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Since the wires are connected in series, they carry the same electric current \(I\). The relationship between drift velocity and current is given by the formula \(I = nAve\), where \(n\) is the number density of conduction electrons, \(A\) is the cross-sectional area, \(v\) is the drift velocity, and \(e\) is the elementary charge. Because both wires are made of copper, \(n\) and \(e\) are the same for both. Therefore, \(A_X v_X = A_Y v_Y\), which gives \(v_X / v_Y = A_Y / A_X\). The cross-sectional area is proportional to the square of the diameter: \(A = \pi d^2 / 4\). Given that the diameter of wire X is twice that of wire Y (\(d_X = 2d_Y\)), the area of wire X is four times larger than wire Y (\(A_X = 4A_Y\)). Hence, the ratio of the drift velocities is \(v_X / v_Y = A_Y / 4A_Y = 0.25\).

1 mark: Correctly identifies that current is identical in series, uses the formula for drift velocity to relate it inversely to cross-sectional area, and calculates the ratio as 0.25.

For undeflected horizontal motion, the magnetic force and the electric force on the electron must be equal in magnitude and opposite in direction. Using Fleming's Left-Hand Rule for a negative charge, the magnetic force on the electron traveling right in a magnetic field directed into the page is directed downwards. Therefore, the electric force must be directed upwards. Since the electron is negatively charged, the electric field must point in the opposite direction to the required electric force, which means the electric field must point downwards. Equating the magnitudes of the electric and magnetic forces gives \(eE = evB\), which simplifies to \(v = E/B\).

1 mark: Identifies that the electric field must point downwards to exert an upward force on the electron, and equates the force equations to obtain the speed \(v = E/B\).

Let the initial speed of the ball be \(v\). Its initial kinetic energy is \(E_k = \frac{1}{2}mv^2\). At the highest point of its trajectory, the vertical component of the velocity is zero, and only the horizontal component of the velocity remains. Since air resistance is negligible, the horizontal component of velocity is constant and is given by \(v_h = v \cos(60^\circ) = 0.5v\). The kinetic energy of the ball at the highest point is \(E_{\text{top}} = \frac{1}{2}m(v_h)^2 = \frac{1}{2}m(0.5v)^2 = 0.25 \left(\frac{1}{2}mv^2\right) = 0.25 E_k\).

1 mark: Recognizes that only horizontal velocity exists at the highest point, applies the horizontal velocity component formula, and correctly determines that the kinetic energy becomes 0.25 of the original kinetic energy.

The fringe spacing \(x\) is given by the formula \(x = \frac{\lambda D}{s}\), where \(s\) is the slit separation. If the new wavelength is \(\lambda' = 1.5\lambda\) and the new slit separation is \(s' = 0.5s\), the new fringe spacing \(x'\) is \(x' = \frac{\lambda' D}{s'} = \frac{1.5\lambda D}{0.5s} = 3 \left(\frac{\lambda D}{s}\right) = 3x\).

1 mark: Uses the double-slit equation to analyze the effect of scaling the wavelength and slit separation, calculating a factor of 3 increase.

Let's check the conservation laws for the interaction. Baryon number: Left side has \(1 \text{ (proton)} + 1 \text{ (neutron)} = 2\). Right side has \(0 \text{ (pion)} + B_X\). Thus, \(B_X\) must be 2. Electric charge: Left side has \(+1 \text{ (proton)} + 0 \text{ (neutron)} = +1\). Right side has \(0 \text{ (pion)} + Q_X\). Thus, \(Q_X\) must be +1. Lepton number: Left side has \(0\). Right side has \(0 + L_X\). Thus, \(L_X\) must be 0. A deuteron (\(^{2}_{1}\text{H}\)) is a nucleus containing one proton and one neutron, giving it a baryon number of 2, a charge of +1, and a lepton number of 0. This matches all criteria.

1 mark: Correctly balances baryon number, charge, and lepton number across the reaction, determining that the final particle must have baryon number 2 and charge +1.

The total mechanical energy \(E\) of the simple harmonic oscillator is given by \(E = E_k + E_p = \frac{1}{2} k A^2\). We are given that \(E_k = 3 E_p\). Substituting this into the total energy equation gives \(3 E_p + E_p = 4 E_p = E\). Since potential energy at displacement \(x\) is \(E_p = \frac{1}{2} k x^2\), we can write \(4 \left(\frac{1}{2} k x^2\right) = \frac{1}{2} k A^2\). This simplifies to \(4 x^2 = A^2\), which yields \(x^2 = A^2 / 4\) and therefore \(x = \pm 0.50 A\).

1 mark: Relates kinetic and potential energies to total energy to find the relationship between potential energy and total energy, solving correctly for \(x = \pm 0.50 A\).

The gravitational potential energy is given by \(U = -\frac{GMm}{r}\). For a circular orbit of radius \(r\), the centripetal force is provided by the gravitational attraction: \(\frac{mv^2}{r} = \frac{GMm}{r^2}\). Rearranging this gives the kinetic energy as \(E_k = \frac{1}{2} mv^2 = \frac{GMm}{2r} = -0.5 U\). The total mechanical energy \(E\) is the sum of the kinetic and potential energies: \(E = E_k + U = -0.5 U + U = 0.5 U\).

1 mark: Shows that kinetic energy equals \(-0.5 U\) and sums the energies to show that the total mechanical energy is \(0.5 U\).

At the top of the vertical circle, both the gravitational force \(mg\) and the normal reaction force \(N\) from the bottom of the bucket on the water act vertically downwards, providing the centripetal force: \(mg + N = \frac{mv^2}{R}\). For no water to spill, the water must maintain contact with the bucket, meaning the normal reaction force must be greater than or equal to zero (\(N \ge 0\)). The minimum speed occurs when \(N = 0\), which simplifies the force equation to \(mg = \frac{mv^2}{R}\). Solving for the minimum speed gives \(v = \sqrt{gR}\).

1 mark: Identifies that the threshold condition is when the normal force is zero, equates centripetal force to weight, and solves to find \(v = \sqrt{gR}\).

The kinetic energy \(E_k\) gained by a charged particle accelerated through a potential difference \(V\) is given by:
\(E_k = qV = \frac{1}{2}mv^2\)

This gives the momentum \(p = mv = \sqrt{2mqV}\).

The radius of the path of a charged particle in a uniform magnetic field is given by:
\(r = \frac{mv}{qB} = \frac{\sqrt{2mqV}}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}\)

Thus, the radius is proportional to \(\sqrt{\frac{m}{q}}\):
\(r \propto \sqrt{\frac{m}{q}}\)

For the proton:
\(r_{\text{proton}} \propto \sqrt{\frac{m_p}{e}}\)

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

Therefore, the ratio of the radii is:
\(\frac{r_{\text{proton}}}{r_{\alpha}} = \frac{\sqrt{\frac{m_p}{e}}}{\sqrt{\frac{2m_p}{e}}} = \frac{1}{\sqrt{2}}\)

So the correct option is B.

[1] B — \(\frac{1}{\sqrt{2}}\)

The potential energy \(E_p\) of a simple harmonic oscillator at displacement \(x\) is given by:
\(E_p = \frac{1}{2}kx^2\)

The total energy \(E\) is given by:
\(E = \frac{1}{2}kA^2\)

At displacement \(x = \frac{A}{2}\):
\(E_p = \frac{1}{2}k\left(\frac{A}{2}\right)^2 = \frac{1}{4}\left(\frac{1}{2}kA^2\right) = \frac{1}{4}E\)

The kinetic energy \(E_k\) is the difference between the total energy and the potential energy:
\(E_k = E - E_p = E - \frac{1}{4}E = \frac{3}{4}E\)

Thus, the ratio of kinetic energy to potential energy is:
\(\frac{E_k}{E_p} = \frac{\frac{3}{4}E}{\frac{1}{4}E} = 3\)

This is equivalent to the ratio 3:1.

So the correct option is C.

[1] C — 3:1

(a) Using Ohm's law for the complete circuit:
\(\varepsilon = I(R_{\text{total}})\)
\(R_{\text{total}} = \frac{\varepsilon}{I} = \frac{6.0\text{ V}}{0.20\text{ A}} = 30.0\ \Omega\)
Since \(R_{\text{total}} = R + R_{\text{th}} + r\):
\(30.0 = 12 + R_{\text{th}} + 1.5\)
\(R_{\text{th}} = 30.0 - 13.5 = 16.5\ \Omega\)

(b) The potential difference across the fixed resistor is:
\(V_R = I R = 0.20\text{ A} \times 12\ \Omega = 2.4\text{ V}\)

(c) As the temperature of an NTC thermistor increases, the charge carriers (electrons) gain thermal energy.
This energy allows more valence electrons to escape their atoms and enter the conduction band, increasing the number density \(n\) of charge carriers.
This increase in charge carriers greatly increases the conductivity, thereby decreasing the resistance of the thermistor.

(d) When the temperature increases, the resistance of the thermistor decreases.
This decreases the total resistance of the circuit, which increases the circuit current \(I\).
Since \(V_R = I R\) and \(R\) remains constant, the potential difference across the fixed resistor must increase.

(a) [3 Marks]
- Use of \(R = V/I\) to find total resistance of \(30\ \Omega\) (1)
- Subtracting internal resistance and fixed resistance from total resistance (1)
- Correct calculation of thermistor resistance to give \(16.5\ \Omega\) (1)

(b) [2 Marks]
- Use of \(V = IR\) with current and fixed resistance (1)
- Correct calculation to give \(2.4\text{ V}\) (1)

(c) [3 Marks]
- Stating that thermal energy is supplied to the thermistor (1)
- Explaining that more charge carriers (conduction electrons) are liberated/released into the conduction band / number density \(n\) increases (1)
- Concluding that this causes resistance to decrease (1)

(d) [2 Marks]
- Stating that the potential difference across the fixed resistor increases (1)
- Explaining that thermistor resistance decreases, lowering total resistance and increasing the current (1)

(a) Work done on the electron equals its kinetic energy:
\(e V = \frac{1}{2} m v^2\)
\(v = \sqrt{\frac{2 e V}{m}} = \sqrt{\frac{2 \times (1.60 \times 10^{-19}\text{ C}) \times 1500\text{ V}}{9.11 \times 10^{-31}\text{ kg}}}\)
\(v = \sqrt{\frac{4.80 \times 10^{-16}}{9.11 \times 10^{-31}}} = \sqrt{5.269 \times 10^{14}} = 2.295 \times 10^7\text{ m s}^{-1} \approx 2.3 \times 10^7\text{ m s}^{-1}\)

(b) For the electron to pass through undeflected, the electric force must balance the magnetic force:
\(F_E = F_B \implies e E = e v B \implies E = v B\)
\(E = (2.295 \times 10^7\text{ m s}^{-1}) \times (4.5 \times 10^{-3}\text{ T}) = 1.033 \times 10^5\text{ V m}^{-1}\) (using \(2.3 \times 10^7\) gives \(1.035 \times 10^5\text{ V m}^{-1}\))

(c) The relation between electric field and potential difference for parallel plates is:
\(E = \frac{V_p}{d} \implies V_p = E d\)
\(V_p = (1.033 \times 10^5\text{ V m}^{-1}) \times (8.0 \times 10^{-3}\text{ m}) = 826\text{ V}\) (using \(2.3 \times 10^7\) gives \(828\text{ V}\))

(d) When the magnetic field is turned off, the only force acting on the electron is the constant vertical electric force. This produces a constant acceleration perpendicular to the initial velocity, resulting in a parabolic trajectory towards the positive plate.

