2024 Edexcel IAS-Level Physics (XPH11) Practice Paper with Answers

The Young modulus is defined as stress divided by strain. Stress has the unit \( \text{N m}^{-2} \) and strain is dimensionless. Therefore, the unit of the Young modulus is the pascal (\( \text{Pa} \)), which is equivalent to \( \text{N m}^{-2} \). Since \( 1 \text{ N} = 1 \text{ kg m s}^{-2} \), we have: \( \text{N m}^{-2} = (\text{kg m s}^{-2}) \text{ m}^{-2} = \text{kg m}^{-1} \text{s}^{-2} \).

1 mark: C is the correct answer.

Using the equation of motion: \( v_y^2 = u_y^2 - 2g y \). At the maximum height \( y = H \), the vertical velocity \( v_y = 0 \), which gives \( u_y^2 = 2gH \), where \( u_y = u \sin \theta \). At a height of \( y = \frac{H}{2} \), the vertical velocity squared is: \( v_y^2 = u_y^2 - 2g \left(\frac{H}{2}\right) = u_y^2 - gH \). Substituting \( gH = \frac{u_y^2}{2} \) into this yields \( v_y^2 = u_y^2 - \frac{u_y^2}{2} = \frac{u_y^2}{2} \). Taking the square root gives: \( v_y = \frac{u_y}{\sqrt{2}} = \frac{u \sin \theta}{\sqrt{2}} \).

1 mark: C is the correct answer.

The Young modulus \( E \) is given by \( E = \frac{F L}{A x} \), which can be rearranged to express the extension as \( x = \frac{F L}{A E} \). For the second wire made of the same material (same \( E \)), the new extension \( x' \) is: \( x' = \frac{(2F)(2L)}{(2A)E} = \frac{4FL}{2AE} = 2 \left( \frac{FL}{AE} \right) = 2x \).

1 mark: C is the correct answer.

For the beam in equilibrium, the sum of the moments about the pivot must be zero. The weight \( W \) of the uniform beam acts at its center of gravity, which is at a distance of \( \frac{L}{2} \) from the pivot. Taking moments about the pivot: clockwise moment = \( W \times \frac{L}{2} \), and anticlockwise moment = \( F \times \frac{3}{4}L \). Equating the moments: \( W \frac{L}{2} = F \frac{3}{4}L \implies \frac{W}{2} = \frac{3F}{4} \implies F = \frac{2}{3}W \).

1 mark: C is the correct answer.

The useful output power \( P_{\text{out}} \) required to lift the load at constant speed is given by: \( P_{\text{out}} = F v = m g v = 150 \text{ kg} \times 9.81 \text{ m s}^{-2} \times 2.0 \text{ m s}^{-1} = 2943 \text{ W} \). The efficiency of the motor is \( \eta = 0.60 \). The electrical power input \( P_{\text{in}} \) is: \( P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{2943 \text{ W}}{0.60} = 4905 \text{ W} \approx 4.9 \text{ kW} \).

1 mark: C is the correct answer.

At terminal velocity, the upward forces (upthrust \( U \) and viscous drag \( F_D \)) equal the downward weight \( W \) of the sphere: \( U + F_D = W \implies F_D = W - U \). According to Stokes' Law, \( F_D = 6 \pi \eta r v \). The weight is \( W = V \rho_m g = \frac{4}{3} \pi r^3 \rho_m g \), and the upthrust is \( U = V \rho_l g = \frac{4}{3} \pi r^3 \rho_l g \). Thus, \( 6 \pi \eta r v = \frac{4}{3} \pi r^3 g (\rho_m - \rho_l) \). Solving for \( v \) gives: \( v = \frac{4 \pi r^3 g (\rho_m - \rho_l)}{18 \pi \eta r} = \frac{2 r^2 g (\rho_m - \rho_l)}{9 \eta} \).

1 mark: A is the correct answer.

The change in momentum \( \Delta p \) of the ball is given by: \( \Delta p = p_f - p_i = -m \left(\frac{v}{2}\right) - mv = -\frac{3}{2}mv \). By Newton's second law, the magnitude of the average force \( F \) is: \( F = \frac{|\Delta p|}{\Delta t} = \frac{3mv}{2\Delta t} \). By Newton's third law, the force exerted on the wall has the same magnitude.

1 mark: C is the correct answer.

Applying Newton's second law to the passenger of mass \( m \) moving downwards with acceleration \( a \): \( F_{\text{net}} = m a \implies m g - R = m a \). Rearranging for the normal contact force \( R \) yields: \( R = m g - m a = m(g - a) \).

1 mark: C is the correct answer.

The block moves at a constant velocity, which means the acceleration is zero and the net force along the slope is zero. Resolving forces parallel to the slope: \(F - f - mg\sin\theta = 0\), where \(f\) is the frictional force. Rearranging this gives \(f = F - mg\sin\theta\). The rate of work done against friction is the power dissipated by the frictional force, which is given by \(P = fv = (F - mg\sin\theta)v\).

C is the correct answer. 1 mark for selecting the correct option.

The extension of a wire under tension is given by \(\Delta x = \frac{FL}{AE}\). Since both wires are made of the same material, they have the same Young modulus \(E\). The cross-sectional area \(A\) is proportional to the square of the diameter \(d^2\). For wire X, the extension is \(\Delta x_X \propto \frac{L}{d^2}\). For wire Y, the extension is \(\Delta x_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\), which means \(\Delta x_Y = 0.5\Delta x_X\). The elastic strain energy stored is given by \(E_{\text{elastic}} = \frac{1}{2}F\Delta x\). Since both wires experience the same tensile force \(F\), the ratio of the strain energies is equal to the ratio of their extensions: \(\frac{E_X}{E_Y} = \frac{\Delta x_X}{\Delta x_Y} = \frac{\Delta x_X}{0.5\Delta x_X} = 2\).

C is the correct answer. 1 mark for selecting the correct option.

Using Newton's Second Law: \( F_{\text{net}} = mg - D = ma \) where \( D \) is the air resistance (drag). Rearranging for \( D \): \( D = m(g - a) \). Substituting the given values: \( D = 120 \text{ kg} \times (9.81 \text{ m s}^{-2} - 3.20 \text{ m s}^{-2}) \). \( D = 120 \times 6.61 = 793.2 \text{ N} \). Expressing to 2 or 3 significant figures: \( D = 790 \text{ N} \) or \( 793 \text{ N} \).

M1: Use of \( F = ma \) or \( W - D = ma \) with correct substitution of mass and acceleration. A1: Correct calculation of air resistance to give \( 790 \text{ N} \) or \( 793 \text{ N} \) (accept \( 793.2 \text{ N} \)).

The Young modulus is given by: \( E = \frac{\text{Stress}}{\text{Strain}} = \frac{F L}{A \Delta L} \). Rearranging for the tension force \( F \): \( F = \frac{E A \Delta L}{L} \). Substituting the given values: \( F = \frac{1.1 \times 10^{11} \text{ Pa} \times 1.5 \times 10^{-7} \text{ m}^2 \times 1.8 \times 10^{-3} \text{ m}}{2.4 \text{ m}} \). \( F = \frac{29.7}{2.4} = 12.375 \text{ N} \). Since the mass is suspended vertically: \( F = mg \implies m = \frac{F}{g} \). \( m = \frac{12.375 \text{ N}}{9.81 \text{ m s}^{-2}} = 1.261 \text{ kg} \). Rounding to 2 significant figures (consistent with given data): \( m = 1.3 \text{ kg} \) (or \( 1.26 \text{ kg} \) to 3 significant figures).

