MetadataPastPaper.sampleTitle

1. The pion-plus (\(\pi^+\)) is a meson, consisting of a quark-antiquark pair. Its quark structure is \(u\bar{d}\).
2. Identify the lepton numbers:
- Before decay: The pion-plus (\(\pi^+\)) is a meson, so its lepton number \(L = 0\).
- After decay: The positive muon (\(\mu^+\)) is an antilepton, which has a lepton number \(L = -1\). The muon neutrino (\(\nu_{\mu}\)) is a lepton, which has a lepton number \(L = +1\).
3. Sum the lepton numbers after decay: \(-1 + 1 = 0\). Since the total lepton number is 0 both before and after the decay, lepton number is conserved.

- State quark structure of \(\pi^+\) as \(u\bar{d}\) [1 mark].
- Correctly identify lepton numbers of the decay products: \(L(\mu^+) = -1\) and \(L(\nu_{\mu}) = +1\) [1 mark].
- State that total lepton number is conserved because total lepton number remains zero before and after the decay [0.3 marks].

1. Change in momentum of the sphere:
Take upwards as positive.
u = -4.2\text{ m s}^{-1}\) (downwards) and \(v = +3.1\text{ m s}^{-1}\) (upwards).
\(\Delta p = m(v - u) = 0.15 \times (3.1 - (-4.2)) = 0.15 \times 7.3 = 1.095\text{ kg m s}^{-1}\).
2. Average net force:
\(F_{\text{net}} = \frac{\Delta p}{\Delta t} = \frac{1.095}{0.085} \approx 12.88\text{ N}\).
3. Average force exerted by the floor \(F_{\text{floor}}\):
\(F_{\text{floor}} - mg = F_{\text{net}} \Rightarrow F_{\text{floor}} = 12.88 + (0.15 \times 9.81) = 12.88 + 1.47 = 14.35\text{ N}\).
Rounding to 2 significant figures gives \(14\text{ N}\). (Accept \(13\text{ N}\) if weight is neglected).

- Calculate momentum change \(\Delta p = 1.1\text{ kg m s}^{-1}\) [1 mark].
- Calculate net average force as \(12.9\text{ N}\) [1 mark].
- Account for gravity to get the final force exerted by the floor as \(14\text{ N}\) (accept \(14.3\text{ N}\) or \(14.4\text{ N}\); accept \(13\text{ N}\) if weight is neglected) [0.3 marks].

1. Maximum power is transferred to the external resistor when the external resistance \(R\) is equal to the internal resistance \(r\) of the source: \(R = r = 1.5\ \Omega\).
2. Calculate the total circuit resistance:
\(R_{\text{total}} = R + r = 1.5 + 1.5 = 3.0\ \Omega\).
3. Calculate the circuit current:
\(I = \frac{\varepsilon}{R_{\text{total}}} = \frac{6.0}{3.0} = 2.0\text{ A}\).
4. Calculate the power dissipated in \(R\):
\(P = I^2 R = (2.0)^2 \times 1.5 = 4.0 \times 1.5 = 6.0\text{ W}\).

- State or apply the maximum power condition \(R = r = 1.5\ \Omega\) [1 mark].
- Find total resistance of \(3.0\ \Omega\) and current of \(2.0\text{ A}\) [1 mark].
- Calculate maximum power as \(6.0\text{ W}\) with unit [0.3 marks].

1. Calculate grating spacing \(d\):
\(d = \frac{1 \times 10^{-3}\text{ m}}{500} = 2.00 \times 10^{-6}\text{ m}\).
2. Use the grating equation: \(d \sin \theta = n \lambda\). For the maximum possible order of diffraction, \(\theta \le 90^{\circ}\), so \(\sin \theta \le 1\).
\(n \le \frac{d}{\lambda} = \frac{2.00 \times 10^{-6}}{632.8 \times 10^{-9}} = 3.16\).
3. Since the order \(n\) must be an integer, the maximum observable order is \(n = 3\).
4. The total number of maxima is given by \(2n + 1\) (for the central maximum, plus 3 orders on each side):
\(2(3) + 1 = 7\).

- Calculate grating spacing \(d = 2.00 \times 10^{-6}\text{ m}\) [1 mark].
- Calculate the maximum order as \(n = 3\) [1 mark].
- Calculate the total number of maxima as \(7\) [0.3 marks].

1. Calculate tension \(F\) in the wire:
\(F = mg = 8.0 \times 9.81 = 78.48\text{ N}\).
2. Calculate extension \(\Delta L\) from Young modulus definition:
\(E = \frac{F L}{A \Delta L} \Rightarrow \Delta L = \frac{F L}{A E}\)
\(\Delta L = \frac{78.48 \times 2.4}{1.2 \times 10^{-6} \times 2.0 \times 10^{11}} = 7.848 \times 10^{-4}\text{ m}\).
3. Calculate stored elastic strain energy \(E_{\text{s}}\):
\(E_{\text{s}} = \frac{1}{2} F \Delta L = 0.5 \times 78.48 \times 7.848 \times 10^{-4} = 0.03079\text{ J} \approx 0.031\text{ J}\).

- Calculate the tension force \(78.5\text{ N}\) and the wire extension \(\Delta L \approx 7.85 \times 10^{-4}\text{ m}\) [1 mark].
- Recall and use formula \(E_{\text{s}} = \frac{1}{2} F \Delta L\) or equivalent [1 mark].
- State final energy as \(0.031\text{ J}\) (or \(0.0308\text{ J}\)) [0.3 marks].

1. Find the time period \(T\) of the simple pendulum:
\(T = 2\pi \sqrt{\frac{L}{g}} = 2\pi \sqrt{\frac{1.25}{9.81}} = 2\pi \times 0.3568 = 2.242\text{ s}\).
2. Find total time for 20 oscillations:
\(t = 20 \times T = 20 \times 2.242 = 44.84\text{ s}\).
Rounding to 2 significant figures gives \(45\text{ s}\) (or \(44.9\text{ s}\) to 3 s.f.).

- Calculate the time period \(T \approx 2.24\text{ s}\) [1 mark].
- State that total time is \(20 T\) [1 mark].
- Obtain final time of \(45\text{ s}\) or \(44.9\text{ s}\) with correct units [0.3 marks].

1. The frequency of the third harmonic is three times the fundamental frequency:
\(f_3 = 3 \times f_1 = 3 \times 120 = 360\text{ Hz}\).
2. The third harmonic has three half-wavelength loops along the length of the string.
3. This creates 4 nodes in total (including the two fixed ends of the string).

- Calculate third harmonic frequency as \(360\text{ Hz}\) [1 mark].
- State that there are 4 nodes [1 mark].
- Express answer with correct unit for frequency [0.3 marks].

1. Calculate photon energy \(E = hf\):
\(E = 6.63 \times 10^{-34}\text{ J s} \times 7.5 \times 10^{14}\text{ Hz} = 4.9725 \times 10^{-19}\text{ J}\).
2. Convert work function \(\Phi\) to joules:
\(\Phi = 2.1\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 3.36 \times 10^{-19}\text{ J}\).
3. Use Einstein's photoelectric equation to find \(E_{\text{k,max}}\):
\(E_{\text{k,max}} = hf - \Phi = 4.9725 \times 10^{-19} - 3.36 \times 10^{-19} = 1.6125 \times 10^{-19}\text{ J}\).
Rounding to 2 significant figures gives \(1.6 \times 10^{-19}\text{ J}\).

- Calculate incident photon energy in joules [1 mark].
- Convert work function to joules [1 mark].
- Subtract work function from photon energy to yield \(1.6 \times 10^{-19}\text{ J}\) (or \(1.61 \times 10^{-19}\text{ J}\)) [0.3 marks].

The neutral kaon \(K^0\) has a quark structure of \(d\bar{s}\) and therefore has a strangeness of \(+1\). The pions \(\pi^+\) (\(u\bar{d}\)) and \(\pi^-\) (\(d\bar{u}\)) both have a strangeness of \(0\).

Since the total initial strangeness is \(+1\) and the total final strangeness is \(0\), strangeness is not conserved (it changes by 1).

In strong and electromagnetic interactions, strangeness must be conserved. However, the weak interaction does not conserve strangeness and can change it by \(\pm 1\), allowing the strange quark to change flavor and decay.

• [1 mark] States that strangeness is not conserved in this decay / strangeness changes by 1 (from +1 to 0).
• [1 mark] Explains that the weak interaction can violate strangeness conservation (or allows quark flavor to change), permitting the decay.

The weight of the uniform beam acts at its center of mass, which is at the midpoint: \(d = 1.20\text{ m}\) from the pivot.
Weight of the beam: \(W = m g = 15.0 \times 9.81 = 147.15\text{ N}\).

Taking moments about the pivot in equilibrium:
\(\text{Clockwise moment} = \text{Anticlockwise moment}\)
\(W \times 1.20\text{ m} = T \sin(35.0^\circ) \times 2.40\text{ m}\)

Substitute the values:
\(147.15 \times 1.20 = T \sin(35.0^\circ) \times 2.40\)
\(176.58 = 1.3766 \times T\)
\(T = \frac{176.58}{2.40 \sin(35.0^\circ)} = 128.27\text{ N}\).

To 3 significant figures, the tension is \(128\text{ N}\) (or \(128\text{ N}\) using \(g = 9.81\text{ m s}^{-2}\); if \(g = 9.8\text{ m s}^{-2}\) is used, \(T = 128\text{ N}\)).

• [1 mark] Applies the principle of moments about the pivot, yielding a correct algebraic expression such as \(15.0 \times g \times 1.20 = T \sin(35.0^\circ) \times 2.40\).
• [1 mark] Evaluates the calculation to obtain \(128\text{ N}\) (allow range \(128\text{ N}\) to \(129\text{ N}\)).

Let the initial length be \(L_0\) and the initial cross-sectional area be \(A_0\). The initial resistance is given by:
\(R_0 = \rho \frac{L_0}{A_0}\)

Since the volume \(V = A_0 L_0\) is constant, the area at any length \(L\) is \(A = \frac{V}{L}\). Substituting this into the resistance formula gives:
\(R = \rho \frac{L^2}{V}\)

Since resistivity \(\rho\) and volume \(V\) are constant, \(R \propto L^2\).

The new length is \(L = 1.015 L_0\). Therefore, the new resistance is:
\(R_{\text{new}} = R_0 \times (1.015)^2 = 1.030225 R_0\)

The fractional increase in resistance is:
\(\frac{R_{\text{new}} - R_0}{R_0} = 0.030225\)

This corresponds to a percentage increase of \(3.02\%\), which to two significant figures is \(3.0\%\).

• [1 mark] Shows or states that resistance is proportional to the square of the length (\(R \propto L^2\)) because volume is conserved and area decreases.
• [1 mark] Calculates the percentage increase as \(3.0\%\) (accept \(3.02\%\)).

First, find the grating spacing \(d\):
\(d = \frac{1}{N} = \frac{1}{5.00 \times 10^5\text{ m}^{-1}} = 2.00 \times 10^{-6}\text{ m}\).

Use the diffraction grating equation:
\(d \sin\theta = n \lambda\)

For the second-order maximum, \(n = 2\) and \(\theta = 38.5^\circ\):
\(2.00 \times 10^{-6} \times \sin(38.5^\circ) = 2 \lambda\)

Evaluate \(\sin(38.5^\circ) \approx 0.6225\):
\(\lambda = 1.00 \times 10^{-6} \times 0.6225 = 6.225 \times 10^{-7}\text{ m}\).

To three significant figures, \(\lambda = 6.23 \times 10^{-7}\text{ m}\) (or \(623\text{ nm}\)).

• [1 mark] Calculates grating spacing \(d = 2.00 \times 10^{-6}\text{ m}\) and substitutes values correctly into \(d \sin\theta = n \lambda\).
• [1 mark] Calculates wavelength correctly to obtain \(6.22 \times 10^{-7}\text{ m}\) or \(6.23 \times 10^{-7}\text{ m}\) (accept \(6.2 \times 10^{-7}\text{ m}\) to \(6.3 \times 10^{-7}\text{ m}\)).

The elastic strain energy \(E_s\) stored in a stretched wire is given by:
\(E_s = \frac{1}{2} F \Delta L\)

First, calculate the extension \(\Delta L\) using the Young modulus formula:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F L}{A \Delta L}\)
\(\Delta L = \frac{F L}{A E}\)

Substitute the values:
\(\Delta L = \frac{240 \times 3.2}{1.5 \times 10^{-6} \times 2.0 \times 10^{11}} = \frac{768}{3.0 \times 10^5} = 2.56 \times 10^{-3}\text{ m}\)

Now, calculate the stored energy:
\(E_s = \frac{1}{2} \times 240 \times 2.56 \times 10^{-3} = 120 \times 2.56 \times 10^{-3} = 0.3072\text{ J}\).

To 2 significant figures, the elastic strain energy is \(0.31\text{ J}\).

• [1 mark] Calculates the extension \(\Delta L = 2.56 \times 10^{-3}\text{ m}\) or correctly combines formulas to find \(E_s = \frac{F^2 L}{2 A E}\).
• [1 mark] Obtains a correct final energy of \(0.31\text{ J}\) (accept \(0.30\text{ J}\) to \(0.31\text{ J}\)).

At the highest point of the bridge, the forces acting on the car are its weight \(mg\) acting downwards and the normal contact force \(R\) acting upwards.

The resultant force towards the center of the circular path is the centripetal force:
\(F_c = mg - R = \frac{m v^2}{r}\)

Rearranging for \(R\):
\(R = mg - \frac{m v^2}{r} = m \left(g - \frac{v^2}{r}\right)\)

Using \(g = 9.81\text{ m s}^{-2}\):
\(R = 1200 \left(9.81 - \frac{15.0^2}{25.0}\right)\)
\(R = 1200 \left(9.81 - 9.00\right) = 1200 \times 0.81 = 972\text{ N}\).

(If \(g = 9.8\text{ m s}^{-2}\) is used, \(R = 1200 \times 0.8 = 960\text{ N}\)).

• [1 mark] Identifies that centripetal force is the resultant of weight and normal reaction, writing the correct equation \(mg - R = \frac{mv^2}{r}\).
• [1 mark] Calculates normal reaction \(R = 972\text{ N}\) (accept range \(960\text{ N}\) to \(972\text{ N}\)).

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

Convert temperature to Kelvin:
\(T = 20.0 + 273.15 = 293.15\text{ K}\)

Rearrange to solve for the number of moles \(n\):
\(n = \frac{p V}{R T}\)

Using the molar gas constant \(R = 8.31\text{ J mol}^{-1}\text{ K}^{-1}\):
\(n = \frac{1.20 \times 10^5 \times 0.0450}{8.31 \times 293.15} = \frac{5400}{2436.08} \approx 2.217\text{ mol}\).

To 3 significant figures, this is \(2.22\text{ mol}\), or to 2 significant figures, \(2.2\text{ mol}\).

• [1 mark] Converts temperature to Kelvin (\(293\text{ K}\)) and substitutes values correctly into \(p V = n R T\).
• [1 mark] Obtains a correct value of \(2.2\text{ mol}\) (accept range \(2.2\text{ mol}\) to \(2.22\text{ mol}\)).

