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The resistivity is given by the formula: \(\rho = \frac{R A}{L} = \frac{\pi R d^2}{4 L}\). The fractional uncertainty in resistivity is calculated by adding the fractional uncertainties of each component, multiplying the fractional uncertainty of diameter by 2 because it is squared: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2\frac{\Delta d}{d} + \frac{\Delta L}{L}\). Calculating the percentage uncertainties: \(\frac{\Delta R}{R} \times 100\% = \frac{0.05}{4.50} \times 100\% \approx 1.11\%\), \(\frac{\Delta d}{d} \times 100\% = \frac{0.01}{0.38} \times 100\% \approx 2.63\%\), \(\frac{\Delta L}{L} \times 100\% = \frac{0.002}{1.200} \times 100\% \approx 0.17\%\). Total percentage uncertainty = \(1.11\% + 2 \times 2.63\% + 0.17\% = 1.11\% + 5.26\% + 0.17\% = 6.54\%\). This rounds to \(6.5\%\).

1 mark for the correct option C. [Method: Recall formula for percentage uncertainty of a compound quantity, correctly double the uncertainty of diameter, and calculate the sum: 1.11% + 2*(2.63%) + 0.17% = 6.5%]

First, calculate the value of kinetic energy: \(E_k = \frac{1}{2} m v^2 = 0.5 \times 2.50 \times (12.0)^2 = 180\ \text{J}\). Next, calculate the fractional uncertainty in \(E_k\): \(\frac{\Delta E_k}{E_k} = \frac{\Delta m}{m} + 2\frac{\Delta v}{v}\). Substitute the values: \(\frac{\Delta E_k}{E_k} = \frac{0.05}{2.50} + 2 \times \frac{0.3}{12.0} = 0.020 + 2 \times 0.025 = 0.020 + 0.050 = 0.070\). Therefore, the absolute uncertainty in kinetic energy is: \\Delta E_k = 0.070 \times 180\ \text{J} = 12.6\ \text{J} \approx 13\ \text{J}.

1 mark for the correct option C. [Method: Calculate kinetic energy as 180 J. Determine fractional uncertainty as 0.02 + 2*(0.025) = 0.07. Multiply by 180 J to find the absolute uncertainty of 12.6 J, which rounds to 13 J]

The intensity \(I\) of a wave is directly proportional to the square of its amplitude \(A\) (\(I \propto A^2\)). If the new amplitude is \(0.40 A\), the new intensity is proportional to \((0.40 A)^2 = 0.16 A^2\). Hence, the new intensity is \(0.16 I\).

1 mark for the correct option A. [Method: Recall intensity-amplitude relationship: I is proportional to A^2. Square the amplitude scale factor 0.40^2 = 0.16]

For a tube closed at one end, the fundamental mode of resonance occurs when the length \(L\) of the tube is equal to a quarter of the wavelength (\(L = \frac{\lambda}{4}\)). Given \(L = 20.0\ \text{cm} = 0.200\ \text{m}\), the wavelength is \(\lambda = 4 \times 0.200\ \text{m} = 0.800\ \text{m}\). Using the wave equation \(v = f \lambda\), the speed of sound is \(v = 425\ \text{Hz} \times 0.800\ \text{m} = 340\ \text{m s}^{-1}\).

1 mark for the correct option C. [Method: Determine fundamental wavelength lambda = 4 * L = 0.80 m. Calculate speed using v = f * lambda = 425 * 0.80 = 340 m/s]

For the plank to be in equilibrium, the sum of vertical forces must equal zero. Hence, the tension at end \(B\) is \(T_B = 120\ \text{N} - 45\ \text{N} = 75\ \text{N}\). Let \(x\) be the distance from end \(A\) to the centre of gravity (where the weight acts downwards). Taking moments about end \(A\): Clockwise moment = \(120\ \text{N} \times x\). Anticlockwise moment = \(T_B \times 4.0\ \text{m} = 75\ \text{N} \times 4.0\ \text{m} = 300\ \text{N m}\). Equating moments: \(120 x = 300\), which gives \(x = \frac{300}{120} = 2.5\ \text{m}\).

1 mark for the correct option C. [Method: Calculate tension at B as 75 N by vertical equilibrium. Set up moments equation about A: 120 * x = 75 * 4.0 to find x = 2.5 m]

For static equilibrium, the vertical forces must balance. The vertical component of the tension \(T\) in the inclined string is \(T \cos(35^\circ)\). The downward force is the weight \(W = mg = 5.0 \times 9.81 = 49.05\ \text{N}\). Therefore, \(T \cos(35^\circ) = 49.05\), which gives \(T = \frac{49.05}{\cos(35^\circ)} \approx 59.9\ \text{N}\), which rounds to \(60\ \text{N}\).

1 mark for the correct option D. [Method: Set up vertical force balance: T * cos(35) = m * g. Solve for T = 49.05 / cos(35) = 59.9 N]

The total displacement is the sum of the displacements in the acceleration and deceleration phases. In Phase 1 (acceleration), displacement \(s_1\) is the area under the velocity-time graph: \(s_1 = \frac{1}{2} \times \text{base} \times \text{height} = 0.5 \times 4.0\ \text{s} \times 6.0\ \text{m s}^{-1} = 12.0\ \text{m}\). In Phase 2 (deceleration), the car decelerates from \(6.0\ \text{m s}^{-1}\) to \(0\) at \(2.0\ \text{m s}^{-2}\). The time taken \(t_2\) is \(t_2 = \frac{v - u}{a} = \frac{0 - 6.0}{-2.0} = 3.0\ \text{s}\). The displacement in this phase is \(s_2 = 0.5 \times 3.0\ \text{s} \times 6.0\ \text{m s}^{-1} = 9.0\ \text{m}\). The total displacement is \(s = s_1 + s_2 = 12.0\ \text{m} + 9.0\ \text{m} = 21.0\ \text{m}\).

1 mark for the correct option C. [Method: Calculate first phase displacement (12 m). Find time for second phase (3 s) and its displacement (9 m). Sum both displacements to get 21 m]

For horizontal projectile motion, the horizontal component of velocity remains constant at \(u_x = 15\ \text{m s}^{-1}\). The time of flight \(t\) is determined by the horizontal distance: \(x = u_x t \implies 45 = 15 t \implies t = 3.0\ \text{s}\). The vertical motion is governed by \(h = \frac{1}{2} g t^2\) (since initial vertical velocity is zero). Substituting the values: \(h = 0.5 \times 9.81 \times (3.0)^2 = 0.5 \times 9.81 \times 9.0 = 44.145\ \text{m}\). This is approximately \(44\ \text{m}\).