M1: Use of \( E = \frac{FL}{A\Delta L} \) to find the force \( F \). M1: Use of \( F = mg \) to relate tension force to mass. A1: Correct final value of mass in the range \( 1.25 \text{ kg} \) to \( 1.30 \text{ kg} \) with appropriate unit.

Calculate the useful energy output (work done against gravity): \( W = mgh = 45 \text{ kg} \times 9.81 \text{ m s}^{-2} \times 8.0 \text{ m} = 3531.6 \text{ J} \). Calculate the total electrical energy input: \( E_{\text{in}} = P_{\text{in}} \times t = 420 \text{ W} \times 12 \text{ s} = 5040 \text{ J} \). Calculate the efficiency: \( \text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} = \frac{3531.6}{5040} = 0.7007 \). Alternatively, calculate the useful power output: \( P_{\text{out}} = \frac{W}{t} = \frac{3531.6}{12} = 294.3 \text{ W} \). \( \text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{294.3}{420} = 0.7007 \). As a percentage: \( \text{Efficiency} = 70\% \) or as a decimal fraction: \( 0.70 \).

M1: Use of \( \Delta E_{\text{grav}} = mgh \) or \( P_{\text{out}} = \frac{mgh}{t} \). M1: Use of \( E = P \times t \) to find input energy, or use of power ratio. A1: Correct calculation of efficiency as \( 0.70 \), \( 70\% \), or \( 70.1\% \).

(a) For the vertical motion, using \(s = ut + \frac{1}{2}at^2\) with \(u = 0\text{ m s}^{-1}\), \(s = 45.0\text{ m}\), and \(a = g = 9.81\text{ m s}^{-2}\): \(45.0 = 0.5 \times 9.81 \times t^2\) which gives \(t^2 = 9.174\), so \(t = 3.03\text{ s}\) (which is about \(3\text{ s}\)). (b) The horizontal distance is calculated using \(s_x = v_x \times t\). Substituting the values: \(s_x = 12.0\text{ m s}^{-1} \times 3.03\text{ s} = 36.36\text{ m}\) which rounds to \(36.4\text{ m}\). (c) The vertical velocity component just before impact is \(v_y = u_y + at = 0 + 9.81 \times 3.03 = 29.72\text{ m s}^{-1}\). The horizontal component is \(v_x = 12.0\text{ m s}^{-1}\). Using \(\tan\theta = v_y / v_x\) gives \(\tan\theta = 29.72 / 12.0 = 2.477\), so \(\theta = \arctan(2.477) = 68.0^\circ\) to the horizontal. (d) With air resistance: The horizontal distance is less because a horizontal drag force causes horizontal deceleration, reducing the horizontal velocity over time. The angle \(\theta\) to the horizontal is larger (steeper) because horizontal velocity is reduced significantly by drag, making the trajectory more vertical at impact.

(a) [3 Marks] Selects \(s = ut + \frac{1}{2}at^2\) for vertical motion (1) Correct substitution of values: \(45.0 = 0.5 \times 9.81 \times t^2\) (1) Calculates \(t = 3.03\text{ s}\) showing 3 s.f. to justify about 3 s (1) (b) [2 Marks] Selects \(s_x = v_x t\) (1) Calculates distance = \(36.4\text{ m}\) (accept \(36\text{ m}\) to \(36.4\text{ m}\)) (1) (c) [3 Marks] Calculates vertical component of velocity \(v_y = 29.7\text{ m s}^{-1}\) (1) Uses \(\tan\theta = v_y / v_x\) (1) Calculates angle \(\theta = 68.0^\circ\) (1) (d) [4 Marks] States actual horizontal distance is smaller (1) Explains that drag opposes horizontal motion, reducing horizontal velocity (1) States angle \(\theta\) is larger/steeper (1) Explains that horizontal velocity decreases more significantly than vertical velocity near impact (1)

(a) Young Modulus is given by \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F L}{A \Delta L}\). Rearranging for extension: \(\Delta L = \frac{F L}{A E} = \frac{150 \times 2.50}{1.20 \times 10^{-6} \times 2.00 \times 10^{11}} = \frac{375}{2.40 \times 10^5} = 1.56 \times 10^{-3}\text{ m}\) (or \(1.56\text{ mm}\)). (b) Elastic strain energy is \(E_{el} = \frac{1}{2} F \Delta L = 0.5 \times 150 \times 1.5625 \times 10^{-3} = 0.117\text{ J}\). (c)(i) Stress is \(\sigma = \frac{F}{A}\), so the maximum force is \(F = \text{Stress} \times A = 3.50 \times 10^8 \times 1.20 \times 10^{-6} = 420\text{ N}\). (c)(ii) The elastic limit is the maximum stress or force a material can withstand and still return to its original length when the load is removed. If exceeded, plastic deformation occurs as planes of atoms slide past each other permanently via the movement of dislocations. (d) The Young Modulus is determined from the gradient of the initial linear region of the stress-strain graph. The ultimate tensile strength is the maximum value of stress on the graph.

(a) [3 Marks] Recalls or uses \(E = \frac{F L}{A \Delta L}\) (1) Correct substitution of values (1) Calculates \(\Delta L = 1.56 \times 10^{-3}\text{ m}\) (1) (b) [2 Marks] Recalls or uses \(E_{el} = \frac{1}{2} F \Delta L\) (1) Calculates energy = \(0.117\text{ J}\) (1) (c)(i) [2 Marks] Recalls or uses \(\text{Stress} = F / A\) (1) Calculates force = \(420\text{ N}\) (1) (c)(ii) [3 Marks] Definition of elastic limit: maximum stress before permanent/plastic deformation (1) Mentions permanent/plastic deformation occurs beyond this point (1) Explanation of atomic rearrangement/sliding of layers/movement of dislocations (1) (d) [2 Marks] States Young Modulus is the gradient of the straight-line/linear section (1) States ultimate tensile strength is the maximum stress reached on the graph (1)

(a) Total momentum is conserved: \(m_A u_A + m_B u_B = m_A v_A + m_B v_B\). Substituting values: \(0.450 \times 3.20 + 0 = 0.450 \times 1.10 + 0.250 \times v_B\) which simplifies to \(1.440 = 0.495 + 0.250 v_B\). Thus, \(0.250 v_B = 0.945\), leading to \(v_B = 3.78\text{ m s}^{-1}\) (approximately \(3.8\text{ m s}^{-1}\)). (b) Total initial kinetic energy is \(E_{ki} = \frac{1}{2} m_A u_A^2 = 0.5 \times 0.450 \times 3.20^2 = 2.304\text{ J}\). Total final kinetic energy is \(E_{kf} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = 0.5 \times 0.450 \times 1.10^2 + 0.5 \times 0.250 \times 3.78^2 = 0.27225 + 1.78605 = 2.0583\text{ J}\). Since \(E_{kf} < E_{ki}\), kinetic energy is not conserved, meaning the collision is inelastic. (c) Impulse equals change in momentum: \(\Delta p_B = m_B v_B - 0 = 0.250 \times 3.78 = 0.945\text{ N s}\). Using \(F_{\text{avg}} \Delta t = \Delta p\), we get \(15.0 \times \Delta t = 0.945\), so \(\Delta t = 0.0630\text{ s}\) (or \(63.0\text{ ms}\)). (d) According to Newton's third law, the force exerted by glider A on glider B is equal in magnitude and opposite in direction to the force exerted by glider B on glider A. These forces act for the same time duration.

