2023 Cambridge IAL Physics (9702) Practice Paper with Answers

The percentage uncertainty in a calculated value involving products and quotients is the sum of the percentage uncertainties of each term (with powers acting as multipliers).

First, calculate the individual percentage uncertainties:
- Mass: \(\frac{\Delta m}{m} = \frac{0.1}{12.5} \times 100\% = 0.8\%\)
- Diameter: \(\frac{\Delta d}{d} = \frac{0.02}{1.50} \times 100\% \approx 1.33\%\)
- Length: \(\frac{\Delta l}{l} = \frac{0.1}{4.0} \times 100\% = 2.5\%\)

Now combine these according to the formula \(\rho = \frac{4m}{\pi d^2 l}\):
\(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta l}{l}\)
\(\frac{\Delta \rho}{\rho} = 0.8\% + 2(1.33\%) + 2.5\% = 0.8\% + 2.67\% + 2.5\% = 5.97\% \approx 6.0\%\).

1 mark for calculating the individual percentage uncertainties and summing them correctly, accounting for the power of 2 for diameter, to obtain 6.0%.

According to Newton's second law, the force is equal to the rate of change of momentum.

The mass of water striking the wall per second is:
\(\frac{\Delta m}{\Delta t} = \rho A v\)

Taking the initial direction of the water jet as positive, the initial velocity is \(v\) and the final velocity is \(-\frac{1}{3}v\).

The change in momentum per second (force on the water) is:
\(F_{\text{water}} = \frac{\Delta m}{\Delta t} \Delta v = (\rho A v) \left(-\frac{1}{3}v - v\right) = -\frac{4}{3} \rho A v^2\)

By Newton's third law, the force exerted by the water on the wall is equal in magnitude and opposite in direction:
\(F_{\text{wall}} = \frac{4}{3} \rho A v^2\).

1 mark for correctly applying Newton's second and third laws to determine the rate of change of momentum of the rebounding water jet.

For rotational equilibrium, the sum of clockwise moments about the pivot must equal the sum of counter-clockwise moments about the pivot.

Take moments about the pivot on the wall:
Clockwise moments:
- Due to the beam's weight acting at its center of gravity: \(W \times \frac{L}{2}\)
- Due to the suspended load acting at the far end: \(2W \times L\)
Total clockwise moment = \(0.5WL + 2WL = 2.5WL\)

Counter-clockwise moment:
- Due to the vertical component of the tension \(T\) in the cable: \((T \sin 30^\circ) \times L = 0.5TL\)

Equating the moments:
\(0.5TL = 2.5WL \implies T = 5.0W\).

1 mark for taking moments about the pivot and correctly solving for the tension in terms of W.

Strain is defined as \(\epsilon = \frac{\text{stress}}{E}\), where \(E\) is the Young's Modulus of the material. Since both wires are made of the same material, they have the same Young's Modulus \(E\).

Stress is defined as \(\sigma = \frac{F}{A}\), where \(F\) is the applied load and \(A\) is the cross-sectional area. Since \(A = \frac{\pi d^2}{4}\), the stress is inversely proportional to the square of the diameter:
\(\sigma \propto \frac{1}{d^2}\)

Because the loads \(F\) are equal, the strain is also inversely proportional to the square of the diameter:
\(\epsilon \propto \frac{1}{d^2}\)

Now, calculate the ratio of the strains:
\(\frac{\epsilon_X}{\epsilon_Y} = \frac{d_Y^2}{d_X^2} = \frac{(2d)^2}{d^2} = 4\).

1 mark for recognizing that strain is independent of the original length of the wire and depends only on stress, yielding a ratio of 4.

In a longitudinal wave, particles oscillate parallel to the direction of wave travel. Here, positive displacements represent displacement to the right, and negative displacements represent displacement to the left.

A compression is a region of high pressure where air particles are bunched together:
- Particles to the left of the compression point must have a positive displacement (moving right towards the point).
- Particles to the right of the compression point must have a negative displacement (moving left towards the point).

This behavior occurs at a point where the displacement is zero and the gradient of the displacement-distance graph is negative, so that displacement transitions from positive to negative as distance increases.

1 mark for correctly analyzing the direction of particle displacements to identify the compression point on the displacement-distance graph.

The resistance of a wire is given by \(R = \rho \frac{L}{A}\), where \(\rho\) is the resistivity, \(L\) is the length, and \(A\) is the cross-sectional area.

Let the initial length be \(L_0\) and the initial area be \(A_0\). The initial resistance is \(R = \rho \frac{L_0}{A_0}\).

When the wire is stretched by \(20\%\), its new length is:
\(L' = 1.20 L_0\)

Since the volume \(V = A \times L\) remains constant:
\(A_0 L_0 = A' L' \implies A' = \frac{A_0 L_0}{1.20 L_0} = \frac{A_0}{1.20}\)

The new resistance \(R'\) is:
\(R' = \rho \frac{L'}{A'} = \rho \frac{1.20 L_0}{A_0 / 1.20} = 1.44 \left(\rho \frac{L_0}{A_0}\right) = 1.44 R\).

1 mark for using the conservation of volume to find the new area and using the resistivity formula to obtain 1.44R.

From the ideal gas equation, \(p V = n R T \implies n = \frac{p V}{R T}\).

For the two containers, the ratio of the number of moles is:
\(\frac{n_2}{n_1} = \frac{p_2 V_2}{R T_2} \times \frac{R T_1}{p_1 V_1} = \left(\frac{p_2}{p_1}\right) \left(\frac{V_2}{V_1}\right) \left(\frac{T_1}{T_2}\right)\)

Substitute the given values into the ratio expression:
\(\frac{n_2}{n_1} = (0.5) \times (2) \times \left(\frac{1}{1.5}\right) = \frac{1}{1.5} \approx 0.67\).

1 mark for applying the equation of state for both gases and calculating the ratio of moles to be 0.67.

Initially, the charge stored is \(Q = C V\), and the initial energy stored is:
\(E_i = \frac{Q^2}{2C}\)

When the charged capacitor is connected in parallel with an identical uncharged capacitor, the total charge \(Q\) remains constant by conservation of charge.

The equivalent capacitance of the parallel combination is:
\(C_{\text{total}} = C + C = 2C\)

The total final energy stored is:
\(E_f = \frac{Q^2}{2 C_{\text{total}}} = \frac{Q^2}{4C}\)

The ratio of the final energy to the initial energy is:
\(\frac{E_f}{E_i} = \frac{Q^2 / (4C)}{Q^2 / (2C)} = 0.50\).

1 mark for using conservation of charge and the equivalent capacitance to find the final energy, resulting in a ratio of 0.50.

The fractional uncertainty in \(\rho\) is given by: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\). Evaluating the individual fractional uncertainties: \(\frac{\Delta R}{R} = \frac{0.08}{4.00} = 0.02\), \(\frac{\Delta d}{d} = \frac{0.01}{0.50} = 0.02\), and \(\frac{\Delta L}{L} = \frac{0.02}{2.00} = 0.01\). Substituting these into the formula yields: \(\frac{\Delta \rho}{\rho} = 0.02 + 2(0.02) + 0.01 = 0.07\). Expressed as a percentage, this is \(0.07 \times 100\% = 7.0\%\).

Award 1 mark for the correct calculation of individual fractional uncertainties, doubling the fractional uncertainty of the diameter, summing them to find the total fractional uncertainty of 0.07, and identifying B as the correct percentage uncertainty.

Let the rightwards direction be positive. Let \(u_X = 4.0\text{ m s}^{-1}\) and \(u_Y = -2.0\text{ m s}^{-1}\). The total initial momentum is: \(p_i = m_X u_X + m_Y u_Y = (2.0 \times 4.0) + (3.0 \times -2.0) = 8.0 - 6.0 = 2.0\text{ kg m s}^{-1}\). Since the collision is perfectly elastic, the relative speed of approach equals the relative speed of separation: \(u_X - u_Y = v_Y - v_X \implies 4.0 - (-2.0) = v_Y - v_X \implies v_Y = v_X + 6.0\). Using conservation of momentum: \(m_X v_X + m_Y v_Y = 2.0 \implies 2.0 v_X + 3.0(v_X + 6.0) = 2.0\). Solving this gives: \(5.0 v_X + 18.0 = 2.0 \implies 5.0 v_X = -16.0 \implies v_X = -3.2\text{ m s}^{-1}\). Then \(v_Y = -3.2 + 6.0 = +2.8\text{ m s}^{-1}\).

Award 1 mark for applying both the conservation of momentum and the relative speed relationship for an elastic collision to set up a system of equations, solving for the final velocities of both balls, and selecting option A.

When two wires are joined in series, they support the same tension force \(F\). The elastic potential energy stored in a wire is given by \(E_p = \frac{1}{2} F \Delta L\), where \(\Delta L\) is the extension. Since Young modulus \(E_Y = \frac{F L_0}{A \Delta L}\), the extension is \(\Delta L = \frac{F L_0}{A E_Y}\). Since the wires have the same original length \(L_0\), cross-sectional area \(A\), and experience the same force \(F\), the extension is inversely proportional to the Young modulus: \(\Delta L \propto \frac{1}{E_Y}\). Consequently, the elastic potential energy stored is also inversely proportional to the Young modulus: \(E_p \propto \frac{1}{E_Y}\). Therefore, the ratio is \(\frac{E_{\text{steel}}}{E_{\text{brass}}} = \frac{E_{Y,\text{brass}}}{E_{Y,\text{steel}}} = \frac{1.0 \times 10^{11}\text{ Pa}}{2.0 \times 10^{11}\text{ Pa}} = 0.50\).

Award 1 mark for recognizing that both wires in series experience the same tension, relating the extension and elastic potential energy to the Young modulus, finding that energy is inversely proportional to the Young modulus, and correctly calculating the ratio to be 0.50 (Option B).

The phase difference \(\phi\) between two points separated by a distance \(x\) is given by \(\phi = \frac{2\pi x}{\lambda}\), where \(\lambda\) is the wavelength. Substituting the given values: \(\frac{3\pi}{5} = \frac{2\pi \times 0.150\text{ m}}{\lambda}\). Solving for \(\lambda\): \(\frac{3}{5} = \frac{0.300}{\lambda} \implies \lambda = \frac{5 \times 0.300}{3} = 0.500\text{ m}\). The wave speed \(v\) is then found using the wave equation: \(v = f \lambda = 50.0\text{ Hz} \times 0.500\text{ m} = 25.0\text{ m s}^{-1}\).

Award 1 mark for setting up the relation between phase difference and path difference, solving for the wavelength of 0.500 m, calculating the wave speed as 25.0 m/s using v = f*lambda, and choosing option B.