1 mark for the correct option C. [Method: Find time of flight t = 45 / 15 = 3.0 s. Apply vertical motion formula h = 0.5 * g * t^2 = 0.5 * 9.81 * 9 = 44 m]

First, express the resistivity formula in terms of the measured quantities \(R\), \(d\), and \(L\): \(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\). Since \(\pi\) and 4 are constants with no uncertainty, the fractional uncertainty in \(\rho\) is given by: \(\frac{\Delta \rho}{\rho} = \frac{\Delta R}{R} + 2 \frac{\Delta d}{d} + \frac{\Delta L}{L}\). Now calculate each percentage uncertainty: (1) Percentage uncertainty in \(R\): \(\frac{0.05}{4.20} \times 100\% \approx 1.19\%\). (2) Percentage uncertainty in \(d\): \(\frac{0.02}{0.50} \times 100\% = 4.0\%\), so \(2 \frac{\Delta d}{d} = 8.0\%\). (3) Percentage uncertainty in \(L\): \(\frac{0.002}{1.250} \times 100\% = 0.16\%\). Summing these gives the total percentage uncertainty in \(\rho\): \(1.19\% + 8.0\% + 0.16\% = 9.35\%\), which rounds to \(9.4\%\).

[1] A - Correct calculation of individual uncertainties and total percentage uncertainty.

Young modulus is defined as \(E = \frac{\text{stress}}{\text{strain}}\). Stress has the formula \(\sigma = \frac{F}{A}\), so its units are \(\text{N m}^{-2}\). Strain is a ratio of lengths \(\varepsilon = \frac{\Delta L}{L}\) and is dimensionless. Therefore, the units of Young modulus are those of stress, \(\text{N m}^{-2}\). Converting Newtons into S.I. base units using \(F = ma\) gives \(1\text{ N} = 1\text{ kg m s}^{-2}\). Substituting this into the units for stress yields \(\text{kg m s}^{-2} \times \text{m}^{-2} = \text{kg m}^{-1}\text{ s}^{-2}\).

[1] B - Correct derivation of the S.I. base units for Young modulus from first principles.

The relationship between phase difference \(\Delta \phi\), distance \(\Delta x\), and wavelength \(\lambda\) is \(\Delta \phi = \frac{2\pi}{\lambda} \Delta x\). Substituting the values gives \(\frac{2\pi}{5} = \frac{2\pi}{\lambda} \times 0.12\), which simplifies to \(\frac{1}{5} = \frac{0.12}{\lambda}\), giving \(\lambda = 0.60\text{ m}\). The wave speed is then calculated using \(v = f \lambda = 5.0\text{ Hz} \times 0.60\text{ m} = 3.0\text{ m s}^{-1}\).

[1] C - Correct calculation of wavelength followed by calculation of wave speed using the wave equation.

The frequency of an electromagnetic wave is determined by its source and does not change when transitioning between media, so it remains \(6.0 \times 10^{14}\text{ Hz}\). The speed of light in the glass is \(v = \frac{c}{n} = \frac{3.00 \times 10^8}{1.5} = 2.00 \times 10^8\text{ m s}^{-1}\). Using \(v = f \lambda\), the wavelength in the glass is \(\lambda = \frac{2.00 \times 10^8}{6.0 \times 10^{14}} \approx 3.33 \times 10^{-7}\text{ m} = 333\text{ nm}\).

[1] D - Identifies that frequency is unchanged and correctly applies the refractive index to calculate the new wavelength.

Let the plank stretch from \(x = 0\text{ m}\) to \(x = 2.4\text{ m}\). The left rope is at \(x = 0\text{ m}\) with tension \(T_1\). The right rope is at \(x = 1.8\text{ m}\) with tension \(T_2\). The weight of the uniform plank, \(W = mg = 15 \times 9.8 = 147\text{ N}\), acts downward at its center of gravity \(x = 1.2\text{ m}\). Taking moments about \(x = 1.8\text{ m}\) (the right support) to eliminate \(T_2\) gives: \(T_1 \times 1.8 = W \times (1.8 - 1.2)\). Substituting \(W = 147\text{ N}\) gives \(T_1 \times 1.8 = 147 \times 0.6\), which simplifies to \(T_1 = \frac{147 \times 0.6}{1.8} = 49\text{ N}\).

[1] A - Correct choice of pivot and application of rotational equilibrium principles.

The aircraft's velocity vector points North with magnitude \(45\text{ m s}^{-1}\). The wind velocity vector points East (since it blows from the West) with magnitude \(15\text{ m s}^{-1}\). These vectors are perpendicular, so the magnitude of the resultant velocity is \(v = \sqrt{45^2 + 15^2} = \sqrt{2250} \approx 47.4\text{ m s}^{-1}\). The angle East of North is \(\theta = \arctan\left(\frac{15}{45}\right) \approx 18.4^\circ\). Thus, the bearing (measured clockwise from North) is \(018^\circ\).

[1] A - Correct magnitude using Pythagoras' theorem and correct direction using trigonometry.

Let upwards be the positive direction. The initial velocity \(u = +12.0\text{ m s}^{-1}\), the acceleration \(a = -9.81\text{ m s}^{-2}\), and the time of flight \(t = 4.50\text{ s}\). Using \(s = ut + \frac{1}{2}at^2\), the displacement of the stone is \(s = (12.0 \times 4.50) + \frac{1}{2}(-9.81)(4.50)^2 = 54.0 - 99.326 = -45.326\text{ m}\). The negative sign indicates that the final position is below the starting point, so the height of the cliff is \(45.3\text{ m}\).

[1] A - Correct calculation of vertical displacement using equations of motion with consistent signs.

