the molecules is zero. The internal energy consists entirely of random translational kinetic energy. Since the average kinetic energy of the molecules is directly proportional to the thermodynamic temperature, an increase in internal energy results in an increase in temperature.

(c)(iii) With the piston fixed, volume change \(\Delta V = 0\), so the work done is \(w = 0\).
Since \(150\text{ J}\) of thermal energy is removed, \(q = -150\text{ J}\).
Change in internal energy during this step is \(\Delta U_2 = -150\text{ J}\).
Total overall change in internal energy from the start is:
\(\Delta U_{\text{total}} = +270\text{ J} - 150\text{ J} = +120\text{ J}\).

(a)
- Sum of random distribution of kinetic and potential energies [1]
- Of the molecules/atoms of the system [1]

(b)
- Formula \(\Delta U = q + w\) (or alternative sign convention if correctly defined) [1]
- Correct definition of all symbols [1]

(c)(i)
- Recognition of work done by gas as negative and thermal energy absorbed as positive [1]
- Calculation: \(+270\text{ J}\) [1]

(c)(ii)
- Mention that for an ideal gas, potential energy is zero, so internal energy is only kinetic energy [1]
- Average kinetic energy of molecules is proportional to thermodynamic temperature [1]

(c)(iii)
- Recognition that work done is zero because volume is constant [1]
- Net overall calculation: \(270 - 150 = +120\text{ J}\) [1]

(a) The two conditions for simple harmonic motion are:
1. Acceleration is directly proportional to displacement from the equilibrium position.
2. Acceleration is always directed towards the equilibrium position.

(b)(i) \(\omega = 2\pi f = 2 \pi \times 2.5 = 15.7\text{ rad s}^{-1}\) (or \(16\text{ rad s}^{-1}\)).

(b)(ii) \(v_{\text{max}} = \omega x_0 = 15.7 \times 0.040 = 0.628\text{ m s}^{-1}\) (or \(0.63\text{ m s}^{-1}\)).

(b)(iii) \(E_{\text{total}} = \frac{1}{2} m v_{\text{max}}^2 = 0.5 \times 0.35 \times (0.628)^2 = 0.069\text{ J}\) (or \(6.9 \times 10^{-2}\text{ J}\)).

(b)(iv) The kinetic energy varies parabolically with displacement, reaching a maximum of \(0.069\text{ J}\) at \(x = 0\), and falling to zero at the amplitudes \(x = \pm 4.0\text{ cm}\). The sketch is a downward-opening parabola symmetric about the y-axis.

(a)
- Acceleration proportional to displacement [1]
- Acceleration (always) directed towards equilibrium position / in opposite direction to displacement [1]

(b)(i)
- Formula \(\omega = 2\pi f\) used [1]
- Answer: \(16\text{ rad s}^{-1}\) or \(15.7\text{ rad s}^{-1}\) [1]

(b)(ii)
- Formula \(v_{\text{max}} = \omega x_0\) used [1]
- Answer: \(0.63\text{ m s}^{-1}\) (allow \(0.628\text{ m s}^{-1}\)) [1]

(b)(iii)
- Formula \(E_{\text{total}} = \frac{1}{2} m \omega^2 x_0^2\) or \(\frac{1}{2} m v_{\text{max}}^2\) used [1]
- Answer: \(0.069\text{ J}\) (allow \(6.9 \times 10^{-2}\text{ J}\) or \(0.070\text{ J}\) from rounded values) [1]

(b)(iv)
- Symmetrical inverted parabola with vertex on y-axis [1]
- Peak labeled as \(0.069\text{ J}\) (or ECF) at \(x = 0\) and intercepts on x-axis labeled at \(\pm 4.0\text{ cm}\) (or \(\pm 0.040\text{ m}\)) [1]

(a) Capacitance is defined as charge per unit potential difference: \(C = Q/V\).

(b)(i) The decay of potential difference during discharge is given by:
\(V = V_0 e^{-\frac{t}{RC}}\)
Substitute the given values:
\(3.0 = 9.0 e^{-\frac{6.0}{47000 C}}\)
\(\frac{1}{3} = e^{-\frac{6.0}{47000 C}}\)
Taking natural logarithms:
\(\ln\left(\frac{1}{3}\right) = -\frac{6.0}{47000 C}\)
\(-1.0986 = -\frac{6.0}{47000 C}\)
\(C = \frac{6.0}{47000 \times 1.0986} = 1.162 \times 10^{-4}\text{ F} \approx 1.2 \times 10^{-4}\text{ F}\).