Let \(A_0\) be the initial activity and \(A\) be the activity after time \(t\).
\(\frac{A}{A_0} = \frac{4.50 \times 10^3}{3.60 \times 10^4} = 0.125\)

Recognizing that \(0.125 = \frac{1}{8} = \left(\frac{1}{2}\right)^3\), we can deduce that exactly 3 half-lives have elapsed.

Therefore, if \(n\) is the number of half-lives:
\(3 \times T_{1/2} = 8.00\text{ hours}\)
\(T_{1/2} = \frac{8.00}{3} = 2.67\text{ hours}\).

Alternatively, using the exponential decay formula:
\(A = A_0 e^{-\lambda t} \implies 0.125 = e^{-\lambda \times 8.00}\)
\(\ln(0.125) = -8.00 \lambda \implies \lambda = 0.2599\text{ h}^{-1}\)
\(T_{1/2} = \frac{\ln(2)}{\lambda} = \frac{0.693}{0.2599} = 2.67\text{ hours}\).

• [1 mark] Identifies that the activity has fallen to \(1/8\) of its original value, indicating 3 half-lives have elapsed (or calculates decay constant \(\lambda = 0.26\text{ h}^{-1}\)).
• [1 mark] Obtains the correct half-life of \(2.67\text{ hours}\) (accept \(2.7\text{ hours}\)).

First, find the weight of the rocket: \( W = mg = 0.250\text{ kg} \times 9.81\text{ m s}^{-2} = 2.453\text{ N} \). Next, calculate the resultant force during thrust: \( F_{\text{net}} = F_{\text{thrust}} - W = 6.50\text{ N} - 2.453\text{ N} = 4.047\text{ N} \). Use Newton's second law to find the acceleration: \( a = \frac{F_{\text{net}}}{m} = \frac{4.047\text{ N}}{0.250\text{ kg}} = 16.19\text{ m s}^{-2} \). Using the SUVAT equation \( v = u + at \) with \( u = 0 \): \( v = 0 + (16.19\text{ m s}^{-2} \times 2.00\text{ s}) = 32.38\text{ m s}^{-1} \). Rounding to three significant figures gives \( 32.4\text{ m s}^{-1} \).

- 1 mark for finding the weight (\(2.45\text{ N}\)) or the resultant force (\(4.05\text{ N}\)).
- 1 mark for calculating the acceleration (\(16.2\text{ m s}^{-2}\)) or using the impulse-momentum equation (\(F_{\text{net}} \Delta t = m \Delta v\)).
- 0.3 marks for the correct final velocity of \(32.4\text{ m s}^{-1}\) (accept \(32\text{ m s}^{-1}\) to \(32.4\text{ m s}^{-1}\)).

Use the equation for a complete circuit: \( \varepsilon = I(R + r) \). This gives two equations for the two scenarios: 1) \( \varepsilon = 1.20(4.50 + r) = 5.40 + 1.20r \) and 2) \( \varepsilon = 0.50(11.5 + r) = 5.75 + 0.50r \). Equating the two expressions for \( \varepsilon \): \( 5.40 + 1.20r = 5.75 + 0.50r \). Rearranging to solve for \( r \): \( 0.70r = 0.35 \Rightarrow r = 0.50\ \Omega \).

- 1 mark for writing down both circuit equations in terms of emf, current, external resistance, and internal resistance.
- 1 mark for equating the two expressions and demonstrating a correct algebraic method to solve for \( r \).
- 0.3 marks for obtaining the correct internal resistance of \( 0.50\ \Omega \) (or \( 0.5\ \Omega \)).

Use the diffraction grating equation: \( d \sin\theta = n\lambda \), where \( n = 3 \), \( \lambda = 633 \times 10^{-9}\text{ m} \), and \( \theta = 48.2^\circ \). Solve for the grating spacing \( d \): \( d = \frac{n\lambda}{\sin\theta} = \frac{3 \times 633 \times 10^{-9}\text{ m}}{\sin(48.2^\circ)} = \frac{1.899 \times 10^{-6}\text{ m}}{0.7455} = 2.547 \times 10^{-6}\text{ m} \). The number of lines per metre is \( N = \frac{1}{d} = \frac{1}{2.547 \times 10^{-6}\text{ m}} = 3.926 \times 10^5\text{ m}^{-1} \). Convert this to lines per millimetre: \( N_{\text{mm}} = \frac{3.926 \times 10^5}{1000} = 392.6 \approx 393\text{ lines mm}^{-1} \).

- 1 mark for correct substitution into \( d \sin\theta = n\lambda \) to find \( d = 2.55 \times 10^{-6}\text{ m} \).
- 1 mark for calculating \( N = 1/d \) and attempting the conversion from lines per metre to lines per millimetre.
- 0.3 marks for the correct value of \( 393 \) (accept in the range of \( 392 \) to \( 394 \)).

First, determine the extension \( \Delta L \) of the wire using the Young modulus formula: \( E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L / L} \Rightarrow \Delta L = \frac{FL}{AE} \). Substitute the given values: \( \Delta L = \frac{180\text{ N} \times 2.40\text{ m}}{1.50 \times 10^{-6}\text{ m}^2 \times 2.00 \times 10^{11}\text{ Pa}} = \frac{432}{3.00 \times 10^5} = 1.44 \times 10^{-3}\text{ m} \). The elastic strain energy \( E_{\text{str}} \) stored is: \( E_{\text{str}} = \frac{1}{2} F \Delta L = \frac{1}{2} \times 180\text{ N} \times 1.44 \times 10^{-3}\text{ m} = 0.1296\text{ J} \). Rounding to three significant figures gives \( 0.130\text{ J} \).

- 1 mark for calculating the extension of the wire (\( 1.44 \times 10^{-3}\text{ m} \)) or writing the combined equation for work done \( E_{\text{str}} = \frac{F^2 L}{2AE} \).
- 1 mark for substitute values into the correct work equation.
- 0.3 marks for the correct final value of \( 0.130\text{ J} \) (accept \( 0.13\text{ J} \)).

The time period of a simple pendulum is given by \( T = 2\pi\sqrt{\frac{L}{g}} \). Since the length \( L \) of the pendulum is constant, \( T^2 g = 4\pi^2 L = \text{constant} \). Therefore, \( T_{\text{Earth}}^2 g_{\text{Earth}} = T_{\text{Moon}}^2 g_{\text{Moon}} \). Rearranging for \( g_{\text{Moon}} \): \( g_{\text{Moon}} = g_{\text{Earth}} \left( \frac{T_{\text{Earth}}}{T_{\text{Moon}}} \right)^2 = 9.81\text{ m s}^{-2} \times \left( \frac{1.65\text{ s}}{4.05\text{ s}} \right)^2 = 9.81 \times 0.16598 = 1.628\text{ m s}^{-2} \). Rounding to three significant figures gives \( 1.63\text{ m s}^{-2} \).

- 1 mark for showing that \( T^2 \propto \frac{1}{g} \) or calculating the length of the pendulum \( L \approx 0.677\text{ m} \).
- 1 mark for substituting values into the ratio equation or calculating \( g_{\text{Moon}} \) from the pendulum length.
- 0.3 marks for the correct final value of \( 1.63\text{ m s}^{-2} \) (accept in the range of \( 1.62 \) to \( 1.64 \)).

In the annihilation process, the total mass of the electron and positron is converted into the energy of two identical photons. The minimum energy of one photon equals the rest energy of a single electron: \( E_0 = m_e c^2 \). Calculate the energy: \( E_0 = 9.11 \times 10^{-31}\text{ kg} \times (3.00 \times 10^8\text{ m s}^{-1})^2 = 8.199 \times 10^{-14}\text{ J} \). Using the relation \( E_0 = hf \), the minimum frequency \( f \) is: \( f = \frac{E_0}{h} = \frac{8.199 \times 10^{-14}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} = 1.237 \times 10^{20}\text{ Hz} \). Rounding to three significant figures gives \( 1.24 \times 10^{20}\text{ Hz} \).

- 1 mark for calculating the rest energy of one electron/positron as \( 8.20 \times 10^{-14}\text{ J} \).
- 1 mark for using \( E = hf \) with correct substitution of Planck's constant.
- 0.3 marks for the correct frequency of \( 1.24 \times 10^{20}\text{ Hz} \) (accept \( 1.23 \times 10^{20} \) to \( 1.25 \times 10^{20} \)).

For a string fixed at both ends, the third harmonic consists of three half-wavelength loops. Therefore, the length \( L \) of the string is related to the wavelength \( \lambda \) by: \( L = \frac{3}{2}\lambda \Rightarrow \lambda = \frac{2}{3}L \). Substitute the given length: \( \lambda = \frac{2}{3} \times 1.20\text{ m} = 0.800\text{ m} \). Using the wave equation \( v = f\lambda \): \( v = 180\text{ Hz} \times 0.800\text{ m} = 144\text{ m s}^{-1} \).

- 1 mark for finding the wavelength of the wave (\( 0.80\text{ m} \)).
- 1 mark for substituting the frequency and wavelength into the wave speed equation \( v = f\lambda \).
- 0.3 marks for obtaining the correct speed of \( 144\text{ m s}^{-1} \).

Convert the work function \( \phi \) from electronvolts to Joules: \( \phi = 2.24 \times 1.60 \times 10^{-19}\text{ J} = 3.584 \times 10^{-19}\text{ J} \). Use Einstein's photoelectric equation: \( E_{\text{photon}} = \phi + E_{k,\text{max}} \). Calculate the photon energy: \( E_{\text{photon}} = 3.584 \times 10^{-19}\text{ J} + 1.45 \times 10^{-19}\text{ J} = 5.034 \times 10^{-19}\text{ J} \). Using \( E_{\text{photon}} = \frac{hc}{\lambda} \), rearrange to solve for wavelength \( \lambda \): \( \lambda = \frac{hc}{E_{\text{photon}}} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{5.034 \times 10^{-19}\text{ J}} = 3.951 \times 10^{-7}\text{ m} \). This is equivalent to \( 3.95 \times 10^{-7}\text{ m} \).

- 1 mark for converting the work function to Joules (\( 3.58 \times 10^{-19}\text{ J} \)).
- 1 mark for calculating the total photon energy (\( 5.03 \times 10^{-19}\text{ J} \)) and applying \( E = hc/\lambda \).
- 0.3 marks for the correct wavelength value of \( 3.95 \times 10^{-7}\text{ m} \) (accept range \( 3.9 \times 10^{-7}\text{ m} \) to \( 4.0 \times 10^{-7}\text{ m} \)).

First, identify the drift velocity equation: \( I = nAve \). Rearrange the equation to solve for the drift velocity \( v \): \( v = \frac{I}{nAe} \). The charge of an electron \( e \) is \( 1.60 \times 10^{-19}\text{ C} \). Substitute the given values into the rearranged equation: \( v = \frac{2.4}{8.5 \times 10^{28} \times 4.5 \times 10^{-7} \times 1.60 \times 10^{-19}} \). Calculate the denominator: \( 8.5 \times 10^{28} \times 4.5 \times 10^{-7} \times 1.60 \times 10^{-19} = 6120\text{ C m}^{-1} \). Finally, calculate \( v \): \( v = \frac{2.4}{6120} \approx 3.92 \times 10^{-4}\text{ m s}^{-1} \). Rounding to 2 significant figures gives \( 3.9 \times 10^{-4}\text{ m s}^{-1} \).

1 mark: Correctly rearranging the formula to \( v = \frac{I}{nAe} \) and substituting the values with \( e = 1.60 \times 10^{-19}\text{ C} \). 1 mark: Calculating the correct value of \( 3.9 \times 10^{-4}\text{ m s}^{-1} \) (or \( 3.92 \times 10^{-4}\text{ m s}^{-1} \)) with appropriate unit.

Use the principle of conservation of linear momentum: Total initial momentum = Total final momentum. Let the initial direction of motion be the positive direction. Mass of rocket, \( m = 0.85\text{ kg} \), and initial velocity, \( u = 12\text{ m s}^{-1} \). Initial momentum, \( p_i = m \times u = 0.85 \times 12 = 10.2\text{ kg m s}^{-1} \). After the explosion: Mass of part A, \( m_A = 0.35\text{ kg} \), and velocity of part A, \( v_A = -5.0\text{ m s}^{-1} \) (negative since it moves backwards). Mass of part B, \( m_B = 0.85 - 0.35 = 0.50\text{ kg} \). Let the velocity of part B be \( v_B \). Final momentum, \( p_f = m_A v_A + m_B v_B = (0.35 \times -5.0) + (0.50 \times v_B) = -1.75 + 0.50 v_B \). Equating initial and final momentum: \( 10.2 = -1.75 + 0.50 v_B \). Rearranging gives: \( 0.50 v_B = 10.2 + 1.75 = 11.95 \). Therefore, \( v_B = \frac{11.95}{0.50} = 23.9\text{ m s}^{-1} \). Rounding to 2 significant figures gives \( 24\text{ m s}^{-1} \) (in the original direction of travel).

1 mark: Correct calculation of the mass of B as \( 0.50\text{ kg} \) and calculating the initial momentum as \( 10.2\text{ kg m s}^{-1} \) (or showing \( 0.85 \times 12 \)). 1 mark: Setting up a correct conservation of momentum equation that accounts for the opposing direction of part A, e.g., \( 10.2 = (0.35 \times -5.0) + 0.50 v_B \). 1 mark: Calculating the correct velocity \( 24\text{ m s}^{-1} \) (accept \( 23.9\text{ m s}^{-1} \) or \( 24\text{ m s}^{-1} \) in the forward direction) with units.

During the weak decay \(\Sigma^- \rightarrow n + \pi^-\), the quark compositions are: \(\Sigma^- = dds\), \(n = udd\), and \(\pi^- = d\bar{u}\). Looking at the initial and final states, we have \(dds \rightarrow udd + d\bar{u}\). This corresponds to an \(s\) quark decaying into a \(u\) quark while emitting a virtual \(W^-\)-boson, which then decays to a \(d\bar{u}\) pair (the \(\pi^-\) meson). Thus, the fundamental quark change is \(s \rightarrow u\).

1 mark for identifying the correct quark change from s to u.

First, define upwards as the positive direction. The initial velocity of the ball is \(u = -20\text{ m s}^{-1}\) and its final velocity is \(v = +15\text{ m s}^{-1}\). The change in momentum is \(\Delta p = m(v - u) = 0.20 \times (15 - (-20)) = 0.20 \times 35 = 7.0\text{ N s}\). The average net force is \(F_{\text{net}} = \frac{\Delta p}{\Delta t} = \frac{7.0}{0.010} = 700\text{ N}\). This net force is the result of the upward contact force from the floor \(F_{\text{floor}}\) and the downward weight of the ball \(W = mg\): \(F_{\text{net}} = F_{\text{floor}} - mg\). Therefore, \(F_{\text{floor}} = F_{\text{net}} + mg = 700 + (0.20 \times 9.81) = 701.96\text{ N} \approx 702\text{ N}\).

