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(a) The initial gravitational potential energy \(E_p\) of the parcel is: \(E_p = mgh\) where \(h = L \sin\theta = 4.2 \sin(28^\circ) = 1.972\text{ m}\). \(E_p = 3.5 \times 9.81 \times 1.972 = 67.70\text{ J}\). The final kinetic energy \(E_k\) is: \(E_k = \frac{1}{2} m v^2 = 0.5 \times 3.5 \times 4.8^2 = 40.32\text{ J}\). The work done against friction \(W_f\) is the loss of mechanical energy: \(W_f = E_p - E_k = 67.70 - 40.32 = 27.38\text{ J} \approx 27.4\text{ J}\).

(b) The average frictional force \(F_f\) is: \(F_f = \frac{W_f}{L} = \frac{27.38}{4.2} = 6.519\text{ N} \approx 6.5\text{ N}\).

(c) If the ramp angle \(\theta\) is increased to \(35^\circ\): 1) The vertical height \(h = L \sin\theta\) increases, so the initial gravitational potential energy increases. 2) The normal force \(R = mg \cos\theta\) decreases, which reduces the frictional force \(F_f = \mu R\) and therefore reduces the energy lost to friction over the same length. 3) With greater initial potential energy and less energy lost to friction, the final kinetic energy, and thus the final velocity, must be greater.

(a) [4 marks total]:
- Correct calculation of vertical height \(h\) (1 mark)
- Correct calculation of initial GPE (1 mark)
- Correct calculation of final KE (1 mark)
- Correct subtraction to get work done against friction: \(27.4\text{ J}\) (allow \(27.3\text{ - }27.5\text{ J}\)) (1 mark)

(b) [3 marks total]:
- Use of \(W = F \times d\) (1 mark)
- Correct substitution of their \(W_f\) and \(L = 4.2\text{ m}\) (1 mark)
- Correct final answer: \(6.5\text{ N}\) (allow \(6.4\text{ - }6.6\text{ N}\)) (1 mark)

(c) [4.6 marks total]:
- Stating that the vertical height increases, leading to a greater initial potential energy (1.5 marks)
- Stating that the normal contact force decreases, leading to a lower frictional force and less work done against friction (1.5 marks)
- Concluding that the final kinetic energy and thus the velocity at the bottom will be greater (1.6 marks)

(a) First find the grating spacing \(d\): \(d = \frac{1 \times 10^{-3}\text{ m}}{500} = 2.00 \times 10^{-6}\text{ m}\). For the second-order maximum (\(n = 2\)): \(d \sin\theta = n \lambda\), hence \(\sin\theta = \frac{2 \times 633 \times 10^{-9}}{2.00 \times 10^{-6}} = 0.633\). \(\theta = \arcsin(0.633) = 39.27^\circ \approx 39.3^\circ\).

(b) For the maximum number of orders, \(\sin\theta \le 1\). Therefore, \(n \le \frac{d}{\lambda} = \frac{2.00 \times 10^{-6}}{633 \times 10^{-9}} = 3.16\). Since \(n\) must be an integer, the maximum observable order is \(n = 3\).

(c) Violet light has a shorter wavelength than red light (\(\lambda_{\text{violet}} < \lambda_{\text{red}}\)). Since \(n \propto \frac{1}{\lambda}\) for a given angle limit, a shorter wavelength increases the maximum order \(n\) that can be observed, thus increasing the total number of maxima visible on the screen.

(a) [4 marks total]:
- Correct calculation of \(d = 2.00 \times 10^{-6}\text{ m}\) (1 mark)
- Correct formula usage \(d \sin\theta = n\lambda\) (1 mark)
- Correct substitution for \(n=2\) and \(\lambda\) (1 mark)
- Correct angle \(39.3^\circ\) (accept \(39.2^\circ - 39.4^\circ\) or \(0.686\text{ rad}\)) (1 mark)

(b) [4 marks total]:
- Setting \(\sin\theta = 1\) as the limiting condition (1 mark)
- Substituting values to find \(n \approx 3.16\) (2 marks)
- Rounding down to the nearest integer to get \(n = 3\) (1 mark)

(c) [3.6 marks total]:
- Stating that violet light has a shorter wavelength than red light (1 mark)
- Explaining that a shorter wavelength results in smaller diffraction angles for each order, or that the limit \(n < d/\lambda\) increases (1.5 marks)
- Concluding that the total number of maxima increases (1.1 marks)

(a) The relationship is: \(V = \varepsilon - Ir\).

(b) Setting up simultaneous equations using the data:
1) \(1.38 = \varepsilon - 0.40r\)
2) \(1.14 = \varepsilon - 1.20r\)
Subtracting equation (2) from (1):
\(0.24 = 0.80r \implies r = 0.30\ \Omega\).
Substituting \(r = 0.30\ \Omega\) back into equation (1):
\(1.38 = \varepsilon - 0.40(0.30) \implies 1.38 = \varepsilon - 0.12 \implies \varepsilon = 1.50\text{ V}\).

(c) With \(R = 2.5\ \Omega\), the total resistance of the circuit is \(R_{\text{total}} = R + r = 2.5 + 0.30 = 2.8\ \Omega\).
The current \(I\) is: \(I = \frac{\varepsilon}{R_{\text{total}}} = \frac{1.50}{2.8} = 0.5357\text{ A}\).
The power dissipated in the internal resistance \(P_r\) is: \(P_r = I^2 r = (0.5357)^2 \times 0.30 = 0.0861\text{ W} \approx 0.086\text{ W}\).

(a) [1.6 marks total]:
- Correct equation: \(V = \varepsilon - Ir\) (or equivalent rearrangement) (1.6 marks)

(b) [6 marks total]:
- Setting up two correct simultaneous equations (2 marks)
- Correctly eliminating \(\varepsilon\) to find \(r\) (1.5 marks)
- Finding \(r = 0.30\ \Omega\) (1 mark)
- Substituting back to find \(\varepsilon\) (1 mark)
- Finding \(\varepsilon = 1.50\text{ V}\) (0.5 marks)

(c) [4 marks total]:
- Correctly calculating total resistance \(R_{\text{total}} = 2.8\ \Omega\) (1 mark)
- Calculating new current \(I = 0.54\text{ A}\) (1.5 marks)
- Using \(P = I^2 r\) to calculate power (1 mark)
- Correct answer: \(0.086\text{ W}\) (accept \(0.086\text{ - }0.087\text{ W}\)) (0.5 marks)

(a) The nuclear equation is: \(^{1}_{0}\text{n} \rightarrow ^{1}_{1}\text{p} + ^{0}_{-1}\text{e}^- + \bar{\nu}_{\text{e}}\).