(b)(ii) Charge \(Q = C V\) at \(t = 6.0\text{ s}\):
Using the calculated value \(C = 1.16 \times 10^{-4}\text{ F}\):
\(Q = 1.162 \times 10^{-4} \times 3.0 = 3.49 \times 10^{-4}\text{ C}\) (or \(3.5 \times 10^{-4}\text{ C}\)).
Using \(1.2 \times 10^{-4}\text{ F}\): \(Q = 3.6 \times 10^{-4}\text{ C}\).

(b)(iii) Initial energy stored in capacitor at \(t = 0\):
\(E_0 = \frac{1}{2} C V_0^2 = 0.5 \times 1.162 \times 10^{-4} \times 9.0^2 = 4.71 \times 10^{-3}\text{ J}\).
Final energy stored in capacitor at \(t = 6.0\text{ s}\):
\(E_f = \frac{1}{2} C V^2 = 0.5 \times 1.162 \times 10^{-4} \times 3.0^2 = 5.23 \times 10^{-4}\text{ J}\).
Energy dissipated in resistor:
\(E_{\text{dissipated}} = E_0 - E_f = 4.71 \times 10^{-3} - 5.23 \times 10^{-4} = 4.19 \times 10^{-3}\text{ J}\) (or \(4.2 \times 10^{-3}\text{ J}\)).

(a)
- Charge per (unit) potential difference / ratio of charge to potential difference [1]
- Correct symbols defined or phrase written clearly [1]

(b)(i)
- Use of \(V = V_0 e^{-t/RC}\) [1]
- Correct substitution of values: \(3.0 = 9.0 e^{-6/(47000 C)}\) [1]
- Finding \(C = 1.16 \times 10^{-4}\text{ F}\) showing clearly how the approximation \(1.2 \times 10^{-4}\text{ F}\) is reached [1]

(b)(ii)
- Use of \(Q = CV\) [1]
- Answer: \(3.5 \times 10^{-4}\text{ C}\) (accept \(3.6 \times 10^{-4}\text{ C}\) if using \(1.2 \times 10^{-4}\text{ F}\)) [1]

(b)(iii)
- Calculation of initial energy \(E_0 = 4.7 \times 10^{-3}\text{ J}\) or final energy \(E_f = 5.2 \times 10^{-4}\text{ J}\) [1]
- Subtraction of final energy from initial energy [1]
- Answer: \(4.2 \times 10^{-3}\text{ J}\) (accept \(4.1 \times 10^{-3}\text{ J}\) to \(4.3 \times 10^{-3}\text{ J}\)) [1]

(a) Magnetic flux linkage is the product of the magnetic flux and the number of turns of the coil, i.e., \(N\Phi = BAN\cos\theta\), where \(B\) is flux density, \(A\) is area, and \(N\) is number of turns.

(b) Area \(A = \pi r^2 = \pi \times (0.032)^2 = 3.217 \times 10^{-3}\text{ m}^2\).
Initially, \(\theta = 0\), so:
\(\text{Flux linkage} = BAN = 0.085 \times 3.217 \times 10^{-3} \times 250 = 0.06836\text{ Wb-turns}\) (or \(6.8 \times 10^{-2}\text{ Wb-turns}\)).

(c)(i) According to Faraday's law, an e.m.f. is induced when there is a change of magnetic flux linkage. During the rotation, the angle between the magnetic field and the area vector changes, causing the flux linkage to drop to zero, hence inducing an e.m.f.

(c)(ii) Average induced e.m.f. \(E = \frac{\Delta(BAN)}{\Delta t} = \frac{0.06836 - 0}{0.15} = 0.456\text{ V} \approx 0.46\text{ V}\).