1 mark for calculating the correct average contact force including the effect of gravity.

The angular position of the \(n\)-th maximum is given by \(d \sin \theta_n = n \lambda\). For small angles, the angular separation between successive orders is \(\Delta \theta \approx \frac{\lambda}{d}\). To decrease this angular separation, we must either decrease \(\lambda\) or increase \(d\). Increasing \(d\) (the slit spacing) is achieved by replacing the grating with one that has fewer lines per millimetre.

1 mark for identifying that fewer lines per millimetre increases d and thus decreases the angular separation.

Use the potential divider formula: \(V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{fixed}}}{R_{\text{LDR}} + R_{\text{fixed}}}\). In the dark: \(2.4 = 12 \times \frac{3.0}{R_{\text{LDR, dark}} + 3.0} \implies R_{\text{LDR, dark}} + 3.0 = 15.0 \implies R_{\text{LDR, dark}} = 12.0\text{ k}\Omega\). In the light: \(8.0 = 12 \times \frac{3.0}{R_{\text{LDR, light}} + 3.0} \implies \frac{2}{3} = \frac{3.0}{R_{\text{LDR, light}} + 3.0} \implies R_{\text{LDR, light}} = 1.5\text{ k}\Omega\). The change in resistance is \(12.0 - 1.5 = 10.5\text{ k}\Omega\).

1 mark for calculating the change in resistance correctly.

The extension is given by \(\Delta L = \frac{F L}{A E}\). Since they are in series, the tension \(F\) is the same in both. Since they are the same material, \(E\) is the same. The cross-sectional area \(A\) is proportional to \(d^2\). Thus, \(\Delta L \propto \frac{L}{d^2}\). For P: \(\Delta L_{\text{P}} \propto \frac{L}{d^2}\). For Q: \(\Delta L_{\text{Q}} \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\). The ratio is \(\frac{\Delta L_{\text{P}}}{\Delta L_{\text{Q}}} = \frac{L/d^2}{L/(2d^2)} = 2\).

1 mark for setting up the ratio of extensions using the Young Modulus equation and finding the correct value.

The potential energy is \(E_p = \frac{1}{2} k x^2\) and the total energy is \(E_{\text{total}} = \frac{1}{2} k A^2\). The kinetic energy is \(E_k = E_{\text{total}} - E_p = \frac{1}{2} k (A^2 - x^2)\). Setting \(E_k = 3 E_p\), we get \(\frac{1}{2} k (A^2 - x^2) = 3 \times \frac{1}{2} k x^2 \implies A^2 - x^2 = 3 x^2 \implies A^2 = 4 x^2 \implies x = \frac{A}{2}\).

1 mark for establishing the relationship between energy terms and solving for x.

The activity decreases from \(360\text{ Bq}\) to \(45\text{ Bq}\), which is a factor of \(\frac{45}{360} = \frac{1}{8}\). Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly 3 half-lives have elapsed in \(12.0\text{ hours}\). Therefore, one half-life is \(T_{1/2} = \frac{12.0\text{ hours}}{3} = 4.0\text{ hours} = 4.0 \times 3600\text{ s} = 14400\text{ s}\). The decay constant is \(\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{14400\text{ s}} \approx 4.8 \times 10^{-5}\text{ s}^{-1}\).

1 mark for calculating the correct decay constant in inverse seconds.

During discharging, the potential difference across the capacitor decays according to \(V(t) = V_0 e^{-t / RC}\). At \(t = RC \ln 2\), the potential difference is \(V = V_0 e^{-RC \ln 2 / RC} = V_0 e^{-\ln 2} = \frac{V_0}{2}\). The energy stored in the capacitor is given by \(E = \frac{1}{2} C V^2\). Substituting \(V = \frac{V_0}{2}\) gives \(E = \frac{1}{2} C \left(\frac{V_0}{2}\right)^2 = \frac{1}{8} C V_0^2\).

1 mark for calculating the energy in terms of capacitance and initial voltage correctly.

First, identify the strangeness \(S\) of each particle involved in the decay:
- The \(\Sigma^0\) baryon has a strangeness of \(S = -1\).
- The proton \(\mathrm{p}\) is a non-strange baryon, so \(S = 0\).
- The negative pion \(\pi^-\) is a non-strange meson, so \(S = 0\).

The total initial strangeness is \(S_{\text{initial}} = -1\).
The total final strangeness is \(S_{\text{final}} = 0 + 0 = 0\).

Therefore, the change in strangeness is:
\(\Delta S = S_{\text{final}} - S_{\text{initial}} = 0 - (-1) = +1\).

Since strangeness is not conserved (it changes by \(+1\)), this decay cannot proceed via the strong or electromagnetic interactions. It must proceed via the weak interaction.

1 mark for the correct option (B).
- Reject A, C, D as they either state the wrong interaction or an incorrect value for \(\Delta S\).

Let downwards be the negative direction and upwards be the positive direction.

1. Calculate the velocity \(v_1\) of the ball just before impact:
Using \(v^2 = u^2 + 2as\):
\(v_1^2 = 0 + 2gH \implies v_1 = -\sqrt{2gH}\) (downwards)

2. Calculate the velocity \(v_2\) of the ball just after impact:
\(v_2^2 = 2g(0.64H) = 1.28gH = 0.64 \times 2gH\)
\(v_2 = +0.8\sqrt{2gH}\) (upwards)

3. Calculate the impulse \(I\) which equals the change in momentum \(\Delta p\):
\(I = p_f - p_i = mv_2 - mv_1\)
\(I = m(0.8\sqrt{2gH}) - m(-\sqrt{2gH}) = 1.8m\sqrt{2gH}\).

Thus, the magnitude of the impulse is \(1.8m\sqrt{2gH}\).

1 mark for the correct option (D).
- Incorrect options represent sign errors in adding initial and final velocities, or failing to convert height ratio correctly.

1. **Power dissipated in \(R\)**:
Using the formula for power: \(P = I^2 R = \left(\frac{\varepsilon}{R+r}\right)^2 R\).
- When \(R \to 0\), \(P \to 0\).
- When \(R = r\), \(P\) reaches its maximum value of \(\frac{\varepsilon^2}{4r}\) (according to the maximum power transfer theorem).
- When \(R \to \infty\), \(P \to 0\).
Therefore, \(P\) first increases and then decreases.

2. **Terminal potential difference \(V\)**:
Using the potential divider equation: \(V = \varepsilon \frac{R}{R+r}\).
As \(R\) increases, the fraction \(\frac{R}{R+r}\) increases continuously from \(0\) towards \(1\).
Thus, \(V\) increases continuously from \(0\) towards \(\varepsilon\).

1 mark for the correct option (B).
- Reject A because power does not increase continuously.
- Reject C and D because terminal PD increases continuously as load resistance increases.

The formula for the fringe spacing is given by:
\(w = \frac{\lambda D}{d}\)

Let the new values be \(\lambda'\), \(D'\), and \(d'\).
- The frequency of the light is doubled: \(f' = 2f\). Since \(c = f\lambda\), the wavelength is halved: \(\lambda' = \frac{\lambda}{2}\).
- The screen-to-slit distance is doubled: \(D' = 2D\).
- The slit separation is halved: \(d' = \frac{d}{2}\).

Substitute these into the fringe width formula:
\(w' = \frac{\lambda' D'}{d'} = \frac{\left(\frac{\lambda}{2}\right) (2D)}{\left(\frac{d}{2}\right)} = \frac{\lambda D}{\frac{d}{2}} = 2 \frac{\lambda D}{d} = 2w\).

1 mark for the correct option (C).
- Common mistake is neglecting the relationship between frequency and wavelength, leading to an incorrect factor of 4 (option D).

The Young modulus \(E\) of the material is given by:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L} = \frac{F L}{A \Delta L}\)

Since \(A = \frac{\pi d^2}{4}\), we have:
\(E = \frac{4 F L}{\pi d^2 \Delta L} \implies \Delta L = \frac{4 F L}{\pi d^2 E}\)

For the first wire with mass \(M\) (tension force \(F_1 = Mg\)):
\(\Delta L_1 = \frac{4 M g L}{\pi d^2 E}\)

For the second wire with mass \(M'\) (tension force \(F_2 = M'g\)):
\(\Delta L_2 = \frac{4 M' g (2L)}{\pi (2d)^2 E} = \frac{8 M' g L}{4 \pi d^2 E} = \frac{2 M' g L}{\pi d^2 E}\)

We require \(\Delta L_1 = \Delta L_2\):
\(\frac{4 M g L}{\pi d^2 E} = \frac{2 M' g L}{\pi d^2 E}\)

Simplifying this yields:
\(4 M = 2 M' \implies M' = 2M\).

1 mark for the correct option (C).
- Reject other options arising from incorrect squaring of the diameter (e.g., forgetting that area scales with \(d^2\)).

Let \(N\) be the normal contact force acting on the car from the banked track, which is directed at an angle \(\theta\) to the vertical.

Resolving the forces:
- Vertically, there is no acceleration: \(N \cos\theta = mg\)
- Horizontally, the component of the normal force provides the centripetal force: \(N \sin\theta = \frac{mv^2}{R}\)

Dividing the horizontal equation by the vertical equation:
\(\frac{N \sin\theta}{N \cos\theta} = \frac{mv^2 / R}{mg}\)
\(\tan\theta = \frac{v^2}{gR}\)

Solving for \(\theta\):
\(\theta = \arctan\left(\frac{v^2}{gR}\right)\).

1 mark for the correct option (C).
- Reject A and B as they use incorrect trigonometric ratios.
- Reject D as it is the reciprocal function.

According to Einstein's photoelectric equation:
\(h f = \Phi + E_{k,\text{max}}\)

For the first scenario:
\(h f = \Phi + E_k\) --- (Equation 1)

For the second scenario:
\(h(1.5f) = \Phi + 2.5E_k\) --- (Equation 2)

Multiply Equation 1 by 1.5:
\(1.5hf = 1.5\Phi + 1.5E_k\) --- (Equation 3)

Since the left-hand sides of Equation 2 and Equation 3 are both equal to \(1.5hf\), equate their right-hand sides:
\(\Phi + 2.5E_k = 1.5\Phi + 1.5E_k\)

Rearranging to solve for \(\Phi\):
\(2.5E_k - 1.5E_k = 1.5\Phi - \Phi\)
\(E_k = 0.5\Phi \implies \Phi = 2E_k\).

1 mark for the correct option (D).
- Incorrect options represent algebra errors when eliminating \(hf\).

The fundamental frequency \(f_0\) of a stretched string fixed at both ends is given by:
\(f_0 = \frac{v}{2L} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\)
where \(\mu\) is the mass per unit length.

Let the new tension be \(T' = 1.44T\) and the new length be \(L' = 0.8L\).
The new fundamental frequency \(f'\) is:
\(f' = \frac{1}{2L'}\sqrt{\frac{T'}{\mu}} = \frac{1}{2(0.8L)}\sqrt{\frac{1.44T}{\mu}}\)

Extracting the constant factors:
\(f' = \frac{\sqrt{1.44}}{0.8} \left( \frac{1}{2L}\sqrt{\frac{T}{\mu}} \right) = \frac{1.2}{0.8} f_0 = 1.5 f_0\).

1 mark for the correct option (C).
- Common mistake: forgetting that speed of wave is proportional to the square root of tension (leading to wrong options like A or B).

To determine which decay is forbidden, we examine the conservation of electron lepton number \(L_e\) and muon lepton number \(L_{\mu}\) for each decay:

* **Option A**: \( p + e^- \rightarrow n + \nu_e \)
Before: \(L_e = +1\)
After: \(L_e = +1\)
(Conserved)

* **Option B**: \( \mu^- \rightarrow e^- + \bar{\nu}_e + \nu_{\mu} \)
Before: \(L_{\mu} = +1\), \(L_e = 0\)
After: \(L_e = +1 - 1 = 0\), \(L_{\mu} = +1\)
(Conserved)

* **Option C**: \( \pi^+ \rightarrow \mu^+ + \nu_{\mu} \)
Before: \(L_{\mu} = 0\)
After: \(L_{\mu} = -1 + 1 = 0\)
(Conserved)

* **Option D**: \( n \rightarrow p + e^- + \nu_e \)
Before: \(L_e = 0\)
After: \(L_e = +1 + 1 = +2\)
(Not conserved, as \(L_e\) changes from 0 to +2. The correct decay requires an electron antineutrino, \(\bar{\nu}_e\), instead of an electron neutrino, \(\nu_e\)).

1 mark for identifying option D as the forbidden decay with correct justification of lepton number non-conservation.

Using the conservation of momentum:
\[ m v + 0 = (m + 3m) v_f \]
\[ m v = 4m v_f \implies v_f = \frac{v}{4} \]

Now, calculate the initial and final kinetic energies of the system:
* Initial kinetic energy:
\[ E_i = \frac{1}{2} m v^2 \]
* Final kinetic energy:
\[ E_f = \frac{1}{2} (4m) \left(\frac{v}{4}\right)^2 = \frac{1}{2} (4m) \frac{v^2}{16} = \frac{1}{4} \left(\frac{1}{2} m v^2\right) = \frac{1}{4} E_i \]

The kinetic energy remaining in the system is \( \frac{1}{4} \) of the initial kinetic energy. Therefore, the fraction of the initial kinetic energy transferred to other forms is:
\[ 1 - \frac{1}{4} = \frac{3}{4} \]

1 mark for selecting option D, derived by applying the principle of conservation of momentum and comparing initial and final kinetic energies.

Let's analyze all possible combinations of three identical resistors of resistance \( R \):

1. **All three in series**:
\[ R_{eq} = R + R + R = 3R \]

2. **All three in parallel**:
\[ \frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{3}{R} \implies R_{eq} = \frac{1}{3} R \]

3. **Two in parallel, connected in series with the third**:
\[ R_{eq} = R + \left(\frac{R \times R}{R + R}\right) = R + \frac{R}{2} = \frac{3}{2} R \]

4. **Two in series, connected in parallel with the third**:
\[ R_{eq} = \frac{(2R) \times R}{2R + R} = \frac{2R^2}{3R} = \frac{2}{3} R \]

Comparing these values with the options, we see that \( \frac{4}{3} R \) (Option C) is not a possible equivalent resistance.

1 mark for identifying option C as the impossible value of equivalent resistance.

Using the diffraction grating equation:
\[ d \sin \theta = n \lambda \]
For \( n = 2 \), the angle is \( \theta = 38^\circ \). Therefore:
\[ d \sin 38^\circ = 2\lambda \implies \frac{\lambda}{d} = \frac{\sin 38^\circ}{2} \]
Using \( \sin 38^\circ \approx 0.6157 \):
\[ \frac{\lambda}{d} \approx \frac{0.6157}{2} = 0.3078 \]

For a maximum to be observable on a screen, the angle of diffraction \( \theta \) must satisfy \( \sin \theta \le 1 \). Thus:
\[ n \frac{\lambda}{d} \le 1 \implies n \le \frac{1}{\lambda/d} \]
\[ n \le \frac{1}{0.3078} \approx 3.25 \]

Since the order \( n \) must be an integer, the highest observable order of maximum is \( n_{max} = 3 \).

