2025 AQA IAL Physics (9630) Practice Paper with Answers

First, we find the extension \(\Delta L\) of the wire using the formula: \(\Delta L = \frac{F L_0}{A E}\). Substituting the values: \(\Delta L = \frac{45.0 \times 2.40}{1.50 \times 10^{-7} \times 2.10 \times 10^{11}} = 3.429 \times 10^{-3} \text{ m}\). Next, calculate the strain energy stored, \(E_s\), which is equal to the work done on the wire: \(E_s = \frac{1}{2} F \Delta L\). Substituting the force and extension: \(E_s = \frac{1}{2} \times 45.0 \times 3.429 \times 10^{-3} = 0.0771 \text{ J}\).

M1: Use of \(\Delta L = \frac{F L_0}{A E}\). A1: Correct calculation of extension as \(3.43 \times 10^{-3} \text{ m}\). M1: Use of elastic strain energy formula \(E_s = \frac{1}{2} F \Delta L\). A1: Correct substitution of values into energy formula. A1.22: Correct final answer to 3 significant figures with appropriate unit: \(0.0771 \text{ J}\) (or \(7.71 \times 10^{-2} \text{ J}\)).

First, calculate the extension of wire X (copper): \(\Delta L_{\text{X}} = \frac{F L}{A E_{\text{X}}} = \frac{60.0 \times 1.80}{2.50 \times 10^{-7} \times 1.10 \times 10^{11}} = 3.927 \times 10^{-3} \text{ m}\). Second, calculate the extension of wire Y (brass): \(\Delta L_{\text{Y}} = \frac{F L}{A E_{\text{Y}}} = \frac{60.0 \times 1.80}{2.50 \times 10^{-7} \times 9.00 \times 10^{10}} = 4.800 \times 10^{-3} \text{ m}\). Third, since they are connected end-to-end, they both experience the same tension of 60.0 N. Thus, the total extension is the sum of the individual extensions: \(\Delta L_{\text{total}} = \Delta L_{\text{X}} + \Delta L_{\text{Y}} = 3.927 \times 10^{-3} + 4.800 \times 10^{-3} = 8.727 \times 10^{-3} \text{ m}\), which is 8.73 mm.

M1: Recall of the extension formula. A1: Correct extension for wire X. A1: Correct extension for wire Y. M1: Statement or implication that total extension is the sum of both individual extensions. A1.22: Final correct calculation of 8.73 mm (or 8.73 * 10^-3 m) with correct unit.

First, resolve the initial velocity into horizontal and vertical components: \(u_x = 25.0 \cos(30.0^\circ) = 21.65 \text{ m s}^{-1}\) and \(u_y = 25.0 \sin(30.0^\circ) = 12.50 \text{ m s}^{-1}\). Second, set up the vertical displacement equation with upwards as positive: \(s_y = u_y t - \frac{1}{2} g t^2\). Since the ground is 45.0 m below the launch point, \(s_y = -45.0 \text{ m}\). This gives \(-45.0 = 12.50 t - 4.905 t^2\), or \(4.905 t^2 - 12.50 t - 45.0 = 0\). Solving this quadratic equation for positive time \(t\) gives \(t \approx 4.560 \text{ s}\). Third, calculate horizontal distance: \(x = u_x \times t = 21.65 \times 4.560 = 98.7 \text{ m}\).

M1: Resolves velocity into components. M1: Sets up vertical equation of motion with displacement of -45 m. A1: Correct time of flight of approximately 4.56 s. M1: Formula for horizontal distance. A1.22: Correct final distance of 98.7 m with unit (accept range 98.5 m to 99.0 m).

