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The strain \(\varepsilon\) in a wire is given by \(\varepsilon = \frac{\sigma}{E} = \frac{F}{A E}\), where \(F\) is the applied force, \(A\) is the cross-sectional area, and \(E\) is the Young modulus. Since the wires are made of the same material, they have the same Young modulus \(E\). Since the same force \(F\) is applied to both wires, the strain is inversely proportional to the cross-sectional area: \(\varepsilon \propto \frac{1}{A} \propto \frac{1}{d^2}\). Therefore, the ratio of the strain in X to the strain in Y is: \(\frac{\varepsilon_X}{\varepsilon_Y} = \frac{d_Y^2}{d_X^2} = \frac{(2d)^2}{d^2} = 4\).

1 mark for the correct calculation of the ratio of strains, demonstrating that strain is inversely proportional to the square of the diameter.

First, calculate the voltmeter reading (potential difference across the LDR) in bright light: \(V_{\text{bright}} = \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}} \times V_{\text{in}} = \frac{3.0}{6.0 + 3.0} \times 12 = 4.0\text{ V}\). Next, calculate the voltmeter reading in darkness: \(V_{\text{dark}} = \frac{18}{6.0 + 18} \times 12 = 9.0\text{ V}\). The change in the reading of the voltmeter is \(9.0\text{ V} - 4.0\text{ V} = +5.0\text{ V}\), which represents an increase of \(5.0\text{ V}\).

1 mark for calculating the correct change in potential difference showing an increase of 5.0 V.

The total energy \(E\) of the oscillation is the sum of the kinetic energy \(E_k\) and the potential energy \(E_p\): \(E = E_k + E_p\). We are given that \(E_k = 3E_p\). Substituting this into the total energy equation gives: \(E = 3E_p + E_p = 4E_p\). The total energy is given by \(E = \frac{1}{2} k A^2\) and the potential energy at displacement \(x\) is given by \(E_p = \frac{1}{2} k x^2\). Thus: \(\frac{1}{2} k A^2 = 4 \left( \frac{1}{2} k x^2 \right) \implies A^2 = 4x^2 \implies x = \pm 0.50A\).

1 mark for establishing the relationship between total energy and potential energy, leading to the correct displacement of 0.50 A.

Let the original length be \(L\) and the original cross-sectional area be \(A\). The original resistance is \(R = \rho \frac{L}{A}\). The new length is \(L' = 1.10 L\). Since the volume \(V = A L\) remains constant, the new cross-sectional area \(A'\) satisfies \(A' L' = A L \implies A' = \frac{A L}{1.10 L} = \frac{A}{1.10}\). The new resistance \(R'\) is: \(R' = \rho \frac{L'}{A'} = \rho \frac{1.10 L}{A / 1.10} = 1.10^2 \left( \rho \frac{L}{A} \right) = 1.21 R\).

1 mark for relating resistance to the square of the length under constant volume, resulting in 1.21 R.

Using the grating equation \(d \sin \theta = n \lambda\) for the third-order maximum (\(n = 3\)): \(d \sin 30^\circ = 3 \lambda\). Since \(\sin 30^\circ = 0.5\), we get \(0.5 d = 3 \lambda \implies d = 6 \lambda\). The maximum possible order of diffraction occurs when \(\sin \theta < 1\), which gives \(n < \frac{d}{\lambda} = 6\). Thus, the highest observable order is \(n = 5\) (since \(n = 6\) would correspond to \(\theta = 90^\circ\), which is parallel to the grating and cannot be projected or observed). The observable orders are \(n = 0, \pm 1, \pm 2, \pm 3, \pm 4, \pm 5\). The total number of observable maxima is \(2 \times 5 + 1 = 11\).

1 mark for calculating the grating spacing, finding the maximum order of 5, and correctly determining the total of 11 maxima.

Let \(I\) be the total current flowing from the source. The single third resistor is in series with the parallel combination, so the entire current \(I\) flows through it. The power dissipated in this third resistor is \(P_3 = I^2 R\). The parallel combination consists of two identical resistors of resistance \(R\) in parallel, so its equivalent resistance is \(R_{\text{p}} = \frac{R}{2}\). The total current flowing through this combination is also \(I\). The total power dissipated in the parallel combination is \(P_{\text{p}} = I^2 R_{\text{p}} = I^2 \left(\frac{R}{2}\right) = 0.5 I^2 R\). The ratio is \(\frac{P_{\text{p}}}{P_3} = \frac{0.5 I^2 R}{I^2 R} = 0.5 = \frac{1}{2}\).

1 mark for finding the equivalent resistance of the parallel pair and calculating the correct ratio of power.

For simple harmonic motion, the maximum speed is \(v_{\text{max}} = \omega x_0\) and the maximum acceleration is \(a_{\text{max}} = \omega^2 x_0\), where \(\omega\) is the angular frequency. Dividing the two expressions gives \(\frac{a_{\text{max}}}{v_{\text{max}}} = \omega\). Since \(\omega = 2 \pi f\), we have \(2 \pi f = \frac{a_{\text{max}}}{v_{\text{max}}} \implies f = \frac{a_{\text{max}}}{2 \pi v_{\text{max}}}\).

1 mark for expressing frequency in terms of maximum acceleration and maximum velocity correctly.

The work done in stretching a wire is given by the integral of force with respect to extension, which corresponds to the area under the tension-extension graph. Within the limit of proportionality, this work is fully stored as strain energy (or elastic potential energy) in the wire. Therefore, the area under the graph represents the elastic potential energy stored in the wire.

1 mark for identifying the area under a force-extension graph as the work done or elastic potential energy.

The terminal potential difference \(V\) is related to the e.m.f. \(E\), current \(I\), and internal resistance \(r\) by the equation:
\(V = E - Ir\)

Rearranging this into the standard equation of a straight line, \(y = mx + c\):
\(V = -rI + E\)

Comparing this to \(y = mx + c\) where \(V\) is on the vertical axis and \(I\) is on the horizontal axis:
- The gradient \(m\) is equal to \(-r\).
- The vertical intercept \(c\) is equal to \(E\).

1 mark for identifying the correct graphical equation and matching its gradient and vertical intercept to the physical variables.

The resistance of a wire is given by \(R = \rho \frac{L}{A}\), where the cross-sectional area \(A = \frac{\pi d^2}{4}\).
This can be written as:
\(R = \frac{4\rho L}{\pi d^2}\)

For the second wire:
\(R_2 = \frac{4\rho (2L)}{\pi (2d)^2} = \frac{8\rho L}{4\pi d^2} = 0.5 \left( \frac{4\rho L}{\pi d^2} \right) = 0.5 R\).

1 mark for correctly applying the resistivity formula and accounting for the changes in both dimensions.

The extension \(e\) is given by:
\(e = \frac{F L}{A E}\)

Since the wires are in series, the tensile force \(F\) is the same in both wires. Let subscripts \(s\) and \(b\) denote steel and brass respectively:
- Length ratio: \(\frac{L_s}{L_b} = 2\)
- Diameter ratio: \(\frac{d_s}{d_b} = 0.5\), which means area ratio \(\frac{A_s}{A_b} = (0.5)^2 = 0.25 \implies \frac{A_b}{A_s} = 4\)
- Young modulus ratio: \(\frac{E_s}{E_b} = \frac{2.0 \times 10^{11}}{1.0 \times 10^{11}} = 2 \implies \frac{E_b}{E_s} = 0.5\)

Calculating the ratio of extensions:
\(\frac{e_s}{e_b} = \frac{L_s}{L_b} \times \frac{A_b}{A_s} \times \frac{E_b}{E_s} = 2 \times 4 \times 0.5 = 4\).