The total number of observed maxima includes the central maximum (\( n = 0 \)) plus three orders on each side of the central maximum:
\[ \text{Total maxima} = 2 n_{max} + 1 = 2(3) + 1 = 7 \]

1 mark for correctly calculating the maximum order as 3, and hence the total number of maxima as 7.

Let's find the effective spring constant of the combination:
1. The two identical springs in parallel have an equivalent spring constant:
\[ k_p = k + k = 2k \]
2. This parallel combination is in series with a third spring of constant \( k \). The overall effective spring constant \( k_{eq} \) is given by:
\[ \frac{1}{k_{eq}} = \frac{1}{k_p} + \frac{1}{k} = \frac{1}{2k} + \frac{1}{k} = \frac{3}{2k} \]
\[ k_{eq} = \frac{2}{3}k \]

Using Hooke's law (\( W = k_{eq} x \)), the total extension \( x \) is:
\[ x = \frac{W}{k_{eq}} = \frac{W}{\frac{2}{3}k} = \frac{3W}{2k} \]

1 mark for calculating the combined spring constant and obtaining the correct total extension expression (Option C).

The displacement of an object starting from the equilibrium position (\( x = 0 \) at \( t = 0 \)) is described by:
\[ x(t) = A \sin(\omega t) \]
where \( \omega = \frac{2\pi}{T} \).

To find the time \( t \) to reach a displacement of \( \frac{A}{2} \):
\[ \frac{A}{2} = A \sin(\omega t) \implies \sin(\omega t) = \frac{1}{2} \]

The smallest positive angle is:
\[ \omega t = \frac{\pi}{6} \]

Substitute \( \omega = \frac{2\pi}{T} \):
\[ \frac{2\pi}{T} t = \frac{\pi}{6} \implies t = \frac{T}{12} \]

1 mark for utilizing the SHM displacement equation and solving for time to obtain Option A.

First, express the unit of work/energy (joule, \(\text{J}\)) in SI base units:
\[ \text{Work} = \text{Force} \times \text{distance} \]
Since \(\text{Force} = \text{mass} \times \text{acceleration}\), the SI unit of force (Newton) is:
\[ \text{N} = \text{kg m s}^{-2} \]
Therefore, the SI unit of energy (Joule) is:
\[ \text{J} = \text{N m} = \text{kg m}^2 \text{s}^{-2} \]

Now, multiply by seconds (\(\text{s}\)) to find the unit of the Planck constant:
\[ [h] = \text{J s} = (\text{kg m}^2 \text{s}^{-2}) \times \text{s} = \text{kg m}^2 \text{s}^{-1} \]

1 mark for correctly expressing Joules in SI base units and multiplying by seconds to find the unit of h (Option B).

Einstein's photoelectric equation for the first case is:
\[ E_k = \frac{hc}{\lambda} - \Phi \implies \frac{hc}{\lambda} = E_k + \Phi \]

For the second case with wavelength \( \frac{\lambda}{2} \):
\[ E_k' = \frac{hc}{\lambda/2} - \Phi = 2 \left(\frac{hc}{\lambda}\right) - \Phi \]

Substitute the expression for \( \frac{hc}{\lambda} \) from the first case into the second equation:
\[ E_k' = 2(E_k + \Phi) - \Phi \]
\[ E_k' = 2E_k + 2\Phi - \Phi \]
\[ E_k' = 2E_k + \Phi \]

1 mark for setting up both photoelectric equations and algebraically solving for the relation to get Option B.

The elastic strain energy stored in a stretched wire is given by: \(E = \frac{1}{2} F \Delta L\). Using the definition of Young's Modulus, \(Y = \frac{F L}{A \Delta L}\), we can express the extension as \(\Delta L = \frac{F L}{A Y}\). Substituting this expression back into the energy equation gives: \(E = \frac{F^2 L}{2 A Y}\). Because both wires are made of the same metal (same \(Y\)) and support the same load (same \(F\)), the energy stored depends only on the dimensions: \(E \propto \frac{L}{A}\). Since cross-sectional area \(A = \frac{\pi d^2}{4}\), the energy stored is proportional to: \(E \propto \frac{L}{d^2}\). For wire \(P\): \(E_P \propto \frac{L}{d^2}\). For wire \(Q\): \(E_Q \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\). Comparing the two values: \(\frac{E_P}{E_Q} = \frac{L / d^2}{L / (2d^2)} = 2\).

1 mark for the correct answer C. Correctly identifying that the stored energy is proportional to the ratio of length to the square of the diameter, leading to a ratio of 2.0.

The work done \(W\) by a gas during an isobaric expansion is given by \(W = P\Delta V\). Since the volume of the gas doubles, the final volume is \(0.080\text{ m}^3\) and the change in volume is \(\Delta V = 0.080 - 0.040 = 0.040\text{ m}^3\). Substituting the values, we get \(W = (1.5 \times 10^5) \times 0.040 = 6000\text{ J}\) (or \(6.0 \times 10^3\text{ J}\)).

1 mark for identifying the correct change in volume \(\Delta V = 0.040\text{ m}^3\) and using \(W = P\Delta V\). 1.14 marks for obtaining the final correct answer of \(6000\text{ J}\) (or \(6.0 \times 10^3\text{ J}\)).

The gravitational force provides the necessary centripetal force for circular orbit: \(\frac{G M m}{r^2} = \frac{m v^2}{r}\), which simplifies to \(v = \sqrt{\frac{GM}{r}}\). Substituting the given values: \(v = \sqrt{\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{1.5 \times 10^7}} = \sqrt{2.668 \times 10^7} \approx 5165\text{ m s}^{-1}\). Rounding to 2 significant figures gives \(5.2 \times 10^3\text{ m s}^{-1}\).

1 mark for using \(v = \sqrt{\frac{GM}{r}}\). 1.14 marks for the correct evaluation to \(5.2 \times 10^3\text{ m s}^{-1}\) (allow 5170 or 5200).

First, calculate the electric field strength \(E\) between the plates using \(E = \frac{V}{d}\): \(E = \frac{400}{8.0 \times 10^{-3}} = 5.0 \times 10^4\text{ V m}^{-1}\). Next, calculate the force \(F\) on the electron using \(F = E Q\): \(F = (5.0 \times 10^4) \times (1.60 \times 10^{-19}) = 8.0 \times 10^{-15}\text{ N}\).

1 mark for calculating the correct electric field strength of \(5.0 \times 10^4\text{ V m}^{-1}\). 1.14 marks for calculating the correct force of \(8.0 \times 10^{-15}\text{ N}\).

The initial charge is \(Q_0 = C V = 47 \times 10^{-6}\text{ F} \times 12\text{ V} = 5.64 \times 10^{-4}\text{ C}\). The time constant of the circuit is \(\tau = R C = 150 \times 10^3\\ \Omega \times 47 \times 10^{-6}\text{ F} = 7.05\text{ s}\). The formula for discharging charge is \(Q = Q_0 e^{-t / \tau}\). Substituting the values: \(Q = 5.64 \times 10^{-4} \times e^{-5.0 / 7.05} = 5.64 \times 10^{-4} \times 0.492 = 2.78 \times 10^{-4}\text{ C}\\. To 2 significant figures, this is \)2.8 \\times 10^{-4}\\text{ C}\).

1 mark for calculating the initial charge \(Q_0 = 5.64 \times 10^{-4}\text{ C}\) or the time constant \(\tau = 7.05\text{ s}\). 1.14 marks for the correct final charge of \(2.8 \times 10^{-4}\text{ C}\) (accept range \(2.7 \times 10^{-4}\text{ C}\) to \(2.8 \times 10^{-4}\text{ C}\)).

The magnetic force \(F\) on a current-carrying conductor in a magnetic field is given by \(F = B I L \sin \theta\). Substituting the given values: \(F = 0.80\text{ T} \times 4.5\text{ A} \times 0.15\text{ m} \times \sin 30^\circ = 0.54 \times 0.5 = 0.27\text{ N}\).

1 mark for using \(F = B I L \sin \theta\) with correct values substituted. 1.14 marks for obtaining the correct answer of \(0.27\text{ N}\).

The activity decreases from \(8.0 \times 10^5\text{ Bq}\) to \(1.0 \times 10^5\text{ Bq}\), which is a reduction by a factor of \(\frac{1.0 \times 10^5}{8.0 \times 10^5} = \frac{1}{8}\). Since \(\frac{1}{8} = \left(\frac{1}{2}\right)^3\), exactly 3 half-lives have elapsed in the 12-hour period. Therefore, the half-life is \(T_{1/2} = \frac{12\text{ hours}}{3} = 4.0\text{ hours}\).

1 mark for identifying that the activity has halved 3 times (or that 3 half-lives have passed). 1.14 marks for calculating the half-life as \(4.0\text{ hours}\).

In thermal equilibrium, the thermal energy lost by the metal equals the thermal energy gained by the water. Let \(T\) be the final temperature in \(^\circ\text{C}\). Energy lost by metal: \(Q_{\text{lost}} = m_{\text{m}} c_{\text{m}} (100 - T) = 0.20 \times 400 \times (100 - T) = 80(100 - T)\). Energy gained by water: \(Q_{\text{gained}} = m_{\text{w}} c_{\text{w}} (T - 20) = 0.10 \times 4200 \times (T - 20) = 420(T - 20)\\. Setting them equal: \)80(100 - T) = 420(T - 20) \\implies 8000 - 80T = 420T - 8400 \\implies 16400 = 500T \\implies T = 32.8^\\circ\\text{C}\). Rounding to 2 significant figures gives \(33^\circ\text{C}\).

1 mark for setting up the heat exchange equation \(m_{\text{m}} c_{\text{m}} \Delta T_{\text{m}} = m_{\text{w}} c_{\text{w}} \Delta T_{\text{w}}\). 1.14 marks for obtaining the final correct temperature of \(32.8^\circ\text{C}\) or \(33^\circ\text{C}\).

First, calculate the total mass of the individual constituent nucleons: \(2 \times 1.00728\text{ u} + 2 \times 1.00866\text{ u} = 2.01456\text{ u} + 2.01732\text{ u} = 4.03188\text{ u}\). Next, find the mass defect \(\Delta m\): \(\Delta m = 4.03188\text{ u} - 4.00150\text{ u} = 0.03038\text{ u}\). The total binding energy is \(0.03038 \times 931.5\text{ MeV} = 28.29897\text{ MeV}\). The binding energy per nucleon is this total divided by the nucleon number (4): \(\frac{28.29897\text{ MeV}}{4} \approx 7.07\text{ MeV}\).

1 mark for finding the mass defect \(\Delta m = 0.03038\text{ u}\). 1.14 marks for dividing the total binding energy by 4 to get \(7.07\text{ MeV}\) (accept \(7.1\text{ MeV}\)).

First, convert the temperature from Celsius to Kelvin:
\(T = 27 + 273.15 = 300.15\text{ K}\).

Using the ideal gas equation:
\(pV = NkT\)

Rearrange to solve for the number of molecules, \(N\):
\(N = \frac{pV}{kT}\)

Substitute the values into the equation:
\(N = \frac{1.2 \times 10^5\text{ Pa} \times 3.5 \times 10^{-3}\text{ m}^3}{1.38 \times 10^{-23}\text{ J K}^{-1} \times 300.15\text{ K}}\)
\(N \approx 1.01 \times 10^{23}\text{ molecules}\).

To two significant figures, \(N = 1.0 \times 10^{23}\).

1 mark for converting temperature to Kelvin and correctly rearranging the ideal gas equation \(pV = NkT\).
1.14 marks for obtaining the final correct value of \(1.0 \times 10^{23}\) (accept \(1.01 \times 10^{23}\)).

The escape velocity \(v\) from a planet is given by the formula:
\(v = \sqrt{\frac{2GM}{R}}\)

Where:
- \(G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}\)
- \(M = 4.8 \times 10^{24}\text{ kg}\)
- \(R = 5.5 \times 10^6\text{ m}\)

Substitute the given values into the formula:
\(v = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 4.8 \times 10^{24}}{5.5 \times 10^6}}\)
\(v = \sqrt{\frac{6.4032 \times 10^{14}}{5.5 \times 10^6}}\)
\(v = \sqrt{1.1642 \times 10^8} \approx 1.079 \times 10^4\text{ m s}^{-1}\).

To two significant figures, this is \(1.1 \times 10^4\text{ m s}^{-1}\) (or \(11\text{ km s}^{-1}\)).

1 mark for the correct substitution of terms into the escape velocity formula \(v = \sqrt{\frac{2GM}{R}}\).
1.14 marks for the final correct answer with appropriate units (accept \(1.08 \times 10^4\text{ m s}^{-1}\) or \(1.1 \times 10^4\text{ m s}^{-1}\)).

The electric field strength \(E\) between the parallel plates is:
\(E = \frac{V}{d} = \frac{2400\text{ V}}{8.0 \times 10^{-3}\text{ m}} = 3.0 \times 10^5\text{ V m}^{-1}\)

The electrostatic force \(F\) acting on the electron is:
\(F = Eq = 3.0 \times 10^5\text{ V m}^{-1} \times 1.60 \times 10^{-19}\text{ C}\)
\(F = 4.8 \times 10^{-14}\text{ N}\)

1 mark for calculating the electric field strength \(E\) or showing a combined formula \(F = \frac{eV}{d}\).
1.14 marks for the final force calculation of \(4.8 \times 10^{-14}\text{ N}\).

The energy stored in a capacitor is given by \(E = \frac{1}{2} C V^2\).

Initial energy stored, \(E_i\):
\(E_i = \frac{1}{2} \times 470 \times 10^{-6}\text{ F} \times (12\text{ V})^2 = 0.03384\text{ J}\)

Final energy stored, \(E_f\):
\(E_f = \frac{1}{2} \times 470 \times 10^{-6}\text{ F} \times (3.0\text{ V})^2 = 0.002115\text{ J}\)

Energy lost, \(\Delta E\):
\(\Delta E = E_i - E_f = 0.03384 - 0.002115 = 0.031725\text{ J}\)

To two significant figures, this is \(3.2 \times 10^{-2}\text{ J}\) (or \(0.032\text{ J}\)).

1 mark for using the energy formula \(E = \frac{1}{2}CV^2\) to find at least one energy value or for using the correct difference formula \(\Delta E = \frac{1}{2}C(V_i^2 - V_f^2)\).
1.14 marks for the correct value of \(3.2 \times 10^{-2}\text{ J}\) (or \(0.032\text{ J}\)).

The centripetal force keeping the proton in its circular path is provided by the magnetic force:
\(\frac{m v^2}{r} = B q v\)

Solving for speed, \(v\):
\(v = \frac{B q r}{m}\)

Substitute the given values into this equation:
\(v = \frac{0.15\text{ T} \times 1.60 \times 10^{-19}\text{ C} \times 0.24\text{ m}}{1.67 \times 10^{-27}\text{ kg}}\)
\(v = \frac{5.76 \times 10^{-21}}{1.67 \times 10^{-27}} \approx 3.45 \times 10^6\text{ m s}^{-1}\)

To two significant figures, the speed of the proton is \(3.5 \times 10^6\text{ m s}^{-1}\).