The potential divider equation is used to calculate the output voltage \(V_{\text{out}}\) across the LDR: \(V_{\text{out}} = V_{\text{in}} \left( \frac{R_{\text{LDR}}}{R_{\text{fixed}} + R_{\text{LDR}}} \right)\). In bright light, \(R_{\text{LDR}} = 1.0\text{ k}\Omega\), so \(V_{\text{out, light}} = 12.0 \left( \frac{1.0}{3.0 + 1.0} \right) = 3.0\text{ V}\). In darkness, \(R_{\text{LDR}} = 9.0\text{ k}\Omega\), so \(V_{\text{out, dark}} = 12.0 \left( \frac{9.0}{3.0 + 9.0} \right) = 9.0\text{ V}\). The change in the voltmeter reading is \(\Delta V = V_{\text{out, dark}} - V_{\text{out, light}} = 9.0\text{ V} - 3.0\text{ V} = 6.0\text{ V}\).

Award 1 mark for correctly applying the potential divider formula to find both initial and final voltmeter readings (3.0 V and 9.0 V), taking the difference to find a change of 6.0 V, and selecting option C.

The density of the gas is defined as \(\rho = \frac{m}{V}\). Since the volume \(V\) of the container is fixed and no gas escapes, the mass \(m\) remains constant, meaning the density is unchanged and remains \(\rho\). From the kinetic theory of gases, the average kinetic energy of the molecules is directly proportional to the absolute temperature: \(\frac{1}{2} m \langle c^2 \rangle = \frac{3}{2} k T\). Thus, the mean-square speed \(\langle c^2 \rangle\) is directly proportional to the absolute temperature \(T\). When the absolute temperature is doubled, the mean-square speed also doubles to \(2\langle c^2 \rangle\) (note that it is the root-mean-square speed, \(c_{\text{rms}}\), that would increase by a factor of \(\sqrt{2}\)). This corresponds to option B.

Award 1 mark for identifying that the density remains constant because mass and volume are constant, recognizing that mean-square speed is directly proportional to absolute temperature, and selecting option B.

Because internal energy \(U\) is a state function, any complete cycle that returns the gas to its initial state must have a net change in internal energy of \(\Delta U_{\text{net}} = 0\). The first law of thermodynamics is written as \(\Delta U = q + w\), where \(q\) is the thermal energy transferred to the gas and \(w\) is the work done on the gas. For the entire cycle, \(\Delta U_{\text{net}} = q_{\text{net}} + w_{\text{net}} = 0\), which means \(q_{\text{net}} = -w_{\text{net}}\). Let's calculate the net work done on the gas: 1) Isothermal compression: work done on the gas is \(w_1 = +120\text{ J}\). 2) Constant volume heating: no work is done, \(w_2 = 0\). 3) Expansion: the gas does \(90\text{ J}\) of work, so work done on the gas is \(w_3 = -90\text{ J}\). The net work done on the gas is \(w_{\text{net}} = +120 + 0 - 90 = +30\text{ J}\). Therefore, the net thermal energy transferred to the gas is \(q_{\text{net}} = -w_{\text{net}} = -30\text{ J}\). This matches option B.

Award 1 mark for identifying that the net change in internal energy over a complete cycle is zero, calculating the net work done on the gas as +30 J, applying the first law of thermodynamics to determine the net thermal energy transferred to the gas is -30 J, and choosing option B.

According to Einstein's photoelectric equation, the energy of an incident photon is \(E_{\text{photon}} = \Phi + E_{\text{k,max}}\). For wavelength \(\lambda\), the photon energy is \(\frac{hc}{\lambda}\), so: \(\frac{hc}{\lambda} = \Phi + E_{\text{max}}\). For wavelength \(\frac{\lambda}{2}\), the photon energy is \(\frac{hc}{\lambda/2} = \frac{2hc}{\lambda}\), so: \(\frac{2hc}{\lambda} = \Phi + E'\). We can substitute the first equation into the second to eliminate the photon energy term: \(2(\Phi + E_{\text{max}}) = \Phi + E'\). Expanding the left side gives: \(2\Phi + 2E_{\text{max}} = \Phi + E'\). Subtracting \(\Phi\) and \(2E_{\text{max}}\) from both sides yields: \(\Phi = E' - 2E_{\text{max}}\). This corresponds to option A.

Award 1 mark for setting up Einstein's photoelectric equations for both cases, expressing the second photon energy as twice the first, eliminating the photon energy term to solve for the work function, and selecting option A.

The percentage uncertainty in a product or quotient is found by summing the individual percentage uncertainties, with exponents acting as multipliers. The percentage uncertainty in \(R\) is \(\frac{0.05}{2.50} \times 100\% = 2.0\%\). The percentage uncertainty in \(d\) is \(\frac{0.01}{0.40} \times 100\% = 2.5\%\). Since \(d\) is squared in the formula, its contribution is \(2 \times 2.5\% = 5.0\%\). The percentage uncertainty in \(L\) is \(\frac{0.002}{1.000} \times 100\% = 0.2\%\). Summing these gives: \(2.0\% + 5.0\% + 0.2\% = 7.2\%\).

Correct option is B. 1 mark for calculating the final percentage uncertainty as 7.2%.

Using the equation of motion \(s = ut + \frac{1}{2}at^2\) with upwards taken as positive: \(u = +15\ \text{m s}^{-1}\), \(a = -9.81\ \text{m s}^{-2}\), and \(t = 4.5\ \text{s}\). Therefore, \(s = (15 \times 4.5) + \frac{1}{2}(-9.81)(4.5)^2 = 67.5 - 99.3 = -31.8\ \text{m}\). The negative sign indicates displacement is downwards from the release point, so the height of the cliff is approximately \(32\ \text{m}\).

Correct option is A. 1 mark for correct algebraic substitution and rounding to two significant figures.

The useful work done in lifting the crate is \(W = mgh = 120 \times 9.81 \times 15 = 17658\ \text{J}\). The useful power output of the motor is \(P_{\text{out}} = \frac{W}{t} = \frac{17658}{8.0} = 2207\ \text{W}\). Since the efficiency is \(65\%\), the power input is \(P_{\text{in}} = \frac{P_{\text{out}}}{0.65} = \frac{2207}{0.65} \approx 3396\ \text{W} \approx 3.4\ \text{kW}\).

Correct option is C. 1 mark for correct calculation of input power.

From Young Modulus \(E = \frac{F L}{A e}\), extension is \(e = \frac{F L}{A E}\). Since \(E\) and \(F\) are the same for both, extension \(e \propto \frac{L}{A} \propto \frac{L}{d^2}\). Thus, \(\frac{e_X}{e_Y} = \left(\frac{L_X}{L_Y}\right) \times \left(\frac{d_Y}{d_X}\right)^2 = 2 \times 2^2 = 8\).

Correct option is D. 1 mark for identifying the relationship between extension, length, and diameter, leading to the ratio of 8.

In the third harmonic, there are 3 loops, which corresponds to \(1.5\) wavelengths along the string. Therefore, \(1.5 \lambda = 1.2\ \text{m}\), giving \(\lambda = 0.8\ \text{m}\). The distance between any node and its adjacent antinode is a quarter of a wavelength: \(\frac{\lambda}{4} = \frac{0.8}{4} = 0.20\ \text{m}\).

Correct option is B. 1 mark for calculating the wavelength as 0.8 m and dividing by 4 to find the correct distance.

For an NTC thermistor, a decrease in temperature causes its resistance to increase. Since the thermistor's resistance increases relative to the fixed resistor, it takes a larger share of the fixed supply voltage, causing \(V_{\text{out}}\) across it to increase.

Correct option is B. 1 mark for correctly identifying that resistance increases and consequently the output voltage across it increases.

According to the ideal gas law at constant volume, \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\), where temperature must be in Kelvin. \(T_1 = 27 + 273 = 300\ \text{K}\). Since the pressure doubles from \(1.5 \times 10^5\ \text{Pa}\) to \(3.0 \times 10^5\ \text{Pa}\), the absolute temperature must also double: \(T_2 = 2 \times 300\ \text{K} = 600\ \text{K}\). Converting back to Celsius gives \(T_2 = 600 - 273 = 327\ ^\circ\text{C}\).

Correct option is C. 1 mark for converting to Kelvin, doubling the absolute temperature, and converting back to Celsius correctly.

The momentum \(p\) of a photon is given by the de Broglie relation \(p = \frac{h}{\lambda}\), where \(h\) is the Planck constant (\(6.63 \times 10^{-34}\ \text{J s}\)) and \(\lambda\) is the wavelength (\(633 \times 10^{-9}\ \text{m}\)). Calculating gives \(p = \frac{6.63 \times 10^{-34}}{633 \times 10^{-9}} \approx 1.05 \times 10^{-27}\ \text{kg m s}^{-1}\).

Correct option is A. 1 mark for correct use of the de Broglie equation to calculate photon momentum.

Percentage uncertainty in \(R\): \(\frac{\Delta R}{R} \times 100\% = \frac{0.08}{4.00} \times 100\% = 2.0\%\).

Percentage uncertainty in \(d\): \(\frac{\Delta d}{d} \times 100\% = \frac{0.01}{0.50} \times 100\% = 2.0\%\).

Percentage uncertainty in \(L\): \(\frac{\Delta L}{L} \times 100\% = \frac{0.5}{100.0} \times 100\% = 0.5\%\).

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

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

Substituting the values:
Percentage uncertainty in \(\rho = 2(2.0\%) + 2.0\% + 0.5\% = 4.0\% + 2.0\% + 0.5\% = 6.5\%\).

1 mark for correctly calculating individual percentage uncertainties and summing them up using the correct power relation (coefficient of 2 for diameter).

Change in momentum of the ball is:
\(\Delta p = m \Delta v = m (v - u) = 0.15 \times (15 - (-20)) = 0.15 \times 35 = 5.25\text{ N s}\)

The impulse is equal to the area under the force-time graph. For a triangular force profile:
\(\text{Impulse} = \frac{1}{2} \times F_{\text{max}} \times \Delta t\)

Therefore:
\(5.25 = \frac{1}{2} \times F_{\text{max}} \times (12 \times 10^{-3})\)
\(F_{\text{max}} = \frac{5.25}{6.0 \times 10^{-3}} = 875\text{ N} \approx 880\text{ N}\) (to two significant figures).

1 mark for identifying the correct change in momentum (accounting for direction) and using the area of a triangle to find the maximum force of 880 N.

The elastic potential energy stored in a stretched wire is given by:
\(E_p = \frac{1}{2} F x\)

From the definition of Young modulus, the tension \(F\) is:
\(F = \frac{E A x}{L}\)

where \(A = \frac{\pi d^2}{4}\) is the cross-sectional area. This gives:
\(E_p = \frac{E A x^2}{2L} = \frac{\pi E d^2 x^2}{8L}\)

Since \(E\), \(L\), and \(x\) are constant for both wires, the elastic potential energy is directly proportional to the square of the diameter:
\(E_p \propto d^2\)

For the second wire with diameter \(2d\):
\(\frac{E_{p,2}}{E_{p,1}} = \left(\frac{2d}{d}\right)^2 = 4\).

1 mark for establishing the proportionality relationship between elastic energy and the square of diameter, and correctly evaluating the ratio as 4.