(a) [3 Marks] Recalls/uses conservation of linear momentum: \(m_A u_A = m_A v_A + m_B v_B\) (1) Correct substitution of values (1) Calculates \(v_B = 3.78\text{ m s}^{-1}\) to 3 s.f. to justify approximately 3.8 (1) (b) [4 Marks] Uses \(E_k = \frac{1}{2}mv^2\) to find initial kinetic energy as \(2.30\text{ J}\) (1) Calculates final kinetic energy as \(2.06\text{ J}\) (1) Compares initial and final kinetic energies (1) Concludes collision is inelastic because kinetic energy is not conserved (1) (c) [3 Marks] Calculates momentum change of B as \(0.945\text{ N s}\) (1) Uses \(F = \Delta p / \Delta t\) (1) Calculates \(\Delta t = 0.063\text{ s}\) (1) (d) [2 Marks] States that glider B exerts an equal and opposite force on glider A (1) Mentions that the forces act for the same duration (resulting in equal and opposite impulses) (1)

(a) The forces are: Weight \(W = mg\) acting vertically downwards; Normal contact force \(R\) acting perpendicular to the slope; Resistive force \(f\) acting down the slope, parallel to the road; Driving force \(D\) acting up the slope, parallel to the road. (b) Since speed is constant, the forces parallel to the slope are in equilibrium: \(D = f + W \sin\theta\). The weight component is \(W \sin\theta = mg \sin\theta = 1200 \times 9.81 \times \sin(6.5^\circ) = 11772 \times 0.1132 = 1332.6\text{ N}\). Adding the resistive force: \(D = 450 + 1332.6 = 1782.6\text{ N}\), which is about \(1800\text{ N}\). (c) Useful power output is \(P = D \times v = 1782.6\text{ N} \times 18\text{ m s}^{-1} = 32087\text{ W}\) (or \(32.1\text{ kW}\)). If using the rounded \(1800\text{ N}\), \(P = 32.4\text{ kW}\). (d) Efficiency is \(\eta = P_{\text{out}} / P_{\text{in}}\). Rearranging gives \(P_{\text{in}} = P_{\text{out}} / \eta = 32087 / 0.32 = 100271\text{ W}\) (or \(100\text{ kW}\)). If using \(32.4\text{ kW}\), \(P_{\text{in}} = 101250\text{ W}\) (or \(101\text{ kW}\)).

(a) [3 Marks] Weight shown acting vertically downwards (1) Normal contact force shown perpendicular to the slope (1) Driving force shown up-slope and resistive force shown down-slope (1) (b) [4 Marks] Identifies equilibrium of forces parallel to slope (1) Correctly uses weight component as \(mg\sin\theta\) (1) Computes \(1200 \times 9.81 \times \sin(6.5^\circ) = 1333\text{ N}\) (1) Adds resistive force to obtain \(1783\text{ N}\) (must show to at least 4 s.f. to support "about 1800 N") (1) (c) [2 Marks] Recalls or uses \(P = F v\) (1) Calculates useful power = \(32.1\text{ kW}\) (or \(32.4\text{ kW}\)) (1) (d) [3 Marks] Recalls or uses \(\text{efficiency} = P_{\text{out}} / P_{\text{in}}\) (1) Correct substitution of values (1) Calculates input power = \(1.00 \times 10^5\text{ W}\) (or \(100\text{ kW}\)) (1)

(a) Viscosity is a measure of a fluid's resistance to flow. Conditions for Stokes' law to apply: the object must be spherical, the flow must be laminar (low speed, no turbulence), and there should be no wall/boundary effects. (b) When falling at terminal velocity, the forces are in equilibrium: Upward forces = Downward force, which is \(U + F_D = W\), where \(U\) is the Upthrust, \(F_D\) is the Drag force (viscous drag), and \(W\) is the Weight. (c) At terminal velocity, \(W - U = F_D\). This is written as \(\rho_{s} V g - \rho_{o} V g = 6 \pi \eta r v\). Since \(V = \frac{4}{3}\pi r^3\), this simplifies to \(\frac{4}{3}\pi r^3 g (\rho_s - \rho_o) = 6\pi\eta r v\). Solving for \(v\): \(v = \frac{2 r^2 g (\rho_s - \rho_o)}{9 \eta}\). Substituting the values: \(v = \frac{2 \times (2.00 \times 10^{-3})^2 \times 9.81 \times (7800 - 900)}{9 \times 0.290} = \frac{2 \times 4.00 \times 10^{-6} \times 9.81 \times 6900}{2.61} = \frac{0.541512}{2.61} = 0.2075\text{ m s}^{-1}\), which is approximately \(0.21\text{ m s}^{-1}\). (d) An increase in temperature reduces the viscosity of the engine oil. Since terminal velocity is inversely proportional to viscosity (\(v \propto 1/\eta\)), the terminal velocity of the sphere will increase.

(a) [3 Marks] Definition of viscosity as fluid's resistance to flow (1) State any two conditions: spherical object, laminar flow, infinite fluid/no wall effects (2) (b) [2 Marks] Identifies three forces: Weight (downward), Upthrust (upward), Drag (upward) (1) States forces are in equilibrium: \(W = U + F_D\) (1) (c) [5 Marks] Recalls equations for Weight, Upthrust, and Drag: \(W = \rho_{s} V g\), \(U = \rho_{o} V g\), \(F_D = 6\pi\eta r v\) (1) Equates forces: \(\frac{4}{3}\pi r^3 g (\rho_s - \rho_o) = 6\pi\eta r v\) (1) Correctly rearranges for terminal velocity: \(v = \frac{2 r^2 g (\rho_s - \rho_o)}{9 \eta}\) (1) Correct substitution of values (1) Calculates terminal velocity = \(0.208\text{ m s}^{-1}\) or \(0.21\text{ m s}^{-1}\) (1) (d) [2 Marks] States that viscosity decreases when temperature increases (1) Concludes that terminal velocity increases (1)

The total resistance of the circuit is \(R + r\). As \(R\) increases, the total resistance increases, which causes the current \(I = E / (R + r)\) to decrease. The terminal potential difference is given by \(V = E - Ir\). Since the current \(I\) decreases, the lost volts \(Ir\) decrease, and therefore the terminal potential difference \(V\) increases.

1 mark for the correct option A.