(c)(iii) The flux linkage at any instant is \(N\Phi = BAN \cos(\omega t)\). The induced e.m.f. is the rate of change of flux linkage: \(e = -\frac{d(N\Phi)}{dt} = BAN\omega \sin(\omega t)\).
As the coil rotates from \(0^\circ\) to \(90^\circ\), the term \(\sin(\omega t)\) increases from \(0\) to \(1\). Therefore, the magnitude of the induced e.m.f. increases from zero to its maximum value at the end of the rotation.

(a)
- Product of magnetic flux and number of turns [1]
- Where magnetic flux is the product of magnetic flux density and area normal to the field [1]

(b)
- Area calculated correctly: \(A = 3.22 \times 10^{-3}\text{ m}^2\) [1]
- Answer: \(6.8 \times 10^{-2}\text{ Wb}\) (or \(\text{Wb-turns}\)) [1]

(c)(i)
- Reference to Faraday's law: induced e.m.f. is proportional to the rate of change of magnetic flux linkage [1]
- Rotation changes the angle, causing a change in flux linkage over time [1]

(c)(ii)
- Use of \(E = \Delta \Phi / \Delta t\) [1]
- Answer: \(0.46\text{ V}\) (allow \(0.45\text{ V}\) to \(0.46\text{ V}\)) [1]

(c)(iii)
- Statement that e.m.f. increases during the rotation [1]
- Explanation using \(\sin\theta\) dependency or rates of cutting flux lines being maximum when plane is parallel to field [1]

(a)
- A tracer containing a positron-emitting radionuclide (like Fluorine-18) is injected into the patient's body.
- The tracer accumulates in high-energy-demand tissues (such as tumor cells).
- The radionuclide decays by emitting positrons.
- These positrons travel a very short distance before colliding with electrons in the tissue.
- Annihilation occurs, producing two gamma-ray photons traveling in opposite directions.
- These photons are detected by a ring of detectors surrounding the patient, and a computer processes the arrival times to locate the tumor.

(b)(i) To conserve momentum. Before annihilation, the total momentum of the stationary electron-positron system is zero. If only one photon were produced, it would carry momentum, violating conservation of linear momentum. Therefore, two photons must be emitted in opposite directions with equal and opposite momenta.

(b)(ii) The total rest mass of one particle (either electron or positron) is converted entirely into the energy of one of the two photons.
Using Einstein's mass-energy equation:
\(E = m c^2\)
\(E = 9.11 \times 10^{-31} \times (3.00 \times 10^8)^2\)
\(E = 8.199 \times 10^{-14}\text{ J}\)
To convert Joules to MeV:
\(E = \frac{8.199 \times 10^{-14}}{1.60 \times 10^{-19}}\text{ eV} = 5.124 \times 10^5\text{ eV} = 0.512\text{ MeV}\) (or \(0.51\text{ MeV}\)).

(a)
- Tracer injected into patient and accumulates in target tissue/tumor [1]
- Positrons emitted during decay of radionuclide [1]
- Positron annihilates with an electron in the tissue [1]
- Two gamma photons emitted in opposite directions and detected by external detectors [1]

(b)(i)
- Total momentum before annihilation is zero [1]
- Two photons traveling in opposite directions are required to conserve momentum [1]

(b)(ii)
- Use of \(E = mc^2\) with electron mass [1]
- Value in Joules: \(8.2 \times 10^{-14}\text{ J}\) [1]
- Division by \(1.6 \times 10^{-19}\) to convert to eV [1]
- Final answer: \(0.51\text{ MeV}\) (allow \(0.512\text{ MeV}\)) [1]

(a) Hubble's law states that the recession speed \(v\) of a galaxy is directly proportional to its distance \(d\) from Earth: \(v = H_0 d\), where \(H_0\) is the Hubble constant.

(b)(i) Redshift \(z\) is defined as:
\(z = \frac{\Delta \lambda}{\lambda} = \frac{682.1 - 656.3}{656.3} = \frac{25.8}{656.3} = 0.0393 \approx 0.039\) (shown).

(b)(ii) The recession velocity \(v\) is given by:
\(v = z c = 0.0393 \times 3.00 \times 10^8 = 1.18 \times 10^7\text{ m s}^{-1}\) (or \(1.2 \times 10^7\text{ m s}^{-1}\)).

(b)(iii) From Hubble's law \(v = H_0 d\):
\(d = \frac{v}{H_0} = \frac{1.179 \times 10^7}{2.2 \times 10^{-18}} = 5.36 \times 10^{24}\text{ m}\) (or \(5.4 \times 10^{24}\text{ m}\)).