(a) [3 Marks]
- Equating electrical work done to kinetic energy: \(eV = \frac{1}{2}mv^2\) (1)
- Rearranging and substituting correct values for \(e\), \(m\), and \(V\) (1)
- Correct calculation to show \(2.3 \times 10^7\text{ m s}^{-1}\) (must show at least 3 sig figs: \(2.295 \times 10^7\text{ m s}^{-1}\)) (1)

(b) [3 Marks]
- Equating forces: \(eE = evB\) or stating \(E = vB\) (1)
- Correct substitution of values (1)
- Calculation to give \(1.03 \times 10^5\text{ V m}^{-1}\) (or \(1.04 \times 10^5\text{ V m}^{-1}\) if using \(2.3\times 10^7\)) (1)

(c) [2 Marks]
- Use of \(V_p = Ed\) with plate separation converted to meters (1)
- Correct calculation of potential difference in the range \(824\text{ V}\) to \(828\text{ V}\) (1)

(d) [2 Marks]
- Description of the path as parabolic / curving (towards the positive plate) (1)
- Explanation that there is a constant force (and constant acceleration) perpendicular to the horizontal velocity (1)

(a) Convert the half-life from years to seconds:
\(T_{1/2} = 4.47 \times 10^9\text{ years} \times 365.25\text{ days/year} \times 24\text{ hours/day} \times 3600\text{ seconds/hour} = 1.411 \times 10^{17}\text{ s}\)
Now calculate the decay constant:
\(\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.69315}{1.411 \times 10^{17}\text{ s}} = 4.912 \times 10^{-18}\text{ s}^{-1} \approx 4.9 \times 10^{-18}\text{ s}^{-1}\)

(b) Number of moles of Uranium-238:
\(n = \frac{\text{mass}}{\text{molar mass}} = \frac{1.50 \times 10^{-3}\text{ g}}{238\text{ g mol}^{-1}} = 6.303 \times 10^{-6}\text{ mol}\)
Number of atoms \(N\):
\(N = n \times N_A = 6.303 \times 10^{-6}\text{ mol} \times 6.02 \times 10^{23}\text{ mol}^{-1} = 3.794 \times 10^{18}\text{ atoms}\)
Activity \(A\):
\(A = \lambda N = 4.912 \times 10^{-18}\text{ s}^{-1} \times 3.794 \times 10^{18} = 18.6\text{ Bq}\) (or \(18.7\text{ Bq}\) if using \(4.9 \times 10^{-18}\text{ s}^{-1}\))

(c) Let \(N_{\text{U}}\) be the number of Uranium atoms left and \(N_{\text{Pb}}\) be the number of Lead atoms formed.
\(\frac{N_{\text{Pb}}}{N_{\text{U}}} = 0.45\)
Since every Lead atom came from a decayed Uranium atom, the initial number of Uranium atoms was:
\(N_0 = N_{\text{U}} + N_{\text{Pb}}\)
Dividing by \(N_{\text{U}}\):
\(\frac{N_0}{N_{\text{U}}} = 1 + \frac{N_{\text{Pb}}}{N_{\text{U}}} = 1 + 0.45 = 1.45\)
Using the decay equation:
\(N_{\text{U}} = N_0 e^{-\lambda t} \implies \frac{N_0}{N_{\text{U}}} = e^{\lambda t} = 1.45\)
\(\lambda t = \ln(1.45)\)
\(t = \frac{\ln(1.45)}{\lambda}\)
Using half-life directly in years:
\(t = \frac{\ln(1.45)}{\ln 2} \times T_{1/2} = \frac{0.3716}{0.69315} \times 4.47 \times 10^9\text{ years} = 2.396 \times 10^9\text{ years} \approx 2.40 \times 10^9\text{ years}\)

(a) [3 Marks]
- Conversion of half-life from years to seconds (1)
- Use of \(\lambda = \frac{\ln 2}{T_{1/2}}\) (1)
- Correct calculation leading to \(4.9 \times 10^{-18}\text{ s}^{-1}\) (minimum 2 sig figs shown in calculation step) (1)

(b) [3 Marks]
- Finding the number of atoms \(N\) using mass and Avogadro's constant (1)
- Use of \(A = \lambda N\) (1)
- Correct calculation of activity to give \(18.6\text{ Bq}\) or \(18.7\text{ Bq}\) (1)

(c) [4 Marks]
- Identifying that \(N_0 = N_{\text{U}} + N_{\text{Pb}}\) so that \(N_0/N_{\text{U}} = 1.45\) (1)
- Use of \(N = N_0 e^{-\lambda t}\) or equivalent logarithmic decay expression (1)
- Correct substitution of either \(\lambda\) or half-life (1)
- Correct age calculation to give \(2.4 \times 10^9\text{ years}\) (accept \(2.38 \times 10^9\) to \(2.41 \times 10^9\)) (1)

(a) The diagram must show:
- Weight, \(W = mg\), acting vertically downwards from the centre of mass.
- Normal contact force, \(R\) (or \(N\)), acting perpendicular to the banked surface (upwards and inwards).
- No friction force parallel to the slope (since it is travelling at the design speed).

(b) Resolving forces:
- Vertically: \(R \cos\theta = mg\)
- Horizontally (towards the centre of the circular path): \(R \sin\theta = \frac{m v_d^2}{r}\)
Dividing the horizontal equation by the vertical equation:
\(\frac{R \sin\theta}{R \cos\theta} = \frac{m v_d^2 / r}{mg}\)
\(\tan\theta = \frac{v_d^2}{gr}\)
Rearranging for \(v_d\):
\(v_d = \sqrt{gr\tan\theta}\)

(c) Substituting the values into the derived equation:
\(v_d = \sqrt{9.81\text{ m s}^{-2} \times 120\text{ m} \times \tan(25^\circ)}\)
\(v_d = \sqrt{1177.2 \times 0.4663} = \sqrt{548.9} = 23.4\text{ m s}^{-1}\)

(d) If the speed is much greater than \(v_d\), a larger centripetal force is required because \(F_c = \frac{mv^2}{r}\).
Since there is no friction on the icy track, the only inward force is the horizontal component of the normal contact force, which is fixed by the car's weight.
Because this force is insufficient to keep the car on its circular path of radius \(r\), the car will slide outwards/up the bank to a larger radius.

(a) [3 Marks]
- Arrow showing weight acting vertically downwards (1)
- Arrow showing normal contact force perpendicular to the banked track (1)
- No other force arrows shown (e.g. no friction) and forces labelled correctly (1)

(b) [3 Marks]
- Resolving vertically to give \(R \cos\theta = mg\) (1)
- Resolving horizontally to give \(R \sin\theta = \frac{mv^2}{r}\) (1)
- Dividing equations to obtain \(v_d = \sqrt{gr\tan\theta}\) (1)

(c) [2 Marks]
- Correct substitution of numbers into the equation (1)
- Correct calculation to give \(23.4\text{ m s}^{-1}\) (or \(23\text{ m s}^{-1}\)) (1)

(d) [2 Marks]
- Stating that the car will slide outwards / up the bank (1)
- Explaining that the horizontal component of the normal force is too small to provide the required centripetal force at this higher speed (1)

(a) A cyclotron consists of two D-shaped electrodes (dees) in a vacuum chamber:
- The magnetic field acts perpendicularly to the plane of the dees. It provides a centripetal force \(F = Bqv\) that causes the protons to move in circular paths within the dees, but does not change their speed.
- The electric field acts only across the gap between the dees. It accelerates the protons as they cross the gap, increasing their kinetic energy.
- The frequency of the alternating electric field must match the frequency of the proton's circular motion so that the electric field reverses direction every time the protons reach the gap, always accelerating them.

(b) The magnetic force provides the necessary centripetal force:
\(Bqv = \frac{m v^2}{r} \implies r = \frac{mv}{Bq}\)
Let \(T\) be the period of one complete circular orbit:
\(T = \frac{2 \pi r}{v}\)
Substitute \(r\):
\(T = \frac{2 \pi \left(\frac{mv}{Bq}\right)}{v} = \frac{2 \pi m}{Bq}\)
The frequency is the reciprocal of the period:
\(f = \frac{1}{T} = \frac{Bq}{2\pi m}\)
Since \(B\), \(q\), and \(m\) are constants, this frequency is independent of the speed \(v\).

(c) Using the derived formula:
\(f = \frac{Bq}{2\pi m}\)
For a proton: \(q = 1.60 \times 10^{-19}\text{ C}\) and \(m = 1.67 \times 10^{-27}\text{ kg}\).
\(f = \frac{1.4\text{ T} \times 1.60 \times 10^{-19}\text{ C}}{2 \pi \times 1.67 \times 10^{-27}\text{ kg}}\)
\(f = \frac{2.24 \times 10^{-19}}{1.0493 \times 10^{-26}} = 2.135 \times 10^7\text{ Hz} = 21.3\text{ MHz}\) (or \(21\text{ MHz}\))

(a) [4 Marks]
- Magnetic field provides centripetal force / causes circular motion (1)
- Electric field exists only in the gap and accelerates protons (increases their speed) (1)
- Speed increases, so radius of orbit increases in the dees (1)
- Alternating electric field frequency matches the orbital frequency of the protons (to ensure acceleration in each gap crossing) (1)

(b) [3 Marks]
- Equating centripetal and magnetic forces: \(Bqv = mv^2/r\) (1)
- Substituting \(v = 2\pi r/T\) or using \(\omega = v/r = Bq/m\) (1)
- Deriving \(f = \frac{Bq}{2\pi m}\) and stating that because there is no \(v\) term, \(f\) is independent of speed (1)

(c) [3 Marks]
- Correct substitution of proton charge and mass (1)
- Correct calculation to give \(2.1 \times 10^7\text{ Hz}\) (accept \(2.10\text{ MHz}\) to \(2.14\text{ MHz}\)) (1)
- Correct unit of Hz or MHz (1)

(a) Energy of a single photon in Joules:
\(E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34}\text{ J s}) \times (3.00 \times 10^8\text{ m s}^{-1})}{240 \times 10^{-9}\text{ m}} = 8.288 \times 10^{-19}\text{ J}\)
Converting to eV:
\(E_{\text{eV}} = \frac{8.288 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ J eV}^{-1}} = 5.18\text{ eV}\)

(b) Using Einstein's photoelectric equation:
\(h f = \phi + E_{k,\text{max}}\)
\(E_{k,\text{max}} = h f - \phi = 5.18\text{ eV} - 2.36\text{ eV} = 2.82\text{ eV}\)
Converting back to Joules:
\(E_{k,\text{max}} = 2.82 \times 1.60 \times 10^{-19}\text{ J} = 4.512 \times 10^{-19}\text{ J}\)

(c) (i) The maximum kinetic energy remains unchanged.
This is because the maximum kinetic energy depends only on the frequency (or wavelength) of the incident photons and the work function of the metal. Doubling the intensity increases the number of photons but does not change the energy of individual photons.

(ii) The rate of emission of photoelectrons doubles.
Intensity is proportional to the number of photons incident per second. Since the photoelectric effect is a one-to-one interaction between a photon and an electron, doubling the number of photons per second doubles the number of photoelectrons emitted per second.