The path difference at point \(P\) is \(36\text{ cm} - 24\text{ cm} = 12\text{ cm}\). Since the wavelength \(\lambda\) is 8 cm, the path difference is \(12 / 8 = 1.5\lambda\). A path difference of \(1\lambda\) corresponds to a phase difference of \(2\pi\text{ rad}\). Therefore, a path difference of \(1.5\lambda\) corresponds to a phase difference of \(1.5 \times 2\pi = 3\pi\text{ rad}\).

1 mark for the correct option C.

According to Einstein's photoelectric equation, \(hf = \Phi + E_{k,max}\), which gives \(hf - \Phi = E_{k,max}\). When the frequency is doubled, the new maximum kinetic energy \(E'_{k,max}\) is given by \(E'_{k,max} = h(2f) - \Phi = 2hf - \Phi\). Substituting \(hf = E_{k,max} + \Phi\) into this equation yields \(E'_{k,max} = 2(E_{k,max} + \Phi) - \Phi = 2E_{k,max} + \Phi\).

1 mark for the correct option B.

The resistance of a wire is given by \(R = \rho L / A\), where \(A = \pi d^2 / 4\). Thus, \(R\) is proportional to \(L / d^2\). For the second wire, the length is \(2L\) and the diameter is \(2d\), so its resistance is proportional to \(2L / (2d)^2 = 2L / 4d^2 = 0.5 (L / d^2)\). Therefore, the new resistance is \(R / 2\).

1 mark for the correct option B.

Using Snell's Law: \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\). Substituting the given values: \(1.6 \sin(30^\circ) = 1.2 \sin(\theta_2)\). Since \(\sin(30^\circ) = 0.5\), we have \(0.8 = 1.2 \sin(\theta_2)\), which gives \(\sin(\theta_2) = 0.8 / 1.2 = 2/3\). Therefore, \(\theta_2 = \sin^{-1}(2/3) \approx 41.8^\circ\). Since the angle of incidence 30 degrees is less than the critical angle \(\theta_c = \sin^{-1}(1.2/1.6) \approx 48.6^\circ\), refraction occurs.

1 mark for the correct option C.

When unpolarised light passes through the first filter, its intensity is halved, so \(I_1 = 0.5 I_0\). According to Malus's Law, when this polarized light passes through the second filter, the transmitted intensity is \(I_2 = I_1 \cos^2(\theta)\). Substituting \(\theta = 60^\circ\) gives \(I_2 = 0.5 I_0 \cos^2(60^\circ) = 0.5 I_0 \times (0.5)^2 = 0.125 I_0\).

1 mark for the correct option A.

An NTC thermistor's resistance decreases as its temperature increases. In a series potential divider circuit, the voltage across a component is proportional to its resistance. As the resistance of the thermistor decreases relative to the fixed resistor, the proportion of the total voltage across the thermistor also decreases, leading to a lower output voltage \(V_{out}\).

1 mark for the correct option D.

For a string of length \(L\) fixed at both ends vibrating at its third harmonic, there are three loops (half-wavelengths) along the length of the string. Thus, \(L = 3 \times (\lambda / 2)\), which means the wavelength \(\lambda = 2L / 3\). The distance between any two adjacent nodes is always equal to half a wavelength, \(\lambda / 2\). Therefore, the distance is \((2L / 3) / 2 = L / 3\).

1 mark for the correct option B.

Since the two wires are connected in series, the current I is the same in both. The current is given by I = nAve, where n is the number density of conduction electrons, A is the cross-sectional area, v is the drift velocity, and e is the charge of an electron. Since both wires are made of copper, n is the same. Therefore, the product of the cross-sectional area and drift velocity (Av) is constant. This means the drift velocity is inversely proportional to the cross-sectional area. Since cross-sectional area is proportional to the square of the diameter, the ratio of the drift velocities v_X / v_Y = (diameter_Y / diameter_X)^2 = (2d / d)^2 = 4.

1 mark for selecting correct option D. Reject all other options.

Using Einstein's photoelectric equation, hf = \Phi + E, which rearranged gives hf = E + \Phi. When the frequency is doubled, the photon energy becomes 2hf. The new maximum kinetic energy E' is given by E' = 2hf - \Phi. Substituting hf = E + \Phi into this equation yields E' = 2(E + \Phi) - \Phi = 2E + 2\Phi - \Phi = 2E + \Phi.

1 mark for selecting correct option B. Reject all other options.

First, calculate the cross-sectional area of the wire: \(A = \frac{\pi d^2}{4} = \frac{\pi \times (1.2 \times 10^{-3}\text{ m})^2}{4} = 1.13 \times 10^{-6}\text{ m}^2\). Next, rearrange the transport equation \(I = nAvq\) to solve for drift velocity \(v\): \(v = \frac{I}{nAq}\). Substituting the known values: \(v = \frac{2.5\text{ A}}{(8.5 \times 10^{28}\text{ m}^{-3}) \times (1.13 \times 10^{-6}\text{ m}^2) \times (1.60 \times 10^{-19}\text{ C})} = 1.63 \times 10^{-4}\text{ m s}^{-1}\).

MP1: Calculation of cross-sectional area \(A = 1.13 \times 10^{-6}\text{ m}^2\) (1) MP2: Correct rearrangement of transport equation \(v = \frac{I}{nAq}\) (1) MP3: Correct calculation of drift velocity to give \(1.6 \times 10^{-4}\text{ m s}^{-1}\) (or \(1.63 \times 10^{-4}\text{ m s}^{-1}\)) with correct unit (1)

Calculate photon energy: \(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{240 \times 10^{-9}\text{ m}} = 8.29 \times 10^{-19}\text{ J}\). Convert the work function into joules: \(\phi = 2.36\text{ eV} \times 1.60 \times 10^{-19}\text{ J eV}^{-1} = 3.78 \times 10^{-19}\text{ J}\). Use the photoelectric equation: \(E_{k,\max} = E - \phi = 8.29 \times 10^{-19}\text{ J} - 3.78 \times 10^{-19}\text{ J} = 4.51 \times 10^{-19}\text{ J}\).

MP1: Calculate energy of the incident photon \(E = 8.3 \times 10^{-19}\text{ J}\) (1) MP2: Convert work function to joules \(\phi = 3.8 \times 10^{-19}\text{ J}\) (1) MP3: Correct application of photoelectric equation to find \(E_{k,\max} = 4.5 \times 10^{-19}\text{ J}\) (accept \(4.51 \times 10^{-19}\text{ J}\)) (1)

First determine the voltage across the fixed resistor: \(V_{\text{fixed}} = V_{\text{total}} - V_{\text{thermistor}} = 9.0\text{ V} - 3.6\text{ V} = 5.4\text{ V}\). Calculate the current in the circuit: \(I = \frac{V_{\text{fixed}}}{R_{\text{fixed}}} = \frac{5.4\text{ V}}{1500\ \Omega} = 3.6 \times 10^{-3}\text{ A}\). Calculate the resistance of the thermistor: \(R_{\text{thermistor}} = \frac{V_{\text{thermistor}}}{I} = \frac{3.6\text{ V}}{3.6 \times 10^{-3}\text{ A}} = 1000\ \Omega\) (or \(1.0\text{ k}\Omega\)).