(c) The Big Bang theory predicts that the early Universe was extremely hot and dense, filled with high-energy short-wavelength radiation. As the Universe expanded, this radiation was redshifted (stretched) to longer wavelengths. Today, we detect this as a highly isotropic microwave background radiation corresponding to a blackbody temperature of about \(2.7\text{ K}\), which strongly supports the expansion of the Universe from a hot origin.

(a)
- Recession speed of a galaxy is proportional to its distance from Earth [1]
- Equation \(v = H_0 d\) with symbols defined [1]

(b)(i)
- Use of \(z = \Delta \lambda / \lambda\) [1]
- Substitution yielding \(0.0393\) (must show calculation with sufficient precision to confirm \(0.039\)) [1]

(b)(ii)
- Use of \(v = zc\) [1]
- Answer: \(1.2 \times 10^7\text{ m s}^{-1}\) (allow \(1.18 \times 10^7\text{ m s}^{-1}\)) [1]

(b)(iii)
- Rearrangement \(d = v/H_0\) [1]
- Answer: \(5.4 \times 10^{24}\text{ m}\) (allow \(5.36 \times 10^{24}\text{ m}\)) [1]

(c)
- Early Universe was very hot/dense and expansion stretched the wavelength of primordial radiation [1]
- Observed today in microwave region / at \(2.7\text{ K}\) / from all directions uniformly [1]

(a) The decay constant \(\lambda\) is the probability of decay per unit time of a nucleus.

(b)(i) Activity is related to the number of nuclei by \(A = \lambda N\).
\(\lambda = \frac{A}{N} = \frac{3.5 \times 10^5}{4.2 \times 10^{11}} = 8.33 \times 10^{-7}\text{ s}^{-1}\) (or \(8.3 \times 10^{-7}\text{ s}^{-1}\)).

(b)(ii) The half-life is given by:
\(T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{8.33 \times 10^{-7}} = 8.32 \times 10^5\text{ s}\).
To convert seconds into days:
\(T_{1/2} = \frac{8.32 \times 10^5}{3600 \times 24} = 9.63\text{ days}\) (or \(9.6\text{ days}\)).

(c) The activity as a function of time is \(A = A_0 e^{-\lambda t}\).
\(1.0 \times 10^4 = 3.5 \times 10^5 e^{-\lambda t}\)
\(\frac{1.0 \times 10^4}{3.5 \times 10^5} = e^{-\lambda t}\)
\(\frac{1}{35} = e^{-\lambda t}\)
\(\ln(35) = \lambda t\)
\(t = \frac{3.555}{\lambda} = \frac{3.555}{8.33 \times 10^{-7}} = 4.268 \times 10^6\text{ s}\).
In days:
\(t = \frac{4.268 \times 10^6}{86400} = 49.4\text{ days}\) (or \(49\text{ days}\)).

(a)
- Probability of decay of a nucleus [1]
- Per unit time [1]

(b)(i)
- Use of \(A = \lambda N\) [1]
- Answer: \(8.3 \times 10^{-7}\text{ s}^{-1}\) (allow \(8.33 \times 10^{-7}\text{ s}^{-1}\)) [1]

(b)(ii)
- Use of \(T_{1/2} = \ln 2 / \lambda\) [1]
- Division by \(86400\) to convert seconds to days [1]
- Answer: \(9.6\text{ days}\) (allow \(9.63\text{ days}\)) [1]

(c)
- Use of \(A = A_0 e^{-\lambda t}\) (or fractional method) [1]
- Calculation of time in seconds: \(4.3 \times 10^6\text{ s}\) [1]
- Conversion to days: \(49\text{ days}\) (allow \(49.4\text{ days}\)) [1]