The motion has three stages. Stage 1 (acceleration): \(u = 0\), \(a = 2.0\text{ m s}^{-2}\), \(t = 5.0\text{ s}\). The final velocity is \(v = u + at = 10\text{ m s}^{-1}\), and the distance is \(s_1 = \frac{u+v}{2}t = 25\text{ m}\). Stage 2 (constant velocity): \(v = 10\text{ m s}^{-1}\), \(t = 8.0\text{ s}\). The distance is \(s_2 = v t = 80\text{ m}\). Stage 3 (deceleration): \(u = 10\text{ m s}^{-1}\), \(v = 0\), \(a = -4.0\text{ m s}^{-2}\). The time to stop is \(t = 2.5\text{ s}\), and the distance is \(s_3 = \frac{u+v}{2}t = 12.5\text{ m}\). The total distance is \(s_1 + s_2 + s_3 = 25 + 80 + 12.5 = 117.5\text{ m}\), which rounds to \(118\text{ m}\).

[1] B - Correct calculations of distance for the three phases of motion and summing them together.

The volume \(V\) of a cylinder is given by the formula \(V = \frac{\pi d^2 h}{4}\). The percentage uncertainty in \(V\) is calculated by adding the percentage uncertainties of the independent variables, multiplying by their powers: \%\Delta V = 2 \times (\%\Delta d) + (\%\Delta h). First, calculate the percentage uncertainty in each measurement: \%\Delta d = \frac{0.02}{2.00} \times 100\% = 1.0\%, and \%\Delta h = \frac{0.1}{5.0} \times 100\% = 2.0\%. Therefore, \%\Delta V = 2 \times (1.0\%) + 2.0\% = 4.0\%.

Correct option is C. 1 mark awarded for the correct percentage uncertainty calculated with clear steps.

The intensity \(I\) of a wave is directly proportional to the square of its amplitude \(A\) (\(I \propto A^2\)). The reflected wave has \(16\%\) of the original wave's energy, meaning its intensity is \(0.16 I\). Since \(A \propto \sqrt{I}\), the reflected amplitude \(A_r\) is given by \(A_r = \sqrt{0.16} A = 0.40 A\).

Correct option is B. 1 mark for applying the relationship between intensity and amplitude to find the correct reflected amplitude.

Since the plank is in equilibrium, the sum of the vertical forces must equal zero. The total upward force is \(T_A + T_B = 200\text{ N}\). Given \(T_A = 80\text{ N}\), the tension at \(B\) is \(T_B = 200\text{ N} - 80\text{ N} = 120\text{ N}\). Taking moments about end \(A\): clockwise moment of weight equals counter-clockwise moment of tension at \(B\). Thus, \(W \times x = T_B \times 4.0\text{ m}\), where \(x\) is the distance from \(A\) to the center of gravity. This gives \(200 \times x = 120 \times 4.0 = 480\text{ N m}\), which yields \(x = 2.4\text{ m}\).

Correct option is C. 1 mark for utilizing translational and rotational equilibrium to solve for the position of the center of gravity.

Using the kinematic equation for displacement \(s = ut + \frac{1}{2}at^2\), taking upwards as positive: \(u = +12\text{ m s}^{-1}\), \(a = -9.81\text{ m s}^{-2}\), and \(t = 4.0\text{ s}\). Substituting these values gives: \(s = (12)(4.0) + \frac{1}{2}(-9.81)(4.0)^2 = 48 - 78.48 = -30.48\text{ m}\). The negative sign indicates that the displacement is downwards from the release point, which means the base of the cliff is \(30.5\text{ m}\) below the edge. To two significant figures, this is \(30\text{ m}\).

Correct option is A. 1 mark for applying the equations of motion with consistent signs to find the height of the cliff.

First, calculate the volume of the cylinder \(V\):
\(V = \pi r^2 l = \pi \times (1.20\text{ cm})^2 \times 8.54\text{ cm} = 38.636\text{ cm}^3 = 3.8636 \times 10^{-5}\text{ m}^3\)

(a) The percentage uncertainty in volume \(V = \pi r^2 l\) is:
\(\%\Delta V = 2 \times \%\Delta r + \%\Delta l\)
\(\%\Delta r = \frac{0.02}{1.20} \times 100\% = 1.667\%\)
\(\%\Delta l = \frac{0.05}{8.54} \times 100\% = 0.585\%\)
\(\%\Delta V = 2 \times 1.667\% + 0.585\% = 3.333\% + 0.585\% = 3.918\% \approx 3.9\%\)
(Note: percentage uncertainty in mass is \(\%\Delta m = \frac{0.1}{124.5} \times 100\% = 0.080\%\))

(b) Calculate the density \(\rho\):
\(\rho = \frac{m}{V} = \frac{0.1245\text{ kg}}{3.8636 \times 10^{-5}\text{ m}^3} = 3222.3\text{ kg m}^{-3}\)

The percentage uncertainty in density \(\rho\) is:
\(\%\Delta \rho = \%\Delta m + \%\Delta V = 0.080\% + 3.918\% = 3.998\%\)

Now, calculate the absolute uncertainty in density \(\Delta \rho\):
\(\Delta \rho = 3.998\% \times 3222.3\text{ kg m}^{-3} = 128.8\text{ kg m}^{-3} \approx 100\text{ kg m}^{-3}\text{ (to 1 s.f.)}\)

Therefore, writing density to match the precision of the uncertainty:
\(\rho = (3.2 \pm 0.1) \times 10^3\text{ kg m}^{-3}\) (or \(3200 \pm 100\text{ kg m}^{-3}\))

Part (a): [3 marks]
- Correct calculation of \(\%\Delta r = 1.67\%\) (1 mark)
- Correct calculation of \(\%\Delta l = 0.59\%\) (1 mark)
- Adding \(2 \times \%\Delta r + \%\Delta l\) to show \(3.9\%\) or \(4.0\%\) (1 mark)

Part (b): [5.33 marks]
- Correct calculation of density value as \(3222\text{ kg m}^{-3}\) or \(3.22\text{ g cm}^{-3}\) (1.33 marks)
- Correct calculation of \(\%\Delta \rho \approx 4.0\%\) (1 mark)
- Correct absolute uncertainty \(\Delta \rho \approx 100\text{ kg m}^{-3}\) or \(130\text{ kg m}^{-3}\) (1.5 marks)
- Final value and uncertainty expressed to 2 s.f.: \((3.2 \pm 0.1) \times 10^3\text{ kg m}^{-3}\) (1.5 marks)