(b) Initially, the neutron has a lepton number of \(0\). Among the products, the proton has a lepton number of \(0\), and the electron has a lepton number of \(+1\). To conserve the total lepton number at \(0\), another product must carry a lepton number of \(-1\). An electron antineutrino has a lepton number of \(-1\), whereas an electron neutrino has a lepton number of \(+1\). Therefore, an antineutrino is required to maintain lepton conservation.

(c) Use Einstein's mass-energy equivalence equation: \(E = \Delta m c^2\).
\(E = (1.39 \times 10^{-30}\text{ kg}) \times (3.00 \times 10^8\text{ m s}^{-1})^2 = 1.251 \times 10^{-13}\text{ J}\).
Convert this energy into \(\text{MeV}\):
\(E = \frac{1.251 \times 10^{-13}\text{ J}}{1.60 \times 10^{-13}\text{ J MeV}^{-1}} = 0.7819\text{ MeV} \approx 0.78\text{ MeV}\).

(a) [3 marks total]:
- Correct symbol and notation for neutron and proton (1 mark)
- Correct electron notation (1 mark)
- Correct electron antineutrino symbol (with bar) (1 mark)

(b) [3.6 marks total]:
- Identifying initial lepton number of neutron as 0 (1 mark)
- Stating that the electron has a lepton number of +1 (1 mark)
- Explaining that the antineutrino has a lepton number of -1, making the total lepton number on the right-hand side equal to 0, which conserves lepton number (1.6 marks)

(c) [5 marks total]:
- Recalling and using \(E = \Delta m c^2\) (1 mark)
- Correct substitution of values into \(E = \Delta m c^2\) (1 mark)
- Finding energy in Joules: \(1.25 \times 10^{-13}\text{ J}\) (1 mark)
- Dividing by \(1.60 \times 10^{-13}\text{ J MeV}^{-1}\) to convert to \(\text{MeV}\) (1 mark)
- Correct final answer: \(0.78\text{ MeV}\) (accept \(0.78\text{ - }0.79\text{ MeV}\)) (1 mark)

(a) A stationary wave with three antinodes (3rd harmonic) has a pattern starting and ending with a node (N) because both ends are clamped. There are three loops, so the order of features along the wire is: Node (at \(0\)), Antinode, Node, Antinode, Node, Antinode, Node (at \(L\)). The nodes are located at \(x = 0\), \(x = 0.28\text{ m}\), \(x = 0.57\text{ m}\), and \(x = 0.85\text{ m}\).

(b) For the third harmonic, the length of the wire is equal to \(1.5\) wavelengths: \(L = \frac{3}{2}\lambda\).
\(\lambda = \frac{2}{3} L = \frac{2}{3} \times 0.85 = 0.567\text{ m} \approx 0.57\text{ m}\).

(c) First find the speed \(v\) of the progressive waves: \(v = f \lambda = 240 \times 0.5667 = 136\text{ m s}^{-1}\).
Using the formula for the speed of a wave on a stretched string: \(v = \sqrt{\frac{T}{\mu}}\).
Squaring both sides gives \(v^2 = \frac{T}{\mu} \implies T = v^2 \mu\).
\(T = (136)^2 \times 4.2 \times 10^{-3} = 18496 \times 4.2 \times 10^{-3} = 77.68\text{ N} \approx 78\text{ N}\).

(a) [3.6 marks total]:
- Identifying that both ends are nodes (1 mark)
- Identifying that there are 3 antinodes and 4 nodes in total (1.5 marks)
- Describing or sketching the alternating N-A-N-A-N-A-N structure correctly (1.1 marks)

(b) [3 marks total]:
- Using the correct relationship between length and wavelength for the third harmonic: \(L = 1.5\lambda\) (1.5 marks)
- Correct final calculation of \(\lambda = 0.57\text{ m}\) (accept \(0.567\text{ m}\)) (1.5 marks)

(c) [5 marks total]:
- Using \(v = f\lambda\) to find the wave speed (1.5 marks)
- Correct calculation of wave speed \(v = 136\text{ m s}^{-1}\) (0.5 marks)
- Recalling and rearranging \(v = \sqrt{\frac{T}{\mu}}\) to make \(T\) the subject (1.5 marks)
- Substituting values to get \(T = 78\text{ N}\) (accept \(77.6\text{ - }78.0\text{ N}\)) (1.5 marks)

(a) Rearranging \(s = \frac{1}{2} g t^2\) for \(g\):
\(g = \frac{2s}{t^2} = \frac{2 \times 1.500}{(0.553)^2} = \frac{3.000}{0.3058} = 9.810\text{ m s}^{-2} \approx 9.8\text{ m s}^{-2}\).

(b) (i) Percentage uncertainty in \(s\):
\(\%\text{ uncertainty in } s = \frac{0.002}{1.500} \times 100\% = 0.133\% \approx 0.13\%\).
(ii) Percentage uncertainty in \(t\):
\(\%\text{ uncertainty in } t = \frac{0.005}{0.553} \times 100\% = 0.904\% \approx 0.90\%\).

(c) The formula for \(g\) is \(g = \frac{2s}{t^2}\).
Using the rules for combining uncertainties:
\(\%\text{ uncertainty in } g = \%\text{ uncertainty in } s + 2 \times (\%\text{ uncertainty in } t)\)
\(\%\text{ uncertainty in } g = 0.133\% + 2 \times 0.904\% = 0.133\% + 1.808\% = 1.941\%\).
Absolute uncertainty in \(g\):
\(\Delta g = g \times \frac{1.941}{100} = 9.810 \times 0.01941 = 0.1904\text{ m s}^{-2} \approx 0.19\text{ m s}^{-2}\) (or \(0.2\text{ m s}^{-2}\) to 1 s.f.).

