2023 AQA IAL Physics (9630) 模擬試題連答案詳解

(a) Taking multiple readings at different orientations and calculating a mean reduces random error. To correct for zero error, the student must close the jaws of the micrometer fully (without any object) and record the reading. This zero reading must then be subtracted from all subsequent diameter measurements.

(b) First, calculate the cross-sectional area \(A\):
\(A = \frac{\pi d^2}{4} = \frac{\pi (0.38 \times 10^{-3}\text{ m})^2}{4} = 1.1341 \times 10^{-7}\text{ m}^2\)

Calculate the central value for resistivity \(\rho\):
\(\rho = \frac{R A}{L} = \frac{4.25 \times 1.1341 \times 10^{-7}}{1.240} = 3.887 \times 10^{-7}\ \Omega\text{ m}\)

Now determine percentage uncertainties:
- For resistance \(R\): \(\frac{\Delta R}{R} = \frac{0.05}{4.25} \times 100\% = 1.176\%\)
- For length \(L\): \(\frac{\Delta L}{L} = \frac{0.002}{1.240} \times 100\% = 0.161\%\)
- For diameter \(d\): \(\frac{\Delta d}{d} = \frac{0.01}{0.38} \times 100\% = 2.632\%\)

Since \(A \propto d^2\), the percentage uncertainty in \(A\) is \(2 \times 2.632\% = 5.264\%\).

Total percentage uncertainty in \(\rho\):
\(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + \frac{\Delta L}{L} + 2\frac{\Delta d}{d} = 1.176\% + 0.161\% + 5.264\% = 6.601\%\)

Now calculate the absolute uncertainty in \(\rho\):
\(\Delta \rho = 3.887 \times 10^{-7} \times 0.06601 = 0.2566 \times 10^{-7}\ \Omega\text{ m} \approx 0.3 \times 10^{-7}\ \Omega\text{ m}\)

Rounding \(\rho\) to match the decimal place of the uncertainty:
\(\rho = (3.9 \pm 0.3) \times 10^{-7}\ \Omega\text{ m}\)

Part (a):
- Identifies random error (1 mark)
- Explains zero-error correction: close micrometer without wire, record reading, subtract from subsequent diameter measurements (1.25 marks)

Part (b):
- Correct calculation of area \(A = 1.13 \times 10^{-7}\text{ m}^2\) (1 mark)
- Correct calculation of \(\rho = 3.89 \times 10^{-7}\ \Omega\text{ m}\) (1 mark)
- Correct percentage uncertainty in \(d\) (\(2.6\%\)) or area (\(5.3\%\)) (1 mark)
- Correct total percentage uncertainty in \(\rho\) of \(6.6\%\) (1 mark)
- Correct absolute uncertainty \(\Delta \rho = 0.26 \times 10^{-7}\ \Omega\text{ m}\) (1 mark)
- Final answer stated correctly with consistent decimal places, appropriate significant figures and unit (1 mark)

(a) The line of best fit is the line drawn through the plotted points that minimises the distances of all points from the line. The worst acceptable line of fit is the steepest or shallowest possible line that still passes through all the error bars of the data points. The absolute uncertainty in the gradient is given by the difference between these two gradients, \(\Delta m = |m_{\text{worst}} - m_{\text{best}}|\).

(b) Squaring the formula: \(T^2 = \frac{4\pi^2}{g} L\). Hence, the gradient \(m = \frac{4\pi^2}{g}\), which rearranges to \(g = \frac{4\pi^2}{m}\).

Calculate the best-fit value of \(g\):
\(g_{\text{best}} = \frac{4\pi^2}{4.02} = 9.820\text{ m s}^{-2}\)

Determine the absolute uncertainty in gradient \(m\):
\(\Delta m = 4.18 - 4.02 = 0.16\text{ s}^2\text{ m}^{-1}\)

The fractional uncertainty in the gradient is:
\(\frac{\Delta m}{m_{\text{best}}} = \frac{0.16}{4.02} = 0.0398\ (3.98\%)\)

Since \(g \propto \frac{1}{m}\), the fractional uncertainty in \(g\) is equal to the fractional uncertainty in \(m\):
\(\Delta g = g_{\text{best}} \times 0.0398 = 9.820 \times 0.0398 = 0.391\text{ m s}^{-2}\)

Alternatively, calculating \(g_{\text{worst}} = \frac{4\pi^2}{4.18} = 9.445\text{ m s}^{-2}\), giving \(\Delta g = 9.820 - 9.445 = 0.375\text{ m s}^{-2}\).

Both methods yield an absolute uncertainty of \(0.4\text{ m s}^{-2}\) when rounded to one significant figure.

Rounding \(g\) to match this, we get:
\(g = 9.8 \pm 0.4\text{ m s}^{-2}\)

Part (a):
- Explains difference between line of best fit and worst acceptable line of fit (e.g., steepest/shallowest passing through error bars) (1.25 marks)
- Identifies how uncertainty is found by taking the difference between the gradients of the two lines (1 mark)

Part (b):
- Correctly identifies \(g = \frac{4\pi^2}{m}\) (1 mark)
- Computes \(g_{\text{best}} = 9.82\text{ m s}^{-2}\) (1 mark)
- Computes uncertainty in gradient \(\Delta m = 0.16\text{ s}^2\text{ m}^{-1}\) (1 mark)
- Uses fractional uncertainty of \(m\) or direct calculation of \(g_{\text{worst}}\) to find absolute uncertainty in \(g\) (1 mark)
- Obtains absolute uncertainty in range \(0.38 - 0.40\text{ m s}^{-2}\) (1 mark)
- Final answer stated as \(9.8 \pm 0.4\text{ m s}^{-2}\) with units (1 mark)

(a) The resolution of the stopwatch is \(0.01\text{ s}\). However, the manual start/stop operation depends on human reaction time, which introduces a random uncertainty of approximately \(0.2\text{ s}\). Since human reaction time is much larger than the instrument's resolution, human reaction time is the dominant and limiting source of uncertainty.

(b) First, calculate the mean total time \(t\) for 20 oscillations:
\(t_{\text{mean}} = \frac{14.23 + 14.19 + 14.25}{3} = 14.2233\text{ s}\)

Find the range of the trials:
\(\text{Range} = 14.25 - 14.19 = 0.06\text{ s}\)

The absolute uncertainty in the total time \(\Delta t\) is estimated as half the range:
\(\Delta t = \frac{0.06}{2} = 0.03\text{ s}\)

The time period for one oscillation \(T\) is:
\(T = \frac{t_{\text{mean}}}{20} = \frac{14.2233}{20} = 0.71117\text{ s}\)

The absolute uncertainty in the time period \(\Delta T\) is:
\[\Delta T = \frac{\Delta t}{20} = \frac{0.03}{20} = 0.0015\text{ s}\]

Rounding the uncertainty to 1 significant figure (or keeping 2 significant figures as \(0.0015\text{ s}\)) gives \(0.002\text{ s}\). Stating the period to match the precision of the uncertainty:
\(T = 0.711 \pm 0.002\text{ s}\) (or \(T = 0.7112 \pm 0.0015\text{ s}\))

Part (a):
- States that resolution is not the limiting factor (1 mark)
- Identifies human reaction time as the dominant source and estimates it to be around \(0.1\text{ s}\) to \(0.2\text{ s}\) (1.25 marks)

Part (b):
- Correctly calculates mean total time \(t_{\text{mean}} = 14.22\text{ s}\) (1 mark)
- Correctly calculates period \(T = 0.711\text{ s}\) (1 mark)
- Finds range \(= 0.06\text{ s}\) (1 mark)
- Finds uncertainty in total time \(\Delta t = 0.03\text{ s}\) (1 mark)
- Finds uncertainty in period \(\Delta T = 0.0015\text{ s}\) (1 mark)
- Formulates final response as \(0.711 \pm 0.002\text{ s}\) or \(0.711 \pm 0.0015\text{ s}\) with units (1 mark)

(a) Precision refers to the degree of agreement between repeated measurements of height or time (governed by the spread of random errors). Accuracy refers to how close the calculated value of \(g\) is to the true accepted value (9.81 m s\(^{-2}\)), which is affected by systematic errors such as air resistance or calibration delays.

(b) Rearrange the equation for \(g\):
\(g = \frac{2h}{t^2}\)

Substitute the central values:
\(g = \frac{2 \times 2.45}{0.71^2} = \frac{4.90}{0.5041} = 9.7203\text{ m s}^{-2}\)

Calculate individual percentage uncertainties:
- Height: \(\frac{\Delta h}{h} \times 100\% = \frac{0.02}{2.45} \times 100\% = 0.816\%\)
- Time: \(\frac{\Delta t}{t} \times 100\% = \frac{0.02}{0.71} \times 100\% = 2.817\%\)

Calculate total percentage uncertainty in \(g\):
\(\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t} = 0.816\% + 2 \times 2.817\% = 6.45\%\)

Now calculate the absolute uncertainty in \(g\):
\(\Delta g = 9.7203 \times 0.0645 = 0.627\text{ m s}^{-2} \approx 0.6\text{ m s}^{-2}\)

Stating the value with the appropriate decimal places matching the uncertainty:
\(g = 9.7 \pm 0.6\text{ m s}^{-2}\)

Part (a):
- Precision definition based on closeness of measurements to each other / size of random errors (1 mark)
- Accuracy definition based on closeness to true value / presence of systematic errors (1.25 marks)

Part (b):
- Correctly calculates \(g = 9.72\text{ m s}^{-2}\) (1 mark)
- Correctly determines percentage uncertainty in \(h\) (\(0.82\%\)) and \(t\) (\(2.82\%\)) (1 mark)
- Doubles the percentage uncertainty of \(t\) in summing (1 mark)
- Finds total percentage uncertainty in \(g = 6.45\%\) (or \(6.5\%\)) (1 mark)
- Computes absolute uncertainty \(\Delta g = 0.63\text{ m s}^{-2}\) (1 mark)
- Expresses final value as \(9.7 \pm 0.6\text{ m s}^{-2}\) with unit (1 mark)

(a) Slow-moving (thermal) neutrons are much more likely to be captured by Uranium-235 nuclei to cause fission than fast neutrons. A moderator (such as water or graphite) is used in the reactor core to slow down these fast fission neutrons through elastic collisions with nuclei of the moderator.

(b) Calculate the total initial binding energy of Uranium-235 (neutrons have zero binding energy):
\(E_{\text{initial}} = 235 \times 7.59\text{ MeV} = 1783.65\text{ MeV}\)

Calculate the total binding energy of the fission products:
\(E_{\text{final}} = (141 \times 8.33\text{ MeV}) + (92 \times 8.51\text{ MeV})\)
\(E_{\text{final}} = 1174.53\text{ MeV} + 782.92\text{ MeV} = 1957.45\text{ MeV}\)

Calculate the energy released (\(\Delta E\)):
\(\Delta E = E_{\text{final}} - E_{\text{initial}} = 1957.45 - 1783.65 = 173.8\text{ MeV}\)

Convert the energy released to Joules (\(1\text{ eV} = 1.60 \times 10^{-19}\text{ J}\)):
\(\Delta E_{\text{Joules}} = 173.8 \times 10^6 \times 1.60 \times 10^{-19}\text{ J} = 2.7808 \times 10^{-11}\text{ J} \approx 2.78 \times 10^{-11}\text{ J}\)

Part (a):
- Explains that slower neutrons have a higher probability of inducing fission (1 mark)
- Identifies the moderator as the component responsible for slowing them down (1.25 marks)

Part (b):
- Calculates total binding energy of U-235 reactant: \(1783.65\text{ MeV}\) (1 mark)
- Calculates total binding energy of products: \(1957.45\text{ MeV}\) (1 mark)
- Subtracts reactant binding energy from product binding energy to get \(173.8\text{ MeV}\) (1 mark)
- Correctly applies the conversion factor \(1.60 \times 10^{-19}\text{ J/eV}\) (1 mark)
- Includes factor of \(10^6\) for Mega-eV (1 mark)
- Obtains final answer of \(2.78 \times 10^{-11}\text{ J}\) (or \(2.8 \times 10^{-11}\text{ J}\)) (1 mark)

(a) Newton's law of gravitation states that the attractive force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Mathematically, \(F = \frac{G M m}{r^2}\), where \(F\) is the gravitational force, \(G\) is the universal gravitational constant, \(M\) and \(m\) are the masses of the two interacting bodies, and \(r\) is the distance between their centers.

(b) Let \(x\) be the distance from the center of the Earth to the point where the net force is zero. The distance from this point to the center of the Moon is \((d - x)\).
At this point, the magnitude of the Earth's gravitational pull equals that of the Moon:
\[ \frac{G M_E m}{x^2} = \frac{G M_M m}{(d - x)^2} \]

Cancel out \(G\) and the spacecraft mass \(m\):
\[ \frac{M_E}{x^2} = \frac{M_M}{(d - x)^2} \]

Take the square root of both sides to simplify solving:
\[ \frac{\sqrt{M_E}}{x} = \frac{\sqrt{M_M}}{d - x} \]
\[ \sqrt{M_E} (d - x) = \sqrt{M_M} x \]
\[ \sqrt{M_E} d - \sqrt{M_E} x = \sqrt{M_M} x \]
\[ x = d \left( \frac{\sqrt{M_E}}{\sqrt{M_E} + \sqrt{M_M}} \right) \]

Substitute the given values:
\[ x = 3.84 \times 10^8 \left( \frac{\sqrt{5.97 \times 10^{24}}}{\sqrt{5.97 \times 10^{24}} + \sqrt{7.35 \times 10^{22}}} \right) \]
\[ \sqrt{5.97 \times 10^{24}} = 2.4434 \times 10^{12} \]
\[ \sqrt{7.35 \times 10^{22}} = 2.7111 \times 10^{11} \]
\[ x = 3.84 \times 10^8 \left( \frac{2.4434 \times 10^{12}}{2.4434 \times 10^{12} + 0.27111 \times 10^{12}} \right) \]
\[ x = 3.84 \times 10^8 \left( \frac{2.4434}{2.7145} \right) = 3.456 \times 10^8\text{ m} \approx 3.46 \times 10^8\text{ m} \]

Part (a):
- States the proportional relationship between force, mass product and separation squared (1.25 marks)
- Defines all symbols correctly: \(F\), \(G\), \(M\), \(m\), \(r\) (1 mark)

Part (b):
- Sets up initial force equilibrium equation: \(\frac{G M_E m}{x^2} = \frac{G M_M m}{(d-x)^2}\) (1 mark)
- Eliminates common constants \(G\) and \(m\) (1 mark)
- Simplifies by taking square roots of both sides (1 mark)
- Rearranges equation to make \(x\) the subject (1 mark)
- Substitutes masses and distance correctly (1 mark)
- Resolves the expression to get \(3.46 \times 10^8\text{ m}\) (1 mark)

(a) Gravitational field strength at a point is defined as the gravitational force exerted per unit mass on a small test mass placed at that point. It is a vector because it has both magnitude and a defined direction, which is always directed radially inwards towards the center of mass creating the field.

(b) The formula for gravitational field strength at a planet's surface is:
\[ g = \frac{G M}{R^2} \]

Substitute the given parameters (with \(G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}\)):
\[ g = \frac{6.67 \times 10^{-11} \times 4.87 \times 10^{24}}{(6.05 \times 10^6)^2} \]
\[ g = \frac{3.2483 \times 10^{14}}{3.6603 \times 10^{13}} = 8.8745\text{ N kg}^{-1} \]

Hence, \(g = 8.87\text{ N kg}^{-1}\) (or \(\text{m s}^{-2}\)).

Now, calculate the weight of the lander:
\[ W = m g \]
\[ W = 75.0 \times 8.8745 = 665.59\text{ N} \]

Rounding to 3 significant figures: \(W = 666\text{ N}\).

Part (a):
- Defines gravitational field strength as force per unit mass (1.25 marks)
- Explains that it is a vector because it has direction (directed towards center of mass) (1 mark)

Part (b):
- Identifies correct formula \(g = \frac{G M}{R^2}\) (1 mark)
- Substitutes numerical values correctly (1 mark)
- Obtains surface gravity \(g = 8.87\text{ N kg}^{-1}\) (or \(\text{m s}^{-2}\)) (2 marks)
- Uses weight formula \(W = m g\) with lander mass (1 mark)
- Obtains final weight \(W = 666\text{ N}\) (accept \(665\text{ N}\) or \(665.6\text{ N}\) depending on rounding of intermediate \(g\)) (1 mark)

(a) Three key assumptions of an ideal gas are:
1. The volume occupied by the gas molecules is completely negligible compared to the total volume of the container.
2. There are no electrostatic forces of attraction or repulsion between the molecules (except during collisions).
3. All collisions between molecules and between the molecules and the container walls are perfectly elastic.