(a) Calculate resistance \(R\):
\(R = \frac{V}{I} = \frac{2.80}{1.45} = 1.931\ \Omega\)

Percentage uncertainty in \(R\):
\(\%\Delta R = \%\Delta V + \%\Delta I = \left(\frac{0.10}{2.80} \times 100\%\right) + \left(\frac{0.05}{1.45} \times 100\%\right) = 3.571\% + 3.448\% = 7.019\%\)

(b) Calculate resistivity \(\rho\):
\(\rho = \frac{R A}{L} = \frac{R \pi d^2}{4 L}\)
\(\rho = \frac{1.931 \times \pi \times (0.42 \times 10^{-3})^2}{4 \times 1.500} = 1.784 \times 10^{-7}\ \Omega\text{ m}\)

Percentage uncertainty in \(\rho\):
\(\%\Delta \rho = \%\Delta R + 2 \times \%\Delta d + \%\Delta L\)
\(\%\Delta d = \frac{0.01}{0.42} \times 100\% = 2.381\%\)
\(\%\Delta L = \frac{0.002}{1.500} \times 100\% = 0.133\%\)
\(\%\Delta \rho = 7.019\% + 2 \times 2.381\% + 0.133\% = 11.914\% \approx 12\%\)

Absolute uncertainty in \(\rho\):
\(\Delta \rho = 11.914\% \times 1.784 \times 10^{-7} = 2.125 \times 10^{-8}\ \Omega\text{ m} \approx 0.2 \times 10^{-7}\ \Omega\text{ m}\)

Therefore, writing the final answer to 2 s.f.:
\(\rho = (1.8 \pm 0.2) \times 10^{-7}\ \Omega\text{ m}\)

Part (a): [3 marks]
- Correct calculation of \(R = 1.93\ \Omega\) (1 mark)
- Calculation of \(\%\Delta V = 3.57\%\) and \(\%\Delta I = 3.45\%\) (1 mark)
- Addition of percentage uncertainties to give \(\%\Delta R \approx 7.0\%\) (1 mark)

Part (b): [5.33 marks]
- Correct calculation of area \(A = 1.39 \times 10^{-7}\text{ m}^2\) or direct calculation of \(\rho = 1.78 \times 10^{-7}\ \Omega\text{ m}\) (1.33 marks)
- Calculation of percentage uncertainty in \(d\) as \(2.38\%\) (1 mark)
- Calculation of total percentage uncertainty in \(\rho\) as \(\approx 12\%\) (1 mark)
- Calculation of absolute uncertainty as \(0.2 \times 10^{-7}\ \Omega\text{ m}\) (1 mark)
- Final value and uncertainty expressed to 2 s.f. with correct unit (1 mark)

First, identify the values given:
\(\lambda_1 = 600 \times 10^{-9}\text{ m}\)
\(a = 0.40 \times 10^{-3}\text{ m}\)
\(D = 1.80\text{ m}\)

(a) Calculate the fringe spacing \(x_1\) for the first laser:
\(x_1 = \frac{\lambda_1 D}{a} = \frac{600 \times 10^{-9} \times 1.80}{0.40 \times 10^{-3}} = 2.70 \times 10^{-3}\text{ m} = 2.70\text{ mm}\)

The distance from the central maximum to the 4th bright fringe is:
\(y_4 = 4 \times x_1 = 4 \times 2.70\text{ mm} = 10.8\text{ mm}\) (or \(1.08 \times 10^{-2}\text{ m}\)).

(b) For the second laser, the 3rd dark fringe occurs at a path difference of \(2.5 \lambda_2\).
The position of the 3rd dark fringe from the central maximum is given by:
\(y_{\text{dark},3} = 2.5 \times x_2 = 2.5 \times \frac{\lambda_2 D}{a}\)

We are given \(y_{\text{dark},3} = 6.075\text{ mm} = 6.075 \times 10^{-3}\text{ m}\).

Rearranging to solve for \(\lambda_2\):
\(\lambda_2 = \frac{y_{\text{dark},3} \times a}{2.5 \times D} = \frac{6.075 \times 10^{-3} \times 0.40 \times 10^{-3}}{2.5 \times 1.80}\)
\(\lambda_2 = \frac{2.43 \times 10^{-6}}{4.50} = 5.40 \times 10^{-7}\text{ m} = 540\text{ nm}\).

Part (a): [4 marks]
- Recall and use of double-slit formula \(x = \frac{\lambda D}{a}\) (1 mark)
- Correct calculation of fringe spacing \(x_1 = 2.70\text{ mm}\) (1 mark)
- Identifying that the 4th bright fringe is at \(4x\) (1 mark)
- Correct calculation of distance as \(10.8\text{ mm}\) (1 mark)

Part (b): [4.33 marks]
- Recalling that the 3rd dark fringe is at a distance of \(2.5 x\) (1.33 marks)
- Setting up the equation \(6.075 \times 10^{-3} = 2.5 \frac{\lambda_2 D}{a}\) (1 mark)
- Rearranging to make \(\lambda_2\) the subject (1 mark)
- Correct calculation of \(\lambda_2 = 540\text{ nm}\) (or \(5.40 \times 10^{-7}\text{ m}\)) (1 mark)

(a) Using Snell's law at the air-glass boundary:
\(n_{\text{air}} \sin \theta_{\text{air}} = n_{\text{glass}} \sin \theta_{\text{glass}}\)
Given \(n_{\text{air}} = 1.00\), \(\theta_{\text{air}} = 42.0^\circ\), and \(n_{\text{glass}} = 1.55\):
\(1.00 \times \sin(42.0^\circ) = 1.55 \times \sin \theta_{\text{glass}}\)
\(\sin \theta_{\text{glass}} = \frac{\sin(42.0^\circ)}{1.55} = \frac{0.6691}{1.55} = 0.4317\)
\(\theta_{\text{glass}} = \sin^{-1}(0.4317) = 25.6^\circ\)

(b)
(i) The critical angle \(\theta_c\) at the glass-liquid boundary is given by:
\(\sin \theta_c = \frac{n_{\text{liquid}}}{n_{\text{glass}}}\)
Given \(n_{\text{liquid}} = 1.33\) and \(n_{\text{glass}} = 1.55\):
\(\sin \theta_c = \frac{1.33}{1.55} = 0.8581\)
\(\theta_c = \sin^{-1}(0.8581) = 59.1^\circ\)

(ii) For total internal reflection to occur, two conditions must be met:
1. The light must travel from a medium of higher refractive index to a medium of lower refractive index (satisfied, as \(1.55 > 1.33\)).
2. The angle of incidence must exceed the critical angle (\(\theta_i > \theta_c\)).
Since the angle of incidence is \(58.0^\circ\) and the critical angle is \(59.1^\circ\), the angle of incidence is less than the critical angle (\(58.0^\circ < 59.1^\circ\)).
Therefore, total internal reflection will NOT occur.