(a) [2.6 marks total]:
- Correctly rearranging formula to \(g = \frac{2s}{t^2}\) (1 mark)
- Correct substitution of numbers (1 mark)
- Showing result of calculation is \(9.81\text{ m s}^{-2}\) (0.6 marks)

(b) [4 marks total]:
- (i) Correct formula and calculation for percentage uncertainty in \(s = 0.13\%\) (2 marks)
- (ii) Correct formula and calculation for percentage uncertainty in \(t = 0.90\%\) (2 marks)

(c) [5 marks total]:
- Identifying that percentage uncertainty of \(t\) must be doubled because of the \(t^2\) term (1.5 marks)
- Summing the percentage uncertainties to find total percentage uncertainty in \(g\) as \(1.94\%\) (1.5 marks)
- Applying this percentage to their calculated value of \(g\) (1 mark)
- Correct final absolute uncertainty: \(0.19\text{ m s}^{-2}\) (accept \(0.2\text{ m s}^{-2}\)) (1 mark)

\( (a)\ A = \frac{\pi d^2}{4} = \frac{\pi (0.38 \times 10^{-3})^2}{4} = 1.134 \times 10^{-7} \text{ m}^2 \). The percentage uncertainty in \( d \) is \( \frac{0.01}{0.38} \times 100\% = 2.63\% \). The percentage uncertainty in \( A \) is \( 2 \times 2.63\% = 5.26\% \). Absolute uncertainty in \( A = 1.134 \times 10^{-7} \times 0.0526 = 5.97 \times 10^{-9} \text{ m}^2 \approx 0.06 \times 10^{-7} \text{ m}^2 \). Therefore, \( A = (1.13 \pm 0.06) \times 10^{-7} \text{ m}^2 \). \( (b)\ \text{Gradient} = \frac{\Delta F}{\Delta (\Delta L)} = \frac{25.0 - 5.0}{(4.50 - 0.90) \times 10^{-3}} = \frac{20.0}{3.60 \times 10^{-3}} = 5.56 \times 10^3 \text{ N m}^{-1} \). \( (c)\ E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L_0} = \text{Gradient} \times \frac{L_0}{A} = 5.556 \times 10^3 \times \frac{2.450}{1.134 \times 10^{-7}} = 1.20 \times 10^{11} \text{ Pa} \). \( (d)\ \text{Percentage uncertainty in } E = \text{percentage uncertainty in gradient} + \text{percentage uncertainty in } L_0 + \text{percentage uncertainty in } A = 1.5\% + \left( \frac{0.002}{2.450} \times 100\% \right) + 5.26\% = 1.5\% + 0.082\% + 5.26\% = 6.84\% \approx 6.8\% \).

\( (a)\ \) [1 mark] for calculating \( A = 1.13 \times 10^{-7} \text{ m}^2 \). [1 mark] for doubling the percentage uncertainty of \( d \) to find the absolute uncertainty as \( \pm 0.06 \times 10^{-7} \text{ m}^2 \). \( (b)\ \) [1 mark] for correct substitution of values to find gradient. [1 mark] for converting \( \text{mm} \) to \( \text{m} \). [1 mark] for gradient value in the range \( 5.5 \times 10^3 \) to \( 5.6 \times 10^3 \text{ N m}^{-1} \). \( (c)\ \) [1 mark] for equating gradient to \( \frac{EA}{L_0} \). [1 mark] for correct substitution. [1 mark] for final Young modulus value \( 1.20 \times 10^{11} \text{ Pa} \). \( (d)\ \) [1 mark] for identifying that percentage uncertainties must be added. [1 mark] for final answer \( 6.8\% \) (accept \( 6.8\% \) to \( 6.9\% \)).

\( (a)\ \) The sphere needs time and distance to accelerate from rest until the viscous drag force plus upthrust balances its weight, ensuring a constant terminal velocity is established. \( (b)\ \) Converting values to SI units: for \( r = 1.0 \times 10^{-3} \text{ m} \), \( r^2 = 1.0 \times 10^{-6} \text{ m}^2 \) and \( v = 0.0142 \text{ m s}^{-1} \). For \( r = 3.0 \times 10^{-3} \text{ m} \), \( r^2 = 9.0 \times 10^{-6} \text{ m}^2 \) and \( v = 0.1278 \text{ m s}^{-1} \). Gradient \( = \frac{\Delta v}{\Delta (r^2)} = \frac{0.1278 - 0.0142}{(9.0 - 1.0) \times 10^{-6}} = 1.42 \times 10^4 \text{ s}^{-1} \). \( (c)\ \) From \( v = \left[ \frac{2 (\rho_s - \rho_f) g}{9 \eta} \right] r^2 \), the gradient \( m = \frac{2 (\rho_s - \rho_f) g}{9 \eta} \). Rearranging gives \( \eta = \frac{2 (\rho_s - \rho_f) g}{9 m} = \frac{2 \times (7800 - 1260) \times 9.81}{9 \times 1.42 \times 10^4} = 1.004 \text{ Pa s} \approx 1.00 \text{ Pa s} \). \( (d)\ \) The student should position their line of sight directly level with the graduation marks on the cylinder so that they do not view the markers from an upward or downward angle.

\( (a)\ \) [2 marks]: 1 mark for stating that the sphere must reach terminal velocity. 1 mark for explaining that forces (weight, upthrust, and drag) must be balanced / acceleration must be zero. \( (b)\ \) [3 marks]: 1 mark for converting \( r^2 \) values to \( \text{m}^2 \) and \( v \) values to \( \text{m s}^{-1} \). 1 mark for a correct gradient calculation method. 1 mark for correct gradient value \( (1.40 \text{ to } 1.44) \times 10^4 \text{ s}^{-1} \). \( (c)\ \) [3 marks]: 1 mark for equating the gradient expression to \( \frac{2 (\rho_s - \rho_f) g}{9 \eta} \). 1 mark for substituting values correctly. 1 mark for \( \eta = 1.0 \text{ Pa s} \) (accept range \( 0.98 \text{ to } 1.02 \text{ Pa s} \)). \( (d)\ \) [2 marks]: 1 mark for viewing the marks at eye level. 1 mark for explaining that this ensures the sight line is perpendicular to the cylinder axis to eliminate parallax angle.