(b) Convert temperatures to Kelvin:
\(T_1 = 22.0 + 273 = 295\text{ K}\)
\(T_2 = 85.0 + 273 = 358\text{ K}\)

Apply the ideal gas equation \(P V = n R T\) to calculate the number of moles \(n\):
\[ n = \frac{P_1 V}{R T_1} = \frac{1.50 \times 10^5 \times 0.0450}{8.31 \times 295} = \frac{6750}{2451.45} = 2.7534\text{ mol} \]

Calculate the mass of the helium:
\[ m = n \times M = 2.7534 \times 4.00 \times 10^{-3}\text{ kg} = 0.01101\text{ kg} = 1.10 \times 10^{-2}\text{ kg} \]

Since the cylinder is rigid, its volume \(V\) is constant. According to Pressure Law, \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\):
\[ P_2 = P_1 \left( \frac{T_2}{T_1} \right) \]
\[ P_2 = 1.50 \times 10^5 \times \left( \frac{358}{295} \right) = 1.8203 \times 10^5\text{ Pa} \approx 1.82 \times 10^5\text{ Pa} \]

Part (a):
- Identifies three valid assumptions (e.g., elastic collisions, negligible volume of particles, negligible collision time, random rapid motion, zero intermolecular forces) (0.75 marks each, max 2.25 marks)

Part (b):
- Correctly converts both Celsius temperatures to Kelvin (1 mark)
- Uses \(PV = nRT\) to correctly calculate \(n = 2.75\text{ mol}\) (1 mark)
- Calculates correct mass \(m = 1.10 \times 10^{-2}\text{ kg}\) (or \(11.0\text{ g}\)) (1 mark)
- Recognises that pressure is directly proportional to absolute temperature when volume is constant (1 mark)
- Substitutes values correctly into pressure ratio equation (1 mark)
- Calculates new pressure \(P_2 = 1.82 \times 10^5\text{ Pa}\) with unit (1 mark)

The density of the wire is given by the formula \( \rho = \frac{4m}{\pi d^2 L} \). The percentage uncertainty in density is the sum of the individual percentage uncertainties: \( \frac{\Delta \rho}{\rho} \times 100 = \left( \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L} \right) \times 100 \). Calculating each percentage uncertainty: \( \frac{\Delta m}{m} \times 100 = \frac{0.01}{0.35} \times 100 \approx 2.86\% \), \( \frac{\Delta d}{d} \times 100 = \frac{0.01}{0.42} \times 100 \approx 2.38\% \), \( \frac{\Delta L}{L} \times 100 = \frac{0.2}{15.0} \times 100 \approx 1.33\% \). Combining these gives: \( 2.86\% + 2(2.38\%) + 1.33\% = 8.95\% \), which rounds to \( 9.0\% \).

1 mark for the correct answer D. Reject all other options.

Correcting each reading by subtracting the systematic zero error (+0.03 mm) removes the systematic error, which brings the mean of the measurements closer to the true value, thereby improving accuracy. However, subtracting a constant from every reading does not affect the spread of the data or the random variations, so the precision remains unchanged.

1 mark for the correct answer B. Reject all other options.

The relationship for a simple pendulum is \( T = 2\pi \sqrt{\frac{L}{g}} \), which gives \( T^2 = \left(\frac{4\pi^2}{g}\right) L \). Therefore, the gradient of the graph is \( m = \frac{4\pi^2}{g} \), so \( g = \frac{4\pi^2}{m} \). Since \( g \) is inversely proportional to \( m \), the percentage uncertainty in \( g \) is equal to the percentage uncertainty in the gradient \( m \). The absolute uncertainty in the gradient is \( \Delta m = 4.18 - 4.02 = 0.16 \text{ s}^2 \text{ m}^{-1} \). The percentage uncertainty is \( \frac{\Delta m}{m_{\text{best}}} \times 100 = \frac{0.16}{4.02} \times 100 \approx 3.98\% \), which rounds to \( 4.0\% \).

1 mark for the correct answer B. Reject all other options.

The energy released is the difference between the total binding energy of the products and the reactants. The total binding energy before fission is: \( 235 \times 7.59 \text{ MeV} = 1783.65 \text{ MeV} \). The total binding energy after fission is: \( (141 \times 8.33) + (92 \times 8.51) = 1174.53 + 782.92 = 1957.45 \text{ MeV} \). The energy released is: \( 1957.45 - 1783.65 = 173.8 \text{ MeV} \).

1 mark for the correct answer A. Reject all other options.

The mass \( M \) of a uniform sphere is proportional to its volume, so \( M \propto R^3 \). If the radius is doubled to \( 2R \), the mass of each sphere increases by a factor of \( 2^3 = 8 \). The distance between the centers of two contacting spheres of radius \( R \) is \( d = 2R \). For spheres of radius \( 2R \), this distance becomes \( d' = 4R \) (which is doubled). Using Newton's law of gravitation, \( F = \frac{G M^2}{d^2} \). For the new spheres: \( F' = \frac{G (8M)^2}{(2d)^2} = \frac{64 G M^2}{4 d^2} = 16 F \).

1 mark for the correct answer D. Reject all other options.

Let \( x \) be the distance from the center of the planet. At this point, the gravitational field strength of the planet equals that of the moon: \( \frac{G M}{x^2} = \frac{G (0.040M)}{(d-x)^2} \). This simplifies to \( \frac{1}{x^2} = \frac{0.040}{(d-x)^2} \). Taking the square root of both sides gives \( \frac{1}{x} = \frac{0.20}{d-x} \). Rearranging yields \( d - x = 0.20 x \), which means \( d = 1.20 x \). Thus, \( x = \frac{d}{1.20} \approx 0.83 d \).

1 mark for the correct answer C. Reject all other options.

The root-mean-square speed \( c_{\text{rms}} \) of gas molecules is directly proportional to the square root of the absolute temperature \( T \) of the gas (\( c_{\text{rms}} = \sqrt{\frac{3RT}{M}} \)). Therefore, \( \frac{v_{\text{new}}}{v} = \sqrt{\frac{T_{\text{new}}}{T_{\text{initial}}}} = \sqrt{\frac{450}{300}} = \sqrt{1.5} \approx 1.22 \). The new speed is \( 1.22 v \).

1 mark for the correct answer A. Reject all other options.

The total number of molecules is \( N = n N_A = 3.5 \times 6.02 \times 10^{23} = 2.107 \times 10^{24} \). The total internal energy of the ideal gas is \( E_{\text{total}} = \frac{3}{2} p V = 1.5 \times 1.2 \times 10^5 \times 0.080 = 14400 \text{ J} \). The average kinetic energy per molecule is \( E_k = \frac{E_{\text{total}}}{N} = \frac{14400}{2.107 \times 10^{24}} \approx 6.83 \times 10^{-21} \text{ J} \).

1 mark for the correct answer C. Reject all other options.

The resistivity is given by the formula \(\rho = \frac{R A}{L} = \frac{\pi R d^2}{4 L}\). The percentage uncertainties for each measured quantity are: For \(R\), \(\frac{0.08}{4.00} \times 100\% = 2.0\%\). For \(L\), \(\frac{0.005}{1.250} \times 100\% = 0.4\%\). For \(d\), \(\frac{0.01}{0.40} \times 100\% = 2.5\%\). Since the diameter \(d\) is squared in the formula, its percentage uncertainty is doubled: \(2 \times 2.5\% = 5.0\%\). Combining these to find the total percentage uncertainty in \(\rho\): \(\%\Delta \rho = \%\Delta R + \%\Delta L + 2 \times \%\Delta d = 2.0\% + 0.4\% + 5.0\% = 7.4\%\).

C is correct (1 mark). Incorrect options: A is a calculation error; B forgets to double the percentage uncertainty of the diameter; D incorrectly doubles the uncertainty of multiple terms.

First, calculate the total binding energy of the reactant nucleus (U-235): \(235 \times 7.59\text{ MeV} = 1783.65\text{ MeV}\). Next, calculate the total binding energy of the product nuclei: For Ba-141: \(141 \times 8.33\text{ MeV} = 1174.53\text{ MeV}\). For Kr-92: \(92 \times 8.51\text{ MeV} = 782.92\text{ MeV}\). Total binding energy of the products: \(1174.53 + 782.92 = 1957.45\text{ MeV}\). The energy released is the difference: \(1957.45\text{ MeV} - 1783.65\text{ MeV} = 173.80\text{ MeV} \approx 174\text{ MeV}\).

B is correct (1 mark). Incorrect options: A calculates the change in binding energy per nucleon; C is a calculation error; D is the total binding energy of the products.

The mass of a uniform sphere of density \(\rho\) is \(M = \frac{4}{3}\pi R^3\rho\). Thus, \(R \propto M^{1/3}\). For the larger spheres of mass \(8M\), the radius is \(R' = 2R\). When in contact, the separation of their centers is \(d = 2R\) for the small spheres and \(d' = 2R' = 4R\) for the large spheres. Using Newton's law of gravitation, \(F = G\frac{M^2}{(2R)^2} = G\frac{M^2}{4R^2}\). For the larger spheres: \(F' = G\frac{(8M)^2}{(4R)^2} = G\frac{64M^2}{16R^2} = 4\left(G\frac{M^2}{R^2}\right)\). Since \(G\frac{M^2}{R^2} = 4F\), we find \(F' = 4 \times (4F) = 16F\).

D is correct (1 mark). Incorrect options: A assumes the force scales with the radius; B assumes the separation distance is kept constant; C assumes linear scaling with mass of a single sphere.

From the ideal gas relation \(\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}\), we can rearrange for the final pressure: \(p_2 = p_1 \times \left(\frac{V_1}{V_2}\right) \times \left(\frac{T_2}{T_1}\right)\). Given that \(p_1 = p\), \(\frac{T_2}{T_1} = 2\), and \(\frac{V_1}{V_2} = 3\), substituting these values yields \(p_2 = p \times 3 \times 2 = 6p\).

D is correct (1 mark). Incorrect options: A multiplies by the volume factor instead of dividing; B is an incorrect ratio rearrangement; C adds the multipliers instead of multiplying them.

First, find the central value: \(g = \frac{2 \times 2.00}{0.64^2} \approx 9.77\text{ m s}^{-2}\). The fractional uncertainty in \(h\) is \(\frac{0.02}{2.00} = 0.01\). The fractional uncertainty in \(t\) is \(\frac{0.04}{0.64} = 0.0625\). Since \(g \propto \frac{h}{t^2}\), the total fractional uncertainty is \(\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t} = 0.01 + 2(0.0625) = 0.135\). The absolute uncertainty is \(\Delta g = 9.77 \times 0.135 \approx 1.32\text{ m s}^{-2}\), which rounds to \(\pm 1.3\text{ m s}^{-2}\).

C is correct (1 mark). Incorrect options: A forgets to double the fractional uncertainty of time; B is a direct addition of absolute values or a rounding error; D uses incorrect propagation formulas.

The root-mean-square speed \(c_{\text{rms}}\) is given by \(c_{\text{rms}} = \sqrt{\frac{3RT}{M}}\), meaning \(c_{\text{rms}} \propto \sqrt{T}\) where \(T\) is in kelvins. Converting the temperatures: \(T_1 = 27 + 273 = 300\text{ K}\) and \(T_2 = 327 + 273 = 600\text{ K}\). The ratio is \(\frac{c_{\text{rms, final}}}{c_{\text{rms, initial}}} = \sqrt{\frac{600}{300}} = \sqrt{2} \approx 1.41\).

B is correct (1 mark). Incorrect options: A is a calculation error; C forgets the square root; D fails to convert temperatures from Celsius to Kelvin.

(a) The time period of a single oscillation is \(T = \frac{T_{20}}{20} = \frac{35.8}{20} = 1.79\text{ s}\).
Since the number of oscillations is a constant, the percentage uncertainty in \(T\) is identical to the percentage uncertainty in \(T_{20}\):
\%\Delta T = \frac{0.2}{35.8} \times 100\% \approx 0.559\% \approx 0.56\%\).

(b) Using the pendulum equation:
\(T = 2\pi \sqrt{\frac{L}{g}} \implies g = \frac{4\pi^2 L}{T^2}\)
\(g = \frac{4\pi^2 \times 0.800}{1.79^2} \approx 9.857\text{ m s}^{-2}\)

Next, determine the percentage uncertainty in \(g\):
\%\Delta L = \frac{0.005}{0.800} \times 100\% = 0.625\%
\%\Delta g = \%\Delta L + 2(\%\Delta T) = 0.625\% + 2(0.559\%) = 1.743\%

The absolute uncertainty in \(g\) is:
\Delta g = 9.857 \times 0.01743 \approx 0.17\text{ m s}^{-2}

Since the absolute uncertainty is expressed to 2 significant figures (or 1 s.f. as \(0.2\)), the value of \(g\) should match this level of precision:
\(g = 9.9 \pm 0.2\text{ m s}^{-2}\) (or \(9.86 \pm 0.17\text{ m s}^{-2}\)).

Part (a):
- Calculates single time period as \(1.79\text{ s}\) OR recognizes percentage uncertainty is unchanged [1 mark]
- Calculates percentage uncertainty as \(0.56\%\) (accept \(0.6\%\)) [1 mark]

Part (b):
- Correct calculation of \(g = 9.86\text{ m s}^{-2}\) (or \(9.9\text{ m s}^{-2}\)) [1 mark]
- Correct calculation of percentage uncertainty in \(L = 0.625\%\) [1 mark]
- Combines percentage uncertainties correctly to get \(\%\Delta g = 1.7\%\) [1 mark]
- Expresses absolute uncertainty and final value matching in decimal places (e.g., \(9.9 \pm 0.2\text{ m s}^{-2}\) or \(9.86 \pm 0.17\text{ m s}^{-2}\)) [1 mark]

(a) The resistivity \(\rho\) is given by:
\(\rho = \frac{R A}{l} = \frac{\pi d^2 R}{4l}\)
\(\rho = \frac{\pi \times (0.38 \times 10^{-3}\text{ m})^2 \times 12.4\ \Omega}{4 \times 1.54\text{ m}}\)
\(\rho = \frac{3.1416 \times 1.444 \times 10^{-7} \times 12.4}{6.16} = 9.13 \times 10^{-7}\ \Omega\text{ m}\) (which is approximately \(9.1 \times 10^{-7}\ \Omega\text{ m}\)).

(b) The formula for percentage uncertainty in \(\rho\) is:
\(\%\Delta \rho = \%\Delta R + 2(\%\Delta d) + \%\Delta l\)
\(\%\Delta R = \frac{0.4}{12.4} \times 100\% \approx 3.23\%\)
\(\%\Delta d = \frac{0.01}{0.38} \times 100\% \approx 2.63\%\)
\(\%\Delta l = \frac{0.02}{1.54} \times 100\% \approx 1.30\%\)

\(\%\Delta \rho = 3.23\% + 2(2.63\%) + 1.30\% = 9.79\% \approx 9.8\%\) (or \(10\%\)).

(c) The absolute uncertainty in \(\rho\) is:
\(\Delta \rho = 9.13 \times 10^{-7} \times 0.0979 \approx 0.89 \times 10^{-7}\ \Omega\text{ m} \approx 0.9 \times 10^{-7}\ \Omega\text{ m}\).

Therefore, the final value with absolute uncertainty is:
\(\rho = (9.1 \pm 0.9) \times 10^{-7}\ \Omega\text{ m}\).

Part (a):
- Recalls and uses formula \(\rho = \frac{\pi d^2 R}{4l}\) with correct units [1 mark]
- Arrives at \(9.13 \times 10^{-7}\ \Omega\text{ m}\) [1 mark]

Part (b):
- Uses addition of percentage uncertainties including the factor of 2 for diameter [1 mark]
- Obtains \(9.8\%\) (accept range \(9.7\%\) to \(10\%\)) [1 mark]

Part (c):
- Correctly calculates absolute uncertainty as \(0.9 \times 10^{-7}\ \Omega\text{ m}\) (or \(0.89 \times 10^{-7}\)) [1 mark]
- Formats final value with uncertainty correctly as \((9.1 \pm 0.9) \times 10^{-7}\ \Omega\text{ m}\) (or \((9 \pm 1) \times 10^{-7}\ \Omega\text{ m}\)) [1 mark]

(a) The difference in distance between two consecutive resonance positions corresponds to half a wavelength:
\(\frac{\lambda}{2} = L_2 - L_1\)
Thus, \(\lambda = 2(L_2 - L_1)\).
\(\lambda = 2(49.4\text{ cm} - 16.2\text{ cm}) = 2(33.2\text{ cm}) = 66.4\text{ cm} = 0.664\text{ m}\).