(a) Capacitance is defined as the charge per unit potential difference: \( C = \frac{Q}{V} \).
(b) When the potential difference decreases to half of its initial value, \( V = 0.5 V_0 \). Substituting this into the discharge equation: \( 0.5 V_0 = V_0 e^{-\frac{t_{1/2}}{RC}} \) which simplifies to \( 0.5 = e^{-\frac{t_{1/2}}{RC}} \). Taking the natural logarithm of both sides: \( \ln(0.5) = -\frac{t_{1/2}}{RC} \). Since \( \ln(0.5) = -\ln 2 \), we have: \( -\ln 2 = -\frac{t_{1/2}}{RC} \) which yields \( t_{1/2} = RC \ln 2 \).
(c)(i) The time constant \( \tau \) is given by: \( \tau = RC \). Substituting the values: \( \tau = 15 \times 10^3\ \Omega \times 470 \times 10^{-6}\text{ F} = 7.05\text{ s} \approx 7.1\text{ s} \).
(c)(ii) The initial charge is \( Q_0 = C V_0 = 470 \times 10^{-6}\text{ F} \times 9.0\text{ V} = 4.23 \times 10^{-3}\text{ C} \). The charge \( Q \) at time \( t = 5.0\text{ s} \) is: \( Q = Q_0 e^{-\frac{t}{RC}} = 4.23 \times 10^{-3} \times e^{-\frac{5.0}{7.05}} = 4.23 \times 10^{-3} \times 0.492 = 2.08 \times 10^{-3}\text{ C} \approx 2.1\text{ mC} \).
(c)(iii) The initial energy stored is: \( E_0 = \frac{1}{2} C V_0^2 = 0.5 \times 470 \times 10^{-6} \times 9.0^2 = 1.90 \times 10^{-2}\text{ J} \). The final energy stored at \( t = 5.0\text{ s} \) is: \( E = \frac{1}{2} C V^2 \). Since \( V = 9.0 \times e^{-\frac{5.0}{7.05}} = 4.43\text{ V} \), \( E = 0.5 \times 470 \times 10^{-6} \times 4.43^2 = 4.61 \times 10^{-3}\text{ J} \). The energy lost is: \( \Delta E = E_0 - E = 1.90 \times 10^{-2}\text{ J} - 4.61 \times 10^{-3}\text{ J} = 1.44 \times 10^{-2}\text{ J} \approx 1.4 \times 10^{-2}\text{ J} \) (or \( 14\text{ mJ} \)).

(a) C1: charge / potential difference (or ratio of charge to potential difference, do not accept product or work). [1]
(b) M1: Substitution of \( V = 0.5 V_0 \) into the equation: \( 0.5 = e^{-\frac{t_{1/2}}{RC}} \). A1: Correct algebraic working shown to arrive at \( t_{1/2} = RC \ln 2 \). [2]
(c)(i) C1: Recall of formula \( \tau = RC \) and substitution of values. A1: \( 7.1\text{ s} \) (accept \( 7.05\text{ s} \)). [2]
(c)(ii) C1: Formula \( Q = Q_0 e^{-t/RC} \) or equivalent, with substitution of \( Q_0 = 4.23 \times 10^{-3}\text{ C} \). A1: \( 2.1 \times 10^{-3}\text{ C} \) (or \( 2.1\text{ mC} \) or \( 2.08 \times 10^{-3}\text{ C} \)). [2]
(c)(iii) C1: Correct calculation of initial energy stored \( E_0 = 1.90 \times 10^{-2}\text{ J} \) (or \( 1.9 \times 10^{-2}\text{ J} \)). C1: Correct calculation of final energy stored \( E = 4.61 \times 10^{-3}\text{ J} \) (or using \( E = \frac{Q^2}{2C} = 4.60 \times 10^{-3}\text{ J} \)). A1: \( 1.4 \times 10^{-2}\text{ J} \) or \( 14\text{ mJ} \) (accept \( 1.44 \times 10^{-2}\text{ J} \)). [3]