\( (a)\ \) The diagram should show a cell (with internal resistance), a variable resistor, and an ammeter connected in a single series loop. A voltmeter must be connected in parallel across the terminals of the cell. \( (b)\ \) The relationship is \( V = \varepsilon - I r = -r I + \varepsilon \). Thus, a graph of \( V \) against \( I \) is a straight line with gradient equal to \( -r \). Gradient \( = \frac{\Delta V}{\Delta I} = \frac{0.90 - 1.38}{0.75 - 0.15} = \frac{-0.48}{0.60} = -0.80 \ \Omega \). Since \( \text{gradient} = -r \), the internal resistance \( r = 0.80 \ \Omega \). \( (c)\ \) Using the equation \( V = \varepsilon - I r \) at \( I = 0.15 \text{ A} \): \( 1.38 = \varepsilon - (0.15 \times 0.80) \implies \varepsilon = 1.38 + 0.12 = 1.50 \text{ V} \). Alternatively, the y-intercept of the graph of \( V \) vs \( I \) represents \( \varepsilon \). Extrapolating to \( I = 0 \) yields \( V = 1.50 \text{ V} \). \( (d)\ \) As the temperature of the cell rises, its internal chemical reaction properties change, altering the internal resistance. If \( r \) changes during the experiment, the relationship between \( V \) and \( I \) is no longer linear, so the data points will deviate from a straight line, reducing the reliability of the determined values of \( \varepsilon \) and \( r \).

\( (a)\ \) [2 marks]: 1 mark for correct symbols and series loop (cell, variable resistor, ammeter). 1 mark for voltmeter connected in parallel across the cell. \( (b)\ \) [3 marks]: 1 mark for calculating the gradient correctly \( (-0.80 \text{ V A}^{-1}) \). 1 mark for relating the gradient to \( -r \). 1 mark for stating \( r = 0.80 \ \Omega \). \( (c)\ \) [2 marks]: 1 mark for using the intercept or substituting points into \( V = \varepsilon - I r \). 1 mark for \( \varepsilon = 1.50 \text{ V} \) (accept range \( 1.48 \text{ to } 1.52 \text{ V} \)). \( (d)\ \) [3 marks]: 1 mark for stating that internal resistance changes as the cell heats up. 1 mark for stating that \( r \) is no longer constant. 1 mark for explaining that this causes a non-linear relationship / systematic error, reducing reliability.

\( (a)\ \) The grating has \( 300 \text{ lines per mm} \), which is \( 300,000 \text{ lines per metre} \). Thus, \( d = \frac{1}{300 \times 10^3} \text{ m} = 3.333 \times 10^{-6} \text{ m} \). \( (b)\ \) The grating equation is \( d \sin\theta = n \lambda \). For \( n=1 \), \( d \sin\theta = \lambda \). For small angles, \( \sin\theta \approx \tan\theta = \frac{x}{D} \). Therefore, \( d \left(\frac{x}{D}\right) = \lambda \implies x = \left(\frac{\lambda}{d}\right) D \). This is in the form \( y = m x \), where the gradient \( m = \frac{\lambda}{d} \). From the data, the gradient is: \( \text{Gradient} = \frac{\Delta x}{\Delta D} = \frac{0.326 - 0.096}{1.70 - 0.50} = \frac{0.230}{1.20} = 0.1917 \approx 0.192 \). \( (c)\ \) Using \( \text{gradient} = \frac{\lambda}{d} \): \( \lambda = \text{gradient} \times d = 0.1917 \times 3.333 \times 10^{-6} \text{ m} = 6.39 \times 10^{-7} \text{ m} \approx 640 \text{ nm} \). \( (d)\ \) Measuring a larger distance (twice the size of \( x \)) with the same absolute uncertainty of the ruler significantly reduces the percentage uncertainty. It also eliminates the systematic error involved in trying to precisely locate the exact center of the very bright zero-order maximum.

\( (a)\ \) [2 marks]: 1 mark for recognizing \( d = \frac{1}{N} \). 1 mark for correct conversion to metres: \( d = 3.33 \times 10^{-6} \text{ m} \). \( (b)\ \) [3 marks]: 1 mark for deriving \( x = \left(\frac{\lambda}{d}\right) D \) using \( \sin\theta \approx \frac{x}{D} \). 1 mark for a valid gradient calculation method. 1 mark for gradient value in the range \( 0.191 \text{ to } 0.193 \). \( (c)\ \) [3 marks]: 1 mark for relating wavelength to the product of gradient and \( d \). 1 mark for substituting values. 1 mark for final answer \( 6.4 \times 10^{-7} \text{ m} \) (accept range \( 6.35 \times 10^{-7} \text{ m} \) to \( 6.45 \times 10^{-7} \text{ m} \)). \( (d)\ \) [2 marks]: 1 mark for stating that a larger measured value reduces the percentage uncertainty. 1 mark for mentioning that it eliminates the difficulty or systematic error of locating the exact center of the central spot.

In option A: the proton \( (\text{p}) \) has a baryon number of \( +1 \). The pions \( (\pi^+ \text{ and } \pi^0) \) are mesons and have a baryon number of \( 0 \). Therefore, the total baryon number before the decay is \( +1 \) and after the decay is \( 0 \), which violates the conservation of baryon number. In all other options, baryon number is conserved (all leptons have a baryon number of \( 0 \), and neutrons and protons have a baryon number of \( +1 \)).

1 mark for the correct option A.

First, apply conservation of linear momentum: \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \). Substituting the values: \( (0.20 \times 4.0) + 0 = (0.20 \times v_1) + (0.60 \times 1.5) \implies 0.80 = 0.20 v_1 + 0.90 \implies 0.20 v_1 = -0.10 \implies v_1 = -0.50 \text{ m s}^{-1} \). The negative sign indicates the car moves in the opposite direction to its initial motion. Next, check the kinetic energies to determine elasticity: Initial Kinetic Energy: \( E_{k,\text{initial}} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 0.20 \times (4.0)^2 = 1.6 \text{ J} \). Final Kinetic Energy: \( E_{k,\text{final}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} \times 0.20 \times (-0.50)^2 + \frac{1}{2} \times 0.60 \times (1.5)^2 = 0.025 + 0.675 = 0.70 \text{ J} \). Since \( E_{k,\text{final}} < E_{k,\text{initial}} \), kinetic energy is not conserved, meaning the collision is inelastic.