(b) The speed of sound is:
\(v = f \lambda = 512 \times 0.664 = 339.97\text{ m s}^{-1} \approx 340\text{ m s}^{-1}\).

To find the percentage uncertainty in \(v\), we first need the absolute uncertainty in \((L_2 - L_1)\):
\(\Delta(L_2 - L_1) = \Delta L_1 + \Delta L_2 = 0.2\text{ cm} + 0.2\text{ cm} = 0.4\text{ cm}\).

Now find percentage uncertainties:
\(\%\Delta \lambda = \%\Delta (L_2 - L_1) = \frac{0.4}{33.2} \times 100\% \approx 1.20\%\).
\(\%\Delta f = \frac{2}{512} \times 100\% \approx 0.39\%\).

Adding these values:
\(\%\Delta v = \%\Delta f + \%\Delta \lambda = 0.39\% + 1.20\% = 1.59\% \approx 1.6\%\).

Part (a):
- Explains that the distance between consecutive resonance points is half a wavelength [1 mark]
- Links this to the formula \(\lambda = 2(L_2 - L_1)\) [1 mark]
- Calculates \(\lambda = 0.664\text{ m}\) [1 mark]

Part (b):
- Calculates \(v = 340\text{ m s}^{-1}\) (or \(339.97\text{ m s}^{-1}\)) [1 mark]
- Finds absolute uncertainty in \((L_2 - L_1)\) is \(0.4\text{ cm}\) or percentage uncertainty in \(\lambda = 1.2\%\) [1 mark]
- Combines uncertainties correctly to find percentage uncertainty in speed \(v = 1.6\%\) (accept range \(1.59\%\) to \(1.6\%\)) [1 mark]

(a) To conserve nucleon number:
\(1 + 235 = 141 + Y + 3(1)\)
\(236 = 144 + Y \implies Y = 236 - 144 = 92\).

(b) Fast neutrons are released with high kinetic energies during fission. However, Uranium-235 has a much larger cross-section (higher probability) for capture/fission when interacting with slow-moving (thermal) neutrons. A moderator (such as water or graphite) is used. The fast neutrons collide elastically with the moderator atoms, transferring kinetic energy and slowing down to thermal equilibrium.

Part (a):
- Equates nucleon numbers on both sides: \(1 + 235 = 141 + Y + 3\) [1 mark]
- Deduces \(Y = 92\) [1 mark]

Part (b):
- Identifies that neutrons produced in fission are fast-moving/high energy [1 mark]
- Explains that Uranium-235 fission is highly probable only with slow/thermal neutrons [1 mark]
- Identifies the 'moderator' as the component used to slow down neutrons [1 mark]
- Explains that slowing down occurs via collisions between neutrons and moderator atoms/nuclei [1 mark]

First, calculate the total mass on the reactant side:
\(m_{\text{react}} = m(\text{Pu}-239) + m(\text{n}) = 239.05216\text{ u} + 1.00866\text{ u} = 240.06082\text{ u}\).

Next, calculate the total mass on the product side:
\(m_{\text{prod}} = m(\text{Xe}-137) + m(\text{Zr}-100) + 3 \times m(\text{n})\)
\(m_{\text{prod}} = 136.91157 + 99.91776 + 3(1.00866) = 236.82933 + 3.02598 = 239.85531\text{ u}\).

Now, calculate the mass defect \(\Delta m\):
\(\Delta m = m_{\text{react}} - m_{\text{prod}} = 240.06082 - 239.85531 = 0.20551\text{ u}\).

Convert this mass defect into energy in MeV:
\(E = 0.20551 \times 931.5\text{ MeV} = 191.43\text{ MeV}\).

Convert energy in MeV to Joules:
\(E = 191.4325 \times 10^6 \times 1.60 \times 10^{-19}\text{ J} = 3.067 \times 10^{-11}\text{ J} \approx 3.07 \times 10^{-11}\text{ J}\).

- Calculates total reactant mass as \(240.06082\text{ u}\) [1 mark]
- Calculates total product mass (including 3 neutrons) as \(239.85531\text{ u}\) [1 mark]
- Calculates mass defect \(\Delta m = 0.20551\text{ u}\) [1 mark]
- Converts mass defect to energy in MeV (\(191.4\text{ MeV}\)) [1 mark]
- Recalls/uses conversion factor \(1.60 \times 10^{-19}\text{ J/eV}\) [1 mark]
- Correctly evaluates energy as \(3.07 \times 10^{-11}\text{ J}\) (accept \(3.06 \times 10^{-11}\text{ J}\) to \(3.07 \times 10^{-11}\text{ J}\)) [1 mark]

(a) Using Newton's Law of Gravitation:
\(F = \frac{G M_A M_B}{r^2}\)
\(F = \frac{6.67 \times 10^{-11} \times 2.4 \times 10^{30} \times 4.8 \times 10^{30}}{(1.2 \times 10^{11})^2}\)
\(F = \frac{7.684 \times 10^{50}}{1.44 \times 10^{22}} = 5.336 \times 10^{28}\text{ N}\) (which is approximately \(5.3 \times 10^{28}\text{ N}\)).

(b) Let \(x\) be the distance from the center of Star \(A\) to the neutral point.
The gravitational field strength due to Star \(A\) is \(g_A = \frac{G M_A}{x^2}\).
The gravitational field strength due to Star \(B\) is \(g_B = \frac{G M_B}{(d-x)^2}\).
At the point where the net field strength is zero, \(g_A = g_B\):
\(\frac{G M_A}{x^2} = \frac{G M_B}{(d-x)^2} \implies \frac{M_A}{x^2} = \frac{M_B}{(d-x)^2}\)
Take the square root of both sides:
\(\frac{d-x}{x} = \sqrt{\frac{M_B}{M_A}} = \sqrt{\frac{4.8 \times 10^{30}}{2.4 \times 10^{30}}} = \sqrt{2} \approx 1.414\)
\(d - x = 1.414x \implies d = 2.414x\)
\(x = \frac{d}{2.414} = \frac{1.2 \times 10^{11}}{2.414} \approx 4.97 \times 10^{10}\text{ m}\) (or \(5.0 \times 10^{10}\text{ m}\)).

Part (a):
- Substitutes into \(F = \frac{G M_A M_B}{r^2}\) with \(G = 6.67 \times 10^{-11}\) [1 mark]
- Arrives at \(5.3 \times 10^{28}\text{ N}\) (minimum 3 sig figs shown in calculation step) [1 mark]

Part (b):
- Formulates the condition for zero field: \(\frac{G M_A}{x^2} = \frac{G M_B}{(d-x)^2}\) [1 mark]
- Correctly simplifies to linear algebraic equation (e.g., \(\frac{d-x}{x} = \sqrt{2}\)) [1 mark]
- Solves for \(x\) algebraically [1 mark]
- Correctly calculates \(5.0 \times 10^{10}\text{ m}\) (accept \(4.97 \times 10^{10}\text{ m}\)) [1 mark]

(a) Convert temperature to Kelvin:
\(T_1 = 27 + 273 = 300\text{ K}\).
Using the ideal gas law \(PV = nRT\):
\(n = \frac{PV}{RT} = \frac{3.0 \times 10^5 \times 0.025}{8.31 \times 300} = \frac{7500}{2493} \approx 3.01\text{ moles}\) (or \(3.0\text{ mol}\)).

(b) Since volume is constant, \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\):
\(T_2 = T_1 \frac{P_2}{P_1} = 300 \times \frac{4.0 \times 10^5}{3.0 \times 10^5} = 400\text{ K}\).
Convert back to degrees Celsius:
\(T_2 = 400 - 273 = 127\ ^\circ\text{C}\).

(c) The internal energy of a monatomic ideal gas is \(U = \frac{3}{2}nRT = \frac{3}{2}PV\).
The increase in internal energy is:
\(\Delta U = \frac{3}{2} V \Delta P = 1.5 \times 0.025 \times (4.0 \times 10^5 - 3.0 \times 10^5)\)
\(\Delta U = 1.5 \times 0.025 \times 1.0 \times 10^5 = 3750\text{ J}\) (or \(3800\text{ J}\) to 2 significant figures).

Part (a):
- Converts temperature to Kelvin (\(300\text{ K}\)) [1 mark]
- Correctly calculates \(n = 3.0\text{ mol}\) (accept \(3.01\text{ mol}\)) [1 mark]

Part (b):
- Applies pressure-temperature law with temperatures in Kelvin [1 mark]
- Obtains new temperature \(T_2 = 127\ ^\circ\text{C}\) [1 mark]

Part (c):
- Recalls and uses formula \(\Delta U = \frac{3}{2} n R \Delta T\) or \(\Delta U = \frac{3}{2} V \Delta P\) [1 mark]
- Obtains \(3750\text{ J}\) (or \(3800\text{ J}\)) [1 mark]

(a) Using the combined gas law equation \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\):
\(T_2 = T_1 \frac{P_2 V_2}{P_1 V_1}\)
\(T_2 = 290 \times \frac{4.8 \times 10^5 \times 1.2 \times 10^{-4}}{1.0 \times 10^5 \times 4.5 \times 10^{-4}}\)
\(T_2 = 290 \times \frac{57.6}{45.0} = 371.2\text{ K} \approx 371\text{ K}\) (or \(370\text{ K}\)).

(b) The mean kinetic energy of a molecule is given by:
\(E_k = \frac{3}{2} k T\)
Where Boltzmann constant \(k = 1.38 \times 10^{-23}\text{ J K}^{-1}\).
\(E_k = 1.5 \times 1.38 \times 10^{-23} \times 371.2 = 7.68 \times 10^{-21}\text{ J}\) (or \(7.7 \times 10^{-21}\text{ J}\) using 3 sig figs).

Part (a):
- Selects and uses the relationship \(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\) [1 mark]
- Rearranges equation correctly for \(T_2\) [1 mark]
- Calculates temperature \(T_2 = 371\text{ K}\) (accept range \(370\text{ K}\) to \(371\text{ K}\)) [1 mark]

Part (b):
- Recalls and uses formula \(E_k = \frac{3}{2} k T\) [1 mark]
- Substitutes Boltzmann constant \(k = 1.38 \times 10^{-23}\text{ J K}^{-1}\) [1 mark]
- Correctly calculates energy as \(7.7 \times 10^{-21}\text{ J}\) (accept \(7.68 \times 10^{-21}\text{ J}\) to \(7.70 \times 10^{-21}\text{ J}\)) [1 mark]

For part (a): The cross-sectional area of the wire is \(A = \frac{\pi d^2}{4} = \frac{\pi \times (0.38 \times 10^{-3})^2}{4} = 1.134 \times 10^{-7}\,\text{m}^2\). The resistivity is \(\rho = \frac{R A}{L} = \frac{4.50 \times 1.134 \times 10^{-7}}{1.200} = 4.25 \times 10^{-7}\,\Omega\,\text{m}\). For part (b): Percentage uncertainty in \(R\) is \(\frac{0.05}{4.50} \times 100\% = 1.11\%\). Percentage uncertainty in \(L\) is \(\frac{0.002}{1.200} \times 100\% = 0.17\%\). Percentage uncertainty in \(d\) is \(\frac{0.01}{0.38} \times 100\% = 2.63\%\). Total percentage uncertainty in \(\rho\) is \(\%\Delta \rho = \%\Delta R + \%\Delta L + 2\%\Delta d = 1.11\% + 0.17\% + 2(2.63\%) = 6.54\%\). The absolute uncertainty is \(\Delta \rho = 6.54\% \times 4.25 \times 10^{-7}\,\Omega\,\text{m} = 0.28 \times 10^{-7}\,\Omega\,\text{m}\). Quoted to 1 significant figure, this is \(0.3 \times 10^{-7}\,\Omega\,\text{m}\). Thus, the final value is \((4.3 \pm 0.3) \times 10^{-7}\,\Omega\,\text{m}\).

Part (a) [2 marks]: 1 mark for calculating the area correctly (\(1.13 \times 10^{-7}\,\text{m}^2\)). 1 mark for correct resistivity value (\(4.25 \times 10^{-7}\,\Omega\,\text{m}\) or \(4.3 \times 10^{-7}\,\Omega\,\text{m}\)). Part (b) [4 marks]: 1 mark for calculating individual percentage uncertainties in \(R\), \(L\), and \(d\). 1 mark for doubling the percentage uncertainty of diameter when summing. 1 mark for finding the total percentage uncertainty (\(6.5\%\)). 1 mark for finding the absolute uncertainty with correct significant figures (\(0.3 \times 10^{-7}\,\Omega\,\text{m}\) or \(0.28 \times 10^{-7}\,\Omega\,\text{m}\)).

For part (a): Mass of reactants = \(235.0439 + 1.0087 = 236.0526\,\text{u}\). Mass of products = \(140.9144 + 91.9262 + 3 \times 1.0087 = 235.8667\,\text{u}\). Mass defect \(\Delta m = 236.0526 - 235.8667 = 0.1859\,\text{u}\). Energy released = \(0.1859 \times 931.5\,\text{MeV} = 173.2\,\text{MeV}\). For part (b): Energy per fission in Joules = \(173.2 \times 10^6 \times 1.60 \times 10^{-19}\,\text{J} = 2.77 \times 10^{-11}\,\text{J}\). Total thermal energy per day = \(150 \times 10^6\,\text{W} \times 24 \times 3600\,\text{s} = 1.296 \times 10^{13}\,\text{J}\). Total number of fissions per day = \(\frac{1.296 \times 10^{13}}{2.77 \times 10^{-11}} = 4.68 \times 10^{23}\). Number of moles of U-235 = \(\frac{4.68 \times 10^{23}}{6.02 \times 10^{23}\,\text{mol}^{-1}} = 0.777\,\text{mol}\). Mass consumed = \(0.777 \times 235\,\text{g} = 183\,\text{g}\) or \(0.183\,\text{kg}\).

Part (a) [3 marks]: 1 mark for finding the correct mass of reactants and products. 1 mark for calculating the mass defect (\(0.1859\,\text{u}\)). 1 mark for converting to energy in MeV (\(173\,\text{MeV}\)). Part (b) [3 marks]: 1 mark for calculating the daily energy requirements (\(1.30 \times 10^{13}\,\text{J}\)). 1 mark for determining the number of fissions (\(4.7 \times 10^{23}\)). 1 mark for converting fissions to total mass of Uranium-235 (\(0.18\,\text{kg}\) or \(180\,\text{g}\)).

For part (a): Setting gravitational force equal to centripetal force: \(\frac{G M m}{r^2} = m \left(\frac{2\pi}{T}\right)^2 r\), which simplifies to \(M = \frac{4\pi^2 r^3}{G T^2}\). Converting the period into seconds: \(T = 29 \times 24 \times 3600 = 2.506 \times 10^6\,\text{s}\). Substituting the known values: \(M = \frac{4\pi^2 \times (2.5 \times 10^8)^3}{6.67 \times 10^{-11} \times (2.506 \times 10^6)^2} = 1.47 \times 10^{24}\,\text{kg}\), which is approximately \(1.5 \times 10^{24}\,\text{kg}\). For part (b): Gravitational field strength at the surface is given by \(g = \frac{G M}{R^2}\). Using \(M = 1.47 \times 10^{24}\,\text{kg}\): \(g = \frac{6.67 \times 10^{-11} \times 1.47 \times 10^{24}}{(4.8 \times 10^6)^2} = 4.26\,\text{N\,kg}^{-1}\). (Using the show-that value of \(1.5 \times 10^{24}\,\text{kg}\) yields \(g = 4.34\,\text{N\,kg}^{-1}\). Both values round to \(4.3\,\text{N\,kg}^{-1}\)).

Part (a) [3 marks]: 1 mark for Equating gravitational force to centripetal force and rearranging to make \(M\) the subject. 1 mark for converting 29 days correctly to seconds (\(2.51 \times 10^6\,\text{s}\)). 1 mark for substitution yielding \(1.47 \times 10^{24}\,\text{kg}\). Part (b) [3 marks]: 1 mark for using the correct formula \(g = \frac{G M}{R^2}\). 1 mark for substituting values (including exoplanet radius squared). 1 mark for final calculated value of \(4.3\,\text{N\,kg}^{-1}\) (allow range \(4.2\) to \(4.4\)).

The tensile stress is given by \(\sigma = \frac{F}{A} = \frac{4F}{\pi d^2}\). The percentage uncertainty in force is \(\frac{0.9}{45.0} \times 100\% = 2.0\%\). The percentage uncertainty in diameter is \(\frac{0.01}{0.50} \times 100\% = 2.0\%\). Since the diameter is squared in the formula, its contribution to the uncertainty is \(2 \times 2.0\% = 4.0\%\). Therefore, the total percentage uncertainty in the tensile stress is \(2.0\% + 4.0\% = 6.0\%\).