Part (a): [3 marks]
- Recall and use of Snell's Law: \(n_1 \sin \theta_1 = n_2 \sin \theta_2\) (1 mark)
- Correct substitution of values (1 mark)
- Correct calculation of angle of refraction as \(25.6^\circ\) (1 mark)

Part (b)(i): [3 marks]
- Recall and use of \(
u_c = \sin^{-1}(\frac{n_2}{n_1})\) or \(\sin \theta_c = \frac{n_2}{n_1}\) (1 mark)
- Correct substitution of values (1 mark)
- Correct calculation of critical angle as \(59.1^\circ\) (1 mark)

Part (b)(ii): [2.33 marks]
- Explicit comparison between the angle of incidence (\(58.0^\circ\)) and the critical angle (\(59.1^\circ\)) (1 mark)
- Correct conclusion that total internal reflection does not occur (1.33 marks)

Identify the forces and their distances from point A:
- Weight of uniform shelf, \(W_{\text{shelf}} = m_{\text{shelf}} g = 4.5\text{ kg} \times 9.81\text{ m s}^{-2} = 44.15\text{ N}\). Since the shelf is uniform, this force acts at the midpoint, \(0.60\text{ m}\) from A.
- Weight of flowerpot, \(W_{\text{pot}} = m_{\text{pot}} g = 8.0\text{ kg} \times 9.81\text{ m s}^{-2} = 78.48\text{ N}\). This acts at \(0.90\text{ m}\) from A.
- Tension \(T\) in the wire acts at point B, which is \(1.2\text{ m}\) from A, at an angle of \(35^\circ\) above the shelf.

(a) Taking moments about point A for rotational equilibrium:
Sum of clockwise moments = Sum of anticlockwise moments
\((W_{\text{shelf}} \times 0.60) + (W_{\text{pot}} \times 0.90) = T \sin(35^\circ) \times 1.2\)

Calculate the clockwise moments:
Clockwise moment = \((44.15 \times 0.60) + (78.48 \times 0.90)\)
Clockwise moment = \(26.49 + 70.63 = 97.12\text{ N m}\)

Equate to the anticlockwise moment:
\(1.2 \times T \sin(35^\circ) = 97.12\)
\(0.6883 T = 97.12 \implies T = 141.1\text{ N} \approx 141\text{ N}\)

(b) For horizontal translational equilibrium, the net horizontal force must be zero.
The wire pulls point B to the left with a horizontal component of tension:
\(T_x = T \cos(35^\circ) = 141.1 \times \cos(35^\circ) = 115.6\text{ N}\)

The only other horizontal force on the shelf is the horizontal component of the force exerted by the hinge at point A, \(H\).
Therefore, for horizontal equilibrium:
\(H = T_x = 115.6\text{ N} \approx 116\text{ N}\) (to 3 s.f.).

Part (a): [5 marks]
- Calculation of the weight of the shelf (\(44.15\text{ N}\)) and the pot (\(78.48\text{ N}\)) (1 mark)
- Identifying that the shelf's weight acts at its midpoint, \(0.60\text{ m}\) (1 mark)
- Correct moment equation about A: \((44.15 \times 0.60) + (78.48 \times 0.90) = T \sin(35^\circ) \times 1.2\) (1 mark)
- Correct calculation of clockwise moment as \(97.1\text{ N m}\) (1 mark)
- Correct calculation of tension \(T = 141\text{ N}\) (1 mark)

Part (b): [3.33 marks]
- Identifying that the horizontal force at the hinge balances the horizontal component of the tension (1 mark)
- Expressing the horizontal component of tension as \(T \cos(35^\circ)\) (1 mark)
- Correct calculation of horizontal hinge force as \(116\text{ N}\) (1.33 marks)

Let us define the upward direction as positive and downward as negative.

Initial velocity components:
\(u_x = 15.0 \cos(30.0^\circ) = 12.99\text{ m s}^{-1}\)
\(u_y = 15.0 \sin(30.0^\circ) = 7.50\text{ m s}^{-1}\)

(a) Using the kinematic equation for vertical displacement \(s_y\):
\(s_y = u_y t + \frac{1}{2} a_y t^2\)
where \(a_y = -9.81\text{ m s}^{-2}\) and \(t = 3.40\text{ s}\):
\(s_y = 7.50 \times 3.40 + \frac{1}{2} (-9.81) \times (3.40)^2\)
\(s_y = 25.50 - 4.905 \times 11.56\)
\(s_y = 25.50 - 56.70 = -31.20\text{ m}\)

Since displacement is \(31.2\text{ m}\) downwards, the height of the cliff \(h\) is \(31.2\text{ m}\) (or \(31\text{ m}\) to 2 s.f.).

(b) Determine the velocity components just before impact:
- Horizontal velocity remains constant because there is no air resistance:
\(v_x = u_x = 12.99\text{ m s}^{-1}\)

- Vertical velocity component \(v_y\):
\(v_y = u_y + a_y t\)
\(v_y = 7.50 + (-9.81 \times 3.40) = 7.50 - 33.35 = -25.85\text{ m s}^{-1}\)

Calculate the magnitude of the final velocity \(v\):
\(v = \sqrt{v_x^2 + v_y^2} = \sqrt{(12.99)^2 + (-25.85)^2}\)
\(v = \sqrt{168.74 + 668.22} = \sqrt{836.96} = 28.93\text{ m s}^{-1} \approx 28.9\text{ m s}^{-1}\) (or \(29\text{ m s}^{-1}\) to 2 s.f.).