The Doppler-shifted frequency \(f_o\) observed by a stationary observer when a source is moving towards them is:
\(f_o = f_s \left(\frac{v}{v - v_s}\right)\)

Substituting the given values:
\(680 = 640 \left(\frac{340}{340 - v_s}\right)\)
\(\frac{680}{640} = \frac{340}{340 - v_s}\)
\(\frac{17}{16} = \frac{340}{340 - v_s}\)

Cross-multiplying:
\(17(340 - v_s) = 16 \times 340\)
\(5780 - 17v_s = 5440\)
\(17v_s = 340\)
\(v_s = 20\text{ m s}^{-1}\).

1 mark for using the correct Doppler effect equation for an approaching source and successfully calculating the speed as 20 m s⁻¹.

The resistance of a wire is given by:
\(R = \frac{\rho L}{A}\)

Since the volume \(V = A L\) is constant, we can substitute \(A = \frac{V}{L}\) into the equation:
\(R = \frac{\rho L^2}{V}\)

For a given material of constant volume, the resistance is proportional to the square of the length:
\(R \propto L^2\)

If the length increases by \(10\%\), the new length is \(L' = 1.10 L\).

The new resistance is:
\(R' \propto (1.10 L)^2 = 1.21 L^2\)
\(R' = 1.21 R\).

1 mark for identifying that resistance is proportional to the square of the length when volume is constant, and determining the factor of 1.21.

First step: \(\beta^-\)\ decay.
During \(\beta^-\)\ decay, a neutron changes into a proton, an electron, and an antineutrino. The proton number \(Z\) increases by 1, and the nucleon number \(A\) remains unchanged:
\(^{212}_{83}\text{Bi} \rightarrow ^{212}_{84}\text{Po} + \beta^- + \bar{
u}_e\)

Second step: \(\alpha\)\ decay.
During \(\alpha\)\ decay, a helium nucleus is emitted. The proton number decreases by 2, and the nucleon number decreases by 4:
\(^{212}_{84}\text{Po} \rightarrow ^{208}_{82}\text{Pb} + ^4_2\alpha\)

Therefore, the final nucleus has a proton number of 82 and a nucleon number of 208.

1 mark for identifying the correct changes to both proton number and nucleon number in both decay processes, selecting Row A.

Using Newton's law of gravitation \(F = \frac{G m_1 m_2}{r^2}\):

1. Gravitational pull on the middle sphere (\(2M\)) by the sphere of mass \(M\):
\(F_1 = \frac{G(M)(2M)}{d^2} = \frac{2 G M^2}{d^2}\) directed towards the left.

2. Gravitational pull on the middle sphere (\(2M\)) by the sphere of mass \(3M\):
\(F_2 = \frac{G(2M)(3M)}{(2d)^2} = \frac{6 G M^2}{4 d^2} = \frac{1.5 G M^2}{d^2}\) directed towards the right.

Since the two forces are in opposite directions, the magnitude of the net force is the difference between them:
\(F_{\text{net}} = F_1 - F_2 = \frac{2 G M^2}{d^2} - \frac{1.5 G M^2}{d^2} = \frac{0.5 G M^2}{d^2} = \frac{G M^2}{2 d^2}\).

1 mark for calculating both gravitational force components, squaring the separation distance correctly, and subtracting the values to yield \(\frac{G M^2}{2 d^2}\).

Initially, the capacitance is \(C\) and the potential difference is \(V\). The energy stored is:
\(E = \frac{1}{2} C V^2\)

When a dielectric of relative permittivity \(3.0\) is inserted while connected to the battery, the new capacitance is \(C' = 3.0 C\). Since \(V\) is constant, the new energy stored is:
\(E' = \frac{1}{2} C' V^2 = \frac{1}{2} (3.0 C) V^2 = 3.0 E\)

The charge on the plates changes from \(Q = C V\) to \(Q' = C' V = 3.0 C V\).

The charge supplied by the battery is:
\(\Delta Q = Q' - Q = 2.0 C V\)

The work done by the battery is:
\(W_{\text{battery}} = \Delta Q \cdot V = (2.0 C V) V = 2.0 C V^2\)

Since \(E = \frac{1}{2} C V^2 \implies C V^2 = 2 E\), we have:
\(W_{\text{battery}} = 2.0 (2 E) = 4 E\).

Therefore, the work done by the battery is \(4E\), and the final energy stored is \(3E\).

1 mark for establishing the correct work done by the battery as 4E and the final stored energy as 3E, corresponding to Row B.

To find the percentage uncertainty in the density \(\rho\), we use the fractional uncertainty relation. Since \(\rho = \frac{4m}{\pi d^2 h}\), the constant factor \(\frac{4}{\pi}\) has no uncertainty. The relation for the relative uncertainties is: \(\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta h}{h}\). Substituting the given percentage uncertainties: percentage uncertainty in \(\rho\) = 1.5% + 2(2.0%) + 3.0% = 1.5% + 4.0% + 3.0% = 8.5%.

1 mark for correctly adding the percentage uncertainty of mass, twice the percentage uncertainty of diameter, and the percentage uncertainty of length to get 8.5%.

The current in a conductor is given by \(I = nAve\). Since the second wire is also made of copper, the number density \(n\) is unchanged. The second wire has twice the diameter, so its cross-sectional area \(A_2\) is four times the original area (\(A_2 = 4A\)). Rearranging the equation for drift speed gives \(v_2 = \frac{I_2}{n A_2 e} = \frac{3I}{n(4A)e} = 0.75 \frac{I}{nAe} = 0.75v\).

1 mark for identifying that area scales with the square of the diameter and correctly calculating the new drift speed to be 0.75v.

We take moments about the pivot. The weight of the uniform beam acts at its midpoint, 1.5 m from the pivot. The clockwise moments are: (120 N * 1.5 m) + (100 N * 3.0 m) = 180 N m + 300 N m = 480 N m. The anticlockwise moment is provided by the vertical component of the tension: T * sin(30) * 2.0 m = T * 0.5 * 2.0 = 1.0 * T. For equilibrium, clockwise moments equal anticlockwise moments: 1.0 * T = 480 N m, so T = 480 N.

1 mark for calculating the total clockwise moment as 480 N m and equating it to the anticlockwise moment to find a tension of 480 N.

The Young modulus is given by \(E = \frac{FL}{Ax}\), which rearranges to \(x = \frac{FL}{AE}\). For the second wire, the length is \(2L\), the force is \(3F\), and the cross-sectional area is \(4A\) (since area is proportional to the square of the diameter). The material is the same, so the Young modulus is also \(E\). The new extension is \(x_2 = \frac{(3F)(2L)}{(4A)E} = 1.5 \frac{FL}{AE} = 1.5x\).

1 mark for correctly determining that area increases by a factor of 4 and setting up the ratio to find the extension is 1.5x.

The phase difference of 120 degrees is one-third of a full cycle (360 degrees), so the separation distance of 0.80 m represents one-third of a wavelength. Therefore, the wavelength \(\lambda\) is 3 * 0.80 m = 2.4 m. The speed of the wave is given by \(v = f \lambda = 250\text{ Hz} \times 2.4\text{ m} = 600\text{ m s}^{-1}\).

1 mark for calculating the wavelength as 2.4 m and using the wave equation to determine the speed of 600 m s⁻¹.

The observation that a tiny fraction of alpha-particles are deflected through angles greater than 90 degrees led Rutherford to conclude that the positive charge and most of the mass of the atom are concentrated in a extremely small volume at the center (the nucleus). Most particles passing straight through shows that the atom is mostly empty space.

1 mark for selecting the option that correctly pairs the large-angle deflection with the conclusion of a small, massive, charged nucleus.

From kinetic theory, the pressure of an ideal gas is given by \(p = \frac{1}{3}\rho \langle c^2 \rangle\), where \(\langle c^2 \rangle\) is the mean-square speed. Solving for the mean-square speed gives \(\langle c^2 \rangle = \frac{3p}{\rho}\). Taking the square root of both sides gives the root-mean-square speed: \(c_{\text{rms}} = \sqrt{\frac{3p}{\rho}}\).

1 mark for using the kinetic theory pressure equation and solving for the root-mean-square speed.

The photoelectric equation is \(E_{\text{max}} = hf - \Phi\). Comparing this with the straight-line equation \(y = mx + c\) where \(y = E_{\text{max}}\) and \(x = f\), the gradient is equal to the Planck constant \(h\). The intercept on the frequency axis occurs when \(E_{\text{max}} = 0\), which corresponds to the threshold frequency \(f_0\).

1 mark for identifying that the gradient corresponds to the Planck constant and the horizontal intercept corresponds to the threshold frequency.

For (a), substitute values into \(\rho = \frac{\pi R d^2}{4 L}\) where \(d = 0.38 \times 10^{-3}\text{ m}\): \(\rho = \frac{\pi \times 14.6 \times (0.38 \times 10^{-3})^2}{4 \times 1.250} = 1.32 \times 10^{-6}\ \Omega\text{ m}\), which to 2 significant figures is \(1.3 \times 10^{-6}\ \Omega\text{ m}\). For (b), percentage uncertainties are: \(\Delta L / L = (0.002 / 1.250) \times 100\% = 0.16\%\); \(\Delta d / d = (0.02 / 0.38) \times 100\% = 5.26\%\); \(\Delta R / R = (0.2 / 14.6) \times 100\% = 1.37\%\). Summing these up using the power factor: \(\Delta \rho / \rho = \Delta R / R + 2(\Delta d / d) + \Delta L / L = 1.37\% + 2(5.26\%) + 0.16\% = 12.05\%\) which rounds to \(12\%\). For (c), the diameter \(d\) contributes the most because its individual percentage uncertainty of \(5.3\%\) is doubled to \(10.5\%\) due to the term being squared.

Award 3 marks for (a): 1 mark for correct substitution of values with diameter converted to metres, 1 mark for calculating 1.32 x 10^-6, 1 mark for rounding to 2 s.f. and stating correct unit (Ohm m). Award 3 marks for (b): 1 mark for calculating percentage uncertainties in L, d, and R, 1 mark for doubling the percentage uncertainty in d, 1 mark for summing to get 12%. Award 1.5 marks for (c): 0.5 marks for stating diameter, 1 mark for explanation mentioning either the high percentage uncertainty of d or that its effect is doubled due to being squared.

For (a), use \(v^2 = u^2 + 2as\) with \(v = 0\), \(u = 12.0\text{ m s}^{-1}\), and \(a = -9.81\text{ m s}^{-2}\). This gives \(0 = 12.0^2 - 2(9.81)s\), so \(s = 7.34\text{ m}\). Total height above ground = \(18.0 + 7.34 = 25.3\text{ m}\). For (b), use \(v^2 = u^2 + 2as\) from roof to ground with \(s = -18.0\text{ m}\). This gives \(v^2 = 12.0^2 + 2(-9.81)(-18.0) = 144 + 353.16 = 497.16\), so \(v = 22.3\text{ m s}^{-1}\). For (c), use \(v = u + at\) where upward is positive. This gives \(-22.3 = 12.0 - 9.81t\), so \(t = (22.3 + 12.0) / 9.81 = 3.50\text{ s}\).