1 mark for the correct answer D. (Method: Calculate percentage uncertainty of force = 2.0%, percentage uncertainty of diameter = 2.0%, compound them as % uncertainty in stress = % uncertainty in F + 2 * (% uncertainty in d) = 6.0%)

First, convert the temperature to kelvin: \(T_1 = 27 + 273.15 \approx 300\text{ K}\). For an ideal gas at constant volume, \(P_1 / T_1 = P_2 / T_2\). Since \(P_2 = 2P_1\), the final temperature is \(T_2 = 2 T_1 = 600\text{ K}\). Converting back to degrees Celsius gives \(T_2 = 600 - 273 = 327^\circ\text{C}\).

1 mark for the correct answer C. (Method: Convert 27 degrees Celsius to 300 K, double the absolute temperature to 600 K, and convert back to 327 degrees Celsius)

The initial gravitational force is given by Newton's law: \(F = \frac{G M m}{r^2}\). When both masses are doubled and the distance is halved, the new force is \(F' = \frac{G (2M) (2m)}{(r/2)^2} = \frac{4 G M m}{r^2 / 4} = 16 \frac{G M m}{r^2} = 16F\).

1 mark for the correct answer C. (Method: Express new force in terms of old force by substituting 2M, 2m, and 0.5r into Newton's law of gravitation, yielding a factor of 16)

Using the conservation of nucleon number (mass number): \(235 + 1 = 141 + X + 3(1)\). This simplifies to \(236 = 144 + X\), which gives \(X = 236 - 144 = 92\).

1 mark for the correct answer B. (Method: Apply conservation of mass number where LHS total nucleons is 236, RHS total nucleons is 141 + X + 3, and solve for X)

Taking multiple measurements at different positions and orientations and calculating their mean reduces the effect of random fluctuations (random errors) and any physical variations in the wire's thickness. It does not eliminate systematic errors (such as zero error) or change the instrument's intrinsic resolution.

1 mark for the correct answer C. (Method: Recall that taking multiple readings and finding the average reduces the impact of random errors but has no effect on systematic errors or instrument resolution)

From the ideal gas equation: \(P V = N k_B T\). Rearranging for the number of molecules: \(N = \frac{P V}{k_B T}\). Substituting the given values: \(N = \frac{1.5 \times 10^5 \text{ Pa} \times 2.4 \times 10^{-3} \text{ m}^3}{1.38 \times 10^{-23} \text{ J K}^{-1} \times 290 \text{ K}} = \frac{360}{4.002 \times 10^{-21}} \approx 8.9955 \times 10^{22} \approx 9.0 \times 10^{22}\).

1 mark for the correct answer C. (Method: Rearrange ideal gas equation PV = N k_B T to solve for N, substitute values correctly to find N = 9.0 * 10^22)

For a circular orbit, the centripetal force is provided by the gravitational force: \(\frac{m v^2}{r} = \frac{G M m}{r^2}\). This simplifies to \(v = \sqrt{\frac{G M}{r}}\). Therefore, orbital speed is inversely proportional to the square root of the orbital radius, \(v \propto \frac{1}{\sqrt{r}}\). When \(r\) increases from \(R\) to \(4R\), the speed changes by a factor of \(\frac{1}{\sqrt{4}} = 0.50\).

1 mark for the correct answer B. (Method: Equate centripetal force to gravitational force to find the relationship v proportional to r^(-0.5), and determine that increasing r by a factor of 4 reduces v by a factor of 2)

First, calculate \(0.5\%\) of the reading: \(0.005 \times 12.00\text{ V} = 0.06\text{ V}\). The term '2 digits' refers to twice the resolution of the instrument: \(2 \times 0.01\text{ V} = 0.02\text{ V}\). Adding these two contributions together gives the total absolute uncertainty: \(0.06\text{ V} + 0.02\text{ V} = 0.08\text{ V}\).

1 mark for the correct answer B. (Method: Calculate 0.5% of 12.00 V = 0.06 V, determine 2 digits = 0.02 V, sum them to get 0.08 V)

First, identify the anomaly in the dataset: 0.46 mm is significantly higher than the other values and should be discarded. The remaining uncorrected values are 0.38 mm, 0.37 mm, 0.39 mm, 0.38 mm, and 0.37 mm. To correct for the systematic zero error of +0.01 mm, subtract 0.01 mm from each reading. The corrected values are 0.37 mm, 0.36 mm, 0.38 mm, 0.37 mm, and 0.36 mm. The mean of these corrected values is (0.37 + 0.36 + 0.38 + 0.37 + 0.36) / 5 = 0.368 mm, which rounds to 0.37 mm. The absolute uncertainty is half the range of these corrected values: (0.38 - 0.36) / 2 = 0.01 mm. Therefore, the corrected mean diameter is \((0.37 \pm 0.01)\text{ mm}\).

1 mark for the correct answer A. Discarding the anomalous value (0.46 mm), subtracting the zero error of 0.01 mm from the remaining values, and calculating the uncertainty as half the range.

First, calculate the percentage uncertainty in the measured time \(t\): \(\% \Delta t = \frac{0.2}{30.0} \times 100\% = 0.67\%\). Since the period \(T\) is \(t/20\), its percentage uncertainty is also \(0.67\%\). In the formula for \(g\), the variable \(T\) is raised to the power of 2, so its percentage uncertainty contribution is doubled: \(2 \times 0.67\% = 1.33\%\). The overall percentage uncertainty in \(g\) is found by summing the individual percentage uncertainties: \(\% \Delta g = \% \Delta I + \% \Delta m + \% \Delta h + 2 \times \% \Delta T = 3.0\% + 1.5\% + 2.0\% + 1.33\% = 7.83\%\), which rounds to \(7.8\%\).

1 mark for the correct answer C. Correct calculation of the percentage uncertainty of T, doubling it for the T^2 term, and summing all component percentage uncertainties.

Using the conservation laws: for atomic number (proton number), \(92 + 0 = 56 + X + 3(0)\) which gives \(X = 92 - 56 = 36\). For mass number (nucleon number), \(235 + 1 = 141 + Y + 3(1)\) which gives \(236 = 144 + Y\), so \(Y = 92\). The energy released is calculated using the mass defect: \(E = 0.186\text{ u} \times 931.5\text{ MeV/u} = 173.3\text{ MeV}\), which rounds to \(173\text{ MeV}\).

1 mark for the correct answer A. Verifying nucleon number conservation, proton number conservation, and calculating the energy released using the MeV conversion factor.

The gravitational field strength at the surface of a spherical planet of mass \(M\) and radius \(R\) is given by \(g = \frac{GM}{R^2}\). Since mass is density times volume, \(M = \rho \times \frac{4}{3}\pi R^3\). Substituting this into the formula for \(g\) yields \(g = \frac{G (\rho \times \frac{4}{3}\pi R^3)}{R^2} = \frac{4}{3}\pi G \rho R\). Since both planets have the same uniform density, the surface gravitational field strength is directly proportional to the radius of the planet. Therefore, if planet Q has twice the radius of planet P, its surface gravitational field strength is \(2 g_{\text{P}}\).

1 mark for the correct answer C. Deducing that surface gravity is proportional to radius when density is constant, and scaling it accordingly.

According to kinetic theory, the mean square speed of ideal gas molecules is directly proportional to the absolute temperature: \(c^2 \propto T\). From the ideal gas equation, \(p V = n R T\), which means \(T \propto p V\). Therefore, the mean square speed is directly proportional to the product of pressure and volume: \(c^2 \propto p V\). The new product of pressure and volume is \(0.8p \times 1.5V = 1.2 pV\). Since the product \(pV\) has increased by a factor of 1.2, the absolute temperature and therefore the mean square speed of the molecules also increase by a factor of 1.2.

1 mark for the correct answer B. Linking mean square speed to absolute temperature and using the ideal gas equation to find the new temperature factor of 1.2.

Let's calculate the current percentage uncertainties in each variable: \(\% \Delta R = \frac{0.1}{4.5} \times 100\% = 2.22\%\). \(\% \Delta d = \frac{0.01}{0.35} \times 100\% = 2.86\%\), which contributes \(2 \times 2.86\% = 5.72\%\) to the total percentage uncertainty because \(d\) is squared. \(\% \Delta L = \frac{0.002}{1.250} \times 100\% = 0.16\%\). The total percentage uncertainty is approximately \(8.10\%\). Analyzing the options: Option A reduces the \(L\) uncertainty contribution by 0.08\%. Option B reduces the \(R\) uncertainty contribution by 1.11\% (from 2.22\% to 1.11\%). Option C reduces the \(d\) uncertainty contribution by 2.86\% (from 5.72\% to 2.86\%). Option D only reduces random error but does not reduce instrumental limits. Therefore, Option C gives the most significant reduction.

1 mark for the correct answer C. Comparing the percentage uncertainty contributions of each parameter, identifying the diameter's squared contribution as dominant, and calculating the effect of each improvement.

(a) The resistivity is given by \(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\). Substituting the measured values:
\(A = \frac{\pi (0.46 \times 10^{-3}\text{ m})^2}{4} = 1.662 \times 10^{-7}\text{ m}^2\)
\(\rho = \frac{3.12 \times 1.662 \times 10^{-7}}{0.850} = 6.10 \times 10^{-7}\text{ }\Omega\text{ m}\)

(b) Percentage uncertainties:
\%\Delta L = \frac{0.002}{0.850} \times 100 = 0.235\%
\%\Delta R = \frac{0.05}{3.12} \times 100 = 1.603\%
\%\Delta d = \frac{0.01}{0.46} \times 100 = 2.174\%
Since \(\rho \propto d^2\), the percentage uncertainty in \(d^2\) is \(2 \times 2.174\% = 4.348\%\).
Total percentage uncertainty in \(\rho\):
\%\Delta \rho = \%\Delta R + 2(\%\Delta d) + \%\Delta L = 1.603\% + 4.348\% + 0.235\% = 6.186\%
Absolute uncertainty:
\(\Delta \rho = 6.10 \times 10^{-7} \times 0.06186 = 0.377 \times 10^{-7}\text{ }\Omega\text{ m}\).
Rounding to 1 significant figure gives \(0.4 \times 10^{-7}\text{ }\Omega\text{ m}\).

(c) Take diameter measurements at multiple different positions and orientations along the wire, and then calculate an average diameter.

(a) [3 marks]
- Formula: \(\rho = \frac{R \pi d^2}{4 L}\) [1 mark]
- Calculation of cross-sectional area: \(1.66 \times 10^{-7}\text{ m}^2\) [1 mark]
- Correct value of resistivity: \(6.1 \times 10^{-7}\text{ }\Omega\text{ m}\) [1 mark]

(b) [4.5 marks]
- Calculation of percentage uncertainties in L (0.24%) and R (1.60%) [1 mark]
- Doubling percentage uncertainty in d: \(2 \times 2.17\% = 4.35\%\) [1.5 marks]
- Summing percentage uncertainties to get 6.19% (or 6.2%) [1 mark]
- Correct calculation of absolute uncertainty as \(0.38 \times 10^{-7}\text{ }\Omega\text{ m}\) or \(0.4 \times 10^{-7}\text{ }\Omega\text{ m}\) [1 mark]

(c) [1 mark]
- Measure at different points along the length and at different orientations and average the readings [1 mark]

(a) Mean time \(t = \frac{31.2 + 31.4 + 31.0}{3} = 31.2\text{ s}\). Range of time measurements = \(31.4 - 31.0 = 0.4\text{ s}\). Absolute uncertainty = \(\frac{\text{Range}}{2} = \frac{0.4}{2} = 0.2\text{ s}\). Since the half-range (0.2 s) is larger than the stopwatch's resolution (0.1 s), the absolute uncertainty is 0.2 s.

(b) Period \(T = \frac{31.2}{20} = 1.56\text{ s}\). Therefore, \(T^2 = (1.56)^2 = 2.43\text{ s}^2\). The percentage uncertainty in \(t\) is \(\frac{0.2}{31.2} \times 100\% = 0.641\%\). Since \(T^2 = (t/20)^2\), the percentage uncertainty in \(T^2\) is twice the percentage uncertainty in \(t\): \(2 \times 0.641\% = 1.28\% \approx 1.3\%\).

(c) For \(T^2\), the absolute uncertainty is \(2.43 \times 0.0128 \approx 0.03\text{ s}^2\). The student should draw a vertical line through the point extending \(0.03\text{ s}^2\) above and below it. For \(L\), the absolute uncertainty is \(0.005\text{ m}\). The student should draw a horizontal line through the point extending \(0.005\text{ m}\) to the left and right.

(a) [2.5 marks]
- Mean time calculated as 31.2 s [0.5 marks]
- Absolute uncertainty based on half-range: \(\Delta t = \frac{31.4 - 31.0}{2} = 0.2\text{ s}\) [1 mark]
- Explanation: the half-range value (0.2 s) is used because it is greater than the instrument resolution (0.1 s) [1 mark]

(b) [4 marks]
- Correct calculation of \(T^2 = 2.43\text{ s}^2\) [1 mark]
- Percentage uncertainty of the time measurement = \(0.64\%\) [1 mark]
- Recognises that percentage uncertainty of \(T^2\) is double that of \(T\) [1 mark]
- Correct final percentage uncertainty: \(1.3\%\) (accept 1.28%) [1 mark]

(c) [2 marks]
- Vertical error bar of length \(\pm 0.03\text{ s}^2\) [1 mark]
- Horizontal error bar of length \(\pm 0.005\text{ m}\) [1 mark]

(a) Random errors cause successive readings to vary unpredictably around the true value; a cause is parallax error in reading the meniscus of the acid bead against the scale. Systematic errors cause all readings to differ from the true value by a constant amount; a cause is an offset where the ruler's zero mark does not align with the closed end of the capillary tube.

(b) Since \(V = A h\), if the cross-sectional area \(A\) varies along the tube, the length \(h\) is no longer directly proportional to volume \(V\). This introduces a systematic error because the volume calculated assuming constant \(A\) will be consistently off. The student can calibrate the tube by placing a small mercury or oil drop of known volume inside and measuring its length at various positions along the tube, creating a calibration curve of local cross-sectional area against position.

(c) A higher resolution sensor only reduces readability uncertainty, not systematic errors. If the digital sensor has not been calibrated, it may have a larger systematic error than the thermometer. Furthermore, temperature gradients in the water bath may mean the sensor is not at the exact temperature of the gas, meaning the measurement is limited by thermal equilibrium rather than sensor resolution.

(a) [3.5 marks]
- Unpredictable variations vs constant offset definition [1.5 marks]
- Valid example of random error (e.g., parallax in reading meniscus, temperature fluctuations) [1 mark]
- Valid example of systematic error (e.g., zero error on scale, ruler starting above base) [1 mark]

(b) [3 marks]
- Explanation that varying area means length is not proportional to volume [1 mark]
- Method of calibration: introduction of a small droplet of known volume [1 mark]
- Recording its length at different positions to map the local area [1 mark]

(c) [2 marks]
- Higher resolution does not guarantee higher accuracy due to potential systematic calibration errors [1 mark]
- The temperature of the water around the sensor may differ from the actual temperature of the gas (thermal gradient/lack of thermal equilibrium) [1 mark]

(a) Using \(h = \frac{1}{2} g t^2\), we find \(g = \frac{2h}{t^2} = \frac{2 \times 1.250}{(0.504)^2} = 9.8419\text{ m s}^{-2}\).

(b) Percentage uncertainties:
\%\Delta h = \frac{0.002}{1.250} \times 100 = 0.16\%
\%\Delta t = \frac{0.003}{0.504} \times 100 = 0.595\%
\%\Delta g = \%\Delta h + 2(\%\Delta t) = 0.16\% + 2(0.595\%) = 1.35\%
Rounding to two significant figures, this is \(1.4\%\).

(c) Releasing the ball bearing by hand may impart a small downward initial velocity \(u > 0\) rather than starting from rest. This means the actual time of fall is less than the true free-fall time from rest. Since \(g \propto \frac{1}{t^2}\), a smaller recorded time \(t\) leads to an calculated value of \(g\) that is larger than the true value, resulting in an overestimate.