1 mark for setting up the ratio equation and calculating the correct numerical factor of 4.

First, calculate the grating spacing \(d\):
\(d = \frac{1}{6.0 \times 10^5\text{ m}^{-1}} = 1.67 \times 10^{-6}\text{ m} = 1670\text{ nm}\)

The diffraction grating equation is:
\(d \sin\theta = n\lambda\)

For the maximum order \(n\) visible, \(\sin\theta \le 1\):
\(n \le \frac{d}{\lambda} = \frac{1670}{540} \approx 3.09\)

Since \(n\) must be an integer, the highest observable order of diffraction is \(n = 3\).

The total number of visible bright fringes is the central maximum (zero-order) plus three orders on each side:
\(\text{Total fringes} = 2n + 1 = 2(3) + 1 = 7\).

1 mark for calculating the maximum diffraction order to be 3, and correctly determining the total number of fringes as 7.

First, calculate the output voltage in bright light:
\(V_{\text{out, bright}} = 12.0\text{ V} \times \frac{1.5\text{ k}\Omega}{3.0\text{ k}\Omega + 1.5\text{ k}\Omega} = 12.0 \times \frac{1.5}{4.5} = 4.0\text{ V}\)

Next, calculate the output voltage in the dark:
\(V_{\text{out, dark}} = 12.0\text{ V} \times \frac{9.0\text{ k}\Omega}{3.0\text{ k}\Omega + 9.0\text{ k}\Omega} = 12.0 \times \frac{9.0}{12.0} = 9.0\text{ V}\)

The change in the output voltage is:
\(\Delta V_{\text{out}} = V_{\text{out, dark}} - V_{\text{out, bright}} = 9.0\text{ V} - 4.0\text{ V} = 5.0\text{ V}\).

1 mark for computing both initial and final output voltages and subtracting them to obtain 5.0 V.

First, find the extension of the spring under a \(24\text{ N}\) load:
\(e_1 = 16.0\text{ cm} - 12.0\text{ cm} = 4.0\text{ cm} = 0.040\text{ m}\)

The spring constant \(k\) is:
\(k = \frac{F_1}{e_1} = \frac{24\text{ N}}{0.040\text{ m}} = 600\text{ N m}^{-1}\)

When the spring has a length of \(20.0\text{ cm}\), its extension \(e_2\) is:
\(e_2 = 20.0\text{ cm} - 12.0\text{ cm} = 8.0\text{ cm} = 0.080\text{ m}\)

The elastic potential energy \(E_p\) stored is:
\(E_p = \frac{1}{2} k e_2^2 = \frac{1}{2} \times 600 \times (0.080)^2 = 300 \times 0.0064 = 1.92\text{ J}\).

1 mark for calculating the spring constant (or scaling) and correctly calculating the energy using the appropriate extension of 0.080 m.

Let us evaluate the four possible configurations of three identical resistors of resistance \(R\):
1. All in series: \(R + R + R = 3R\)
2. All in parallel: \(\left(\frac{1}{R} + \frac{1}{R} + \frac{1}{R}\right)^{-1} = \frac{R}{3} \approx 0.33 R\)
3. Two in parallel, in series with the third: \(\frac{R}{2} + R = 1.5 R\)
4. Two in series, in parallel with the third: \(\frac{2R \times R}{2R + R} = \frac{2}{3}R \approx 0.67 R\)

Comparing with the options, \(0.75 R\) is not a mathematically possible combination of all three resistors.

1 mark for analyzing possible configurations of three resistors and identifying 0.75 R as impossible.

The electric current \(I\) in a conductor is related to the drift speed \(v\) by the equation:
\(I = n A v e\)
where \(n\) is the number density of conduction electrons and \(e\) is the elementary charge.

Since both wires are made of copper at the same temperature, \(n\) remains unchanged. Rearranging for drift speed:
\(v \propto \frac{I}{A}\)

For the second wire:
\(v_2 \propto \frac{2I}{3A} \implies v_2 = \frac{2}{3} v\).

1 mark for utilizing the current equation and finding the correct ratio of the drift speed.

The resistance \(R\) of a wire of length \(L\), resistivity \(\rho\), and diameter \(d\) is given by:

\(R = \rho \frac{L}{A} = \rho \frac{L}{\pi d^2 / 4} \propto \frac{L}{d^2}

Let the resistance of wire X be \)R_X \propto \frac{L}{d^2}\).
Then the resistance of wire Y is:

\(R_Y \propto \frac{3L}{(2d)^2} = \frac{3L}{4d^2} = 0.75 R_X\)

Since the wires are connected in parallel, the potential difference \(V\) across each wire is the same. The power dissipated \(P\) is given by \(P = \frac{V^2}{R}\).

Thus, the ratio of power dissipated is:

\(\frac{P_X}{P_Y} = \frac{V^2 / R_X}{V^2 / R_Y} = \frac{R_Y}{R_X} = 0.75\)

1 mark for the correct answer B.
- Award 1 mark for calculating the resistance ratio \(R_Y/R_X = 0.75\) and recognizing that power is inversely proportional to resistance for parallel connections, leading to \(\frac{P_X}{P_Y} = 0.75\).

First, calculate the tensile force \(F\) acting on the wire due to the hanging mass:
\(F = mg = 12 \times 9.81 = 117.72\text{ N}

The extension \)x\) of the wire is given by:
\(x = \frac{FL}{AE} = \frac{117.72 \times 2.5}{1.5 \times 10^{-6} \times 1.2 \times 10^{11}} = \frac{294.3}{1.8 \times 10^5} = 1.635 \times 10^{-3}\text{ m}

The elastic potential energy \)E_p\) stored in the wire is:
\(E_p = \frac{1}{2} F x = 0.5 \times 117.72 \times 1.635 \times 10^{-3} \approx 0.0962\text{ J}

Rounding to two significant figures gives \)0.096\text{ J}\).

1 mark for the correct answer B.
- Award 1 mark for calculating the correct force \(F\), determining the extension \(x\), and using \(E_p = \frac{1}{2}Fx\) to obtain \(0.096\text{ J}\).

The diffraction grating equation is given by:
\(d \sin \theta = n \lambda

where:
\)n = 3
\(\lambda = 540 \times 10^{-9}\text{ m}
\)\theta = 35.0^\circ

Calculate the grating spacing \(d\):
\(d = \frac{n \lambda}{\sin \theta} = \frac{3 \times 540 \times 10^{-9}}{\sin(35.0^\circ)} = \frac{1.62 \times 10^{-6}}{0.5736} \approx 2.824 \times 10^{-6}\text{ m}

The number of lines per metre \)N\) is:
\(N = \frac{1}{d} = \frac{1}{2.824 \times 10^{-6}} \approx 3.54 \times 10^5\text{ m}^{-1}

To find the number of lines per millimetre, divide by 1000:
\)N_{\text{mm}} = \frac{3.54 \times 10^5}{1000} \approx 354\text{ lines mm}^{-1}\).

1 mark for the correct answer B.
- Award 1 mark for correctly applying the grating formula to find \(d\), converting it to lines per metre, and then to lines per millimetre.

The relationship between terminal potential difference \(V\), electromotive force \(E\), current \(I\), and internal resistance \(r\) is:
\(V = E - Ir

Rearranging into the standard linear equation form \)y = mx + c\):
\(V = -r I + E

This is a straight-line graph where:
- The vertical intercept (y-intercept) is \)E\), which is a non-zero positive value.
- The gradient is \(-r\), which is a constant negative value.

Therefore, the graph is a straight line with a constant negative gradient and a non-zero positive vertical intercept.

1 mark for the correct answer A.
- Award 1 mark for identifying the linear relation \(V = E - Ir\) and correctly interpreting its gradient and intercept.