MP1: Determine the voltage across the fixed resistor is \(5.4\text{ V}\) or set up a valid ratio equation (1) MP2: Calculate the circuit current as \(3.6\text{ mA}\) or correct substitution into the potential divider equation (1) MP3: Correct final value of thermistor resistance as \(1000\ \Omega\) or \(1.0\text{ k}\Omega\) (1)

First, calculate the grating spacing \(d\): \(d = \frac{1 \times 10^{-3}\text{ m}}{500} = 2.00 \times 10^{-6}\text{ m}\). Using the diffraction grating formula \(d \sin\theta = n\lambda\), rearrange for wavelength \(\lambda\) with \(n = 2\): \(\lambda = \frac{d \sin\theta}{n} = \frac{2.00 \times 10^{-6}\text{ m} \times \sin(38.0^\circ)}{2} = 6.1566 \times 10^{-7}\text{ m}\). Rounding to 3 significant figures gives \(6.16 \times 10^{-7}\text{ m}\).

MP1: Calculate the grating spacing \(d = 2.00 \times 10^{-6}\text{ m}\) (1) MP2: Correct substitution into the formula \(d \sin\theta = n\lambda\) using \(n = 2\) (1) MP3: Final wavelength calculated to be \(6.16 \times 10^{-7}\text{ m}\) (or \(6.2 \times 10^{-7}\text{ m}\)) (1)

**(a)**
1. The total resistance of the circuit is \(R_{\text{total}} = R + r\).
2. Using Ohm's Law, the current \(I\) in the circuit is:
\(I = \frac{\varepsilon}{R_{\text{total}}} = \frac{\varepsilon}{R+r}\)
3. The power \(P\) dissipated in the external resistor \(R\) is:
\(P = I^2 R = \left(\frac{\varepsilon}{R+r}\right)^2 R = \frac{\varepsilon^2 R}{(R+r)^2}\) (as required).

**(b)**
1. Calculate the current:
\(I = \frac{\varepsilon}{R+r} = \frac{1.50\text{ V}}{2.40\ \Omega + 0.85\ \Omega} = \frac{1.50}{3.25} = 0.4615\text{ A} \approx 0.46\text{ A}\)
2. Calculate the power:
\(P = I^2 R = (0.4615\text{ A})^2 \times 2.40\ \Omega = 0.511\text{ W} \approx 0.51\text{ W}\)

**(c)**
1. Find the total resistance at \(50^\circ\text{C}\) with the fixed resistor:
\(R_{\text{total}} = R_T + R_F + r = 0.55\ \Omega + 1.20\ \Omega + 0.85\ \Omega = 2.60\ \Omega\)
2. Calculate the new current \(I_2\):
\(I_2 = \frac{\varepsilon}{R_{\text{total}}} = \frac{1.50\text{ V}}{2.60\ \Omega} \approx 0.577\text{ A}\)
3. Calculate the terminal potential difference \(V_{\text{term}}\):
\(V_{\text{term}} = \varepsilon - I_2 r = 1.50\text{ V} - (0.577\text{ A} \times 0.85\ \Omega) = 1.50\text{ V} - 0.49\text{ V} = 1.01\text{ V}\)
(Alternatively: \(V_{\text{term}} = I_2(R_T + R_F) = 0.577 \times (0.55 + 1.20) = 1.01\text{ V}\))
4. Calculate the potential difference across the thermistor \(V_{\text{thermistor}}\):
\(V_{\text{thermistor}} = I_2 R_T = 0.577\text{ A} \times 0.55\ \Omega = 0.317\text{ V} \approx 0.32\text{ V}\)

**(a)**
* **(1 mark)** Expresses total resistance as \(R + r\).
* **(1 mark)** Expresses current as \(I = \frac{\varepsilon}{R+r}\).
* **(1 mark)** Substitutes \(I\) into \(P = I^2 R\) to show the given formula clearly.

**(b)**
* **(1 mark)** Uses total resistance to find \(I = 0.46\text{ A}\) (accept \(0.462\text{ A}\)).
* **(1 mark)** Uses \(P = I^2 R\) or \(P = V I\) with appropriate substitutions.
* **(1 mark)** Calculates \(P = 0.51\text{ W}\) (accept \(0.511\text{ W}\)).
* **(1 mark)** Correct units shown for both current (A) and power (W).

**(c)**
* **(1 mark)** Calculates new total circuit resistance as \(2.60\ \Omega\).
* **(1 mark)** Calculates new current \(I_2 = 0.577\text{ A}\) (accept \(0.58\text{ A}\)).
* **(1 mark)** Uses \(V_{\text{term}} = \varepsilon - I r\) or \(I(R_T + R_F)\).
* **(1 mark)** Calculates terminal potential difference as \(1.01\text{ V}\) (accept \(1.00\text{ V}\) to \(1.02\text{ V}\)).
* **(1 mark)** Calculates potential difference across the thermistor as \(0.32\text{ V}\) (accept \(0.317\text{ V}\)).

**(a)**
1. Calculate the photon energy in Joules:
\(E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34}\text{ J s} \times 3.00 \times 10^8\text{ m s}^{-1}}{240 \times 10^{-9}\text{ m}} = 8.2875 \times 10^{-19}\text{ J}\)
2. Convert the work function to Joules:
\(\Phi = 2.36\text{ eV} \times 1.60 \times 10^{-19}\text{ J/eV} = 3.776 \times 10^{-19}\text{ J}\)
3. Use Einstein's photoelectric equation:
\(E_{k,\text{max}} = E - \Phi = 8.2875 \times 10^{-19}\text{ J} - 3.776 \times 10^{-19}\text{ J} = 4.5115 \times 10^{-19}\text{ J} \approx 4.51 \times 10^{-19}\text{ J}\).

**(b)**
1. Kinetic energy gained by the electron:
\(E_k = e V = 1.60 \times 10^{-19}\text{ C} \times 150\text{ V} = 2.40 \times 10^{-17}\text{ J}\)
2. Calculate the momentum \(p\) of the electron:
\(p = \sqrt{2 m_e E_k} = \sqrt{2 \times 9.11 \times 10^{-31}\text{ kg} \times 2.40 \times 10^{-17}\text{ J}} = 6.613 \times 10^{-24}\text{ kg m s}^{-1}\)
3. Calculate the de Broglie wavelength \(\lambda_{\text{dB}}\):
\(\lambda_{\text{dB}} = \frac{h}{p} = \frac{6.63 \times 10^{-34}\text{ J s}}{6.613 \times 10^{-24}\text{ kg m s}^{-1}} = 1.0026 \times 10^{-10}\text{ m} \approx 1.00 \times 10^{-10}\text{ m}\).

**(c)**
1. Doubling the intensity of the light source doubles the number of photons arriving per second. Since the photoelectric effect is a one-to-one interaction, the rate of emission of photoelectrons doubles.
2. The maximum kinetic energy depends only on the photon energy (frequency) and the work function. Since the wavelength remains constant, the maximum kinetic energy is unchanged.