First, find the time taken to fall to the ground from vertical motion: \(h = \frac{1}{2} g t^2 \Rightarrow 120 = 4.905 t^2 \Rightarrow t = 4.946 \text{ s}\). Second, find the vertical component of the velocity at impact: \(v_y = g t = 9.81 \times 4.946 = 48.52 \text{ m s}^{-1}\). Third, find the horizontal component of the velocity at impact, accounting for deceleration: \(v_x = u_x - a_x t = 68.0 - (0.550 \times 4.946) = 65.28 \text{ m s}^{-1}\). Finally, find the magnitude of the final velocity: \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{65.28^2 + 48.52^2} = \sqrt{4261.5 + 2354.2} = 81.33 \text{ m s}^{-1}\).

M1: Calculate flight time or final vertical velocity. A1: Correct vertical velocity of 48.5 m/s. M1: Equation for horizontal velocity with deceleration. A1: Correct final horizontal velocity of 65.3 m/s. A1.22: Uses Pythagoras to find final velocity magnitude of 81.3 m/s with correct unit (accept 81.1 to 81.5 m/s).

First, convert the half-life to seconds: \(T_{1/2} = 8.50 \text{ days} = 8.50 \times 24 \times 3600 = 7.344 \times 10^5 \text{ s}\). Second, calculate the decay constant \(\lambda\): \(\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.69315}{7.344 \times 10^5} = 9.438 \times 10^{-7} \text{ s}^{-1}\). Third, calculate the activity: \(A = \lambda N = 9.438 \times 10^{-7} \times 3.20 \times 10^{15} = 3.02 \times 10^9 \text{ Bq}\).

M1: Converts half-life from days to seconds. A1: Obtains decay constant lambda = 9.44 * 10^-7 s^-1. M1: Use of activity formula A = lambda * N. A1: Correct substitution of values. A1.22: Correct final answer of 3.02 * 10^9 Bq with correct unit.

By conservation of linear momentum, the alpha particle and radon nucleus must have equal and opposite momenta: \(p_{\alpha} = p_{\text{Rn}}\). The relationship between kinetic energy and momentum is \(E_k = \frac{p^2}{2m}\). Therefore, the kinetic energies are inversely proportional to their masses: \(\frac{E_{\alpha}}{E_{\text{Rn}}} = \frac{m_{\text{Rn}}}{m_{\alpha}} = \frac{222}{4}\). Expressing \(E_{\text{Rn}}\) in terms of \(E_{\alpha}\) gives \(E_{\text{Rn}} = \frac{4}{222} E_{\alpha}\). Since the total energy is the sum of both: \(E_{\alpha} + \frac{4}{222} E_{\alpha} = 4.87 \text{ MeV}\), so \(E_{\alpha} \left(1 + \frac{4}{222}\right) = 4.87 \text{ MeV}\). This simplifies to \(E_{\alpha} \left(\frac{226}{222}\right) = 4.87 \text{ MeV}\). Solving for \(E_{\alpha}\): \(E_{\alpha} = 4.87 \times \frac{222}{226} = 4.78 \text{ MeV}\).

M1: States or applies momentum conservation showing equal momentum. A1: Establishes inverse relationship between kinetic energy and mass. M1: Sets up the total energy equation using mass ratios. A1: Substitutes the values 4 and 222 (or 226) correctly. A1.22: Correct final answer of 4.78 MeV with unit (accept 4.78 to 4.79 MeV).

First, find the net charge of the oxygen-18 ion. Since it has a charge of 2-, it has two excess electrons, so its charge \(q = 2 e = 2 \times 1.60 \times 10^{-19} \text{ C} = 3.20 \times 10^{-19} \text{ C}\). Second, find the mass of the ion using the number of nucleons (18): \(m = 18 \times 1.67 \times 10^{-27} \text{ kg} = 3.006 \times 10^{-26} \text{ kg}\). Third, calculate specific charge: \(\text{Specific Charge} = \frac{q}{m} = \frac{3.20 \times 10^{-19}}{3.006 \times 10^{-26}} = 1.0645 \times 10^7 \text{ C kg}^{-1}\).