The angular frequency \(\omega\) is given by:
\(\omega = \frac{2\pi}{T} = \frac{2\pi}{2.0} = \pi\text{ rad s}^{-1}

The maximum acceleration \)a_{\text{max}}\) of an object in simple harmonic motion is given by:
\(a_{\text{max}} = \omega^2 A

where \)A\) is the amplitude in metres (\(4.0\text{ cm} = 0.040\text{ m}\)):
\(a_{\text{max}} = \pi^2 \times 0.040 \approx 9.87 \times 0.040 = 0.395\text{ m s}^{-2}

Rounding to two significant figures, we get \)0.40\text{ m s}^{-2}\).

1 mark for the correct answer C.
- Award 1 mark for finding \(\omega = \pi\text{ rad s}^{-1}\) and calculating \(a_{\text{max}} = \omega^2 A\) to get \(0.40\text{ m s}^{-2}\).

The voltmeter reading \(V_{\text{out}}\) is the potential difference across the fixed resistor \(R = 2.0\text{ k}\Omega\).

Initially, the resistance of the thermistor is \(R_{\text{th}} = 4.0\text{ k}\Omega\). The initial voltmeter reading is:
\(V_{\text{initial}} = V_{\text{supply}} \times \frac{R}{R + R_{\text{th}}} = 9.0 \times \frac{2.0}{2.0 + 4.0} = 9.0 \times \frac{2.0}{6.0} = 3.0\text{ V}

Finally, the resistance of the thermistor decreases to \)R_{\text{th}} = 1.0\text{ k}\Omega\). The final voltmeter reading is:
\(V_{\text{final}} = V_{\text{supply}} \times \frac{R}{R + R_{\text{th}}} = 9.0 \times \frac{2.0}{2.0 + 1.0} = 9.0 \times \frac{2.0}{3.0} = 6.0\text{ V}

The change in the reading is:
\)V_{\text{final}} - V_{\text{initial}} = 6.0 - 3.0 = +3.0\text{ V}\) (an increase of \(3.0\text{ V}\)).

1 mark for the correct answer D.
- Award 1 mark for calculating the initial and final voltage across the fixed resistor and finding the difference of 3.0 V (increase).

When damping is increased in a forced oscillation system:
1. The resonance curve becomes flatter, and the peak (maximum) amplitude decreases.
2. The peak of the curve shifts slightly to the left, which means the resonant frequency (the frequency at which the maximum amplitude occurs) decreases.

Therefore, both the maximum amplitude and the frequency at which it occurs decrease.

1 mark for the correct answer A.
- Award 1 mark for correctly identifying both effects of increased damping on the resonance curve (amplitude decreases, peak shifts to a lower frequency).

The strain energy \(E_s\) stored in a wire stretched by force \(F\) is given by:
\(E_s = \frac{1}{2} F x

where the extension \)x\) is given by:
\(x = \frac{FL}{AE}

Substituting \)x\) into the energy formula:
\(E_s = \frac{F^2 L}{2AE}

Since both wires are made of the same material (copper), they have the same Young modulus \)E\). They are also subjected to the same force \(F\). Thus:
\(E_s \propto \frac{L}{A}

Since cross-sectional area \)A \propto d^2\) (where \(d\) is the diameter):
\(E_s \propto \frac{L}{d^2}

For wire P:
\)L_P = 2L_Q
\(d_P = 0.5d_Q

Therefore, the ratio of strain energy stored in P to Q is:
\)\frac{E_{s,P}}{E_{s,Q}} = \frac{L_P / d_P^2}{L_Q / d_Q^2} = \frac{2L_Q / (0.5d_Q)^2}{L_Q / d_Q^2} = \frac{2}{0.25} = 8\)

1 mark for the correct answer C.
- Award 1 mark for identifying the relationship between strain energy, length, and diameter, and calculating the ratio of 8.

We use the equation for terminal potential difference: \(V = E - Ir\), where \(I = \frac{V}{R}\). Thus, \(E = V\left(1 + \frac{r}{R}\right)\). For the first case: \(E = 8.0\left(1 + \frac{r}{4.0}\right) = 8.0 + 2.0r\). For the second case: \(E = 10.0\left(1 + \frac{r}{10.0}\right) = 10.0 + r\). Equating the two expressions for \(E\): \(8.0 + 2.0r = 10.0 + r\), which gives \(r = 2.0\ \Omega\). Substituting \(r\) back into one of the equations: \(E = 10.0 + 2.0 = 12\text{ V}\).

1 mark for the correct combination of E and r. No partial marks.

The resistance of a wire is given by \(R = \rho \frac{L}{A} = \rho \frac{4L}{\pi d^2}\). Thus, resistance is proportional to \(\frac{L}{d^2}\). For wire X: \(R_X \propto \frac{L}{d^2}\). For wire Y: \(R_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2} \frac{L}{d^2}\). This means \(R_Y = 0.5 R_X\). Since the wires are connected in parallel, they have the same potential difference \(V\) across them. The power dissipated is \(P = \frac{V^2}{R}\). Therefore, the ratio of power is: \(\frac{P_Y}{P_X} = \frac{V^2 / R_Y}{V^2 / R_X} = \frac{R_X}{R_Y} = \frac{R_X}{0.5 R_X} = 2.0\).

1 mark for the correct ratio of 2.0.

The relationship between Young modulus \(E\), force \(F\), original length \(L\), cross-sectional area \(A\), and extension \(\Delta L\) is: \(E = \frac{F L}{A \Delta L}\). Rearranging for extension \(\Delta L\): \(\Delta L = \frac{F L}{A E}\). The force \(F\) is due to gravity on the mass: \(F = m g = 15 \times 9.81 = 147.15\text{ N}\). Substituting the values: \(\Delta L = \frac{147.15 \times 2.0}{(1.5 \times 10^{-6}) \times (2.0 \times 10^{11})} = \frac{294.3}{3.0 \times 10^5} = 9.81 \times 10^{-4}\text{ m} = 0.98\text{ mm}\).

1 mark for the correct extension in mm.

First, find the grating spacing \(d\): \(d = \frac{1 \times 10^{-3}\text{ m}}{500} = 2.0 \times 10^{-6}\text{ m}\). The condition for diffraction maxima is \(d \sin\theta = n \lambda\). For the maximum order \(n\), we set \(\sin\theta \le 1\): \(n \le \frac{d}{\lambda} = \frac{2.0 \times 10^{-6}\text{ m}}{600 \times 10^{-9}\text{ m}} \approx 3.33\). Since \(n\) must be an integer, the highest observable order of bright fringe on either side of the central maximum is \(n = 3\). The total number of bright fringes includes the central maximum (\(n = 0\)) and the three orders on each side: \(\text{Total} = 2n + 1 = 2(3) + 1 = 7\).

1 mark for the correct calculation of total visible maxima.

The total resistance in the driver circuit is: \(R_{\text{total}} = R_{AB} + R = 5.0 + 15.0 = 20.0\ \Omega\). The current \(I\) in the driver circuit is: \(I = \frac{2.0\text{ V}}{20.0\ \Omega} = 0.10\text{ A}\). The potential difference \(V_{AB}\) across the entire potentiometer wire \(AB\) is: \(V_{AB} = I \times R_{AB} = 0.10\text{ A} \times 5.0\ \Omega = 0.50\text{ V}\). The potential gradient along the wire is: \(\frac{V_{AB}}{L} = \frac{0.50\text{ V}}{1.00\text{ m}} = 0.50\text{ V m}^{-1}\). At the balance point, the e.m.f. of the test cell is equal to the potential difference across the balance length (\(60.0\text{ cm} = 0.60\text{ m}\)): \(E_{\text{test}} = 0.50\text{ V m}^{-1} \times 0.60\text{ m} = 0.30\text{ V}\).