1 mark for the correct option A.

The terminal potential difference across the cell is given by: \( V = \varepsilon - Ir \). This equation can be rearranged into the standard straight-line equation \( y = mx + c \): \( V = (-r)I + \varepsilon \). If \( V \) is plotted on the vertical axis (\( y \)) and \( I \) on the horizontal axis (\( x \)), the gradient \( m \) is \( -r \) and the vertical intercept \( c \) is \( \varepsilon \).

1 mark for the correct option A.

For a string of length \( L = 1.2 \text{ m} \) fixed at both ends vibrating in the third harmonic (\( n = 3 \)), the wavelength is \( \lambda = \frac{2L}{3} = \frac{2 \times 1.2}{3} = 0.8 \text{ m} \). The distance between adjacent nodes is half a wavelength: \( \frac{\lambda}{2} = \frac{0.8}{2} = 0.4 \text{ m} \). The nodes are located at \( x = 0 \text{ m} \), \( 0.4 \text{ m} \), \( 0.8 \text{ m} \), and \( 1.2 \text{ m} \). The first loop is between \( 0 \text{ m} \) and \( 0.4 \text{ m} \), so the point at \( 0.2 \text{ m} \) lies in this loop. The second loop is between \( 0.4 \text{ m} \) and \( 0.8 \text{ m} \), so the point at \( 0.6 \text{ m} \) lies in this adjacent loop. Since adjacent loops in a stationary wave vibrate in antiphase, the phase difference between these two points is \( \pi \text{ rad} \).

1 mark for the correct option A.

Using the grating equation: \( d \sin \theta = n \lambda \). For \( n = 2 \), we have \( d \sin(30.0^\circ) = 2 \lambda \implies d (0.500) = 2 \times 600 \times 10^{-9} \text{ m} \implies d = 2.40 \times 10^{-6} \text{ m} \). To find the maximum order of diffraction, we use \( \theta < 90^\circ \) (since at \( \theta = 90^\circ \), the light would be parallel to the grating surface and could not be projected onto a screen behind the grating). Thus, \( n < \frac{d}{\lambda} = \frac{2.40 \times 10^{-6}}{600 \times 10^{-9}} = 4 \). Since \( n \) must be an integer, the highest observable order is \( n = 3 \). The total number of maxima is therefore \( 2 \times 3 + 1 = 7 \) (corresponding to \( n = -3, -2, -1, 0, 1, 2, 3 \)).

1 mark for the correct option C.

First, calculate the energy of the incident photons in electronvolts: \( E = \frac{hc}{\lambda} \). Substituting the values: \( E = \frac{(6.63 \times 10^{-34} \text{ J s}) \times (3.00 \times 10^8 \text{ m s}^{-1})}{350 \times 10^{-9} \text{ m}} = 5.68 \times 10^{-19} \text{ J} \). Convert this energy to electronvolts: \( E = \frac{5.68 \times 10^{-19} \text{ J}}{1.60 \times 10^{-19} \text{ J eV}^{-1}} \approx 3.55 \text{ eV} \). Using Einstein's photoelectric equation: \( E_{k, \text{max}} = hf - \phi = E - \phi = 3.55 \text{ eV} - 2.2 \text{ eV} = 1.35 \text{ eV} \approx 1.4 \text{ eV} \).

1 mark for the correct option A.

The extension \( \Delta L \) is calculated from the Young modulus equation: \( E = \frac{F L}{A \Delta L} \implies \Delta L = \frac{F L}{A E} \). Substituting the given values: \( \Delta L = \frac{180 \times 2.0}{(1.5 \times 10^{-6}) \times (2.0 \times 10^{11})} = \frac{360}{3.0 \times 10^5} = 1.2 \times 10^{-3} \text{ m} \). The elastic strain energy \( E_s \) stored in the wire within its limit of proportionality is: \( E_s = \frac{1}{2} F \Delta L = \frac{1}{2} \times 180 \times (1.2 \times 10^{-3}) = 0.108 \text{ J} \approx 0.11 \text{ J} \).

1 mark for the correct option B.

First, identify the anomalous reading. The value \( 0.55 \text{ mm} \) is significantly far from the others (\( 0.41 - 0.43 \text{ mm} \)) and is discarded. The remaining values are: \( 0.42 \text{ mm} \), \( 0.43 \text{ mm} \), \( 0.41 \text{ mm} \), and \( 0.42 \text{ mm} \). The mean of these values is: \( \text{Mean} = \frac{0.42 + 0.43 + 0.41 + 0.42}{4} = 0.42 \text{ mm} \). The absolute uncertainty in a set of repeated measurements is half of the range: \( \text{Uncertainty} = \frac{\text{Range}}{2} = \frac{0.43 - 0.41}{2} = 0.01 \text{ mm} \). The percentage uncertainty is: \( \frac{\text{Absolute Uncertainty}}{\text{Mean}} \times 100\% = \frac{0.01}{0.42} \times 100\% \approx 2.38\% \), which to two significant figures is \( 2.4\% \).

1 mark for the correct option B.

Let upwards be the positive direction. Just before impact, the velocity of the ball is \(v_i = -\sqrt{2gH}\). Just after impact, its velocity is \(v_f = +\sqrt{2gh}\). The change in momentum of the ball is given by: \(\Delta p = m v_f - m v_i = m(\sqrt{2gh} - (-\sqrt{2gH})) = m(\sqrt{2gH} + \sqrt{2gh})\). According to Newton's second law, the average net force acting on the ball during the collision is: \(F_{\text{net}} = \frac{\Delta p}{\Delta t} = \frac{m}{\Delta t}(\sqrt{2gH} + \sqrt{2gh})\). The forces acting on the ball during the collision are the upward average force from the floor \(F_{\text{floor}}\) and the downward force of gravity \(mg\). Therefore: \(F_{\text{net}} = F_{\text{floor}} - mg \implies F_{\text{floor}} = F_{\text{net}} + mg = \frac{m}{\Delta t}(\sqrt{2gH} + \sqrt{2gh}) + mg\).

1 mark for the correct option A. Method: Identify initial and final velocities using kinematics. Write the equation for change of momentum. Apply Newton's second law to find net force. Relate net force to the force from the floor and weight to find the final force expression.