**(a)**
* **(1 mark)** Calculates photon energy \(E = 8.29 \times 10^{-19}\text{ J}\).
* **(1 mark)** Converts work function to Joules: \(\Phi = 3.78 \times 10^{-19}\text{ J}\).
* **(1 mark)** Uses Einstein's photoelectric equation \(E_{k,\text{max}} = hf - \Phi\).
* **(1 mark)** Shows final answer of \(4.51 \times 10^{-19}\text{ J}\) (clearly showing conversion/steps to at least 2 significant figures).

**(b)**
* **(1 mark)** Calculates kinetic energy \(E_k = eV = 2.40 \times 10^{-17}\text{ J}\).
* **(1 mark)** Uses momentum formula \(p = \sqrt{2 m_e E_k}\) or calculates velocity \(v = \sqrt{\frac{2E_k}{m_e}} = 7.26 \times 10^6\text{ m s}^{-1}\).
* **(1 mark)** Calculates momentum \(p = 6.61 \times 10^{-24}\text{ kg m s}^{-1}\).
* **(1 mark)** Uses \(\lambda_{\text{dB}} = \frac{h}{p}\).
* **(1 mark)** Obtains \(1.00 \times 10^{-10}\text{ m}\) (accept range \(1.00 \times 10^{-10}\text{ m}\) to \(1.01 \times 10^{-10}\text{ m}\)).

**(c)**
* **(1 mark)** States that the number of electrons emitted per second doubles.
* **(1 mark)** Explains this in terms of one-to-one interaction (each photon can release one electron, and photon flux has doubled).
* **(1 mark)** States that the maximum kinetic energy is unchanged because photon energy / frequency is unchanged.

**(a)**
1. Use the double-slit formula: \(w = \frac{\lambda D}{d}\)
Where \(\lambda = 633 \times 10^{-9}\text{ m}\), \(D = 1.80\text{ m}\), and \(d = 0.250 \times 10^{-3}\text{ m}\).
2. Substitute the values:
\(w = \frac{633 \times 10^{-9}\text{ m} \times 1.80\text{ m}}{0.250 \times 10^{-3}\text{ m}} = 4.5576 \times 10^{-3}\text{ m} \approx 4.56\text{ mm}\).

**(b)(i)**
1. Use the diffraction grating formula: \(d \sin\theta = n\lambda\)
2. Rearrange for slit spacing \(d\) with \(n = 2\) and \(\theta = 15.0^\circ\):
\(d = \frac{2 \lambda}{\sin(15.0^\circ)} = \frac{2 \times 633 \times 10^{-9}\text{ m}}{\sin(15.0^\circ)} = 4.8914 \times 10^{-6}\text{ m}\)
3. Calculate the number of lines per metre \(N_{\text{m}}\):
\(N_{\text{m}} = \frac{1}{d} = \frac{1}{4.8914 \times 10^{-6}\text{ m}} = 2.0444 \times 10^5\text{ lines/m}\)
4. Convert to lines per millimetre:
\(N_{\text{mm}} = \frac{2.0444 \times 10^5}{1000} = 204.4\text{ lines/mm} \approx 204\text{ lines/mm}\).

**(b)(ii)**
1. For the maximum possible order, \(\theta \le 90^\circ\), so \(\sin\theta \le 1\).
2. \(n_{\text{max}} = \frac{d}{\lambda} = \frac{4.8914 \times 10^{-6}\text{ m}}{633 \times 10^{-9}\text{ m}} = 7.73\)
3. Since \(n\) must be an integer, the highest observable order is \(n = 7\).
4. The total number of maxima is \(2n_{\text{max}} + 1 = 2(7) + 1 = 15\) (consisting of the central maximum, 7 on the left, and 7 on the right).

**(a)**
* **(1 mark)** Recalls or uses \(w = \frac{\lambda D}{d}\).
* **(1 mark)** Correctly substitutes values with matching units.
* **(1 mark)** Calculates fringe width as \(4.56\text{ mm}\) (or \(4.6 \times 10^{-3}\text{ m}\)).

**(b)(i)**
* **(1 mark)** Recalls or uses \(d \sin\theta = n\lambda\).
* **(1 mark)** Substitutes \(n=2\), \(\theta = 15.0^\circ\), and \(\lambda = 633\text{ nm}\) to find \(d\).
* **(1 mark)** Calculates \(d = 4.89 \times 10^{-6}\text{ m}\).
* **(1 mark)** Recognizes \(N = \frac{1}{d}\).
* **(1 mark)** Calculates \(N = 204\text{ lines/mm}\) (accept \(204 - 205\text{ lines/mm}\)).

**(b)(ii)**
* **(1 mark)** Sets \(\theta = 90^\circ\) or \(\sin\theta = 1\).
* **(1 mark)** Calculates maximum theoretical order \(n = 7.73\).
* **(1 mark)** Identifies the highest visible integer order as \(n = 7\).
* **(1 mark)** Calculates total number of maxima as \(15\).

**(a)**
1. Use the resistance formula: \(R = \rho \frac{L}{A}\)
2. Substitute the given values:
\(R = 1.72 \times 10^{-8}\ \Omega\text{ m} \times \frac{2.50\text{ m}}{4.50 \times 10^{-7}\text{ m}^2} = 0.09556\ \Omega\)
This is approximately \(0.10\ \Omega\) (as required).

**(b)(i)**
1. Use Ohm's Law: \(I = \frac{V}{R}\)
2. Substitute \(V = 12.0\text{ V}\) and \(R = 0.09556\ \Omega\):
\(I = \frac{12.0\text{ V}}{0.09556\ \Omega} = 125.6\text{ A} \approx 126\text{ A}\).

**(b)(ii)**
1. Use the transport equation: \(I = n A v e\)
Where \(n = 8.50 \times 10^{28}\text{ m}^{-3}\), \(A = 4.50 \times 10^{-7}\text{ m}^2\), \(e = 1.60 \times 10^{-19}\text{ C}\), and \(I = 125.6\text{ A}\).
2. Rearrange for drift velocity \(v\):
\(v = \frac{I}{n A e} = \frac{125.6\text{ A}}{(8.50 \times 10^{28}\text{ m}^{-3}) \times (4.50 \times 10^{-7}\text{ m}^2) \times (1.60 \times 10^{-19}\text{ C})}\)
3. Calculate the denominator:
\(n A e = 8.50 \times 10^{28} \times 4.50 \times 10^{-7} \times 1.60 \times 10^{-19} = 6120\text{ C m}^{-1}\)
4. Solve for \(v\):
\(v = \frac{125.6}{6120} = 0.0205\text{ m s}^{-1} = 2.05 \times 10^{-2}\text{ m s}^{-1}\).

**(c)**
1. As temperature increases, the positive metal lattice ions vibrate with greater amplitude and frequency.
2. This increases the frequency of collisions between the conduction electrons and the vibrating lattice ions.
3. Consequently, the resistance of the wire increases.
4. Since the potential difference remains constant, the increased resistance causes the current \(I\) to decrease. Since \(I = nAve\) and \(n\), \(A\), and \(e\) are constant, the drift velocity \(v\) must decrease.

**(a)**
* **(1 mark)** Recalls and uses \(R = \rho \frac{L}{A}\).
* **(1 mark)** Shows calculation to get \(0.096\ \Omega\) (demonstrates proper substitution and rounding).