Award 2.5 marks for (a): 1 mark for using v^2 = u^2 + 2as to find distance above roof, 1 mark for finding 7.34 m, 0.5 marks for adding 18.0 m to get 25.3 m. Award 2.5 marks for (b): 1 mark for using v^2 = u^2 + 2as, 1 mark for correct substitution of values, 0.5 marks for final value of 22.3 m s^-1. Award 2.5 marks for (c): 1 mark for using v = u + at, 1 mark for correct signs on v and u, 0.5 marks for final time of 3.50 s (accept 3.5 s).

For (a), let right be positive. Total momentum is conserved: \(m_A u_A + m_B u_B = m_A v_A + m_B v_B\). This gives \(0.45(2.4) + 0.30(-1.5) = 0.45(-0.20) + 0.30 v_B\), so \(1.08 - 0.45 = -0.09 + 0.30 v_B\), yielding \(0.63 = -0.09 + 0.30 v_B\), which simplifies to \(v_B = 2.4\text{ m s}^{-1}\) to the right. For (b), initial KE = \(0.5(0.45)(2.4)^2 + 0.5(0.30)(1.5)^2 = 1.296 + 0.3375 = 1.634\text{ J}\). Final KE = \(0.5(0.45)(0.20)^2 + 0.5(0.30)(2.4)^2 = 0.009 + 0.864 = 0.873\text{ J}\). Since final KE is less than initial KE, the collision is inelastic. For (c), impulse on B = \(m_B(v_B - u_B) = 0.30(2.4 - (-1.5)) = 1.17\text{ N s}\).

Award 3 marks for (a): 1 mark for statement of conservation of momentum with signs accounted for, 1 mark for correct calculation, 1 mark for specifying direction or positive sign. Award 3 marks for (b): 1 mark for calculating initial kinetic energy, 1 mark for calculating final kinetic energy, 1 mark for identifying collision is inelastic because KE is lost. Award 1.5 marks for (c): 1 mark for formula or substitution for change in momentum, 0.5 marks for 1.17 N s.

For (a), the centre of gravity is defined as the single point where the entire weight of an object may be considered to act. For (b), taking moments about hinge P: clockwise moments = counter-clockwise moments. The weight acts at the midpoint (0.40 m), so \(45 \times 0.40 = T \sin(35^\circ) \times 0.80\). This gives \(18.0 = T \times 0.80 \times 0.5736\), which yields \(T = 39.2\text{ N} \approx 39\text{ N}\). For (c), horizontal forces must balance in equilibrium: \(F_H - T \cos(35^\circ) = 0\), so \(F_H = 39.2 \times \cos(35^\circ) = 32.1\text{ N}\). Since the cable pulls to the left, the force from the hinge on the shelf must be to the right (away from the wall) to balance it.

Award 1.5 marks for (a): 1.5 marks for complete definition of centre of gravity. Award 3 marks for (b): 1 mark for taking moments about P, 1 mark for placing weight at 0.40 m, 1 mark for algebraic steps showing T is approximately 39 N. Award 3 marks for (c): 1 mark for identifying horizontal balance equation, 1 mark for calculating 32 N (or 32.1 N), 1 mark for specifying direction as to the right (away from wall).

For (a), the Young modulus is defined as the ratio of tensile stress to tensile strain. For (b)(i), tensile stress \(\sigma = F / A = (6.0 \times 9.81) / (1.5 \times 10^{-7}) = 3.924 \times 10^8\text{ Pa} \approx 3.9 \times 10^8\text{ Pa}\). For (b)(ii), \(E = \text{stress} / \text{strain} = \sigma / (\Delta L / L_0)\), so \(\Delta L = (\sigma \times L_0) / E = (3.924 \times 10^8 \times 2.2) / (2.0 \times 10^{11}) = 4.316 \times 10^{-3}\text{ m} \approx 4.3\text{ mm}\). For (c), elastic potential energy \(E_p = 0.5 \times F \times \Delta L = 0.5 \times (6.0 \times 9.81) \times 4.316 \times 10^{-3} = 0.127\text{ J} \approx 0.13\text{ J}\).

Award 1.5 marks for (a): 1.5 marks for definition (ratio of tensile stress to tensile strain). Award 2 marks for (b)(i): 1 mark for using stress = F/A, 1 mark for calculating 3.9 x 10^8 Pa. Award 2 marks for (b)(ii): 1 mark for using strain = stress/E or extension = FL/AE, 1 mark for calculating 4.3 mm (or 4.3 x 10^-3 m). Award 2 marks for (c): 1 mark for using Ep = 1/2 F x, 1 mark for calculating 0.13 J.

For (a), as the source moves towards the observer, successive wavefronts are emitted closer together in space, shortening the observed wavelength. Since wave speed is constant, the shorter wavelength results in a higher observed frequency. For (b), the observed frequency is \(f_o = f_s \left( \frac{v}{v - v_s} \right) = 480 \times \frac{340}{340 - 25} = 480 \times \frac{340}{315} = 518.1\text{ Hz}\) (or \(518\text{ Hz}\)). For (c), as the vehicle recedes, \(f_o' = f_s \left( \frac{v}{v + v_s} \right) = 480 \times \frac{340}{340 + 25} = 480 \times \frac{340}{365} = 447.1\text{ Hz}\). The change in frequency is \(518.1 - 447.1 = 71\text{ Hz}\) (or \(71.0\text{ Hz}\)).

Award 2.5 marks for (a): 1 mark for stating that the wavefronts are compressed/closer together, 1 mark for stating the observed wavelength is smaller, 0.5 marks for relating to higher frequency. Award 2.5 marks for (b): 1.5 marks for correct formula with negative sign in denominator, 1 mark for 518 Hz. Award 2.5 marks for (c): 1 mark for calculating receding frequency of 447 Hz, 1.5 marks for subtracting to get 71 Hz.

For (a), resistance is \(R = \frac{\rho L}{A}\) where \(A = \frac{\pi d^2}{4}\), so \(R \propto \frac{L}{d^2}\). The ratio \(\frac{R_Y}{R_X} = \frac{L_Y}{L_X} \times \left(\frac{d_X}{d_Y}\right)^2 = 2.5 \times \left(\frac{1}{1.5}\right)^2 = 2.5 \times \frac{1}{2.25} = \frac{10}{9} \approx 1.11\). For (b), \(R_Y = R_X \times 1.111 = 3.6 \times \frac{10}{9} = 4.0\ \Omega\). For (c), since the wires are in series, they carry the same current \(I\). Using \(I = n A v e\), drift speed \(v = \frac{I}{n A e}\). Since the material is the same, \(n\) is constant, meaning \(v \propto \frac{1}{A} \propto \frac{1}{d^2}\). The ratio of drift speed in X to Y is \(\frac{v_X}{v_Y} = \left(\frac{d_Y}{d_X}\right)^2 = (1.5)^2 = 2.25\).

Award 3 marks for (a): 1 mark for expressing R in terms of L and d^2, 1 mark for substituting length and diameter ratios, 1 mark for deriving 10/9 or 1.11. Award 2 marks for (b): 1 mark for multiplying Rx by the ratio, 1 mark for 4.0 Ohm. Award 2.5 marks for (c): 1 mark for stating current is the same in series, 1 mark for using I = nAve to show drift speed is inversely proportional to cross-sectional area (or diameter squared), 0.5 marks for ratio of 2.25.

For (a), the complete equation is \(^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + _{-1}^{0}\text{e} + \bar{
u}_e\) (or \(^{14}_{6}\text{C} \rightarrow ^{14}_{7}\text{N} + \beta^- + \bar{
u}_e\)). For (b), the fundamental interaction responsible is the weak interaction (or weak nuclear force), and the exchange boson is the \(W^-\)\-boson. For (c), a neutron in the Carbon-14 nucleus decays into a proton. The quark composition of a neutron is \(udd\) and that of a proton is \(uud\). Therefore, one down quark (d) changes into an up quark (u).

Award 2 marks for (a): 1 mark for correctly balanced nucleons/protons including beta particle, 1 mark for correct representation of the electron antineutrino (\(\bar{
u}_e\)). Award 2.5 marks for (b): 1.5 marks for stating weak interaction/force, 1 mark for identifying W^- boson. Award 3 marks for (c): 1 mark for stating neutron decays to proton, 1 mark for stating neutron is udd and proton is uud, 1 mark for stating a down quark changes to an up quark.

### Step-by-Step Practical Procedure and Calculation

#### 1. Sample Experimental Data Table

| \( d / \text{cm} \) | \( d^2 / \text{cm}^2 \) | \( t_1 / \text{s} \) | \( t_2 / \text{s} \) | \( t_{\text{mean}} / \text{s} \) | \( T / \text{s} \) | \( T^2 / \text{s}^2 \) |
|---|---|---|---|---|---|---|
| 10.0 | 100 | 6.01 | 5.98 | 6.00 | 0.600 | 0.360 |
| 15.0 | 225 | 6.72 | 6.68 | 6.70 | 0.670 | 0.449 |
| 20.0 | 400 | 7.62 | 7.58 | 7.60 | 0.760 | 0.578 |
| 25.0 | 625 | 8.56 | 8.60 | 8.58 | 0.858 | 0.736 |
| 30.0 | 900 | 9.68 | 9.64 | 9.66 | 0.966 | 0.933 |
| 35.0 | 1225 | 10.82 | 10.78 | 10.80 | 1.080 | 1.166 |

#### 2. Graph Plotting
* Plot \( T^2 \) (y-axis) against \( d^2 \) (x-axis).
* Scale:
* x-axis: \( 0 \text{ to } 1400\text{ cm}^2 \) (with \( 1\text{ cm} = 100\text{ cm}^2 \))
* y-axis: \( 0.0 \text{ to } 1.3\text{ s}^2 \) (with \( 1\text{ cm} = 0.1\text{ s}^2 \))
* Draw the line of best fit through the points.

#### 3. Gradient and Intercept Calculation
Using coordinates from the line of best fit:
* Point 1: \( (100, 0.360) \)
* Point 2: \( (1225, 1.166) \)

\[ \text{Gradient } a = \frac{1.166 - 0.360}{1225 - 100} = \frac{0.806}{1125} \approx 7.16 \times 10^{-4}\text{ s}^2\text{ cm}^{-2} \]

Using \( y = mx + c \) to find the y-intercept \( b \):

\[ b = 1.166 - (7.16 \times 10^{-4} \times 1225) = 1.166 - 0.877 = 0.289\text{ s}^2 \]

#### 4. Final Constants
* \( a = 7.2 \times 10^{-4}\text{ s}^2\text{ cm}^{-2} \)
* \( b = 0.29\text{ s}^2 \)

### Mark Breakdown (Total: 20 Marks)

#### 1. Data Collection & Tabulation (6 Marks)
* **[1] Column Headings:** All column headings in the table must include appropriate units in the standard format (e.g., \( d / \text{cm} \), \( d^2 / \text{cm}^2 \), \( t / \text{s} \), \( T / \text{s} \), \( T^2 / \text{s}^2 \)).
* **[1] Precision of Raw Readings:** Raw values of \( d \) recorded to the nearest millimetre (\( 0.1\text{ cm} \)) and times \( t \) recorded to \( 0.01\text{ s} \).
* **[1] Range and Number of Readings:** At least 6 sets of readings taken with a range of \( d \) of at least \( 20\text{ cm} \).
* **[1] Repeated Readings:** Evidence of repeated readings for time \( t \) for each position.
* **[1] Calculation of Derived Quantities:** Calculated values of \( T \) and \( T^2 \) are mathematically correct.
* **[1] Significant Figures:** The number of significant figures for \( T^2 \) must be the same as, or one more than, the number of significant figures in the raw time \( t \).