(a) [2.5 marks]
- Relates formula: \(g = \frac{2h}{t^2}\) [1 mark]
- Correct substitution: \(g = \frac{2 \times 1.250}{(0.504)^2}\) [0.5 marks]
- Final value and units: \(9.84\text{ m s}^{-2}\) [1 mark]

(b) [3 marks]
- \%\Delta h = 0.16\% and \%\Delta t = 0.60\% [1 mark]
- Sums uncertainties correctly: \(\%\Delta g = \%\Delta h + 2 \times \%\Delta t\) [1 mark]
- Final percentage uncertainty: \(1.4\%\) (accept 1.35%) [1 mark]

(c) [3 marks]
- Identifies that manual release may impart an initial downward velocity (or force) [1 mark]
- Explains that this causes the ball to travel distance \(h\) in less time (shorter \(t\)) [1 mark]
- Concludes that because \(g \propto \frac{1}{t^2}\), a smaller \(t\) leads to an overestimate of \(g\) [1 mark]

(a) The balanced equation is:
\({}^{235}_{92}\text{U} + {}^{1}_{0}\text{n} \rightarrow {}^{141}_{56}\text{Ba} + {}^{92}_{36}\text{Kr} + 3\, {}^{1}_{0}\text{n}\)

(b) Initial mass: \(M_i = 234.9935 + 1.00867 = 236.00217\text{ u}\)
Final mass: \(M_f = 140.8831 + 91.9064 + 3(1.00867) = 235.81551\text{ u}\)
Mass defect: \(\Delta m = M_i - M_f = 236.00217 - 235.81551 = 0.18666\text{ u}\)
Energy released: \(E = 0.18666 \times 931.5\text{ MeV} = 173.87\text{ MeV} \approx 173.9\text{ MeV}\).
In Joules: \(E = 173.87 \times 10^6 \times 1.602 \times 10^{-19}\text{ J} = 2.785 \times 10^{-11}\text{ J} \approx 2.79 \times 10^{-11}\text{ J}\).

(c) Control rods are made of materials (such as boron or cadmium) that readily absorb neutrons. By inserting them deeper into the core, they absorb more neutrons, preventing them from inducing further fission reactions, which reduces the rate of energy release. Pulling them out increases the fission rate.

(a) [2.5 marks]
- Correct left side: \({}^{235}_{92}\text{U} + {}^{1}_{0}\text{n}\) [1 mark]
- Correct products including 3 neutrons: \({}^{141}_{56}\text{Ba} + {}^{92}_{36}\text{Kr} + 3\, {}^{1}_{0}\text{n}\) [1 mark]
- Fully balanced proton and mass numbers [0.5 marks]

(b) [4 marks]
- Calculate mass defect \(\Delta m = 0.18666\text{ u}\) [1 mark]
- Energy in MeV: \(173.9\text{ MeV}\) [1 mark]
- Usage of conversion factor \(1.60 \times 10^{-19}\text{ J/eV}\) [1 mark]
- Energy in Joules: \(2.79 \times 10^{-11}\text{ J}\) [1 mark]

(c) [2 marks]
- States that control rods absorb neutrons [1 mark]
- Explains that insertion depth regulates the rate of the chain reaction by controlling the neutron population [1 mark]

(a) Newton's law of gravitation states that the attractive force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

(b) Using \(F = \frac{G M_1 M_2}{r^2}\):
\(F = \frac{6.67 \times 10^{-11} \times 3.20 \times 10^{30} \times 1.60 \times 10^{30}}{(4.50 \times 10^{10})^2} = 1.6875 \times 10^{29}\text{ N} \approx 1.69 \times 10^{29}\text{ N}\).

(c) Let \(x\) be the distance from Star 1. At the zero-field point:
\(\frac{G M_1}{x^2} = \frac{G M_2}{(r-x)^2}\)
\(\frac{M_1}{x^2} = \frac{M_2}{(r-x)^2} \implies \frac{r-x}{x} = \sqrt{\frac{M_2}{M_1}}\)
Since \(\frac{M_2}{M_1} = \frac{1.60}{3.20} = 0.500\):
\(\frac{r-x}{x} = \sqrt{0.500} \approx 0.7071\)
\(r - x = 0.7071 x \implies r = 1.7071 x\)
\(x = \frac{4.50 \times 10^{10}}{1.7071} = 2.636 \times 10^{10}\text{ m} \approx 2.64 \times 10^{10}\text{ m}\).

(a) [2 marks]
- Force proportional to product of masses [1 mark]
- Force inversely proportional to distance squared [1 mark]

(b) [2.5 marks]
- Use of \(F = \frac{G M_1 M_2}{r^2}\) [1 mark]
- Substitution of correct values [0.5 marks]
- Correct calculation with units: \(1.69 \times 10^{29}\text{ N}\) [1 mark]

(c) [4 marks]
- Equates gravitational field strengths: \(\frac{G M_1}{x^2} = \frac{G M_2}{(r-x)^2}\) [1 mark]
- Takes square root: \(\frac{r-x}{x} = \sqrt{0.50}\) [1 mark]
- Solves algebraically for \(x\) [1 mark]
- Correct final answer: \(2.64 \times 10^{10}\text{ m}\) (accept \(2.6 \times 10^{10}\text{ m}\)) [1 mark]

(a) Using the ideal gas equation \(P V = n R T\):
\(T = 22.0 + 273.15 = 295.15\text{ K}\)
\(P = \frac{n R T}{V} = \frac{1.40 \times 8.31 \times 295.15}{0.0350} = 9.811 \times 10^4\text{ Pa} \approx 9.81 \times 10^4\text{ Pa}\).

(b) The new temperature is \(T = 115 + 273.15 = 388.15\text{ K}\).
The mean kinetic energy of a single atom is given by:
\(E_k = \frac{3}{2} k_B T\)
\(E_k = 1.5 \times 1.38 \times 10^{-23} \times 388.15 = 8.035 \times 10^{-21}\text{ J} \approx 8.04 \times 10^{-21}\text{ J}\).

(c) Raising the temperature increases the mean kinetic energy and thus the average speed of the helium atoms. This has two effects:
1. The atoms collide with the container walls more frequently.
2. Each collision involves a larger change in momentum, meaning a larger average force is exerted per collision.
Since pressure is force per unit area, both factors contribute to an increase in the pressure exerted on the walls.

(a) [2.5 marks]
- Temperature conversion to Kelvin: \(295.15\text{ K}\) [0.5 marks]
- Rearranging and substituting into \(P V = n R T\) [1 mark]
- Correct pressure: \(9.81 \times 10^4\text{ Pa}\) [1 mark]

(b) [3 marks]
- Temperature conversion: \(388.15\text{ K}\) [0.5 marks]
- Use of \(E_k = \frac{3}{2} k_B T\) with correct Boltzmann constant [1 mark]
- Correct calculation: \(8.04 \times 10^{-21}\text{ J}\) (accept \(8.0 \times 10^{-21}\text{ J}\)) [1.5 marks]

(c) [3 marks]
- Higher temperature means higher mean speed/kinetic energy of molecules [1 mark]
- More frequent collisions with the walls [1 mark]
- Greater momentum change per collision, hence greater force exerted on the walls [1 mark]

The volume is given by \(V = \pi r^2 L\). Using the rules of combining uncertainties for products and powers: \(\frac{\Delta V}{V} = 2 \frac{\Delta r}{r} + \frac{\Delta L}{L}\). The fractional uncertainty in \(r\) is \(\frac{\Delta r}{r} = \frac{0.05}{2.00} = 0.025\) (or \(2.5\%\)). The fractional uncertainty in \(L\) is \(\frac{\Delta L}{L} = \frac{0.2}{10.0} = 0.02\) (or \(2.0\%\)). Combining these terms gives: \(\frac{\Delta V}{V} = 2 \times 2.5\% + 2.0\% = 5.0\% + 2.0\% = 7.0\%\).

1 mark for correctly identifying that the percentage uncertainty of the radius must be doubled and added to the percentage uncertainty of the length, resulting in 7.0% (choice C).

Precision is indicated by how close the measurements are to one another. The range is small (4.56 V - 4.54 V = 0.02 V), which represents high precision. Accuracy is indicated by how close the average measurement is to the true value. The average value is around 4.55 V, which is far from the true value of 4.80 V, representing low accuracy due to a systematic error.

1 mark for identifying high precision and low accuracy (choice C).

Since \(y = k x^2\), the gradient is equal to \(k\). The absolute uncertainty in \(k\) is equal to the absolute uncertainty in the gradient, which is \(|m_{\text{best}} - m_{\text{worst}}| = |4.25 - 3.95| = 0.30\text{ m}^{-1}\).

1 mark for calculating the absolute difference as 0.30 m\(^{-1}\) (choice B).

The mean time is \(t_{\text{mean}} = \frac{28.4 + 28.1 + 28.3 + 28.5 + 28.2}{5} = 28.3\text{ s}\). The uncertainty is half the range: \(\Delta t = \frac{\text{Range}}{2} = \frac{28.5 - 28.1}{2} = 0.2\text{ s}\). Thus, the value is \((28.3 \pm 0.2)\text{ s}\).

1 mark for calculating the mean as 28.3 s and the uncertainty as 0.2 s (choice B).

The temperature rise is \(\Delta \theta = 61.5 - 21.5 = 40.0\text{ }^\circ\text{C}\). For addition and subtraction, absolute uncertainties are added: \(\Delta(\Delta \theta) = 0.5 + 0.5 = 1.0\text{ }^\circ\text{C}\). The percentage uncertainty is \(\frac{1.0}{40.0} \times 100\% = 2.5\%\).

1 mark for calculating absolute uncertainty of 1.0 °C and percentage uncertainty of 2.5% (choice C).

Conservation of nucleon number (A) gives: \(235 + 1 = 141 + A + 3 \times 1 \implies 236 = 144 + A \implies A = 92\). Conservation of proton number (Z) gives: \(92 + 0 = 56 + Z + 3 \times 0 \implies 92 = 56 + Z \implies Z = 36\). This corresponds to krypton-92 (\(^{92}_{36}\text{Kr}\)).

1 mark for calculating both nucleon number and proton number correctly to identify choice A.

Let \(x\) be the distance of R from P. The distance from R to Q is \(d - x\). Equating the gravitational forces: \(\frac{G M m}{x^2} = \frac{G (4M) m}{(d - x)^2}\). Simplifying: \(\frac{1}{x^2} = \frac{4}{(d - x)^2}\). Taking the square root of both sides: \(\frac{1}{x} = \frac{2}{d - x}\). Rearranging gives: \(d - x = 2x \implies 3x = d \implies x = \frac{d}{3}\).

1 mark for setting up the force equality and solving for \(x\) to get \(\frac{d}{3}\) (choice C).

The ideal gas law indicates \(P \propto T\) at a constant volume. Converting temperatures to Kelvin: \(T_1 = 27 + 273.15 = 300.15\text{ K}\) and \(T_2 = 327 + 273.15 = 600.15\text{ K}\). Since the temperature in Kelvin doubles, the pressure also doubles: \(P_2 = 2 P_0\).

1 mark for using Kelvin temperatures to find that the pressure doubles (choice C).

The density is given by \(\rho = \frac{m}{V} = \frac{4m}{\pi d^2 l}\). The percentage uncertainty in \(\rho\) is the sum of the percentage uncertainties of the contributing quantities (with the diameter's percentage uncertainty multiplied by 2 due to its exponent of 2): \%\Delta \rho = \%\Delta m + 2\%\Delta d + \%\Delta l. The percentage uncertainties are: \%\Delta m = \frac{0.002}{0.400} \times 100 = 0.5\%, \%\Delta d = \frac{0.02}{0.80} \times 100 = 2.5\%, and \%\Delta l = \frac{0.5}{120.0} \times 100 = 0.417\%. Adding these together gives: 0.5\% + 2(2.5\%) + 0.417\% = 5.917\%, which rounds to 5.9\%.

1 mark for the correct answer C. Method: Calculate individual percentage uncertainties (0.5% for mass, 2.5% for diameter, 0.42% for length), double the diameter's percentage uncertainty, and sum them up.

The binding energy of a nucleus is equal to its nucleon number multiplied by its binding energy per nucleon. Total initial binding energy = \(235 \times 7.59\text{ MeV} = 1783.65\text{ MeV}\) (the free neutron has zero binding energy). Total final binding energy = \((141 \times 8.33) + (92 \times 8.51)\text{ MeV} = 1174.53 + 782.92 = 1957.45\text{ MeV}\). Energy released = final binding energy - initial binding energy = \(1957.45 - 1783.65 = 173.8\text{ MeV}\), which is approximately 174 MeV.

1 mark for the correct answer A. Method: Calculate initial total binding energy and final total binding energy, then find the difference.

The gravitational field strength at the surface is given by \(g = \frac{GM}{R^2}\). Since mass \(M = \rho V = \rho \frac{4}{3}\pi R^3\), substituting \(M\) into the equation for \(g\) gives: \(g = \frac{G \rho \frac{4}{3}\pi R^3}{R^2} = \frac{4}{3}\pi G \rho R\). Since both planets have the same average density, \(g\) is directly proportional to the radius \(R\). Since Planet Y has three times the radius of Planet X, the gravitational field strength at its surface is three times greater.

1 mark for the correct answer C. Method: Relate mass to density and radius, show that \(g\) is proportional to \(R\) for constant density, and apply the scale factor of 3.

Using the pressure law for an ideal gas at constant volume: \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\), where temperature must be in Kelvin. Initial temperature in Kelvin: \(T_1 = 27 + 273 = 300\text{ K}\). Solving for \(T_2\): \(T_2 = T_1 \times \frac{P_2}{P_1} = 300\text{ K} \times \frac{2.0 \times 10^5\text{ Pa}}{1.2 \times 10^5\text{ Pa}} = 500\text{ K}\). Converting back to Celsius: \(T_2 = 500 - 273 = 227^\circ\text{C}\).

1 mark for the correct answer B. Method: Convert initial temperature to Kelvin, apply Gay-Lussac's Law to find the final temperature in Kelvin, and convert back to Celsius.

Rearranging the equation for \(g\) gives \(g = \frac{2h}{t^2}\). The percentage uncertainty in \(g\) is given by: \%\Delta g = \%\Delta h + 2\%\Delta t = 1.5\% + 2(2.0\%) = 1.5\% + 4.0\% = 5.5\%.

1 mark for the correct answer C. Method: Identify the formula for combining uncertainties, recognizing that the power of 2 of \(t\) doubles its percentage uncertainty contribution.

An effective moderator needs to slow down fast neutrons through elastic collisions. In elastic collisions, the maximum kinetic energy is transferred when the target nucleus has a mass similar to that of the incoming neutron (i.e. a very low nucleon number, such as hydrogen or carbon). If the nucleon number were high, the neutron would simply bounce off with little loss of kinetic energy.

1 mark for correct answer B. Explanation: Low mass nuclei optimize kinetic energy transfer per elastic collision, slowing neutrons down rapidly without absorbing them.

The gravitational force provides the centripetal force for a circular orbit: \(\frac{GMm}{r^2} = \frac{mv^2}{r}\), which simplifies to \(v = \sqrt{\frac{GM}{r}}\). Therefore, the orbital speed is inversely proportional to the square root of the orbital radius: \(v \propto \frac{1}{\sqrt{r}}\). Taking the ratio: \(\frac{v_P}{v_Q} = \sqrt{\frac{r_Q}{r_P}} = \sqrt{\frac{4r_P}{r_P}} = \sqrt{4} = 2.0\).

1 mark for the correct answer C. Method: Derive the relationship \(v \propto \frac{1}{\sqrt{r}}\), substitute the given ratio of radii, and find the speed ratio.

(a) Two assumptions:
1. All kinetic energy of the wind passing through the swept area is assumed to be fully extractable / the wind is brought to a complete standstill behind the turbine (or that wind speed is completely uniform across the blade swept area).
2. No energy is lost due to friction in the bearings, turbulence around the blades, or electrical inefficiencies in the generator.

(b)(i) Swept area:
\(A = \pi R^2 = \pi \times (35)^2 = 3848.5\text{ m}^2\)

Kinetic energy per second (theoretical power available in the wind):
\(P_{\text{wind}} = \frac{1}{2} \rho A v^3\)
\(P_{\text{wind}} = 0.5 \times 1.2 \times 3848.5 \times 12^3 = 3.99 \times 10^6\text{ W}\) (or \(3.99\text{ MJ s}^{-1}\))

(b)(ii) Efficiency:
\(\eta = \frac{P_{\text{out}}}{P_{\text{wind}}} = \frac{1.5 \times 10^6\text{ W}}{3.99 \times 10^6\text{ W}} = 0.376\)
Efficiency \(= 37.6\% \approx 38\%\)

(c) If the downstream wind speed were zero, air would pile up behind the turbine and block subsequent airflow, preventing any continuous transfer of kinetic energy. The air must retain some velocity to move out of the way, which theoretically limits the maximum proportion of kinetic energy that can be extracted to approximately \(59\%\) (the Betz limit).

(a) 1 mark for each valid assumption up to a maximum of 2 marks.
- Accept: Wind is uniform / air is incompressible / no mechanical frictional losses / wind is brought to rest.