(a) Hubble's law states that the recession speed \( v \) of a distant galaxy is directly proportional to its distance \( d \) from the observer. Mathematically: \( v = H_0 d \) where \( H_0 \) is the Hubble constant, \( v \) is the velocity of recession, and \( d \) is the distance.
(b)(i) The phenomenon is redshift. It indicates that the galaxy is moving away from the Earth (receding).
(b)(ii) Redshift is calculated using: \( z = \frac{\Delta \lambda}{\lambda} \). Here, \( \Delta \lambda = 675.4\text{ nm} - 656.3\text{ nm} = 19.1\text{ nm} \). Thus, \( z = \frac{19.1}{656.3} = 0.02911 \approx 0.0291 \).
(b)(iii) The recession speed \( v \) is related to redshift by: \( v = z c \) where \( c = 3.00 \times 10^8\text{ m s}^{-1} \). Substituting the values: \( v = 0.02911 \times 3.00 \times 10^8 = 8.73 \times 10^6\text{ m s}^{-1} \approx 8.7 \times 10^6\text{ m s}^{-1} \).
(b)(iv) Using Hubble's law: \( d = \frac{v}{H_0} \). Substituting the values: \( d = \frac{8.73 \times 10^6}{2.3 \times 10^{-18}} = 3.80 \times 10^{24}\text{ m} \approx 3.8 \times 10^{24}\text{ m} \).

(a) B1: States that recession speed (of a galaxy) is directly proportional to its distance (from Earth). B1: Identifies symbols in formula \( v = H_0 d \) (where \( v \) is recession speed, \( d \) is distance, and \( H_0 \) is the Hubble constant). [2]
(b)(i) B1: (Cosmological) Redshift. B1: Identifies that the galaxy is moving away from Earth (or receding). [2]
(b)(ii) C1: Recall and substitution of \( z = \frac{\Delta \lambda}{\lambda} \). A1: \( 0.0291 \) (accept \( 0.029 \) or \( 2.9 \times 10^{-2} \)). [2]
(b)(iii) C1: Recall of \( v = z c \) (or \( \frac{\Delta \lambda}{\lambda} = \frac{v}{c} \)). A1: \( 8.7 \times 10^6\text{ m s}^{-1} \) (or \( 8.73 \times 10^6\text{ m s}^{-1} \)). [2]
(b)(iv) C1: Recall of Hubble's law formula \( v = H_0 d \) and substitution. A1: \( 3.8 \times 10^{24}\text{ m} \) (accept \( 3.79 \times 10^{24}\text{ m} \) or \( 3.8 \times 10^{24}\text{ m} \) based on earlier rounding). [2]

### 1. Diagram and Circuit Design
- Draw a diagram showing two parallel metal plates fixed at a distance \(d\) apart using insulating spacers.
- Show the plastic sheet partially inserted between the plates by a distance \(x\).
- Include a circuit diagram showing the parallel plates connected in parallel with a digital voltmeter (or data logger / cathode-ray oscilloscope) and a discharging resistor \(R\). A switch should allow charging the plates from a DC power supply and then discharging them through \(R\).

### 2. Experimental Procedure & Control of Variables
- Keep the plate separation \(d\) constant by using non-conducting spacers at the edges.
- Keep the plate width \(w\) and length \(L\) constant.
- Measure the plate separation \(d\) using a micrometer screw gauge and the width \(w\) using a vernier caliper.
- Slide the plastic sheet into the gap between the plates by a distance \(x\). Measure \(x\) using a ruler or a scale marked directly on the plastic sheet.

### 3. Measuring the Time Constant \(\tau\)
- For each value of \(x\), close the switch to charge the capacitor to an initial potential difference \(V_0\).
- Open the charging circuit and close the discharge circuit. Record the discharge of the potential difference \(V\) across the capacitor over time \(t\) using a storage oscilloscope or a fast-sampling data logger.
- Determine \(\tau\) by plotting a graph of \(\ln(V)\) against \(t\). Since \(V = V_0 e^{-t/\tau}\), the relationship is \(\ln(V) = -\frac{t}{\tau} + \ln(V_0)\). The gradient of this graph is equal to \(-\frac{1}{\tau}\), from which \(\tau\) is calculated. Repeat this process for at least six different values of \(x\).

### 4. Analysis and Determining \(\epsilon_r\)
- Plot a graph of \(\tau\) against \(x\).
- If the relationship is valid, the graph of \(\tau\) against \(x\) will be a straight line with a non-zero y-intercept.
- Determine the gradient \(A\) of the line of best fit.
- Calculate the dielectric constant \(\epsilon_r\) using the relationship:
$$A = \frac{R \epsilon_0 w (\epsilon_r - 1)}{d} \implies \epsilon_r = 1 + \frac{A d}{R \epsilon_0 w}$$

### 5. Safety & Quality of Results
- Avoid touching the capacitor terminals during charging to prevent electric shock.
- Ensure the plastic sheet is flat and in complete contact with both plates to avoid air gaps.
- Clean the plates and dielectric sheet with a dry cloth to prevent leakage currents due to dust or moisture.