M1: Finds charge of the ion (3.20 * 10^-19 C). A1: Finds mass of the ion (3.01 * 10^-26 kg). M1: States formula for specific charge = charge / mass. A1: Performs substitution correctly. A1.22: Correct final answer of 1.06 * 10^7 C kg^-1 with appropriate units.

First, calculate the weight of the beam: \(W = m g = 24.0 \times 9.81 = 235.44 \text{ N}\). Second, identify the positions of forces relative to pivot A: weight acts at 1.40 m from A; support C is located at \(3.60 - 1.00 = 2.60 \text{ m}\) from A. Third, take moments about pivot A for equilibrium: clockwise moments = counter-clockwise moments, so \(W \times 1.40 = F_{\text{C}} \times 2.60\). Substituting the weight: \(235.44 \times 1.40 = F_{\text{C}} \times 2.60 \Rightarrow 329.616 = 2.60 F_{\text{C}}\). Solving for \(F_{\text{C}}\): \(F_{\text{C}} = \frac{329.616}{2.60} = 126.78 \text{ N}\), which rounds to 127 N.

M1: Correctly calculates position of support C as 2.60 m from A. A1: Correctly calculates weight of beam as 235 N. M1: Writes principle of moments equation about A. A1: Correctly substitutes values into moments equation. A1.22: Correct final answer of 127 N with correct unit (accept 126 N to 127 N).

(a) Using the Young modulus formula: \(E = \frac{F L}{A \Delta L}\). Rearranging for tension \(F\): \(F = \frac{E A \Delta L}{L}\). Substituting the values: \(F = \frac{(2.0 \times 10^{11}\text{ Pa}) \times (3.5 \times 10^{-7}\text{ m}^2) \times (1.8 \times 10^{-3}\text{ m})}{3.0\text{ m}}\). \(F = \frac{126}{3.0} = 42\text{ N}\). (b) The elastic strain energy stored \(E_s\) is given by: \(E_s = \frac{1}{2} F \Delta L\). \(E_s = 0.5 \times 42\text{ N} \times 1.8 \times 10^{-3}\text{ m} = 3.78 \times 10^{-2}\text{ J}\) (or \(0.038\text{ J}\)). (c) Plastic deformation means that the wire does not return to its original length when the applied load is removed (it undergoes permanent extension).

(a) [2 marks] 1 mark for rearranging the Young modulus equation to make tension the subject: \(F = \frac{E A \Delta L}{L}\) (or correct substitution of values into the formula). 1 mark for showing the final value is exactly \(42\text{ N}\) with clear working. (b) [2 marks] 1 mark for using the formula for elastic strain energy: \(E_s = \frac{1}{2} F \Delta L\). 1 mark for the correct calculation: \(3.8 \times 10^{-2}\text{ J}\) or \(3.78 \times 10^{-2}\text{ J}\) (allow ECF from (a)). (c) [1 mark] 1 mark for stating that the material is permanently deformed / does not return to its original length when the force is removed.

(a) To measure the diameter accurately:
- Use a micrometer screw gauge.
- To minimize systematic error: Check for zero error when the jaws of the micrometer are closed and subtract/add this correction to all readings.
- To minimize random error: Measure the diameter at several different positions along the wire and in multiple perpendicular orientations at each position to account for non-circularity, then calculate the average.

(b) First, calculate the cross-sectional area \(A\):
\(A = \frac{\pi d^2}{4} = \frac{\pi (0.46 \times 10^{-3}\text{ m})^2}{4} = 1.6619 \times 10^{-7}\text{ m}^2\)

Using the Young modulus formula:
\(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{e / L} = \frac{F L}{A e}\)

Substitute the values:
\(E = \frac{35.0\text{ N} \times 2.58\text{ m}}{1.6619 \times 10^{-7}\text{ m}^2 \times 1.4 \times 10^{-3}\text{ m}} = 3.881 \times 10^{11}\text{ Pa} \approx 3.9 \times 10^{11}\text{ Pa}\) (or \(390\text{ GPa}\)).