**(b)(i)**
* **(1 mark)** Recalls or uses \(I = \frac{V}{R}\).
* **(1 mark)** Calculates \(I = 126\text{ A}\) (accept range \(120\text{ A}\) to \(126\text{ A}\) depending on rounding of \(R\)).

**(b)(ii)**
* **(1 mark)** Recalls or uses \(I = n A v e\).
* **(1 mark)** Rearranges equation to make \(v\) the subject.
* **(1 mark)** Substitutes values correctly, including electronic charge \(e = 1.60 \times 10^{-19}\text{ C}\).
* **(1 mark)** Calculates \(v = 2.05 \times 10^{-2}\text{ m s}^{-1}\) (accept \(2.1 \times 10^{-2}\text{ m s}^{-1}\)).

**(c)**
* **(1 mark)** States that lattice ions vibrate with greater amplitude/energy as temperature rises.
* **(1 mark)** States that this leads to more frequent collisions between conduction electrons and lattice ions.
* **(1 mark)** Explains that this increases resistance (and thus decreases current for constant voltage).
* **(1 mark)** Concludes that drift velocity decreases as a result of the decreased current.

**(a)**
1. Use the critical angle formula: \(\sin\theta_c = \frac{n_2}{n_1}\)
2. Substitute the refractive indices:
\(\sin\theta_c = \frac{1.45}{1.52} = 0.9539\)
3. Calculate \(\theta_c\):
\(\theta_c = \sin^{-1}(0.9539) = 72.54^\circ \approx 72.5^\circ\).

**(b)(i)**
1. Use Snell's Law at the air-core boundary:
\(n_{\text{air}} \sin\theta_{\text{air}} = n_1 \sin\theta_r\)
2. Substitute \(n_{\text{air}} = 1.00\), \(\theta_{\text{air}} = 12.0^\circ\), and \(n_1 = 1.52\):
\(1.00 \times \sin(12.0^\circ) = 1.52 \sin\theta_r\)
3. Solve for \(\sin\theta_r\):
\(\sin\theta_r = \frac{\sin(12.0^\circ)}{1.52} = \frac{0.2079}{1.52} = 0.1368\)
4. Calculate \(\theta_r\):
\(\theta_r = \sin^{-1}(0.1368) = 7.86^\circ\).

**(b)(ii)**
1. Consider the geometry of the fibre. Since the end face is perpendicular to the core-cladding boundary, the angle of incidence \(\phi\) at the boundary is given by:
\(\phi = 90^\circ - \theta_r\)
2. Substitute \(\theta_r = 7.86^\circ\):
\(\phi = 90^\circ - 7.86^\circ = 82.14^\circ\).
3. Compare the angle of incidence \(\phi\) with the critical angle \(\theta_c\):
Since \(\phi = 82.14^\circ\) is greater than \(\theta_c = 72.5^\circ\), the condition for total internal reflection is met.
4. Therefore, the ray of light will undergo total internal reflection.

**(c)**
1. Advantage: Confining light to the core and preventing 'cross-talk' (light leakage between adjacent fibres).
2. Explanation: If two bare fibres touch, light can pass from one to another because there is no boundary with lower refractive index. Cladding ensures total internal reflection occurs inside each fibre even if they are in contact. (Alternative: Cladding protects the core from scratches/damage, which would otherwise alter the boundary and cause light to scatter out).

**(a)**
* **(1 mark)** Recalls or uses \(\sin\theta_c = \frac{n_2}{n_1}\).
* **(1 mark)** Substitutes values correctly (showing \(1.45/1.52\)).
* **(1 mark)** Calculates \(72.5^\circ\) (accept \(72.5^\circ\) to \(73^\circ\)).

**(b)(i)**
* **(1 mark)** Recalls or uses \(n_1 \sin\theta_1 = n_2 \sin\theta_2\).
* **(1 mark)** Substitutes values correctly (\(1.00\sin(12.0^\circ) = 1.52\sin\theta_r\)).
* **(1 mark)** Calculates \(7.86^\circ\) (accept \(7.9^\circ\)).

**(b)(ii)**
* **(1 mark)** Recognizes that the angle of incidence at the core-cladding boundary is \(90^\regular - \theta_r\).
* **(1 mark)** Calculates the boundary angle of incidence as \(82.1^\circ\) (or \(82.14^\circ\)).
* **(1 mark)** Evaluates by comparing boundary angle with critical angle (\(82.1^\circ > 72.5^\circ\)).
* **(1 mark)** Clearly states the conclusion that total internal reflection will occur.

**(c)**
* **(1 mark)** Identifies a valid advantage (e.g., prevents cross-talk/light leakage, protects core from physical damage/contamination, reduces signal attenuation from dirt/scratches).
* **(1 mark)** Gives a clear physical explanation supporting the advantage (e.g., because total internal reflection occurs at the core-cladding interface rather than core-air interface, meaning external contact does not affect the light confinement).

(a) To measure the diameter: Measure at several different positions along the wire to check for uniformity. Measure at different orientations at each position to account for non-circularity. Calculate a mean value. Check for and subtract any zero error on the micrometer. (b) The formula for resistance is \( R = \rho L / A \). Comparing this to the equation of a straight line \( y = mx + c \) where \( y = R \) and \( x = L \), the gradient \( m \) of the graph is given by \( m = \rho / A \). Therefore, resistivity is calculated as \( \rho = m \times A \). (c) The percentage uncertainty in diameter \( d \) is: \( \% \Delta d = (0.02 / 0.46) \times 100 = 4.348\% \). Since the area is given by \( A = \pi d^2 / 4 \), the percentage uncertainty in area \( A \) is twice the percentage uncertainty in \( d \): \( \% \Delta A = 2 \times 4.348\% = 8.696\% \), which rounds to \( 8.7\% \). (d) Keeping the current flowing causes heating of the wire. An increase in temperature would increase the resistivity and resistance of the wire, introducing a systematic error. Disconnecting the power supply allows the temperature to remain constant.

Part (a): [1 mark] Measure diameter at multiple positions along the wire. [1 mark] Measure at different orientations/angles at each position. [1 mark] Check for zero error and calculate mean diameter. Part (b): [1 mark] Correctly state formula \( R = \rho L / A \). [1 mark] Identify gradient \( m = \rho / A \). [1 mark] Rearrange to show \( \rho = m \times A \). Part (c): [1 mark] Calculation of percentage uncertainty in \( d \): \( 4.35\% \). [1 mark] Recall that uncertainty is doubled for \( d^2 \) term. [1.5 marks] Correct calculation showing \( 8.7\% \) (accept 8.69% to 8.70%). Part (d): [1 mark] Current causes a temperature rise in the wire. [1 mark] Resistance of the wire increases with temperature. [1 mark] Disconnecting the current maintains a constant temperature / ensures a fair test.