(a) [3 Marks]
- Use of \(E = hc/\lambda\) (1)
- Division by \(1.60 \times 10^{-19}\) to convert to eV (1)
- Correct calculation to give \(5.18\text{ eV}\) (accept \(5.17\text{ eV}\) to \(5.20\text{ eV}\)) (1)

(b) [3 Marks]
- Use of \(E_{k,\text{max}} = hf - \phi\) (1)
- Subtraction of work function from photon energy (1)
- Correct calculation of KE to give \(4.5 \times 10^{-19}\text{ J}\) (accept \(4.45 \times 10^{-19}\) to \(4.55 \times 10^{-19}\text{ J}\)) (1)

(c) (i) [2 Marks]
- Stating that maximum kinetic energy is unchanged (1)
- Explaining that photon energy depends on frequency/wavelength, which is constant (or interaction is one-to-one) (1)

(c) (ii) [2 Marks]
- Stating that rate of emission of photoelectrons doubles (1)
- Explaining that intensity is proportional to photon rate, resulting in twice as many interactions per second (1)

(a) The vertical component of the initial velocity is:
\(v_y = u \sin\theta = 22.0\text{ m s}^{-1} \times \sin(35.0^\circ)\)
\(v_y = 22.0 \times 0.5736 = 12.62\text{ m s}^{-1} \approx 12.6\text{ m s}^{-1}\)

(b) Taking upwards as positive, the vertical displacement \(s_y\) when the ball returns to the ground is \(0\).
\(s_y = v_y t - \frac{1}{2} g t^2\)
\(0 = 12.62 t - 4.905 t^2\)
Since \(t \neq 0\):
\(t = \frac{2 \times 12.62}{9.81} = 2.573\text{ s} \approx 2.57\text{ s}\)

(c) The horizontal component of velocity is constant:
\(v_x = u \cos\theta = 22.0 \times \cos(35.0^\circ) = 22.0 \times 0.8192 = 18.02\text{ m s}^{-1}\)
The range is:
\(R = v_x \times t = 18.02\text{ m s}^{-1} \times 2.573\text{ s} = 46.37\text{ m} \approx 46.4\text{ m}\)

(d) If air resistance is not negligible, the range would be significantly less.
This is because air resistance acts in the direction opposite to motion, creating a horizontal decelerating force that continuously reduces the horizontal component of velocity throughout the flight. It also reduces the maximum height and time of flight.

(a) [2 Marks]
- Identifying sine component as vertical: \(22.0 \sin(35.0^\circ)\) (1)
- Correct calculation showing at least 3 sig figs: \(12.62\text{ m s}^{-1}\) (1)

(b) [3 Marks]
- Using suvat equation \(s = ut + \frac{1}{2}at^2\) vertically with \(s = 0\) (or finding time to maximum height where \(v = 0\)) (1)
- Correct substitution of values (1)
- Correct calculation to give \(2.57\text{ s}\) (1)

(c) [3 Marks]
- Calculating horizontal component of velocity: \(18.0\text{ m s}^{-1}\) (1)
- Use of \(\text{distance} = \text{speed} \times \text{time}\) horizontally (1)
- Correct calculation of range to give \(46.4\text{ m}\) (accept \(46.0\text{ m}\) to \(47.0\text{ m}\)) (1)

(d) [2 Marks]
- Stating that the range would be less (1)
- Explaining that air resistance opposes motion / reduces horizontal speed (1)

(a) The period of a mass-spring system is:
\(T = 2 \pi \sqrt{\frac{m}{k}}\)
\(T^2 = 4 \pi^2 \frac{m}{k}\)
\(k = \frac{4 \pi^2 m}{T^2}\)
\(k = \frac{4 \pi^2 \times 0.45\text{ kg}}{(1.5\text{ s})^2} = \frac{17.765}{2.25} = 7.896\text{ N m}^{-1} \approx 7.9\text{ N m}^{-1}\)

(b) First find the angular frequency \(\omega\):
\(\omega = \frac{2 \pi}{T} = \frac{2 \pi}{1.5\text{ s}} = 4.189\text{ rad s}^{-1}\)
The maximum velocity \(v_{\text{max}}\) is at the equilibrium position:
\(v_{\text{max}} = \omega A\)
\(v_{\text{max}} = 4.189\text{ rad s}^{-1} \times 0.12\text{ m} = 0.503\text{ m s}^{-1}\)

(c) The maximum acceleration \(a_{\text{max}}\) is at maximum displacement:
\(a_{\text{max}} = \omega^2 A\)
\(a_{\text{max}} = (4.189)^2 \times 0.12 = 17.55 \times 0.12 = 2.11\text{ m s}^{-2}\)

(d) The graph of kinetic energy \(E_k\) against displacement \(x\) is an inverted parabola.
Key features include:
- Maximum kinetic energy occurs at the equilibrium position (\(x = 0\)).
- Kinetic energy is zero at maximum positive and negative displacements (\(x = +0.12\text{ m}\) and \(x = -0.12\text{ m}\)).
- The curve is smooth and symmetrical.

(a) [3 Marks]
- Recall and use of \(T = 2\pi\sqrt{\frac{m}{k}}\) (1)
- Rearranging for \(k = \frac{4\pi^2 m}{T^2}\) and correct substitution (1)
- Correct calculation showing at least 3 sig figs: \(7.90\text{ N m}^{-1}\) (1)

(b) [3 Marks]
- Finding angular frequency \(\omega = 4.19\text{ rad s}^{-1}\) (1)
- Use of \(v_{\text{max}} = \omega A\) (1)
- Correct calculation to give \(0.50\text{ m s}^{-1}\) (accept \(0.50\text{ m s}^{-1}\) to \(0.51\text{ m s}^{-1}\)) (1)

(c) [2 Marks]
- Use of \(a_{\text{max}} = \omega^2 A\) (or use of \(a_{\text{max}} = F_{\text{max}}/m = kA/m\)) (1)
- Correct calculation to give \(2.1\text{ m s}^{-2}\) (accept \(2.10\text{ m s}^{-2}\) to \(2.12\text{ m s}^{-2}\)) (1)

(d) [2 Marks]
- Graph sketched as an inverted parabola / symmetrical loop with its peak on the vertical axis (1)
- Peak correctly identified at \(x = 0\) and zeros correctly identified at \(x = \pm A\) (or \(\pm 0.12\text{ m}\)) (1)

The root mean square speed of ideal gas molecules is given by \( v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}} \). This means \( v_{\text{rms}} \propto \sqrt{T} \). Since the absolute temperature increases from \( T \) to \( 1.44 T \), the new root mean square speed is \( \sqrt{1.44} v = 1.20 v \).

1 mark for selecting the correct option (A).

According to the Stefan-Boltzmann law, the luminosity of a star is given by \( L = 4 \pi R^2 \sigma T^4 \). For the second star, the luminosity is \( L' = 4 \pi (2R)^2 \sigma \left(\frac{T}{2}\right)^4 = 4 \pi (4R^2) \sigma \frac{T^4}{16} = \frac{4}{16} \left(4 \pi R^2 \sigma T^4\right) = 0.25 L \).

1 mark for selecting the correct option (A).

After 24 hours, the number of half-lives elapsed for X is \( \frac{24}{4} = 6 \), so the remaining number of nuclei is \( N_X = N_0 \left(\frac{1}{2}\right)^6 = \frac{N_0}{64} \). The number of half-lives elapsed for Y is \( \frac{24}{8} = 3 \), so the remaining number of nuclei is \( N_Y = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8} \). The ratio is \( \frac{N_X}{N_Y} = \frac{N_0 / 64}{N_0 / 8} = \frac{8}{64} = 0.125 \).

1 mark for selecting the correct option (D).

The gravitational potential energy of the satellite is \( E_p = -\frac{G M m}{r} \). For a circular orbit, the centripetal force is provided by the gravitational force: \( \frac{G M m}{r^2} = \frac{m v^2}{r} \implies m v^2 = \frac{G M m}{r} \). Thus, the kinetic energy is \( E_k = \frac{1}{2} m v^2 = \frac{G M m}{2r} \). The total energy is \( E_{\text{total}} = E_k + E_p = \frac{G M m}{2r} - \frac{G M m}{r} = -\frac{G M m}{2r} \).

1 mark for selecting the correct option (A).

The displacement of an object in simple harmonic motion can be represented as \( x = A \cos(\omega t) \), so its velocity is \( v = -A \omega \sin(\omega t) \). The maximum velocity is \( v_{\text{max}} = \omega A \). Therefore, the maximum kinetic energy is \( E_{k, \text{max}} = \frac{1}{2} m v_{\text{max}}^2 = \frac{1}{2} m \omega^2 A^2 \).

1 mark for selecting the correct option (A).

The grating spacing \( d \) in meters is the distance between adjacent lines. Since there are \( N \) lines per millimeter, there are \( 10^3 N \) lines per meter. Therefore, \( d = \frac{1}{10^3 N} \) meters. Using the grating equation \( d \sin\theta = n \lambda \) for the second-order maximum (\( n = 2 \)): \( \frac{1}{10^3 N} \sin\theta = 2 \lambda \implies \sin\theta = 2 \times 10^3 N \lambda \).

1 mark for selecting the correct option (A).

The work done in stretching the wire is given by the area under the force-extension graph: \( W = \frac{1}{2} F x \). From the definition of Young modulus, \( E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{x/L} = \frac{F L}{A x} \implies F = \frac{E A x}{L} \). Substituting this force expression into the work equation gives: \( W = \frac{1}{2} \left(\frac{E A x}{L}\right) x = \frac{E A x^2}{2 L} \).

1 mark for selecting the correct option (A).

The redshift \( z \) is given by: \( z = \frac{\Delta \lambda}{\lambda} = \frac{682 \text{ nm} - 656 \text{ nm}}{656 \text{ nm}} = \frac{26 \text{ nm}}{656 \text{ nm}} \approx 0.03963 \). Using the Doppler relationship for non-relativistic speed: \( v = z c = 0.03963 \times 3.00 \times 10^8 \text{ m s}^{-1} \approx 1.19 \times 10^7 \text{ m s}^{-1} \).

1 mark for selecting the correct option (A).

The average kinetic energy of a molecule in an ideal gas is given by the equation \(\frac{1}{2}m\langle c^2 \rangle = \frac{3}{2}kT\) where \(m\) is the mass of a single molecule, \(\langle c^2 \rangle\) is the mean square speed, \(k\) is the Boltzmann constant, and \(T\) is the absolute temperature. Rearranging this expression gives \(\langle c^2 \rangle = \frac{3kT}{m}\). Since the gas is of a fixed mass, the molecular mass \(m\) is constant. Therefore, the mean square speed is directly proportional to the absolute temperature: \(\langle c^2 \rangle \propto T\). The volume change (halving) affects the gas pressure but does not affect the mean square speed, which depends only on the temperature. Since the absolute temperature is doubled, the mean square speed is also doubled, giving a ratio of 2.

C is the correct answer. Award 1 mark for the correct ratio of 2. Reject option A because the speed must change due to the temperature increase. Reject option B because this is the ratio of the root-mean-square speeds, not the mean square speeds. Reject option D because the volume change does not directly scale the molecular speeds.

According to Wien's displacement law, the peak radiation wavelength is inversely proportional to the absolute temperature: \(\lambda_{\text{max}} \propto \frac{1}{T}\). Since Star Y has twice the peak wavelength of Star X, its temperature is half that of Star X: \(T_Y = \frac{1}{2} T_X\). According to the Stefan-Boltzmann law, the luminosity is given by \(L = 4\pi R^2 \sigma T^4\), which means \(L \propto R^2 T^4\). Comparing Star Y to Star X, we have a radius ratio of \(\frac{R_Y}{R_X} = 2\) and a temperature ratio of \(\frac{T_Y}{T_X} = \frac{1}{2}\). Therefore, the ratio of their luminosities is \(\frac{L_Y}{L_X} = \left(\frac{R_Y}{R_X}\right)^2 \left(\frac{T_Y}{T_X}\right)^4 = 2^2 \times \left(\frac{1}{2}\right)^4 = 4 \times \frac{1}{16} = \frac{1}{4}\).

B is the correct answer. Award 1 mark for the correct ratio of 1/4. Reject option A as it incorrectly scales the temperature relation. Reject option C as it assumes the radius and temperature effects cancel exactly. Reject option D as it fails to use the inverse relationship in Wien's law.