Part (a): [4.33 marks]
- Correct resolution of initial velocity into vertical component \(u_y = 7.50\text{ m s}^{-1}\) (1 mark)
- Recall and use of \(s = ut + \frac{1}{2}at^2\) with appropriate signs (1.33 marks)
- Correct calculation of vertical displacement as \(-31.2\text{ m}\) (1 mark)
- Height of cliff stated as positive value \(31.2\text{ m}\) (1 mark)

Part (b): [4 marks]
- Correct statement that horizontal velocity \(v_x = 13.0\text{ m s}^{-1}\) is constant (1 mark)
- Correct calculation of final vertical velocity \(v_y = -25.9\text{ m s}^{-1}\) (1 mark)
- Use of Pythagoras' theorem to combine components (1 mark)
- Correct calculation of final velocity magnitude as \(28.9\text{ m s}^{-1}\) or \(29\text{ m s}^{-1}\) (1 mark)

(a) Convert measurements to SI units: \(d = 4.6 \times 10^{-4}\text{ m}\), \(L = 0.750\text{ m}\), \(m = 1.12 \times 10^{-3}\text{ kg}\). Volume \(V = \frac{\pi d^2 L}{4} = \frac{\pi \times (4.6 \times 10^{-4})^2 \times 0.750}{4} = 1.246 \times 10^{-7}\text{ m}^3\). Density \(\rho = \frac{m}{V} = \frac{1.12 \times 10^{-3}}{1.246 \times 10^{-7}} = 8986\text{ kg m}^{-3} \approx 9.0 \times 10^3\text{ kg m}^{-3}\).
(b) Percentage uncertainties: \%\Delta m = \frac{0.02}{1.12} \times 100\% = 1.79\%; \%\Delta L = \frac{0.2}{75.0} \times 100\% = 0.27\%; \%\Delta d = \frac{0.01}{0.46} \times 100\% = 2.17\%. Since \(\rho \propto \frac{m}{d^2 L}\), the percentage uncertainty in \(\rho\) is: \%\Delta \rho = \%\Delta m + 2(\%\Delta d) + \%\Delta L = 1.79\% + 2(2.17\%) + 0.27\% = 6.4\%.
(c) The diameter \(d\) contributes the most because its percentage uncertainty is doubled in the density calculation (to 4.35\%), representing the largest portion of the total uncertainty. It can be improved by measuring the diameter at several different positions and orientations along the wire and calculating a mean, or by using a thicker wire of the same material to reduce percentage uncertainty.

(a) C1 for volume calculation: \(1.25 \times 10^{-7}\text{ m}^3\) or correct formula substitution. C1 for density calculation: \(8986\text{ kg m}^{-3}\). A1 for final answer rounded to 2 significant figures: \(9.0 \times 10^3\text{ kg m}^{-3}\) (accept \(9000\text{ kg m}^{-3}\)).
(b) C1 for calculating individual percentage uncertainties (\(m\): 1.79\%, \(L\): 0.27\%, \(d\): 2.17\%). C1 for doubling the percentage uncertainty of \(d\) (4.35\%) and summing all three: \%\Delta \rho = \%\Delta m + 2 \%\Delta d + \%\Delta L. A1 for final answer 6.4\% (accept 6.40\% or 6.43\%).
(c) B1 for stating diameter \(d\). B1 for explaining that its percentage uncertainty is the largest / is doubled in the formula. B0.75 for suggesting measuring at multiple points and taking a mean / using a higher-precision micrometer / using a thicker wire.

(a) A plane-polarized wave is a transverse wave in which the oscillations are restricted to a single plane that is perpendicular to the direction of energy transfer (or wave propagation).
(b)(i) Using Malus's Law: \(I = I_0 \cos^2\theta\). Substitute values: \(I = 4.8 \times \cos^2(35^\circ) = 4.8 \times (0.8192)^2 = 4.8 \times 0.6710 = 3.22\text{ W m}^{-2} \approx 3.2\text{ W m}^{-2}\).
(ii) \(\theta = 90^\circ\) (or \(\frac{\pi}{2}\text{ rad}\)). Explanation: Since \(\cos(90^\circ) = 0\), the component of the electric field vector along the transmission axis of the filter is zero, so no wave energy is transmitted.
(c) Intensity is directly proportional to the square of the amplitude (\(I \propto A^2\)). Therefore, doubling the amplitude increases the incident intensity by a factor of \(2^2 = 4\). The new transmitted intensity will also be 4 times the original transmitted intensity: \(3.22 \times 4 = 12.9\text{ W m}^{-2}\) (or \(13\text{ W m}^{-2}\)).

(a) B1 for stating oscillations are in one direction / plane only. B1 for specifying this plane is perpendicular to the direction of propagation / energy transfer.
(b)(i) C1 for stating Malus's law \(I = I_0 \cos^2\theta\). C1 for substitution: \(4.8 \times \cos^2(35^\circ)\). A0.75 for correct answer to 2 or 3 s.f.: \(3.2\text{ W m}^{-2}\) or \(3.22\text{ W m}^{-2}\).
(b)(ii) B1 for \(\theta = 90^\circ\). B1 for explaining that the electric field component along the transmission axis is zero / \(\cos(90^\circ) = 0\).
(c) B1 for stating that \(I \propto A^2\) hence incident intensity increases by a factor of 4. B1 for calculating the new transmitted intensity as \(12.9\text{ W m}^{-2}\) (accept \(13\text{ W m}^{-2}\) or \(12.88\text{ W m}^{-2}\)).

(a) The moment of a force is defined as the force multiplied by the perpendicular distance from the pivot to the line of action of the force.
(b) The forces are: 1. Weight of \(45\text{ N}\) acting vertically downwards at the midpoint of the shelf (\(0.60\text{ m}\) from the hinge). 2. Tension \(T\) in the cable acting at the far end (\(1.2\text{ m}\) from the hinge) at an angle of \(30^\circ\) above the horizontal. 3. Hinge force acting at the hinge (having both vertical and horizontal components to balance the other forces).
(c) Taking moments about the hinge (pivot) for rotational equilibrium: Clockwise moment = Weight \(\times 0.60\text{ m} = 45 \times 0.60 = 27\text{ N m}\). Anticlockwise moment = perpendicular component of Tension \(\times 1.2\text{ m} = (T \sin 30^\circ) \times 1.2 = 0.60 T\). Setting moments equal: \(0.60 T = 27 \implies T = 45\text{ N}\).
(d) For horizontal equilibrium, the horizontal force from the hinge \(H_h\) must balance the horizontal component of the tension: \(H_h = T \cos 30^\circ = 45 \times \cos 30^\circ = 38.97\text{ N} \approx 39\text{ N}\).