1 mark for the correct value of the test cell's e.m.f.

As temperature increases, the amplitude of the vibrations of the metal lattice ions increases. This increases the rate of collisions between the conduction electrons and the lattice ions, which increases the resistance \(R\) of the conductor. Since the potential difference \(V\) is kept constant and \(R\) increases, the current \(I = \frac{V}{R}\) must decrease. From the equation \(I = n A v q\), because the number density of free electrons \(n\), the cross-sectional area \(A\), and the elementary charge \(q\) are constant, the decrease in current \(I\) results in a decrease in the drift velocity \(v\).

1 mark for identifying that drift velocity decreases and resistance increases.

The total work done is the area under the force-extension graph. We can divide this area into a triangle (for the linear elastic region) and a rectangle (for the plastic region): 1. Area of the triangle (extension from \(0\) to \(2.0 \times 10^{-3}\text{ m}\)): \(W_1 = \frac{1}{2} \times \text{Force} \times \text{Extension} = \frac{1}{2} \times 200\text{ N} \times 2.0 \times 10^{-3}\text{ m} = 0.20\text{ J}\). 2. Area of the rectangle (extension from \(2.0 \times 10^{-3}\text{ m}\) to \(5.0 \times 10^{-3}\text{ m}\)): \(W_2 = \text{Force} \times \Delta x = 200\text{ N} \times (5.0 - 2.0) \times 10^{-3}\text{ m} = 200\text{ N} \times 3.0 \times 10^{-3}\text{ m} = 0.60\text{ J}\). Total work done: \(W_{\text{total}} = 0.20\text{ J} + 0.60\text{ J} = 0.80\text{ J}\).

1 mark for the correct total work done of 0.80 J.

First, calculate the grating spacing \(d\): \(d = \frac{1}{4.0 \times 10^5\text{ m}^{-1}} = 2.5 \times 10^{-6}\text{ m}\). For the first-order spectrum (\(n = 1\)), use the diffraction grating formula \(d \sin\theta = n \lambda\): For violet light (\(\lambda_V = 400 \times 10^{-9}\text{ m}\)): \(\sin\theta_V = \frac{\lambda_V}{d} = \frac{400 \times 10^{-9}}{2.5 \times 10^{-6}} = 0.160\), which gives \(\theta_V = \sin^{-1}(0.160) \approx 9.20^\circ\). For red light (\(\lambda_R = 700 \times 10^{-9}\text{ m}\)): \(\sin\theta_R = \frac{\lambda_R}{d} = \frac{700 \times 10^{-9}}{2.5 \times 10^{-6}} = 0.280\), which gives \(\theta_R = \sin^{-1}(0.280) \approx 16.26^\circ\). Calculate the angular separation: \(\Delta \theta = \theta_R - \theta_V = 16.26^\circ - 9.20^\circ = 7.06^\circ \approx 7.1^\circ\).

1 mark for calculating the correct angles and finding the difference of 7.1 degrees.

Let the resistance of the external resistor be \(R = 3r\).

**Case 1: Single cell**
The total resistance in the circuit is:
\(R_{\text{total, 1}} = r + R = r + 3r = 4r\)

The current \(I_1\) in the circuit is:
\(I_1 = \frac{E}{4r\)}

The power \(P_1\) dissipated in the external resistor is:
\(P_1 = I_1^2 R = \left(\frac{E}{4r}\right)^2 (3r) = \frac{3E^2}{16r}\)

**Case 2: Two cells in parallel**
The two identical cells in parallel are equivalent to a single cell with:
- Equivalent e.m.f. \(E_{\text{eq}} = E\)
- Equivalent internal resistance \(r_{\text{eq}} = \frac{r}{2}\)

The total resistance in the circuit is:
\(R_{\text{total, 2}} = \frac{r}{2} + 3r = 3.5r = \frac{7r}{2}\)

The current \(I_2\) in the circuit is:
\(I_2 = \frac{E}{3.5r} = \frac{2E}{7r}\)

The power \(P_2\) dissipated in the external resistor is:
\(P_2 = I_2^2 R = \left(\frac{2E}{7r}\right)^2 (3r) = \frac{12E^2}{49r}\)

**Ratio of power:**
\(\frac{P_2}{P_1} = \frac{\frac{12E^2}{49r}}{\frac{3E^2}{16r}} = \frac{12}{49} \times \frac{16}{3} = \frac{64}{49}\)

1 mark for the correct calculation of the ratio of power, showing clear steps for both circuit configurations (single cell vs parallel cells).

For simple harmonic motion, the speed \(v\) of an object at displacement \(x\) is given by:
\(v = \omega \sqrt{x_0^2 - x^2}\)

The maximum speed \(v_0\) occurs at the equilibrium position (\(x=0\)):
\(v_0 = \omega x_0\)

We are given that \(v = \frac{\sqrt{3}}{2} v_0\):
\(\omega \sqrt{x_0^2 - x^2} = \frac{\sqrt{3}}{2} \omega x_0\)

Divide both sides by \(\omega\):
\(\sqrt{x_0^2 - x^2} = \frac{\sqrt{3}}{2} x_0\)

Square both sides:
\(x_0^2 - x^2 = \frac{3}{4} x_0^2\)
\(x^2 = x_0^2 - \frac{3}{4} x_0^2 = \frac{1}{4} x_0^2\)
\(x = \pm 0.50 x_0\)

Thus, the magnitude of the displacement is \(0.50 x_0\).

1 mark for correctly relating speed to displacement in simple harmonic motion and solving for \(x\) in terms of \(x_0\).

Let the first wire have mass \(M_1\), length \(L_1 = L\), cross-sectional area \(A_1\), diameter \(d_1 = d\), and resistance \(R_1 = R\).
Let the second wire have mass \(M_2 = 2M_1\), length \(L_2\), cross-sectional area \(A_2\), diameter \(d_2 = 0.5d\), and resistance \(R_2\).

Since both wires are made of the same material, they have the same density \(D\):
\(M = D \times A \times L\)

For the second wire, the diameter is halved, so its cross-sectional area is:
\(A_2 = \frac{\pi d_2^2}{4} = \frac{\pi (0.5d)^2}{4} = 0.25 A_1\)

Using the relation for mass:
\(M_2 = 2M_1 \implies A_2 L_2 = 2 A_1 L_1\)
\((0.25 A_1) L_2 = 2 A_1 L_1 \implies L_2 = 8 L_1\)

The resistance of a wire is given by \(R = \rho \frac{L}{A}\):
\(R_2 = \rho \frac{L_2}{A_2} = \rho \frac{8 L_1}{0.25 A_1} = 32 \left(\rho \frac{L_1}{A_1}\right) = 32 R\)

1 mark for finding the correct relationship between length, area, mass, and resistance, yielding \(32R\).

The Young modulus is defined as:
\(E = \frac{\text{stress}}{\text{strain}} = \frac{F / A}{x / L} = \frac{F L}{A x}\)

Rearranging for extension \(x\):
\(x = \frac{F L}{A E}\)

Since the wires are joined in series, the tensile force \(F\) is the same in both wires.

For the steel wire:
\(x_s = \frac{F L}{\left(\frac{\pi d^2}{4}\right) E_s}\)

For the brass wire:
\(x_b = \frac{F (2L)}{\left(\frac{\pi (2d)^2}{4}\right) E_b} = \frac{2 F L}{4 \left(\frac{\pi d^2}{4}\right) E_b} = \frac{F L}{2 \left(\frac{\pi d^2}{4}\right) E_b}\)

Taking the ratio of the extensions:
\(\frac{x_s}{x_b} = \frac{\frac{F L}{\left(\frac{\pi d^2}{4}\right) E_s}}{\frac{F L}{2 \left(\frac{\pi d^2}{4}\right) E_b}} = \frac{1 / E_s}{1 / (2 E_b)} = \frac{2 E_b}{E_s}\)

1 mark for correctly applying the Young modulus formula to both series-connected wires and evaluating the ratio of extensions.