(a) For undeflected passage, the electric and magnetic forces must balance: \(qE = qvB\). Rearranging gives: \(v = \frac{E}{B} = \frac{1.2 \times 10^5\text{ V m}^{-1}}{0.40\text{ T}} = 3.0 \times 10^5\text{ m s}^{-1}\).

(b) In the deflection chamber, the magnetic force acts as the centripetal force: \(qvB = \frac{mv^2}{r}\). Rearranging for radius \(r\): \(r = \frac{mv}{qB}\). Substituting the given values: \(r = \frac{3.8 \times 10^{-26}\text{ kg} \times 3.0 \times 10^5\text{ m s}^{-1}}{1.60 \times 10^{-19}\text{ C} \times 0.40\text{ T}} = 0.178\text{ m} \approx 0.18\text{ m}\).

(c) The radius of the path is inversely proportional to the charge of the ion since \(r = \frac{mv}{qB}\). If the charge doubles to \(+2e\) while the mass and velocity remain essentially the same, the radius of the circular path will be halved. This means the path curves more sharply.

(a) [1 mark] Equates electric and magnetic forces: \(qE = qvB\) or writes \(v = E/B\).
[1 mark] Substitutes values and shows the correct calculation leading to \(3.0 \times 10^5\text{ m s}^{-1}\).

(b) [1 mark] Recalls or uses \(qvB = mv^2/r\) or \(r = mv/qB\).
[1 mark] Substitutes correct values including \(q = 1.60 \times 10^{-19}\text{ C}\).
[1 mark] Obtains correct radius of \(0.18\text{ m}\) (accept \(0.178\text{ m}\)).

(c) [1 mark] Recognises that \(r \propto 1/q\) (radius is inversely proportional to charge).
[1 mark] Concludes that the radius of the path will be halved (or path is more curved).

(a) According to Wien's displacement law: \(\lambda_{\text{max}} T = 2.898 \times 10^{-3}\text{ m K}\). Rearranging for \(T\): \(T = \frac{2.898 \times 10^{-3}\text{ m K}}{725 \times 10^{-9}\text{ m}} = 3997\text{ K} \approx 4000\text{ K}\).

(b) According to Stefan's law: \(L = 4\pi R^2 \sigma T^4\). Rearranging for \(R\): \(R = \sqrt{\frac{L}{4\pi \sigma T^4}}\). Substituting \(T = 4000\text{ K}\) (or \(3997\text{ K}\)): \(R = \sqrt{\frac{4.5 \times 10^{29}}{4\pi \times 5.67 \times 10^{-8} \times (4000)^4}} = \sqrt{\frac{4.5 \times 10^{29}}{1.824 \times 10^8}} = 4.97 \times 10^{10}\text{ m} \approx 5.0 \times 10^{10}\text{ m}\).

(c) Since peak wavelength is inversely proportional to temperature (\(\lambda_{\text{max}} \propto 1/T\)), a higher temperature means the peak wavelength will decrease. Despite the much higher temperature, a white dwarf has an extremely small surface area, which dominates Stefan's law equation, causing the overall luminosity to decrease significantly.

(a) [1 mark] Uses Wien's displacement law formula \(\lambda_{\text{max}} T = 2.898 \times 10^{-3}\).
[1 mark] Calculates temperature as \(4000\text{ K}\) (or \(3997\text{ K}\)).

(b) [1 mark] Recalls or uses \(L = 4\pi R^2 \sigma T^4\).
[1 mark] Substitutes values into rearranged equation for \(R\).
[1 mark] Correctly computes \(R\) as \(4.97 \times 10^{10}\text{ m}\) or \(5.0 \times 10^{10}\text{ m}\).

(c) [1 mark] States that peak wavelength decreases (or shifts to shorter wavelengths/blue-shifted).
[1 mark] States that luminosity decreases due to the vastly reduced surface area.

(a) Using the ideal gas law: \(pV = nRT\). Convert temperature to Kelvin: \(T = 20.0 + 273.15 = 293.15\text{ K}\). Rearranging for \(n\): \(n = \frac{pV}{RT} = \frac{101 \times 10^3\text{ Pa} \times 15.0\text{ m}^3}{8.31\text{ J mol}^{-1}\text{ K}^{-1} \times 293.15\text{ K}} = 621.5\text{ moles} \approx 622\text{ moles}\).

(b) The average kinetic energy of a gas atom is given by \(E_k = \frac{3}{2}kT\). Convert temperature to Kelvin: \(T = -40.0 + 273.15 = 233.15\text{ K}\). Substitute Boltzmann's constant \(k = 1.38 \times 10^{-23}\text{ J K}^{-1}\): \(E_k = 1.5 \times 1.38 \times 10^{-23} \times 233.15 = 4.83 \times 10^{-21}\text{ J}\).

(c) Any two standard assumptions of an ideal gas: (1) The helium atoms are in continuous, rapid, random motion. (2) The volume occupied by the helium atoms is negligible compared to the volume of the balloon. (3) There are no intermolecular forces acting between the helium atoms except during collisions. (4) All collisions between atoms and with the balloon wall are perfectly elastic.

(a) [1 mark] Converts temperature to Kelvin (\(293.15\text{ K}\)).
[1 mark] Rearranges ideal gas equation: \(n = pV/RT\).
[1 mark] Correctly evaluates \(n = 622\text{ moles}\) (accept \(620\text{ moles}\) if \(273\text{ K}\) is used).

(b) [1 mark] Recalls or uses \(E_k = \frac{3}{2}kT\) with temperature in Kelvin (\(233.15\text{ K}\)).
[1 mark] Calculates kinetic energy as \(4.8 \times 10^{-21}\text{ J}\) (accept \(4.83 \times 10^{-21}\text{ J}\)).

(c) [2 marks] States any two correct assumptions of the kinetic theory of gases (1 mark per assumption).

(a) When there is a current in the circuit, some of the chemical energy converted per unit charge (e.m.f.) is dissipated as thermal energy within the internal resistance of the cell. This potential drop is known as 'lost volts' (\(v = Ir\)). Since the voltmeter measures terminal potential difference \(V = \mathcal{E} - Ir\), the reading \(V\) is less than the e.m.f. \(\mathcal{E}\).

(b) The terminal potential difference is given by \(V = IR\). Therefore, we can find the current in each case:
Case 1: \(I_1 = \frac{1.2\text{ V}}{6.0\ \Omega} = 0.20\text{ A}\)
Case 2: \(I_2 = \frac{1.5\text{ V}}{15.0\ \Omega} = 0.10\text{ A}\)
Using \(\mathcal{E} = V + Ir\):
From Case 1: \(\mathcal{E} = 1.2 + 0.20r\)
From Case 2: \(\mathcal{E} = 1.5 + 0.10r\)
Equating these two expressions:
\(1.2 + 0.20r = 1.5 + 0.10r\)
\(0.10r = 0.3\)
\(r = 3.0\ \Omega\)
Substitute \(r\) back into either expression:
\(\mathcal{E} = 1.2 + 0.20(3.0) = 1.8\text{ V\).

(a) [1 mark] Mentions that current flows through the internal resistance of the cell.
[1 mark] Explains that this causes a potential drop (or lost volts) within the cell, reducing terminal p.d. below e.m.f.

(b) [1 mark] Calculates the currents in both cases correctly: \(I_1 = 0.20\text{ A}\) and \(I_2 = 0.10\text{ A}\).
[1 mark] Sets up two simultaneous equations using \(\mathcal{E} = V + Ir\).
[1 mark] Solves for internal resistance to get \(r = 3.0\ \Omega\).
[1 mark] Solves for e.m.f. to get \(\mathcal{E} = 1.8\text{ V}\).

(a) Activity is the rate of decay of unstable nuclei, measured as the number of disintegrations (or decays) per second.

(b) The relationship between decay constant \(\lambda\) and half-life \(T_{1/2}\) is given by: \(\lambda = \frac{\ln 2}{T_{1/2}}\). First, convert half-life into seconds:
\(T_{1/2} = 5730\text{ years} \times 365.25\text{ days/year} \times 24\text{ hours/day} \times 3600\text{ seconds/hour} = 1.808 \times 10^{11}\text{ s}\).
Now calculate \(\lambda\):
\(\lambda = \frac{\ln 2}{1.808 \times 10^{11}\text{ s}} = 3.83 \times 10^{-12}\text{ s}^{-1}\).

(c) Use the radioactive decay law \(A = A_0 e^{-\lambda t}\), where \(A = 0.145\text{ Bq}\) and \(A_0 = 0.230\text{ Bq}\):
\(\frac{0.145}{0.230} = e^{-\lambda t} \implies 0.6304 = e^{-\lambda t}\).
Taking the natural logarithm of both sides:
\(\ln(0.6304) = -\lambda_{\text{years}} t \implies -0.4613 = -\lambda_{\text{years}} t\).
Find \(\lambda\) in \(\text{year}^{-1}\):
\(\lambda_{\text{years}} = \frac{\ln 2}{5730\text{ years}} = 1.21 \times 10^{-4}\text{ year}^{-1}\).
Therefore: \(t = \frac{0.4613}{1.21 \times 10^{-4}\text{ year}^{-1}} = 3812\text{ years} \approx 3800\text{ years}\).

(a) [1 mark] Defines activity as the number of nuclear decays per unit time.

(b) [1 mark] Recalls or uses \(\lambda = \ln 2 / T_{1/2}\).
[1 mark] Correctly converts years to seconds (\(1.81 \times 10^{11}\text{ s}\)).
[1 mark] Correctly evaluates decay constant \(\lambda = 3.83 \times 10^{-12}\text{ s}^{-1}\).

(c) [1 mark] Recalls or uses the exponential decay formula \(A = A_0 e^{-\lambda t}\).
[1 mark] Performs algebraic steps to isolate \(t\) using logs correctly.
[1 mark] Obtains a final age of approximately \(3800\text{ years}\) (accept range \(3790 - 3830\text{ years}\)).

(a) The three main forces are: (1) The weight of the car acting vertically downwards. (2) The normal contact force from the road surface acting perpendicular to the slope. (3) The friction (braking force) acting up the slope parallel to the road, opposing the motion of the car.

(b) The vertical height lost by the car as it travels \(45\text{ m}\) along the slope is:
\(\Delta h = 45 \sin(8.0^\circ) = 6.26\text{ m}\).
The loss in gravitational potential energy is:
\(\Delta E_p = m g \Delta h = 1200\text{ kg} \times 9.81\text{ m s}^{-2} \times 6.26\text{ m} = 7.37 \times 10^4\text{ J} \approx 7.4 \times 10^4\text{ J}\).

(c) Using the principle of conservation of energy, the work done by the braking force \(F\) over the distance \(d = 45\text{ m}\) must equal the total energy change (loss of kinetic energy + loss of potential energy):
\(\Delta E_k = \frac{1}{2} m v^2 = 0.5 \times 1200\text{ kg} \times (18\text{ m s}^{-1})^2 = 1.944 \times 10^5\text{ J}\).
Total energy to be dissipated \(W = \Delta E_k + \Delta E_p = 1.944 \times 10^5\text{ J} + 7.37 \times 10^4\text{ J} = 2.681 \times 10^5\text{ J}\).
Work done \(W = F \times d\), so:
\(F = \frac{W}{d} = \frac{2.681 \times 10^5\text{ J}}{45\text{ m}} = 5.96 \times 10^3\text{ N} \approx 6.0 \times 10^3\text{ N}\).