**Defining the Problem (3 marks):**
- [1] Identify \(x\) as the independent variable and \(\tau\) as the dependent variable.
- [1] State that the plate separation \(d\), plate width \(w\), and discharging resistance \(R\) must be kept constant.
- [1] State that the temperature/environmental conditions should be kept constant to prevent leakage current variations.

**Methods of Data Collection (5 marks):**
- [1] Detailed diagram of the physical arrangement showing parallel plates, the dielectric sheet, and the circuit with a charging supply, switch, and discharge resistor \(R\).
- [1] Describe how the insertion depth \(x\) is measured using a ruler or markings on the sheet.
- [1] Describe how the plate separation \(d\) is measured using a micrometer screw gauge at multiple points, and plate width \(w\) using a vernier caliper.
- [1] Describe how a storage oscilloscope or data logger is used to record the rapid discharge of voltage against time.
- [1] Explain how \(\tau\) is found: timing the decay to \(0.37 V_0\) or by plotting \(\ln(V)\) against \(t\) and finding \(\tau = -1/\text{gradient}\).

**Method of Analysis (2 marks):**
- [1] State that a graph of \(\tau\) against \(x\) should be plotted.
- [1] State that a straight line of best fit with a positive y-intercept confirms the proposed equation, and show how to calculate \(\epsilon_r = 1 + \frac{A d}{R \epsilon_0 w}\) from the gradient \(A\).

**Additional Details (5 marks for any 5 of the following):**
- [1] Use of high-value resistor \(R\) (e.g. \(M\Omega\) range) to slow down the discharge and make manual timing possible if an oscilloscope is unavailable.
- [1] Repeat measurements of \(x\) and \(\tau\) to average.
- [1] Safety precaution: Switch off power supply before adjusting the dielectric sheet; use insulated wires.
- [1] Ensure the dielectric sheet is tightly fitting to eliminate air gaps which lower effective capacitance.
- [1] Clean the metal plates with isopropyl alcohol to prevent leakage current across dirty surfaces.
- [1] Use non-conducting clamps/spacers to hold the plates firmly at a uniform distance.

### Part (a)
We calculate the values using:
- \(f^2\) divided by \(10^4\).
- \(1/I_0^2 = 1 / (I_0 \times 10^{-3})^2\).
- Absolute uncertainty: \(\Delta(1/I_0^2) \approx 2 \times \frac{\Delta I_0}{I_0} \times \frac{1}{I_0^2}\).

$$\begin{array}{|c|c|c|c|}
\hline
f \text{ / Hz} & f^2 \text{ / } 10^4\text{ s}^{-2} & I_0 \text{ / mA} & (1/I_0^2) \text{ / A}^{-2} & \Delta (1/I_0^2) \text{ / A}^{-2} \\ \hline
100 & 1.00 & 57.0 \pm 1.0 & 308 & 11 \\ \hline
120 & 1.44 & 49.0 \pm 1.0 & 416 & 17 \\ \hline
140 & 1.96 & 42.9 \pm 1.0 & 543 & 25 \\ \hline
160 & 2.56 & 38.0 \pm 1.0 & 693 & 36 \\ \hline
180 & 3.24 & 34.1 \pm 1.0 & 860 & 50 \\ \hline
200 & 4.00 & 30.9 \pm 1.0 & 1047 & 68 \\ \hline
\end{array}$$

### Part (b)
Plot \((1/I_0^2)\) against \(f^2\). Plot error bars vertically for each point. Draw the best-fit line through the points. Draw the worst acceptable line (the steepest or shallowest line passing through all error bars).