Let us check the conservation laws for each option: Option A: \(K^+ \rightarrow e^- + \bar{\nu}_e\). Charge: \(+1 \rightarrow -1 + 0 = -1\) (not conserved, so impossible). Option B: \(K^+ \rightarrow \pi^+ + \pi^0\). Charge: \(+1 \rightarrow +1 + 0 = +1\) (conserved). Baryon number: \(0 \rightarrow 0 + 0 = 0\) (conserved). Lepton number: \(0 \rightarrow 0 + 0 = 0\) (conserved). Strangeness: \(+1 \rightarrow 0 + 0 = 0\) (strangeness changes by \(1\), which is allowed in weak interactions). This is a possible weak decay. Option C: \(K^- \rightarrow \pi^- + e^- + \bar{\nu}_e\). Charge: \(-1 \rightarrow -1 - 1 + 0 = -2\) (not conserved). Option D: \(K^0 \rightarrow p + \pi^-\). Baryon number: \(0 \rightarrow 1 + 0 = 1\) (not conserved).

1 mark for the correct option B. Method: Systematically verify conservation of charge, baryon number, lepton number, and strangeness for each option, keeping in mind that strangeness can change by 1 in weak decays.

The diffraction grating equation is given by: \(d \sin\theta = n \lambda\). We are given that \(n = 3\) and \(\theta = 30^\circ\). The grating has \(N\) lines per millimetre, which means there are \(10^3 N\) lines per metre. Therefore, the grating spacing \(d\) in metres is: \(d = \frac{1}{10^3 N} = \frac{10^{-3}}{N}\text{ m}\). Substitute \(d\) and \(\theta\) into the grating equation: \(\left(\frac{10^{-3}}{N}\right) \sin(30^\circ) = 3 \lambda\). Since \(\sin(30^\circ) = 0.5\), this simplifies to: \(\frac{10^{-3}}{2N} = 3 \lambda \implies \lambda = \frac{10^{-3}}{6N} = \frac{1}{6000N}\text{ m\).

1 mark for the correct option C. Method: State the grating formula \(d \sin\theta = n\lambda\). Express the grating spacing \(d\) in metres in terms of \(N\). Substitute the given values to solve for \(\lambda\).

The total resistance of the series circuit is \(R_{\text{total}} = R + r\). The current flowing through the circuit is \(I = \frac{\varepsilon}{R + r\). As the resistance of the variable resistor \(R\) is decreased, the total resistance of the circuit decreases, which causes the current \(I\) to increase. The voltmeter measures the terminal potential difference across the battery, given by \(V = \varepsilon - I r\). Since the current \(I\) increases, the internal potential drop (lost volts) \(I r\) increases, which causes the terminal potential difference \(V\) to decrease. The power dissipated in the internal resistance of the battery is given by \(P_r = I^2 r\). Since \(I\) increases and \(r\) remains constant, \(P_r\) must increase. Thus, \(V\) decreases and \(P_r\) increases.

1 mark for the correct option D. Method: Analyze how decreasing the load resistance affects the total circuit current, then use the terminal PD equation and internal power dissipation equation to determine the changes.

For a string fixed at both ends, the \(n\)-th harmonic has \(n\) loops, which means it has \(n + 1\) nodes and \(n\) antinodes. For the third harmonic: the number of nodes is \(3 + 1 = 4\), and the number of antinodes is \(3\) (making option A incorrect). The length \(L\) consists of \(3\) half-wavelengths: \(L = \frac{3\lambda}{2} \implies \lambda = \frac{2L}{3}\). The distance between two adjacent nodes is half a wavelength, which is \(\frac{\lambda}{2} = \frac{L}{3}\) (making options B and C incorrect). The frequency of the \(n\)-th harmonic is given by \(f_n = n f_1\), where \(f_1\) is the fundamental frequency. Therefore, \(f = 3 f_1 \implies f_1 = \frac{f}{3}\), which makes option D correct.

1 mark for the correct option D. Method: Deduce the wave properties of a third-harmonic stationary wave on a fixed string and use harmonic relations to find the fundamental frequency.

The elastic strain energy stored in a stretched wire within its limit of proportionality is given by: \(E_s = \frac{1}{2} F \Delta L\). By definition, the Young modulus is: \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} = \frac{F L}{A \Delta L}\). Rearranging this for the extension \(\Delta L\) gives: \(\Delta L = \frac{F L}{A E}\). Substituting this expression for \(\Delta L\) back into the equation for the strain energy yields: \(E_s = \frac{1}{2} F \left(\frac{F L}{A E}\right) = \frac{F^2 L}{2 A E}\).

1 mark for the correct option B. Method: Express the elastic strain energy as \(E_s = \frac{1}{2} F \Delta L\), then express \(\Delta L\) in terms of Young modulus and substitute back to obtain the final equation.

Einstein's photoelectric equation is: \(h f = \phi + E_{k,\text{max}}\), where the maximum kinetic energy is related to the stopping potential by \(E_{k,\text{max}} = e V_s\). Thus: \(e V_s = h f - \phi \implies h f = e V_s + \phi\). When the frequency is doubled to \(2f\), the new stopping potential \(V'_s\) satisfies: \(e V'_s = h (2f) - \phi = 2(h f) - \phi\). Substituting the first expression for \(h f\) into this equation gives: \(e V'_s = 2(e V_s + \phi) - \phi = 2 e V_s + \phi\). Dividing both sides by the elementary charge \(e\) gives: \(V'_s = 2 V_s + \frac{\phi}{e}\).

1 mark for the correct option C. Method: State Einstein's photoelectric equation. Set up equations for frequency \(f\) and frequency \(2f\). Eliminate the photon energy \(hf\) to express the new stopping potential in terms of \(V_s\) and \(\phi\).

The formula for resistivity is: \(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\). The percentage uncertainty in \(\rho\) is given by the sum of the percentage uncertainties of the independent variables (with the power of \(d\) being \(2\)): \(\%\Delta \rho = \%\Delta R + 2(\%\Delta d) + \%\Delta L\). Calculate the individual percentage uncertainties:
- \(\%\Delta R = \frac{0.1}{4.0} \times 100\% = 2.5\%\)
- \(\%\Delta d = \frac{0.01}{0.50} \times 100\% = 2.0\%\)
- \(\%\Delta L = \frac{0.01}{1.00} \times 100\% = 1.0\%\)
Substitute these values into the total uncertainty equation:
\(\%\Delta \rho = 2.5\% + 2(2.0\%) + 1.0\% = 2.5\% + 4.0\% + 1.0\% = 7.5\%\).