The diffraction grating equation is:
\(d \sin \theta = n \lambda\)

where:
- \(d\) is the slit separation,
- \(\theta = 36.9^\circ\) is the angle of diffraction,
- \(n = 3\) is the order of the maximum,
- \(\lambda = 600 \times 10^{-9}\text{ m}\) is the wavelength.

Rearranging to solve for \(d\):
\(d = \frac{n \lambda}{\sin \theta} = \frac{3 \times 600 \times 10^{-9}\text{ m}}{\sin(36.9^\circ)}

Using \)\sin(36.9^\circ) \approx 0.60\):
\(d = \frac{1.80 \times 10^{-6}\text{ m}}{0.60} = 3.0 \times 10^{-6}\text{ m}

The number of lines per metre \)N\) is:
\(N = \frac{1}{d} = \frac{1}{3.0 \times 10^{-6}\text{ m}} = 3.33 \times 10^5\text{ lines per m}

To find the number of lines per millimetre:
\)N_{\text{mm}} = \frac{3.33 \times 10^5}{1000} \approx 330\text{ lines per mm}\)

1 mark for calculating the slit separation and converting it correctly to the number of lines per millimetre.

**Initial State:**
- Resistance of fixed resistor \(R = 4.0\text{ k}\Omega\)
- Resistance of thermistor \(R_T = 8.0\text{ k}\Omega\)
- Total resistance \(R_{\text{total}} = 4.0\text{ k}\Omega + 8.0\text{ k}\Omega = 12.0\text{ k}\Omega\)

The initial potential difference \(V_{\text{initial}}\) across the fixed resistor is:
\(V_{\text{initial}} = 12.0\text{ V} \times \frac{4.0\text{ k}\Omega}{12.0\text{ k}\Omega} = 4.0\text{ V}

**Final State:**
- Resistance of fixed resistor \)R = 4.0\text{ k}\Omega\)
- Resistance of thermistor \(R_T = 2.0\text{ k}\Omega\)
- Total resistance \(R_{\text{total}} = 4.0\text{ k}\Omega + 2.0\text{ k}\Omega = 6.0\text{ k}\Omega\)

The final potential difference \(V_{\text{final}}\) across the fixed resistor is:
\(V_{\text{final}} = 12.0\text{ V} \times \frac{4.0\text{ k}\Omega}{6.0\text{ k}\Omega} = 8.0\text{ V}

**Change in potential difference:**
\)\Delta V = V_{\text{final}} - V_{\text{initial}} = 8.0\text{ V} - 4.0\text{ V} = 4.0\text{ V}\) (increase)

1 mark for determining both initial and final potential differences and finding the correct change of +4.0 V.

Since the wire remains within its limit of proportionality, the force-extension graph is a straight line.

The increase in elastic potential energy is represented by the area under the force-extension graph between the two states. This area is a trapezoid:
\(\Delta E = \frac{1}{2} (F_1 + F_2) \Delta x\)

where:
- \(F_1 = 80\text{ N}\)
- \(F_2 = 120\text{ N}\)
- \(\Delta x = 0.75\text{ mm} = 0.75 \times 10^{-3}\text{ m}\)

Calculating \(\Delta E\):
\(\Delta E = \frac{1}{2} \times (80\text{ N} + 120\text{ N}) \times (0.75 \times 10^{-3}\text{ m})\)
\(\Delta E = \frac{1}{2} \times 200\text{ N} \times 0.75 \times 10^{-3}\text{ m}\)
\(\Delta E = 100 \times 0.75 \times 10^{-3}\text{ m} = 0.075\text{ J}\)

1 mark for calculating the change in stored energy using the area under the force-extension graph (trapezoid formula) or by subtracting initial from final energy.

For forced oscillations:
1. Increasing the degree of damping always reduces the amplitude of the oscillations at all frequencies, so the maximum amplitude (at resonance) decreases.
2. Increasing the damping also shifts the peak of the resonance curve (the resonance frequency) slightly to a lower frequency.

Therefore, the amplitude at resonance decreases, and the resonance frequency shifts to a lower frequency.

1 mark for correctly identifying both the reduction in peak amplitude and the shift to a lower frequency due to increased damping.

(a) Electromotive force (e.m.f.) is defined as the total energy converted from other forms (e.g. chemical) into electrical energy per unit charge that passes through the cell/source (or the work done per unit charge in driving charge around a complete circuit).

(b) For the two parallel lamps of resistance \( R_L = 6.0\ \Omega \):
\( \frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{6.0} = \frac{2}{6.0} \implies R_p = 3.0\ \Omega \)
Total external resistance:
\( R_{\text{ext}} = R_1 + R_p = 4.5 + 3.0 = 7.5\ \Omega \).

(c) Total resistance of the circuit:
\( R_{\text{total}} = R_{\text{ext}} + r = 7.5 + 1.5 = 9.0\ \Omega \)
Total circuit current:
\( I = \frac{E}{R_{\text{total}}} = \frac{9.0\text{ V}}{9.0\ \Omega} = 1.0\text{ A} \)
Potential difference across the parallel combination:
\( V_p = I \times R_p = 1.0\text{ A} \times 3.0\ \Omega = 3.0\text{ V} \).

(d) When one filament breaks, the parallel network is replaced by a single lamp of \( 6.0\ \Omega \).
New total external resistance:
\( R_{\text{ext}}' = 4.5 + 6.0 = 10.5\ \Omega \)
New total circuit resistance:
\( R_{\text{total}}' = 10.5 + 1.5 = 12.0\ \Omega \)
New total current:
\( I' = \frac{9.0\text{ V}}{12.0\ \Omega} = 0.75\text{ A} \)
Previously, the total current of \( 1.0\text{ A} \) split equally between the two lamps, so the current through the remaining lamp was only \( 0.50\text{ A} \).
Now, the entire current of \( 0.75\text{ A} \) flows through the single remaining lamp.
Since the current through the remaining lamp has increased (from \( 0.50\text{ A} \) to \( 0.75\text{ A} \)), the power dissipated in it increases, and its brightness increases.

(a)
- Energy converted from other forms to electrical energy per unit charge [1]
- In the whole circuit / through the source [1]

(b)
- Calculation of parallel resistance: \( R_p = (1/6.0 + 1/6.0)^{-1} = 3.0\ \Omega \) [1]
- Summing with series resistor: \( 4.5 + 3.0 = 7.5\ \Omega \) with clear working [1]

(c)
- Correct calculation of total resistance \( 9.0\ \Omega \) [1]
- Calculation of total current \( 1.0\text{ A} \) [1]
- Calculation of p.d. across parallel combination \( 3.0\text{ V} \) (accept alternative potential divider methods) [1]

(d)
- Identification that the total circuit resistance increases, reducing total current to \( 0.75\text{ A} \) [1]
- Explanation that the current in the remaining lamp was originally \( 0.50\text{ A} \) and is now \( 0.75\text{ A} \) (current increased) [1]
- Conclusion that power/brightness increases [1]

(a) Potential difference is the work done (or energy transferred) per unit charge between two points.

(b) As light intensity increases, the resistance of the LDR decreases (or vice versa: as light intensity decreases, the resistance increases).