(a) B1 for 'force \(\times\) distance from pivot'. B1 for specifying 'perpendicular' distance.
(b) B1 for identifying weight acting downwards at the midpoint (\(0.60\text{ m}\) from hinge). B1 for identifying tension acting at the end at \(30^\circ\) and a force at the hinge.
(c) C1 for calculating clockwise moment as \(27\text{ N m}\). C1 for writing anticlockwise moment as \(T \sin(30^\circ) \times 1.2\). A1 for \(T = 45\text{ N}\).
(d) C1 for identifying that horizontal hinge force equals \(T \cos 30^\circ\). A0.75 for \(39\text{ N}\) (accept \(38.97\text{ N}\) or \(39.0\text{ N}\)).

(a) Initially, when the skydiver jumps, the only force acting is the weight acting vertically downwards. The initial acceleration is equal to \(g = 9.81\text{ m s}^{-2}\). As the speed of the skydiver increases, the upward drag force (air resistance) increases. The resultant downward force is \(W - F_D\). Since \(F_D\) increases, the resultant force decreases, which causes the acceleration of the skydiver to decrease. However, since there is still a resultant downward force, the speed continues to increase. Eventually, the drag force increases to a value equal to the weight of the skydiver. At this point, the resultant force becomes zero, and the acceleration becomes zero. The skydiver now falls at a constant speed, which is the terminal velocity.
(b)(i) Rearrange the equation for \(k\): \(k = \frac{F_D}{v^2}\). The unit of force (Newton, \(\text{N}\)) in SI base units is \(\text{kg m s}^{-2}\) (from \(F = ma\)). The unit of \(v^2\) is \(\text{m}^2\text{s}^{-2}\). Substituting these base units: \([k] = \frac{\text{kg m s}^{-2}}{\text{m}^2\text{s}^{-2}} = \text{kg m}^{-1}\).
(b)(ii) At terminal velocity, the drag force equals the weight of the skydiver: \(F_D = W = mg\). So, \(k v^2 = mg \implies k = \frac{mg}{v^2}\). Substituting the values: \(k = \frac{80 \times 9.81}{52^2} = \frac{784.8}{2704} = 0.290\text{ kg m}^{-1}\).

(a) This is an extended response question.
Level 3 (4-4.75 marks): Detailed description of motion (increasing speed, decreasing acceleration, constant terminal velocity) with complete explanation in terms of forces (weight, drag, resultant force, balanced forces). Quality of written communication is high.
Level 2 (2-3 marks): Describes main stages of motion but explanation of forces is incomplete (e.g., omits mention of resultant force or how drag changes with speed).
Level 1 (1 mark): Identifies some forces or basic motion description.
(b)(i) C1 for stating \(\text{N} = \text{kg m s}^{-2}\) or showing \(F = ma\) units. A1 for correct algebraic simplification leading to \(\text{kg m}^{-1}\).
(b)(ii) C1 for setting \(k v^2 = mg\) at terminal velocity. A1 for correct calculation of \(k = 0.29\text{ kg m}^{-1}\) (accept \(0.290\text{ kg m}^{-1}\)).

(a) Mean time for 20 oscillations: \(t_{\text{mean}} = \frac{37.2 + 37.6 + 37.4}{3} = 37.4\text{ s}\). Uncertainty in time: \(\Delta t = \frac{\text{range}}{2} = \frac{37.6 - 37.2}{2} = 0.2\text{ s}\). Mean period of one oscillation: \(T = \frac{37.4}{20} = 1.87\text{ s}\). Absolute uncertainty in period: \(\Delta T = \frac{\Delta t}{20} = \frac{0.2}{20} = 0.01\text{ s}\). Thus, \(T = 1.87 \pm 0.01\text{ s}\).
(b)(i) Square both sides: \(T^2 = 4\pi^2 \frac{L}{g}\). Rearrange for \(g\): \(g = \frac{4 \pi^2 L}{T^2}\).
(b)(ii) Substitute values: \(g = \frac{4 \pi^2 \times 0.850}{1.87^2} = \frac{33.558}{3.4969} = 9.596\text{ m s}^{-2} \approx 9.60\text{ m s}^{-2}\).
(c) Percentage uncertainty in \(L\): \(\%\Delta L = \frac{0.005}{0.850} \times 100\% = 0.588\%\). Percentage uncertainty in \(T\): \(\%\Delta T = \frac{0.01}{1.87} \times 100\% = 0.535\%\). Since \(g = \frac{4\pi^2 L}{T^2}\): \(\%\Delta g = \%\Delta L + 2(\%\Delta T) = 0.588\% + 2(0.535\%) = 0.588\% + 1.07\% = 1.658\% \approx 1.7\%\).

(a) C1 for calculating mean time \(37.4\text{ s}\) and time uncertainty \(0.2\text{ s}\). C1 for dividing mean time by 20 to get \(T = 1.87\text{ s}\). A1 for dividing uncertainty by 20 to get \(\Delta T = 0.01\text{ s}\).
(b)(i) B1 for showing clear algebraic steps from \(T = 2\pi\sqrt{L/g}\) to \(g = 4\pi^2 L / T^2\).
(b)(ii) C1 for correct substitution of \(L = 0.850\) and \(T = 1.87\). A0.75 for \(9.60\text{ m s}^{-2}\) (accept \(9.6\text{ m s}^{-2}\)).
(c) C1 for calculating percentage uncertainty in \(L\) (0.59\%) or \(T\) (0.53\%). C1 for combining them using \(\%\Delta g = \%\Delta L + 2\%\Delta T\). A1 for final answer of 1.7\% (accept 1.66\% or 1.65\%).