(b)(i)
- Use of \(A = \pi R^2\) to get \(3850\text{ m}^2\) [1 mark]
- Correct substitution into \(\frac{1}{2}\rho A v^3\) [1 mark]
- Correct final value \(3.99 \times 10^6\text{ W}\) (accept range \(3.98 \times 10^6\) to \(4.00 \times 10^6\text{ W}\)) [1 mark]

(b)(ii)
- Use of \(\text{efficiency} = \frac{\text{useful power out}}{\text{total power in}}\) [1 mark]
- Substitution: \(\frac{1.5 \times 10^6}{3.99 \times 10^6}\) [1 mark]
- Correct answer \(38\%\) (or \(0.38\)) [1 mark]

(c)
- Correct statement that if velocity was zero, air would pile up / block airflow [1 mark]
- Statement that air must retain some kinetic energy to escape [1 mark]
- Reference to Betz's limit (limiting efficiency to ~59%) [1 mark]

(a) The solar constant is the solar electromagnetic radiation intensity received per unit area perpendicular to the rays, at the top of the Earth's atmosphere, when the Earth is at its mean distance from the Sun. Its approximate value is \(1.36\text{ kW m}^{-2}\) (accept values in the range \(1.3\text{ to }1.4\text{ kW m}^{-2}\)).

(b)(i) Rate of solar energy incident:
\(P_{\text{in}} = I \times A = 850\text{ W m}^{-2} \times 4.5\text{ m}^2 = 3825\text{ W} = 3.825\text{ kW}\)

(b)(ii) Thermal energy required to heat the water:
\(Q = m c \Delta T = 180\text{ kg} \times 4200\text{ J kg}^{-1}\text{ K}^{-1} \times (55 - 15)\text{ K}\)
\(Q = 180 \times 4200 \times 40 = 3.024 \times 10^7\text{ J}\)

Time interval:
\(t = 4.0\text{ hours} = 4.0 \times 3600\text{ s} = 14400\text{ s}\)

Average rate of heat transfer:
\(P_{\text{out}} = \frac{Q}{t} = \frac{3.024 \times 10^7\text{ J}}{14400\text{ s}} = 2100\text{ W} = 2.1\text{ kW}\)

(b)(iii) Efficiency:
\(\eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{2100}{3825} = 0.549 \approx 55\%\)

(c) Two factors:
1. Cloud cover / air pollution: Atmospheric particles scatter and absorb solar rays, decreasing the incident light intensity.
2. Tilt angle / orientation relative to the sun: Panels not mounted at the optimal tilt angle relative to the local latitude collect less energy per unit area because the light is not normal to the surface.

(a)
- Definition: Solar energy incident per unit time per unit area normal to the radiation outside Earth's atmosphere [1 mark]
- Correct value: \(1.3\) to \(1.4\text{ kW m}^{-2}\) [1 mark]

(b)(i)
- Correct equation: \(P = I A\) [1 mark]
- Correct substitution and result: \(3825\text{ W}\) or \(3.8\text{ kW}\) [1 mark]

(b)(ii)
- Calculate temperature difference \(\Delta T = 40\text{ K}\) and time in seconds \(14400\text{ s}\) [1 mark]
- Correct calculation of energy \(Q = 3.024 \times 10^7\text{ J}\) [1 mark]
- Correct calculation of power \(2100\text{ W}\) [1 mark]

(b)(iii)
- Recall and use of efficiency formula [1 mark]
- Correct calculation: \(0.55\) or \(55\%\) (allow ecf from b(i) and b(ii)) [1 mark]

(c)
- 1 mark for each valid factor with explanation up to 2 marks (e.g., latitude/tilt, clouds, shading, surface cleanliness).

(a) Run-of-river schemes generate electricity continuously using the natural flow of a river, without large-scale storage capacity [1 mark]. Pumped storage schemes act as energy storage systems, using electricity during off-peak times to pump water up to a high reservoir, and releasing it to generate electricity during peak-demand periods [2 marks].

(b)(i) Mass flow rate:
\(\frac{\Delta m}{\Delta t} = \rho \frac{\Delta V}{\Delta t} = 1000\text{ kg m}^{-3} \times 85\text{ m}^3\text{ s}^{-1} = 85000\text{ kg s}^{-1}\)

Maximum theoretical power (gravitational potential energy lost per second):
\(P_{\text{th}} = \frac{\Delta E_p}{\Delta t} = \frac{\Delta m}{\Delta t} g h = 85000\text{ kg s}^{-1} \times 9.81\text{ m s}^{-2} \times 360\text{ m}\)
\(P_{\text{th}} = 3.002 \times 10^8\text{ W} \approx 300\text{ MW}\) (or \(3.0 \times 10^8\text{ W}\))

(b)(ii) Generation efficiency:
\(\eta = \frac{240\text{ MW}}{300\text{ MW}} = 0.80 = 80\%\)

(b)(iii) Gravitational potential energy gained by the water:
\(\Delta E_p = m g h = (1.0 \times 10^7\text{ kg}) \times 9.81\text{ m s}^{-2} \times 360\text{ m} = 3.53 \times 10^{10}\text{ J}\)

With a pumping efficiency of \(85\%\), the electrical energy required is:
\(E_{\text{input}} = \frac{\Delta E_p}{\eta_{\text{pump}}} = \frac{3.53 \times 10^{10}\text{ J}}{0.85} = 4.15 \times 10^{10}\text{ J} \approx 4.2 \times 10^{10}\text{ J}\)

(a)
- Run-of-river: continuous/base-load generation using natural river flow without large reservoir [1 mark]
- Pumped storage: energy storage role, uses electricity off-peak to pump water up [1 mark]
- Generates electricity on demand during peak times by releasing water down [1 mark]

(b)(i)
- Determine mass flow rate as \(85000\text{ kg s}^{-1}\) [1 mark]
- Substitution into \(P = \frac{\Delta m}{\Delta t} g h\) [1 mark]
- Correct calculation: \(3.0 \times 10^8\text{ W}\) or \(300\text{ MW}\) [1 mark]

(b)(ii)
- Formulates ratio of actual power to theoretical power [1 mark]
- Correct answer: \(0.80\) or \(80\%\) (allow ecf from b(i)) [1 mark]

(b)(iii)
- Calculate potential energy gain: \(3.53 \times 10^{10}\text{ J}\) [1 mark]
- Use of efficiency equation: \(E = \frac{\Delta E_p}{0.85}\) [1 mark]
- Correct answer: \(4.2 \times 10^{10}\text{ J}\) (or \(4.15 \times 10^{10}\text{ J}\)) [1 mark]

(a)
- Moderator role: Slows down fast-moving neutrons produced during fission to thermal speeds [1 mark]. This is necessary because thermal (slower) neutrons are much more likely to be captured by Uranium-235 nuclei to cause further fission [1 mark].
- Control rods role: Absorb excess neutrons [1 mark]. They are raised or lowered to ensure that, on average, exactly one neutron per fission goes on to cause another fission, maintaining a steady, safe chain reaction [1 mark].

(b)(i) Mass of reactants:
\(m_{\text{reactants}} = 235.0439\text{ u} + 1.0087\text{ u} = 236.0526\text{ u}\)

Mass of products:
\(m_{\text{products}} = 140.9144\text{ u} + 91.9262\text{ u} + 3 \times 1.0087\text{ u}\)
\(m_{\text{products}} = 232.8406 + 3.0261 = 235.8667\text{ u}\)

Mass defect:
\(\Delta m = m_{\text{reactants}} - m_{\text{products}} = 236.0526 - 235.8667 = 0.1859\text{ u}\)

Energy released:
\(E = 0.1859 \times 931.5\text{ MeV} = 173.17\text{ MeV} \approx 173\text{ MeV}\)

(b)(ii) First, convert fission energy into Joules:
\(E = 173.17 \times 10^6 \times 1.60 \times 10^{-19}\text{ J} = 2.771 \times 10^{-11}\text{ J}\)

The thermal power of the reactor is:
\(P_{\text{thermal}} = \frac{P_{\text{electrical}}}{\eta} = \frac{950 \times 10^6\text{ W}}{0.33} = 2.879 \times 10^9\text{ W}\)

Thermal energy required per day (\(1\text{ day} = 86400\text{ s}\)):
\(Q = 2.879 \times 10^9\text{ W} \times 86400\text{ s} = 2.487 \times 10^{14}\text{ J}\)

Number of fissions per day:
\(N = \frac{2.487 \times 10^{14}\text{ J}}{2.771 \times 10^{-11}\text{ J}} = 8.975 \times 10^{24}\text{ fissions}\)

Number of moles of Uranium-235 consumed:
\(n = \frac{8.975 \times 10^{24}}{6.02 \times 10^{23}\text{ mol}^{-1}} = 14.91\text{ mol}\)

Mass of U-235 consumed:
\(m = 14.91\text{ mol} \times 235\text{ g mol}^{-1} = 3504\text{ g} \approx 3.5\text{ kg}\)

(a)
- Moderator slows down fast neutrons [1 mark] to thermal speeds / to match thermal equilibrium of core [1 mark]
- Control rods absorb neutrons [1 mark] to control rate of reaction / prevent supercritical state [1 mark]

(b)(i)
- Calculations of total reactant and product masses [1 mark]
- Correct mass defect \(\Delta m = 0.1859\text{ u}\) [1 mark]
- Final energy: \(173\text{ MeV}\) (accept \(173.1\text{ to }173.3\text{ MeV}\)) [1 mark]

(b)(ii)
- Calculate thermal power: \(2.88 \times 10^9\text{ W}\) [1 mark]
- Convert fission energy to Joules: \(2.77 \times 10^{-11}\text{ J}\) [1 mark]
- Calculate total fissions per day: \(8.98 \times 10^{24}\) [1 mark]
- Convert to mass: \(3.5\text{ kg}\) (accept range \(3.4 \text{ to } 3.6\text{ kg}\)) [1 mark]

(a) Two assumptions of kinetic theory:
1. The volume occupied by the gas molecules is negligible compared to the volume of the container.
2. All collisions between molecules and between molecules and container walls are perfectly elastic.
(Accept other valid assumptions such as: no intermolecular forces except during collisions; motion of molecules is random; time of collisions is negligible compared to time between collisions).

(b)(i) Temperatures in Kelvin:
\(T_1 = 280 + 273.15 = 553.15\text{ K} \approx 553\text{ K}\)
\(T_2 = 650 + 273.15 = 923.15\text{ K} \approx 923\text{ K}\)

(b)(ii) At constant volume, \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\):
\(P_2 = P_1 \left(\frac{T_2}{T_1}\right) = 4.5\text{ MPa} \times \left(\frac{923.15}{553.15}\right) = 7.51\text{ MPa}\)

(b)(iii) Rate of heat transfer:
\(P = \frac{\Delta Q}{\Delta t} = \frac{\Delta m}{\Delta t} c \Delta T\)
\(\Delta T = T_2 - T_1 = 650 - 280 = 370\text{ K}\) (or \(370\text{ }^\circ\text{C}\))
\(P = 180\text{ kg s}^{-1} \times 3.12 \times 10^3\text{ J kg}^{-1}\text{ K}^{-1} \times 370\text{ K}\)
\(P = 2.078 \times 10^8\text{ W} \approx 208\text{ MW}\) (or \(210\text{ MW}\) to two significant figures)

(c) Explanation:
- An increase in temperature increases the average kinetic energy and thus the mean speed of the gas molecules [1 mark].
- Fast-moving molecules collide with the container walls more frequently [1 mark].
- Each collision involves a larger change in momentum, producing a larger force per collision [1 mark]. Combined, this increases the average force per unit area (pressure).

(a) 1 mark for each correct assumption up to 2 marks.

(b)(i) Correctly converts both temperatures to Kelvin (553 K and 923 K) [1 mark].

(b)(ii)
- Uses \(\frac{P}{T} = \text{constant}\) [1 mark]
- Obtains \(7.51\text{ MPa}\) (or \(7.5\text{ MPa}\)) [1 mark]

(b)(iii)
- Calculates temperature change \(\Delta T = 370\text{ K}\) [1 mark]
- Substitutes correctly into \(P = \frac{\Delta m}{\Delta t} c \Delta T\) [1 mark]
- Obtains \(2.1 \times 10^8\text{ W}\) or \(210\text{ MW}\) (accept range \(207\text{ to }210\text{ MW}\)) [1 mark]

(c)
- Higher temperature means higher mean speed/kinetic energy of molecules [1 mark]
- Higher collision rate/frequency with walls [1 mark]
- Larger change in momentum per collision [1 mark]

(a) Newton's law of gravitation states that the attractive force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them [1 mark].
Formula: \(F = \frac{GMm}{r^2}\) [1 mark] where:
- \(F\) is the gravitational force,
- \(G\) is the gravitational constant,
- \(M\) and \(m\) are the masses,
- \(r\) is the separation between their centres [1 mark].

(b)(i) For a geostationary orbit, the period \(T = 24\text{ hours} = 86400\text{ s}\).
The gravitational force provides the centripetal force:
\(\frac{GMm}{r^2} = m \omega^2 r = m \left(\frac{2\pi}{T}\right)^2 r\)
\(r^3 = \frac{GMT^2}{4\pi^2}\)
\(r^3 = \frac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times (86400)^2}{4\pi^2}\)
\(r^3 = 7.537 \times 10^{22}\text{ m}^3\)
\(r = 4.22 \times 10^7\text{ m}\) (which is approximately \(4.2 \times 10^7\text{ m}\))

(b)(ii) Gravitational force:
\(F = \frac{GMm}{r^2} = \frac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times (2.5 \times 10^6)}{(4.22 \times 10^7)^2}\)
\(F = 5.59 \times 10^4\text{ N} \approx 5.6 \times 10^4\text{ N}\)

(c) Required incident solar power:
\(P_{\text{solar}} = \frac{P_{\text{microwave}}}{\eta} = \frac{1.5 \times 10^9\text{ W}}{0.22} = 6.818 \times 10^9\text{ W}\)

Required active area:
\(A = \frac{P_{\text{solar}}}{I} = \frac{6.818 \times 10^9\text{ W}}{1360\text{ W m}^{-2}} = 5.01 \times 10^6\text{ m}^2 \approx 5.0 \times 10^6\text{ m}^2\)

(a)
- Statement of proportionality / word definition [1 mark]
- Correct equation [1 mark]
- Correct definition of all terms including separation of centres [1 mark]

(b)(i)
- Equating gravitational and centripetal forces [1 mark]
- Stating \(T = 86400\text{ s}\) [1 mark]
- Correct substitution and rearrangement to yield \(4.2 \times 10^7\text{ m}\) [1 mark]

(b)(ii)
- Substituting values correctly into gravitational formula [1 mark]
- Calculating \(5.6 \times 10^4\text{ N}\) (accept \(5.5 \times 10^4\text{ to }5.7 \times 10^4\text{ N}\)) [1 mark]

(c)
- Correct calculation of required solar power: \(6.8 \times 10^9\text{ W}\) [1 mark]
- Use of \(A = P/I\) [1 mark]
- Correct area: \(5.0 \times 10^6\text{ m}^2\) (accept \(5.01 \times 10^6\text{ m}^2\)) [1 mark]

The percentage uncertainty in each measurement is calculated as:
\% uncertainty in \(m = \frac{0.3}{15.0} \times 100\% = 2.0\%\n\% uncertainty in \)d = \frac{0.02}{1.20} \times 100\% \approx 1.67\%\n\% uncertainty in \(L = \frac{0.1}{5.0} \times 100\% = 2.0\%\n\nFor the density formula \)\rho = \frac{4m}{\pi d^2 L}\), the fractional uncertainty relationship is:
\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}

Summing the individual percentage uncertainties:
\% uncertainty in \(\rho = 2.0\% + 2(1.67\%) + 2.0\% = 7.33\%\n\nRounding to 2 significant figures, this gives \)7.3\%\).

1 mark for correct option C.
- Method: Identification of the need to double the percentage uncertainty of the diameter because it is squared, and add all percentage uncertainties together.
- Accuracy: Correct computation leading to \(7.3\%\).

First, find the percentage uncertainty in \(L\):
\% uncertainty in \(L = \frac{0.5}{80.0} \times 100\% = 0.625\%\n\nThe time period is \)T = t / 20\). The percentage uncertainty in \(T\) is identical to the percentage uncertainty in the total measured time \(t\):
\% uncertainty in \(T = \frac{0.2}{35.8} \times 100\% \approx 0.559\%\n\nThe formula is \)g = \frac{4\pi^2 L}{T^2}\), so the percentage uncertainty in \(g\) is:
\% uncertainty in \(g = \% \text{ uncertainty in } L + 2 \times (\% \text{ uncertainty in } T) = 0.625\% + 2 \times 0.559\% = 1.74\%\n\nThis rounds to \)1.7\%\).