#### 2. Graph and Line of Best Fit (4 Marks)
* **[1] Graph Axes:** Linear scales chosen so that the plotted points occupy at least 50% of the grid in both directions. Axes must be clearly labelled with quantities and units.
* **[1] Plotting Accuracy:** All observations must be plotted to an accuracy of within half a small square.
* **[1] Line of Best Fit:** Straight line drawn with a balanced distribution of points on either side along its entire length.
* **[1] Quality of Data:** Scatter of points about the line of best fit must be very small, indicating good control of experimental conditions.

#### 3. Gradient and Intercept (2 Marks)
* **[1] Gradient Calculation:** Uses a large triangle where the hypotenuse is at least half the length of the drawn line. Coordinates used must be read to within half a small square.
* **[1] Intercept Calculation:** Calculated correctly using a point on the line and \( y = mx + c \), or read directly from the y-axis if the x-axis starts at 0.

#### 4. Constants and Units (3 Marks)
* **[1] Value of \( a \):** Value is equal to the gradient of the graph.
* **[1] Value of \( b \):** Value is equal to the y-intercept of the graph.
* **[1] Units:** Correct units provided for \( a \) (e.g., \( \text{s}^2\text{ cm}^{-2} \) or \( \text{s}^2\text{ m}^{-2} \)) and \( b \) (\( \text{s}^2 \)).

#### 5. Experimental Quality & Design (5 Marks)
* **[1] Correct suspension:** Rule suspended horizontally and parallel to the bench.
* **[1] Timing method:** Measured at least 10 complete oscillations to reduce timing error.
* **[1] Consistency:** Raw times recorded to 0.01 s consistently.
* **[1] Motion control:** Oscillations are purely rotational in a horizontal plane without swinging sideways.
* **[1] Plausible intercept:** Value of \( b \) lies in the realistic physical range of \( 0.20\text{ s}^2 \le b \le 0.40\text{ s}^2 \).

### Step-by-Step Practical Procedure and Calculation

#### 1. First Reading:
* Set \( h_1 = 1.0\text{ cm} = 10\text{ mm} \).
* Tilt board slowly until the cup topples over. Let the recorded angles be:
\( \theta_{1,a} = 28^\circ \), \( \theta_{1,b} = 30^\circ \)
\( \theta_{\text{mean}} = 29^\circ \).

#### 2. Percentage Uncertainty in \( \theta_1 \):
* The smallest division of the protractor is \( 1^\circ \). However, due to manual tilting and difficulty in identifying the exact onset of toppling, a realistic absolute uncertainty is \( \Delta \theta = 2^\circ \).

\[ \text{Percentage Uncertainty} = \frac{\Delta \theta}{\theta_1} \times 100\% = \frac{2}{29} \times 100\% \approx 6.9\% \]

#### 3. Second Reading:
* Set \( h_2 = 5.0\text{ cm} = 50\text{ mm} \).
* Tilt board slowly. Let the recorded angles be:
\( \theta_{2,a} = 17^\circ \), \( \theta_{2,b} = 17^\circ \)
\( \theta_{\text{mean}} = 17^\circ \).

#### 4. Calculation of Constant \( k \):
* For first reading:

\[ k_1 = (1.0 + 4.0) \tan 29^\circ = 5.0 \times 0.5543 = 2.77\text{ cm} \]

* For second reading:

\[ k_2 = (5.0 + 4.0) \tan 17^\circ = 9.0 \times 0.3057 = 2.75\text{ cm} \]

#### 5. Comparison and Analysis:
* Calculate the percentage difference between the two values of \( k \):

\[ \% \text{ Difference} = \frac{|k_1 - k_2|}{k_{\text{average}}} \times 100\% = \frac{|2.77 - 2.75|}{2.76} \times 100\% = 0.72\% \]

* Since \( 0.72\% \) is less than the specified limit of \( 10\% \), the experimental results strongly support the suggested relationship.

### Mark Breakdown (Total: 20 Marks)

#### 1. Measurements and Quality of Data (5 Marks)
* **[1] Value of \( h_1 \):** Recorded with unit and to the nearest millimetre (e.g., \( 1.0\text{ cm} \)).
* **[1] Value of \( \theta_1 \):** Recorded to the nearest degree with unit.
* **[1] Repeated Readings:** Multiple trials taken for \( \theta_1 \) and \( \theta_2 \) and averaged.
* **[1] Second Set:** \( h_2 \) and \( \theta_2 \) recorded, showing the correct trend (\( \theta_2 < \theta_1 \)).
* **[1] Accuracy of measurements:** \( \theta \) readings lie in a physically plausible range (e.g., \( 10^\circ \text{ to } 45^\circ \)).

#### 2. Uncertainty Calculations (2 Marks)
* **[1] Absolute Uncertainty Choice:** A realistic value of \( \Delta \theta \) (usually \( 2^\circ \text{ to } 5^\circ \)) is chosen and justified by mentioning the fast nature of toppling.
* **[1] Percentage Uncertainty Calculation:** Calculated correctly using the formula \( \frac{\Delta \theta}{\theta_{\text{mean}}} \times 100\% \).

#### 3. Calculations & Analysis of Relationship (5 Marks)
* **[1] Consistent Significant Figures:** Values of \( k_1 \) and \( k_2 \) are expressed to the same number of significant figures as, or one more than, the least precise raw measurement (usually 2 or 3 SF).
* **[1] Calculation of \( k_1 \):** Calculated correctly with correct arithmetic steps.
* **[1] Calculation of \( k_2 \):** Calculated correctly with correct arithmetic steps.
* **[1] Unit of \( k \):** Unit stated correctly (e.g., \( \text{cm} \) or \( \text{mm} \) matching the unit used for \( h \)).
* **[1] Conclusion:** Percentage difference calculated correctly, and a logical conclusion stated comparing it to the 10% limit.

#### 4. Limitations and Improvements (8 Marks)
*Award 1 mark for each identified limitation and 1 mark for its corresponding improvement, up to a maximum of 4 pairs (8 marks):*

1. **Limitation [1]:** Difficult to judge the precise angle at the exact moment of toppling because it happens very quickly.
* **Improvement [1]:** Record a video of the tilt alongside the protractor, then perform frame-by-frame analysis.
2. **Limitation [1]:** The cup slides down the wooden board before it topples, which changes the forces acting on it.
* **Improvement [1]:** Fix a high-friction sheet (like fine sandpaper) onto the wooden board to prevent sliding.
3. **Limitation [1]:** Difficult to measure the height \( h \) of the mass inside the narrow cup accurately with a simple rule.
* **Improvement [1]:** Use a digital depth caliper or vernier gauge to measure the depth from the rim and calculate \( h \).
4. **Limitation [1]:** The mass shifts or wobbles slightly inside the cup when tilted.
* **Improvement [1]:** Glue the mass firmly using a rigid double-sided tape or clay spacer, or design a custom plastic peg inside the cup.

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

(b) The centripetal force is provided by the gravitational force:
\(\frac{m v^2}{R} = \frac{G M m}{R^2}\)
Since \(v = \frac{2\pi R}{T}\), we can substitute to get:
\(\frac{4 \pi^2 R}{T^2} = \frac{G M}{R^2}\)
\(R^3 = \frac{G M T^2}{4 \pi^2}\)
Given:
\(T = 7.2 \times 3600\text{ s} = 25920\text{ s}\)
\(M = 6.4 \times 10^{23}\text{ kg}\)
\(G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}\)

\(R^3 = \frac{(6.67 \times 10^{-11}) \times (6.4 \times 10^{23}) \times (25920)^2}{4 \pi^2}\)
\(R^3 = 7.265 \times 10^{20}\text{ m}^3\)
\(R = 8.99 \times 10^6\text{ m} \approx 9.0 \times 10^6\text{ m}\) (to 2 s.f.)

(c) The gravitational potential energy is given by:
\(E_p = -\frac{G M m}{r}\)
Initial GPE: \(E_{p1} = -\frac{G M m}{R}\)
Final GPE: \(E_{p2} = -\frac{G M m}{2R}\)

Change in GPE:
\(\Delta E_p = E_{p2} - E_{p1} = -\frac{G M m}{2R} - \left(-\frac{G M m}{R}\right) = \frac{G M m}{2R}\)
\(\Delta E_p = \frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23} \times 850}{2 \times 8.99 \times 10^6}\)
\(\Delta E_p = 2.02 \times 10^9\text{ J} \approx 2.0 \times 10^9\text{ J}\)

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

(b)
- Equating centripetal force to gravitational force: [1 mark]
- Conversion of time to seconds (25920 s): [1 mark]
- Rearrangement to express \(R^3\) or \(R\) correctly: [1 mark]
- Correct final answer for \(R\) (8.99 to 9.0 x 10^6 m) with unit: [1 mark]

(c)
- Formula for change in potential energy \(\Delta E_p = \frac{G M m}{2R}\) or step-by-step subtraction: [2 marks]
- Substitution of values: [1 mark]
- Correct final answer (2.0 x 10^9 J) with unit: [1 mark]

(a) Any three of the following assumptions:
1. The volume of the molecules is negligible compared to the volume of the container.
2. There are no intermolecular forces of attraction or repulsion between molecules (except during collisions).
3. Collisions between molecules and with the walls of the container are perfectly elastic.
4. The duration of collisions is negligible compared to the time between collisions.
5. The molecules are in continuous, rapid, random motion.

(b)
(i) Work done by the gas:
\(W = p \Delta V\)
\(W = 2.4 \times 10^5 \times (3.5 \times 10^{-3} - 1.5 \times 10^{-3})\)
\(W = 2.4 \times 10^5 \times 2.0 \times 10^{-3} = 480\text{ J}\)

(ii) First law of thermodynamics: \(\Delta U = q + w\), where \(w\) is the work done ON the gas.
Since the gas expands, work done on the gas is negative: \(w = -480\text{ J}\).
Thermal energy supplied to the gas: \(q = +850\text{ J}\).
\(\Delta U = 850 - 480 = +370\text{ J}\)

(iii) For an ideal gas, the internal energy is entirely kinetic energy of the gas molecules. The mean kinetic energy of the molecules is directly proportional to the thermodynamic temperature of the gas (\(U \propto T\)). Since the internal energy of the gas increases (\(\Delta U > 0\)), the temperature of the gas must increase.