(c) Calculate the percentage uncertainties of each variable:
- \(\% F = \frac{0.5}{35.0} \times 100\% = 1.43\%\)
- \(\% L = \frac{0.01}{2.58} \times 100\% = 0.39\%\)
- \(\% d = \frac{0.01}{0.46} \times 100\% = 2.17\%\)
Since \(A \propto d^2\), \(\% A = 2 \times 2.17\% = 4.35\%\)
- \(\% e = \frac{0.1}{1.4} \times 100\% = 7.14\%\)

Sum the percentage uncertainties to find the percentage uncertainty in \(E\):
\(\% E = \% F + \% L + \% A + \% e = 1.43\% + 0.39\% + 4.35\% + 7.14\% = 13.31\%\)

Now calculate the absolute uncertainty \(\Delta E\):
\(\Delta E = 3.881 \times 10^{11}\text{ Pa} \times 0.1331 = 5.17 \times 10^{10}\text{ Pa} \approx 0.5 \times 10^{11}\text{ Pa}\)

Quoting the absolute uncertainty to 1 significant figure and match the decimal precision of the value:
\(E = (3.9 \pm 0.5) \times 10^{11}\text{ Pa}\) (or \(390 \pm 50\text{ GPa}\)).

(d) Because extension is directly proportional to the original length of the wire (\(e \propto L\)), a longer wire will yield a larger extension for the same applied load. Since the absolute uncertainty in measuring the extension remains constant, a larger absolute extension value decreases the percentage uncertainty in the extension measurement (\(\frac{\Delta e}{e}\)), leading to a lower overall percentage uncertainty in \(E\).

(a) [2 marks]
- 1 mark: Use of a micrometer screw gauge at multiple points and orientations along the wire, averaging the readings.
- 1 mark: Checks for and corrects for zero error.

(b) [3 marks]
- 1 mark: Correct calculation of cross-sectional area \(A = 1.66 \times 10^{-7}\text{ m}^2\).
- 1 mark: Correct substitution into the Young modulus formula \(E = \frac{FL}{Ae}\).
- 1 mark: Correct value calculated to 2 or 3 significant figures: \(3.9 \times 10^{11}\text{ Pa}\) (allow \(3.88 \times 10^{11}\text{ Pa}\) or \(390\text{ GPa}\)).

(c) [3.5 marks]
- 1 mark: Correctly calculates individual percentage uncertainties, including doubling the diameter percentage uncertainty for the area.
- 1 mark: Correctly sums percentage uncertainties to get \(13\%\) (or \(13.3\%\)).
- 1 mark: Correctly calculates absolute uncertainty as \(5 \times 10^{10}\text{ Pa}\) (or \(0.5 \times 10^{11}\text{ Pa}\)).
- 0.5 mark: Expresses final answer with consistent decimal places/significant figures: \((3.9 \pm 0.5) \times 10^{11}\text{ Pa}\).

(d) [1 mark]
- 1 mark: States that a longer wire increases the measured extension, which reduces the percentage uncertainty in the extension measurement.

(a) Circuit diagram description:
- Draw a series circuit loop with a cell, an ammeter, and a variable resistor.
- Connect a voltmeter in parallel across the terminals of the cell.
- State that the variable resistor's resistance is altered to vary the current in the circuit.

(b) By comparing the equation \(V = -r I + \varepsilon\) with the standard straight-line equation \(y = m x + c\):
- The vertical intercept (y-intercept) on the \(V\)-axis is equal to the electromotive force, \(\varepsilon\).
- The gradient of the graph is equal to \(-r\), meaning the internal resistance \(r\) is the absolute value of the gradient.

(c) (i)
- The emf, \(\varepsilon\), is given by the y-intercept: \(\varepsilon = 1.45 \pm 0.03\text{ V}\).
- The internal resistance, \(r\), is given by the negative of the gradient: \(r = 1.24 \pm 0.08\ \Omega\).