1 mark for correct option C.
- Method: Correctly find percentage uncertainties in \(L\) and \(t\), then apply the power rule for combining uncertainties.
- Accuracy: Correctly obtain \(1.7\%\).

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

Initial binding energy (reactants):
\(E_{\text{initial}} = 235 \times 7.59\text{ MeV} = 1783.65\text{ MeV}\) (free neutrons have zero binding energy).

Final binding energy (products):
\(E_{\text{final}} = (141 \times 8.33\text{ MeV}) + (92 \times 8.51\text{ MeV}) = 1174.53\text{ MeV} + 782.92\text{ MeV} = 1957.45\text{ MeV}\).

Energy released:
\(\Delta E = 1957.45\text{ MeV} - 1783.65\text{ MeV} = 173.8\text{ MeV}\).

This is closest to \(174\text{ MeV}\).

1 mark for correct option B.
- Method: Determine the total binding energy for the reactants and products by multiplying binding energy per nucleon by the nucleon number, then find the energy difference.
- Accuracy: Correctly calculate the energy released to be \(174\text{ MeV}\).

The primary role of a moderator in a nuclear reactor is to slow down the high-speed neutrons produced during fission to thermal speeds, making them more likely to initiate further fission in uranium-235. Water and graphite are common and suitable materials. Option A describes the function of control rods (e.g., boron). Option C describes the coolant. Option D describes the biological shielding.

1 mark for correct option B.
- Method: Correctly identify the physical purpose of the moderator (slowing down neutrons to thermal speeds) and a valid material (water).

According to Newton's law of gravitation, the force is:
\(F = G \frac{M_X M_Y}{d^2}\)

Applying the changes:
- New mass of X: \(2 M_X\)
- New mass of Y: \(0.5 M_Y\)
- New distance: \(0.5 d\)

The new force \(F'\) is:
\(F' = G \frac{(2 M_X)(0.5 M_Y)}{(0.5 d)^2} = G \frac{1 \cdot M_X M_Y}{0.25 d^2} = 4 \left( G \frac{M_X M_Y}{d^2} \right) = 4F\).

1 mark for correct option D.
- Method: Apply proportional change analysis using Newton's law of gravitation.
- Accuracy: Correctly determine that the force is quadrupled.

Let \(x\) be the distance from the centre of the planet where the gravitational field strength is zero. The distance from this point to the centre of the moon is \(d - x\).

Equating the magnitudes of the gravitational fields:
\(\frac{G M}{x^2} = \frac{G (0.04M)}{(d - x)^2}\)

Cancel \(G\) and \(M\):
\(\frac{1}{x^2} = \frac{0.04}{(d - x)^2}\)

Take the square root of both sides:
\(\frac{1}{x} = \frac{0.2}{d - x}\)

Cross-multiplying:
\(d - x = 0.2 x \implies d = 1.2 x \implies x = \frac{d}{1.2} \approx 0.83 d\).

1 mark for correct option C.
- Method: Set up field equality equation for zero net field strength, solve algebraically for \(x\) in terms of \(d\).
- Accuracy: Correctly calculate the value of \(0.83 d\).

From the ideal gas equation of state:
\(\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}\)

Substitute the given conditions:
\(\frac{p V}{T} = \frac{p_2 (1.5 V)}{1.2 T}\)

Solving for \(p_2\):
\(p_2 = p \times \frac{1.2}{1.5} = 0.8 p\).

1 mark for correct option B.
- Method: Use the equation of state for a fixed mass of gas to relate the ratios of pressure, volume, and temperature.
- Accuracy: Correct calculation of the final pressure as \(0.8 p\).

The relationship between root-mean-square speed and absolute temperature is:
\(v_{\text{rms}} = \sqrt{\frac{3 k T}{m}}\)

This implies that \(v_{\text{rms}} \propto \sqrt{T}\). To double the root-mean-square speed, the absolute temperature \(T\) in Kelvin must be increased by a factor of \(2^2 = 4\).

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

Required final temperature:
\(T_2 = 4 \times 300\text{ K} = 1200\text{ K}\)

Convert back to Celsius:
\(t_2 = 1200 - 273 = 927^\circ\text{C}\).

1 mark for correct option C.
- Method: Convert temperatures to Kelvin, use the relationship between root-mean-square speed and temperature, and convert the final temperature back to Celsius.
- Accuracy: Correct calculation yielding \(927^\circ\text{C}\).

1. Find the percentage uncertainty in \(L\):
\[\%\Delta L = \left(\frac{0.2\text{ cm}}{80.0\text{ cm}}\right) \times 100\% = 0.25\%\]

2. Find the percentage uncertainty in \(T\):
Since \(T = \frac{t}{50}\), the percentage uncertainty in \(T\) is equal to the percentage uncertainty in the total measured time \(t\):
\[\%\Delta T = \left(\frac{0.5\text{ s}}{90.0\text{ s}}\right) \times 100\% \approx 0.556\%\]

3. Combine the percentage uncertainties for \(g \propto \frac{L}{T^2}\):
\[\%\Delta g = \%\Delta L + 2(\%\Delta T) = 0.25\% + 2(0.556\%) = 1.36\%\]

Rounding to two significant figures gives \(1.4\%\).

1 mark for the correct answer C.

[1] Correctly identifies that the percentage uncertainty in \(g\) is given by \(\%\Delta L + 2(\%\Delta T)\) and calculates this value to be \(1.4\%\).

1. Calculate the total binding energy of the reactant nucleus (Uranium-235):
\[\text{BE}_{\text{reactants}} = 235 \times 7.59\text{ MeV} = 1783.65\text{ MeV}\]
Note: Neutrons have a binding energy of 0.

2. Calculate the total binding energy of the product nuclei:
\[\text{BE}_{\text{products}} = (141 \times 8.33\text{ MeV}) + (92 \times 8.51\text{ MeV}) = 1174.53\text{ MeV} + 782.92\text{ MeV} = 1957.45\text{ MeV}\]

3. Calculate the energy released (\( \Delta E \)):
\[\Delta E = \text{BE}_{\text{products}} - \text{BE}_{\text{reactants}} = 1957.45\text{ MeV} - 1783.65\text{ MeV} = 173.8\text{ MeV} \approx 174\text{ MeV}\]

1 mark for the correct answer B.

[1] Correct calculation of reactant and product binding energies, leading to an energy release of \(174\text{ MeV}\).

1. Since the spheres are made of the same material, they have the same density \(\rho\). Mass is proportional to volume:
\[M = \rho V = \rho \left(\frac{4}{3}\pi r^3\right) \propto r^3\]

2. Let the mass of the first sphere (radius \(R\)) be \(m_1 = k R^3\).
* The initial second sphere (radius \(2R\)) has mass \(m_2 = k (2R)^3 = 8 k R^3\).
* The new second sphere (radius \(3R\)) has mass \(m'_2 = k (3R)^3 = 27 k R^3\).

3. The gravitational force is given by Newton's law: \(F = G \frac{m_1 m_2}{d^2}\). Since \(d = 6R\) is constant for both cases:
\[F \propto m_1 m_2\]

4. Find the ratio of the new force \(F'\) to the original force \(F\):
\[\frac{F'}{F} = \frac{m_1 m'_2}{m_1 m_2} = \frac{m'_2}{m_2} = \frac{27 k R^3}{8 k R^3} = \frac{27}{8} = 3.375\]

Thus, the new force is \(3.375 F\).

1 mark for the correct answer C.

[1] Correctly recognizes that mass scales with \(r^3\) and uses the ratio of the masses of the second sphere (\(27/8 = 3.375\)) to determine the new force.

1. Convert temperatures to Kelvin:
* \(T_1 = 27 + 273.15 = 300\text{ K}\)
* \(T_2 = 177 + 273.15 = 450\text{ K}\)

2. Use the ideal gas equation \(pV = nRT = \frac{m}{M}RT\). Since the volume \(V\) and the molar mass \(M\) are constant:
\[m \propto \frac{p}{T}\]

3. Find the ratio of the final mass \(m_2\) to the initial mass \(m_1\):
\[\frac{m_2}{m_1} = \frac{p_2}{p_1} \times \frac{T_1}{T_2}\]

4. Substitute the values:
\[\frac{m_2}{m_1} = \frac{1.8 \times 10^5\text{ Pa}}{2.4 \times 10^5\text{ Pa}} \times \frac{300\text{ K}}{450\text{ K}} = 0.75 \times \frac{2}{3} = 0.50\]

Therefore, \(0.50\) of the original mass remains.

1 mark for the correct answer A.

[1] Converts temperatures to Kelvin, uses the ideal gas equation to express mass as proportional to \(p/T\), and correctly calculates the ratio as \(0.50\).

1. Find the mean of the raw measurements:
\[\text{Mean measured value} = \frac{0.54 + 0.55 + 0.53 + 0.54 + 0.54}{5} = 0.54\text{ mm}\]

2. Correct for the systematic zero error:
Since the zero error is \(-0.03\text{ mm}\), the actual reading is:
\[\text{Actual value} = \text{Measured value} - \text{Zero error} = 0.54 - (-0.03) = 0.57\text{ mm}\]

3. Determine the absolute uncertainty in the measurements:
Using half the range of the measurements:
\[\text{Uncertainty} = \frac{\text{Maximum} - \text{Minimum}}{2} = \frac{0.55 - 0.53}{2} = 0.01\text{ mm}\]
Since this is equal to the resolution of the instrument (\(0.01\text{ mm}\)), the absolute uncertainty is indeed \(\pm 0.01\text{ mm}\).

Therefore, the final result is \((0.57 \pm 0.01)\text{ mm}\).

1 mark for the correct answer B.

[1] Correctly applies the zero error correction (adds \(0.03\text{ mm}\)) and calculates the uncertainty as half the range (\(0.01\text{ mm}\)).

1. For a satellite in circular orbit, the centripetal force is provided by gravity:
\[\frac{m v^2}{r} = \frac{G M m}{r^2} \Rightarrow v^2 = \frac{G M}{r}\]

2. This implies that the orbit radius \(r\) is related to speed \(v\) by:
\[r = \frac{G M}{v^2} \propto \frac{1}{v^2}\]

3. If the orbital speed is halved (\(v \rightarrow \frac{1}{2}v\)), the new radius \(r'\) becomes:
\[r' = 4r\]

4. The gravitational force \(F\) is given by Newton's law:
\[F = \frac{G M m}{r^2} \propto \frac{1}{r^2}\]

5. Substitute the new radius \(r' = 4r\) into the force equation:
\[F' \propto \frac{1}{(4r)^2} = \frac{1}{16 r^2} \Rightarrow F' = \frac{F}{16}\]

1 mark for the correct answer A.

[1] Derives the orbital relation \(r \propto 1/v^2\), finds that the new radius is four times larger, and correctly concludes that force is reduced by a factor of 16.

1. Since the gases are in thermal equilibrium, they are at the same temperature \(T\). This means their average kinetic energy per atom is equal:
\[\frac{1}{2} m_{\text{He}} c_{\text{rms, He}}^2 = \frac{1}{2} m_{\text{Ne}} c_{\text{rms, Ne}}^2\]

2. Solve for the ratio of the root-mean-square speeds:
\[\frac{c_{\text{rms, He}}}{c_{\text{rms, Ne}}} = \sqrt{\frac{m_{\text{Ne}}}{m_{\text{He}}}}\]

3. The ratio of atomic masses is equal to the ratio of their molar masses:
\[\frac{c_{\text{rms, He}}}{c_{\text{rms, Ne}}} = \sqrt{\frac{20\text{ g mol}^{-1}}{4.0\text{ g mol}^{-1}}} = \sqrt{5.0} \approx 2.236\]

To two significant figures, the ratio is \(2.2\).

1 mark for the correct answer C.

[1] Uses the equality of average kinetic energy at thermal equilibrium to establish that \(c_{\text{rms}} \propto 1/\sqrt{m}\), leading to a ratio of \(\sqrt{5} \approx 2.2\).

[Part a] A solid steel ball has a much larger mass-to-surface-area ratio than a light plastic ball. This makes air resistance negligible compared to the gravitational force, ensuring the ball's acceleration is extremely close to the true free-fall acceleration \(g\). [Part b] Using the equation of motion \(h = \frac{1}{2}gt^2\), we find \(g = \frac{2h}{t^2} = \frac{2 \times 1.250}{0.505^2} = 9.803\text{ m s}^{-2}\). The percentage uncertainty in \(h\) is \(\%\Delta h = (0.002/1.250) \times 100\% = 0.16\%\). The percentage uncertainty in \(t\) is \(\%\Delta t = (0.004/0.505) \times 100\% = 0.792\%\). Since \(g \propto h/t^2\), the total percentage uncertainty in \(g\) is \(\%\Delta g = \%\Delta h + 2 \times \%\Delta t = 0.16\% + 2 \times 0.792\% = 1.74\%\). Absolute uncertainty in \(g = 9.803 \times 0.0174 = 0.17\text{ m s}^{-2}\). [Part c] If the recorded times are systematically larger than the true times, the denominator in \(g = 2h/t^2\) is systematically overestimated. Therefore, the calculated value of \(g\) will be systematically lower than the true acceleration due to gravity. [Part d] By measuring \(h\) at several different heights and recording the corresponding times \(t\), a student can plot a graph of \(2h\) on the vertical axis against \(t^2\) on the horizontal axis. Since \(2h = g t^2\), the line of best fit will be straight and pass through the origin with a gradient equal to \(g\). This graphical method is superior because any systematic timing delay will affect only the intercept on the horizontal axis and not the gradient, and random errors are averaged out across the entire range of measurements.

[Part a] 1 mark: Identifies that air resistance is negligible for the steel ball / motion is purely under gravity. [Part b] 1 mark: Correct rearrangement of formula to \(g = 2h/t^2\). 1 mark: Calculation of \(g = 9.80\text{ m s}^{-2}\). 1 mark: Calculates percentage uncertainties for both variables (\(\%\Delta h = 0.16\%\) and \(\%\Delta t = 0.79\%\)). 1 mark: Correctly doubles \(\%\Delta t\) and adds \(\%\Delta h\) to obtain \(1.74\%\) (or absolute uncertainty of \(0.17\text{ m s}^{-2}\)). [Part c] 1 mark: Explains that measured \(t\) is systematically too large. 1 mark: Explains that because \(g\) is inversely proportional to \(t^2\), the calculated \(g\) is systematically smaller than the true value. [Part d] 1 mark: Identifies that plotting \(2h\) against \(t^2\) gives a straight line. 1 mark: States that the gradient of this line equals \(g\). 1.4 marks: Explains that a graphical approach filters out systematic zero errors (constant delays) which only shift the intercept, and reduces the impact of random variations through a line of best fit.

[Part a] The student should use a micrometer screw gauge because it has a high resolution (typically \(0.01\text{ mm}\)). To ensure the measurement is accurate and representative, the student must: (i) Check for and record any zero error on the micrometer when closed, subtracting this from subsequent readings; (ii) Measure the diameter at multiple points along the length of the wire to account for any non-uniformity; (iii) Take readings at different perpendicular orientations at each point to check for non-circular cross-section; (iv) Calculate a mean diameter. [Part b] The cross-sectional area of the wire is \(A = \frac{\pi d^2}{4} = \frac{\pi (0.38 \times 10^{-3})^2}{4} = 1.134 \times 10^{-7}\text{ m}^2\). Resistivity is given by \(\rho = \frac{RA}{L} = \frac{12.4 \times 1.134 \times 10^{-7}}{1.000} = 1.406 \times 10^{-6}\ \Omega\text{ m}\). Percentage uncertainties: \(\%\Delta d = (0.01/0.38) \times 100\% = 2.63\%\). Since \(A \propto d^2\), the percentage uncertainty in area is \(\%\Delta A = 2 \times \%\Delta d = 5.26\%\). Percentage uncertainty in resistance is \(\%\Delta R = (0.2/12.4) \times 100\% = 1.61\%\). Percentage uncertainty in length is \(\%\Delta L = (0.002/1.000) \times 100\% = 0.20\%\). The total percentage uncertainty in \(\rho\) is \(\%\Delta \rho = \%\Delta R + \%\Delta A + \%\Delta L = 1.61\% + 5.26\% + 0.20\% = 7.07\%\). The absolute uncertainty in \(\rho\) is \(1.406 \times 10^{-6} \times 0.0707 = 0.099 \times 10^{-6}\ \Omega\text{ m}\). Hence, \(\rho = (1.41 \pm 0.10) \times 10^{-6}\ \Omega\text{ m}\). [Part c] Passing a high current through the wire causes a significant temperature rise due to Joule heating. Since the resistivity of a metal increases with temperature, this would cause the measured resistance to be higher than its value at room temperature, introducing a systematic error in the calculated resistivity.