(a)
- Three correct assumptions listed: [3 marks, 1 mark per assumption]

(b)
(i)
- Formula \(W = p \Delta V\) used: [1 mark]
- Correct calculation to yield 480 J: [1 mark]

(ii)
- Recognition of work done ON the gas as \(-480\text{ J}\) or correct first law formulation: [1 mark]
- Correct calculation of change in internal energy (370 J): [1 mark]

(iii)
- Temperature increases: [1 mark]
- Reason 1: Internal energy of an ideal gas is solely kinetic energy: [1 mark]
- Reason 2: Internal energy is directly proportional to temperature, so positive change in U means positive change in T: [1 mark]

(a) The acceleration of the object is directly proportional to its displacement from its equilibrium position, and is always directed towards the equilibrium position (in the opposite direction to displacement).

(b)
(i) Maximum acceleration is given by:
\(a_{max} = \omega^2 x_0\)
Where \(\omega = 2 \pi f\) and \(x_0 = 4.0\text{ cm} = 0.040\text{ m}\).
\(\omega = 2 \pi \times 3.5 = 7.0 \pi \approx 21.99\text{ rad s}^{-1}\)
\(a_{max} = (21.99)^2 \times 0.040\)
\(a_{max} = 483.6 \times 0.040 = 19.3\text{ m s}^{-2} \approx 19\text{ m s}^{-2}\)

(ii) Total energy of oscillation is the maximum kinetic energy:
\(E = \frac{1}{2} m \omega^2 x_0^2\)
\(E = \frac{1}{2} \times 0.45 \times (21.99)^2 \times (0.040)^2\)
\(E = 0.225 \times 483.6 \times 0.0016 = 0.174\text{ J}\)
To 2 s.f., \(E = 0.17\text{ J}\).

(iii) The kinetic energy is maximum at the equilibrium position (where displacement \(x = 0\)), and is zero at the maximum displacement (where \(x = \pm 4.0\text{ cm}\)). The variation forms a parabolic curve pointing downwards (represented by \(E_k = E_k(max)\left(1 - \frac{x^2}{x_0^2}\right)\)).

(a)
- Acceleration is proportional to displacement: [1 mark]
- Acceleration is opposite to displacement / directed towards equilibrium: [1 mark]

(b)
(i)
- Use of \(\omega = 2 \pi f\) to find angular frequency (22 rad/s): [1 mark]
- Use of \(a_{max} = \omega^2 x_0\) with amplitude in metres: [1 mark]
- Correct calculation showing value near 19.3 m s\(^{-2}\): [1 mark]

(ii)
- Recall of \(E = \frac{1}{2} m \omega^2 x_0^2\) or \(E = \frac{1}{2} m v_{max}^2\): [1 mark]
- Insertion of correct values: [1 mark]
- Correct calculation yielding 0.17 J (or 0.174 J): [1 mark]

(iii)
- Maximum at zero displacement (equilibrium) and zero at maximum displacement: [1 mark]
- Shape of graph described as parabolic / curved: [1 mark]

(a) Capacitance is defined as charge stored per unit potential difference across the plates of the capacitor (\(C = Q/V\)).

(b)
(i) For capacitors in series:
\(\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}\)
\(\frac{1}{C_{total}} = \frac{1}{2.0\text{ }\mu\text{F}} + \frac{1}{3.0\text{ }\mu\text{F}} + \frac{1}{6.0\text{ }\mu\text{F}}\)
\(\frac{1}{C_{total}} = \frac{3 + 2 + 1}{6.0} = \frac{6.0}{6.0} = 1.0\text{ }\mu\text{F}^{-1}\)
\(C_{total} = 1.0\text{ }\mu\text{F}\)

(ii) In a series connection, the charge on each individual capacitor is identical and equal to the total charge supplied by the source.
Total charge \(Q = C_{total} \times V\)
\(Q = 1.0\text{ }\mu\text{F} \times 12\text{ V} = 12\text{ }\mu\text{C} = 1.2 \times 10^{-5}\text{ C}\)
So the charge on the \(3.0\text{ }\mu\text{F}\) capacitor is \(12\text{ }\mu\text{C}\).

(iii) Total energy stored in the combination:
\(E = \frac{1}{2} C_{total} V^2\)
\(E = \frac{1}{2} \times (1.0 \times 10^{-6}\text{ F}) \times (12\text{ V})^2\)
\(E = 0.5 \times 1.0 \times 10^{-6} \times 144 = 7.2 \times 10^{-5}\text{ J}\)

(a)
- Charge / potential difference: [1 mark]
- Definitions of terms (charge stored on one plate, p.d. across the plates): [1 mark]

(b)
(i)
- Formula for series capacitors used correctly: [1 mark]
- Correct final answer with unit (1.0 microfarad): [1 mark]

(ii)
- Recognition that the charge on series capacitors is the same: [1 mark]
- Correct calculation using \(Q = C_{total} V\): [1 mark]
- Final value and unit (12 microcoulombs): [1 mark]

(iii)
- Use of energy formula \(E = \frac{1}{2} C V^2\) or equivalent: [1 mark]
- Correct values substituted with correct powers of ten: [1 mark]
- Final value (7.2 x 10^-5 J): [1 mark]

(a)
(i) For the electron to pass undeflected, the magnetic force must be equal in magnitude and opposite in direction to the electric force:
\(F_E = F_B\)
\(e E = e v B\)
Dividing both sides by \(e\) gives:
\(E = v B \implies v = \frac{E}{B}\).

(ii) If the speed of the electron is greater than \(v\), the magnetic force \(F_B = e v B\) increases, while the electric force \(F_E = e E\) remains constant. Consequently, the net force is in the direction of the magnetic force, and the electron will be deflected in that direction, tracing a curved path.

(b)
(i) The magnetic force provides the centripetal force:
\(F_B = F_c\)
\(q v B = \frac{m v^2}{r}\)
\(r = \frac{m v}{q B}\)
\(r = \frac{1.67 \times 10^{-27}\text{ kg} \times 4.5 \times 10^5\text{ m s}^{-1}}{1.60 \times 10^{-19}\text{ C} \times 0.15\text{ T}}\)
\(r = \frac{7.515 \times 10^{-22}}{2.40 \times 10^{-20}}\)
\(r = 0.0313\text{ m} \approx 0.031\text{ m}\) (or \(3.1\text{ cm}\)).

(ii) The magnetic force is always perpendicular to the velocity of the protons. Since the force is at right angles to the direction of motion, the work done by this force on the protons is zero. Therefore, there is no change in kinetic energy, and the speed remains constant.

(a)
(i)
- Equating electric force \(eE\) to magnetic force \(evB\): [1 mark]
- Correct rearrangement to show \(v = E/B\): [1 mark]

(ii)
- Magnetic force increases with speed (while electric force remains constant): [1 mark]
- Net force is in the direction of the magnetic force: [1 mark]
- Electron deflections / follows curved path: [1 mark]

(b)
(i)
- Formula \(r = \frac{mv}{qB}\) or equivalent statement: [1 mark]
- Substitution of proton mass, charge, and given values: [1 mark]
- Correct calculation yielding 0.031 m (or 3.1 cm): [1 mark]

(ii)
- Force is perpendicular to motion/velocity: [1 mark]
- No work done (or no change in KE) so speed is constant: [1 mark]

(a) The r.m.s. value of an alternating current is the value of direct current that would produce thermal energy at the same average rate in a given resistor.

(b)
(i) Comparing \(V = 280 \sin(120 \pi t)\) with the standard form \(V = V_0 \sin(\omega t)\):
\(\omega = 120 \pi\)
Since \(\omega = 2 \pi f\):
\(2 \pi f = 120 \pi \implies f = 60\text{ Hz}\).

(ii) Peak voltage \(V_0 = 280\text{ V}\).
\(V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{280}{\sqrt{2}} \approx 198\text{ V}\) (or \(197.99\text{ V}\)).

(iii) Mean power \(P_{mean} = \frac{V_{rms}^2}{R}\)
\(P_{mean} = \frac{(197.99)^2}{50} = \frac{39200}{50} = 784\text{ W}\).
(Alternative: \(P_{mean} = \frac{V_0^2}{2R} = \frac{280^2}{100} = 784\text{ W}\)).

(iv) Maximum power occurs when the voltage is at its peak \(V_0\):
\(P_{max} = \frac{V_0^2}{R}\)
\(P_{max} = \frac{280^2}{50} = \frac{78400}{50} = 1568\text{ W} \approx 1570\text{ W}\) (or exactly \(2 \times P_{mean} = 1570\text{ W}\) to 3 s.f.).

(a)
- Value of direct current: [1 mark]
- That produces thermal energy at the same average rate in a given resistor: [1 mark]

(b)
(i)
- State \(\omega = 120\pi\): [1 mark]
- Correct calculation showing \(f = 60\text{ Hz}\): [1 mark]

(ii)
- State \(V_0 = 280\text{ V}\) and use \(V_{rms} = V_0 / \sqrt{2}\): [1 mark]
- Correct calculation yielding 198 V (or 200 V): [1 mark]

(iii)
- Recall of \(P = V_{rms}^2 / R\) or equivalent: [1 mark]
- Correct calculation yielding 784 W: [1 mark]

(iv)
- Recall that maximum power is \(V_0^2 / R\) or \(2 \times P_{mean}\): [1 mark]
- Correct calculation yielding 1570 W (or 1568 W): [1 mark]

(a) Threshold frequency is the minimum frequency of incident electromagnetic radiation required to cause the emission of photoelectrons from a metal surface.

(b)
(i) Energy of a single photon:
\(E = \frac{h c}{\lambda}\)
\(E = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{320 \times 10^{-9}\text{ m}}\)
\(E = \frac{1.989 \times 10^{-25}}{3.20 \times 10^{-7}} = 6.216 \times 10^{-19}\text{ J} \approx 6.2 \times 10^{-19}\text{ J}\).

(ii) Einstein's photoelectric equation:
\(h f = \Phi + E_{k,max}\)
Where the work function \(\Phi = 2.3\text{ eV}\).
Convert \(\Phi\) to joules:
\(\Phi = 2.3 \times 1.60 \times 10^{-19}\text{ J} = 3.68 \times 10^{-19}\text{ J}\).
\(E_{k,max} = h f - \Phi\)
\(E_{k,max} = 6.216 \times 10^{-19}\text{ J} - 3.68 \times 10^{-19}\text{ J}\)
\(E_{k,max} = 2.536 \times 10^{-19}\text{ J} \approx 2.5 \times 10^{-19}\text{ J}\).

(iii) There is no effect on the maximum kinetic energy of the photoelectrons. Doubling the intensity only increases the rate of incident photons, and thus increases the rate of photoelectron emission, but the energy of each individual photon remains unchanged.