1 mark for the correct option A. Method: Identify the formula for resistivity in terms of measured variables. Apply the rules for combining uncertainties (add percentage uncertainties, double for squared variables). Perform the calculations accurately.

The initial kinetic energy of the system is \(E_{ki} = \frac{1}{2}(2m)v^2 = mv^2\).

By conservation of linear momentum, the initial momentum equals the final momentum:
\(2mv = (2m + m)V\)
where \(V\) is the common final velocity of the coupled trolleys.

Solving for \(V\):
\(V = \frac{2}{3}v\)

The final kinetic energy of the system is:
\(E_{kf} = \frac{1}{2}(3m)V^2 = \frac{3}{2}m \left(\frac{2}{3}v\right)^2 = \frac{2}{3}mv^2\)

The kinetic energy lost is:
\(\Delta E_k = E_{ki} - E_{kf} = mv^2 - \frac{2}{3}mv^2 = \frac{1}{3}mv^2\)

Therefore, the fraction of initial kinetic energy lost is:
\(\frac{\Delta E_k}{E_{ki}} = \frac{\frac{1}{3}mv^2}{mv^2} = \frac{1}{3}\)

1 mark for the correct option (B).

- Correct application of conservation of momentum to find the final velocity: \(V = \frac{2}{3}v\)
- Correct calculation of final kinetic energy: \(E_{kf} = \frac{2}{3}mv^2\)
- Correct determination of the fraction of energy lost: \(\frac{1}{3}\)

For a negative temperature coefficient (NTC) thermistor, as temperature decreases, the number of charge carriers decreases, causing its resistance \(R_T\) to increase.

The output voltage across the thermistor is given by the potential divider equation:
\(V_{\text{out}} = V \times \frac{R_T}{R + R_T}\)

As \(R_T\) increases, the fraction \(\frac{R_T}{R + R_T}\) increases, which means \(V_{\text{out}}\) increases. Thus, both the resistance of the thermistor and the output voltage increase.

1 mark for the correct option (D).

- Identification that NTC thermistor resistance increases when temperature decreases.
- Identification that output voltage across the thermistor increases as its resistance increases relative to the fixed resistor.

In \(\beta^-\) decay, a down quark inside a neutron changes into an up quark (\(d \to u\)), transforming the neutron into a proton. To conserve charge, the down quark (charge \(-\frac{1}{3}e\)) must emit a virtual \(\mathrm{W}^-\) boson (charge \(-1e\)) to become an up quark (charge \(+\frac{2}{3}e\)).

The \(\mathrm{W}^-\) boson then immediately decays into an electron (\(e^-\)) and an electron antineutrino (\(\bar{\nu}_e\)). Therefore, the mediating exchange particle is the \(\mathrm{W}^-\) boson.

1 mark for the correct option (C).

- Recognition that \(\beta^-\) decay is mediated by the weak force, eliminating photons.
- Conservation of charge at the quark vertex: \(d (-\frac{1}{3}) \to u (+\frac{2}{3}) + \mathrm{W}^- (-1)\).

The formula for the fringe spacing in double-slit interference is:
\(w = \frac{\lambda D}{d}\)

Let the new parameters be:
- \(\lambda' = 2\lambda\)
- \(d' = \frac{d}{2}\)
- \(D' = 2D\)

Substituting these new values into the formula gives the new fringe spacing \(w'\):
\(w' = \frac{\lambda' D'}{d'} = \frac{(2\lambda)(2D)}{\frac{d}{2}} = 8 \frac{\lambda D}{d} = 8w\)

Thus, the new fringe spacing is \(8w\).

1 mark for the correct option (C).

- Recall and correct use of the double-slit equation: \(w = \frac{\lambda D}{d}\).
- Correct algebraic substitution of the scale factors: \(\frac{2 \times 2}{1/2} = 8\).

A wire fixed at both ends vibrating in its third harmonic has three loops (loops are the regions between adjacent nodes).

Therefore, the length of the wire \(L\) contains three half-wavelengths:
\(L = 3 \left(\frac{\lambda}{2}\right)\)

The distance between adjacent nodes is exactly equal to half a wavelength (\(\frac{\lambda}{2}\)).

Rearranging the equation for \(\frac{\lambda}{2}\):
\(\frac{\lambda}{2} = \frac{L}{3}\)

So, the distance between adjacent nodes is \(\frac{L}{3}\).

1 mark for the correct option (B).

- Recognition that the third harmonic has 3 loops (half-wavelengths) across length \(L\).
- Recognition that the distance between adjacent nodes is one half-wavelength (\(\lambda/2\)).
- Rearranging to find \(\lambda/2 = L/3\).

According to Einstein's photoelectric equation, the maximum kinetic energy is:
\(E_{k,\max} = hf - \Phi\)

This can be rewritten to express the photon energy in terms of kinetic energy and work function:
\(hf = E_{k,\max} + \Phi\)

When the frequency is doubled to \(2f\), the new maximum kinetic energy \(E_{k,\max}'\) is:
\(E_{k,\max}' = h(2f) - \Phi = 2(hf) - \Phi\)

Substituting \(hf = E_{k,\max} + \Phi\) into this equation:
\(E_{k,\max}' = 2(E_{k,\max} + \Phi) - \Phi = 2E_{k,\max} + 2\Phi - \Phi = 2E_{k,\max} + \Phi\)

1 mark for the correct option (B).

- Correct statement of Einstein's photoelectric equation: \(E_{k,\max} = hf - \Phi\).
- Correct algebraic substitution to find the relation for doubled frequency: \(2E_{k,\max} + \Phi\).