(ii)
- The x-intercept occurs when terminal potential difference \(V = 0\).
- This represents the maximum current that can be drawn from the cell (short-circuit current).
- Calculation:
\(0 = 1.45 - (1.24 \times I_{\text{max}}) \implies I_{\text{max}} = \frac{1.45}{1.24} = 1.17\text{ A}\) (or \(1.2\text{ A}\)).

(d) Closing the switch only when taking readings prevents continuous current from flowing, which would heat up the cell and circuit components. Heating changes the internal resistance of the cell, causing systematic error in subsequent measurements.

(a) [2 marks]
- 1 mark: Correct circuit diagram with voltmeter in parallel with the cell, and ammeter and variable resistor in series.
- 1 mark: Explicitly states that adjusting the variable resistor varies the current.

(b) [2 marks]
- 1 mark: Identifies that the y-intercept corresponds to the electromotive force \(\varepsilon\).
- 1 mark: Identifies that the internal resistance \(r\) is equal to \(-\text{gradient}\) (or magnitude of gradient).

(c) [4.5 marks]
- 1 mark: \(\varepsilon = 1.45 \pm 0.03\text{ V}\) (must include units).
- 1 mark: \(r = 1.24\ \Omega\).
- 1 mark: Absolute uncertainty of \(r = \pm 0.08\ \Omega\) (must include units).
- 1 mark: Identifies x-intercept as short-circuit current / maximum current and calculates \(1.17\text{ A}\) (allow range \(1.17 - 1.20\text{ A}\)).
- 0.5 mark: Units correct throughout part (c).

(d) [1 mark]
- 1 mark: To prevent temperature increase of the cell/resistors, which would alter the internal resistance.

The Young modulus \(E\) is given by:
\[E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x/L} = \frac{F L}{A x}\]
where the cross-sectional area of the first wire is \(A = \frac{\pi d^2}{4}\).
Thus, the extension \(x\) is:
\[x = \frac{F L}{A E} = \frac{4 F L}{\pi d^2 E}\]

For the second wire, the length is \(2L\), the diameter is \(2d\) (which means the cross-sectional area is \(A' = \frac{\pi (2d)^2}{4} = 4A\)), and the force is \(3F\).

Its extension \(x'\) is:
\[x' = \frac{(3F) (2L)}{(4A) E} = \frac{6 F L}{4 A E} = 1.5 \left(\frac{F L}{A E}\right) = 1.5x\]

1 mark for identifying that area increases by a factor of 4, length by a factor of 2, and force by a factor of 3, leading to a factor of 1.5 change in extension.

The time of flight \(T\) of a projectile on horizontal ground depends only on the vertical component of its initial velocity \(u_y\) and the acceleration due to gravity \(g\):
\[T = \frac{2 u_y}{g}\]
If \(u_y\) is doubled, the new time of flight \(T'\) becomes:
\[T' = \frac{2 (2 u_y)}{g} = 2T\]

The maximum height \(H\) is given by:
\[H = \frac{u_y^2}{2g}\]
If \(u_y\) is doubled, the new maximum height \(H'\) becomes:
\[H' = \frac{(2 u_y)^2}{2g} = 4 \left(\frac{u_y^2}{2g}\right) = 4H\]

1 mark for identifying that time of flight is directly proportional to the vertical velocity component and maximum height is proportional to the square of the vertical velocity component, leading to 2T and 4H.

Using Fleming's Left-Hand Rule:
1. For \(\alpha\) particles (positive charge): The conventional current is in the direction of motion (upwards). The magnetic field is into the page. The force is directed to the left. Since \(\alpha\) particles are much more massive than \(\beta^-\) particles, their charge-to-mass ratio is small, resulting in a much larger radius of curvature (\(r = \frac{mv}{Bq}\)).
2. For \(\beta^-\) particles (negative charge): The conventional current is opposite to the direction of motion (downwards). The force is directed to the right. Due to their very small mass, their radius of curvature is much smaller (greater deflection).
3. For \(\gamma\) rays (neutral): They carry no charge and are completely undeflected.