Alternatively, using equations of motion:
\(v^2 = u^2 + 2as \implies 0 = 18^2 + 2a(45) \implies a = -3.6\text{ m s}^{-2}\).
Net force \(F_{\text{net}} = m a = 1200 \times (-3.6) = -4320\text{ N}\).
Resolving weight along the slope: \(F_{\text{gravity}} = mg \sin(8.0^\circ) = 1200 \times 9.81 \sin(8.0^\circ) = 1638\text{ N}\).
Since net force is up-slope: \(F_{\text{braking}} - F_{\text{gravity}} = 4320\text{ N} \implies F_{\text{braking}} = 4320 + 1638 = 5958\text{ N} \approx 6.0 \times 10^3\text{ N}\).

(a) [2 marks] Mentions weight downwards, normal contact force perpendicular to the slope, and friction/braking force up the slope. (Deduct 1 mark if direction of any force is incorrect or missing).

(b) [1 mark] Calculates vertical height dropped using trigonometry: \(45 \sin(8.0^\circ)\) (\(= 6.26\text{ m}\)).
[1 mark] Calculates \(\Delta E_p\) correctly as \(7.4 \times 10^4\text{ J}\).

(c) [1 mark] Calculates initial kinetic energy as \(1.94 \times 10^5\text{ J}\).
[1 mark] Equates work done by brakes to sum of kinetic energy and potential energy changes, or uses Newton's second law with correct component of gravity resolved.
[1 mark] Correctly calculates braking force as \(6.0 \times 10^3\text{ N}\) (accept \(5.96 \times 10^3\text{ N}\)).

(a) Comparing \(x = 0.085 \cos(12.5 t)\) to the standard SHM equation \(x = A \cos(\omega t)\):
- Amplitude \(A = 0.085\text{ m}\).
- Angular frequency \(\omega = 12.5\text{ rad s}^{-1}\).
Frequency \(f = \frac{\omega}{2\pi} = \frac{12.5}{2\pi} = 1.99\text{ Hz}\).

(b) For a mass-spring system, the angular frequency is related to the spring constant \(k\) and mass \(m\) by \(\omega^2 = \frac{k}{m}\).
Rearranging for \(k\):
\(k = m \omega^2 = 0.350\text{ kg} \times (12.5\text{ rad s}^{-1})^2 = 0.350 \times 156.25 = 54.69\text{ N m}^{-1} \approx 54.7\text{ N m}^{-1}\).

(c) The maximum speed is given by \(v_{\text{max}} = \omega A\):
\(v_{\text{max}} = 12.5\text{ rad s}^{-1} \times 0.085\text{ m} = 1.0625\text{ m s}^{-1}\).
The maximum kinetic energy is:
\(E_{k,\text{max}} = \frac{1}{2} m v_{\text{max}}^2 = 0.5 \times 0.350\text{ kg} \times (1.0625\text{ m s}^{-1})^2 = 0.1975\text{ J} \approx 0.197\text{ J}\).
Alternatively, using potential energy at maximum displacement:
\(E_{k,\text{max}} = \frac{1}{2} k A^2 = 0.5 \times 54.69 \times (0.085)^2 = 0.1975\text{ J} \approx 0.197\text{ J}\).

(a) [1 mark] Identifies amplitude \(A = 0.085\text{ m}\).
[1 mark] Calculates frequency \(f = 1.99\text{ Hz}\) (or \(2.0\text{ Hz}\)) from \(\omega = 12.5\text{ rad s}^{-1}\).

(b) [1 mark] Recalls or uses \(\omega^2 = k/m\) or \(T = 2\pi\sqrt{m/k}\).
[1 mark] Calculates spring constant \(k = 54.7\text{ N m}^{-1}\).

(c) [1 mark] Identifies expression for maximum velocity \(v_{\text{max}} = \omega A\) or states \(E_k = \frac{1}{2}kA^2\).
[1 mark] Substitutes values to find maximum speed (\(1.06\text{ m s}^{-1}\)) or maximum displacement into kinetic energy formula.
[1 mark] Obtains correct maximum kinetic energy of \(0.197\text{ J}\) (accept range \(0.197 - 0.20\text{ J}\)).

(a) The energy of a photon in Joules is given by: \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{380 \times 10^{-9}\text{ m}} = 5.234 \times 10^{-19}\text{ J}\).
To convert this energy to electronvolts, divide by the charge of an electron \(e\):
\(E_{\text{eV}} = \frac{5.234 \times 10^{-19}\text{ J}}{1.60 \times 10^{-19}\text{ C}} = 3.27\text{ eV} \approx 3.3\text{ eV}\).

(b) According to Einstein's photoelectric equation:
\(h\nu = \phi + E_{k,\text{max}}\)
\(E_{k,\text{max}} = h\nu - \phi = 3.27\text{ eV} - 2.28\text{ eV} = 0.99\text{ eV}\).
To convert this back into Joules:
\(E_{k,\text{max}} = 0.99 \times 1.60 \times 10^{-19}\text{ J} = 1.58 \times 10^{-19}\text{ J} \approx 1.6 \times 10^{-19}\text{ J}\).

(c) For a wavelength of \(600\text{ nm}\), the energy of an incident photon is:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^8}{600 \times 10^{-9}} = 3.315 \times 10^{-19}\text{ J} = 2.07\text{ eV}\).
Since the individual photon energy (\(2.07\text{ eV}\)) is less than the work function of sodium (\(2.28\text{ eV}\)), a single photon does not possess enough energy to liberate an electron. Because photoelectric interaction is a one-to-one photon-electron event, no emission can occur.

(a) [1 mark] Recalls or uses \(E = hc/\lambda\).
[1 mark] Calculates energy in Joules correctly (\(5.23 \times 10^{-19}\text{ J}\)).
[1 mark] Correctly divides by \(1.60 \times 10^{-19}\) to obtain \(3.27\text{ eV}\) (shows that it is close to \(3.3\text{ eV}\)).

(b) [1 mark] Subtracts the work function from photon energy to get kinetic energy in eV (\(0.99\text{ eV}\)).
[1 mark] Correctly converts maximum kinetic energy to Joules: \(1.6 \times 10^{-19}\text{ J}\) (accept \(1.58 \times 10^{-19}\text{ J}\)).

(c) [1 mark] Calculates the energy of a \(600\text{ nm}\) photon (\(2.07\text{ eV}\) or \(3.3 \times 10^{-19}\text{ J}\)).
[1 mark] Concludes that since photon energy is less than the work function (\(2.28\text{ eV}\)), no emission occurs because the process is a one-to-one interaction.

(a) Convert temperature to Kelvin: \( T_1 = 20 + 273 = 293\text{ K} \). Using the ideal gas equation \( pV = NkT \), we rearrange to find the number of atoms \( N \): \( N = \frac{pV}{kT} = \frac{1.2 \times 10^5\text{ Pa} \times 0.025\text{ m}^3}{1.38 \times 10^{-23}\text{ J K}^{-1} \times 293\text{ K}} = 7.42 \times 10^{23} \). This is approximately \( 7.4 \times 10^{23} \). (b) Convert final temperature to Kelvin: \( T_2 = 85 + 273 = 358\text{ K} \). The temperature change is \( \Delta T = 358\text{ K} - 293\text{ K} = 65\text{ K} \). The increase in total kinetic energy of the gas is \( \Delta E_k = \frac{3}{2} N k \Delta T \). Substituting the values: \( \Delta E_k = 1.5 \times 7.42 \times 10^{23} \times 1.38 \times 10^{-23}\text{ J K}^{-1} \times 65\text{ K} = 998.4\text{ J} \) (or \( 1000\text{ J} \) to 2 significant figures).

(a) - Conversion of temperature to 293 K: 1 mark. - Use of pV = NkT or pV = nRT with N = n N_A: 1 mark. - Correct calculation showing N = 7.42 * 10^(23): 1 mark. (b) - Recognition that total kinetic energy is 1.5 NkT or 1.5 nRT: 1 mark. - Calculation of temperature change of 65 K: 1 mark. - Correct substitution of values into total kinetic energy formula: 1 mark. - Correct calculation of energy change as 998 J or 1000 J: 1 mark.

(a) First, calculate the angular frequency \( \omega \) using \( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{180}{4.5}} = \sqrt{40} \approx 6.32\text{ rad s}^{-1} \). The maximum velocity \( v_{\max} \) is given by \( v_{\max} = \omega A \), where \( A = 0.12\text{ m} \). Thus, \( v_{\max} = 6.32\text{ rad s}^{-1} \times 0.12\text{ m} = 0.759\text{ m s}^{-1} \approx 0.76\text{ m s}^{-1} \). (b) Since the object starts at the maximum positive displacement at \( t = 0 \), its displacement is described by \( x = A \cos(\omega t) \). Set \( x = 0.060\text{ m} \): \( 0.060 = 0.12 \cos(\omega t) \implies \cos(\omega t) = 0.5 \). The shortest time corresponds to the first angle: \( \omega t = \frac{\pi}{3}\text{ rad} \). Therefore, \( t = \frac{\pi}{3 \omega} = \frac{\pi}{3 \times 6.325} \approx 0.166\text{ s} \approx 0.17\text{ s} \).

(a) - Calculation of angular frequency: 1 mark. - Formula for maximum velocity: 1 mark. - Correct value of 0.76 m/s: 1 mark. (b) - Use of equation x = A cos(wt): 1 mark. - Correct angle in radians (pi/3 or 1.05 rad): 1 mark. - Rearranging for time t: 1 mark. - Correct final value of 0.17 s (allow 0.165 s to 0.17 s): 1 mark.

(a) The electrical energy stored is given by \( E = \frac{1}{2} C V^2 \). Substituting the values: \( E = 0.5 \times 470 \times 10^{-6}\text{ F} \times (9.0\text{ V})^2 = 0.019035\text{ J} \approx 0.019\text{ J} \) (or \( 1.9 \times 10^{-2}\text{ J} \)). (b) The time constant \( \tau = R C = 15 \times 10^3\ \Omega \times 470 \times 10^{-6}\text{ F} = 7.05\text{ s} \). The potential difference at time \( t \) is given by \( V = V_0 e^{-t/RC} \). For \( t = 5.0\text{ s} \): \( V = 9.0 \times e^{-5.0 / 7.05} = 9.0 \times e^{-0.7092} \approx 4.43\text{ V} \approx 4.4\text{ V} \). (c) The current is given by \( I = I_0 e^{-t/RC} \). The ratio of current to initial current is \( \frac{I}{I_0} = e^{-t/RC} \). For \( t = 15\text{ s} \): \( \frac{I}{I_0} = e^{-15 / 7.05} = e^{-2.128} \approx 0.119 \). Expressed as a percentage, the current is \( 11.9\% \) of its initial value. Since \( 11.9\% < 15\% \), this shows the statement is correct.

(a) - Correct use of E = 0.5 C V^2: 1 mark. - Correct answer to 2 s.f.: 1 mark. (b) - Correct calculation of time constant (7.05 s): 1 mark. - Use of V = V_0 e^(-t/RC): 1 mark. - Correct potential difference (4.4 V): 1 mark. (c) - Correct use of ratio equation for current: 1 mark. - Calculation of 11.9% (or 12%) and concluding statement: 1 mark.

(a) Mean diameter \(d_{\text{mean}} = \frac{0.38 + 0.39 + 0.37 + 0.38}{4} = 0.38\text{ mm}\).
Range of diameter = \(0.39 - 0.37 = 0.02\text{ mm}\).
Uncertainty in diameter (half range) = \(\pm 0.01\text{ mm}\).
Percentage uncertainty in diameter = \(\frac{0.01\text{ mm}}{0.38\text{ mm}} \times 100\% \approx 2.63\%\).
Mean cross-sectional area \(A = \frac{\pi d^2}{4} = \frac{\pi (3.8 \times 10^{-4}\text{ m})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\).
Percentage uncertainty in \(A = 2 \times \text{percentage uncertainty in } d = 2 \times 2.63\% = 5.26\%\).