(a) Independent variable: height \( h \). Dependent variable: time \( t \) (or \( t^2 \)). (b) A graph of \( h \) against \( t \) would be a curve, which makes it difficult to determine a precise gradient. A graph of \( h \) against \( t^2 \) is a straight line through the origin, where the gradient is constant and equals \( g/2 \), allowing a much more accurate and straightforward calculation of \( g \). (c) Using \( g = 2h / t^2 \): \( g = (2 \times 1.200) / (0.495)^2 = 2.400 / 0.245025 \approx 9.80 \text{ m s}^{-2} \). Percentage uncertainty in \( h \): \( \% \Delta h = (0.002 / 1.200) \times 100 = 0.167\% \). Percentage uncertainty in \( t \): \( \% \Delta t = (0.005 / 0.495) \times 100 = 1.010\% \). Percentage uncertainty in \( g \): \( \% \Delta g = \% \Delta h + 2 \times \% \Delta t = 0.167\% + 2.020\% = 2.187\% \). Absolute uncertainty in \( g \): \( \Delta g = 9.795 \times 0.02187 \approx 0.21 \text{ m s}^{-2} \). (d) One potential systematic error is the time delay in the electromagnet releasing the ball. This can be minimized by using a double light gate setup where the timer starts when the ball passes the first gate (eliminating release delay) rather than starting when the electromagnet is switched off.

Part (a): [1 mark] Identify height \( h \) as the independent variable. [1 mark] Identify time \( t \) as the dependent variable. Part (b): [1 mark] State that \( h \) vs \( t \) is a curve. [1 mark] State that \( h \) vs \( t^2 \) is a straight line through the origin. [1 mark] Gradient of the straight line is constant and equals \( g/2 \), which is easier to calculate accurately. Part (c): [1 mark] Correct calculation of \( g \approx 9.80 \text{ m s}^{-2} \) (accept 9.79 to 9.80). [1 mark] Percentage uncertainty in \( h \) calculated as \( 0.17\% \). [1 mark] Percentage uncertainty in \( t \) calculated as \( 1.01\% \). [1 mark] Correct addition of uncertainties: \( \% \Delta g = \% \Delta h + 2\% \Delta t = 2.19\% \) (or 2.2%). [1.5 marks] Calculation of absolute uncertainty \( \approx 0.21 \text{ m s}^{-2} \) (accept 0.2 to 0.22). Part (d): [1 mark] Identify a valid systematic error (e.g., air resistance, delay in electromagnet release, parallax error in measuring height). [1 mark] Suggestion to minimize: use a heavy/dense sphere (for air resistance), or start timing with a light gate rather than release switch.

(a) A suitable diagram should show: a long wire clamped at one end to a bench, passing over a pulley at the other end, with a mass hanger attached. A marker (e.g., tape) is attached to the wire, and a metre rule with a vernier scale or travelling microscope is aligned next to the marker to measure its displacement. (b) To avoid slippage errors, the student can use a reference wire suspended from the same support alongside the test wire (Searle's apparatus), or measure the extension from a fixed reference marker on the wire near the clamp rather than measuring from the clamp itself. Any movement at the clamp will then not affect the measured distance between the marker and the pulley. (c) First, calculate the cross-sectional area \( A = \pi d^2 / 4 = \pi (0.32 \times 10^{-3})^2 / 4 = 8.042 \times 10^{-8} \text{ m}^2 \). Stress \( \sigma = F / A = 15.0 / (8.042 \times 10^{-8}) = 1.865 \times 10^8 \text{ Pa} \). Strain \( \epsilon = \Delta x / L = 3.2 \times 10^{-3} / 2.05 = 1.561 \times 10^{-3} \). Young modulus \( E = \sigma / \epsilon = (1.865 \times 10^8) / (1.561 \times 10^{-3}) \approx 1.20 \times 10^{11} \text{ Pa} \). (d) The student can check if the elastic limit has been exceeded by removing the loads after each measurement and checking if the wire returns to its original length \( L \) (i.e., zero permanent extension).

Part (a): [1 mark] Diagram showing wire clamped at one end, passing over a pulley with hanging masses. [1 mark] Diagram shows a marker on the wire. [1 mark] Labelled measuring instrument (ruler, vernier, or microscope) placed near the marker. Part (b): [1 mark] Use of a reference wire (Searle's apparatus) OR measuring from a fixed reference marker on the wire itself. [1 mark] Measure the distance from the reference marker to the pulley as the original length. [1 mark] Explain that any slippage at the support/clamp will affect both wires equally or won't change the position of the marker relative to the ruler. Part (c): [1 mark] Area calculation: \( 8.04 \times 10^{-8} \text{ m}^2 \). [1 mark] Stress calculation: \( 1.87 \times 10^8 \text{ Pa} \). [1 mark] Strain calculation: \( 1.56 \times 10^{-3} \). [1.5 marks] Correct calculation of \( E \approx 1.20 \times 10^{11} \text{ Pa} \) (accept range \( 1.19 \times 10^{11} \) to \( 1.21 \times 10^{11} \text{ Pa} \)). Part (d): [1 mark] Remove the loads (unload the wire). [1 mark] Check if the marker returns to its original position / wire returns to its original length (no plastic deformation).

(a) The ray should enter the flat face at an angle of incidence \( i \) relative to the normal, bend towards the normal inside the glass, making an angle \( r \). It travels along a radius to the curved surface. Since the ray meets the curved boundary along a normal (at \( 90^\circ \) to the tangent), the angle of incidence at this boundary is \( 0^\circ \), so no refraction (bending) occurs. (b) Place the block on a sheet of paper and trace its outline. Use a ray box to direct a thin beam of light. Mark the entry point, the exit point, and the path of the incident and refracted rays. Draw a normal line perpendicular to the flat face at the point of entry. Use a protractor to measure the angles \( i \) and \( r \) relative to this normal line. (c) Using Snell's law: \( n = \sin i / \sin r = \sin 45.0^\circ / \sin 28.0^\circ = 0.7071 / 0.4695 \approx 1.51 \). (d) To maximize \( n \), we use the maximum possible \( i \) and the minimum possible \( r \). \( i_{\text{max}} = 45.0^\circ + 1.0^\circ = 46.0^\circ \), \( r_{\text{min}} = 28.0^\circ - 1.0^\circ = 27.0^\circ \). \( n_{\text{max}} = \sin 46.0^\circ / \sin 27.0^\circ = 0.7193 / 0.4540 \approx 1.58 \).

Part (a): [1 mark] Diagram showing ray bending towards normal at the flat face. [1 mark] Correctly labelled angles \( i \) (in air) and \( r \) (in glass) relative to the normal. [1 mark] Ray exits the curved face along the radius without bending. [1 mark] Explanation: Ray strikes curved face at normal incidence / angle of incidence at curved boundary is \( 0^\circ \). Part (b): [1 mark] Draw around the block and mark ray paths with pencil dots. [1 mark] Draw the normal line at the point of entry. [1 mark] Use a protractor to measure the angles from the normal. Part (c): [1 mark] Recall Snell's law \( n = \sin i / \sin r \). [1.5 marks] Calculate \( n = 1.51 \) (accept 1.50 to 1.51). Part (d): [1 mark] Identify that \( n_{\text{max}} \) occurs when \( i \) is maximized and \( r \) is minimized. [1 mark] Correct values: \( i_{\text{max}} = 46.0^\circ \), \( r_{\text{min}} = 27.0^\circ \). [1 mark] Calculate \( n_{\text{max}} \approx 1.58 \) (accept 1.58 to 1.60).