1 mark for equating centripetal force and magnetic force to find \(v = \frac{Bqr}{m}\).
1.14 marks for calculating the correct speed in the range \(3.4 \times 10^6\text{ m s}^{-1}\) to \(3.5 \times 10^6\text{ m s}^{-1}\).

First, convert the half-life into seconds:
\(T_{1/2} = 8.0\text{ days} = 8.0 \times 24 \times 3600\text{ s} = 691,200\text{ s}\)

Next, calculate the decay constant, \(\lambda\):
\(\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.69315}{691200} \approx 1.003 \times 10^{-6}\text{ s}^{-1}\)

Now, calculate the activity, \(A\), of the sample:
\(A = \lambda N\)
\(A = 1.003 \times 10^{-6}\text{ s}^{-1} \times 3.2 \times 10^{15} \approx 3.21 \times 10^9\text{ Bq}\)

To two significant figures, this is \(3.2 \times 10^9\text{ Bq}\).

1 mark for converting half-life to seconds and calculating the decay constant \(\lambda\).
1.14 marks for obtaining the activity of \(3.2 \times 10^9\text{ Bq}\) (accept values between \(3.2 \times 10^9\) and \(3.21 \times 10^9\)).

The maximum kinetic energy of a simple harmonic oscillator is given by:
\(E_{\text{k,max}} = \frac{1}{2} m v_{\text{max}}^2\)

Where the maximum speed \(v_{\text{max}} = \omega A\).
First, calculate the angular frequency \(\omega\):
\(\omega = \frac{2\pi}{T} = \frac{2\pi}{0.82\text{ s}} \approx 7.662\text{ rad s}^{-1}\)

The amplitude is \(A = 45\text{ mm} = 0.045\text{ m}\).

Now calculate \(v_{\text{max}}\):
\(v_{\text{max}} = 7.662 \times 0.045 = 0.3448\text{ m s}^{-1}\)

Substitute \(v_{\text{max}}\) into the kinetic energy equation:
\(E_{\text{k,max}} = \frac{1}{2} \times 0.35\text{ kg} \times (0.3448\text{ m s}^{-1})^2\)
\(E_{\text{k,max}} = 0.175 \times 0.1189 \approx 0.0208\text{ J}\)

To two significant figures, the maximum kinetic energy is \(2.1 \times 10^{-2}\text{ J}\) (or \(0.021\text{ J}\)).

1 mark for calculating the angular frequency \(\omega\) or finding the equivalent spring constant \(k\).
1.14 marks for obtaining the final correct kinetic energy of \(2.1 \times 10^{-2}\text{ J}\) (or \(0.021\text{ J}\)).

The volume of a cylindrical wire is given by:
\(V = \pi r^2 L = \frac{\pi d^2 L}{4}\)

The equation for combining percentage uncertainties in this relationship is:
\(\%\Delta V = 2 \times \%\Delta d + \%\Delta L\)

First, calculate the percentage uncertainty in \(d\):
\(\%\Delta d = \frac{0.01}{0.46} \times 100 \approx 2.174\%\)

Next, calculate the percentage uncertainty in \(L\):
\(\%\Delta L = \frac{0.02}{1.50} \times 100 \approx 1.333\%\)

Now, sum the percentage uncertainties as required by the formula:
\(\%\Delta V = 2 \times (2.174\%) + 1.333\% = 4.348\% + 1.333\% = 5.681\%\)

To two significant figures, the percentage uncertainty in the volume is \(5.7\%\) (accept \(6\%\)).

1 mark for identifying that the percentage uncertainty of the diameter must be doubled when calculating the total percentage uncertainty.
1.14 marks for obtaining the final correct percentage uncertainty of \(5.7\%\) (or \(6\%\)).

First, calculate the number of moles of the gas: \(n = \frac{m}{M} = \frac{2.4 \times 10^{-3} \text{ kg}}{28 \times 10^{-3} \text{ kg mol}^{-1}} = 0.0857 \text{ mol}\). Next, find the total number of molecules: \(N = n N_A = 0.0857 \times 6.02 \times 10^{23} = 5.16 \times 10^{22}\). The mean kinetic energy of a molecule is given by \(E_k = \frac{3}{2} k_B T\). Since \(p V = N k_B T\), we can rewrite this as \(E_k = \frac{3}{2} \frac{p V}{N} = 1.5 \times \frac{1.5 \times 10^5 \times 1.8 \times 10^{-3}}{5.16 \times 10^{22}} = 7.85 \times 10^{-21} \text{ J}\).

1 Mark: Correct calculation of the number of molecules \(N\) or moles \(n\). 1 Mark: Correct final value of kinetic energy of \(7.9 \times 10^{-21} \text{ J}\) (accept range \(7.8 \times 10^{-21}\) to \(7.9 \times 10^{-21}\)).

The total energy of a satellite in a circular orbit is given by \(E = -\frac{G M m}{2 r}\). The work done to move the satellite from orbit 1 to orbit 2 is the difference in their total energies: \(\Delta E = E_2 - E_1 = \frac{G M m}{2} \left(\frac{1}{r_1} - \frac{1}{r_2}\right)\). Substituting the values: \(\Delta E = 0.5 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 450 \times \left(\frac{1}{8.2 \times 10^6} - \frac{1}{1.2 \times 10^7}\right) = 3.46 \times 10^9 \text{ J}\).

1 Mark: Correct formulation of the total orbit energy difference. 1 Mark: Correct final answer of \(3.5 \times 10^9 \text{ J}\) (accept \(3.4 \times 10^9\) to \(3.5 \times 10^9\)).

By conservation of energy, the electrical potential energy lost equals the kinetic energy gained: \(q V = \frac{1}{2} m v^2\). Rearranging for speed: \(v = \sqrt{\frac{2 q V}{m}} = \sqrt{\frac{2 \times 3.2 \times 10^{-19} \times 120}{6.64 \times 10^{-27}}} = 1.08 \times 10^5 \text{ m s}^{-1}\).

1 Mark: Equating electrostatic potential energy change to kinetic energy gain. 1 Mark: Correct calculation of the final speed as \(1.1 \times 10^5 \text{ m s}^{-1}\) (accept \(1.08 \times 10^5\) to \(1.1 \times 10^5\)).

The charge on a discharging capacitor is given by \(Q = Q_0 e^{-t / RC}\). To find when the charge falls to 15% of its initial value, we set \(0.15 Q_0 = Q_0 e^{-t / RC}\), which simplifies to \(e^{-t / RC} = 0.15\). Taking the natural logarithm: \(t = -RC \ln(0.15) = -(47 \times 10^3 \times 220 \times 10^{-6}) \ln(0.15) = 19.6 \text{ s}\).

1 Mark: Correct substitution of values into the exponential decay relation. 1 Mark: Correct final value of \(20 \text{ s}\) (accept \(19.6\) to \(20\)).

Using the equation \(A = \lambda N\), the decay constant is \(\lambda = \frac{1.8 \times 10^{12}}{4.2 \times 10^{18}} = 4.286 \times 10^{-7} \text{ s}^{-1}\). The half-life in seconds is \(T_{1/2} = \frac{\ln(2)}{\lambda} = \frac{0.693}{4.286 \times 10^{-7}} = 1.617 \times 10^6 \text{ s}\). Converting this to hours: \(T_{1/2} = \frac{1.617 \times 10^6}{3600} = 449.2 \text{ hours}\).

1 Mark: Correct calculation of the decay constant \(\lambda\) or half-life in seconds. 1 Mark: Correct conversion to hours to give \(450 \text{ hours}\) (accept \(449\) to \(450\)).

For the forces to balance, the magnetic force must equal the weight of the wire: \(B I L = m g\). Rearranging for the current: \(I = \frac{m g}{B L} = \frac{18 \times 10^{-3} \text{ kg} \times 9.81 \text{ m s}^{-2}}{0.15 \text{ T} \times 0.35 \text{ m}} = 3.36 \text{ A}\).

1 Mark: Equating the magnetic force formula to weight \(B I L = m g\). 1 Mark: Correct final value of \(3.4 \text{ A}\) (accept \(3.36\) to \(3.4\)).

The angular frequency is \(\omega = \frac{2\pi}{T} = \frac{2\pi}{0.85} = 7.392 \text{ rad s}^{-1}\). The maximum velocity is \(v_{\text{max}} = \omega A = 7.392 \times 0.040 = 0.2957 \text{ m s}^{-1}\). The maximum kinetic energy is then \(E_{k,\text{max}} = \frac{1}{2} m v_{\text{max}}^2 = 0.5 \times 0.12 \times (0.2957)^2 = 5.25 \times 10^{-3} \text{ J}\).

1 Mark: Correct calculation of angular frequency \(\omega\) or maximum velocity \(v_{\text{max}}\). 1 Mark: Correct calculation of maximum kinetic energy to give \(5.3 \times 10^{-3} \text{ J}\) (accept \(5.2 \times 10^{-3}\) to \(5.3 \times 10^{-3}\)).

The thermal energy required is \(\Delta Q = m c \Delta T = 0.75 \times 130 \times (327 - 20) = 29932.5 \text{ J}\). The time taken by the heater of power \(P\) is \(t = \frac{\Delta Q}{P} = \frac{29932.5}{150} = 199.6 \text{ s}\).

1 Mark: Correct calculation of the total thermal energy needed. 1 Mark: Correct final value of \(200 \text{ s}\) (accept \(199\) to \(200\)).

First, find the total mass of the helium gas in the container: \(M = n \times \text{molar mass} = 0.12\text{ mol} \times 4.0 \times 10^{-3}\text{ kg mol}^{-1} = 4.8 \times 10^{-4}\text{ kg}\). Next, use the kinetic theory of gases equation relating pressure and volume to mean square speed: \(PV = \frac{1}{3} M c_{\text{rms}}^2\). Rearranging for the root-mean-square speed: \(c_{\text{rms}} = \sqrt{\frac{3PV}{M}} = \sqrt{\frac{3 \times 1.5 \times 10^5\text{ Pa} \times 3.5 \times 10^{-3}\text{ m}^3}{4.8 \times 10^{-4}\text{ kg}}} = \sqrt{3.281 \times 10^6} \approx 1810\text{ m s}^{-1}\). Alternatively, the temperature of the gas can be calculated first using \(PV = nRT\): \(T = \frac{1.5 \times 10^5 \times 3.5 \times 10^{-3}}{0.12 \times 8.31} \approx 526.5\text{ K}\). Then use \(c_{\text{rms}} = \sqrt{\frac{3RT}{M_{\text{molar}}}} = \sqrt{\frac{3 \times 8.31 \times 526.5}{4.0 \times 10^{-3}}} \approx 1810\text{ m s}^{-1}\).

1 mark: Correct calculation of total mass of gas \(M = 4.8 \times 10^{-4}\text{ kg}\) OR correct calculation of the gas temperature \(T = 526.5\text{ K}\). 1.14 marks: Correct substitution into the root-mean-square speed equation leading to \(1800\text{ m s}^{-1}\) or \(1810\text{ m s}^{-1}\) (accept range \(1800\text{ m s}^{-1}\) to \(1820\text{ m s}^{-1}\)).

The ratio of the activity at time \(t\) to the initial activity is \(\frac{A}{A_0} = \frac{4.5 \times 10^4}{3.6 \times 10^5} = 0.125\). Since \(0.125 = (1/2)^3\), this duration corresponds to exactly 3 half-lives. Thus, the half-life \(T_{1/2} = \frac{15\text{ hours}}{3} = 5.0\text{ hours}\). Converting this to seconds gives \(T_{1/2} = 5.0 \times 3600\text{ s} = 1.8 \times 10^4\text{ s}\). The decay constant is \(\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{1.8 \times 10^4\text{ s}} \approx 3.85 \times 10^{-5}\text{ s}^{-1}\). Alternatively, using \(A = A_0 e^{-\lambda t}\): \(0.125 = e^{-\lambda \times 54000}\) which gives \(\ln(0.125) = -54000 \lambda \implies -2.079 = -54000 \lambda \implies \lambda = 3.85 \times 10^{-5}\text{ s}^{-1}\).

1 mark: Recognition that the activity has decreased to 1/8 of its initial value, determining the half-life to be \(5.0\text{ hours}\) or converting \(15\text{ hours}\) to \(5.4 \times 10^4\text{ s}\). 1.14 marks: Correctly calculating the decay constant \(\lambda\) as \(3.85 \times 10^{-5}\text{ s}^{-1}\) (accept \(3.8 \times 10^{-5}\) to \(3.9 \times 10^{-5}\text{ s}^{-1}\)).

The gravitational potential energy of a mass \(m\) in the gravitational field of a planet of mass \(M\) at distance \(r\) is given by \(E_p = -\frac{GMm}{r}\). The change in gravitational potential energy is \(\Delta E_p = E_{p2} - E_{p1} = -\frac{GMm}{r_2} - \left(-\frac{GMm}{r_1}\right) = GMm \left( \frac{1}{r_1} - \frac{1}{r_2} \right)\). Substituting the given values: \(\Delta E_p = (6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}) \times (6.0 \times 10^{24}\text{ kg}) \times (450\text{ kg}) \times \left( \frac{1}{1.2 \times 10^7\text{ m}} - \frac{1}{2.4 \times 10^7\text{ m}} \right)\). \(\Delta E_p = 1.8009 \times 10^{17} \times (8.333 \times 10^{-8} - 4.167 \times 10^{-8}) = 1.8009 \times 10^{17} \times 4.167 \times 10^{-8} \approx 7.50 \times 10^9\text{ J}\).

1 mark: Evidence of using the correct expression for gravitational potential energy or calculating the potential energy at either or both of the orbits (e.g., \(E_{p1} = -1.50 \times 10^{10}\text{ J}\) or \(E_{p2} = -7.50 \times 10^9\text{ J}\)). 1.14 marks: Correct calculation of the increase in gravitational potential energy, yielding \(7.5 \times 10^9\text{ J}\) (accept \(7.50 \times 10^9\text{ J}\)).

The work done on the proton by the uniform electric field is converted into its kinetic energy. The work done is given by \(W = F \times d = q E d\), where \(q\) is the charge of the proton, \(E\) is the field strength, and \(d\) is the distance travelled. \(W = 1.60 \times 10^{-19}\text{ C} \times 4.5 \times 10^4\text{ V m}^{-1} \times 0.050\text{ m} = 3.60 \times 10^{-16}\text{ J}\). Equating this to the kinetic energy of the proton: \(\frac{1}{2} m v^2 = W \implies v = \sqrt{\frac{2W}{m}} = \sqrt{\frac{2 \times 3.60 \times 10^{-16}\text{ J}}{1.67 \times 10^{-27}\text{ kg}}} = \sqrt{4.311 \times 10^{11}} \approx 6.57 \times 10^5\text{ m s}^{-1}\). Alternatively, the force on the proton is \(F = qE = 7.20 \times 10^{-15}\text{ N}\), and the acceleration is \(a = F/m = 4.31 \times 10^{12}\text{ m s}^{-2}\). Using \(v^2 = u^2 + 2as\) with \(u = 0\): \(v = \sqrt{2 \times 4.31 \times 10^{12} \times 0.050} \approx 6.57 \times 10^5\text{ m s}^{-1}\).