(b) Resistivity \(\rho = A \times \text{gradient} = (1.134 \times 10^{-7}\text{ m}^2) \times 4.15\text{ }\Omega\text{ m}^{-1} = 4.706 \times 10^{-7}\text{ }\Omega\text{ m}\).
Percentage uncertainty in gradient = \(\frac{0.08}{4.15} \times 100\% = 1.93\%\).
Total percentage uncertainty in \(\rho = 5.26\% + 1.93\% = 7.19\%\).
Absolute uncertainty in \(\rho = 7.19\% \times 4.706 \times 10^{-7} = 0.338 \times 10^{-7}\text{ }\Omega\text{ m}\).
Rounding the uncertainty to one significant figure gives \(\pm 0.3 \times 10^{-7}\text{ }\Omega\text{ m}\).
Hence, \(\rho = (4.7 \pm 0.3) \times 10^{-7}\text{ }\Omega\text{ m}\).

1. Mean diameter calculation shown correctly: \(0.38\text{ mm}\) [1 Mark]
2. Half-range used to find uncertainty \(\pm 0.01\text{ mm}\) and percentage uncertainty of diameter calculated as \(2.6\%\) [1 Mark]
3. Area calculation shown with unit: \(1.13 \times 10^{-7}\text{ m}^2\) [1 Mark]
4. Percentage uncertainty in area correctly identified as twice that of diameter (\(5.2\%\) to \(5.3\%\)) [1 Mark]
5. Correct substitution to find resistivity: \(4.71 \times 10^{-7}\text{ }\Omega\text{ m}\) [1 Mark]
6. Percentage uncertainty in resistivity calculated by adding percentage uncertainties of Area and Gradient (\(7.2\%\)) [1 Mark]
7. Absolute uncertainty in resistivity calculated: \(\pm 0.3 \times 10^{-7}\text{ }\Omega\text{ m}\) [1 Mark]
8. Final answer written to 2 s.f. with matching absolute uncertainty and unit: \((4.7 \pm 0.3) \times 10^{-7}\text{ }\Omega\text{ m}\) [2.23 Marks]

(a) Mean time for 20 oscillations \(t_{\text{mean}} = \frac{35.9 + 36.1 + 35.8}{3} = 35.93\text{ s}\).
Range of \(t = 36.1 - 35.8 = 0.3\text{ s}\).
Uncertainty in \(t = \frac{\text{range}}{2} = 0.15\text{ s}\).
Percentage uncertainty in \(t = \frac{0.15}{35.93} \times 100\% = 0.4175\%\).
Mean period for 1 oscillation \(T = \frac{35.93}{20} = 1.797\text{ s}\).
Percentage uncertainty in \(T\) remains \(0.4175\% \approx 0.42\%\) because dividing by the constant 20 does not change the percentage uncertainty.

(b) Rearranging \(T = 2\pi\sqrt{\frac{L}{g}}\) gives \(g = \frac{4\pi^2 L}{T^2}\).
\(g = \frac{4\pi^2 \times 0.800}{(1.797)^2} = 9.780\text{ m s}^{-2}\).

(c) Percentage uncertainty in \(L = \frac{0.002}{0.800} \times 100\% = 0.25\%\).
Percentage uncertainty in \(T^2 = 2 \times \%\text{ uncertainty in } T = 2 \times 0.4175\% = 0.835\%\).
Percentage uncertainty in \(g = \%\text{ uncertainty in } L + \%\text{ uncertainty in } T^2 = 0.25\% + 0.835\% = 1.085\%\).
Absolute uncertainty in \(g = 1.085\% \times 9.780 = 0.106\text{ m s}^{-2} \approx 0.11\text{ m s}^{-2}\).
Thus, \(g = 9.78 \pm 0.11\text{ m s}^{-2}\).

1. Calculates mean time for 20 oscillations (\(35.93\text{ s}\)) and the period \(T = 1.797\text{ s} \approx 1.80\text{ s}\) [1 Mark]
2. Calculates uncertainty in time using half-range (\(\pm 0.15\text{ s}\)) [1 Mark]
3. Calculates percentage uncertainty in \(T\) (\(0.42\%\)) [1 Mark]
4. Correct rearrangement of the pendulum equation to find \(g = \frac{4\pi^2 L}{T^2}\) [1 Mark]
5. Correct calculation of \(g = 9.78\text{ m s}^{-2}\) [1 Mark]
6. Calculates percentage uncertainty in \(L\) (\(0.25\%\)) [1 Mark]
7. Correctly doubles the percentage uncertainty for \(T^2\) to get \(0.84\%\) [1 Mark]
8. Adds percentage uncertainties to find percentage uncertainty in \(g\) (\(1.09\%\)) and absolute uncertainty (\(\pm 0.11\text{ m s}^{-2}\)) [2.23 Marks]

(a) According to Boyle's law, \(p \propto 1/V\). Plotting \(p\) against \(1/V\) produces a straight line through the origin, which is easier to verify and analyze (by calculating a constant gradient) than a curved hyperbola of \(p\) against \(V\).

(b) Using the ideal gas equation: \(pV = nRT \Rightarrow p = nRT \left(\frac{1}{V}\right)\).
Comparing to \(y = mx + c\), the gradient \(m = nRT\).
Given: \(m = 1.42\text{ Pa m}^3\), \(T = 20.0 + 273.15 = 293.15\text{ K}\), \(R = 8.31\text{ J mol}^{-1}\text{ K}^{-1}\).
\(n = \frac{m}{RT} = \frac{1.42}{8.31 \times 293.15} = \frac{1.42}{2436.08} = 5.829 \times 10^{-4}\text{ moles}\).
So the number of moles is \(5.83 \times 10^{-4}\text{ moles}\).

(c) Precautions:
1. Adjust the volume very slowly to allow any heat generated during compression to dissipate, maintaining thermal equilibrium with the surroundings.
2. Wait a short period (e.g., 15-30 seconds) after each volume change before recording the pressure to ensure the temperature has returned to ambient room temperature.
3. Avoid holding the body of the syringe with bare hands to prevent heat transfer from the body to the trapped gas.

1. Explanation: Straight line through origin confirms direct proportionality of \(p\) and \(1/V\) [1 Mark]
2. Explanation: Easier to identify systematic errors or determine constant gradient than a curve of \(p\) vs \(V\) [1 Mark]
3. Identifies gradient \(m = nRT\) from the equation \(p = \frac{nRT}{V}\) [1 Mark]
4. Converts temperature to Kelvin: \(293.15\text{ K}\) [1 Mark]
5. Correct substitution of \(R\) and \(T\) into the rearranged equation \(n = \frac{m}{RT}\) [1 Mark]
6. Obtains correct answer of \(5.83 \times 10^{-4}\text{ mol}\) (accept \(5.8 \times 10^{-4}\text{ mol}\)) [1 Mark]
7. Precautions: mentions slow compression of gas [1 Mark]
8. Precautions: mentions waiting after compression before taking a reading or avoiding touching the syringe [2.23 Marks]

(a) Slit spacing \(d = \frac{1 \times 10^{-3}\text{ m}}{300} = 3.333 \times 10^{-6}\text{ m}\) (or \(3.33\text{ }\mu\text{m}\)).

(b) Angle of diffraction \(\theta\):
\(\tan \theta = \frac{x}{D} = \frac{0.585}{1.50} = 0.390 \Rightarrow \theta = \arctan(0.390) = 21.30^\circ\).
To find the uncertainty in \(\theta\):
\(x_{\text{max}} = 0.590\text{ m}\), \(D_{\text{min}} = 1.49\text{ m} \Rightarrow \tan \theta_{\text{max}} = \frac{0.590}{1.49} = 0.3960 \Rightarrow \theta_{\text{max}} = 21.60^\circ\).
\(x_{\text{min}} = 0.580\text{ m}\), \(D_{\text{max}} = 1.51\text{ m} \Rightarrow \tan \theta_{\text{min}} = \frac{0.580}{1.51} = 0.3841 \Rightarrow \theta_{\text{min}} = 21.01^\circ\).
Range of \(\theta = 21.60^\circ - 21.01^\circ = 0.59^\circ\).
Uncertainty in \(\theta = \frac{0.59^\circ}{2} \approx \pm 0.3^\circ\).
Thus, \(\theta = 21.3 \pm 0.3^\circ\).

(c) Wavelength \(\lambda\):
Using \(d \sin \theta = n \lambda\) for \(n = 2\):
\(\lambda = \frac{d \sin \theta}{2} = \frac{(3.333 \times 10^{-6}\text{ m}) \times \sin(21.30^\circ)}{2} = 6.05 \times 10^{-7}\text{ m}\) (or \(605\text{ nm}\)).
To find the uncertainty in \(\lambda\):
\(\lambda_{\text{max}} = \frac{3.333 \times 10^{-6} \times \sin(21.60^\circ)}{2} = 6.13 \times 10^{-7}\text{ m}\).
\(\lambda_{\text{min}} = \frac{3.333 \times 10^{-6} \times \sin(21.01^\circ)}{2} = 5.97 \times 10^{-7}\text{ m}\).
Range in \(\lambda = 6.13 \times 10^{-7} - 5.97 \times 10^{-7} = 0.16 \times 10^{-7}\text{ m}\).
Absolute uncertainty in \(\lambda = \pm 0.08 \times 10^{-7}\text{ m}\) (or \(\pm 8\text{ nm}\)).
Percentage uncertainty in \(\lambda = \frac{8\text{ nm}}{605\text{ nm}} \times 100\% \approx 1.3\%\).
Therefore, \(\lambda = 6.1 \times 10^{-7}\text{ m} \pm 1.3\%\).

1. Calculates slit spacing \(d = 3.33 \times 10^{-6}\text{ m}\) [1 Mark]
2. Calculates the central angle \(\theta = 21.3^\circ\) [1 Mark]
3. Calculates max angle \(\theta_{\text{max}} = 21.6^\circ\) or min angle \(\theta_{\text{min}} = 21.0^\circ\) [1 Mark]
4. Determines uncertainty in angle \(\theta = \pm 0.3^\circ\) [1 Mark]
5. Correct formula used: \(d \sin \theta = 2 \lambda\) [1 Mark]
6. Calculates \(\lambda = 6.05 \times 10^{-7}\text{ m}\) [1 Mark]
7. Calculates max/min wavelength values: \(\lambda_{\text{max}} = 6.13 \times 10^{-7}\text{ m}\) and \(\lambda_{\text{min}} = 5.97 \times 10^{-7}\text{ m}\) [1 Mark]
8. Obtains absolute uncertainty of \(\pm 8\text{ nm}\) (or \(\pm 0.08 \times 10^{-7}\text{ m}\)) and percentage uncertainty of \(1.3\%\) [2.23 Marks]

(a) The equation is: \(V = \mathcal{E} - Ir\).
Comparing with \(y = mx + c\) (where \(V\) is on the y-axis and \(I\) is on the x-axis):
- The y-intercept represents the e.m.f. \(\mathcal{E}\).
- The gradient of the graph is equal to \(-r\), where \(r\) is the internal resistance.

(b) From the best-fit line, the gradient is \(-0.85\text{ }\Omega\), so \(r = 0.85\text{ }\Omega\).
From the worst acceptable line of fit, the gradient is \(-0.98\text{ }\Omega\), so \(r_{\text{worst}} = 0.98\text{ }\Omega\).
The absolute uncertainty in the internal resistance is: \(\Delta r = |r_{\text{worst}} - r_{\text{best}}| = |0.98 - 0.85| = 0.13\text{ }\Omega\).
Therefore, \(r = 0.85 \pm 0.13\text{ }\Omega\).