(c)(i) Using the potential divider equation:
\( V_{\text{out}} = \frac{R_{\text{LDR}}}{R + R_{\text{LDR}}} \times V_{\text{in}} = \frac{2.0\text{ k}\Omega}{8.0\text{ k}\Omega + 2.0\text{ k}\Omega} \times 12.0\text{ V} = 0.20 \times 12.0\text{ V} = 2.4\text{ V} \)

(c)(ii) With the new LDR resistance:
\( V_{\text{out}} = \frac{40\text{ k}\Omega}{8.0\text{ k}\Omega + 40\text{ k}\Omega} \times 12.0\text{ V} = \frac{40}{48} \times 12.0\text{ V} = 10.0\text{ V} \)

(d)(i) In daylight, the LDR resistance is low, so the p.d. across it is small (\( 2.4\text{ V} \)), which is below the \( 8.0\text{ V} \) threshold required to activate the switch. At night/in the dark, the LDR resistance is high, causing the p.d. across it to increase to a value (e.g. \( 10.0\text{ V} \)) that exceeds the \( 8.0\text{ V} \) threshold, thereby turning the lamp on.

(d)(ii) Set \( V_{\text{out}} = 8.0\text{ V} \):
\( 8.0 = \frac{R_{\text{LDR}}}{8.0\text{ k}\Omega + R_{\text{LDR}}} \times 12.0 \)
\( \frac{8.0}{12.0} = \frac{2}{3} = \frac{R_{\text{LDR}}}{8.0\text{ k}\Omega + R_{\text{LDR}}} \)
\( 2(8.0\text{ k}\Omega + R_{\text{LDR}}) = 3 R_{\text{LDR}} \implies 16.0\text{ k}\Omega + 2 R_{\text{LDR}} = 3 R_{\text{LDR}} \implies R_{\text{LDR}} = 16\text{ k}\Omega \).

(a)
- Energy per unit charge / work done per unit charge [1]

(b)
- Resistance decreases as light intensity increases (or vice versa) [1]

(c)(i)
- Correct potential divider formula or current calculation (\( I = 1.2\text{ mA} \)) [1]
- Correct calculation of voltage: \( 2.4\text{ V} \) [1]

(c)(ii)
- Correct calculation of voltage: \( 10.0\text{ V} \) (or \( 10\text{ V} \)) [2]

(d)(i)
- Link low light/night to high LDR resistance and hence high p.d. [1]
- Identify that daylight p.d. is below the threshold and dark p.d. is above \( 8.0\text{ V} \) to activate the switch [1]

(d)(ii)
- Set up correct ratio equation: \( \frac{R}{R+8} = \frac{8}{12} \) or equivalent [1]
- Correct calculation: \( 16\text{ k}\Omega \) (or \( 1.6 \times 10^4\ \Omega \)) [1]

(a) The two conditions for simple harmonic motion are:
1. Acceleration is directly proportional to displacement from the equilibrium position.
2. Acceleration is always directed towards the equilibrium position (opposite to the direction of displacement).

(b)(i) The maximum acceleration is given by \( a_0 = \omega^2 x_0 \), where \( x_0 = 5.0\text{ cm} = 0.050\text{ m} \).
\( 8.0 = \omega^2 \times 0.050 \)
\( \omega^2 = \frac{8.0}{0.050} = 160\text{ s}^{-2} \)
\( \omega = \sqrt{160} \approx 12.65\text{ rad s}^{-1} \approx 13\text{ rad s}^{-1} \).

(b)(ii) The frequency \( f \) is related to angular frequency by \( \omega = 2\pi f \):
\( f = \frac{\omega}{2\pi} = \frac{12.65}{2\pi} \approx 2.01\text{ Hz} \approx 2.0\text{ Hz} \).

(c) The maximum kinetic energy occurs at the equilibrium position (where \( v = v_0 = \omega x_0 \)):
\( E_{k,\text{max}} = \frac{1}{2} m v_0^2 = \frac{1}{2} m \omega^2 x_0^2 \)
\( E_{k,\text{max}} = \frac{1}{2} \times 0.25 \times 160 \times (0.050)^2 \)
\( E_{k,\text{max}} = 0.125 \times 160 \times 0.0025 = 0.050\text{ J} \).

(d) Total energy is \( E_T = E_k + E_p = \frac{1}{2} m \omega^2 x_0^2 \).
When \( E_k = E_p \), we have:
\( E_p = \frac{1}{2} E_T \implies \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} \left( \frac{1}{2} m \omega^2 x_0^2 \right) \)
\( x^2 = \frac{x_0^2}{2} \implies x = \frac{x_0}{\sqrt{2}} \)
\( x = \frac{5.0\text{ cm}}{\sqrt{2}} \approx 3.54\text{ cm} \approx 3.5\text{ cm} \).

(a)
- Acceleration proportional to displacement [1]
- Directed towards equilibrium / fixed point / opposite to displacement [1]

(b)(i)
- Formula used: \( a_0 = \omega^2 x_0 \) with correct substitution of \( a_0 = 8.0 \) and \( x_0 = 0.050 \) [1]
- Calculated value: \( 12.65\text{ rad s}^{-1} \) leading to \( 13\text{ rad s}^{-1} \) [1]

(b)(ii)
- Formula used: \( f = \omega / (2\pi) \) [1]
- Correct calculation: \( 2.0\text{ Hz} \) (accept \( 2.01\text{ Hz} \)) [1]

(c)
- Formula used: \( E_k = \frac{1}{2} m \omega^2 x_0^2 \) or \( E_k = \frac{1}{2} m v^2 \) [1]
- Correct calculation: \( 0.050\text{ J} \) (or \( 5.0 \times 10^{-2}\text{ J} \)) [1]

(d)
- Equating energies to get \( x = x_0 / \sqrt{2} \) [1]
- Correct calculation: \( 3.5\text{ cm} \) (or \( 0.035\text{ m} \), accept \( 3.54\text{ cm} \)) [1]

(a) Resistance is the ratio of potential difference across a conductor to the current through it.

(b) Resistance of the wire is:
\( R = \frac{V}{I} = \frac{3.0\text{ V}}{0.80\text{ A}} = 3.75\ \Omega \)
Using the resistivity formula \( R = \frac{\rho L}{A} \):
\( \rho = \frac{R A}{L} = \frac{3.75\ \Omega \times 1.5 \times 10^{-7}\text{ m}^2}{1.8\text{ m}} \)
\( \rho = \frac{5.625 \times 10^{-7}}{1.8} = 3.125 \times 10^{-7}\ \Omega\text{ m} \approx 3.1 \times 10^{-7}\ \Omega\text{ m} \).

(c)(i) Number density is the number of free electrons per unit volume.

(c)(ii) Using the relation \( I = n A v e \):
\( v = \frac{I}{n A e} \)
\( v = \frac{0.80}{8.5 \times 10^{28} \times 1.5 \times 10^{-7} \times 1.6 \times 10^{-19}} \)
\( v = \frac{0.80}{2040} \approx 3.92 \times 10^{-4}\text{ m s}^{-1} \approx 3.9 \times 10^{-4}\text{ m s}^{-1} \).

(d) Volume \( V = A L \) remains constant. When the length is doubled (\( L' = 2L \)), the cross-sectional area must be halved (\( A' = A/2 \)).
Since \( R = \frac{\rho L}{A} \):
\( R' = \frac{\rho (2L)}{(A/2)} = 4 \times \left(\frac{\rho L}{A}\right) = 4R \).
Thus, the resistance increases by a factor of 4.