1 mark: Correct calculation of the work done on the proton (\(3.6 \times 10^{-16}\text{ J}\)) OR the acceleration of the proton (\(4.31 \times 10^{12}\text{ m s}^{-2}\)). 1.14 marks: Correct calculation of the final speed, leading to \(6.6 \times 10^5\text{ m s}^{-1}\) (accept \(6.5 \times 10^5\text{ m s}^{-1}\) to \(6.6 \times 10^5\text{ m s}^{-1}\)).

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

For gas \(X\):
\(c_X = \sqrt{\frac{3kT}{m}}\)

For gas \(Y\):
\(c_Y = \sqrt{\frac{3k(2T)}{4m}} = \sqrt{\frac{6kT}{4m}} = \sqrt{\frac{2}{4}} \sqrt{\frac{3kT}{m}} = \frac{\sqrt{2}}{2} c_X\)

Therefore, the ratio \(\frac{c_Y}{c_X} = \frac{\sqrt{2}}{2}\).

1 mark for the correct calculation of the ratio of root-mean-square speeds leading to Option B.

For a stable circular orbit of radius \(r\), the gravitational force provides the centripetal force:
\(\frac{GMm}{r^2} = \frac{mv^2}{r}\)

Multiplying both sides by \(\frac{1}{2} r\) gives the kinetic energy \(E_{\text{k}}\):
\(E_{\text{k}} = \frac{1}{2}mv^2 = \frac{GMm}{2r}\)

The gravitational potential energy \(E_{\text{p}}\) is:
\(E_{\text{p}} = -\frac{GMm}{r} = -2 E_{\text{k}}\)

The total energy \(E\) of the satellite is:
\(E = E_{\text{k}} + E_{\text{p}} = E_{\text{k}} - 2E_{\text{k}} = -E_{\text{k}}\)

When the orbit radius is changed to \(3r\):
- The new kinetic energy is: \(E_{\text{k}}' = \frac{GMm}{2(3r)} = \frac{1}{3} E_{\text{k}}\)
- The new potential energy is: \(E_{\text{p}}' = -\frac{GMm}{3r} = -\frac{2}{3} E_{\text{k}}\)
- The new total energy is: \(E' = E_{\text{k}}' + E_{\text{p}}' = \frac{1}{3} E_{\text{k}} - \frac{2}{3} E_{\text{k}} = -\frac{1}{3} E_{\text{k}}\)

The work done on the satellite is the change in total energy:
\(W = E' - E = -\frac{1}{3} E_{\text{k}} - (-E_{\text{k}}) = \frac{2}{3} E_{\text{k}}\).

1 mark for the correct application of orbital energy formulas and calculation of the change in total energy to yield Option B.

Let the distance from charge \(+Q\) to point \(P\) be \(x\). Since \(P\) lies between the charges, the distance from \(P\) to \(-4Q\) is \(d - x\).

The electric potential \(V\) at \(P\) is the sum of the potentials due to both charges:
\(V = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q}{x} + \frac{-4Q}{d - x} \right) = 0\)

This simplifies to:
\(\frac{Q}{x} = \frac{4Q}{d - x}\)
\(\frac{1}{x} = \frac{4}{d - x}\)
\(d - x = 4x\)
\(5x = d\)
\(x = 0.20d\).

1 mark for setting up the potential equation and solving for \(x\) to obtain Option A.

The potential difference across the discharging capacitor at time \(t\) is given by:
\(V_t = V e^{-\frac{t}{RC}}\)

Substituting \(t = RC \ln 2\):
\(V_t = V e^{-\frac{RC \ln 2}{RC}} = V e^{-\ln 2} = \frac{V}{2}\)

The energy \(E\) stored in a capacitor is proportional to the square of its potential difference:
\(\frac{E_t}{E_{\text{initial}}} = \left(\frac{V_t}{V}\right)^2 = \left(\frac{1}{2}\right)^2 = 0.25\).

1 mark for calculating the potential difference at the specified time and finding the remaining fraction of energy to be 0.25 (Option B).

Let the initial activity of isotope \(Y\) be \(A_0\). The initial activity of isotope \(X\) is \(2A_0\).

After 16 days:
- For isotope \(X\) (half-life of 4 days), the number of half-lives is \(16 / 4 = 4\).
The final activity of \(X\) is:
\(A_X = 2A_0 \times \left(\frac{1}{2}\right)^4 = \frac{2A_0}{16} = \frac{A_0}{8}\)

- For isotope \(Y\) (half-life of 8 days), the number of half-lives is \(16 / 8 = 2\).
The final activity of \(Y\) is:
\(A_Y = A_0 \times \left(\frac{1}{2}\right)^2 = \frac{A_0}{4}\)

The ratio of the activities after 16 days is:
\(\frac{A_X}{A_Y} = \frac{A_0 / 8}{A_0 / 4} = \frac{4}{8} = 0.50\).

1 mark for calculating the activities after 16 days and correctly evaluating their ratio to be 0.50 (Option B).

The radius of the circular path of a charged particle in a magnetic field is given by:
\(r = \frac{mv}{Bq}\)

For the second particle:
- Mass, \(m' = 2m\)
- Charge, \(q' = 0.5q\)
- Velocity, \(v' = 3v\)
- Magnetic flux density is the same, \(B\).

The radius of its path, \(r'\), is:
\(r' = \frac{m'v'}{Bq'} = \frac{(2m)(3v)}{B(0.5q)} = \frac{6mv}{0.5Bq} = 12 \left(\frac{mv}{Bq}\right) = 12r\).

1 mark for scaling the parameters in the magnetic radius formula to obtain \(12r\) (Option C).

The total energy \(E\) of a simple harmonic oscillator is proportional to the square of its amplitude:
\(E = \frac{1}{2} k A^2\)

At displacement \(x\), the potential energy \(E_{\text{p}}\) is given by:
\(E_{\text{p}} = \frac{1}{2} k x^2\)

For \(x = 0.6A\):
\(E_{\text{p}} = \frac{1}{2} k (0.6A)^2 = 0.36 \left(\frac{1}{2} k A^2\right) = 0.36E\)

By conservation of energy, the kinetic energy \(E_{\text{k}}\) is:
\(E_{\text{k}} = E - E_{\text{p}} = E - 0.36E = 0.64E\).

1 mark for using the conservation of energy to determine that the kinetic energy is \(0.64E\) (Option D).

Let us check the electron-lepton number conservation for Option A: \(\text{n} \rightarrow \text{p} + \text{e}^- + \nu_{\text{e}}\)

- LHS: Lepton number \(L_{\text{e}} = 0\) (since the neutron is a baryon).
- RHS: Proton has \(L_{\text{e}} = 0\); electron (\(\text{e}^-\)) has \(L_{\text{e}} = +1\); electron-neutrino (\(\nu_{\text{e}}\)) has \(L_{\text{e}} = +1\).
Total lepton number on RHS = \(+2\).

Since the lepton number is not conserved (\(0 \neq +2\)), this decay cannot occur. The correct decay equation must produce an electron-antineutrino (\(\bar{\nu}_{\text{e}}\)) with \(L_{\text{e}} = -1\).

1 mark for correctly analyzing the lepton numbers in each reaction and identifying Option A as the incorrect decay.

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

Initially, the total internal energy of the two gases is:
\(U_i = U_1 + U_2 = \frac{3}{2} n_1 R T + \frac{3}{2} n_2 R (2T) = \frac{3}{2} R T (n_1 + 2n_2)\)

After thermal equilibrium is reached at temperature \(1.5T\), the total internal energy of the system is:
\(U_f = \frac{3}{2} (n_1 + n_2) R (1.5T) = \frac{9}{4} R T (n_1 + n_2)\)

Since no heat is lost to the surroundings, internal energy is conserved (\(U_i = U_f\)):
\(\frac{3}{2} R T (n_1 + 2n_2) = \frac{9}{4} R T (n_1 + n_2)\)

Dividing both sides by \(\frac{3}{4} R T\):
\(2(n_1 + 2n_2) = 3(n_1 + n_2)\)
\(2n_1 + 4n_2 = 3n_1 + 3n_2\)
\(n_2 = n_1\)

Therefore, the ratio is \(\frac{n_1}{n_2} = 1.0\).

Award 1 mark for the correct option B.
- Correct identification of energy conservation principle.
- Correct algebraic simplification leading to \(n_1 = n_2\).

For any satellite of mass \(m\) in a circular orbit of radius \(r\) around a planet of mass \(M\), the centripetal force is provided by the gravitational force:
\(\frac{G M m}{r^2} = \frac{m v^2}{r}\)

Multiplying both sides by \(\frac{1}{2} r\), we obtain the kinetic energy \(E_k\):
\(E_k = \frac{1}{2} m v^2 = \frac{G M m}{2r}\)

The gravitational potential energy \(E_p\) is given by:
\(E_p = -\frac{G M m}{r}\)

This shows that in a circular orbit, \(E_p = -2 E_k\).

Since the two satellites have equal kinetic energy (\(E_{k,\text{A}} = E_{k,\text{B}}\)), their gravitational potential energies must also be equal:
\(E_{p,\text{A}} = -2 E_{k,\text{A}} = -2 E_{k,\text{B}} = E_{p,\text{B}}\)

Thus, the ratio of their gravitational potential energies is:
\(\frac{E_{p,\text{A}}}{E_{p,\text{B}}} = 1.0\)

Award 1 mark for the correct option C.
- Correct recall of the relationship between potential energy and kinetic energy in a circular orbit.

Let \(+Q\) be located at \(x = 0\) and \(-4Q\) be located at \(x = d\).
At a point \(x\) between the two charges (\(0 < x < d\)), the total potential \(V\) is:
\(V = \frac{1}{4 \pi \epsilon_0} \left( \frac{Q}{x} + \frac{-4Q}{d - x} \right)\)

Setting \(V = 0\):
\(\frac{Q}{x} = \frac{4Q}{d - x} \implies d - x = 4x \implies 5x = d \implies x = \frac{d}{5}\)

At this point \(x = \frac{d}{5}\), the electric field due to \(+Q\) points away from \(+Q\) (towards \(-4Q\)):
\(E_+ = \frac{Q}{4 \pi \epsilon_0 \left(\frac{d}{5}\right)^2} = \frac{25 Q}{4 \pi \epsilon_0 d^2}\)

The electric field due to \(-4Q\) points towards \(-4Q\) (also away from \(+Q\)):
\(E_- = \frac{4Q}{4 \pi \epsilon_0 \left(d - \frac{d}{5}\right)^2} = \frac{4Q}{4 \pi \epsilon_0 \left(\frac{4d}{5}\right)^2} = \frac{25 Q}{16 \pi \epsilon_0 d^2}\)

Since both electric fields point in the same direction, they add together:
\(E = E_+ + E_- = \frac{25 Q}{4 \pi \epsilon_0 d^2} + \frac{25 Q}{16 \pi \epsilon_0 d^2} = \frac{100 Q + 25 Q}{16 \pi \epsilon_0 d^2} = \frac{125 Q}{16 \pi \epsilon_0 d^2}\)

Award 1 mark for the correct option C.
- Correct identification of the position where potential is zero.
- Correct vector addition of the electric field strengths.

The initial electrostatic energy stored in the capacitor \(C\) is:
\(E_i = \frac{1}{2} C V^2\)

The initial charge on this capacitor is \(Q = C V\).
When connected in parallel with an uncharged capacitor of capacitance \(2C\), the total charge is conserved:
\(Q_f = Q = C V\)

The total equivalent capacitance in parallel is:
\(C_{\text{eq}} = C + 2C = 3C\)

The final energy stored in the parallel combination is:
\(E_f = \frac{Q_f^2}{2 C_{\text{eq}}} = \frac{(C V)^2}{2 (3C)} = \frac{1}{6} C V^2\)

The energy dissipated as heat and radiation is:
\(\Delta E = E_i - E_f = \frac{1}{2} C V^2 - \frac{1}{6} C V^2 = \frac{1}{3} C V^2\)

The fraction of the initial energy dissipated is:
\(\frac{\Delta E}{E_i} = \frac{\frac{1}{3} C V^2}{\frac{1}{2} C V^2} = \frac{2}{3}\)

Award 1 mark for the correct option C.
- Correct expression for initial and final energy.
- Correct fraction determination.

Let the initial activity of Y be \(A_0\). The initial activity of X is therefore \(4A_0\).

The activity of a nuclide as a function of time \(t\) is given by:
\(A(t) = A_{\text{initial}} \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}\)

At time \(t\), the activities are:
\(A_{\text{X}}(t) = 4 A_0 \left(\frac{1}{2}\right)^{\frac{t}{3.0}}\)
\(A_{\text{Y}}(t) = A_0 \left(\frac{1}{2}\right)^{\frac{t}{6.0}}\)

For the activities to be equal:
\(4 A_0 \left(\frac{1}{2}\right)^{\frac{t}{3.0}} = A_0 \left(\frac{1}{2}\right)^{\frac{t}{6.0}}\)

Dividing both sides by \(A_0\):
\(4 = \frac{\left(\frac{1}{2}\right)^{\frac{t}{6.0}}}{\left(\frac{1}{2}\right)^{\frac{t}{3.0}}} = \left(\frac{1}{2}\right)^{\frac{t}{6.0} - \frac{t}{3.0}} = \left(\frac{1}{2}\right)^{-\frac{t}{6.0}} = 2^{\frac{t}{6.0}}\)

Since \(4 = 2^2\):
\(2^{\frac{t}{6.0}} = 2^2 \implies \frac{t}{6.0} = 2 \implies t = 12 \text{ hours}\)

Award 1 mark for the correct option C.
- Setting up the correct ratio or equality using the decay law.
- Correct algebraic solution.

A particle entering a uniform magnetic field perpendicularly and undergoing a semi-circular path spends a time interval equal to half the period of its uniform circular motion.

The orbital period \(T\) is given by:
\(T = \frac{2\pi m}{q B}\)

Therefore, the time interval \(\Delta t\) spent in the field is:
\(\Delta t = \frac{T}{2} = \frac{\pi m}{q B}\)

Notice that \(\Delta t\) is independent of the particle's velocity \(v\).

For the second particle:
\(m' = 2m\)
\(q' = 2q\)

Its time interval in the field is:
\(\Delta t' = \frac{\pi m'}{q' B} = \frac{\pi (2m)}{(2q) B} = \frac{\pi m}{q B} = \Delta t\)

Thus, the time interval remains unchanged as \(\Delta t\).

Award 1 mark for the correct option B.
- Recognizing that circular period/half-period is independent of speed.
- Calculating correct substitution of mass and charge.