1 mark for using Fleming's Left-Hand Rule and the relative magnitudes of deflection (radius of curvature) based on charge-to-mass ratios to identify the correct behavior of the three types of radiation.

In \(\beta^-\) decay, a neutron decays into a proton, an electron, and an electron antineutrino:
\[\text{n} \to \text{p} + \text{e}^- + \bar{\nu}_\text{e}\]
The quark structure of a neutron is \(udd\).
The quark structure of a proton is \(uud\).
This means that one down quark (\(d\)) decays into an up quark (\(u\)).
Consequently, the number of up quarks increases by 1 and the number of down quarks decreases by 1.

1 mark for identifying the correct quark-level transition in beta-minus decay (d to u) and stating the corresponding net changes.

Take moments about the pivot in equilibrium:
- The weight of the uniform beam (\(120\text{ N}\)) acts at its center of gravity, which is at the midpoint: \(1.5\text{ m}\) from the pivot.
Clockwise moment of the beam's weight = \(120\text{ N} \times 1.5\text{ m} = 180\text{ N m}\).
- The weight hung from the end (\(80\text{ N}\)) is at \(3.0\text{ m}\) from the pivot.
Clockwise moment of the hung weight = \(80\text{ N} \times 3.0\text{ m} = 240\text{ N m}\).
- The tension \(T\) in the wire acts vertically upwards at \(2.0\text{ m}\) from the pivot.
Anticlockwise moment of the tension = \(T \times 2.0\text{ m}\).

By the principle of moments:
\[2.0 T = 180 + 240\]
\[2.0 T = 420 \implies T = 210\text{ N}\]

1 mark for setting up a correct moment equation about the pivot and solving for the tension.

The density \(\rho\) of a sphere is given by:
\[\rho = \frac{m}{V} = \frac{m}{\frac{4}{3}\pi r^3} = \frac{6m}{\pi d^3}\]
The formula for combining independent uncertainties gives:
\[\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \frac{\Delta d}{d}\]

Calculate the percentage uncertainty in mass:
\[\frac{0.1}{45.2} \times 100\% \approx 0.22\%\]

Calculate the percentage uncertainty in diameter:
\[\frac{0.02}{2.04} \times 100\% \approx 0.98\%\]

Total percentage uncertainty in density:
\[0.22\% + 3 \times 0.98\% = 0.22\% + 2.94\% = 3.16\% \approx 3.2\%\]

1 mark for correctly combining the percentage uncertainty in mass with three times the percentage uncertainty in diameter, obtaining 3.2%.

The useful output power \(P_{\text{out}}\) developed by the motor is:
\[P_{\text{out}} = F v = m g v\]
\[P_{\text{out}} = 200\text{ kg} \times 9.81\text{ m s}^{-2} \times 1.5\text{ m s}^{-1} = 2943\text{ W} = 2.943\text{ kW}\]

The efficiency \(\eta\) is \(75\% = 0.75\). The input electrical power \(P_{\text{in}}\) is:
\[P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{2.943\text{ kW}}{0.75} = 3.924\text{ kW} \approx 3.9\text{ kW}\]

1 mark for calculating the useful power output and dividing by the efficiency (0.75) to obtain 3.9 kW.

Let the direction to the right be positive.
Initial total momentum:
\[p_{\text{initial}} = m_A u_A + m_B u_B = (2.0 \times 6.0) + (4.0 \times 0) = 12.0\text{ kg m s}^{-1}\]

After the collision, trolley A moves to the left, so its velocity is \(v_A = -1.0\text{ m s}^{-1}\).
Final total momentum:
\[p_{\text{final}} = m_A v_A + m_B v_B = (2.0 \times -1.0) + (4.0 \times v_B) = -2.0 + 4.0 v_B\]

By conservation of momentum:
\[12.0 = -2.0 + 4.0 v_B\]
\[14.0 = 4.0 v_B \implies v_B = 3.5\text{ m s}^{-1}\]
Since the value is positive, trolley B moves to the right with a speed of \(3.5\text{ m s}^{-1}\).