(a)
- Ratio of potential difference to current (\( R = V/I \)) [1]

(b)
- Calculation of resistance \( R = 3.75\ \Omega \) [1]
- Rearrangement to \( \rho = R A / L \) and substitution [1]
- Correct calculation showing \( 3.125 \times 10^{-7}\ \Omega\text{ m} \) leading to \( 3.1 \times 10^{-7}\ \Omega\text{ m} \) [1]

(c)(i)
- Number of free electrons per unit volume [1]

(c)(ii)
- Recall of formula \( I = n A v e \) [1]
- Substitution of values including electronic charge \( e = 1.6 \times 10^{-19}\text{ C} \) [1]
- Calculation: \( 3.9 \times 10^{-4}\text{ m s}^{-1} \) (accept \( 3.92 \times 10^{-4}\text{ m s}^{-1} \)) [1]

(d)
- Area is halved when length is doubled [1]
- Resistance is proportional to \( L/A \), so increases by a factor of 4 [1]

(a)(i) Tensile stress is the force per unit cross-sectional area.
(a)(ii) Tensile strain is the extension per unit original length.

(b) The diameter \( d = 0.80\text{ mm} = 8.0 \times 10^{-4}\text{ m} \).
\( A = \frac{\pi d^2}{4} = \frac{\pi \times (8.0 \times 10^{-4})^2}{4} = \frac{\pi \times 6.4 \times 10^{-7}}{4} = 1.6\pi \times 10^{-7} \approx 5.027 \times 10^{-7}\text{ m}^2 \approx 5.0 \times 10^{-7}\text{ m}^2 \).

(c)(i) Tensile stress \( \sigma = \frac{F}{A} = \frac{150\text{ N}}{5.027 \times 10^{-7}\text{ m}^2} \approx 2.98 \times 10^8\text{ Pa} \approx 3.0 \times 10^8\text{ Pa} \) (or \( 3.0 \times 10^8\text{ N m}^{-2} \)).

(c)(ii) Young modulus \( E = \frac{\text{Stress}}{\text{Strain}} \implies \text{Strain} = \frac{\text{Stress}}{E} \)
\( \text{Strain} = \frac{2.98 \times 10^8\text{ Pa}}{2.0 \times 10^{11}\text{ Pa}} = 1.49 \times 10^{-3} \)
Since \( \text{Strain} = \frac{e}{L} \):
\( e = \text{Strain} \times L = 1.49 \times 10^{-3} \times 2.4\text{ m} = 3.58 \times 10^{-3}\text{ m} \approx 3.6\text{ mm} \) (or \( 3.6 \times 10^{-3}\text{ m} \)).

(d) Under this load, the stress developed is \( 3.0 \times 10^8\text{ Pa} \).
Since this stress exceeds the elastic limit of the copper wire (\( 1.2 \times 10^8\text{ Pa} \)), the copper wire would undergo plastic deformation (permanent stretching) and would not return to its original length, or it would break.

(a)(i)
- Force / cross-sectional area [1]
(a)(ii)
- Extension / original length [1]

(b)
- Calculation of area: \( \pi \times (0.40 \times 10^{-3})^2 \) or \( \pi \times (0.80 \times 10^{-3})^2 / 4 \) to show \( 5.0 \times 10^{-7}\text{ m}^2 \) with sufficient precision [1]

(c)(i)
- Formula: \( \text{Stress} = F / A \) [1]
- Calculation: \( 3.0 \times 10^8\text{ Pa} \) (or \( 2.98 \times 10^8\text{ Pa} \)) [1]

(c)(ii)
- Correct formula relating stress, strain and Young Modulus [1]
- Calculation of strain: \( 1.5 \times 10^{-3} \) [1]
- Calculation of extension: \( 3.6\text{ mm} \) (or \( 3.6 \times 10^{-3}\text{ m} \)) [1]

(d)
- Compares stress in wire (\( 3.0 \times 10^8\text{ Pa} \)) to copper's limit (\( 1.2 \times 10^8\text{ Pa} \)) [1]
- Concludes copper will deform plastically / permanently / break [1]

(a) The principle of superposition states that when two or more waves meet at a point, the resultant displacement is the vector sum of the individual displacements of the waves.

(b) Grating spacing \( d \) is the inverse of the number of lines per meter \( N \):
\( d = \frac{1}{N} = \frac{1}{5.0 \times 10^5\text{ m}^{-1}} = 2.0 \times 10^{-6}\text{ m} \).

(c)(i) Using the grating equation \( d \sin\theta = n\lambda \) for \( n = 1 \):
\( \sin\theta_1 = \frac{1 \times 630 \times 10^{-9}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 0.315 \)
\( \theta_1 = \sin^{-1}(0.315) = 18.36^\circ \approx 18.4^\circ \).

(c)(ii) For \( n = 2 \):
\( \sin\theta_2 = \frac{2 \times 630 \times 10^{-9}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 0.630 \)
\( \theta_2 = \sin^{-1}(0.630) = 39.05^\circ \approx 39.1^\circ \).

(d) The maximum possible angle of diffraction is \( 90^\circ \), for which \( \sin\theta = 1 \).
\( n_{\text{max}} \le \frac{d}{\lambda} \)
\( n_{\text{max}} \le \frac{2.0 \times 10^{-6}\text{ m}}{630 \times 10^{-9}\text{ m}} \approx 3.17 \)
Since the order \( n \) must be an integer, the highest observable order is \( n = 3 \). (At \( n = 4 \), \( \sin\theta = 1.26 \) which is impossible).

(a)
- Mention of two (or more) waves meeting / overlapping at a point [1]
- Resultant displacement is the sum of individual displacements [1]

(b)
- Calculation: \( d = 1 / N = 1 / (5.0 \times 10^5) = 2.0 \times 10^{-6}\text{ m} \) [1]

(c)(i)
- Formula \( d \sin\theta = n\lambda \) substituted correctly [1]
- Calculation of angle: \( 18.4^\circ \) (or \( 18^\circ \)) [1]

(c)(ii)
- Correct substitution for \( n = 2 \) [1]
- Calculation of angle: \( 39.1^\circ \) (or \( 39^\circ \)) [1]

(d)
- Statement that the maximum angle of diffraction is \( 90^\circ \) / \( \sin\theta \le 1 \) [1]
- Calculation of maximum \( n = 3.17 \) [1]
- Deduction of highest integer order: \( n = 3 \) [1]