(c) The total resistance of the circuit is \(R_{\text{total}} = R + r = 3.0 + 0.85 = 3.85\text{ }\Omega\).
The current \(I = \frac{\mathcal{E}}{R_{\text{total}}} = \frac{1.48}{3.85} = 0.3844\text{ A}\).
The power dissipated in the external resistor \(P = I^2 R = (0.3844)^2 \times 3.0 = 0.1478 \times 3.0 = 0.443\text{ W}\).
Rounding to an appropriate 2 significant figures gives \(0.44\text{ W}\).

1. States \(V = \mathcal{E} - Ir\) [1 Mark]
2. Explains y-intercept corresponds to \(\mathcal{E}\) [1 Mark]
3. Explains gradient corresponds to \(-r\) [1 Mark]
4. Determines best-fit \(r = 0.85\text{ }\Omega\) [1 Mark]
5. Calculates absolute uncertainty in \(r\) using the worst-fit line: \(\Delta r = 0.13\text{ }\Omega\) [1 Mark]
6. Calculates total resistance of the circuit: \(3.85\text{ }\Omega\) [1 Mark]
7. Calculates circuit current \(I = 0.384\text{ A}\) [1 Mark]
8. Calculates power dissipated in the external resistor: \(P = 0.44\text{ W}\) (accept \(0.443\text{ W}\)) [2.23 Marks]

(a) Taking natural logarithms on both sides of the equation \(R = R_0 e^{-\lambda t}\):
\(\ln(R) = \ln(R_0 e^{-\lambda t}) = \ln(R_0) + \ln(e^{-\lambda t}) = \ln(R_0) - \lambda t\).
Rearranging into the standard linear equation form \(y = mx + c\):
\(\ln(R) = (-\lambda)t + \ln(R_0)\).
Plotting \(\ln(R)\) on the y-axis against \(t\) on the x-axis yields a straight line with gradient \(m = -\lambda\). Thus, the decay constant \(\lambda = -\text{gradient}\).

(b) Using the data provided:
At \(t_1 = 60\text{ s}\), \(R_1 = 320\text{ s}^{-1} \Rightarrow \ln(R_1) = \ln(320) = 5.7683\).
At \(t_2 = 300\text{ s}\), \(R_2 = 80\text{ s}^{-1} \Rightarrow \ln(R_2) = \ln(80) = 4.3820\).
Gradient \(m = \frac{\ln(R_2) - \ln(R_1)}{t_2 - t_1} = \frac{4.3820 - 5.7683}{300 - 60} = \frac{-1.3863}{240} = -0.005776\text{ s}^{-1}\).
Since \(m = -\lambda\), the decay constant \(\lambda = 0.00578\text{ s}^{-1}\) (or \(5.8 \times 10^{-3}\text{ s}^{-1}\)).
Half-life \(t_{1/2} = \frac{\ln(2)}{\lambda} = \frac{0.69315}{0.005776} = 120\text{ s}\).

(c)
- A short-lived source decays significantly over a short period. If we use long counting intervals, the activity changes significantly during the count, which averages out the rapid changes and prevents us from measuring the instantaneous activity at a specific time.
- However, a shorter counting interval means the total number of recorded counts \(N\) is much lower. Since radioactive decay is a random process, the statistical fluctuation (uncertainty) is \(\sqrt{N}\). The percentage uncertainty in the count rate is \(\frac{\sqrt{N}}{N} \times 100\% = \frac{100\%}{\sqrt{N}}\), which increases as \(N\) decreases. Hence, short intervals result in greater statistical uncertainty.

1. Derives \(\ln(R) = -\lambda t + \ln(R_0)\) correctly [1 Mark]
2. Explains that the gradient \(m = -\lambda\) [1 Mark]
3. Calculates \(\ln(R_1)\) and \(\ln(R_2)\) correctly [1 Mark]
4. Correctly calculates gradient \(m = -0.00578\text{ s}^{-1}\) [1 Mark]
5. Identifies decay constant \(\lambda = 5.78 \times 10^{-3}\text{ s}^{-1}\) (or \(5.8 \times 10^{-3}\text{ s}^{-1}\)) [1 Mark]
6. Calculates half-life \(t_{1/2} = 120\text{ s}\) [1 Mark]
7. Explains necessity of short intervals: activity decays rapidly, long counts average out the decay curve [1 Mark]
8. Explains statistical uncertainty: smaller number of counts \(N\) leads to larger relative fluctuation \(\frac{1}{\sqrt{N}}\) [2.23 Marks]

(a) The reference wire is suspended from the same ceiling/support beam and experiences the same thermal conditions as the test wire. It corrects for:
- Thermal expansion or contraction of the wires due to changing room temperature.
- Any sagging or downward deflection of the support beam when heavy loads are added. Since both wires move together, the relative measurement of extension is unaffected.

(b) The formula for the Young Modulus is:
\(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L}\)
First, calculate the cross-sectional area \(A\):
\(A = \frac{\pi d^2}{4} = \frac{\pi (0.56 \times 10^{-3}\text{ m})^2}{4} = 2.463 \times 10^{-7}\text{ m}^2\).
Now substitute the values into the Young Modulus equation:
\(E = \frac{35.0 \times 2.450}{(2.463 \times 10^{-7}) \times (1.8 \times 10^{-3})} = 1.934 \times 10^{11}\text{ Pa}\).
Rounding to 2 significant figures gives \(1.9 \times 10^{11}\text{ Pa}\).

(c) Let's compute the percentage uncertainty for each measurement:
1. Original length \(L\): \(\%\Delta L/L = \frac{0.002}{2.450} \times 100\% = 0.082\%\).
2. Diameter \(d\): \(\%\Delta d/d = \frac{0.02}{0.56} \times 100\% = 3.571\%\).
Since \(A \propto d^2\), the percentage uncertainty in area \(A\) is \(2 \times 3.571\% = 7.14\%\). Thus, the contribution of \(d\) to the uncertainty in \(E\) is \(7.14\%\).
3. Extension \(\Delta L\): \(\%\Delta (\Delta L) / \Delta L = \frac{0.1}{1.8} \times 100\% = 5.56\%\).

Comparing the contributions:
- \(L\): \(0.082\%\)
- \(d\): \(7.14\%\)
- \(\Delta L\): \(5.56\%\)

The diameter \(d\) contributes the most to the overall percentage uncertainty in the Young Modulus, because its percentage uncertainty is doubled in the calculation of the cross-sectional area.

1. Explanation of reference wire: compensates for thermal expansion/contraction of the wire [1 Mark]
2. Explanation of reference wire: compensates for support flexure or sagging [1 Mark]
3. Calculates cross-sectional area \(A = 2.46 \times 10^{-7}\text{ m}^2\) [1 Mark]
4. Substitutes values correctly into the Young Modulus formula [1 Mark]
5. Correct value for Young Modulus: \(1.9 \times 10^{11}\text{ Pa}\) (or \(\text{N m}^{-2}\)) [1 Mark]
6. Calculates percentage uncertainty of original length: \(0.08\%\) [1 Mark]
7. Calculates percentage uncertainty of extension: \(5.6\%\) [1 Mark]
8. Calculates percentage uncertainty of diameter (\(3.6\%\)), doubles it for area (\(7.1\%\)), and concludes \(d\) has the greatest contribution [2.23 Marks]

(a) From the equation \(B_0 = \left(\frac{\mu_0 N}{2R}\right) I\), comparing with \(y = mx + c\) shows that a plot of \(B_0\) on the y-axis against \(I\) on the x-axis will yield a straight line through the origin with a gradient:
\(m = \frac{\mu_0 N}{2R}\).
To find \(\mu_0\), the student must:
- Rearrange to get \(\mu_0 = \frac{2 R m}{N}\).
- Measure the number of turns \(N\) of the coil (usually known/given).
- Measure the radius \(R\) of the coil (which involves measuring the diameter and dividing by 2).

(b) Diameter \(D = 12.0\text{ cm} = 0.120\text{ m}\), so radius \(R = 0.060\text{ m}\).
Gradient \(m = 1.57 \times 10^{-3}\text{ T A}^{-1}\).
Number of turns \(N = 150\).
\(\mu_0 = \frac{2 \times 0.060 \times 1.57 \times 10^{-3}}{150} = \frac{1.884 \times 10^{-4}}{150} = 1.256 \times 10^{-6}\text{ T m A}^{-1}\) (or \(\text{H m}^{-1}\)).
Rounding to 3 significant figures, \(\mu_0 = 1.26 \times 10^{-6}\text{ T m A}^{-1}\).

(c) The Hall voltage is proportional to the component of the magnetic field perpendicular to the probe sensor surface (\(B = B_{\text{total}} \cos \theta\), where \(\theta\) is the angle of deviation from perpendicularity).
- The probe face must be aligned perfectly perpendicular to the axis of the coil (which is the direction of the magnetic field along the axis).
- To achieve this, the student should rotate the probe slightly in three dimensions at each location until the reading on the voltmeter/gaussmeter is maximized. The maximum reading corresponds to \(\theta = 0\) (\(\cos \theta = 1\)), which represents the true field strength.

1. Compares formula to linear form to show gradient \(m = \frac{\mu_0 N}{2R}\) [1 Mark]
2. Rearranges to find \(\mu_0 = \frac{2 R m}{N}\) [1 Mark]
3. Identifies that number of turns \(N\) and radius \(R\) (or diameter) must be measured [1 Mark]
4. Correctly converts diameter to radius in meters: \(R = 0.060\text{ m}\) [1 Mark]
5. Substitutes values into rearranged equation correctly [1 Mark]
6. Obtains \(\mu_0 = 1.26 \times 10^{-6}\text{ T m A}^{-1}\) (or \(\text{H m}^{-1}\) or \(\text{N A}^{-2}\)) [1 Mark]
7. Explains that Hall voltage depends on angle (varies with \(\cos \theta\)) [1 Mark]
8. Explains how to align perpendicularity: rotate probe until maximum reading is achieved [2.23 Marks]

(a) Comparing the equation \( L = \frac{v}{4} \left(\frac{1}{f}\right) - c \) with the general equation for a straight line \( y = mx + c \), we can identify that plotting \( L \) against \( \frac{1}{f} \) yields a straight line where the gradient is \( m = \frac{v}{4} \) and the y-intercept is \( -c \). Thus, the speed of sound is determined by \( v = 4 \times \text{gradient} \), and the end correction \( c \) is equal to the magnitude of the y-intercept.

(b) The speed of sound is:
\( v = 4 \times 85.5 = 342 \text{ m s}^{-1} \)

The percentage uncertainty in the speed of sound is the same as the percentage uncertainty in the gradient, which is \( 2.4\% \). The absolute uncertainty in \( v \) is:
\( \Delta v = 342 \times 0.024 = 8.2 \approx 8 \text{ m s}^{-1} \)

Thus, the speed of sound with its absolute uncertainty is:
\( v = 342 \pm 8 \text{ m s}^{-1} \)

(c) A major source of systematic error is the end correction \( c \) (the position of the antinode slightly outside the tube opening). If only a single length and frequency measurement were taken, neglecting this end correction would systematically skew the calculated speed of sound. By plotting multiple points on a graph, this end correction only shifts the y-intercept of the line downwards and does not affect the gradient. Since the speed of sound is calculated purely from the gradient of the line, this systematic error is eliminated.

Part (a) (3 marks):
- 1 mark for expressing the equation in the form \( y = mx + c \) to demonstrate a linear relationship.
- 1 mark for identifying that the speed of sound \( v = 4 \times \text{gradient} \).
- 1 mark for identifying that the end correction is equal to the magnitude of the vertical intercept (y-intercept).