### Part (c)
Calculate the gradient \(G\) of the graph in terms of the plotted scales:
Using points on the best fit line:
\(G = \frac{1047 - 308}{4.00 - 1.00} = \frac{739}{3.00} = 246.3\text{ A}^{-2}\text{ / } (10^4\text{ s}^{-2})\)
So the true gradient \(m\) is:
\(m = 246.3 \times 10^{-4} = 0.0246\text{ A}^{-2}\text{s}^2\)

From the worst acceptable line, let's say the gradient is \(G_{\text{worst}} = 231.3\text{ A}^{-2}\text{ / } (10^4\text{ s}^{-2})\).
True \(m_{\text{worst}} = 0.0231\text{ A}^{-2}\text{s}^2\).
Uncertainty \(\Delta m = m_{\text{best}} - m_{\text{worst}} = 0.0246 - 0.0231 = 0.0015\text{ A}^{-2}\text{s}^2\).

### Part (d)
Determine the y-intercept \(c\) of the line of best fit:
\(c = y - m f^2 = 308 - 246.3 \times 1.00 = 61.7\text{ A}^{-2}\)
From the worst fit line:
\(c_{\text{worst}} = 308 - 231.3 \times 1.00 = 76.7\text{ A}^{-2}\) (if worst line goes through the top of the first error bar).
Uncertainty \(\Delta c \approx 15\text{ A}^{-2}\).

### Part (e)
1. **Inductance \(L\)**:
\(m = \frac{4\pi^2 L^2}{V_0^2} \implies L = \frac{V_0 \sqrt{m}}{2\pi}\)
\(L = \frac{6.0 \times \sqrt{0.0246}}{2\pi} = 0.150\text{ H}\)

To find the uncertainty in \(L\):
\(\frac{\Delta L}{L} = \frac{\Delta V_0}{V_0} + \frac{1}{2} \frac{\Delta m}{m} = \frac{0.2}{6.0} + 0.5 \times \frac{0.0015}{0.0246} = 0.0333 + 0.0305 = 0.0638\)
\(\Delta L = 0.0638 \times 0.150 = 0.0096\text{ H} \approx 0.010\text{ H}\)
So, \(L = 0.150 \pm 0.010\text{ H}\).

2. **Resistance \(R\)**:
\(c = \frac{R^2}{V_0^2} \implies R = V_0 \sqrt{c}\)
\(R = 6.0 \times \sqrt{61.7} = 47.1\ \Omega\)

To find the uncertainty in \(R\):
\(\frac{\Delta R}{R} = \frac{\Delta V_0}{V_0} + \frac{1}{2} \frac{\Delta c}{c} = \frac{0.2}{6.0} + 0.5 \times \frac{15}{61.7} = 0.0333 + 0.1216 = 0.155\)
\(\Delta R = 0.155 \times 47.1 = 7.3\ \Omega \approx 7\ \Omega\).
So, \(R = 47 \pm 7\ \Omega\).

**Part (a) [3 marks]**
- [1] Table completed with correct values of \(f^2\) (to 3 s.f.) and \(1/I_0^2\) (to 3 s.f.).
- [1] Absolute uncertainties in \(1/I_0^2\) calculated correctly.
- [1] All values recorded to consistent decimal places and units in table headers are correct.

**Part (b) [5 marks]**
- [1] Suitable scales chosen such that the plotted points occupy more than half the grid.
- [1] All six points plotted accurately within half a small square.
- [1] Error bars drawn accurately to scale for all points.
- [1] Best-fit straight line drawn with balanced distribution of points.
- [1] Worst acceptable line drawn passing through the top of the first error bar and the bottom of the last error bar.

**Part (c) [3 marks]**
- [1] Gradient of best-fit line calculated using points that are separated by more than half the line length.
- [1] Value of the gradient converted correctly taking into account the scale factor \(10^4\) (yielding approx. \(0.0246\text{ A}^{-2}\text{s}^2\)).
- [1] Uncertainty in gradient calculated correctly using: \(\Delta m = m_{\text{best}} - m_{\text{worst}}\).

**Part (d) [2 marks]**
- [1] y-intercept \(c\) found correctly by reading from \(x = 0\) or calculation using \(c = y - mx\).
- [1] Uncertainty in intercept determined from \(\Delta c = |c_{\text{best}} - c_{\text{worst}}|\).

**Part (e) [2 marks]**
- [1] Inductance \(L\) calculated correctly with its absolute uncertainty, including unit (H) and consistent s.f.
- [1] Resistance \(R\) calculated correctly with its absolute uncertainty, including unit (\(\Omega\)) and consistent s.f.