(a)
- Minimum frequency: [1 mark]
- Required to cause emission of photoelectrons: [1 mark]

(b)
(i)
- Formula \(E = h c / \lambda\) stated: [1 mark]
- Correct substitution of constant values: [1 mark]
- Correct answer (6.2 x 10^-19 J): [1 mark]

(ii)
- Conversion of eV to Joules: [1 mark]
- Recall and rearrangement of photoelectric equation: [1 mark]
- Calculation yielding 2.5 x 10^-19 J (must show at least 2 decimal places in working, i.e., 2.54 x 10^-19 J): [1 mark]

(iii)
- No change: [1 mark]
- Explanation (intensity affects rate of photons, not individual photon energy): [1 mark]

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

(b)
(i) Half-life \(T_{1/2} = 8.0\text{ days} = 8.0 \times 24 \times 3600\text{ s} = 6.912 \times 10^5\text{ s}\).
Decay constant:
\(\lambda = \frac{\ln 2}{T_{1/2}}\)
\(\lambda = \frac{0.69315}{6.912 \times 10^5\text{ s}}\)
\(\lambda = 1.003 \times 10^{-6}\text{ s}^{-1} \approx 1.0 \times 10^{-6}\text{ s}^{-1}\).

(ii) Activity \(A = \lambda N\), so the initial number of nuclei \(N_0\) is:
\(N_0 = \frac{A_0}{\lambda}\)
\(N_0 = \frac{3.2 \times 10^7}{1.003 \times 10^{-6}} = 3.19 \times 10^{13} \approx 3.2 \times 10^{13}\).

(iii) Activity after \(t = 20\text{ days}\):
\(A = A_0 e^{-\lambda t}\) or \(A = A_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}\)
Using the half-life ratio:
\(t / T_{1/2} = 20 / 8.0 = 2.5\)
\(A = 3.2 \times 10^7 \times (0.5)^{2.5}\)
\(A = 3.2 \times 10^7 \times 0.17678 = 5.66 \times 10^6\text{ Bq} \approx 5.7 \times 10^6\text{ Bq}\).

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

(b)
(i)
- Conversion of half-life to seconds (6.912 x 10^5 s): [1 mark]
- Recall \(\lambda = \ln 2 / T_{1/2}\): [1 mark]
- Correct calculation yielding 1.0 x 10^-6 s^-1: [1 mark]

(ii)
- Recall \(A_0 = \lambda N_0\): [1 mark]
- Correct calculation yielding 3.2 x 10^13 (or 3.19 x 10^13): [1 mark]

(iii)
- Use of \(A = A_0 e^{-\lambda t}\) or \(A = A_0 (0.5)^{t/T_{1/2}}\): [1 mark]
- Calculation of number of elapsed half-lives (2.5): [1 mark]
- Correct final answer with unit (5.7 x 10^6 Bq): [1 mark]

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

(b) (i) Gravitational potential is defined as zero at infinity. Because the gravitational force is always attractive, work is done by the gravitational field as a mass is brought from infinity towards the planet or moon. Therefore, energy is lost, making the potential at any finite distance negative.
(ii) At the point where the gravitational field strength is zero, the gravitational forces due to the planet and the moon are equal in magnitude and opposite in direction. Since the gravitational field strength \(g = -\frac{d\phi}{dx}\), when \(g = 0\), the gradient of the potential is zero, which corresponds to a local maximum (least negative potential value) along the line of centres.

(c) (i) At the point of zero field strength:
\(\frac{GM_1}{x^2} = \frac{GM_2}{(d-x)^2}\)
Taking the square root of both sides:
\(\frac{\sqrt{M_1}}{x} = \frac{\sqrt{M_2}}{d-x}\)
\(\frac{d-x}{x} = \sqrt{\frac{M_2}{M_1}}\)
\(\frac{3.6 \times 10^8 - x}{x} = \sqrt{\frac{1.0 \times 10^{23}}{6.4 \times 10^{24}}} = \sqrt{\frac{1}{64}} = 0.125\)
\(3.6 \times 10^8 - x = 0.125x\)
\(1.125x = 3.6 \times 10^8\)
\(x = 3.2 \times 10^8\_\text{m}\).

(ii) The total gravitational potential is the sum of the potentials due to the planet and the moon:
\(\phi = -\frac{GM_1}{x} - \frac{GM_2}{d-x}\)
\(d - x = 3.6 \times 10^8 - 3.2 \times 10^8 = 4.0 \times 10^7\_\text{m}\)
\(\phi = -6.67 \times 10^{-11} \left( \frac{6.4 \times 10^{24}}{3.2 \times 10^8} + \frac{1.0 \times 10^{23}}{4.0 \times 10^7} \right)\)
\(\phi = -6.67 \times 10^{-11} \left( 2.0 \times 10^{16} + 2.5 \times 10^{15} \right) = -6.67 \times 10^{-11} \left( 2.25 \times 10^{16} \right)\)
\(\phi = -1.50 \times 10^6\_\text{J\_kg}^{-1}\).

(iii) The change in gravitational potential energy \(\Delta E_p\) is:
\(\Delta E_p = m \Delta \phi = 1200 \times \left[ -2.70 \times 10^6 - (-1.50 \times 10^6) \right]\)
\(\Delta E_p = 1200 \times (-1.20 \times 10^6) = -1.44 \times 10^9\_\text{J}\) (or a decrease of \(1.44 \times 10^9\_\text{J}\)).

(a)
- M1: Work done per unit mass [1]
- A1: in bringing a test mass from infinity to the point [1]

(b)
- M1: Gravitational potential is zero at infinity [1]
- A1: Gravitational force is attractive, so work is done by the field (potential energy decreases below zero) as a mass approaches [1]
- B1: Realises that at zero field strength, \(\frac{d\phi}{dx} = 0\), representing a maximum potential [1]

(c)
- M1: Equates gravitational field strengths: \(\frac{M_1}{x^2} = \frac{M_2}{(d-x)^2}\) [1]
- A1: Correctly processes the square root and algebraic steps to show \(x = 3.2 \times 10^8\_\text{m}\) [1]
- M1: Correct expression for total potential: \(\phi = -\frac{GM_1}{x} - \frac{GM_2}{d-x}\) [1]
- A1: Correct calculation of \(\phi = -1.50 \times 10^6\_\text{J\_kg}^{-1}\) (must include unit and negative sign) [1]
- B1: Correct calculation of \(\Delta E_p = -1.44 \times 10^9\_\text{J}\) or states a decrease of \(1.44 \times 10^9\_\text{J}\) [1]

(a) Three basic assumptions of the kinetic theory of ideal gases are:
1. The gas consists of a very large number of molecules in rapid, random motion.
2. The volume of the gas molecules is negligible compared to the total volume occupied by the gas (the volume of the container).
3. There are no intermolecular forces of attraction or repulsion between molecules, except during collisions.
4. All collisions (between molecules, and between molecules and the container walls) are perfectly elastic.
5. The time duration of a collision is negligible compared to the time interval between collisions.

(b) The density of the gas is given by \(\rho = \frac{Nm}{V}\), where \(N\) is the number of molecules, \(m\) is the mass of a single molecule, and \(V\) is the volume of the gas.
Substituting this into the pressure equation:
\(p = \frac{1}{3} \frac{Nm}{V} \langle c^2 \rangle\)
\(pV = \frac{1}{3} Nm \langle c^2 \rangle\)
From the ideal gas equation of state, \(pV = NkT\).
Equating the two expressions for \(pV\):
\(NkT = \frac{1}{3} Nm \langle c^2 \rangle\)
\(kT = \frac{1}{3} m \langle c^2 \rangle\)
The average translational kinetic energy of a molecule is \(E_k = \frac{1}{2} m \langle c^2 \rangle\).
Rearranging the expression: \(\frac{3}{2} k T = \frac{1}{2} m \langle c^2 \rangle\).
Hence, \(E_k = \frac{3}{2} k T\).

(c) (i) First, convert the temperature to kelvin:
\(T = 27 + 273.15 = 300.15\_\text{K}\) (or \(300\_\text{K}\))
Using \(pV = NkT\):
\(N = \frac{pV}{kT} = \frac{2.0 \times 10^5 \times 4.5 \times 10^{-2}}{1.38 \times 10^{-23} \times 300} = \frac{9000}{4.14 \times 10^{-21}} = 2.17 \times 10^{24} \approx 2.2 \times 10^{24}\).

(ii) The mass of one helium atom is:
\(m = \frac{4.0 \times 10^{-3}\_\text{kg\_mol}^{-1}}{6.02 \times 10^{23}\_\text{mol}^{-1}} = 6.64 \times 10^{-27}\_\text{kg}\)
Using \(\langle c^2 \rangle = \frac{3 k T}{m}\):
\(\langle c^2 \rangle = \frac{3 \times 1.38 \times 10^{-23} \times 300}{6.64 \times 10^{-27}} = 1.87 \times 10^6\_\text{m}^2\_\text{s}^{-2}\)
\(c_{\text{rms}} = \sqrt{1.87 \times 10^6} = 1.37 \times 10^3\_\text{m\_s}^{-1}\) (or \(1370\_\text{m\_s}^{-1}\)).
Alternatively, using \(c_{\text{rms}} = \sqrt{\frac{3RT}{M}}\):
\(c_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \times 300}{4.0 \times 10^{-3}}} = \sqrt{1.87 \times 10^6} = 1.37 \times 10^3\_\text{m\_s}^{-1}\).

(a)
- B1: Any one correct assumption (e.g., molecules in rapid random motion) [1]
- B1: Second correct assumption (e.g., volume of molecules is negligible compared to container volume) [1]
- B1: Third correct assumption (e.g., intermolecular forces are negligible / collisions are perfectly elastic) [1]

(b)
- M1: Expresses density as \(\rho = \frac{Nm}{V}\) and writes \(pV = \frac{1}{3} Nm \langle c^2 \rangle\) [1]
- M1: Equates to \(pV = NkT\) to obtain \(kT = \frac{1}{3} m \langle c^2 \rangle\) [1]
- A1: Clearly substitutes the definition \(E_k = \frac{1}{2} m \langle c^2 \rangle\) to arrive at \(E_k = \frac{3}{2} k T\) [1]

(c)
- M1: Correct conversion of temperature to Kelvin (\(300\_\text{K}\)) and uses \(pV = NkT\) (or \(pV = nRT\)) [1]
- A1: Calculates \(N = 2.2 \times 10^{24}\) (accept \(2.17 \times 10^{24}\)) [1]
- M1: Uses a correct formula for r.m.s. speed, e.g., \(c_{\text{rms}} = \sqrt{\frac{3kT}{m}}\) or \(\sqrt{\frac{3RT}{M}}\), with correct values [1]
- A1: Calculates \(c_{\text{rms}} = 1.37 \times 10^3\_\text{m\_s}^{-1}\) (accept \(1.4 \times 10^3\_\text{m\_s}^{-1}\) or \(1370\_\text{m\_s}^{-1}\)) [1]

Diagram:
The setup should show a metal (e.g. aluminium or copper) pendulum suspended from a rigid support. The pendulum sheet oscillates between the pole pieces of an electromagnet. The electromagnet is connected in a series circuit containing a variable DC power supply, a digital ammeter, and a switch. A vertical ruler (or scale) is placed behind the pendulum, and a video camera with a digital timer is positioned to capture the oscillations.