The definition of Young modulus is \(E = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L}\), which gives the extension:
\(\Delta L = \frac{FL}{AE}\)

Since the tensile force \(F\) and the initial length \(L\) are the same for both wires:
\(\Delta L \propto \frac{1}{A E}\)

The cross-sectional area of a wire is \(A = \frac{\pi d^2}{4} \propto d^2\), where \(d\) is the diameter. Thus:
\(\Delta L \propto \frac{1}{d^2 E}\)

Let's compare the steel wire (\(s\)) to the copper wire (\(c\)):
- \(E_s = 2E_c\)
- \(d_s = 0.5d_c\)

Using the proportional relationship:
\(\frac{\Delta L_s}{\Delta L_c} = \frac{d_c^2 E_c}{d_s^2 E_s} = \frac{d_c^2 E_c}{(0.5 d_c)^2 (2 E_c)} = \frac{d_c^2 E_c}{0.25 d_c^2 \times 2 E_c} = \frac{1}{0.5} = 2.0\)

1 mark for the correct option (D).

- Relating extension to Young modulus, area, and diameter: \(\Delta L \propto \frac{1}{d^2 E}\).
- Setting up and simplifying the ratio: \(\frac{1}{(1/2)^2 \times 2} = \frac{1}{0.25 \times 2} = 2.0\).

First, calculate the percentage uncertainty in the independent measurements:

Percentage uncertainty in \(s\):
\(\% \Delta s = \frac{0.02}{2.00} \times 100\% = 1.0\%\)

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

The formula for \(g\) is:
\(g = \frac{2s}{t^2}\)

To find the overall percentage uncertainty in \(g\), we add the percentage uncertainties of each term (the constant \(2\) has no uncertainty). For \(t^2\), the percentage uncertainty is multiplied by its power (2):

\(\% \Delta g = \% \Delta s + 2 \times (\% \Delta t)\)

\(\% \Delta g = 1.0\% + 2 \times 2.0\% = 5.0\%\)

So, the percentage uncertainty in \(g\) is \(5.0\%\).

1 mark for the correct option (B).

- Correct calculation of percentage uncertainty in \(s\) (1.0%) and \(t\) (2.0%).
- Application of the uncertainty combination rule for powers: adding \(\% \Delta s\) and \(2 \times \% \Delta t\) to obtain 5.0%.

To move up the slope at a constant speed, the forward force \(F\) exerted by the motor must balance the components of weight acting down the slope and the resistive force:

\(F = mg \sin\theta + R\)

\(F = 0.50 \times 9.81 \times \sin(30^\circ) + 1.5\)

\(F = 2.4525 + 1.5 = 3.9525\text{ N}\)

Now, calculate the power using \(P = F v\):

\(P = 3.9525 \times 2.0 = 7.905\text{ W} \approx 7.9\text{ W}\).

1 mark for the correct answer C.

- Award 1 mark for correct calculation leading to option C.
- Reject option A (neglected resistive force: \(4.9\text{ W}\)).
- Reject option B (neglected gravity component: \(3.0\text{ W}\)).
- Reject option D (used cosine instead of sine: \(11.5\text{ W}\)).

The formula for the fringe separation is given by:

\(w = \frac{\lambda D}{d}\)

Substituting the new values into the formula:

\(w' = \frac{(1.2\lambda)(0.5D)}{2d} = \frac{0.6 \lambda D}{2d} = 0.3 \frac{\lambda D}{d} = 0.3w\)

1 mark for the correct answer A.

- Award 1 mark for identifying the correct relationship and scaling factors to obtain \(0.3w\).
- Reject other options where algebraic errors are made (e.g., multiplying instead of dividing by the slit separation scaling factor).

Use the equation \(\varepsilon = I(R + r)\) for both scenarios:

1) \(\varepsilon = 1.2(4.0 + r)\)
2) \(\varepsilon = 0.6(10.0 + r)\)

Equating both expressions for \(\varepsilon\):

\(1.2(4.0 + r) = 0.6(10.0 + r)\)

\(2(4.0 + r) = 10.0 + r\)

\(8.0 + 2r = 10.0 + r\)

\(r = 2.0\ \Omega\)

Substitute \(r = 2.0\ \Omega\) back into equation 1:

\(\varepsilon = 1.2(4.0 + 2.0) = 1.2 \times 6.0 = 7.2\text{ V}\)

1 mark for the correct answer B.

- Award 1 mark for solving the simultaneous equations correctly for both \(\varepsilon\) and \(r\).
- Reject incorrect options arising from algebraic mistakes.

The decay involves a meson decaying into leptons, which involves a change of quark flavor (from \(d\bar{u}\) quarks to leptons). This must occur via the weak interaction. Because charge is conserved and the initial pion has a charge of \(-1\), the weak exchange particle responsible is the charged \(W^-\) boson.

1 mark for the correct answer C.

- Reject A: The decay is mediated by the weak interaction, not the strong interaction (gluons).
- Reject B: The muon lepton family number is conserved (initially 0, after is \(+1 - 1 = 0\)).
- Reject D: Pions do not contain strange quarks, so there is no change in strangeness.

The frequency of the first harmonic is given by:

\(f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\)

When the length is changed to \(L' = 0.5L\) and tension to \(T' = 1.44T\):

\(f' = \frac{1}{2(0.5L)} \sqrt{\frac{1.44T}{\mu}} = \frac{1}{0.5} \sqrt{1.44} \times \left( \frac{1}{2L} \sqrt{\frac{T}{\mu}} \right)\)

\(f' = 2 \times 1.2 \times f = 2.4f\)

1 mark for the correct answer C.

- Award 1 mark for correct application of the fundamental frequency formula and scaling factors.
- Reject option A (incorrect inverse scaling for length: \(0.6f\)).
- Reject option B (neglecting the length change scaling: \(1.2f\)).

The formula for density \(\rho\) of a cylinder is:

\(\rho = \frac{m}{V} = \frac{m}{\pi (d/2)^2 h} = \frac{4m}{\pi d^2 h}

To find the percentage uncertainty in \)\rho\), we sum the percentage uncertainties of the independent terms, multiplying by their respective powers:

\(\%\Delta \rho = \%\Delta m + 2(\%\Delta d) + \%\Delta h\)

\(\%\Delta \rho = 1.5\% + 2(1.0\%) + 2.0\% = 1.5\% + 2.0\% + 2.0\% = 5.5\%\)

1 mark for the correct answer B.

- Award 1 mark for doubling the percentage uncertainty of diameter and adding all uncertainties.
- Reject option A (adding without doubling diameter: \(4.5\%\)).
- Reject other options containing computational or power relationship errors.