Step 1: Raw data collection. We set up the apparatus and measure the time for 10 oscillations at different distances \( d \). Let us compile a sample set of results: For \( d = 20.0\text{ cm} \) (0.20 m): \( t_1 = 8.5\text{ s} \), \( t_2 = 8.5\text{ s} \), \( t_{\text{mean}} = 8.5\text{ s} \), \( T = 0.85\text{ s} \), \( T^2 = 0.723\text{ s}^2 \). For \( d = 30.0\text{ cm} \) (0.30 m): \( t_1 = 9.8\text{ s} \), \( t_2 = 9.8\text{ s} \), \( t_{\text{mean}} = 9.8\text{ s} \), \( T = 0.98\text{ s} \), \( T^2 = 0.960\text{ s}^2 \). For \( d = 40.0\text{ cm} \) (0.40 m): \( t_1 = 11.0\text{ s} \), \( t_2 = 11.0\text{ s} \), \( t_{\text{mean}} = 11.0\text{ s} \), \( T = 1.10\text{ s} \), \( T^2 = 1.210\text{ s}^2 \). For \( d = 50.0\text{ cm} \) (0.50 m): \( t_1 = 12.1\text{ s} \), \( t_2 = 12.1\text{ s} \), \( t_{\text{mean}} = 12.1\text{ s} \), \( T = 1.21\text{ s} \), \( T^2 = 1.464\text{ s}^2 \). For \( d = 60.0\text{ cm} \) (0.60 m): \( t_1 = 13.0\text{ s} \), \( t_2 = 13.0\text{ s} \), \( t_{\text{mean}} = 13.0\text{ s} \), \( T = 1.30\text{ s} \), \( T^2 = 1.690\text{ s}^2 \). For \( d = 80.0\text{ cm} \) (0.80 m): \( t_1 = 14.8\text{ s} \), \( t_2 = 14.8\text{ s} \), \( t_{\text{mean}} = 14.8\text{ s} \), \( T = 1.48\text{ s} \), \( T^2 = 2.190\text{ s}^2 \). Step 2: Plotting the graph. A linear graph is plotted with \( T^2 / \text{s}^2 \) on the y-axis and \( d / \text{m} \) on the x-axis. All points are plotted within half a small square and form an excellent straight line. Step 3: Determining gradient and intercept. Two points on the line of best fit are selected: \( (0.20, 0.72) \) and \( (0.80, 2.19) \). Gradient \( m = \frac{2.19 - 0.72}{0.80 - 0.20} = \frac{1.47}{0.60} = 2.45\text{ s}^2\text{ m}^{-1} \). To find the y-intercept \( c \): using \( y = mx + c \) at point \( (0.20, 0.72) \), \( 0.72 = 2.45 \times 0.20 + c \), which gives \( c = 0.72 - 0.49 = 0.23\text{ s}^2 \). Step 4: Finding the constants. Since \( T^2 = \alpha d + \beta \), we have: \( \alpha = \text{gradient} = 2.45\text{ s}^2\text{ m}^{-1} \) and \( \beta = \text{intercept} = 0.23\text{ s}^2 \).

Table of results (6 marks): - 1 mark: Successfully collects 6 sets of data for \( d \) and \( t \). - 1 mark: Range of \( d \) covers at least \( 20.0\text{ cm} \) to \( 80.0\text{ cm} \). - 1 mark: Table column headings show correct quantities and units (e.g., \( d / \text{cm} \), \( t_1 / \text{s} \), \( t_2 / \text{s} \), \( T / \text{s} \), \( T^2 / \text{s}^2 \)). - 1 mark: Raw measurements of \( d \) are recorded to the nearest millimetre. - 1 mark: Time measurements recorded consistently to 0.1 s or 0.01 s. - 1 mark: Calculated values of \( T^2 \) are worked out correctly and recorded to a consistent number of significant figures (usually 3 s.f.). Graph (5 marks): - 1 mark: Linear axes with sensible scales (no awkward scale factors like 3 or 7, points must occupy more than half the grid). - 1 mark: Correct plotting of all points within half a small square. - 1 mark: Good line of best fit drawn with a balanced distribution of points. - 1 mark: Quality of data (no massive outliers, points lie very close to the line). - 1 mark: Large triangle used for gradient calculation (hypotenuse > 50% of the drawn line). Calculations (5 marks): - 1 mark: Correct gradient calculation with clear working. - 1 mark: Correct y-intercept calculation using coordinates from the line. - 1 mark: Constant \( \alpha \) equated to the gradient with correct numerical value and unit (\( \text{s}^2\text{ m}^{-1} \) or \( \text{s}^2\text{ cm}^{-1} \)). - 1 mark: Constant \( \beta \) equated to the y-intercept with correct numerical value and unit (\( \text{s}^2 \)). - 1 mark: Correct calculation of period \( T \) by dividing average time by 10. Experimental Technique (4 marks): - 1 mark: Evidence of repeated raw timing measurements. - 1 mark: Timing technique described (e.g., counting down "3, 2, 1, 0" to start the watch at "0"). - 1 mark: Fiducial marker used at the equilibrium position to improve timing accuracy. - 1 mark: Metre rule clamped firmly to prevent slipping of the pivot during oscillation.

Step 1: Deflection measurements. For \( L_1 = 80.0\text{ cm} \), we measure: \( h_0 = 95.0\text{ cm} \) (undeflected height) and \( h_1 = 81.2\text{ cm} \) (deflected height). Deflection \( s_1 = h_0 - h_1 = 13.8\text{ cm} \). Step 2: Percentage uncertainty. The absolute uncertainty in each height measurement is \( \pm 1\text{ mm} \) (or \( 0.1\text{ cm} \)). Since \( s_1 \) is calculated from the difference of two heights, its absolute uncertainty \( \Delta s_1 = 0.1\text{ cm} + 0.1\text{ cm} = 0.2\text{ cm} \). Percentage uncertainty in \( s_1 = \frac{0.2\text{ cm}}{13.8\text{ cm}} \times 100\% = 1.45\% \). Step 3: Second trial. For \( L_2 = 60.0\text{ cm} \), we measure: \( h_0 = 95.0\text{ cm} \) and \( h_2 = 89.2\text{ cm} \). Deflection \( s_2 = h_0 - h_2 = 5.8\text{ cm} \). Step 4: Testing the relation \( \frac{s}{L^3} = k \). For the first trial: \( k_1 = \frac{13.8}{80.0^3} = 2.70 \times 10^{-5}\text{ cm}^{-2} \). For the second trial: \( k_2 = \frac{5.8}{60.0^3} = 2.69 \times 10^{-5}\text{ cm}^{-2} \). Step 5: Comparing results. The percentage difference between \( k_1 \) and \( k_2 \) is: \( \frac{|2.70 - 2.69|}{2.695} \times 100\% = 0.37\% \). Since this percentage difference is extremely small (much less than the \( 10\% \) benchmark or the estimated percentage uncertainty in our measurements), we conclude that the experimental results strongly support the suggested relationship.

Measurements and calculations (4 marks): - 1 mark: Raw values of \( h_0 \) and \( h_1 \) recorded to the nearest millimetre with units. - 1 mark: Correct calculation of \( s_1 \) with matching unit. - 1 mark: Second trial deflection \( s_2 \) measured and shown to be smaller than \( s_1 \). - 1 mark: Repeated measurements of the heights to ensure accuracy. Uncertainty (2 marks): - 1 mark: Absolute uncertainty of \( s_1 \) estimated reasonably (typically between \( 0.1\text{ cm} \) and \( 0.3\text{ cm} \) based on reading two scales). - 1 mark: Correct calculation of percentage uncertainty in \( s_1 \) with appropriate working shown. Analysis and comparison (4 marks): - 1 mark: Correct calculation of \( k_1 \) with appropriate unit (e.g., \( \text{cm}^{-2} \) or \( \text{m}^{-2} \)). - 1 mark: Correct calculation of \( k_2 \). - 1 mark: Correct calculation of the percentage difference between \( k_1 \) and \( k_2 \). - 1 mark: Clear, logical conclusion stating whether the relationship is supported, directly comparing the percentage difference to a stated percentage criterion (e.g., \( 10\% \)). Evaluation (10 marks - 4 pairs of limitation/improvement, plus 2 floating marks for detail): - 1 mark: Limitation: Cantilever rule twists sideways when loaded. - 1 mark: Improvement: Securely clamp a guide block on either side of the rule at the bench edge. - 1 mark: Limitation: Parallax error reading the vertical scale. - 1 mark: Improvement: Use a mirror behind the scale or use a digital height gauge. - 1 mark: Limitation: Difficult to ensure vertical half-metre rule is perfectly perpendicular to the floor. - 1 mark: Improvement: Use a set-square resting on the floor or a plumb-line to align the rule vertically. - 1 mark: Limitation: Only two sets of readings are taken, which is insufficient to draw a definitive conclusion. - 1 mark: Improvement: Collect multiple readings of \( s \) for various values of \( L \) and plot a graph of \( s \) against \( L^3 \). - 2 marks: Quality of explanation (one for clearly specifying the source of friction or wobble, one for specifying exact details of the apparatus used for improvements).