The total energy of a simple harmonic oscillator is:
\(E_{\text{total}} = \frac{1}{2} k A^2\)

At a displacement of \(x = \frac{1}{3}A\), the potential energy \(E_p\) of the system is:
\(E_p = \frac{1}{2} k x^2 = \frac{1}{2} k \left(\frac{1}{3}A\right)^2 = \frac{1}{9} \left(\frac{1}{2} k A^2\right) = \frac{1}{9} E_{\text{total}}\)

Since energy is conserved, the kinetic energy \(E_k\) at this point is:
\(E_k = E_{\text{total}} - E_p = E_{\text{total}} - \frac{1}{9} E_{\text{total}} = \frac{8}{9} E_{\text{total}}\)

The ratio of kinetic energy to potential energy is:
\(\frac{E_k}{E_p} = \frac{\frac{8}{9} E_{\text{total}}}{\frac{1}{9} E_{\text{total}}} = 8\)

Award 1 mark for the correct option C.
- Finding potential energy in terms of total energy.
- Calculating correct ratio.

Let wire Y have length \(L\), diameter \(D\), and cross-sectional area \(A = \frac{\pi D^2}{4}\).

Wire X has length \(2L\), diameter \(0.5D\), and cross-sectional area:
\(A_{\text{X}} = \frac{\pi (0.5D)^2}{4} = \frac{A}{4}\)

Both wires support the same load \(F\) and have the same Young modulus \(E\). The extension \(\Delta L\) is given by:
\(\Delta L = \frac{F L}{A E}\)

For wire Y:
\(\Delta L_{\text{Y}} = \frac{F L}{A E}\)

For wire X:
\(\Delta L_{\text{X}} = \frac{F (2L)}{\left(\frac{A}{4}\right) E} = 8 \frac{F L}{A E} = 8 \Delta L_{\text{Y}}\)

The elastic strain energy stored is given by \(U = \frac{1}{2} F \Delta L\). Since the force \(F\) is equal for both:
\(\frac{U_{\text{X}}}{U_{\text{Y}}} = \frac{\frac{1}{2} F \Delta L_{\text{X}}}{\frac{1}{2} F \Delta L_{\text{Y}}} = \frac{\Delta L_{\text{X}}}{\Delta L_{\text{Y}}} = 8\)

Award 1 mark for the correct option C.
- Deducing that the area of X is a quarter of the area of Y.
- Deducing that the extension of X is eight times that of Y.
- Calculating correct ratio of stored energy.

Using the ideal gas equation: \(pV = nRT\), which can be written as \(p = \frac{nRT}{V}\).

Let the initial state be \(p_1, V_1, T_1\) and the final state be \(p_2, V_2, T_2\).

We are given that \(T_2 = 2T_1\) and \(V_2 = 0.5V_1\).

Therefore, \(p_2 = \frac{nR(2T_1)}{0.5V_1} = 4 \left(\frac{nRT_1}{V_1}\right) = 4p_1\). The pressure ratio \(\frac{p_2}{p_1} = 4\).

The root mean square speed \(c_{\text{rms}}\) is given by \(c_{\text{rms}} = \sqrt{\frac{3RT}{M}}\), meaning \(c_{\text{rms}} \propto \sqrt{T}\).

Since the absolute temperature is doubled, the ratio of the new rms speed to the original rms speed is \(\sqrt{\frac{T_2}{T_1}} = \sqrt{2}\).

1 mark for selecting correct option B.
- Correctly identifies that pressure increases by a factor of 4.
- Correctly identifies that rms speed increases by a factor of \(\sqrt{2}\).

The nuclear radius \(R\) is given by the formula: \(R = R_0 A^{1/3}\), where \(R_0\) is a constant and \(A\) is the nucleon (mass) number.

For the gold nucleus: \(R_{\text{Au}} = R_0 (197)^{1/3}\).

For the helium nucleus: \(R_{\text{He}} = R_0 (4)^{1/3}\).

Taking the ratio of the two equations:
\(\frac{R_{\text{He}}}{R_{\text{Au}}} = \left(\frac{4}{197}\right)^{1/3}\)

Therefore, \(R_{\text{He}} = R_{\text{Au}} \left(\frac{4}{197}\right)^{1/3}\).

1 mark for selecting correct option B.
- Recognises that nuclear radius is proportional to the cube root of the nucleon number (\(A^{1/3}\)).
- Correctly applies the ratio using nucleon numbers 4 and 197.

Gravitational potential energy \(E_p\) is given by: \(E_p = -\frac{GMm}{r}\).

Initially, the orbit radius is \(2R\):
\(E_{p,\text{initial}} = -\frac{GMm}{2R}\)

Finally, the orbit radius is \(3R\):
\(E_{p,\text{final}} = -\frac{GMm}{3R}\)

Change in potential energy \(\Delta E_p = E_{p,\text{final}} - E_{p,\text{initial}}\):
\(\Delta E_p = \left(-\frac{GMm}{3R}\right) - \left(-\frac{GMm}{2R}\right)\)
\(\Delta E_p = \frac{GMm}{R} \left(\frac{1}{2} - \frac{1}{3}\right) = \frac{GMm}{6R}\).

Since this is positive, it represents an increase of \(\frac{GMm}{6R}\).

1 mark for selecting correct option A.
- Uses the correct potential energy formula with negative signs.
- Subtracts initial GPE from final GPE to find positive increase of \(\frac{GMm}{6R}\).

Initially, the droplet is stationary, which means the electrostatic force upwards balances the weight downwards:
\(F_E = mg \implies qE = mg \implies q\frac{V}{d} = mg\).

When the potential difference is increased to \(1.5V\), the new electric field strength is \(1.5E\), and the new electric force is:
\(F_E' = q\frac{1.5V}{d} = 1.5 \left(q\frac{V}{d}\right) = 1.5mg\).

The net upward force \(F_{\text{net}}\) acting on the droplet is:
\(F_{\text{net}} = F_E' - mg = 1.5mg - mg = 0.5mg\).

Using Newton's second law, \(F_{\text{net}} = ma\):
\(0.5mg = ma \implies a = 0.5g\).

1 mark for selecting correct option A.
- Identifies that \(q\frac{V}{d} = mg\).
- Finds the net force is \(0.5mg\).
- Calculates the acceleration to be \(0.5g\).

The potential difference \(V\) across the capacitor during discharge decays according to:
\(V = V_0 e^{-\frac{t}{RC}}\).

Substituting the given time \(t = RC \ln(4)\):
\(V = V_0 e^{-\frac{RC \ln(4)}{RC}} = V_0 e^{-\ln(4)} = V_0 \left(4^{-1}\right) = \frac{V_0}{4}\).

The energy \(E\) stored in a capacitor is given by \(E = \frac{1}{2} C V^2\).

Therefore, the remaining energy is:
\(E = \frac{1}{2} C \left(\frac{V_0}{4}\right)^2 = \frac{1}{16} \left(\frac{1}{2} C V_0^2\right) = \frac{1}{16} E_0\).

The fraction of the initial energy remaining is \(\frac{1}{16}\).

1 mark for selecting correct option D.
- Correctly determines that the voltage drops to \(\frac{1}{4}\) of its initial value.
- Relates energy to the square of voltage and calculates the fraction as \(\left(\frac{1}{4}\right)^2 = \frac{1}{16}\).

For a charged particle in a magnetic field, the centripetal force is provided by the magnetic force:
\(qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB}\).

We can express momentum \(p = mv\) in terms of kinetic energy \(E_k\):
\(E_k = \frac{p^2}{2m} \implies p = \sqrt{2mE_k}\).

Thus, the radius is:
\(r = \frac{\sqrt{2mE_k}}{qB}\).

Since \(E_k\) and \(B\) are the same for both particles, \(r \propto \frac{\sqrt{m}}{q}\).

For the alpha particle: \(r_{\alpha} \propto \frac{\sqrt{4}}{2} = 1\).

For the proton: \(r_p \propto \frac{\sqrt{1}}{1} = 1\).

Therefore, the ratio \(\frac{r_{\alpha}}{r_p} = 1.0\).

1 mark for selecting correct option B.
- Uses circular motion equation and kinetic energy to show \(r \propto \frac{\sqrt{m}}{q}\).
- Plugs in relative mass and charge values to get a ratio of 1.

Let the initial number of nuclei of each type be \(N_0\).

After 16.0 hours:
- For X, 16.0 hours corresponds to \(\frac{16.0}{4.0} = 4\) half-lives.
The number of remaining X nuclei is \(N_X = N_0 \times \left(\frac{1}{2}\right)^4 = \frac{N_0}{16}\).

- For Y, 16.0 hours corresponds to \(\frac{16.0}{8.0} = 2\) half-lives.
The number of remaining Y nuclei is \(N_Y = N_0 \times \left(\frac{1}{2}\right)^2 = \frac{N_0}{4}\).

Activity is given by \(A = \lambda N = \frac{\ln(2)}{T_{1/2}} N\).

Therefore, the ratio of activities is:
\(\frac{A_X}{A_Y} = \frac{\lambda_X N_X}{\lambda_Y N_Y} = \frac{\frac{\ln(2)}{4.0} \times \frac{N_0}{16}}{\frac{\ln(2)}{8.0} \times \frac{N_0}{4}} = \frac{8.0}{4.0} \times \frac{4}{16} = 2 \times \frac{1}{4} = 0.5\).

1 mark for selecting correct option B.
- Calculates remaining fractions of nuclei correctly for X (1/16) and Y (1/4).
- Correctly applies the definition of activity \(A = \lambda N\) to find the ratio.

Total energy supplied by the heater in time \(t\) is \(P t\).

The energy used to heat the liquid is \(m c (T_2 - T_1)\).

The energy lost to the surroundings in time \(t\) is \(Q_{\text{loss}} t\).

By conservation of energy:
\(P t = m c (T_2 - T_1) + Q_{\text{loss}} t\).

Subtracting \(Q_{\text{loss}} t\) from both sides:
\((P - Q_{\text{loss}}) t = m c (T_2 - T_1)\).

Dividing by \(t\):
\(P - Q_{\text{loss}} = \frac{m c (T_2 - T_1)}{t}\).

1 mark for selecting correct option A.
- Writes correct conservation of energy equation containing power and rate of heat loss.
- Rearranges algebraic terms correctly.

1. Calculate the mean diameter: \(d = \frac{0.32 + 0.34 + 0.31 + 0.33 + 0.35}{5} = 0.33\text{ mm}\). 2. Find the range of the measurements: \(\text{Range} = 0.35 - 0.31 = 0.04\text{ mm}\). 3. Calculate the absolute uncertainty in diameter as half the range: \(\Delta d = \frac{0.04}{2} = 0.02\text{ mm}\). 4. Calculate the percentage uncertainty in the diameter: \(\%\Delta d = \frac{0.02}{0.33} \times 100 \approx 6.06\%\). 5. Since cross-sectional area \(A \propto d^2\), the percentage uncertainty in area is twice the percentage uncertainty in diameter: \(\%\Delta A = 2 \times \%\Delta d = 2 \times 6.06\% = 12.12\%\). Rounded to 2 significant figures, this is \(12\%\).

1 mark: Correct calculation of the mean diameter (\(0.33\text{ mm}\)) and the absolute uncertainty in the diameter (\(0.02\text{ mm}\)). 1 mark: Calculation of percentage uncertainty in the diameter (\(6.06\%\)). 0.37 marks: Final percentage uncertainty in the area (\(12\%\)), rounded to 2 significant figures.

1. Find the time period of one oscillation: \(T = \frac{36.0}{20} = 1.80\text{ s}\). The absolute uncertainty in \(T\) is \(\Delta T = \frac{0.2}{20} = 0.01\text{ s}\). 2. The formula for the acceleration due to gravity is \(g = \frac{4\pi^2 L}{T^2}\). 3. Calculate the percentage uncertainty in \(L\): \(\%\Delta L = \frac{0.002}{0.800} \times 100 = 0.25\%\). 4. Calculate the percentage uncertainty in \(T\): \(\%\Delta T = \frac{0.01}{1.80} \times 100 \approx 0.556\%\). 5. The total percentage uncertainty in \(g\) is \(\%\Delta g = \%\Delta L + 2 \times \%\Delta T = 0.25\% + 2(0.556\%) = 1.361\%\). 6. The calculated value of \(g\) is \(g = \frac{4\pi^2 \times 0.800}{1.80^2} \approx 9.748\text{ m s}^{-2}\). 7. The absolute uncertainty in \(g\) is \(\Delta g = 9.748 \times 0.01361 \approx 0.133\text{ m s}^{-2}\). Rounded to 2 significant figures, this is \(0.13\text{ m s}^{-2}\).

1 mark: Calculation of percentage uncertainties in length (\(0.25\%\)) and time period (\(0.556\%\)), combined to get percentage uncertainty in \(g\) of \(1.36\%\). 1 mark: Calculation of the standard value of \(g = 9.75\text{ m s}^{-2}\). 0.37 marks: Calculation of the absolute uncertainty as \(0.13\text{ m s}^{-2}\) to 2 significant figures.

1. When a voltmeter is connected in parallel with the resistor, the total resistance of the parallel combination is the apparent resistance measured: \(R_{\text{apparent}} = \frac{R \times R_v}{R + R_v}\). 2. Substitute the values: \(R_{\text{apparent}} = \frac{500 \times 20000}{500 + 20000} = \frac{10^7}{20500} \approx 487.8\ \Omega\). 3. The percentage error is \(\frac{|R - R_{\text{apparent}}|}{R} \times 100 = \frac{500 - 487.8}{500} \times 100 = \frac{12.2}{500} \times 100 = 2.44\%\). Rounded to 2 significant figures, this is \(2.4\%\).

1 mark: Formula or calculation of the parallel equivalent resistance of the resistor and voltmeter (\(487.8\ \Omega\)). 1 mark: Formula for percentage error using the nominal value (\(500\ \Omega\)) and equivalent resistance. 0.37 marks: Correct answer of \(2.4\%\) to 2 significant figures.

1. The distance from the central maximum to either second-order maximum is \(x = \frac{76.0}{2} = 38.0\text{ cm} = 0.380\text{ m}\). 2. The angle \(\theta\) for the second-order maximum is given by \(\tan\theta = \frac{x}{D} = \frac{0.380}{1.50} \approx 0.2533\), so \(\theta = \tan^{-1}(0.2533) \approx 14.22^\circ\). 3. Use the grating equation \(d \sin\theta = n \lambda\) with \(n=2\): \(d = \frac{2 \times 630 \times 10^{-9}}{\sin(14.22^\circ)} = \frac{1.26 \times 10^{-6}}{\sin(14.22^\circ)} \approx 5.13 \times 10^{-6}\text{ m}\). 4. The number of lines per millimetre is \(N = \frac{10^{-3}}{d} = \frac{10^{-3}}{5.13 \times 10^{-6}} \approx 195\text{ lines mm}^{-1}\).

1 mark: Correct determination of the diffraction angle \(\theta = 14.2^\circ\) using the half-distance of \(38.0\text{ cm}\). 1 mark: Application of the grating equation to find the slit spacing \(d = 5.13 \times 10^{-6}\text{ m}\). 0.37 marks: Final answer of \(195\) lines per millimetre.