(a) The two light sources must be coherent (meaning they have a constant phase difference and the same frequency/wavelength). Additionally, they should have similar amplitudes to ensure good contrast (high visibility) of the fringes.
(b) First find the fringe spacing \(x\): \(x = \frac{15.6\text{ mm}}{10} = 1.56\text{ mm} = 1.56 \times 10^{-3}\text{ m}\). Using the double-slit equation \(\lambda = \frac{a x}{D}\), where \(a\) is the slit separation \(d\): \(d = \frac{\lambda D}{x} = \frac{650 \times 10^{-9} \times 2.4}{1.56 \times 10^{-3}} = 1.00 \times 10^{-3}\text{ m} = 1.0\text{ mm}\).
(c)(i) Since green light has a shorter wavelength than red light (\(532\text{ nm} < 650\text{ nm}\)), and fringe spacing \(x\) is proportional to wavelength (\(x \propto \lambda\)), the fringe spacing will decrease. The fringes will be green and more closely spaced.
(ii) Since \(x \propto \lambda\), the new fringe spacing \(x_{\text{new}} = x_{\text{old}} \times \frac{\lambda_{\text{new}}}{\lambda_{\text{old}}} = 1.56\text{ mm} \times \frac{532\text{ nm}}{650\text{ nm}} = 1.277\text{ mm}\). The new distance across 10 fringe spacings is \(10 \times 1.277\text{ mm} = 12.8\text{ mm}\) (or \(12.77\text{ mm}\)).

(a) B1 for coherent sources / constant phase difference. B1 for same wavelength/frequency / similar amplitude.
(b) C1 for calculating \(x = 1.56 \times 10^{-3}\text{ m}\). C1 for rearranging \(\lambda = \frac{d x}{D}\) to \(d = \frac{\lambda D}{x}\) and substituting. A1 for \(1.0 \times 10^{-3}\text{ m}\) or \(1.0\text{ mm}\).
(c)(i) B1 for stating that the fringes are closer together (smaller spacing). B0.75 for mentioning the fringes are now green.
(c)(ii) C1 for establishing \(x \propto \lambda\) or calculating new fringe spacing \(1.28\text{ mm}\). A1 for multiplying by 10 to get \(12.8\text{ mm}\) (accept \(12.77\text{ mm}\) or \(13\text{ mm}\) if 2 s.f. used).

(a) The sphere is in equilibrium because the resultant force acting on it is zero (the vector sum of the three coplanar forces—tension, weight, and the horizontal applied force—is zero).
(b) Resolving forces vertically for equilibrium: \(T \cos(25^\circ) = W\). Substitute the given value of \(T = 14\text{ N}\): \(W = 14 \times \cos(25^\circ) = 14 \times 0.9063 = 12.69\text{ N} \approx 12.7\text{ N}\).
(c) Resolving forces horizontally for equilibrium: \(T \sin(25^\circ) = F\). Substitute the given values: \(F = 14 \times \sin(25^\circ) = 14 \times 0.4226 = 5.92\text{ N}\).
(d) The vertical equilibrium equation is \(W = T \cos\theta \implies T = \frac{W}{\cos\theta}\). The weight \(W\) of the sphere remains constant. If the angle \(\theta\) increases, \(\cos\theta\) decreases. Therefore, the tension \(T\) must increase.

(a) B1 for stating that the resultant force is zero. B1 for mentioning that the vector sum of the forces forms a closed triangle / forces balance in all directions.
(b) C1 for stating the vertical equation: \(T \cos(25^\circ) = W\). C1 for substitution: \(14 \times \cos(25^\circ)\). A0.75 for \(12.7\text{ N}\) (accept \(13\text{ N}\) or \(12.69\text{ N}\)).
(c) C1 for stating the horizontal equation: \(T \sin(25^\circ) = F\). A1 for \(5.92\text{ N}\) (accept \(5.9\text{ N}\) or \(5.91\text{ N}\)).
(d) B1 for identifying that vertical balance must still be maintained: \(T = W / \cos\theta\). B1 for explaining that as \(\theta\) increases, \(\cos\theta\) decreases, so \(T\) must increase.

(a) Use the equation of motion for vertical direction: \(s_y = u_y t + \frac{1}{2} a_y t^2\). Here, vertical displacement \(s_y = 35\text{ m}\), initial vertical velocity \(u_y = 0\text{ m s}^{-1}\), and vertical acceleration \(a_y = g = 9.81\text{ m s}^{-2}\). Substituting these: \(35 = 0 + 0.5 \times 9.81 \times t^2 \implies t^2 = \frac{35}{4.905} = 7.136 \implies t = 2.67\text{ s} \approx 2.7\text{ s}\).
(b) In the horizontal direction, there is no acceleration, so speed remains constant. Horizontal distance \(s_x = v_x \times t = 12 \times 2.671 = 32.05\text{ m} \approx 32\text{ m}\) (or \(32.1\text{ m}\)).
(c) Just before impact, the horizontal component of velocity is \(v_x = 12\text{ m s}^{-1}\). The vertical component of velocity is \(v_y = u_y + g t = 0 + 9.81 \times 2.671 = 26.20\text{ m s}^{-1}\). The magnitude of the resultant velocity is \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{12^2 + 26.20^2} = \sqrt{144 + 686.4} = \sqrt{830.4} = 28.8\text{ m s}^{-1} \approx 29\text{ m s}^{-1}\). The direction is given by the angle \(\theta\) below the horizontal: \(\tan\theta = \frac{v_y}{v_x} = \frac{26.20}{12} = 2.183 \implies \theta = \tan^{-1}(2.183) = 65.4^\circ \approx 65^\circ\) below the horizontal.

(a) C1 for using \(s = \frac{1}{2} g t^2\) with appropriate vertical values. A1 for showing calculated value is 2.67 s (which rounds to 2.7 s).
(b) C1 for horizontal distance formula \(s = v t\). A0.75 for \(32\text{ m}\) (or \(32.1\text{ m}\), accept \(32.4\text{ m}\) if using 2.7 s).
(c) C1 for stating horizontal velocity remains \(12\text{ m s}^{-1}\). C1 for calculating vertical velocity \(v_y = 26.2\text{ m s}^{-1}\) (or \(26.5\text{ m s}^{-1}\) using 2.7 s). C1 for using Pythagoras' theorem to find magnitude: \(v = \sqrt{12^2 + 26.2^2}\). A1 for magnitude \(29\text{ m s}^{-1}\) (accept \(28.8\text{ m s}^{-1}\)). A1 for direction angle \(65^\circ\) (accept \(65.4^\circ\) or \(65.6^\circ\)) below the horizontal.