[Part a] 1 mark: Identifies the micrometer screw gauge as the correct instrument. 1 mark: Mentions checking and correcting for zero error. 1 mark: Describes measuring at multiple locations along the wire. 1 mark: Describes measuring at different orientations at each location and calculating the mean. [Part b] 1 mark: Calculates the correct cross-sectional area \(A = 1.13 \times 10^{-7}\text{ m}^2\). 1 mark: Calculates the correct central value of resistivity \(\rho = 1.41 \times 10^{-6}\ \Omega\text{ m}\). 1 mark: Finds percentage uncertainty in area \(\%\Delta A = 5.26\%\) by doubling diameter uncertainty. 1 mark: Adds percentage uncertainties of \(R\), \(A\), and \(L\) to find total percentage uncertainty (\(7.07\%\)). 1.4 marks: Converts to absolute uncertainty (\(0.10 \times 10^{-6}\ \Omega\text{ m}\)) and presents the final value with consistent decimal places. [Part c] 1 mark: States that high current leads to resistive/Joule heating and temperature increase. 1 mark: Connects temperature rise to an increase in resistance, which systematically overestimates resistivity.

[Part a] The temperature of the trapped air must remain constant. To ensure this: (i) The syringe plunger must be compressed or expanded very slowly to allow heat transfer with the surrounding environment; (ii) The student must wait a short period of time after moving the plunger before taking the pressure reading to allow the gas to return to thermal equilibrium with the room. [Part b] From the ideal gas equation, \(p V = n R T\), which expands to \(p(V_s + V_t) = n R T\). Rearranging this equation to match a straight-line format \(y = mx + c\) gives: \(V_s = n R T (\frac{1}{p}) - V_t\). Therefore, when plotting \(V_s\) against \(1/p\): (i) The vertical intercept (y-intercept) is equal to \(-V_t\), so the magnitude of the y-intercept directly gives the volume of the connection tube \(V_t\); (ii) The gradient of the graph is equal to \(n R T\). Since \(R\) is the molar gas constant and \(T\) is the constant room temperature in Kelvin, the number of moles is calculated as \(n = \frac{\text{gradient}}{RT}\). [Part c] Using the ideal gas law \(p V = n R T\), we rearrange for \(n\): \(n = \frac{pV}{RT}\). First convert volume to SI units: \(V = 45.0 \times 10^{-6}\text{ m}^3\). Substituting the central values: \(n = \frac{1.20 \times 10^5 \times 45.0 \times 10^{-6}}{8.31 \times 293} = 2.218 \times 10^{-3}\text{ mol}\). Percentage uncertainties: \(\%\Delta p = (0.05 / 1.20) \times 100\% = 4.17\%\). \(\%\Delta V = (1.0 / 45.0) \times 100\% = 2.22\%\). \(\%\Delta T = (1 / 293) \times 100\% = 0.34\%\). Since \(n = \frac{pV}{RT}\), the percentage uncertainty in \(n\) is the sum of the individual percentage uncertainties: \(\%\Delta n = \%\Delta p + \%\Delta V + \%\Delta T = 4.17\% + 2.22\% + 0.34\% = 6.73\%\). Thus, \(n = 2.22 \times 10^{-3}\text{ mol} \pm 6.7\%\).

[Part a] 1 mark: Identifies temperature as the constant quantity. 1 mark: Recommends moving the plunger slowly. 1 mark: Recommends waiting for thermal equilibrium before taking readings. [Part b] 1 mark: Derives the linear equation \(V_s = nRT(1/p) - V_t\). 1 mark: Correctly identifies that the magnitude of the negative vertical intercept is equal to the tube volume \(V_t\). 1 mark: Identifies the gradient of the plot as equal to \(nRT\). 1 mark: Shows how to solve for \(n\) by dividing the gradient by \(RT\). [Part c] 1 mark: Converts volume correctly to cubic meters (\(45.0 \times 10^{-6}\text{ m}^3\)). 1 mark: Calculates correct central value for \(n = 2.22 \times 10^{-3}\text{ mol}\). 1 mark: Calculates individual percentage uncertainties of \(p\) and \(V\) correctly. 1.4 marks: Calculates total percentage uncertainty by summation to obtain \(6.7\%\) (accept range \(6.7\%\) to \(6.8\%\)).

[Part a] A thermal neutron is a slow-moving neutron that has been slowed down by collisions within the moderator until its mean kinetic energy is in thermal equilibrium with its surroundings (approximately \(0.025\text{ eV}\)). Thermal neutrons have a much higher probability of causing fission (a larger capture cross-section) in Uranium-235 because they remain in close proximity to the nucleus longer, allowing the strong nuclear force to capture them. Fast neutrons are highly energetic and are much more likely to simply scatter off the Uranium-235 nucleus. [Part b] The total binding energy of the initial reactants is: \(E_{\text{initial}} = 235 \times 7.59\text{ MeV} = 1783.65\text{ MeV}\) (since the free neutron has no binding energy). The total binding energy of the products is: \(E_{\text{final}} = (141 \times 8.33) + (92 \times 8.51) = 1174.53 + 782.92 = 1957.45\text{ MeV}\). The energy released in the reaction is the difference in binding energies: \(\Delta E = E_{\text{final}} - E_{\text{initial}} = 1957.45 - 1783.65 = 173.80\text{ MeV}\). [Part c] Control rods are made of neutron-absorbing materials (such as boron, cadmium, or hafnium). In a reactor, they are inserted into the core to absorb excess neutrons, regulating the rate of the chain reaction. To maintain a constant power output, the reaction must be critical, meaning exactly one neutron from each fission event goes on to cause another fission. Moving the rods further into the core absorbs more neutrons, slowing the reaction down. Withdrawing the rods increases the number of available neutrons, speeding up the reaction and increasing power. In an emergency, they can be fully inserted (scrammed) to completely shut down the fission process.

[Part a] 1 mark: Defines a thermal neutron as slow-moving/in thermal equilibrium with the moderator. 1 mark: Explains that slower speed increases the likelihood of capture / increases the fission cross-section. [Part b] 1 mark: Calculates total binding energy of Uranium-235 as \(1783.65\text{ MeV}\). 1 mark: Calculates total binding energy of Barium-141 as \(1174.53\text{ MeV}\) and Krypton-92 as \(782.92\text{ MeV}\). 1 mark: Sums products to get \(1957.45\text{ MeV}\). 1.4 marks: Subtracts initial from final binding energy to get \(173.8\text{ MeV}\) (accept range \(173.3\text{ MeV}\) to \(174.0\text{ MeV}\) due to intermediate rounding). [Part c] 1 mark: Identifies control rods as neutron absorbers. 1 mark: Explains that adjusting depth regulates the neutron population to keep the chain reaction steady/critical. 1 mark: Explains that full insertion stops the reaction quickly for safety.

[Part a] The gravitational force between the star and the planet provides the necessary centripetal force for the circular orbit. Therefore: \(\frac{G M m}{r^2} = m \omega^2 r\). Substituting \(\omega = \frac{2̀̀\pi}{T}\) gives: \(\frac{G M}{r^2} = \left(\frac{2\pi}{T}\right)^2 r = \frac{4\pi^2 r}{T^2}\). Rearranging this yields Kepler's Third Law: \(T^2 = \left(\frac{4\pi^2}{G M}\right) r^3\). Hence, \(k = \frac{4\pi^2}{G M}\). [Part b] To find \(M\), we rearrange the formula: \(M = \frac{4\pi^2 r^3}{G T^2}\). Substituting the values: \(r = 5.6 \times 10^{10}\text{ m}\), \(T = 2.40 \times 10^6\text{ s}\), and \(G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}\): \(M = \frac{4\pi^2 \times (5.6 \times 10^{10})^3}{6.67 \times 10^{-11} \times (2.40 \times 10^6)^2} = \frac{39.4784 \times 1.75616 \times 10^{32}}{6.67 \times 10^{-11} \times 5.76 \times 10^{12}} = \frac{6.9333 \times 10^{33}}{3.8419 \times 10^2} = 1.805 \times 10^{30}\text{ kg}\). Rounded to 3 significant figures, \(M = 1.81 \times 10^{30}\text{ kg}\). [Part c] The percentage uncertainty in the orbital radius is: \(\%\Delta r = (0.2 / 5.6) \times 100\% = 3.57\%\). The percentage uncertainty in the orbital period is: \(\%\Delta T = (0.03 / 2.40) \times 100\% = 1.25\%\). Since the formula for mass is \(M = \frac{4\pi^2 r^3}{G T^2}\), the total percentage uncertainty in \(M\) is: \(\%\Delta M = 3 \times \%\Delta r + 2 \times \%\Delta T = 3 \times 3.57\% + 2 \times 1.25\% = 10.71\% + 2.50\% = 13.21\%\). Therefore, \(\%\Delta M \approx 13.2\%\). The measurement of the orbital radius \(r\) contributes significantly more to the final uncertainty. This is because \(r\) has a larger initial percentage uncertainty (\(3.57\%\) compared to \(1.25\%\) for \(T\)), and its uncertainty is multiplied by 3 due to being cubed in the formula, whereas the uncertainty of \(T\) is only multiplied by 2.

[Part a] 1 mark: Equates gravitational force to centripetal force. 1 mark: Correctly substitutes centripetal acceleration in terms of \(T\). 1 mark: Shows algebraic steps to reach \(T^2 = \frac{4\pi^2 r^3}{GM}\). [Part b] 1 mark: Correctly rearranges the formula to make \(M\) the subject. 1 mark: Correctly substitutes the numbers, including raising them to the correct powers. 1 mark: Calculates the mass \(M = 1.81 \times 10^{30}\text{ kg}\) (accept \(1.80 \times 10^{30}\text{ kg}\) to \(1.82 \times 10^{30}\text{ kg}\)). [Part c] 1 mark: Computes \(\%\Delta r = 3.57\%\) and \(\%\Delta T = 1.25\%\) correctly. 1 mark: Uses the correct absolute uncertainty combination formula (\(3\%\Delta r + 2\%\Delta T\)). 1.4 marks: Correctly calculates \(\%\Delta M \approx 13.2\%\) and justifies that \(r\) is the dominant source because of its larger initial percentage uncertainty and the factor of 3 multiplier.

[Part a] A fiducial marker (such as a thin pin or thread) should be placed horizontally at the exact equilibrium position of the oscillating mass. The student should start and stop the stopwatch when the mass passes this marker moving in a specified direction. This reduces reaction time uncertainty because the mass has its maximum speed at the equilibrium position, making the transit time past the marker extremely sharp and well-defined compared to timing at the turning points where the mass is temporarily stationary and moving slowly. [Part b] First, find the mean time for 20 oscillations: \(\overline{t_{20}} = \frac{25.4 + 25.1 + 25.6 + 25.3}{4} = 25.35\text{ s}\). The range of the measurements is \(25.6 - 25.1 = 0.5\text{ s}\). The absolute uncertainty in \(t_{20}\) is \(\frac{\text{range}}{2} = \frac{0.5}{2} = 0.25\text{ s}\). Therefore, the time period of a single oscillation is \(T = \frac{25.35}{20} = 1.2675\text{ s}\), and its absolute uncertainty is \(\Delta T = \frac{0.25}{20} = 0.0125\text{ s}\). Rounding to a sensible number of decimal places: \(T = 1.27 \pm 0.01\text{ s}\) (or \(1.268 \pm 0.013\text{ s}\)). [Part c] Rearranging \(T = 2\pi \sqrt{m/k}\) for \(k\) gives \(k = \frac{4\pi^2 m}{T^2}\). Substituting the values: \(k = \frac{4\pi^2 \times 0.400}{1.2675^2} = 9.829\text{ N m}^{-1}\). Percentage uncertainty in \(m\) is \(\%\Delta m = (0.005 / 0.400) \times 100\% = 1.25\%\). Percentage uncertainty in \(T\) is \(\%\Delta T = (0.0125 / 1.2675) \times 100\% = 0.986\%\). Since \(k \propto m / T^2\), the percentage uncertainty in \(k\) is \(\%\Delta k = \%\Delta m + 2 \times \%\Delta T = 1.25\% + 2 \times 0.986\% = 3.22\%\). Thus, the spring constant is \(k = 9.81\text{ N m}^{-1}\) with a percentage uncertainty of \(3.2\%\) (or \(3.3\%\) depending on intermediate rounding).

[Part a] 1 mark: Recommends placing the marker at the equilibrium position. 1 mark: Explains that velocity of the mass is maximum at this point. 1 mark: Mentions starting/stopping the clock when the mass passes the marker in a consistent direction. [Part b] 1 mark: Calculates the correct mean for 20 oscillations (\(25.35\text{ s}\)) and divides by 20 to get \(T = 1.27\text{ s}\). 1 mark: Determines the range as \(0.5\text{ s}\) and halves it to find the uncertainty in \(t_{20}\) as \(0.25\text{ s}\). 1 mark: Divides the uncertainty by 20 to find \(\Delta T = 0.013\text{ s}\) (or \(0.01\text{ s}\)). [Part c] 1 mark: Correctly rearranges the formula to \(k = 4\pi^2 m / T^2\). 1 mark: Calculates the central value \(k = 9.81\text{ N m}^{-1}\) (accept \(9.8\text{ N m}^{-1}\) to \(9.83\text{ N m}^{-1}\)). 1 mark: Calculates \(\%\Delta m = 1.25\%\). 1.4 marks: Correctly doubles \(\%\Delta T\) and adds to \(\%\Delta m\) to get total percentage uncertainty of \(3.2\%\) to \(3.3\%\).

[Part a] The student must measure the potential difference \(V\) across the capacitor and the time \(t\). Because the discharge can be extremely fast, manual timing is unsuitable. A data logger or a digital storage oscilloscope (DSO) should be connected across the capacitor, configured to sample the voltage at a high frequency (e.g., hundreds of Hertz). [Part b] Taking the natural logarithm of both sides of the equation \(V = V_0 e^{-t / (RC)}\) yields: \(\ln(V) = \ln(V_0) - \frac{1}{RC} t\). This matches the equation of a straight line, \(y = mx + c\), where the dependent variable \(y\) is \(\ln(V)\) and the independent variable \(x\) is \(t\). The gradient \(m\) of this straight line is equal to \(-\frac{1}{RC}\), and the vertical intercept \(c\) is equal to \(\ln(V_0)\). [Part c] From the gradient formula \(m = -\frac{1}{RC}\), we rearrange for \(C\): \(C = -\frac{1}{mR}\). Substituting the central values: \(C = -\frac{1}{(-0.0245) \times (47.0 \times 10^3)} = 8.684 \times 10^{-4}\text{ F}\) (or \(868.4\ \mu\text{F}\)). Percentage uncertainties: \(\%\Delta m = (0.0008 / 0.0245) \times 100\% = 3.27\%\). \(\%\Delta R = (1.0 / 47.0) \times 100\% = 2.13\%\). Since \(C = (mR)^{-1}\), the total percentage uncertainty in \(C\) is: \(\%\Delta C = \%\Delta m + \%\Delta R = 3.27\% + 2.13\% = 5.40\%\). The absolute uncertainty in \(C\) is: \(\Delta C = 8.684 \times 10^{-4} \times 0.0540 = 0.469 \times 10^{-4}\text{ F}\). Thus, rounding to 1 or 2 significant figures for uncertainty: \(C = (8.7 \pm 0.5) \times 10^{-4}\text{ F}\) (or \(868 \pm 47\ \mu\text{F}\)).

[Part a] 1 mark: Identifies voltage and time as the key variables. 1 mark: Explains that manual stopwatch timing is too slow for rapid discharges. 1 mark: Proposes the use of a data logger or digital storage oscilloscope. [Part b] 1 mark: Derives the logarithmic equation \(\ln(V) = -\frac{t}{RC} + \ln(V_0)\). 1 mark: Explains that gradient equals \(-1/RC\). 1 mark: Explains that vertical intercept equals \(\ln(V_0)\). [Part c] 1 mark: Correctly relates capacitance to gradient and resistance: \(C = -1/(mR)\). 1 mark: Calculates the correct central value of \(C = 8.68 \times 10^{-4}\text{ F}\). 1 mark: Calculates percentage uncertainty in gradient (\(3.27\%\)). 1 mark: Calculates percentage uncertainty in resistance (\(2.13\%\)). 1.4 marks: Correctly sums the percentage uncertainties to find \(5.4\%\) and converts this to an absolute uncertainty of \(0.5 \times 10^{-4}\text{ F}\).