1 mark for setting up a conservation of momentum equation that accounts for direction with opposite signs, yielding 3.5 m s^-1 to the right.

The Young modulus \(E\) is given by \(E = \frac{F L}{A \Delta L}\), where \(A = \frac{\pi d^2}{4}\). Thus, the tension force \(F\) is \(F = \frac{E A \Delta L}{L} = \frac{E \pi d^2 \Delta L}{4 L}\). For the first wire, \(M g = \frac{E \pi d^2 \Delta L}{4 L}\). For the second wire, the mass is \(M_2\), so \(M_2 g = \frac{E \pi (2d)^2 \Delta L}{4 (2L)} = \frac{4 E \pi d^2 \Delta L}{8 L} = 2 \left( \frac{E \pi d^2 \Delta L}{4 L} \right) = 2 M g\). Therefore, \(M_2 = 2M\).

1 mark for selecting the correct answer C.

At the highest point, the vertical component of the velocity is zero. The horizontal component of the velocity remains constant at \(v_x = u \cos(60^\circ) = 0.5 u\), where \(u\) is the initial launch speed. The kinetic energy at the highest point is therefore \(E_{\text{k, top}} = \frac{1}{2} m v_x^2 = \frac{1}{2} m (0.5 u)^2 = 0.25 \left(\frac{1}{2} m u^2\right) = 0.25 E_{\text{k}}\).

1 mark for selecting B.

After a time of \(3T\) (three half-lives), the fraction of the original nuclei that remain undecayed is \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\). The fraction of nuclei that have decayed is \(1 - \frac{1}{8} = \frac{7}{8}\). The ratio of decayed nuclei to remaining undecayed nuclei is therefore \(\frac{7/8}{1/8} = 7\).

1 mark for selecting B.

Specific charge is defined as the ratio of charge to mass (\(Q/m\)). For a singly ionized helium-4 atom (\(^4\text{He}^+\)), the charge is \(+1e\) and the mass is approximately \(4u\), so its specific charge is \(\frac{1e}{4u} = 0.25\,e\,u^{-1}\). For a carbon-12 nucleus (\(^{12}\text{C}^{6+}\)), the charge is \(+6e\) and the mass is \(12u\), so its specific charge is \(\frac{6e}{12u} = 0.50\,e\,u^{-1}\). The ratio of the two specific charges is \(\frac{0.25}{0.50} = 0.50\).

1 mark for selecting B.

The weight \(W\) of the uniform beam acts at its center of gravity, which is at the midpoint (\(0.50L\) from either end). Taking moments about the pivot (located at \(0.25L\) from the left end): The clockwise moment due to the weight is \(W \times (0.50L - 0.25L) = W \times 0.25L\). The anticlockwise moment due to the upward vertical force \(F\) at the right-hand end (at distance \(L - 0.25L = 0.75L\) from the pivot) is \(F \times 0.75L\). For rotational equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments: \(F \times 0.75L = W \times 0.25L \implies F = \frac{0.25}{0.75} W = \frac{1}{3} W\).

1 mark for selecting B.

The percentage uncertainty in a quantity calculated using multiplication and division is found by adding the percentage uncertainties of each term, multiplied by the power of that term. For density \(\rho = \frac{4m}{\pi d^2 h}\): \% uncertainty in \(\rho\) = \% uncertainty in \(m\) + 2 * (\% uncertainty in \(d\)) + \% uncertainty in \(h\) = 1.5% + 2(1.0%) + 2.0% = 1.5% + 2.0% + 2.0% = 5.5%.

1 mark for selecting C.