1. The Young modulus is defined as \(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L_0} = \left(\frac{F}{\Delta L}\right) \frac{L_0}{A}\). 2. The gradient of the graph represents \(\frac{F}{\Delta L} = 1.18 \times 10^4\text{ N m}^{-1}\). 3. The cross-sectional area of the wire is \(A = \frac{\pi d^2}{4} = \frac{\pi (0.50 \times 10^{-3})^2}{4} \approx 1.963 \times 10^{-7}\text{ m}^2\). 4. Calculate the Young modulus: \(E = 1.18 \times 10^4 \times \frac{2.00}{1.963 \times 10^{-7}} \approx 1.20 \times 10^{11}\text{ Pa} = 120\text{ GPa}\). To 2 significant figures, this is \(120\text{ GPa}\).

1 mark: Correct calculation of the cross-sectional area \(A = 1.96 \times 10^{-7}\text{ m}^2\). 1 mark: Correct expression relating the gradient to the Young modulus: \(E = \text{gradient} \times \frac{L_0}{A}\). 0.37 marks: Calculation of the Young modulus as \(120\text{ GPa}\) (accept 120).

1. Set up the simultaneous equations for both trials to account for constant heat transfer rate to the surroundings: \(P_1 t = m_1 L_v + Q_{\text{loss}}\) and \(P_2 t = m_2 L_v + Q_{\text{loss}}\). 2. Subtract the two equations: \((P_2 - P_1) t = (m_2 - m_1) L_v\). 3. Substitute the given values: \((75.0 - 45.0) \times 300 = (8.5 \times 10^{-3} - 4.8 \times 10^{-3}) L_v\). 4. \(30.0 \times 300 = 3.7 \times 10^{-3} L_v \implies 9000 = 3.7 \times 10^{-3} L_v\). 5. Solve for \(L_v\): \(L_v = \frac{9000}{3.7 \times 10^{-3}} \approx 2.43 \times 10^6\text{ J kg}^{-1} = 2.43\text{ MJ kg}^{-1}\). Rounded to 2 significant figures, this is \(2.4\text{ MJ kg}^{-1}\).

1 mark: Setting up the subtraction formula \((P_2 - P_1)t = (m_2 - m_1)L_v\) or showing that subtracting the simultaneous equations eliminates the heat loss term. 1 mark: Correct substitution of values with mass converted to kg (\(3.7 \times 10^{-3}\text{ kg}\)). 0.37 marks: Final answer of \(2.4\text{ MJ kg}^{-1}\) (accept 2.4).

1. The discharge equation is \(V = V_0 e^{-t/RC}\). 2. Taking the natural logarithm of both sides gives \(\ln(V) = \ln(V_0) - \frac{t}{RC}\). 3. Therefore, a plot of \(\ln(V)\) against \(t\) has a gradient of \(-\frac{1}{RC}\). 4. Set the gradient equal to the measured value: \(-0.303 = -\frac{1}{RC} \implies RC = \frac{1}{0.303} \approx 3.300\text{ s}\). 5. Solve for \(C\) using \(R = 15 \times 10^3\ \Omega\): \(C = \frac{3.300}{15 \times 10^3} \approx 2.20 \times 10^{-4}\text{ F} = 220\ \mu\text{F}\). Rounded to 2 significant figures, this is \(220\ \mu\text{F}\).

1 mark: Identifying that the gradient of the graph is equal to \(-\frac{1}{RC}\). 1 mark: Calculation of the time constant \(RC \approx 3.30\text{ s}\). 0.37 marks: Correct calculation of capacitance as \(220\ \mu\text{F}\) (accept 220).

1. Find the peak-to-peak voltage: \(V_{pp} = 4.2\text{ divisions} \times 50\text{ mV/division} = 210\text{ mV} = 0.210\text{ V}\). 2. Find the peak voltage \(V_0\): \(V_0 = \frac{V_{pp}}{2} = 105\text{ mV} = 0.105\text{ V}\). 3. The induced EMF in the search coil is given by \(V_0 = N A B_0 \omega = N A B_0 (2 \pi f)\). 4. Rearrange to solve for \(B_0\): \(B_0 = \frac{V_0}{2 \pi f N A}\). 5. Substitute the values: \(B_0 = \frac{0.105}{2 \pi \times 50 \times 500 \times 1.2 \times 10^{-4}} = \frac{0.105}{18.85} \approx 5.57 \times 10^{-3}\text{ T} = 5.57\text{ mT}\). Rounded to 2 significant figures, this is \(5.6\text{ mT}\).

1 mark: Correct calculation of the peak voltage \(V_0 = 0.105\text{ V}\) from the peak-to-peak reading. 1 mark: Correct recall and rearrangement of the formula \(V_0 = 2\pi f N A B_0\). 0.37 marks: Final answer of \(5.6\text{ mT}\) (accept 5.6).

First, find the mean diameter \(d = \frac{0.45 + 0.46 + 0.44 + 0.45 + 0.45}{5} = 0.45\text{ mm}\). Next, find the absolute uncertainty in \(d\) as half the range: \\text{Uncertainty} = \\frac{0.46 - 0.44}{2} = 0.01\\text{ mm}\). The percentage uncertainty in the diameter is \(\%\Delta d = \frac{0.01}{0.45} \times 100\% \approx 2.22\%\). Since the cross-sectional area is \(A = \frac{\pi d^2}{4}\), the percentage uncertainty in area is twice the percentage uncertainty in diameter: \(\%\Delta A = 2 \times 2.22\% \approx 4.4\%\).

1.00 mark for calculating the mean diameter and absolute uncertainty of 0.01 mm; 1.37 marks for correctly doubling the percentage uncertainty of the diameter to find the area uncertainty (accept answers between 4.4% and 4.44%).

First, calculate the value of \(g = \frac{2 \times 1.250}{0.505^2} \approx 9.803\text{ m s}^{-2}\). Next, calculate the percentage uncertainty in \(h\): \(\%\Delta h = \frac{0.002}{1.250} \times 100\% = 0.16\%\). Calculate the percentage uncertainty in \(t\): \(\%\Delta t = \frac{0.005}{0.505} \times 100\% \approx 0.99\%\). The percentage uncertainty in \(g\) is \(\%\Delta g = \%\Delta h + 2 \times \%\Delta t = 0.16\% + 2 \times 0.99\% = 2.14\%\). Finally, calculate the absolute uncertainty in \(g\): \(\Delta g = 9.803 \times 0.0214 \approx 0.21\text{ m s}^{-2}\).

1.00 mark for finding the total percentage uncertainty of 2.14%; 1.37 marks for converting this to the absolute uncertainty of 0.21 m s^{-2} (accept 0.2 or 0.21 m s^{-2}).

First, calculate the cross-sectional area \(A\) of the wire: \(r = 0.40\text{ mm} = 4.0 \times 10^{-4}\text{ m}\) and \(A = \pi r^2 = \pi \times (4.0 \times 10^{-4})^2 \approx 5.027 \times 10^{-7}\text{ m}^2\). Now use the Young modulus formula: \(E = \frac{F L}{A \Delta L} = \frac{35.0 \times 2.50}{(5.027 \times 10^{-7}) \times (1.50 \times 10^{-3})} \approx 1.16 \times 10^{11}\text{ Pa}\). Convert to GPa: \(1.16 \times 10^{11}\text{ Pa} = 116\text{ GPa}\). Rounding to 2 significant figures gives \(120\text{ GPa}\).

1.00 mark for correct calculation of the cross-sectional area; 1.37 marks for the correct Young modulus calculation in GPa to 2 s.f.

Using Boyle's Law: \(p_1 V_1 = p_2 V_2 \implies 1.01 \times 10^5 \times 3.50 \times 10^{-5} = p_2 \times 2.10 \times 10^{-5}\). Solving for \(p_2\) gives \(p_2 = \frac{1.01 \times 10^5 \times 3.50}{2.10} = 1.68 \times 10^5\text{ Pa}\). The change must be made slowly so that the system remains in thermal equilibrium with its surroundings, ensuring that any heat generated during compression escapes, keeping the temperature constant.

1.00 mark for calculating the pressure of 1.68 x 10^5 Pa; 1.37 marks for explaining that a slow change allows heat transfer to maintain a constant temperature.

First, find the angle of diffraction \(\theta\) for the second-order maximum: \(\tan \theta = \frac{0.82}{1.50} \implies \theta \approx 28.67^\circ\). Using the grating equation: \(d \sin \theta = n \lambda \implies d = \frac{2 \times 632.8 \times 10^{-9}}{\sin(28.67^\circ)} \approx 2.638 \times 10^{-6}\text{ m}\). Calculate the number of lines per millimetre \(N\): \(N = \frac{10^{-3}}{d} = \frac{10^{-3}}{2.638 \times 10^{-6}} \approx 379\text{ lines/mm}\). Rounding to two significant figures gives \(380\text{ lines/mm}\).

1.00 mark for finding the angle of diffraction theta; 1.37 marks for the correct lines per mm calculation (accept 379 or 380).

The discharge equation is \(V = V_0 e^{-t/RC}\). Taking the natural logarithm: \(\ln(V) = \ln(V_0) - \frac{t}{RC}\). Thus, the gradient of the graph of \(\ln(V)\) against \(t\) is equal to \(-\frac{1}{RC}\). Given \\text{gradient} = -0.085 = -\\frac{1}{RC} \\implies RC = \\frac{1}{0.085} \\approx 11.76\\text{ s}\). Now calculate \(C\): \(C = \frac{11.76}{47 \times 10^3} \approx 2.50 \times 10^{-4}\text{ F} = 250\text{ }\mu\text{F}\).

1.00 mark for identifying the gradient as -1/RC; 1.37 marks for the correct calculation of capacitance to 250 microfarads (accept 250 or 250.3).

According to the inverse-square law with a distance offset \(d_0\): \(R = \frac{k}{(d + d_0)^2}\), where \(d\) is the measured distance and \(d_0\) is the distance between the active material inside the casing and the start of the ruler. Taking the square root and reciprocal of both sides: \(R^{-1/2} = \frac{d + d_0}{\sqrt{k}} = \frac{1}{\sqrt{k}}d + \frac{d_0}{\sqrt{k}}\). This is of the form \(y = mx + c\), which is a straight-line equation. When \(R^{-1/2} = 0\), \(d = -d_0\). Therefore, the negative of the \(d\)-intercept represents the distance offset \(d_0\).

1.00 mark for deriving the linear relationship between R^{-1/2} and d; 1.37 marks for identifying the d-intercept as the negative of the offset distance d_0.

The period of oscillation of a mass-spring system is given by \(T = 2\pi\sqrt{\frac{m}{k}}\). Squaring both sides yields: \(T^2 = \frac{4\pi^2}{k} m\). Thus, the gradient of the graph of \(T^2\) against \(m\) is \\text{gradient} = \\frac{4\\pi^2}{k}\). Given the gradient is \(1.62\text{ s}^2\text{ kg}^{-1}\): \(1.62 = \frac{4\pi^2}{k} \implies k = \frac{4\pi^2}{1.62} \approx 24.37\text{ N m}^{-1}\). Rounding to 3 significant figures gives \(24.4\text{ N m}^{-1}\).

1.00 mark for identifying that gradient = 4 pi^2 / k; 1.37 marks for calculating the correct value of 24.4 N m^-1 (accept 24 N m^-1 if 2 s.f. is used).

1. Calculate the mean diameter: \( d_{\text{mean}} = \frac{0.42 + 0.43 + 0.41 + 0.44 + 0.42}{5} = 0.424\text{ mm} \).
2. Calculate the range of the measurements: \( \text{Range} = 0.44 - 0.41 = 0.03\text{ mm} \).
3. Calculate the absolute uncertainty in the diameter: \( \Delta d = \frac{\text{Range}}{2} = 0.015\text{ mm} \).
4. Calculate the percentage uncertainty in the diameter: \( \%\Delta d = \frac{0.015}{0.424} \times 100\% \approx 3.538\% \).
5. Since the cross-sectional area is \( A = \frac{\pi d^2}{4} \), the percentage uncertainty in \( A \) is twice the percentage uncertainty in \( d \): \( \%\Delta A = 2 \times \%\Delta d = 2 \times 3.538\% = 7.08\% \).
Rounding to a sensible number of significant figures gives 7.1%.

Award [1 mark] for calculating the mean diameter of 0.424 mm and its absolute uncertainty of 0.015 mm.
Award [1 mark] for finding the percentage uncertainty of the diameter as 3.54%.
Award [0.37 mark] for doubling the percentage uncertainty of the diameter to find the percentage uncertainty of the area as 7.1% (accept 7.08% or 7%).

1. Calculate the percentage uncertainty in the length \( L \): \( \%\Delta L = \frac{0.002}{0.850} \times 100\% \approx 0.235\% \).
2. Calculate the percentage uncertainty in the period \( T \) (which is equal to the percentage uncertainty in the total time \( t \)): \( \%\Delta T = \frac{0.2}{37.0} \times 100\% \approx 0.541\% \).
3. Use the formula for percentage uncertainty propagation: \( \%\Delta g = \%\Delta L + 2 \times \%\Delta T \).
4. Calculate the result: \( \%\Delta g = 0.235\% + 2 \times 0.541\% = 1.317\% \).
Rounding to two significant figures gives 1.3%.

Award [1 mark] for calculating the percentage uncertainty in L (0.24%) and T (0.54%).
Award [1 mark] for stating or using the correct propagation rule: % uncertainty in g = % uncertainty in L + 2 * % uncertainty in T.
Award [0.37 mark] for the final correct answer of 1.3% (accept 1.32% or 1.3%).

1. Calculate the fringe spacing: \( w = \frac{w_{10}}{10} = 3.71\text{ mm} \).
2. Calculate the central value of wavelength \( \lambda \): \( \lambda = \frac{a w}{D} = \frac{0.35 \times 10^{-3} \times 3.71 \times 10^{-3}}{2.00} = 6.4925 \times 10^{-7}\text{ m} = 649.25\text{ nm} \).
3. Calculate the percentage uncertainties:
- \( \%\Delta w = \frac{0.1}{37.1} \times 100\% \approx 0.270\% \)
- \( \%\Delta a = \frac{0.01}{0.35} \times 100\% \approx 2.857\% \)
- \( \%\Delta D = \frac{0.01}{2.00} \times 100\% = 0.500\% \)
4. Calculate the total percentage uncertainty in \( \lambda \): \( \%\Delta \lambda = 2.857\% + 0.270\% + 0.500\% = 3.627\% \).
5. Calculate the absolute uncertainty in \( \lambda \): \( \Delta \lambda = 649.25\text{ nm} \times 0.03627 \approx 23.55\text{ nm} \).
To 2 significant figures, the absolute uncertainty is 24 nm.

Award [1 mark] for calculating the central value of wavelength as 649 nm (or 6.5 x 10^-7 m).
Award [1 mark] for summing the percentage uncertainties of a, w, and D to get a total of 3.63%.
Award [0.37 mark] for calculating the absolute uncertainty and stating it to 2 significant figures as 24 nm (accept 23.6 nm or 24 nm).