Part (b) (3 marks):
- 1 mark for calculating the correct speed of sound: \( v = 342 \text{ m s}^{-1} \).
- 1 mark for calculating the absolute uncertainty: \( 8.2 \text{ m s}^{-1} \) or \( 8 \text{ m s}^{-1} \).
- 1 mark for expressing the final answer to consistent decimal places or appropriate significant figures, e.g., \( 342 \pm 8 \text{ m s}^{-1} \).

Part (c) (3.23 marks):
- 1 mark for identifying the systematic error associated with the position of the antinode relative to the physical end of the tube.
- 1 mark for explaining that using a single measurement would lead to an over- or under-estimation of \( v \).
- 1.23 marks for explaining that on the graph, this offset affects only the y-intercept, leaving the gradient (and therefore \( v \)) unaffected.

(a) First, find the mean diameter \( d \):
\( d = \frac{0.35 + 0.34 + 0.36 + 0.35 + 0.35}{5} = 0.35 \text{ mm} \)

The range of measurements is \( 0.36 - 0.34 = 0.02 \text{ mm} \).
Using half-range for absolute uncertainty:
\( \Delta d = \frac{0.02}{2} = 0.01 \text{ mm} \)

Percentage uncertainty in \( d \):
\( \frac{0.01}{0.35} \times 100\% \approx 2.86\% \)

The cross-sectional area is given by \( A = \frac{\pi d^2}{4} \). Since \( A \propto d^2 \), the percentage uncertainty in the area is double that of the diameter:
\( \text{Percentage uncertainty in } A = 2 \times 2.86\% = 5.71\% \)

(b) First, calculate the cross-sectional area:
\( A = \frac{\pi \times (0.35 \times 10^{-3} \text{ m})^2}{4} = 9.621 \times 10^{-8} \text{ m}^2 \)

Now, calculate the resistivity:
\( \rho = \frac{R A}{L} = \frac{5.60 \times 9.621 \times 10^{-8}}{1.000} = 5.388 \times 10^{-7} \ \Omega \text{ m} \approx 5.39 \times 10^{-7} \ \Omega \text{ m} \)

Next, calculate individual percentage uncertainties:
- Resistance: \( \frac{0.12}{5.60} \times 100\% = 2.14\% \)
- Length: \( \frac{0.002}{1.000} \times 100\% = 0.20\% \)
- Area: \( 5.71\% \) (from part a)

Sum the percentage uncertainties to find the total percentage uncertainty in resistivity:
\( \text{Total uncertainty in } \rho = 2.14\% + 5.71\% + 0.20\% = 8.05\% \)

Calculate the absolute uncertainty:
\( \Delta \rho = 5.388 \times 10^{-7} \times 0.0805 \approx 4.34 \times 10^{-8} \ \Omega \text{ m} = 0.43 \times 10^{-7} \ \Omega \text{ m} \)

Thus, the resistivity is \( \rho = (5.39 \pm 0.43) \times 10^{-7} \ \Omega \text{ m} \).

Part (a) (4 marks):
- 1 mark for correct mean diameter of \( 0.35 \text{ mm} \).
- 1 mark for finding absolute uncertainty of diameter using half-range as \( 0.01 \text{ mm} \).
- 1 mark for percentage uncertainty of diameter of \( 2.86\% \) (accept \( 2.9\% \)).
- 1 mark for multiplying the diameter percentage uncertainty by 2 to obtain the area uncertainty of \( 5.71\% \) (accept \( 5.7\% \)).

Part (b) (5.23 marks):
- 1 mark for calculating correct cross-sectional area \( A = 9.62 \times 10^{-8} \text{ m}^2 \).
- 1 mark for calculating correct resistivity \( \rho = 5.39 \times 10^{-7} \ \Omega \text{ m} \).
- 1 mark for calculating percentage uncertainty in resistance as \( 2.14\% \).
- 1 mark for calculating total percentage uncertainty in resistivity as \( 8.05\% \) (accept \( 8.1\% \)).
- 1.23 marks for calculating absolute uncertainty as \( 0.43 \times 10^{-7} \ \Omega \text{ m} \) and giving final formatted answer with units.

(a) Taking the natural logarithm of both sides of the relation:
\( \ln T = \ln(k l^n) \implies \ln T = n \ln l + \ln k \)

Comparing this to the equation of a straight line, \( y = mx + c \), where \( y = \ln T \) and \( x = \ln l \):
- The gradient is \( m = n \).
- The y-intercept is \( c = \ln k \).

According to simple pendulum theory, \( T = 2\pi \sqrt{\frac{l}{g}} = \left(\frac{2\pi}{\sqrt{g}}\right) l^{0.5} \). Since \( n = 0.5 \), the gradient is expected to be \( 0.5 \), and the intercept is \( \ln \left(\frac{2\pi}{\sqrt{g}}\right) \).

(b) Timing a single oscillation is highly subject to human reaction time (typically about \( \pm 0.2 \text{ s} \) for starting and stopping a stopwatch). By timing 20 oscillations, the absolute uncertainty in the total measured time remains approximately \( \pm 0.2 \text{ s} \). When dividing by 20 to find the period \( T \), the absolute uncertainty in \( T \) is reduced to \( \frac{0.2 \text{ s}}{20} = 0.01 \text{ s} \). Thus, the percentage uncertainty in the period is drastically reduced.

(c) From the graph, the intercept is \( \ln k = 0.70 \).
\( k = e^{0.70} = 2.0138 \text{ s m}^{-0.5} \)

Using the formula for \( k \):
\( k = \frac{2\pi}{\sqrt{g}} \implies \sqrt{g} = \frac{2\pi}{k} \implies g = \frac{4\pi^2}{k^2} \)
\( g = \frac{4\pi^2}{(2.0138)^2} = 9.73 \text{ m s}^{-2} \)

Part (a) (3 marks):
- 1 mark for correct logarithmic expansion to yield the linear relationship: \( \ln T = n \ln l + \ln k \).
- 1 mark for matching variables to show the gradient is \( n \) and showing that \( n = 0.5 \) based on \( T \propto \sqrt{l} \).
- 1 mark for identifying that the y-intercept is \( \ln k \).

Part (b) (3 marks):
- 1 mark for identifying that human reaction time is the major source of absolute uncertainty in timing.
- 1 mark for stating that timing multiple (20) oscillations keeps the absolute reaction time uncertainty constant while increasing the total measured time.
- 1 mark for explaining that dividing the total time by 20 reduces the absolute uncertainty in the calculated single period (or equivalently, reduces percentage uncertainty).

Part (c) (3.23 marks):
- 1 mark for calculating \( k = e^{0.70} \approx 2.01 \).
- 1 mark for algebraically rearranging to show \( g = \frac{4\pi^2}{k^2} \).
- 1.23 marks for calculating \( g = 9.73 \text{ m s}^{-2} \) (accept a range of 9.70 to 9.80 depending on rounding).

(a) Take the natural logarithm of both sides of the equation:
\( \ln V = \ln(V_0 e^{-t/RC}) \implies \ln V = -\frac{t}{RC} + \ln V_0 \)

This is a linear equation of the form \( y = mx + c \), where:
- \( y = \ln V \) is plotted on the vertical axis.
- \( x = t \) is plotted on the horizontal axis.
- The gradient is \( m = -\frac{1}{RC} \).

Thus, the capacitance can be found using: \( C = -\frac{1}{m R} \).

(b) Substitute the values into the formula:
\( C = -\frac{1}{(-0.0135 \text{ s}^{-1}) \times (150 \times 10^3 \ \Omega)} \)
\( C = 4.938 \times 10^{-4} \text{ F} \approx 494 \ \mu\text{F} \)

(c) The formula for capacitance is \( C = \frac{1}{R \times |m|} \).
For calculations involving multiplication or division, we sum the percentage uncertainties of each component.

Percentage uncertainty in \( R \):
\( \% \Delta R = \frac{3}{150} \times 100\% = 2.0\% \)

Percentage uncertainty in gradient \( m \):
\( \% \Delta m = 4.0\% \)

Total percentage uncertainty in \( C \):
\( \% \Delta C = 2.0\% + 4.0\% = 6.0\% \)

Part (a) (3 marks):
- 1 mark for taking the natural log to find \( \ln V = -\frac{t}{RC} + \ln V_0 \).
- 1 mark for specifying the variables on the axes: \( \ln V \) on the vertical axis and \( t \) on the horizontal axis.
- 1 mark for showing that the gradient \( m = -\frac{1}{RC} \), so \( C = -\frac{1}{mR} \).

Part (b) (3 marks):
- 1 mark for converting resistance correctly to ohms (\( 150,000 \ \Omega \)).
- 1 mark for correct numerical substitution of both gradient and resistance.
- 1 mark for obtaining the value \( 4.94 \times 10^{-4} \text{ F} \) (or \( 494 \ \mu\text{F} \)) to 3 significant figures with units.

Part (c) (3.23 marks):
- 1 mark for determining that the percentage uncertainty of \( R \) is \( 2.0\% \).
- 1 mark for showing that percentage uncertainties of \( R \) and the gradient should be summed.
- 1.23 marks for calculating the final percentage uncertainty as \( 6.0\% \).

(a) Starting from \( P t = m c \Delta\theta \):
\( \Delta\theta = \left(\frac{P}{mc}\right) t \)

Since \( \theta = \theta_0 + \Delta\theta \):
\( \theta = \left(\frac{P}{mc}\right) t + \theta_0 \)

Comparing this with \( y = mx + c \), where temperature \( \theta \) is on the y-axis and time \( t \) is on the x-axis, the gradient is \( \frac{P}{mc} \).

(b) Calculate power \( P \):
\( P = I \times V = 2.50 \times 12.0 = 30.0 \text{ W} \)

Calculate percentage uncertainties:
- For current: \( \% \Delta I = \frac{0.05}{2.50} \times 100\% = 2.0\% \)
- For voltage: \( \% \Delta V = \frac{0.2}{12.0} \times 100\% = 1.67\% \)

Total percentage uncertainty in power:
\( \% \Delta P = 2.0\% + 1.67\% = 3.67\% \approx 3.7\% \)

(c) Using the gradient equation:
\( \text{gradient} = \frac{P}{mc} \implies c = \frac{P}{m \times \text{gradient}} \)
\( c = \frac{30.0}{1.00 \times 0.058} = 517.2 \approx 517 \text{ J kg}^{-1}\text{ K}^{-1} \)

The experimental value is higher than the textbook value because thermal energy is lost to the surroundings. Some of the electrical energy supplied is transferred to the environment rather than warming the block. As a result, the temperature of the block rises slower than it would in an ideal closed system, resulting in a smaller gradient and an artificially elevated value for \( c \).

Part (a) (2 marks):
- 1 mark for expressing temperature \( \theta \) as a function of time \( t \).
- 1 mark for identifying that the gradient of the graph is equal to \( \frac{P}{mc} \).

Part (b) (3.23 marks):
- 1 mark for calculating the correct electrical power \( P = 30.0 \text{ W} \).
- 1 mark for finding percentage uncertainties for current (2%) and voltage (1.67%).
- 1.23 marks for adding the percentage uncertainties together to get \( 3.67\% \) (accept \( 3.7\% \)).

Part (c) (4 marks):
- 1 mark for rearranging the formula for \( c \): \( c = \frac{P}{m \times \text{gradient}} \).
- 1 mark for calculating the specific heat capacity correctly as \( 517 \text{ J kg}^{-1}\text{ K}^{-1} \) (or \( 520 \text{ J kg}^{-1}\text{ K}^{-1} \)).
- 1 mark for stating that thermal energy is lost to the surroundings.
- 1 mark for explaining that heat loss means a lower rate of temperature increase, leading to a smaller gradient and consequently a larger calculated value of \( c \).