Variables:
- Independent variable: Current \(I\) in the electromagnet.
- Dependent variable: Damping constant \(b\).
- Control variables: Length of the pendulum (keeping the natural frequency of oscillation constant), initial displacement/amplitude \(A_0\) of the pendulum, shape and mass of the pendulum sheet, and the distance between the magnetic poles.

Data Collection:
1. Set up the pendulum and align the electromagnet so that the metal sheet oscillates symmetrically between the poles without touching them.
2. Connect the electromagnet to the DC power supply, ammeter, and a variable resistor in series.
3. Set the current \(I\) to a low value (e.g., 0.5 A) and measure it using the ammeter.
4. Pull the pendulum back to a fixed starting mark (initial amplitude \(A_0\)) and release it.
5. Use the video camera to record the entire decay of the oscillations. Play back the video frame-by-frame (or in slow motion) to record the amplitude \(A\) of consecutive swings and the corresponding time \(t\) from the digital timer.
6. Repeat this process for at least five more values of current \(I\) (e.g., 1.0 A, 1.5 A, 2.0 A, 2.5 A, 3.0 A) by adjusting the variable power supply or the rheostat.

Analysis of Data:
1. For a constant current \(I\), the amplitude decays according to \(A = A_0 e^{-bt}\). Taking natural logarithms: \(\ln A = \ln A_0 - bt\).
2. Plot a graph of \(\ln A\) against time \(t\). This will be a straight line with a gradient equal to \(-b\). Determine \(b\) from this gradient.
3. Once \(b\) is obtained for each current \(I\), test the relationship \(b = k I^n\). Taking logarithms of both sides: \(\ln b = \ln k + n \ln I\).
4. Plot a graph of \(\ln b\) (y-axis) against \(\ln I\) (x-axis).
5. If the relationship is valid, a straight line will be obtained. The gradient of this line will be equal to \(n\). The y-intercept of this line will be equal to \(\ln k\), which gives \(k = e^{\text{intercept}}\).

Safety and Reliability:
- Electromagnets carrying high currents can become very hot. Switch off the current between runs to prevent overheating.
- Ensure the pendulum sheet is made of a non-ferromagnetic conductor (like aluminium) so that it experiences electromagnetic damping (eddy currents) rather than magnetic attraction to the poles.
- Use a non-magnetic (wooden or plastic) clamp stand to support the pendulum to avoid magnetic interference.
- Use a fiducial marker or video analysis to improve the accuracy of amplitude and time measurements.

Marking Scheme (15 Marks Total):

Defining the Problem (3 Marks):
- [1] Identify current \(I\) as the independent variable.
- [1] Identify damping constant \(b\) as the dependent variable.
- [1] State that the initial amplitude \(A_0\) or pendulum length is kept constant.

Methods of Data Collection (4 Marks):
- [1] Draw a labeled diagram showing the pendulum oscillating between the poles of an electromagnet, with a circuit diagram showing a DC power supply, ammeter, and electromagnet in series.
- [1] Describe how the current \(I\) is measured using an ammeter and varied using a variable power supply or rheostat.
- [1] Describe how the amplitude \(A\) is measured as a function of time \(t\) (e.g., using a video camera with a scale/timer, or motion sensor with data logger).
- [1] State the need to record \(A\) and \(t\) over several cycles to find \(b\).

Method of Analysis (3 Marks):
- [1] Explain how to find \(b\) by plotting \(\ln A\) against \(t\), where \(\text{gradient} = -b\).
- [1] Explain how to analyze the overall relationship: plot a graph of \(\ln b\) against \(\ln I\).
- [1] State that a straight-line graph confirms the relationship, with \(\text{gradient} = n\) and \(\text{y-intercept} = \ln k\).

Safety and Experimental Details (5 Marks - Any 5 from the following):
- [1] Switch off current between readings to prevent overheating of the electromagnet.
- [1] Use non-magnetic materials (e.g. aluminium sheet) to avoid direct magnetic force attraction.
- [1] Use a non-magnetic support stand (e.g., wood or heavy plastic clamp) for the pendulum.
- [1] Explain the use of video playback with slow motion to accurately record maximum amplitude in each cycle.
- [1] Ensure the pendulum swings in a single plane perpendicular to the magnetic field lines.
- [1] Keep the gap between the electromagnet poles constant throughout the experiment.

Let us process the data first. The thermodynamic temperature is calculated using \(T = \theta + 273.15\).
The uncertainty in \(T\) is \(\pm 0.5\text{K}\).
The uncertainty in \(\frac{1}{T}\) is calculated via fractional uncertainty: \(\Delta(1/T) = \frac{\Delta T}{T^2}\). For \(T = 293.15\text{K}\), \(\Delta(1/T) = 0.5 / (293.15)^2 \approx 0.006 \times 10^{-3} \text{K}^{-1}\). Since this is extremely small, error bars in the x-direction are negligible.
The value on the y-axis is \(\ln(R/\Omega)\). The absolute uncertainty in \(\ln(R/\Omega)\) is \(\Delta(\ln R) \approx \frac{\Delta R}{R}\).

Table of values:
1. \(\theta = 20.0^\circ\text{C}\) \(\implies T = 293.15\text{K}\) \(\implies 10^3/T = 3.411 \text{K}^{-1}\). \(R = 2500 \pm 100 \Omega \implies \ln R = 7.824 \pm 0.040\).
2. \(\theta = 30.0^\circ\text{C}\) \(\implies T = 303.15\text{K}\) \(\implies 10^3/T = 3.299 \text{K}^{-1}\). \(R = 1750 \pm 70 \Omega \implies \ln R = 7.467 \pm 0.040\).
3. \(\theta = 40.0^\circ\text{C}\) \(\implies T = 313.15\text{K}\) \(\implies 10^3/T = 3.193 \text{K}^{-1}\). \(R = 1250 \pm 50 \Omega \implies \ln R = 7.131 \pm 0.040\).
4. \(\theta = 50.0^\circ\text{C}\) \(\implies T = 323.15\text{K}\) \(\implies 10^3/T = 3.095 \text{K}^{-1}\). \(R = 910 \pm 40 \Omega \implies \ln R = 6.813 \pm 0.044\).
5. \(\theta = 60.0^\circ\text{C}\) \(\implies T = 333.15\text{K}\) \(\implies 10^3/T = 3.002 \text{K}^{-1}\). \(R = 670 \pm 30 \Omega \implies \ln R = 6.507 \pm 0.045\).
6. \(\theta = 70.0^\circ\text{C}\) \(\implies T = 343.15\text{K}\) \(\implies 10^3/T = 2.914 \text{K}^{-1}\). \(R = 500 \pm 20 \Omega \implies \ln R = 6.215 \pm 0.040\).

Linearized Relationship:
\(\ln R = \ln R_0 + \frac{B}{T}\)
Comparing with \(y = mx + c\), the graph of \(\ln R\) against \(\frac{1}{T}\) has:
- \(\text{gradient } m = B\)
- \(\text{y-intercept } c = \ln R_0\)

Gradient and Intercept Calculations:
Using the best-fit line through the coordinates \((2.914 \times 10^{-3}, 6.215)\) and \((3.411 \times 10^{-3}, 7.824)\):
\(m = \frac{7.824 - 6.215}{(3.411 - 2.914) \times 10^{-3}} \approx 3237 \text{K}\).
Let us say the best-fit gradient is \(3240 \pm 130 \text{K}\) (with uncertainty obtained from the worst acceptable line gradient).
Substituting back to find the intercept \(c = \ln(R_0)\):
\(7.824 = 3237 \times 3.411 \times 10^{-3} + c \implies c \approx -3.22\).
The uncertainty in the intercept is found from the worst acceptable line intercept, giving \(c = -3.22 \pm 0.40\).

Finding B and R_0:
- \(B = \text{gradient} = 3240 \pm 130\text{ K}\) (or approx \(3200 \pm 130 \text{ K}\)).
- \(R_0 = e^c = e^{-3.22} \approx 0.040 \Omega\).
- Uncertainty in \(R_0\): \(\Delta R_0 = R_0 \times \Delta c = 0.040 \times 0.40 = 0.016 \Omega\).
Therefore, \(R_0 = (4.0 \pm 1.6) \times 10^{-2} \Omega\).

Marking Scheme (15 Marks Total):

Table of Results (2 Marks):
- [1] Columns for \(T\) and \(1/T\) correctly calculated (to 3 or 4 significant figures).
- [1] Column for \(\ln(R/\Omega)\) with correct absolute uncertainties \(\Delta(\ln R) \approx \Delta R / R\) calculated to 2 decimal places (typically 0.04 to 0.05).

Plotting and Graph Work (4 Marks):
- [1] Axes labeled correctly with units: y-axis is \(\ln(R/\Omega)\) and x-axis is \(\frac{1}{T} / 10^{-3}\text{K}^{-1}\). Scale chosen so that plotted points cover more than half the grid.
- [1] Error bars plotted accurately on all points according to \(\Delta(\ln R)\) (size \(\approx \pm 0.04\)).
- [1] Line of best fit drawn as a single, thin, straight line passing through the main body of all error bars.
- [1] Worst acceptable line drawn as the steepest or shallowest possible line that passes through the top of the error bar of the first/last point and the bottom of the error bar of the last/first point.

Gradient and Intercept (4 Marks):
- [1] Calculation of gradient of best-fit line using points on the line separated by more than half the line length.
- [1] Gradient of worst acceptable line calculated, and uncertainty in gradient determined as: \(\Delta m = |m_{\text{best}} - m_{\text{worst}}|\). (Accept range \(3100 \text{ K}\) to \(3300 \text{ K}\), uncertainty around \(\pm 100 \text{ K}\) to \(\pm 150 \text{ K}\)).
- [1] y-intercept calculated by substituting a point on the best-fit line into \(y = mx + c\).
- [1] y-intercept of worst acceptable line calculated, and uncertainty determined as: \(\Delta c = |c_{\text{best}} - c_{\text{worst}}|\). (Accept range \(-3.0\) to \(-3.4\), uncertainty \(\approx \pm 0.3\) to \(\pm 0.5\)).

Analysis and Constants (5 Marks):
- [1] State that \(B = \text{gradient}\) and \(R_0 = e^c\).
- [1] Calculate \(B\) within acceptable range (\(3100\text{ K}\) to \(3300\text{ K}\)) with unit K.
- [1] Calculate \(R_0\) within acceptable range (\(0.03\Omega\) to \(0.05\Omega\)) with unit \(\Omega\).
- [1] Calculate absolute uncertainty in \(B\) from uncertainty in gradient, with 1 or 2 significant figures.
- [1] Calculate absolute uncertainty in \(R_0\) using \(\Delta R_0 = R_0 \times \Delta c\) or \(e^{c_{\text{max}}} - e^{c_{\text{best}}}\).